[{"data":1,"prerenderedAt":2126},["ShallowReactive",2],{"public-task-95":3,"public-task-topics-en":67},{"id":4,"topic_id":5,"code":6,"title":7,"locale":8,"uk_question_id":9,"slug":10,"seo_title":11,"seo_description":12,"seo_text":13,"description":14,"approx_time_min":15,"content":16,"updated_at":66},95,60,"SOM-BEND-04","Simple supported Beam under Forces","en",75,"simple-supported-beam-forces-shear-bending","Simple Supported Beam: Reactions, Shear and Bending","Determine support reactions and construct shear force and bending moment diagrams for a simply supported beam subjected to concentrated loads.","This task focuses on the static analysis of a simply supported beam under concentrated loads. Students calculate the support reactions, determine the shear force in each beam segment, and construct the bending moment diagram from the calculated internal forces. The exercise reinforces equilibrium equations, interpretation of shear and moment distributions, and the relationship between loading, support reactions, and internal force diagrams. It is intended for mechanics of materials and structural analysis courses.","",25,[17,20,24,27,30,33,36,39,42,45,48,51,54,57,60,63],{"type":18,"html":19},"html","\u003Csvg id=\"svg\" xmlns=\"http:\u002F\u002Fwww.w3.org\u002F2000\u002Fsvg\" version=\"1.1\" height=\"200\" width=\"700\">\n\u003Cg transform=\"translate(50,0)\">\n \u003Cimage xlink:href=\"https:\u002F\u002Fsopromat.xyz\u002Fimages\u002Fprojects\u002Fbeams\u002FR0.bmp\" preserveAspectRatio=\"none\" x=\"80\" y=\"103\" width=\"50\" height=\"50\" i=\"0\">\u003C\u002Fimage>\n \u003Cimage xlink:href=\"https:\u002F\u002Fsopromat.xyz\u002Fimages\u002Fprojects\u002Fbeams\u002FR1.bmp\" preserveAspectRatio=\"none\" x=\"480\" y=\"103\" width=\"40\" height=\"50\" i=\"1\">\u003C\u002Fimage>\n \u003Cpath d=\"M 0 100 H 600\" stroke=\"#000000\" stroke-width=\"6\">\u003C\u002Fpath>\n\n \u003Cpath d=\"M 0 0 l 0 -4 L 42 -4 L 42 -8 L60 0 L42 8 L 42 4 L 0,4 L 0 -4\" stroke=\"#ff0000\" fill=\"#ff0000\" transform=\"translate(0,36) rotate(90)\">\u003C\u002Fpath>\n \u003Ctext x=\"0\" y=\"30\" fill=\"#ff0000\">6 kN\u003C\u002Ftext>\n\n \u003Cpath d=\"M 0 0 l 0 -4 L 42 -4 L 42 -8 L60 0 L42 8 L 42 4 L 0,4 L 0 -4\" stroke=\"#ff0000\" fill=\"#ff0000\" transform=\"translate(200,36) rotate(90)\">\u003C\u002Fpath>\n \u003Ctext x=\"200\" y=\"30\" fill=\"#ff0000\">20 kN\u003C\u002Ftext>\n\n \u003Cpath d=\"M 0 0 l 0 -4 L 42 -4 L 42 -8 L60 0 L42 8 L 42 4 L 0,4 L 0 -4\" stroke=\"#ff0000\" fill=\"#ff0000\" transform=\"translate(300,95) rotate(-90)\">\u003C\u002Fpath>\n \u003Ctext x=\"300\" y=\"30\" fill=\"#ff0000\">25 kN\u003C\u002Ftext>\n\n \u003Cpath d=\"M 0 0 l 0 -4 L 42 -4 L 42 -8 L60 0 L42 8 L 42 4 L 0,4 L 0 -4\" stroke=\"#ff0000\" fill=\"#ff0000\" transform=\"translate(400,36) rotate(90)\">\u003C\u002Fpath>\n \u003Ctext x=\"400\" y=\"30\" fill=\"#ff0000\">10 kN\u003C\u002Ftext>\n\n \u003Cpath d=\"M 0 0 l 0 -4 L 42 -4 L 42 -8 L60 0 L42 8 L 42 4 L 0,4 L 0 -4\" stroke=\"#ff0000\" fill=\"#ff0000\" transform=\"translate(600,36) rotate(90)\">\u003C\u002Fpath>\n \u003Ctext x=\"600\" y=\"30\" fill=\"#ff0000\">6 kN\u003C\u002Ftext>\n\n \u003Cpath d=\"M 0 150 v 40 M 100 150 v 40 M 200 150 v 40 M 300 150 v 40 M 400 150 v 40 M 500 150 v 40 M 600 150 v 40 M 0 170 h 600\" stroke=\"#000000\">\u003C\u002Fpath>\n \u003Cpath d=\"M-5 175 l10 -10 M95 175 l10 -10 M195 175 l10 -10 M295 175 l10 -10 M395 175 l10 -10 M495 175 l10 -10 M595 175 l10 -10\" stroke=\"#000000\">\u003C\u002Fpath>\n\n \u003Ctext x=\"30\" y=\"165\" fill=\"#ff0000\">1.1 m\u003C\u002Ftext>\n \u003Ctext x=\"130\" y=\"165\" fill=\"#ff0000\">1.6 m\u003C\u002Ftext>\n \u003Ctext x=\"230\" y=\"165\" fill=\"#ff0000\">3 m\u003C\u002Ftext>\n \u003Ctext x=\"330\" y=\"165\" fill=\"#ff0000\">2.3 m\u003C\u002Ftext>\n \u003Ctext x=\"430\" y=\"165\" fill=\"#ff0000\">2.9 m\u003C\u002Ftext>\n \u003Ctext x=\"530\" y=\"165\" fill=\"#ff0000\">0.7 m\u003C\u002Ftext>\n\u003C\u002Fg>\n\u003C\u002Fsvg>\n\n\u003Cp>For the given beam, determine the support reactions and construct the shear force and bending moment diagrams.\u003C\u002Fp>",{"type":21,"html":22,"name":23},"number","Reaction at the left support","ans1",{"type":21,"html":25,"name":26},"Reaction at the right support","ans2",{"type":21,"html":28,"name":29},"Shear force in segment 1","ans3",{"type":21,"html":31,"name":32},"Shear force in segment 2","ans4",{"type":21,"html":34,"name":35},"Shear force in segment 3","ans5",{"type":21,"html":37,"name":38},"Shear force in segment 4","ans6",{"type":21,"html":40,"name":41},"Shear force in segment 5","ans7",{"type":21,"html":43,"name":44},"Shear force in segment 6","ans8",{"type":21,"html":46,"name":47},"Bending moment at the beginning of segment 1","ans9",{"type":21,"html":49,"name":50},"Bending moment at the beginning of segment 2","ans10",{"type":21,"html":52,"name":53},"Bending moment at the beginning of segment 3","ans11",{"type":21,"html":55,"name":56},"Bending moment at the beginning of segment 4","ans12",{"type":21,"html":58,"name":59},"Bending moment at the beginning of segment 5","ans13",{"type":21,"html":61,"name":62},"Bending moment at the beginning of segment 6","ans14",{"type":21,"html":64,"name":65},"Bending moment at the end of the beam","ans15","2026-08-05 19:40:03",[68,82,90,783,1677,2117],{"id":69,"parent_id":70,"code":70,"slug":71,"name":72,"seo_title":73,"seo_description":74,"seo_text":75,"content":70,"locale":8,"uk_topic_id":76,"show_in_theory_list":77,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":71,"children":81},79,null,"school-knowledge-test","School Knowledge Test","Engineering Mechanics Readiness Test — Physics & Math Quiz","Test your physics and math foundation before studying engineering mechanics: vectors, Newton’s laws, trigonometry, geometry, areas, and centroids.","Mastering strength of materials and engineering mechanics requires a strong foundation in high school physics and geometry. This online quiz allows students to assess vector algebra, Newton's laws, trigonometry, plane area calculations, and centroid locations.",34,0,1,"published","public",[],{"id":83,"parent_id":70,"code":70,"slug":84,"name":85,"seo_title":70,"seo_description":70,"seo_text":70,"content":86,"locale":8,"uk_topic_id":87,"show_in_theory_list":77,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":84,"children":89},386,"test-page","Test Page","\u003Ch2>Test\u003C\u002Fh2>\n\u003Cp>{{chunk:threejs-typical-cross-sections}}\u003C\u002Fp>",385,"draft",[],{"id":91,"parent_id":70,"code":70,"slug":92,"name":93,"seo_title":94,"seo_description":95,"seo_text":96,"content":97,"locale":8,"uk_topic_id":98,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":92,"children":99},81,"theoretical-mechanics","Theoretical Mechanics","Theoretical Mechanics — Statics, Kinematics & Dynamics Course","Learn theoretical mechanics through statics, kinematics, and dynamics: equilibrium, rigid-body motion, forces, work, energy, momentum, formulas, and problems.","Theoretical mechanics provides the mathematical and physical foundation for rigid body motion and equilibrium. The course is divided into three parts: Statics (equilibrium of forces), Kinematics (geometry of motion without forces), and Dynamics (motion caused by forces).","\u003Cp>\u003Cstrong>Theoretical mechanics\u003C\u002Fstrong> is a fundamental engineering discipline that studies the general laws of mechanical motion and equilibrium of material bodies. It provides a mathematical foundation for further study of mechanics of materials, theory of mechanisms and machines, machine dynamics, structural mechanics, and other engineering subjects.\u003C\u002Fp>\u003Ch2>What theoretical mechanics studies\u003C\u002Fh2>\u003Cp>Real bodies and mechanical systems are represented by idealized models such as a particle, a rigid body, or a system of particles. These models make it possible to isolate the essential laws of motion and equilibrium and describe them mathematically.\u003C\u002Fp>\u003Cp>The course consists of three main branches: \u003Cstrong>statics, kinematics, and dynamics\u003C\u002Fstrong>. They are studied in a natural sequence — from forces and equilibrium, through the description of motion, to the causes of that motion.\u003C\u002Fp>\u003Ch2>Statics\u003C\u002Fh2>\u003Cp>\u003Cstrong>Statics\u003C\u002Fstrong> studies force systems and the conditions of equilibrium of material bodies. Topics include forces and their projections, moments of forces, couples, reduction of force systems, constraints and reactions, equilibrium of two- and three-dimensional systems, distributed loads, centers of gravity, and friction.\u003C\u002Fp>\u003Cp>The main practical goal is to learn how to construct a free-body diagram, identify external forces and support reactions correctly, and write independent equilibrium equations.\u003C\u002Fp>\u003Ch2>Kinematics\u003C\u002Fh2>\u003Cp>\u003Cstrong>Kinematics\u003C\u002Fstrong> describes mechanical motion without considering the forces that cause it. Its principal quantities include position, trajectory, velocity, acceleration, angular velocity, and angular acceleration.\u003C\u002Fp>\u003Cp>The section covers particle kinematics, translation and fixed-axis rotation of rigid bodies, plane motion, relative motion of a particle, Coriolis acceleration, spherical motion, and general motion of a free rigid body.\u003C\u002Fp>\u003Ch2>Dynamics\u003C\u002Fh2>\u003Cp>\u003Cstrong>Dynamics\u003C\u002Fstrong> establishes the relationship between motion and the forces that produce it. It uses Newton's laws, differential equations of motion, momentum, angular-momentum and work-energy theorems, and conservation laws.\u003C\u002Fp>\u003Cp>For mechanical systems, the course covers center-of-mass motion, general theorems of dynamics, D'Alembert's principle, virtual displacements, the general equation of dynamics, and Lagrange's equations of the second kind. The course concludes with an introduction to small oscillations of a one-degree-of-freedom system.\u003C\u002Fp>\u003Ch2>How to study the course\u003C\u002Fh2>\u003Cp>A recommended sequence is \u003Cstrong>Statics → Kinematics → Dynamics\u003C\u002Fstrong>. For each topic, first understand the physical meaning of the concepts and the assumptions of the model, then study the governing equations and the procedure for applying them. After that, move to problems, where the key skill is selecting an appropriate mechanical model and solution method.\u003C\u002Fp>\u003Ch2>Notation and units\u003C\u002Fh2>\u003Cp>The formulas use standard vector and scalar notation of engineering mechanics. Unless stated otherwise, quantities should be expressed in a consistent system of units, preferably SI: length in meters, time in seconds, mass in kilograms, force in newtons, and moment of force in N·m.\u003C\u002Fp>",40,[100,324,523],{"id":101,"parent_id":91,"code":70,"slug":102,"name":103,"seo_title":104,"seo_description":105,"seo_text":106,"content":107,"locale":8,"uk_topic_id":108,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":102,"children":109},136,"statics","Statics","Statics — Forces, Moments and Equilibrium | Mechanics","Engineering statics: forces, moments, constraints and equilibrium conditions. Concise theory, formulas and practical mechanics problems.","Statics is the branch of theoretical mechanics concerned with equilibrium of bodies under applied forces. This section covers forces and their projections, moments, constraints and reactions, force systems and equilibrium equations, with practical problems for engineering students.","\u003Cp>Statics studies the conditions under which material bodies remain in equilibrium under applied forces. The section develops forces and their projections, moments, constraints and reactions, force systems, and equilibrium equations.\u003C\u002Fp>\u003Cp>MechClassroom emphasizes application: use the concise theory and formulas as a foundation, then practise setting up mechanical models and equilibrium equations through problems.\u003C\u002Fp>",133,[110,185,245,284],{"id":111,"parent_id":101,"code":70,"slug":112,"name":113,"seo_title":114,"seo_description":115,"seo_text":116,"content":117,"locale":8,"uk_topic_id":118,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":112,"children":119},274,"fundamentals-of-forces-and-moments","Fundamentals of Forces and Moments","Fundamentals of Forces and Moments in Statics","Forces, projections, moments, couples, resultants, and reduction of force systems in engineering statics.","This section covers the foundations of statics: forces and projections, moments, couples, addition of forces, and reduction of force systems to a point.","\u003Cp>\u003Cstrong>Fundamentals of Forces and Moments\u003C\u002Fstrong> introduces the mathematical description of force action on rigid bodies. Because force is a vector, its magnitude, direction, point of application, and coordinate projections are all important in statics.\u003C\u002Fp>\u003Ch2>Force systems and equivalent transformations\u003C\u002Fh2>\u003Cp>Several forces form a force system. For analysis, a system may be replaced by an equivalent resultant or by a force and a couple moment at a selected point, provided the mechanical effect on the rigid body is preserved.\u003C\u002Fp>\u003Ch2>Moments and couples\u003C\u002Fh2>\u003Cp>The moment of a force characterizes its rotational effect. A force couple has zero resultant force but produces a pure moment. These concepts form the basis for reducing force systems and writing equilibrium equations.\u003C\u002Fp>",270,[120,130,141,152,163,174],{"id":121,"parent_id":111,"code":70,"slug":122,"name":123,"seo_title":124,"seo_description":125,"seo_text":126,"content":127,"locale":8,"uk_topic_id":128,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":122,"children":129},182,"basic-concepts-statics-force-force-systems","Basic Concepts of Statics. Force and Force Systems","Basic Concepts of Statics: Force and Force Systems","Basic concepts of statics: force, force systems, resultant force and balanced systems. Core definitions for studying equilibrium of rigid bodies.","This introductory statics topic explains force and force systems, their mechanical meaning and essential terminology. It prepares students to study force projections, moments, resultants and equilibrium conditions for rigid bodies.","\u003Cp>\u003Cstrong>Statics\u003C\u002Fstrong> studies the conditions of equilibrium of material bodies and methods for transforming force systems. Its basic model is the rigid body—an idealized body in which the distance between any two points is assumed to remain unchanged during mechanical interaction.\u003C\u002Fp>\u003Ch2>Force and force systems\u003C\u002Fh2>{{chunk:statics-force-system-basics}}\u003Ch2>How a force is specified\u003C\u002Fh2>\u003Cp>A force is a vector. To describe it unambiguously in a mechanics problem, its \u003Cstrong>magnitude\u003C\u002Fstrong> $F$, \u003Cstrong>direction\u003C\u002Fstrong> and \u003Cstrong>point of application\u003C\u002Fstrong> must be known. The straight line along which the force vector acts is called its \u003Cstrong>line of action\u003C\u002Fstrong>.\u003C\u002Fp>\u003Cp>The SI unit of force is the newton (N). One newton is the force that gives a mass of 1 kg an acceleration of 1 m\u002Fs².\u003C\u002Fp>\u003Ch2>Classification of force systems\u003C\u002Fh2>\u003Cp>Force systems are classified according to the relative positions of their lines of action. Forces may be \u003Cstrong>collinear\u003C\u002Fstrong>, \u003Cstrong>concurrent\u003C\u002Fstrong>, \u003Cstrong>parallel\u003C\u002Fstrong>, or form a general force system. If all lines of action lie in one plane, the system is coplanar; otherwise, it is a three-dimensional force system.\u003C\u002Fp>\u003Ch2>Equivalence and resultant force\u003C\u002Fh2>\u003Cp>In statics, one force system is often replaced by another, simpler but equivalent system. If a system can be replaced by a single force, that force is its resultant. The resultant should not be confused with a simple sum of magnitudes: forces are added as vectors, and the positions of their lines of action also affect their mechanical effect.\u003C\u002Fp>\u003Ch2>Equilibrium\u003C\u002Fh2>\u003Cp>A rigid body is in equilibrium relative to a chosen inertial reference frame if the given force system does not change its state of rest. A force system satisfying this condition is called balanced. Specific equilibrium equations for coplanar and three-dimensional systems are developed in later topics.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose two forces of 100 N act on a ring along the same line in opposite directions. Their vector sum is zero, so the system is balanced. If one of the forces is instead 120 N, the resultant of the collinear system has a magnitude of 20 N and acts in the direction of the larger force.\u003C\u002Fp>\u003Ch2>Key points\u003C\u002Fh2>\u003Cul>\u003Cli>a force has magnitude, direction and a point of application;\u003C\u002Fli>\u003Cli>a force system is a set of forces acting on a body;\u003C\u002Fli>\u003Cli>equivalent systems produce the same mechanical effect;\u003C\u002Fli>\u003Cli>a resultant replaces a system by one force only when such a replacement is possible;\u003C\u002Fli>\u003Cli>mechanics problems require attention not only to force magnitudes but also to their directions and lines of action.\u003C\u002Fli>\u003C\u002Ful>",167,[],{"id":131,"parent_id":111,"code":70,"slug":132,"name":133,"seo_title":134,"seo_description":135,"seo_text":136,"content":137,"locale":8,"uk_topic_id":138,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":132,"children":140},82,"force-projections","Force Projections","Force Projection on an Axis — Formula and Sign Convention","Learn how to calculate the projection of a force on an axis using F cos θ, choose the correct sign, resolve force components, and avoid common angle mistakes.","A force projection onto an axis is a scalar quantity equal to the force magnitude multiplied by the cosine of the angle to that axis. This section details sign conventions for projecting force vectors onto Cartesian axes (X, Y, Z) and setting up equilibrium equations.","\u003Cp\r\n >A \u003Cstrong>projection of a force onto an axis\u003C\u002Fstrong> is an algebraic scalar quantity\r\n representing the component of the force vector along a selected direction. Projections convert\r\n vector relations into scalar equations and are therefore fundamental to analytical solutions in\r\n statics.\u003C\u002Fp\r\n>\r\n{{chunk:force-projections-spatial-components}}\r\n\u003Ch2>Projection onto a coordinate axis\u003C\u002Fh2>{{chunk:statics-force-projection}}\u003Cp\r\n >If the force $\\vec F$ forms an angle $\\alpha$ with the positive $x$ direction, its projection is\r\n $F_x$. The projections $F_y$ and, in three-dimensional problems, $F_z$ are defined in the same\r\n way.\u003C\u002Fp\r\n>\u003Ch2>Signs of projections\u003C\u002Fh2\r\n>\u003Cp\r\n >A projection is positive when the corresponding force component points in the positive direction\r\n of the axis and negative when it points in the opposite direction. A force perpendicular to an\r\n axis has zero projection onto that axis.\u003C\u002Fp\r\n>\u003Ch2>Resolving a force in two dimensions\u003C\u002Fh2\r\n>\u003Cp\r\n >For a force in the $xy$ plane, $\\vec F=F_x\\vec i+F_y\\vec j$. If $\\alpha$ is measured from the\r\n positive $x$ axis, then $F_x=F\\cos\\alpha$ and $F_y=F\\sin\\alpha$, with signs determined by the\r\n actual component directions.\u003C\u002Fp\r\n>\u003Ch2>Checking the force magnitude\u003C\u002Fh2\r\n>\u003Cp\r\n >In a rectangular Cartesian coordinate system, the magnitude of a two-dimensional force satisfies\r\n $F=\\sqrt{F_x^2+F_y^2}$. This relation is useful for checking calculated components.\u003C\u002Fp\r\n>\u003Ch2>Example\u003C\u002Fh2\r\n>\u003Cp\r\n >A force $F=10$ kN acts at $30^\\circ$ above the positive $x$ axis in the first quadrant. Then\r\n $F_x=10\\cos30^\\circ\\approx8.66$ kN and $F_y=10\\sin30^\\circ=5$ kN. Both projections are\r\n positive.\u003C\u002Fp\r\n>\u003Ch2>Common mistakes\u003C\u002Fh2\r\n>\u003Cul\r\n >\u003Cli>confusing an angle measured from the $x$ axis with one measured from the $y$ axis;\u003C\u002Fli\r\n >\u003Cli>using only the magnitude of a projection and ignoring its sign;\u003C\u002Fli\r\n >\u003Cli>using sine instead of cosine without checking which axis the angle is measured from;\u003C\u002Fli\r\n >\u003Cli>adding force magnitudes instead of algebraic force projections.\u003C\u002Fli>\u003C\u002Ful\r\n>\u003Cp\r\n >Equilibrium equations use the algebraic projections of all forces onto the selected coordinate\r\n axes.\u003C\u002Fp\r\n>\r\n",43,2,[],{"id":142,"parent_id":111,"code":70,"slug":143,"name":144,"seo_title":145,"seo_description":146,"seo_text":147,"content":148,"locale":8,"uk_topic_id":149,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":143,"children":151},140,"moment-of-force-about-point-and-axis","Moment of a Force About a Point and an Axis","Moment of a Force About a Point and Axis — Formulas","Moment of a force in statics: moment arm, sign convention, point and axis formulas, coordinate calculation, worked example and practice foundation.","The moment of a force is a fundamental statics quantity describing the rotational effect of a force. This topic covers moments about a point and an axis, the moment arm, sign convention, vector definition, coordinate calculation and their use in equilibrium problems.","\u003Cp>The \u003Cstrong>moment of a force\u003C\u002Fstrong> describes the rotational effect produced by a force. It depends on both the force magnitude and the position of its line of action relative to the point or axis about which rotation is considered.\u003C\u002Fp>\u003Ch2>Moment of a force about a point\u003C\u002Fh2>\u003Cp>Let a force $\\vec F$ act at point $A$, and let $O$ be the point about which the moment is required. The moment vector is defined by the cross product\u003C\u002Fp>\u003Cp>$$\\vec M_O=\\vec r\\times\\vec F,$$\u003C\u002Fp>\u003Cp>where $\\vec r=\\overrightarrow{OA}$. Its direction follows the right-hand rule. Its magnitude can be written as $M_O=rF\\sin\\alpha$, where $\\alpha$ is the angle between $\\vec r$ and $\\vec F$. Since $d=r\\sin\\alpha$ is the perpendicular distance from $O$ to the line of action, the principal practical formula follows.\u003C\u002Fp>{{chunk:moment-of-force-about-point}}\u003Ch2>Sign convention in planar statics\u003C\u002Fh2>\u003Cp>For a planar force system, moments are conveniently treated as algebraic quantities. Counterclockwise rotation is commonly taken as positive and clockwise rotation as negative. Choose a convention once and use it consistently throughout the equilibrium equations.\u003C\u002Fp>\u003Ch2>When the moment is zero\u003C\u002Fh2>\u003Cp>If the force's line of action passes through point $O$, the moment arm is $d=0$ and the moment about that point is zero. Consequently, moving a force anywhere along its line of action does not change its moment about an arbitrary point.\u003C\u002Fp>\u003Ch2>Coordinate calculation\u003C\u002Fh2>\u003Cp>In the $xy$ plane, if $\\vec r=(x,y)$ and $\\vec F=(F_x,F_y)$, the moment about the origin is\u003C\u002Fp>\u003Cp>$$M_O=xF_y-yF_x.$$\u003C\u002Fp>\u003Cp>This form is especially useful when a problem gives the coordinates of the point of application and the Cartesian components of the force.\u003C\u002Fp>\u003Ch2>Moment of a force about an axis\u003C\u002Fh2>\u003Cp>The moment about a specified axis is the projection of the moment vector about any point on that axis onto the axis direction. If $\\vec e$ is a unit vector along the axis,\u003C\u002Fp>\u003Cp>$$M_{axis}=\\vec e\\cdot(\\vec r\\times\\vec F).$$\u003C\u002Fp>\u003Cp>In three-dimensional statics, this measures the tendency of the force to rotate the body specifically about the selected axis.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A force $F=200$ N acts on a lever with a perpendicular distance $d=0.30$ m from the pivot to the force's line of action. The moment magnitude is $M=200\\cdot0.30=60$ N·m. Its sign depends on the resulting sense of rotation and the adopted sign convention.\u003C\u002Fp>\u003Ch2>Skills for solving problems\u003C\u002Fh2>\u003Cp>Before calculating a moment, identify the reference point or axis, locate the force's line of action, determine the perpendicular moment arm or resolve the force into components, and check the sign. For several forces, their moments are summed algebraically. This procedure forms the basis of equilibrium equations in statics.\u003C\u002Fp>",139,3,[],{"id":153,"parent_id":111,"code":70,"slug":154,"name":155,"seo_title":156,"seo_description":157,"seo_text":158,"content":159,"locale":8,"uk_topic_id":160,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":154,"children":162},183,"force-couples-couple-moment","Force Couples and Couple Moment","Force Couple & Couple Moment — Formula, Properties, Examples","Learn force couples in statics: calculate couple moment with M = Fd, understand its direction and free-vector property, combine couples, and see examples.","This topic introduces a force couple as two equal, parallel and oppositely directed forces. It explains the couple moment, its key properties and its use when transforming and simplifying force systems.","\u003Cp>A \u003Cstrong>force couple\u003C\u002Fstrong> consists of two parallel forces that are equal in magnitude, opposite in direction, and separated by a perpendicular distance. The resultant force is zero, but the couple produces a pure rotational effect.\u003C\u002Fp>\n{{chunk:force-couple-interactive-moment}}\n\u003Ch2>Moment of a couple\u003C\u002Fh2>\n\u003Cp>The magnitude of the couple moment is\u003C\u002Fp>\n\u003Cp>\\[M=Fd,\\]\u003C\u002Fp>\n\u003Cp>where \\(F\\) is the magnitude of either force and \\(d\\) is the perpendicular distance between their lines of action, called the \u003Cstrong>couple arm\u003C\u002Fstrong>.\u003C\u002Fp>\n\u003Cp>In a planar problem, the sign of \\(M\\) is assigned according to the sense of rotation. A common convention takes counterclockwise moments as positive and clockwise moments as negative.\u003C\u002Fp>\n\u003Ch2>Why the couple moment is independent of the reference point\u003C\u002Fh2>\n\u003Cp>Let the forces be \\(\\vec F\\) and \\(-\\vec F\\), applied at position vectors \\(\\vec r_1\\) and \\(\\vec r_2\\) measured from an arbitrary point \\(O\\). Their total moment about \\(O\\) is\u003C\u002Fp>\n\u003Cp>\\[\\vec M_O=\\vec r_1\\times\\vec F+\\vec r_2\\times(-\\vec F)=(\\vec r_1-\\vec r_2)\\times\\vec F.\\]\u003C\u002Fp>\n\u003Cp>The result depends only on the relative position of the two lines of action, not on the location of \\(O\\). Therefore, the moment of a couple is the same about any point.\u003C\u002Fp>\n\u003Ch2>Vector representation\u003C\u002Fh2>\n\u003Cp>In three dimensions, the couple moment is represented by a vector \\(\\vec M\\) perpendicular to the plane containing the two forces. Its direction is determined by the right-hand rule.\u003C\u002Fp>\n\u003Cp>Because the couple moment is independent of the reference point, \\(\\vec M\\) is a \u003Cstrong>free vector\u003C\u002Fstrong>: it may be translated parallel to itself without changing the external effect on a rigid body.\u003C\u002Fp>\n\u003Ch2>Equivalent couples\u003C\u002Fh2>\n\u003Cp>Two couples are mechanically equivalent if their moment vectors are equal:\u003C\u002Fp>\n\u003Cp>\\[\\vec M_1=\\vec M_2.\\]\u003C\u002Fp>\n\u003Cp>Thus, the force magnitude and arm may change while the product \\(Fd\\) and the moment direction remain the same.\u003C\u002Fp>\n\u003Ch2>Addition of couples\u003C\u002Fh2>\n\u003Cp>Several couples acting on a rigid body can be replaced by a single resultant couple whose moment is the vector sum\u003C\u002Fp>\n\u003Cp>\\[\\vec M_R=\\sum_i\\vec M_i.\\]\u003C\u002Fp>\n\u003Cp>For coplanar couples, this becomes an algebraic sum of signed moments.\u003C\u002Fp>\n\u003Ch2>Example\u003C\u002Fh2>\n\u003Cp>Two opposite parallel forces of magnitude \\(F=100\\,\\text{N}\\) are separated by \\(d=0.20\\,\\text{m}\\). The couple moment magnitude is\u003C\u002Fp>\n\u003Cp>\\[M=100\\cdot0.20=20\\,\\text{N}\\cdot\\text{m}.\\]\u003C\u002Fp>\n\u003Cp>A pair of \\(50\\,\\text{N}\\) forces separated by \\(0.40\\,\\text{m}\\) creates the same couple moment, provided the sense of rotation is unchanged.\u003C\u002Fp>",168,4,[],{"id":164,"parent_id":111,"code":70,"slug":165,"name":166,"seo_title":167,"seo_description":168,"seo_text":169,"content":170,"locale":8,"uk_topic_id":171,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":165,"children":173},184,"addition-forces-resultant-force-system","Addition of Forces. Resultant of a Force System","Addition of Forces & Resultant — Vector and Component Methods","Learn how to add forces and find a resultant using the parallelogram rule, vector addition, Cartesian components, and analytical methods for statics problems.","This topic explains addition of forces and the resultant of a force system. Geometric and analytical approaches provide the foundation for reducing force systems and formulating equilibrium conditions.","\u003Cp>\u003Cstrong>Forces are added\u003C\u002Fstrong> according to the rules of vector addition. The result of this vector addition is the resultant vector of the force system. For a concurrent force system, whose lines of action pass through a common point, this vector can be applied at the point of concurrency and is the resultant force of the system.\u003C\u002Fp>\u003Ch2>Geometric addition\u003C\u002Fh2>\u003Cp>Two forces can be added using the parallelogram or triangle rule. For several forces, construct a force polygon by placing each successive vector at the end of the previous one. The vector from the start of the first force to the end of the last is their geometric sum.\u003C\u002Fp>\u003Ch2>Analytical addition in two dimensions\u003C\u002Fh2>\u003Cp>In Cartesian coordinates, first sum the components algebraically:\u003C\u002Fp>\u003Cp>$$R_x=\\sum_i F_{ix},\\qquad R_y=\\sum_i F_{iy}.$$\u003C\u002Fp>{{chunk:statics-resultant-two-components}}\u003Cp>The direction of the resultant vector is determined from its components while accounting for the correct quadrant, for example with $\\operatorname{atan2}(R_y,R_x)$.\u003C\u002Fp>\u003Ch2>Resultant force and resultant vector\u003C\u002Fh2>\u003Cp>The vector sum of all forces always defines the \u003Cstrong>resultant vector\u003C\u002Fstrong> $\\vec R=\\sum\\vec F_i$. For a general force system, however, this vector alone is not sufficient to replace the entire system by one force because the moment effect must also be preserved. The term \u003Cstrong>resultant force\u003C\u002Fstrong> should therefore be used when a single force is actually equivalent to the original system.\u003C\u002Fp>\u003Ch2>Collinear forces\u003C\u002Fh2>\u003Cp>For forces acting along one line, choose a positive direction and add the forces algebraically. For example, forces of 8 kN and 5 kN acting in the same direction give 13 kN. If they act in opposite directions, the magnitude of the sum is 3 kN and its direction is that of the larger force.\u003C\u002Fp>\u003Ch2>Example with perpendicular forces\u003C\u002Fh2>\u003Cp>Suppose forces of 3 kN along $+x$ and 4 kN along $+y$ act at a common point. Then $R_x=3$ kN and $R_y=4$ kN, so the resultant magnitude of the concurrent system is $R=\\sqrt{3^2+4^2}=5$ kN. Its direction is $\\alpha=\\arctan(4\u002F3)\\approx53.1^\\circ$ from the positive $x$ axis.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>adding magnitudes of non-collinear forces as ordinary numbers;\u003C\u002Fli>\u003Cli>ignoring the signs of force components;\u003C\u002Fli>\u003Cli>using only $\\arctan(R_y\u002FR_x)$ without checking the quadrant;\u003C\u002Fli>\u003Cli>treating the resultant vector of a general force system as a single equivalent force without checking moments.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>The next step in statics is reduction of a general force system to a specified point, where the resultant moment is considered together with the resultant vector.\u003C\u002Fp>",169,5,[],{"id":175,"parent_id":111,"code":70,"slug":176,"name":177,"seo_title":178,"seo_description":179,"seo_text":180,"content":181,"locale":8,"uk_topic_id":182,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":176,"children":184},185,"reduction-force-system-given-point","Reduction of a Force System to a Given Point","Reduce a Force System to a Point — Resultant Force & Moment","Learn how to reduce a general force system to a chosen point using the resultant force vector and resultant moment, and identify equilibrium or a pure couple.","This topic explains how a general force system is reduced to a specified point. It introduces the resultant force vector and resultant moment used to analyze equivalent force systems and equilibrium.","\u003Cp>\u003Cstrong>Reduction of a force system to a specified point\u003C\u002Fstrong> replaces a general set of forces and couples by a simpler equivalent system: one force applied at the selected point and one couple moment. This representation is fundamental for subsequent equilibrium analysis.\u003C\u002Fp>\u003Ch2>Moving a force to a specified point\u003C\u002Fh2>\u003Cp>A force $\\vec F$ applied at point $A$ can be moved parallel to itself to point $O$ if a couple with moment $\\vec M_O=\\vec r\\times\\vec F$ is added at the same time, where $\\vec r=\\overrightarrow{OA}$. The resulting force-and-couple system is equivalent to the original force.\u003C\u002Fp>\u003Ch2>Resultant force vector and resultant moment\u003C\u002Fh2>{{chunk:statics-force-system-reduction}}\u003Cp>For a planar problem, the resultant force vector is described by $R_x=\\sum F_{ix}$ and $R_y=\\sum F_{iy}$, while the resultant moment about $O$ is the algebraic sum of the moments of all forces and applied couples.\u003C\u002Fp>\u003Ch2>Changing the reduction point\u003C\u002Fh2>\u003Cp>If a system has been reduced to point $O$, moving the reduction point to $A$ does not change the resultant force vector. The resultant moment transforms according to\u003C\u002Fp>\u003Cp>$$\\vec M_A=\\vec M_O+\\overrightarrow{AO}\\times\\vec R.$$\u003C\u002Fp>\u003Cp>Thus, the same force system has the same resultant force vector at every reduction point, but generally different resultant moments when $\\vec R\\ne0$.\u003C\u002Fp>\u003Ch2>Main cases after reduction\u003C\u002Fh2>\u003Cul>\u003Cli>if $\\vec R=0$ and $\\vec M_O=0$, the system is balanced;\u003C\u002Fli>\u003Cli>if $\\vec R=0$ but $\\vec M_O\\ne0$, the system is equivalent to a force couple;\u003C\u002Fli>\u003Cli>if $\\vec R\\ne0$, whether the system can be reduced further to a single resultant force depends on the relation between the resultant force vector and resultant moment.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Planar example\u003C\u002Fh2>\u003Cp>Suppose a 10 kN force acts in the $+y$ direction at a point 2 m to the right of $O$. When the force is moved to $O$, an additional moment $M_O=10\\cdot2=20$ kN·m must be introduced. With the usual planar sign convention, this moment is positive because the original force tends to rotate the body counterclockwise about $O$.\u003C\u002Fp>\u003Ch2>Reduction procedure\u003C\u002Fh2>\u003Col>\u003Cli>choose the reduction point $O$;\u003C\u002Fli>\u003Cli>calculate the components of all forces and form the resultant vector $\\vec R$;\u003C\u002Fli>\u003Cli>calculate the moment of every force about $O$;\u003C\u002Fli>\u003Cli>add all applied couple moments;\u003C\u002Fli>\u003Cli>write the equivalent system $\\vec R$ and $\\vec M_O$;\u003C\u002Fli>\u003Cli>if required, determine whether the system can be simplified further.\u003C\u002Fli>\u003C\u002Fol>\u003Cp>The conditions $\\vec R=0$ and $\\vec M_O=0$ lead directly to the general equilibrium equations for a rigid body.\u003C\u002Fp>",170,6,[],{"id":186,"parent_id":101,"code":70,"slug":187,"name":188,"seo_title":189,"seo_description":190,"seo_text":191,"content":192,"locale":8,"uk_topic_id":193,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":187,"children":194},275,"equilibrium-of-coplanar-systems","Equilibrium of Coplanar Systems","Equilibrium of Coplanar Force Systems","Equilibrium equations, constraints and reactions, systems of bodies, and distributed loads in two-dimensional statics.","This section covers equilibrium of coplanar force systems, constraint reactions, equilibrium of bodies and systems of bodies, and distributed loading.","\u003Cp>\u003Cstrong>Equilibrium of coplanar systems\u003C\u002Fstrong> covers problems in which forces act in a single plane. The central task is to replace the real object by an appropriate free-body model and determine the conditions under which a rigid body or a system of bodies remains in equilibrium.\u003C\u002Fp>\u003Ch2>Constraints and reactions\u003C\u002Fh2>\u003Cp>Supports, hinges, cables, and other constraints restrict possible motion. In a free-body diagram their action is replaced by reaction forces or moments whose unknown components depend on the type of constraint.\u003C\u002Fp>\u003Ch2>Equilibrium equations\u003C\u002Fh2>\u003Cp>For a general coplanar force system, equilibrium requires zero sums of force projections and zero sum of moments. Systems consisting of several connected bodies can be separated and analyzed while retaining the interaction forces between their parts.\u003C\u002Fp>\u003Ch2>Distributed loads\u003C\u002Fh2>\u003Cp>Loads distributed over a length or surface are often replaced by an equivalent resultant for equilibrium calculations. The magnitude and line of action of that resultant follow from the distribution law.\u003C\u002Fp>",271,[195,205,215,225,235],{"id":196,"parent_id":186,"code":70,"slug":197,"name":198,"seo_title":199,"seo_description":200,"seo_text":201,"content":202,"locale":8,"uk_topic_id":203,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":197,"children":204},186,"equilibrium-conditions-coplanar-force-systems","Equilibrium Conditions for Coplanar Force Systems","2D Equilibrium Equations — Coplanar Force Systems","Learn the equilibrium equations for coplanar force systems: ΣFx = 0, ΣFy = 0, and ΣM = 0, with alternative equation sets and independence conditions.","This topic organizes the equilibrium conditions for coplanar force systems. It covers force-component and moment equations and their use in determining unknown forces and reactions in statics problems.","\u003Cp>\u003Cstrong>Equilibrium of a planar force system\u003C\u002Fstrong> requires both the resultant force vector and the resultant moment of the system to be zero. For a rigid body in two dimensions, this produces three independent scalar equations used to determine unknown forces and constraint reactions.\u003C\u002Fp>\u003Ch2>Standard equilibrium equations\u003C\u002Fh2>{{chunk:statics-planar-equilibrium-equations}}\u003Cp>The first two equations eliminate the translational effect of the system along the coordinate axes, while the third eliminates its rotational effect. When all three conditions are satisfied, the body has no tendency toward translation or rotation under the forces considered.\u003C\u002Fp>\u003Ch2>Choosing coordinate axes\u003C\u002Fh2>\u003Cp>The $x$ and $y$ axes may be chosen freely, but a convenient orientation can greatly simplify the calculations. One axis is often aligned with an inclined surface, a member, or the direction of several known forces. Once the axes are selected, component signs must be used consistently.\u003C\u002Fp>\u003Ch2>Choosing the moment point\u003C\u002Fh2>\u003Cp>It is usually advantageous to choose point $O$ where the lines of action of one or more unknown reactions intersect. Their moments about that point are then zero, reducing the number of unknowns in the moment equation.\u003C\u002Fp>\u003Ch2>Equivalent forms\u003C\u002Fh2>\u003Cp>For a general planar force system, other independent sets of three equilibrium equations may be used instead of two component equations and one moment equation. For example, two moment equations about different points and one force-component equation can be valid when the selected equations remain independent. The important requirement is independence, not merely writing three equations.\u003C\u002Fp>\u003Ch2>Special force systems\u003C\u002Fh2>\u003Cp>For a concurrent planar force system, all lines of action intersect at one point, so equilibrium is described by the two independent conditions $\\sum F_x=0$ and $\\sum F_y=0$. For a parallel force system, an equation for force components along the common force direction together with one independent moment equation is generally sufficient.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A body is subjected to a horizontal force of 8 kN to the right and an unknown force $P$ to the left, together with vertical forces of 5 kN upward and 5 kN downward. From $\\sum F_x=0$, $8-P=0$, giving $P=8$ kN. The vertical condition is already satisfied. Complete equilibrium still requires $\\sum M_O=0$ because a zero vector sum of forces can still leave a nonzero couple moment.\u003C\u002Fp>\u003Ch2>Solution procedure\u003C\u002Fh2>\u003Col>\u003Cli>isolate the body or system of bodies being analyzed;\u003C\u002Fli>\u003Cli>show all applied forces and constraint reactions;\u003C\u002Fli>\u003Cli>choose coordinate axes and a moment sign convention;\u003C\u002Fli>\u003Cli>write the force components;\u003C\u002Fli>\u003Cli>choose a convenient point for the moment equation;\u003C\u002Fli>\u003Cli>form independent equilibrium equations and solve for the unknowns;\u003C\u002Fli>\u003Cli>check signs, units and the physical meaning of the result.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>omitting a support reaction;\u003C\u002Fli>\u003Cli>confusing the sign of a force with the sign of its component;\u003C\u002Fli>\u003Cli>using an incorrect moment arm;\u003C\u002Fli>\u003Cli>omitting applied couple moments;\u003C\u002Fli>\u003Cli>writing dependent equations instead of independent ones.\u003C\u002Fli>\u003C\u002Ful>",171,[],{"id":206,"parent_id":186,"code":70,"slug":207,"name":208,"seo_title":209,"seo_description":210,"seo_text":211,"content":212,"locale":8,"uk_topic_id":213,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":207,"children":214},187,"constraints-constraint-reactions","Constraints and Constraint Reactions","Support Reactions in Statics — Roller, Pin, Fixed Support & Cable","Learn how to model support reactions for rollers, pins, fixed supports, cables, links, and smooth contacts when drawing free-body diagrams in statics.","This topic examines mechanical constraints and the reaction forces that replace them in a free-body diagram. It covers common supports, hinges, cables and contacts and the directions of their reactions.","\u003Cp>\u003Cstrong>Constraints\u003C\u002Fstrong> are bodies or devices that restrict the possible motion of the body being analyzed. In statics, each constraint is replaced by its \u003Cstrong>reaction\u003C\u002Fstrong>, after which the body is treated as free under the action of the applied forces and constraint reactions.\u003C\u002Fp>\u003Ch2>Principle of releasing constraints\u003C\u002Fh2>\u003Cp>To write equilibrium equations, conceptually remove the constraints from the body. Replace each removed constraint by a reaction force or a system of forces and moments that reproduces its mechanical action. The number and directions of unknown reaction components are determined by the motions that the constraint prevents.\u003C\u002Fp>\u003Ch2>Common constraints in two dimensions\u003C\u002Fh2>{{chunk:statics-constraint-reactions}}\u003Ch2>Smooth contact\u003C\u002Fh2>\u003Cp>For an ideally smooth surface, friction is neglected. The contact reaction acts along the common normal to the surfaces at the contact point. For a flat surface, the reaction direction is known in advance and only its magnitude is unknown.\u003C\u002Fp>\u003Ch2>Cables and two-force members\u003C\u002Fh2>\u003Cp>An ideal flexible cable or rope carries tension only, so the tension force acts along the cable. A straight member acted on only by forces at two pin-connected ends is a two-force member: the end forces are collinear with the member axis, equal in magnitude, and opposite in direction.\u003C\u002Fp>\u003Ch2>Pin and roller supports\u003C\u002Fh2>\u003Cp>A roller or movable support in a planar model produces one reaction in the direction in which it prevents motion. A pin support prevents two independent translations, so its reaction is usually represented by two unknown components $R_x$ and $R_y$. An ideal pin does not transmit a reaction moment.\u003C\u002Fp>\u003Ch2>Fixed support\u003C\u002Fh2>\u003Cp>A fixed support in two dimensions prevents two translations and rotation. Its action is therefore represented by two reaction components $R_x$, $R_y$ and a reaction moment $M$.\u003C\u002Fp>\u003Ch2>Free-body diagram\u003C\u002Fh2>\u003Cp>After releasing the constraints, construct a free-body diagram. Show all applied forces, applied couple moments, the body weight when relevant, and every constraint reaction. The equilibrium equations are applied to this isolated diagram.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A beam supported by a pin at $A$ and a roller at $B$ has three unknown reactions in a typical planar arrangement: $A_x$, $A_y$, and $B_y$ when the roller reaction is vertical. This matches the three independent equilibrium equations available for a planar rigid body.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>leaving a support on the free-body diagram while also drawing its reactions;\u003C\u002Fli>\u003Cli>adding a reaction moment at an ideal pin;\u003C\u002Fli>\u003Cli>assigning an arbitrary direction to a smooth-contact reaction instead of the normal direction;\u003C\u002Fli>\u003Cli>assuming a cable can carry compression;\u003C\u002Fli>\u003Cli>omitting one of the reaction components of a fixed support.\u003C\u002Fli>\u003C\u002Ful>",172,[],{"id":216,"parent_id":186,"code":70,"slug":217,"name":218,"seo_title":219,"seo_description":220,"seo_text":221,"content":222,"locale":8,"uk_topic_id":223,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":217,"children":224},188,"equilibrium-rigid-body-two-dimensions","Equilibrium of a Rigid Body in Two Dimensions","2D Rigid-Body Equilibrium — Support Reactions Step by Step","Solve 2D rigid-body equilibrium problems step by step using free-body diagrams, support reactions, ΣFx = 0, ΣFy = 0, and ΣM = 0.","This topic applies statics equations to rigid bodies in two dimensions. It covers free-body diagrams, replacement of constraints by reactions and calculation of unknown forces and support reactions.","\u003Cp>In practical statics problems, \u003Cstrong>equilibrium of a rigid body in two dimensions\u003C\u002Fstrong> is analyzed using a free-body diagram. Real supports and contacts are replaced by their reactions, and the equilibrium equations are then applied to the isolated body.\u003C\u002Fp>\u003Ch2>Free-body diagram\u003C\u002Fh2>\u003Cp>A free-body diagram should contain only the body being analyzed and all external forces and moments acting on it. The physical constraints are removed from the diagram and replaced by their corresponding reactions. Internal forces within the isolated rigid body are not shown.\u003C\u002Fp>\u003Ch2>Solution procedure\u003C\u002Fh2>{{chunk:statics-planar-rigid-body-procedure}}\u003Ch2>Static determinacy\u003C\u002Fh2>\u003Cp>For one rigid body subjected to a general planar force system, three independent equilibrium equations are available. If the correctly modeled constraints introduce no more than three independent unknown reaction components and those equations determine them uniquely, the problem may be statically determinate. Additional reaction unknowns generally require deformation relations beyond rigid-body statics.\u003C\u002Fp>\u003Ch2>Efficient order of equations\u003C\u002Fh2>\u003Cp>It is often best to begin with a moment equation about a point through which the lines of action of several unknown reactions pass. Those reactions then have zero moment about that point and disappear from the equation. After one unknown is found, force-component equations can be used for the others.\u003C\u002Fp>\u003Ch2>Example: simply supported beam\u003C\u002Fh2>\u003Cp>A 6 m beam is supported by a pin at $A$ and a roller at $B$. A vertical force $P=12$ kN acts at midspan. For this vertical loading, $A_x=0$. Taking moments about $A$ gives $B_y\\cdot6-12\\cdot3=0$, so $B_y=6$ kN. From $\\sum F_y=0$, $A_y+B_y-12=0$, giving $A_y=6$ kN.\u003C\u002Fp>\u003Ch2>Meaning of a negative reaction\u003C\u002Fh2>\u003Cp>The initial direction of an unknown reaction may be assumed. If the calculated value is negative, this is not automatically an error: the actual reaction acts opposite to the assumed direction. The result should still be checked against the physical contact model.\u003C\u002Fp>\u003Ch2>Checking the solution\u003C\u002Fh2>\u003Cp>After calculating the reactions, verify the force and moment sums again, preferably by taking moments about a different point. This helps reveal errors in signs, moment arms, or arithmetic.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>writing equations before constructing a free-body diagram;\u003C\u002Fli>\u003Cli>modeling a support reaction incorrectly;\u003C\u002Fli>\u003Cli>including forces that act on another body rather than the isolated body;\u003C\u002Fli>\u003Cli>omitting an applied couple moment;\u003C\u002Fli>\u003Cli>using the distance to the force application point as the moment arm without checking perpendicularity.\u003C\u002Fli>\u003C\u002Ful>",173,[],{"id":226,"parent_id":186,"code":70,"slug":227,"name":228,"seo_title":229,"seo_description":230,"seo_text":231,"content":232,"locale":8,"uk_topic_id":233,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":227,"children":234},189,"equilibrium-systems-bodies-method-disassembly","Equilibrium of Systems of Bodies. Method of Disassembly","Equilibrium of Systems of Bodies — Statics","Equilibrium of connected bodies using the method of disassembly: interaction forces, separate free-body diagrams and equilibrium equations.","This topic explains equilibrium of systems containing several connected bodies. The method of disassembly treats each body separately, introduces interaction forces and provides the equations needed for equilibrium analysis.","\u003Cp>A \u003Cstrong>system of bodies\u003C\u002Fstrong> consists of several bodies connected by pins, contacts, members, or other constraints. To determine not only external reactions but also interaction forces between the parts, the system often has to be separated and the equilibrium of each body analyzed individually.\u003C\u002Fp>\u003Ch2>External and internal forces\u003C\u002Fh2>\u003Cp>For the complete system, forces of interaction between its bodies are internal. They occur in action-reaction pairs and cancel when the system is considered as a whole. Reactions from external supports and applied loads are external forces.\u003C\u002Fp>\u003Ch2>Method of disassembly\u003C\u002Fh2>{{chunk:statics-system-of-bodies-procedure}}\u003Ch2>Forces at an internal pin\u003C\u002Fh2>\u003Cp>In a planar model, an ideal internal pin can transmit two force components but no moment. If the pin forces on the first body are denoted by $H_x$ and $H_y$, the second body is subjected to $-H_x$ and $-H_y$. These internal forces should not also be included as external forces on the free-body diagram of the complete system.\u003C\u002Fp>\u003Ch2>Why analyze the complete system first\u003C\u002Fh2>\u003Cp>The equilibrium equations of the entire structure often determine some external reactions without introducing internal forces. The system can then be separated to obtain additional equations for pin and connection reactions. This order usually reduces the number of unknowns in each equation.\u003C\u002Fp>\u003Ch2>Example: two beams connected by an internal pin\u003C\u002Fh2>\u003Cp>Suppose beams $AC$ and $CB$ are connected by an internal pin at $C$ and have external supports at $A$ and $B$. After analyzing the complete system, separate the beams at $C$. The beam $AC$ is subjected to pin-force components $C_x$ and $C_y$, while beam $CB$ is subjected to equal and opposite components $-C_x$ and $-C_y$. Separate equilibrium equations are then written for each beam.\u003C\u002Fp>\u003Ch2>Two-force members\u003C\u002Fh2>\u003Cp>If a straight member is pin-connected only at two points and carries no other loads, it is a two-force member. The forces at its ends must be collinear, equal in magnitude, and opposite in direction. This allows two unknown Cartesian components to be replaced by one unknown axial force.\u003C\u002Fp>\u003Ch2>Checking the solution\u003C\u002Fh2>\u003Cp>After solving, verify that interaction forces on paired free-body diagrams have equal magnitudes and opposite directions. It is also useful to substitute the calculated external reactions back into the equilibrium equations of the complete system.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating internal forces as external when analyzing the complete system;\u003C\u002Fli>\u003Cli>drawing the same direction for an internal pin force on both isolated bodies;\u003C\u002Fli>\u003Cli>adding a reaction moment at an ideal internal pin;\u003C\u002Fli>\u003Cli>failing to use the two-force-member property when applicable;\u003C\u002Fli>\u003Cli>writing equations for a separated part without a complete free-body diagram.\u003C\u002Fli>\u003C\u002Ful>",174,[],{"id":236,"parent_id":186,"code":70,"slug":237,"name":238,"seo_title":239,"seo_description":240,"seo_text":241,"content":242,"locale":8,"uk_topic_id":243,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":237,"children":244},190,"distributed-loads-resultants","Distributed Loads and Their Resultants","Distributed Load Resultant — How to Find Magnitude & Location","Replace uniform, triangular, trapezoidal, or variable distributed loads with an equivalent resultant using the load-diagram area and centroid location.","This topic covers distributed loads and their replacement by equivalent concentrated forces. It explains how to determine the magnitude and point of application of the resultant for common load distributions.","\u003Cp>A \u003Cstrong>distributed load\u003C\u002Fstrong> acts continuously over a length or area instead of being applied at one single point. In beam and statics problems, a load distributed along a length is usually described by its intensity \\(q\\), measured in force per unit length, for example N\u002Fm or kN\u002Fm.\u003C\u002Fp>\u003Cp>Distributed loads are commonly used to model effects such as the weight of a beam, pressure acting along a member, or many closely spaced forces that are more conveniently represented as one continuous load.\u003C\u002Fp>\u003Ch2>Types of distributed loads\u003C\u002Fh2>\u003Cp>The load intensity may remain constant or vary from point to point. The most common cases are:\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>Uniformly distributed load (UDL):\u003C\u002Fstrong> the intensity is constant over the loaded length. Its load diagram is a rectangle.\u003C\u002Fli>\u003Cli>\u003Cstrong>Triangular distributed load:\u003C\u002Fstrong> the intensity changes linearly from zero to a maximum value, or from a maximum value to zero.\u003C\u002Fli>\u003Cli>\u003Cstrong>Trapezoidal distributed load:\u003C\u002Fstrong> the intensity changes linearly between two nonzero values.\u003C\u002Fli>\u003Cli>\u003Cstrong>General nonuniform load:\u003C\u002Fstrong> the intensity varies according to a function \\(q(x)\\) and may have a curved load diagram.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>In equilibrium calculations, it is often convenient to replace a distributed load by a single \u003Cstrong>equivalent concentrated force\u003C\u002Fstrong>, called its resultant. The replacement is equivalent when it produces the same total force and the same moment.\u003C\u002Fp>\u003Ch2>Uniform load\u003C\u002Fh2>\u003Cp>Start with the simplest case. If a constant load intensity \\(q\\) acts over a length \\(L\\), its resultant is\u003C\u002Fp>\u003Cp>\\[R=qL.\\]\u003C\u002Fp>\u003Cp>The resultant acts at the middle of the loaded length:\u003C\u002Fp>\u003Cp>\\[x_R=\\frac{L}{2}.\\]\u003C\u002Fp>\u003Cp>This follows directly from the rectangular load diagram: its area is \\(qL\\), and its centroid is at the center.\u003C\u002Fp>\u003Ch2>Triangular load\u003C\u002Fh2>\u003Cp>For a load that changes linearly from zero to \\(q_{max}\\), the load diagram is a triangle. Its resultant equals the area of that triangle:\u003C\u002Fp>\u003Cp>\\[R=\\frac{q_{max}L}{2}.\\]\u003C\u002Fp>\u003Cp>The resultant passes through the centroid of the triangle: \\(2L\u002F3\\) from the zero-intensity end, or \\(L\u002F3\\) from the maximum-intensity end.\u003C\u002Fp>\u003Ch2>Trapezoidal load\u003C\u002Fh2>\u003Cp>A trapezoidal load is often easiest to understand by splitting it into two simple loads: a uniform rectangular part and a triangular part. Find the resultant of each part, place each force at the centroid of its diagram, and then combine the forces and their moments.\u003C\u002Fp>\u003Ch2>General rule: area and centroid\u003C\u002Fh2>\u003Cp>The previous cases reveal a useful geometric rule: the magnitude of the resultant equals the \u003Cstrong>area under the load-intensity diagram\u003C\u002Fstrong>, and its line of action passes through the \u003Cstrong>centroid of that area\u003C\u002Fstrong>.\u003C\u002Fp>{{chunk:distributed-load-resultant-builder}}\u003Cp>For rectangles, triangles, and combinations of simple shapes, this rule usually lets you solve the problem without integration.\u003C\u002Fp>\u003Ch2>General nonuniform load\u003C\u002Fh2>\u003Cp>When the load varies according to an arbitrary function \\(q(x)\\), the same area-and-centroid idea is written mathematically using integration. For a load acting from \\(x=a\\) to \\(x=b\\),\u003C\u002Fp>\u003Cp>\\[R=\\int_a^b q(x)\\,dx.\\]\u003C\u002Fp>\u003Cp>The location of the resultant follows from equality of moments:\u003C\u002Fp>\u003Cp>\\[x_R=\\frac{\\int_a^b xq(x)\\,dx}{\\int_a^b q(x)\\,dx}.\\]\u003C\u002Fp>\u003Cp>Thus, the integral formulas are not a separate rule: they are the general form of the same area-and-centroid method used for rectangular, triangular, and trapezoidal load diagrams.\u003C\u002Fp>\u003Ch2>Sign-changing loads\u003C\u002Fh2>\u003Cp>If \\(q(x)\\) changes direction, positive and negative parts of the load diagram must be treated algebraically using the chosen sign convention. The expression for \\(x_R\\) applies when \\(R\\ne0\\). If the net resultant force is zero, the distributed system may instead reduce to a pure couple.\u003C\u002Fp>",175,[],{"id":246,"parent_id":101,"code":70,"slug":247,"name":248,"seo_title":249,"seo_description":250,"seo_text":251,"content":252,"locale":8,"uk_topic_id":253,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":247,"children":254},276,"three-dimensional-statics","Three-Dimensional Statics","Three-Dimensional Statics and Equilibrium","Three-dimensional force systems, equilibrium equations, centers of parallel forces, and centers of gravity.","This section covers three-dimensional force systems, their equilibrium conditions, and the determination of centers of parallel forces and centers of gravity.","\u003Cp>\u003Cstrong>Three-dimensional statics\u003C\u002Fstrong> extends the methods of statics to spatial force systems. Forces and moments may have three components, so complete equilibrium requires both translational and rotational effects to be considered about the three coordinate axes.\u003C\u002Fp>\u003Ch2>Spatial force systems\u003C\u002Fh2>\u003Cp>A general three-dimensional force system can be characterized by a resultant force and a resultant moment about a selected point. For equilibrium, both quantities must vanish.\u003C\u002Fp>\u003Ch2>Equilibrium equations\u003C\u002Fh2>\u003Cp>In the general case there are six scalar equilibrium conditions: three force-component equations and three moment equations. A suitable choice of axes and moment centers can simplify the calculation considerably.\u003C\u002Fp>\u003Ch2>Centers of parallel forces and gravity\u003C\u002Fh2>\u003Cp>Parallel force systems form an important special case. Their resultant passes through the center of parallel forces; for gravitational forces in a uniform field, the corresponding concept leads to the center of gravity.\u003C\u002Fp>",272,[255,265,275],{"id":256,"parent_id":246,"code":70,"slug":257,"name":258,"seo_title":259,"seo_description":260,"seo_text":261,"content":262,"locale":8,"uk_topic_id":263,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":257,"children":264},191,"three-dimensional-force-systems","Three-Dimensional Force Systems","Three-Dimensional Force Systems — Statics","Three-dimensional force systems in statics: force components, moments about axes, resultant force vector and resultant moment in 3D problems.","This topic extends the main concepts of statics to three-dimensional force systems. It covers force components in space, moments and system characteristics required for spatial equilibrium analysis.","\u003Cp>A \u003Cstrong>three-dimensional force system\u003C\u002Fstrong> is one whose force lines of action are not confined to a single plane. Its analytical description requires three coordinate axes, vector moments, and the rules of the vector cross product.\u003C\u002Fp>\u003Ch2>Force in three dimensions\u003C\u002Fh2>\u003Cp>In Cartesian coordinates, a force is written as $\\vec F=F_x\\vec i+F_y\\vec j+F_z\\vec k$. The three components uniquely define the force vector, and their signs determine the component directions along the coordinate axes.\u003C\u002Fp>{{chunk:statics-force-magnitude-3d}}\u003Ch2>Direction cosines\u003C\u002Fh2>\u003Cp>If $\\alpha$, $\\beta$, and $\\gamma$ are the angles between the force vector and the positive $x$, $y$, and $z$ axes, then $F_x=F\\cos\\alpha$, $F_y=F\\cos\\beta$, and $F_z=F\\cos\\gamma$. The direction cosines satisfy $\\cos^2\\alpha+\\cos^2\\beta+\\cos^2\\gamma=1$.\u003C\u002Fp>\u003Ch2>Force along a specified line\u003C\u002Fh2>\u003Cp>If a force is directed from point $A(x_A,y_A,z_A)$ toward point $B(x_B,y_B,z_B)$, first form the vector $\\overrightarrow{AB}$. Normalize it to obtain $\\vec e_{AB}=\\overrightarrow{AB}\u002F|\\overrightarrow{AB}|$, then write the force as $\\vec F=F\\vec e_{AB}$.\u003C\u002Fp>\u003Ch2>Moment of a force about a point\u003C\u002Fh2>\u003Cp>The moment of a force $\\vec F$ applied at point $A$ about point $O$ is $\\vec M_O=\\vec r\\times\\vec F$, where $\\vec r=\\overrightarrow{OA}$. The moment vector is perpendicular to the plane formed by $\\vec r$ and $\\vec F$, with its direction given by the right-hand rule.\u003C\u002Fp>\u003Ch2>Moment of a force about an axis\u003C\u002Fh2>\u003Cp>The moment about a coordinate or arbitrary axis equals the projection of the force-moment vector onto that axis. If $\\vec e$ is a unit vector along the axis, then $M_{axis}=\\vec e\\cdot(\\vec r\\times\\vec F)$.\u003C\u002Fp>\u003Ch2>Resultant force vector and resultant moment\u003C\u002Fh2>\u003Cp>For a three-dimensional force system, the resultant force vector is $\\vec R=\\sum\\vec F_i$, and the resultant moment about point $O$ is $\\vec M_O=\\sum(\\vec r_i\\times\\vec F_i)+\\sum\\vec M_j$. In a 3D problem, each of these vectors has three Cartesian components.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose a force has components $F_x=3$ kN, $F_y=4$ kN, and $F_z=12$ kN. Its magnitude is $F=\\sqrt{3^2+4^2+12^2}=13$ kN. The force vector can be written as $\\vec F=(3\\vec i+4\\vec j+12\\vec k)$ kN.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using a planar moment sign convention for a three-dimensional moment vector;\u003C\u002Fli>\u003Cli>failing to normalize the vector between two points before multiplying it by a specified force magnitude;\u003C\u002Fli>\u003Cli>confusing the moment about a point with its projection onto an axis;\u003C\u002Fli>\u003Cli>reversing the order in $\\vec r\\times\\vec F$, which reverses the moment direction;\u003C\u002Fli>\u003Cli>omitting one of the three force or moment components.\u003C\u002Fli>\u003C\u002Ful>",176,[],{"id":266,"parent_id":246,"code":70,"slug":267,"name":268,"seo_title":269,"seo_description":270,"seo_text":271,"content":272,"locale":8,"uk_topic_id":273,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":267,"children":274},192,"equilibrium-conditions-three-dimensional-force-systems","Equilibrium Conditions for Three-Dimensional Force Systems","Equilibrium of Three-Dimensional Force Systems","Equilibrium equations for three-dimensional force systems using sums of force components and moments about coordinate axes in statics problems.","This topic organizes the equilibrium conditions for general three-dimensional force systems. It covers force-component and moment equations used to determine reactions and unknown loads in spatial statics problems.","\u003Cp>\u003Cstrong>Equilibrium of a rigid body in three dimensions\u003C\u002Fstrong> requires the simultaneous absence of translational and rotational effects of the force system. In vector form, this means $\\sum\\vec F=0$ and $\\sum\\vec M_O=0$. Projection onto the three coordinate axes gives six scalar conditions.\u003C\u002Fp>\u003Ch2>General equilibrium equations\u003C\u002Fh2>{{chunk:statics-spatial-equilibrium-equations}}\u003Cp>For a single rigid body, these six independent equations are the basic tool for determining unknown reactions of three-dimensional supports and unknown loads.\u003C\u002Fp>\u003Ch2>Force components\u003C\u002Fh2>\u003Cp>Before writing the equations, each three-dimensional force is expressed through its components $F_x$, $F_y$, and $F_z$. If a force direction is specified by two points, first determine the unit vector along its line of action and then multiply it by the force magnitude.\u003C\u002Fp>\u003Ch2>Moment components\u003C\u002Fh2>\u003Cp>The moment of a force about point $O$ is calculated as $\\vec M_O=\\vec r\\times\\vec F$. The resulting moment vector is then resolved into $M_x$, $M_y$, and $M_z$. Applied couple moments are added directly to the corresponding components of the total moment.\u003C\u002Fp>\u003Ch2>Choosing the moment reference point\u003C\u002Fh2>\u003Cp>As in planar statics, the moment reference point can be selected for convenience. If the lines of action of several unknown forces pass through the chosen point, their moments about that point vanish. In three dimensions, however, all three components of the moment vector must be handled consistently.\u003C\u002Fp>\u003Ch2>Reactions of three-dimensional constraints\u003C\u002Fh2>\u003Cp>The number of unknown reactions depends on the motions prevented by the constraint. For example, an idealized fixed support in three dimensions can transmit three force components and three moment components. The specific support model must be established before equilibrium equations are written.\u003C\u002Fp>\u003Ch2>Special force systems\u003C\u002Fh2>\u003Cp>For a concurrent three-dimensional force system, all force lines of action pass through one point, so the three force-component equations are the independent equilibrium conditions. Other special geometries may reduce the number of required equations, but the general set of six equations remains the fundamental form.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose three forces acting at a point have components $\\vec F_1=(3,0,-4)$ kN, $\\vec F_2=(-3,5,0)$ kN, and $\\vec F_3=(0,-5,4)$ kN. The sums of all three components are zero, so this concurrent force system is balanced. For a general nonconcurrent system, the three moment equations must also be checked.\u003C\u002Fp>\u003Ch2>Solution procedure\u003C\u002Fh2>\u003Col>\u003Cli>construct a three-dimensional free-body diagram;\u003C\u002Fli>\u003Cli>replace all constraints by the appropriate reactions;\u003C\u002Fli>\u003Cli>express forces in Cartesian components;\u003C\u002Fli>\u003Cli>choose a reference point $O$ and calculate force moments;\u003C\u002Fli>\u003Cli>write $\\sum F_x=\\sum F_y=\\sum F_z=0$;\u003C\u002Fli>\u003Cli>write $\\sum M_x=\\sum M_y=\\sum M_z=0$;\u003C\u002Fli>\u003Cli>solve the equations and check the physical meaning of the calculated reactions.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>omitting one of the three force or moment components;\u003C\u002Fli>\u003Cli>determining the force unit vector in the wrong direction;\u003C\u002Fli>\u003Cli>confusing a moment about a point with a moment about an axis;\u003C\u002Fli>\u003Cli>reversing the order of the cross product $\\vec r\\times\\vec F$;\u003C\u002Fli>\u003Cli>using six equations without first modeling the three-dimensional constraints correctly.\u003C\u002Fli>\u003C\u002Ful>",177,[],{"id":276,"parent_id":246,"code":70,"slug":277,"name":278,"seo_title":278,"seo_description":279,"seo_text":280,"content":281,"locale":8,"uk_topic_id":282,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":277,"children":283},193,"center-parallel-forces-center-gravity","Center of Parallel Forces and Center of Gravity","Center of parallel forces and center of gravity: coordinates, symmetry and methods for locating the center of gravity of bodies and plane areas.","This topic explains the center of parallel forces and center of gravity. It covers coordinate methods, use of symmetry and applications to composite bodies and plane areas in engineering statics.","\u003Cp>The \u003Cstrong>center of parallel forces\u003C\u002Fstrong> is the point through which the line of action of the resultant of a parallel-force system passes when the relative positions of the force application points remain unchanged. The center of gravity is an important physical application of this concept.\u003C\u002Fp>\u003Ch2>Coordinates of the center of parallel forces\u003C\u002Fh2>\u003Cp>For parallel forces $F_i$ acting along a common direction, the center coordinates follow from moment equivalence. For example, $x_C=\\sum F_i x_i\u002F\\sum F_i$ and $y_C=\\sum F_i y_i\u002F\\sum F_i$, provided the denominator is nonzero. Oppositely directed forces are included with their algebraic signs.\u003C\u002Fp>\u003Ch2>Center of gravity\u003C\u002Fh2>\u003Cp>The gravitational forces acting on the particles of a body are effectively parallel in a uniform gravitational field. The point of application of their resultant is the \u003Cstrong>center of gravity\u003C\u002Fstrong>. For a homogeneous body it coincides with the center of mass, and for a homogeneous thin plate it coincides with the geometric centroid of its area.\u003C\u002Fp>\u003Ch2>Composite plane areas\u003C\u002Fh2>{{chunk:statics-composite-area-centroid}}\u003Cp>This method is especially convenient for sections that can be decomposed into rectangles, triangles, circles, and other simple regions. Holes and cutouts are assigned negative areas.\u003C\u002Fp>\u003Ch2>Using symmetry\u003C\u002Fh2>\u003Cp>If a homogeneous area has an axis of symmetry, its centroid lies on that axis. With two axes of symmetry, the centroid is at their intersection. Symmetry can therefore determine one or both coordinates without calculation.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>An area consists of two rectangles with $A_1=6$ cm² and $A_2=4$ cm² whose centroid coordinates are $x_1=2$ cm and $x_2=7$ cm. Then $x_C=(6\\cdot2+4\\cdot7)\u002F(6+4)=4$ cm.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>mixing coordinates measured from different origins;\u003C\u002Fli>\u003Cli>failing to treat holes as negative areas;\u003C\u002Fli>\u003Cli>assuming the center of gravity of a nonhomogeneous body is its geometric volume centroid;\u003C\u002Fli>\u003Cli>using an arithmetic mean of coordinates instead of a weighted mean.\u003C\u002Fli>\u003C\u002Ful>",178,[],{"id":285,"parent_id":101,"code":70,"slug":286,"name":287,"seo_title":288,"seo_description":289,"seo_text":290,"content":291,"locale":8,"uk_topic_id":292,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":286,"children":293},277,"friction","Friction","Friction in Statics — Sliding and Rolling","Coulomb sliding friction, equilibrium with friction, and rolling resistance in engineering statics.","This section covers dry sliding friction, Coulomb's laws, equilibrium in the presence of friction, and rolling resistance.","\u003Cp>\u003Cstrong>Friction\u003C\u002Fstrong> accounts for resistance forces that arise at contacting surfaces and oppose relative sliding or rolling. Unlike an ideal smooth contact, a real contact can transmit tangential forces.\u003C\u002Fp>\u003Ch2>Sliding friction\u003C\u002Fh2>\u003Cp>Dry friction acts against the tendency of relative sliding. During static contact its magnitude adjusts to the applied loading up to a limiting value related to the normal reaction and the coefficient of friction.\u003C\u002Fp>\u003Ch2>Equilibrium with friction\u003C\u002Fh2>\u003Cp>In equilibrium problems, the direction of friction is determined from the impending or possible relative motion. It is important to distinguish ordinary equilibrium from limiting equilibrium, where slipping is about to begin.\u003C\u002Fp>\u003Ch2>Rolling resistance\u003C\u002Fh2>\u003Cp>Rolling resistance is associated with deformation in the contact region and displacement of the resultant contact reaction. Its mechanical model differs from the Coulomb model of dry sliding friction.\u003C\u002Fp>",273,[294,304,314],{"id":295,"parent_id":285,"code":70,"slug":296,"name":297,"seo_title":298,"seo_description":299,"seo_text":300,"content":301,"locale":8,"uk_topic_id":302,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":296,"children":303},194,"sliding-friction-coulombs-laws","Sliding Friction. Coulomb's Laws","Sliding Friction — Coulomb Law, F ≤ μN & Friction Angle","Learn dry sliding friction in statics: F ≤ μN, limiting friction, coefficient μ, friction angle and cone, and Coulomb’s laws for rough-contact equilibrium.","This topic covers dry sliding friction and Coulomb's laws. It introduces friction force and coefficient, limiting friction, the angle and cone of friction used to analyze equilibrium with rough contacts.","\u003Cp>\u003Cstrong>Dry sliding friction\u003C\u002Fstrong> arises at the contact between two rough bodies and opposes their relative sliding or tendency to slide. In the simplest Coulomb model, the tangential friction force is related to the normal contact reaction.\u003C\u002Fp>\u003Ch2>Friction during static equilibrium\u003C\u002Fh2>\u003Cp>As long as the body does not slide, the friction force adjusts to the external loading within the range required for equilibrium. Therefore, $F_{fr}=\\mu N$ must not automatically be used for every static condition.\u003C\u002Fp>\u003Ch2>Limiting friction\u003C\u002Fh2>{{chunk:statics-coulomb-friction-limit}}\u003Cp>The equality applies at impending sliding. The friction force acts opposite to the direction of the impending relative motion.\u003C\u002Fp>\u003Ch2>Coefficient of friction\u003C\u002Fh2>\u003Cp>The dry-friction coefficient $\\mu$ is a dimensionless property of the contacting pair within the adopted model. Its value depends on the materials and surface condition. In the elementary Coulomb model, the limiting friction force is proportional to the normal reaction.\u003C\u002Fp>\u003Ch2>Angle of friction\u003C\u002Fh2>\u003Cp>The total reaction of a rough surface is the vector sum of the normal reaction $N$ and the friction force. At impending sliding it is inclined from the normal by the angle of friction $\\varphi$, for which $\\tan\\varphi=\\mu$.\u003C\u002Fp>\u003Ch2>Cone of friction\u003C\u002Fh2>\u003Cp>In a three-dimensional problem, the possible direction of the limiting total reaction forms a cone around the contact normal. If the resultant contact reaction lies inside the friction cone, sticking equilibrium may be possible; a reaction on the cone surface corresponds to impending sliding.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A block is pressed against a horizontal surface with normal reaction $N=500$ N and coefficient of friction $\\mu=0.30$. The maximum static-friction force is $F_{fr,max}=0.30\\cdot500=150$ N. If the applied horizontal force is only 80 N and there are no other horizontal forces, the equilibrium friction force is 80 N, not 150 N.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>always setting $F_{fr}=\\mu N$ instead of using the static-friction inequality;\u003C\u002Fli>\u003Cli>assigning the friction direction without considering impending motion;\u003C\u002Fli>\u003Cli>confusing the normal reaction with the total rough-contact reaction;\u003C\u002Fli>\u003Cli>treating the coefficient of friction as a dimensional quantity;\u003C\u002Fli>\u003Cli>failing to check whether the friction force found from equilibrium exceeds its limiting value.\u003C\u002Fli>\u003C\u002Ful>",179,[],{"id":305,"parent_id":285,"code":70,"slug":306,"name":307,"seo_title":308,"seo_description":309,"seo_text":310,"content":311,"locale":8,"uk_topic_id":312,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":306,"children":313},195,"equilibrium-with-friction","Equilibrium with Friction","Equilibrium with Friction — How to Solve Statics Problems","Solve dry-friction equilibrium problems by choosing the friction direction, applying equilibrium equations, checking F ≤ μN, and identifying impending sliding.","This topic focuses on equilibrium problems involving dry friction. It explains how to select the friction-force direction, check limiting conditions and combine friction laws with statics equilibrium equations.","\u003Cp>In \u003Cstrong>equilibrium problems with friction\u003C\u002Fstrong>, the ordinary equations of statics are supplemented by dry-friction conditions. Unlike a smooth contact, a rough-surface reaction has both normal and tangential components.\u003C\u002Fp>\u003Ch2>Free-body diagram\u003C\u002Fh2>\u003Cp>It is convenient to resolve the rough-contact reaction into a normal reaction $N$ and a friction force $F_{fr}$. The normal component is perpendicular to the surface, while friction acts tangentially and opposes possible relative sliding.\u003C\u002Fp>\u003Ch2>Equilibrium check procedure\u003C\u002Fh2>{{chunk:statics-friction-equilibrium-check}}\u003Ch2>Inclined plane\u003C\u002Fh2>\u003Cp>For a body on an inclined plane, its weight resolves into $G\\sin\\alpha$ along the plane and $G\\cos\\alpha$ normal to it. With no other forces, rest requires $G\\sin\\alpha\\le\\mu G\\cos\\alpha$, or $\\tan\\alpha\\le\\mu$. The limiting inclination corresponds to the angle of friction.\u003C\u002Fp>\u003Ch2>Direction of friction\u003C\u002Fh2>\u003Cp>In an equilibrium problem, friction direction is determined from the tendency of relative motion that would occur without friction, not from actual motion. If an assumed friction direction is opposite to the required one, the calculated friction value will be negative.\u003C\u002Fp>\u003Ch2>Impending equilibrium\u003C\u002Fh2>\u003Cp>When a problem states that a body is on the verge of motion, is about to slide, or asks for a minimum or maximum force immediately before motion, the limiting condition $|F_{fr}|=\\mu N$ is generally used. Ordinary static equilibrium requires the inequality instead.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 1 kN block rests on a horizontal surface with $\\mu=0.25$. A horizontal force of 180 N requires 180 N of friction. Since $F_{fr,max}=0.25\\cdot1000=250$ N, equilibrium is possible. If the horizontal force rises to 300 N, static friction can no longer maintain rest.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>immediately setting $F_{fr}=\\mu N$ without an impending-motion condition;\u003C\u002Fli>\u003Cli>failing to check the calculated friction against its limiting value;\u003C\u002Fli>\u003Cli>directing friction with, rather than against, the tendency to slide;\u003C\u002Fli>\u003Cli>ignoring changes in the normal reaction caused by inclined external forces;\u003C\u002Fli>\u003Cli>using the dry-friction model when the problem specifies a different contact model.\u003C\u002Fli>\u003C\u002Ful>",180,[],{"id":315,"parent_id":285,"code":70,"slug":316,"name":317,"seo_title":318,"seo_description":319,"seo_text":320,"content":321,"locale":8,"uk_topic_id":322,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":316,"children":323},196,"rolling-resistance","Rolling Resistance","Rolling Resistance — Engineering Statics","Rolling resistance in engineering statics: physical origin, resistance moment and equilibrium conditions for wheels and cylindrical bodies.","This topic explains rolling resistance and how it differs from sliding friction. It introduces the resistance moment and its use in equilibrium problems involving wheels and cylindrical bodies.","\u003Cp>\u003Cstrong>Rolling resistance\u003C\u002Fstrong> occurs when a wheel, cylinder, or other rounded body rolls over a real surface. Unlike ideal point contact, the contact region deforms, so the resultant normal reaction may be offset from the geometric vertical through the body center and produce a moment opposing rolling.\u003C\u002Fp>\u003Ch2>Resistance-moment model\u003C\u002Fh2>{{chunk:statics-rolling-resistance-moment}}\u003Cp>The coefficient $\\delta$ in this model has units of length, such as metres or millimetres. This is fundamentally different from the dimensionless sliding-friction coefficient $\\mu$.\u003C\u002Fp>\u003Ch2>Physical meaning\u003C\u002Fh2>\u003Cp>For a perfectly rigid wheel on a perfectly rigid surface, idealized rolling resistance is absent. In a real contact, deformation of the wheel and supporting surface, material hysteresis, and other losses create resistance. A simple statics model replaces these effects by a resistance moment or an equivalent offset of the normal reaction.\u003C\u002Fp>\u003Ch2>Condition for the onset of rolling\u003C\u002Fh2>\u003Cp>If external forces create a driving moment about the wheel center, rolling begins when that moment exceeds the maximum available rolling-resistance moment in the adopted model. The no-slip condition must be checked separately when dry friction is also included in the problem.\u003C\u002Fp>\u003Ch2>Rolling versus sliding\u003C\u002Fh2>\u003Cp>Rolling resistance and sliding friction are different phenomena. Contact friction may be required for rolling without slipping, whereas rolling resistance describes losses that oppose the rolling motion itself. The coefficients $\\mu$ and $\\delta$ are therefore not interchangeable.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a wheel with normal reaction $N=2$ kN and rolling-resistance coefficient $\\delta=5$ mm, $M_{rr}=2000\\cdot0.005=10$ N·m. In the simplified model, the applied driving moment must overcome this resistance for rolling to begin.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating $\\delta$ as dimensionless;\u003C\u002Fli>\u003Cli>confusing rolling resistance with the sliding-friction force $\\mu N$;\u003C\u002Fli>\u003Cli>failing to convert millimetres to metres when calculating a moment in N·m;\u003C\u002Fli>\u003Cli>ignoring the possibility of slipping while analyzing rolling;\u003C\u002Fli>\u003Cli>using the simple model $M_{rr}=N\\delta$ without checking which rolling-resistance model the problem specifies.\u003C\u002Fli>\u003C\u002Ful>",181,[],{"id":325,"parent_id":91,"code":70,"slug":326,"name":327,"seo_title":328,"seo_description":329,"seo_text":330,"content":331,"locale":8,"uk_topic_id":332,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":326,"children":333},137,"kinematics","Kinematics","Kinematics — Motion, Velocity and Acceleration | Mechanics","Engineering kinematics: position, trajectory, velocity and acceleration of particles and rigid bodies, with formulas and practical problems.","Kinematics describes mechanical motion without considering the forces that cause it. This section covers methods of describing motion, trajectories, particle velocity and acceleration, and fundamental types of rigid-body motion, supported by practical engineering problems.","\u003Cp>Kinematics describes mechanical motion without considering the forces that cause it. Its fundamental quantities include position, displacement, trajectory, velocity and acceleration.\u003C\u002Fp>\u003Cp>The material is organized around problem solving, from finding particle motion parameters to analysing translation, rotation and plane motion of rigid bodies.\u003C\u002Fp>",134,[334,383,431,466,494],{"id":335,"parent_id":325,"code":70,"slug":336,"name":337,"seo_title":338,"seo_description":339,"seo_text":340,"content":341,"locale":8,"uk_topic_id":342,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":336,"children":343},284,"particle-kinematics","Particle Kinematics","Particle Kinematics — Position, Velocity & Acceleration","Learn particle kinematics through position and trajectory, methods of describing motion, velocity, acceleration, and common special cases of particle motion.","This section covers methods of describing particle motion, particle velocity and acceleration, and important special cases of motion.","\u003Cp>\u003Cstrong>Particle kinematics\u003C\u002Fstrong> studies how the motion of a point is described without considering the forces that cause it. The main objective is to determine position, trajectory, velocity, and acceleration from a given law of motion.\u003C\u002Fp>\u003Ch2>Describing motion\u003C\u002Fh2>\u003Cp>Motion may be specified by Cartesian coordinates, a position vector, or natural coordinates along a trajectory. The most convenient representation depends on the geometry of the problem and the quantities to be determined.\u003C\u002Fp>\u003Ch2>Velocity and acceleration\u003C\u002Fh2>\u003Cp>Velocity describes the rate of change of position and is tangent to the trajectory. Acceleration describes the rate of change of the velocity vector and can be resolved into components associated with changes in speed and direction.\u003C\u002Fp>\u003Ch2>Typical laws of motion\u003C\u002Fh2>\u003Cp>Rectilinear, uniform, uniformly accelerated, and curvilinear motions provide basic models for understanding the relationships among position, velocity, and acceleration.\u003C\u002Fp>",278,[344,354,364,374],{"id":345,"parent_id":335,"code":70,"slug":346,"name":347,"seo_title":348,"seo_description":349,"seo_text":350,"content":351,"locale":8,"uk_topic_id":352,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":346,"children":353},212,"particle-kinematics-methods-describing-motion","Particle Kinematics. Methods of Describing Motion","Particle Kinematics and Methods of Describing Motion","Particle kinematics: vector, Cartesian and path-coordinate descriptions of motion, trajectory and equations of motion.","This topic introduces particle kinematics and the main methods of describing motion: vector, Cartesian-coordinate and path-coordinate descriptions, including trajectories and equations of motion.","\u003Cp>\u003Cstrong>Particle kinematics\u003C\u002Fstrong> describes the motion of a particle without considering the forces that cause it. The basic objective is to specify particle position as a function of time and determine its trajectory, velocity, and acceleration relative to a selected reference frame.\u003C\u002Fp>\u003Ch2>Reference frame and equations of motion\u003C\u002Fh2>\u003Cp>A motion description requires a reference body, an associated coordinate system, and a measure of time. The equations of motion must determine the particle position at any instant within the interval being studied.\u003C\u002Fp>\u003Ch2>Main methods of describing motion\u003C\u002Fh2>{{chunk:kinematics-particle-motion-description}}\u003Ch2>Vector description\u003C\u002Fh2>\u003Cp>The position vector $\\vec r(t)$ extends from the coordinate origin to the moving particle. As time varies, the endpoints of $\\vec r(t)$ trace the trajectory. In a Cartesian basis, $\\vec r=x\\vec i+y\\vec j+z\\vec k$.\u003C\u002Fp>\u003Ch2>Cartesian-coordinate description\u003C\u002Fh2>\u003Cp>The relations $x=x(t)$, $y=y(t)$, and $z=z(t)$ are the kinematic equations of motion. An equation of the trajectory can be obtained by eliminating time from these relations. Two coordinates are sufficient for planar motion.\u003C\u002Fp>\u003Ch2>Path-coordinate description\u003C\u002Fh2>\u003Cp>If the trajectory is already known, choose an origin $O_1$ on the curve, assign a positive direction, and specify $s=s(t)$. The sign of $s$ locates the particle relative to the path-coordinate origin, while the sign of $\\dot s$ indicates its direction of motion along the trajectory.\u003C\u002Fp>\u003Ch2>Trajectory, distance traveled, and displacement\u003C\u002Fh2>\u003Cp>The \u003Cstrong>trajectory\u003C\u002Fstrong> is the geometric locus of successive particle positions. \u003Cstrong>Distance traveled\u003C\u002Fstrong> is the length accumulated along the trajectory and does not decrease as the particle moves. \u003Cstrong>Displacement\u003C\u002Fstrong> is the vector from the initial to the final position; its magnitude is generally not equal to the distance traveled.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose planar motion is given by $x=2t$ and $y=t^2$ in metres. Eliminating time with $t=x\u002F2$ gives the trajectory $y=x^2\u002F4$, a parabola. The parametric equations also specify where the particle is on that trajectory at every instant.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing equations of motion with the equation of the trajectory;\u003C\u002Fli>\u003Cli>treating distance traveled as the magnitude of displacement for arbitrary curved motion;\u003C\u002Fli>\u003Cli>eliminating time and losing information about motion direction or the valid time interval;\u003C\u002Fli>\u003Cli>using a path coordinate without specifying the trajectory, origin, and positive direction.\u003C\u002Fli>\u003C\u002Ful>",197,[],{"id":355,"parent_id":335,"code":70,"slug":356,"name":357,"seo_title":358,"seo_description":359,"seo_text":360,"content":361,"locale":8,"uk_topic_id":362,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":356,"children":363},213,"particle-velocity","Particle Velocity","Particle Velocity — v = dr\u002Fdt, Components & Path Coordinates","Learn how to calculate particle velocity from v = dr\u002Fdt, Cartesian components, magnitude and direction, and the path-coordinate relation v = ds\u002Fdt.","This topic explains how to determine particle velocity from vector, Cartesian-coordinate and path-coordinate descriptions of motion and discusses the geometric meaning of the velocity vector.","\u003Cp>\u003Cstrong>Particle velocity\u003C\u002Fstrong> describes the rate of change of particle position and its instantaneous direction of motion. Average velocity describes a finite change of position over a time interval, while instantaneous velocity is obtained by a limiting process and equals the time derivative of the position vector.\u003C\u002Fp>\u003Ch2>Velocity vector\u003C\u002Fh2>\u003Cp>For $\\vec r=\\vec r(t)$, instantaneous velocity is $\\vec v=d\\vec r\u002Fdt$. The vector $\\vec v$ is tangent to the trajectory and points in the direction of particle motion.\u003C\u002Fp>\u003Ch2>Cartesian-coordinate description\u003C\u002Fh2>{{chunk:kinematics-particle-velocity-cartesian}}\u003Cp>The signs of $v_x$, $v_y$, and $v_z$ indicate how the corresponding coordinates are changing. A zero value of one component does not imply that the particle is at rest.\u003C\u002Fp>\u003Ch2>Path-coordinate description\u003C\u002Fh2>\u003Cp>If position is specified by a path coordinate $s=s(t)$, the algebraic velocity along the trajectory is $v_s=ds\u002Fdt$. In vector form, $\\vec v=(ds\u002Fdt)\\vec\\tau$, where $\\vec\\tau$ is the unit tangent vector in the positive $s$ direction.\u003C\u002Fp>\u003Ch2>Average and instantaneous velocity\u003C\u002Fh2>\u003Cp>The average vector velocity over $\\Delta t$ is $\\Delta\\vec r\u002F\\Delta t$. It depends on displacement rather than the length of the path traveled. As $\\Delta t\\to0$, average velocity approaches instantaneous velocity.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For $x=3t^2$ and $y=4t$, the components are $v_x=6t$ and $v_y=4$. At $t=1$ s, the speed is $v=\\sqrt{6^2+4^2}=\\sqrt{52}\\approx7.21$ m\u002Fs. The velocity direction coincides with the tangent to the trajectory at that point.\u003C\u002Fp>\u003Ch2>Stopping and reversal\u003C\u002Fh2>\u003Cp>A particle is instantaneously at rest only when its entire velocity vector is zero. In rectilinear motion, a change in the sign of algebraic velocity indicates reversal of direction; an instant with $v=0$ should be interpreted together with the equation of motion.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing average vector velocity with distance traveled divided by time;\u003C\u002Fli>\u003Cli>calculating speed by adding the magnitudes of velocity components;\u003C\u002Fli>\u003Cli>ignoring component signs when determining direction;\u003C\u002Fli>\u003Cli>assuming that $v_x=0$ means the particle is completely at rest.\u003C\u002Fli>\u003C\u002Ful>",198,[],{"id":365,"parent_id":335,"code":70,"slug":366,"name":367,"seo_title":368,"seo_description":369,"seo_text":370,"content":371,"locale":8,"uk_topic_id":372,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":366,"children":373},214,"particle-acceleration","Particle Acceleration","Particle Acceleration — Tangential & Normal Components, Formulas","Calculate particle acceleration from a = dv\u002Fdt, Cartesian components, tangential acceleration at and normal acceleration an = v²\u002Fρ for curvilinear motion.","This topic treats particle acceleration as the time derivative of velocity, including Cartesian components and tangential-normal components for curvilinear motion.","\u003Cp>\u003Cstrong>Particle acceleration\u003C\u002Fstrong> describes the rate of change of the velocity vector. Acceleration can arise from a change in speed, a change in velocity direction, or both.\u003C\u002Fp>\u003Ch2>Acceleration vector\u003C\u002Fh2>\u003Cp>Instantaneous acceleration is $\\vec a=d\\vec v\u002Fdt=d^2\\vec r\u002Fdt^2$. Unlike velocity, the acceleration vector is not generally tangent to the trajectory.\u003C\u002Fp>\u003Ch2>Cartesian and path components\u003C\u002Fh2>{{chunk:kinematics-particle-acceleration-components}}\u003Ch2>Tangential acceleration\u003C\u002Fh2>\u003Cp>The component $a_\\tau=dv\u002Fdt$ describes the change in speed. When the tangential acceleration points with the velocity, speed increases; when it points opposite the velocity, speed decreases.\u003C\u002Fp>\u003Ch2>Normal acceleration\u003C\u002Fh2>\u003Cp>The component $a_n=v^2\u002F\\rho$ results from a change in velocity direction. It always points toward the center of curvature and vanishes for rectilinear motion, for which the radius of curvature is formally infinite.\u003C\u002Fp>\u003Ch2>Total acceleration\u003C\u002Fh2>\u003Cp>Because tangential and normal components are perpendicular, the acceleration magnitude is $a=\\sqrt{a_\\tau^2+a_n^2}$. Its direction follows from $\\vec a=a_\\tau\\vec\\tau+a_n\\vec n$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A particle moves on a circle of radius $\\rho=2$ m at a speed of 6 m\u002Fs that is increasing at 3 m\u002Fs². Then $a_\\tau=3$ m\u002Fs², $a_n=6^2\u002F2=18$ m\u002Fs², and $a=\\sqrt{3^2+18^2}\\approx18.25$ m\u002Fs².\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>assuming acceleration is zero whenever speed is constant;\u003C\u002Fli>\u003Cli>directing normal acceleration along the tangent;\u003C\u002Fli>\u003Cli>confusing the radius of curvature with distance to an arbitrary coordinate origin;\u003C\u002Fli>\u003Cli>adding $a_\\tau$ and $a_n$ algebraically when calculating total acceleration magnitude.\u003C\u002Fli>\u003C\u002Ful>",199,[],{"id":375,"parent_id":335,"code":70,"slug":376,"name":377,"seo_title":377,"seo_description":378,"seo_text":379,"content":380,"locale":8,"uk_topic_id":381,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":376,"children":382},215,"special-cases-particle-motion","Special Cases of Particle Motion","Uniform and uniformly accelerated rectilinear motion, circular motion and basic kinematic relations for a particle.","This topic organizes common particle-motion laws, including uniform motion, uniformly accelerated rectilinear motion and circular motion, with the main relations among position, velocity and acceleration.","\u003Cp>Many kinematics problems reduce to several \u003Cstrong>standard cases of particle motion\u003C\u002Fstrong>. They are conveniently classified by trajectory shape and by how velocity changes.\u003C\u002Fp>\u003Ch2>Uniform rectilinear motion\u003C\u002Fh2>\u003Cp>If a particle moves along a straight line with constant algebraic velocity $v$, its coordinate follows $s=s_0+vt$. Acceleration is zero.\u003C\u002Fp>\u003Ch2>Uniformly accelerated rectilinear motion\u003C\u002Fh2>{{chunk:kinematics-uniformly-accelerated-motion}}\u003Cp>If $a$ and $v_0$ have the same sign, speed initially increases. If their signs are opposite, the particle may slow to rest and then reverse direction.\u003C\u002Fp>\u003Ch2>Free fall as a special case\u003C\u002Fh2>\u003Cp>If air resistance and variation of gravitational acceleration with altitude are neglected, vertical motion near Earth’s surface is uniformly accelerated with $\\vec g$ directed downward. Signs in the scalar equations depend on the selected positive vertical direction.\u003C\u002Fp>\u003Ch2>Uniform circular motion\u003C\u002Fh2>\u003Cp>For constant speed $v$ on a circle of radius $R$, tangential acceleration is zero but normal acceleration is not: $a_n=v^2\u002FR$. It points toward the circle center, so the velocity vector continuously changes direction.\u003C\u002Fp>\u003Ch2>Nonuniform circular motion\u003C\u002Fh2>\u003Cp>If speed changes, the particle has both tangential acceleration $a_\\tau=dv\u002Fdt$ and normal acceleration $a_n=v^2\u002FR$. Total acceleration is their vector sum.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A car moves along a straight line with $v_0=5$ m\u002Fs and constant acceleration $a=2$ m\u002Fs². After 4 s, its velocity is $v=5+2\\cdot4=13$ m\u002Fs and its displacement from the initial position is $5\\cdot4+2\\cdot4^2\u002F2=36$ m.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using constant-acceleration formulas when $a$ varies with time;\u003C\u002Fli>\u003Cli>assuming acceleration is zero in uniform circular motion;\u003C\u002Fli>\u003Cli>substituting $g$ without matching its sign to the chosen axis direction;\u003C\u002Fli>\u003Cli>confusing coordinate $s$ with distance traveled when the particle reverses direction.\u003C\u002Fli>\u003C\u002Ful>",200,[],{"id":384,"parent_id":325,"code":70,"slug":385,"name":386,"seo_title":387,"seo_description":388,"seo_text":389,"content":390,"locale":8,"uk_topic_id":391,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":385,"children":392},285,"basic-motions-of-a-rigid-body","Basic Motions of a Rigid Body","Translation and Fixed-Axis Rotation of a Rigid Body","Rigid-body translation, fixed-axis rotation, point velocities and accelerations, and transmission of rotational motion.","This section covers rigid-body translation, rotation about a fixed axis, kinematics of body points, and transmission of rotational motion.","\u003Cp>\u003Cstrong>Basic motions of a rigid body\u003C\u002Fstrong> are translation and rotation about a fixed axis. They are fundamental rigid-body kinematic models and also serve as components of more general motion.\u003C\u002Fp>\u003Ch2>Translation\u003C\u002Fh2>\u003Cp>In translation, every line fixed in the body remains parallel to its original direction. At any instant all points of the body have identical velocity vectors and identical acceleration vectors.\u003C\u002Fp>\u003Ch2>Fixed-axis rotation\u003C\u002Fh2>\u003Cp>The body configuration is described by an angular coordinate. Its time derivatives give angular velocity and angular acceleration, while the linear velocities and accelerations of body points depend on their distance from the axis.\u003C\u002Fp>\u003Ch2>Transmission of rotation\u003C\u002Fh2>\u003Cp>Mechanisms transmit rotational motion through gears, belts, friction drives, and other arrangements. Their kinematic relations connect the angular velocities of components through the geometry of the transmission.\u003C\u002Fp>",279,[393,402,411,421],{"id":394,"parent_id":384,"code":70,"slug":395,"name":396,"seo_title":396,"seo_description":397,"seo_text":398,"content":399,"locale":8,"uk_topic_id":400,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":395,"children":401},216,"translation-rigid-body","Translation of a Rigid Body","Rigid-body translation: trajectories, velocities and accelerations of points and the fundamental properties of translational motion.","This topic explains rigid-body translation and shows that at any instant all points of a translating rigid body have identical velocity and acceleration vectors.","\u003Cp>\u003Cstrong>Translation of a rigid body\u003C\u002Fstrong> is motion in which every line fixed in the body remains parallel to its initial orientation. The body therefore does not rotate relative to the selected reference frame.\u003C\u002Fp>\u003Ch2>Fundamental property\u003C\u002Fh2>{{chunk:kinematics-rigid-body-translation}}\u003Cp>Consequently, the kinematics of a translating rigid body can be determined by studying any one of its points. The velocity and acceleration found for that point at an instant apply to every other point at the same instant.\u003C\u002Fp>\u003Ch2>Point trajectories\u003C\u002Fh2>\u003Cp>Let $\\vec r_B=\\vec r_A+\\vec r_{AB}$, where $\\vec r_{AB}$ is constant because the body is rigid and its orientation does not change. The trajectory of point $B$ is therefore the trajectory of point $A$ shifted by the constant vector $\\vec r_{AB}$.\u003C\u002Fp>\u003Ch2>Velocities\u003C\u002Fh2>\u003Cp>Differentiating the position relation gives $\\vec v_B=\\vec v_A$ because $d\\vec r_{AB}\u002Fdt=0$. All points have identical velocity vectors at each instant even though their trajectories occupy different locations in space.\u003C\u002Fp>\u003Ch2>Accelerations\u003C\u002Fh2>\u003Cp>Differentiating again gives $\\vec a_B=\\vec a_A$. Thus, for velocity and acceleration analysis, rigid-body translation is kinematically equivalent to the motion of a single particle.\u003C\u002Fp>\u003Ch2>Translation need not be rectilinear\u003C\u002Fh2>\u003Cp>The term “translation” describes unchanged body orientation, not the shape of its trajectory. For example, with a suitable suspension, a Ferris-wheel cabin can undergo curvilinear translation while remaining vertically oriented.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If one point of a translating platform has velocity $\\vec v=(2\\vec i+3\\vec j)$ m\u002Fs at an instant, every other point has the same velocity vector at that instant. Their acceleration vectors are likewise identical.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>assuming translation must be rectilinear;\u003C\u002Fli>\u003Cli>confusing translation with motion at zero acceleration;\u003C\u002Fli>\u003Cli>assigning different angular velocities to different points of a translating body;\u003C\u002Fli>\u003Cli>assuming identical trajectory shapes must occupy the same geometric curve.\u003C\u002Fli>\u003C\u002Ful>",201,[],{"id":403,"parent_id":384,"code":70,"slug":404,"name":405,"seo_title":405,"seo_description":406,"seo_text":407,"content":408,"locale":8,"uk_topic_id":409,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":404,"children":410},217,"rotation-rigid-body-fixed-axis","Rotation of a Rigid Body About a Fixed Axis","Angular position, angular velocity and angular acceleration of a rigid body rotating about a fixed axis.","This topic covers the rotation law of a rigid body about a fixed axis, angular velocity, angular acceleration and the kinematic characteristics of points of the body.","\u003Cp>\u003Cstrong>Rotation of a rigid body about a fixed axis\u003C\u002Fstrong> is motion in which two points of the body, and therefore the line through them, remain fixed. This line is the axis of rotation. Every other point moves on a circle whose center lies on the axis.\u003C\u002Fp>\u003Ch2>Rotation law\u003C\u002Fh2>\u003Cp>Body orientation is specified by the angular position $\\varphi=\\varphi(t)$. A positive angular direction is selected in advance and determines the signs of angular velocity and angular acceleration.\u003C\u002Fp>\u003Ch2>Angular velocity and angular acceleration\u003C\u002Fh2>{{chunk:kinematics-angular-velocity-acceleration}}\u003Cp>In SI, angular position is measured in radians. Angular velocity is commonly expressed in rad\u002Fs and angular acceleration in rad\u002Fs²; the radian is dimensionless in SI, but retaining its symbol is useful for identifying angular quantities.\u003C\u002Fp>\u003Ch2>Vector description\u003C\u002Fh2>\u003Cp>The vector $\\vec\\omega$ lies along the rotation axis according to the right-hand rule. For a fixed axis, $\\vec\\varepsilon=d\\vec\\omega\u002Fdt$ also lies along that axis; its direction relative to $\\vec\\omega$ indicates whether the angular-speed magnitude is increasing or decreasing.\u003C\u002Fp>\u003Ch2>Uniform rotation\u003C\u002Fh2>\u003Cp>If $\\omega=const$, then $\\varepsilon=0$ and $\\varphi=\\varphi_0+\\omega t$. One complete revolution corresponds to an angular change of magnitude $2\\pi$ rad.\u003C\u002Fp>\u003Ch2>Constant angular acceleration\u003C\u002Fh2>\u003Cp>If $\\varepsilon=const$, then $\\omega=\\omega_0+\\varepsilon t$ and $\\varphi=\\varphi_0+\\omega_0t+\\varepsilon t^2\u002F2$. These relations are analogous to those for uniformly accelerated rectilinear motion.\u003C\u002Fp>\u003Ch2>Period and frequency\u003C\u002Fh2>\u003Cp>For uniform rotation, the period $T$ is the time for one revolution and the frequency $f=1\u002FT$ is the number of revolutions per unit time. Angular velocity is related by $\\omega=2\\pi\u002FT=2\\pi f$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A disk rotates uniformly at frequency $f=5$ Hz. Its angular velocity is $\\omega=2\\pi f=10\\pi\\approx31.4$ rad\u002Fs. In 2 s, the disk completes 10 revolutions.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing angular velocity with the linear velocity of a point on the body;\u003C\u002Fli>\u003Cli>using degrees in formulas intended for radians;\u003C\u002Fli>\u003Cli>ignoring the sign of $\\omega$ or $\\varepsilon$;\u003C\u002Fli>\u003Cli>assuming points on the rotation axis have nonzero linear velocity.\u003C\u002Fli>\u003C\u002Ful>",202,[],{"id":412,"parent_id":384,"code":70,"slug":413,"name":414,"seo_title":415,"seo_description":416,"seo_text":417,"content":418,"locale":8,"uk_topic_id":419,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":413,"children":420},218,"point-velocities-accelerations-fixed-axis-rotation","Velocities and Accelerations of Points in Fixed-Axis Rotation","Fixed-Axis Rotation — v = ωr, Tangential & Normal Acceleration","Calculate point velocity and acceleration in fixed-axis rotation using v = ωr, tangential acceleration at = αr, and normal acceleration an = ω²r.","This topic relates rigid-body angular velocity and angular acceleration to the linear velocity, tangential acceleration, normal acceleration and total acceleration of its points.","\u003Cp>During \u003Cstrong>rotation of a rigid body about a fixed axis\u003C\u002Fstrong>, all points share the same angular velocity $\\omega$ and angular acceleration $\\varepsilon$, but their linear velocities and accelerations depend on perpendicular distance from the axis.\u003C\u002Fp>\u003Ch2>Linear velocity\u003C\u002Fh2>{{chunk:kinematics-fixed-axis-point-velocity}}\u003Cp>The farther a point is from the axis, the greater its speed for the same $\\omega$. Points located directly on the axis have $r=0$ and remain fixed.\u003C\u002Fp>\u003Ch2>Vector velocity relation\u003C\u002Fh2>\u003Cp>For a point whose position from the axis is represented by $\\vec r$, velocity can be written as $\\vec v=\\vec\\omega\\times\\vec r$. The cross product automatically gives the tangential direction according to the right-hand rule.\u003C\u002Fp>\u003Ch2>Tangential acceleration\u003C\u002Fh2>\u003Cp>A change in speed produces the tangential component $a_\\tau=\\varepsilon r$. It is tangent to the circular path, with direction determined by the sign of angular acceleration.\u003C\u002Fp>\u003Ch2>Normal acceleration\u003C\u002Fh2>\u003Cp>The change in velocity direction produces the normal component $a_n=\\omega^2r=v^2\u002Fr$, directed from the point toward the rotation axis.\u003C\u002Fp>\u003Ch2>Total acceleration\u003C\u002Fh2>\u003Cp>The tangential and normal components are perpendicular, so $a=\\sqrt{a_\\tau^2+a_n^2}=r\\sqrt{\\varepsilon^2+\\omega^4}$. In vector form, $\\vec a=\\vec\\varepsilon\\times\\vec r+\\vec\\omega\\times(\\vec\\omega\\times\\vec r)$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A point on a disk is $r=0.20$ m from the axis. With $\\omega=10$ rad\u002Fs and $\\varepsilon=4$ rad\u002Fs², $v=2$ m\u002Fs, $a_\\tau=0.8$ m\u002Fs², $a_n=20$ m\u002Fs², and $a\\approx20.02$ m\u002Fs².\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using distance from the body center instead of perpendicular distance from the axis;\u003C\u002Fli>\u003Cli>confusing $a_\\tau=\\varepsilon r$ with $a_n=\\omega^2r$;\u003C\u002Fli>\u003Cli>assuming point acceleration is zero when $\\omega$ is constant;\u003C\u002Fli>\u003Cli>adding tangential and normal accelerations algebraically instead of vectorially.\u003C\u002Fli>\u003C\u002Ful>",203,[],{"id":422,"parent_id":384,"code":70,"slug":423,"name":424,"seo_title":425,"seo_description":426,"seo_text":427,"content":428,"locale":8,"uk_topic_id":429,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":423,"children":430},219,"transmission-rotational-motion","Transmission of Rotational Motion","Transmission of Rotational Motion in Kinematics","Kinematic relations for gears, belts and friction drives: angular velocities, radii and transmission ratio.","This topic covers transmission of rotational motion between bodies without slipping and the relationships among angular velocities, radii and transmission ratio.","\u003Cp>\u003Cstrong>Rotational-motion transmissions\u003C\u002Fstrong> transfer motion from an input member to an output member. In kinematics, the main objective is to relate angular velocities, rotational frequencies, and geometric parameters of the members.\u003C\u002Fp>\u003Ch2>No-slip condition\u003C\u002Fh2>{{chunk:kinematics-rotation-transmission-ratio}}\u003Cp>Equality of tangential velocities at contact is the basis for analyzing simple friction, belt, and gear transmissions under the ideal no-slip assumption.\u003C\u002Fp>\u003Ch2>Friction wheels\u003C\u002Fh2>\u003Cp>For two externally contacting wheels without slip, $|\\omega_1|r_1=|\\omega_2|r_2$. They rotate in opposite directions. With internal contact, their rotation directions are the same.\u003C\u002Fp>\u003Ch2>Belt drives\u003C\u002Fh2>\u003Cp>For an open belt drive without slip, belt speed is the same at both pulleys, so $\\omega_1r_1=\\omega_2r_2$ in magnitude. An open belt gives the pulleys the same rotation direction; a crossed belt gives opposite directions.\u003C\u002Fp>\u003Ch2>Gear drives\u003C\u002Fh2>\u003Cp>For an external gear pair, $|\\omega_1|\u002F|\\omega_2|=z_2\u002Fz_1$, where $z_1$ and $z_2$ are tooth numbers. External meshing reverses rotation direction, while internal meshing preserves it.\u003C\u002Fp>\u003Ch2>Transmission ratio\u003C\u002Fh2>\u003Cp>A transmission ratio must always be interpreted with its definition. Here $i=\\omega_1\u002F\\omega_2$, where member 1 is the input and member 2 is the output. If only the magnitude is required, rotation direction is handled separately.\u003C\u002Fp>\u003Ch2>Multistage transmissions\u003C\u002Fh2>\u003Cp>For successive stages, the overall transmission ratio is the product of the individual stage ratios. Output rotation direction follows from the number of external gear meshes or crossed-belt stages.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>An input gear has $z_1=20$ teeth and rotates at $\\omega_1=60$ rad\u002Fs. The output gear has $z_2=60$. Then $|\\omega_2|=60\\cdot20\u002F60=20$ rad\u002Fs, with opposite direction for external meshing.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>reversing the radius ratio $r_2\u002Fr_1$;\u003C\u002Fli>\u003Cli>defining $i$ without stating which member is input;\u003C\u002Fli>\u003Cli>ignoring rotation direction in external gear meshing;\u003C\u002Fli>\u003Cli>using equal contact speeds when the problem explicitly includes slip;\u003C\u002Fli>\u003Cli>considering only one pair in a multistage transmission.\u003C\u002Fli>\u003C\u002Ful>",204,[],{"id":432,"parent_id":325,"code":70,"slug":433,"name":434,"seo_title":434,"seo_description":435,"seo_text":436,"content":437,"locale":8,"uk_topic_id":438,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":433,"children":439},286,"rigid-body-plane-motion","Plane Motion of a Rigid Body","Plane motion of a rigid body and methods for determining velocities and accelerations of its points.","This section covers plane motion of a rigid body and methods for determining velocities and accelerations of body points.","\u003Cp>\u003Cstrong>Plane motion of a rigid body\u003C\u002Fstrong> occurs when all body points move in planes parallel to a fixed plane. This type of motion is common in planar mechanisms.\u003C\u002Fp>\u003Ch2>Decomposition of motion\u003C\u002Fh2>\u003Cp>Plane motion can be represented as translation of a selected reference point combined with rotation about an axis perpendicular to the plane of motion. The choice of reference point does not change the physical motion but may simplify the analysis.\u003C\u002Fp>\u003Ch2>Velocities of points\u003C\u002Fh2>\u003Cp>The velocity of any point is obtained from the velocity of the reference point plus the relative velocity caused by body rotation. The instantaneous center of zero velocity is also useful for velocity analysis.\u003C\u002Fp>\u003Ch2>Accelerations of points\u003C\u002Fh2>\u003Cp>Point acceleration consists of the acceleration of the reference point and tangential and normal components associated with rotation. The instantaneous center used for velocities is not, in general, a universal center for acceleration calculations.\u003C\u002Fp>",281,[440,448,457],{"id":441,"parent_id":432,"code":70,"slug":442,"name":434,"seo_title":434,"seo_description":443,"seo_text":444,"content":445,"locale":8,"uk_topic_id":446,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":442,"children":447},220,"plane-motion-rigid-body","Plane motion of a rigid body: decomposition into translation of a reference point and rotation about that point.","This topic introduces plane motion of a rigid body and its representation as translation of a reference point combined with rotation of the body about that point.","\u003Cp>\u003Cstrong>Plane motion of a rigid body\u003C\u002Fstrong> is motion in which all points of the body move in planes parallel to a fixed plane. Its kinematics can be studied through the motion of a plane figure representing a section of the body parallel to the plane of motion.\u003C\u002Fp>\u003Ch2>Kinematic decomposition\u003C\u002Fh2>{{chunk:kinematics-plane-motion-decomposition}}\u003Cp>The reference point can be selected arbitrarily. The translational component depends on that choice, but the angular velocity and angular acceleration of the plane figure at an instant do not.\u003C\u002Fp>\u003Ch2>Equations of motion of a plane figure\u003C\u002Fh2>\u003Cp>The three functions $x_A=x_A(t)$, $y_A=y_A(t)$, and $\\varphi=\\varphi(t)$ completely specify the figure position in the plane. The first two describe translation of the reference point and the third describes change of body orientation.\u003C\u002Fp>\u003Ch2>Translational and rotational components\u003C\u002Fh2>\u003Cp>Decomposing the motion into translation and rotation does not mean the body physically performs them one after another. It is a kinematic representation of one actual motion: the reference point moves while the figure simultaneously changes orientation.\u003C\u002Fp>\u003Ch2>Angular characteristics\u003C\u002Fh2>\u003Cp>Angular velocity is $\\omega=\\dot\\varphi$ and angular acceleration is $\\varepsilon=\\ddot\\varphi$. For plane motion, their vectors are perpendicular to the plane of motion.\u003C\u002Fp>\u003Ch2>Special cases\u003C\u002Fh2>\u003Cp>If $\\omega=0$ throughout an interval, the motion reduces to translation. If the reference point is fixed, the motion reduces to rotation about a fixed axis perpendicular to the plane.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A wheel rolling along a straight path undergoes plane motion: its center translates while the wheel simultaneously rotates. The motion of any point on the rim combines these two components.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating the reference point as a physically fixed point;\u003C\u002Fli>\u003Cli>assuming $\\omega$ depends on the selected reference point;\u003C\u002Fli>\u003Cli>describing plane-figure position only by one point’s coordinates and omitting $\\varphi$;\u003C\u002Fli>\u003Cli>confusing plane motion with pure translation.\u003C\u002Fli>\u003C\u002Ful>",205,[],{"id":449,"parent_id":432,"code":70,"slug":450,"name":451,"seo_title":451,"seo_description":452,"seo_text":453,"content":454,"locale":8,"uk_topic_id":455,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":450,"children":456},221,"point-velocities-plane-motion","Velocities of Points in Plane Motion","Relative velocity relation for a plane rigid body, instantaneous center of zero velocity and determination of point velocities.","This topic covers point velocities in plane rigid-body motion using a reference point, angular velocity and the instantaneous center of zero velocity.","\u003Cp>In \u003Cstrong>plane rigid-body motion\u003C\u002Fstrong>, different points generally have different velocities. Their velocities are related by the rigid-body relative-velocity equation.\u003C\u002Fp>\u003Ch2>Velocity relation\u003C\u002Fh2>{{chunk:kinematics-plane-motion-velocity-theorem}}\u003Cp>Choosing a point $A$ with known velocity as the reference point allows the velocity of any other point $B$ to be determined. The vector $\\vec\\omega\\times\\vec r_{B\u002FA}$ is always perpendicular to $AB$.\u003C\u002Fp>\u003Ch2>Velocity projections\u003C\u002Fh2>\u003Cp>Because the relative velocity $\\vec v_{B\u002FA}$ is perpendicular to $AB$, the velocity components of two points of a rigid body projected onto the line joining them are equal. This property often determines unknown components without a complete vector construction.\u003C\u002Fp>\u003Ch2>Instantaneous center of zero velocity\u003C\u002Fh2>\u003Cp>The \u003Cstrong>instantaneous center of zero velocity (IC)\u003C\u002Fstrong> is a point in the plane whose velocity is zero at the instant considered. If a finite IC $P$ exists, the velocity field of the figure at that instant is equivalent to instantaneous rotation about $P$.\u003C\u002Fp>\u003Ch2>Locating the IC\u003C\u002Fh2>\u003Cp>If the velocity directions of two points are known, draw through each point a line perpendicular to its velocity. Their intersection is the IC when the lines meet at a finite point. For pure translation, the IC is regarded as lying at infinity.\u003C\u002Fp>\u003Ch2>Velocity from the IC\u003C\u002Fh2>\u003Cp>For a point $A$ and known IC $P$, $v_A=|\\omega|PA$. Thus $v_A\u002Fv_B=PA\u002FPB$. Velocity directions are perpendicular to $PA$ and $PB$ and must correspond to one consistent sense of instantaneous rotation.\u003C\u002Fp>\u003Ch2>Rolling without slipping\u003C\u002Fh2>\u003Cp>For a wheel rolling without slip on a fixed surface, the contact point has zero instantaneous velocity and is the IC. Therefore the wheel-center speed satisfies $v_C=\\omega R$ in magnitude.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If the IC of a plane link is at $P$, $PA=0.2$ m, $PB=0.5$ m, and $v_A=1$ m\u002Fs, then $|\\omega|=1\u002F0.2=5$ rad\u002Fs and $v_B=5\\cdot0.5=2.5$ m\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating the IC as one material point fixed for a finite time interval;\u003C\u002Fli>\u003Cli>locating the IC along velocity directions instead of along perpendiculars to them;\u003C\u002Fli>\u003Cli>using $v_A\u002Fv_B=PA\u002FPB$ without a common IC;\u003C\u002Fli>\u003Cli>assuming zero contact-point velocity when rolling includes slip.\u003C\u002Fli>\u003C\u002Ful>",206,[],{"id":458,"parent_id":432,"code":70,"slug":459,"name":460,"seo_title":460,"seo_description":461,"seo_text":462,"content":463,"locale":8,"uk_topic_id":464,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":459,"children":465},222,"point-accelerations-plane-motion","Accelerations of Points in Plane Motion","Acceleration relation for points of a plane rigid body: reference-point, tangential and normal components.","This topic explains how to determine accelerations of points in plane rigid-body motion from reference-point acceleration, angular velocity and angular acceleration.","\u003Cp>In \u003Cstrong>plane rigid-body motion\u003C\u002Fstrong>, accelerations of different points are related through the acceleration of a selected reference point, the angular velocity, and the angular acceleration of the body.\u003C\u002Fp>\u003Ch2>Acceleration relation\u003C\u002Fh2>{{chunk:kinematics-plane-motion-acceleration-theorem}}\u003Cp>Unlike the velocity relation, the relative acceleration contains two components: a tangential component caused by changing angular velocity and a normal component caused by changing direction of relative velocity.\u003C\u002Fp>\u003Ch2>Tangential component\u003C\u002Fh2>\u003Cp>The vector $\\vec a^\\tau_{B\u002FA}=\\vec\\varepsilon\\times\\vec r_{B\u002FA}$ is perpendicular to $AB$. Its magnitude is $|\\varepsilon|AB$, and its direction follows from the sign of angular acceleration.\u003C\u002Fp>\u003Ch2>Normal component\u003C\u002Fh2>\u003Cp>The vector $\\vec a^n_{B\u002FA}=\\vec\\omega\\times(\\vec\\omega\\times\\vec r_{B\u002FA})$ points from $B$ toward reference point $A$ and has magnitude $\\omega^2AB$.\u003C\u002Fp>\u003Ch2>Choosing a reference point\u003C\u002Fh2>\u003Cp>A useful reference point is one whose acceleration is known or easily obtained from the constraints. The angular quantities $\\omega$ and $\\varepsilon$ do not depend on which reference point is selected.\u003C\u002Fp>\u003Ch2>Instantaneous center of acceleration\u003C\u002Fh2>\u003Cp>In some cases a point of the plane figure may have zero acceleration, but it must not be confused with the instantaneous center of zero velocity. Zero velocity at an instant does not imply zero acceleration.\u003C\u002Fp>\u003Ch2>Rolling without slipping\u003C\u002Fh2>\u003Cp>The contact point of a wheel rolling without slip on a fixed surface has zero instantaneous velocity, but generally nonzero acceleration. It therefore cannot be used as a fixed center for acceleration analysis in the same way that the IC is used for velocities.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a link $AB=0.4$ m with $\\omega=5$ rad\u002Fs and $\\varepsilon=3$ rad\u002Fs², the relative acceleration components of $B$ with respect to $A$ have magnitudes $a^\\tau_{B\u002FA}=1.2$ m\u002Fs² and $a^n_{B\u002FA}=10$ m\u002Fs². They must be added vectorially to $\\vec a_A$.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>including only the tangential or only the normal component;\u003C\u002Fli>\u003Cli>directing the normal component from the reference point toward the other point;\u003C\u002Fli>\u003Cli>assuming the instantaneous center of zero velocity has zero acceleration;\u003C\u002Fli>\u003Cli>adding component magnitudes instead of vectors;\u003C\u002Fli>\u003Cli>using different $\\omega$ or $\\varepsilon$ values for different points of the same rigid body.\u003C\u002Fli>\u003C\u002Ful>",207,[],{"id":467,"parent_id":325,"code":70,"slug":468,"name":469,"seo_title":469,"seo_description":470,"seo_text":471,"content":472,"locale":8,"uk_topic_id":473,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":468,"children":474},287,"relative-motion-of-a-particle-section","Relative Motion of a Particle","Relative, transport and absolute motion, acceleration addition, and Coriolis acceleration.","This section covers relative motion of a particle, addition of velocities and accelerations, and Coriolis acceleration.","\u003Cp>\u003Cstrong>Relative motion of a particle\u003C\u002Fstrong> is considered when a particle moves with respect to a reference frame that itself moves relative to another frame. The motion is separated into relative, transport, and absolute components.\u003C\u002Fp>\u003Ch2>Addition of velocities\u003C\u002Fh2>\u003Cp>The absolute velocity equals the vector sum of relative and transport velocities. This relation is fundamental in the analysis of particles moving in guides or mechanisms whose supporting frame is itself in motion.\u003C\u002Fp>\u003Ch2>Addition of accelerations\u003C\u002Fh2>\u003Cp>The acceleration relation is more involved. When the moving frame rotates, the absolute acceleration includes the Coriolis acceleration in addition to the relative and transport terms.\u003C\u002Fp>\u003Ch2>Choice of reference frames\u003C\u002Fh2>\u003Cp>A reliable solution begins by defining the fixed and moving frames clearly and assigning every velocity and acceleration term to the correct component of motion.\u003C\u002Fp>",282,[475,484],{"id":476,"parent_id":467,"code":70,"slug":477,"name":469,"seo_title":478,"seo_description":479,"seo_text":480,"content":481,"locale":8,"uk_topic_id":482,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":477,"children":483},223,"relative-motion-particle","Relative Motion of a Particle — Velocity Addition Theorem","Learn absolute, relative, and transport motion of a particle and calculate velocity in moving reference frames using the velocity-addition theorem.","This topic introduces absolute, relative and transport motion of a particle and shows how absolute velocity is obtained from relative and transport velocity components.","\u003Cp>\u003Cstrong>Relative motion of a particle\u003C\u002Fstrong> is considered when a particle moves with respect to a reference frame that itself moves relative to another frame treated as fixed. This description separates the motion into absolute, relative, and transport components.\u003C\u002Fp>\u003Ch2>Three types of motion\u003C\u002Fh2>\u003Cp>\u003Cstrong>Absolute motion\u003C\u002Fstrong> is particle motion relative to the fixed frame. \u003Cstrong>Relative motion\u003C\u002Fstrong> is motion relative to the moving frame. \u003Cstrong>Transport motion\u003C\u002Fstrong> is the motion of the point of the moving frame that instantaneously coincides with the particle.\u003C\u002Fp>\u003Ch2>Velocity addition\u003C\u002Fh2>{{chunk:kinematics-relative-motion-velocity-addition}}\u003Cp>This is a vector relation, so velocity magnitudes generally cannot simply be added algebraically; their directions must be taken into account.\u003C\u002Fp>\u003Ch2>Transport velocity\u003C\u002Fh2>\u003Cp>If the moving frame undergoes translation only, all its points have the same transport velocity. If it also rotates, then for a point with position vector $\\vec r$ from a selected moving-frame origin $O'$, $\\vec v_e=\\vec v_{O'}+\\vec\\omega_e\\times\\vec r$.\u003C\u002Fp>\u003Ch2>Relative velocity\u003C\u002Fh2>\u003Cp>Relative velocity is measured by an observer moving with the moving coordinate frame. For example, for a slider moving in a slot fixed to a moving link, relative velocity is directed along the slot.\u003C\u002Fp>\u003Ch2>Vector addition\u003C\u002Fh2>\u003Cp>The equation $\\vec v_a=\\vec v_r+\\vec v_e$ can be solved by resolving vectors into components or by constructing a velocity triangle. The convenient method depends on which directions and magnitudes are known.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A person walks along a train car at 1.5 m\u002Fs relative to the car while the train moves along a straight track at 12 m\u002Fs in the same direction. The person’s absolute velocity is 13.5 m\u002Fs. If the person walks in the opposite direction, using the same sign convention the absolute velocity is 10.5 m\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing relative and transport velocities;\u003C\u002Fli>\u003Cli>adding velocity magnitudes without considering directions;\u003C\u002Fli>\u003Cli>for a rotating moving frame, using only the velocity of its origin as the transport velocity;\u003C\u002Fli>\u003Cli>failing to state the reference frame relative to which each velocity is defined.\u003C\u002Fli>\u003C\u002Ful>",208,[],{"id":485,"parent_id":467,"code":70,"slug":486,"name":487,"seo_title":488,"seo_description":489,"seo_text":490,"content":491,"locale":8,"uk_topic_id":492,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":486,"children":493},224,"acceleration-addition-coriolis-acceleration","Acceleration Addition. Coriolis Acceleration","Coriolis Acceleration — Formula & Acceleration Addition Theorem","Calculate absolute acceleration in relative motion using relative, transport, and Coriolis components, including the formula aC = 2ω × vrel.","This topic presents the acceleration-addition theorem for a particle observed from a moving frame and explains the geometric and physical meaning of Coriolis acceleration.","\u003Cp>In relative particle motion with a rotating reference frame, absolute acceleration is not merely the sum of relative and transport accelerations. An additional term appears: \u003Cstrong>Coriolis acceleration\u003C\u002Fstrong>.\u003C\u002Fp>\u003Ch2>Acceleration-addition theorem\u003C\u002Fh2>\u003Cp>For a particle moving relative to a moving reference frame, absolute acceleration is $\\vec a_a=\\vec a_r+\\vec a_e+\\vec a_C$, where $\\vec a_r$ is relative acceleration, $\\vec a_e$ is transport acceleration, and $\\vec a_C$ is Coriolis acceleration.\u003C\u002Fp>\u003Ch2>Coriolis acceleration\u003C\u002Fh2>{{chunk:kinematics-coriolis-acceleration}}\u003Ch2>When Coriolis acceleration is zero\u003C\u002Fh2>\u003Cp>The term $\\vec a_C$ vanishes if the moving frame does not rotate ($\\omega_e=0$), if the particle has no relative velocity ($v_r=0$), or if $\\vec v_r$ is parallel or antiparallel to $\\vec\\omega_e$.\u003C\u002Fp>\u003Ch2>Transport acceleration\u003C\u002Fh2>\u003Cp>For a moving frame with origin $O'$, the transport acceleration of the coincident frame point contains origin acceleration, tangential rotational acceleration, and centripetal acceleration: $\\vec a_e=\\vec a_{O'}+\\vec\\varepsilon_e\\times\\vec r+\\vec\\omega_e\\times(\\vec\\omega_e\\times\\vec r)$.\u003C\u002Fp>\u003Ch2>Direction of Coriolis acceleration\u003C\u002Fh2>\u003Cp>The direction of $\\vec a_C$ follows from $2\\vec\\omega_e\\times\\vec v_r$. In planar problems, first determine the direction of $\\vec\\omega_e$ by the right-hand rule and then take its cross product with $\\vec v_r$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A disk rotates with $\\omega_e=4$ rad\u002Fs while a slider moves radially in a slot at $v_r=0.5$ m\u002Fs. Here $\\theta=90^\\circ$, so $a_C=2\\cdot4\\cdot0.5=4$ m\u002Fs². Its direction lies in the disk plane and is perpendicular to the radial relative-velocity direction.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>omitting the factor 2 in the Coriolis formula;\u003C\u002Fli>\u003Cli>using absolute velocity instead of relative velocity $\\vec v_r$;\u003C\u002Fli>\u003Cli>adding acceleration magnitudes without accounting for direction;\u003C\u002Fli>\u003Cli>including a Coriolis term when the moving frame undergoes pure translation;\u003C\u002Fli>\u003Cli>reversing the cross-product order $\\vec\\omega_e\\times\\vec v_r$.\u003C\u002Fli>\u003C\u002Ful>",209,[],{"id":495,"parent_id":325,"code":70,"slug":496,"name":497,"seo_title":497,"seo_description":498,"seo_text":499,"content":500,"locale":8,"uk_topic_id":501,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":496,"children":502},288,"spatial-motion-of-a-rigid-body","Spatial Motion of a Rigid Body","Spherical motion and general motion of a free rigid body in three-dimensional space.","This section covers spherical rigid-body motion and general motion of a free rigid body in three-dimensional space.","\u003Cp>\u003Cstrong>Spatial motion of a rigid body\u003C\u002Fstrong> covers motions that cannot be reduced to planar kinematics. The orientation of the body changes in three-dimensional space, so angular velocity and angular acceleration are treated as vectors.\u003C\u002Fp>\u003Ch2>Spherical motion\u003C\u002Fh2>\u003Cp>In spherical motion, one point of the rigid body remains fixed while all other points move on spherical surfaces centered at that point. Instantaneously, the motion may be characterized as rotation about an instantaneous axis.\u003C\u002Fp>\u003Ch2>General rigid-body motion\u003C\u002Fh2>\u003Cp>The general motion of a free rigid body can be decomposed into translation of a selected reference point and rotation of the body relative to that point. This is the three-dimensional counterpart of the decomposition used in plane motion.\u003C\u002Fp>\u003Ch2>Kinematic quantities\u003C\u002Fh2>\u003Cp>Velocities and accelerations of body points are determined from the translational component together with the body's angular velocity and angular acceleration. Vector notation provides a unified description independent of body orientation.\u003C\u002Fp>",283,[503,513],{"id":504,"parent_id":495,"code":70,"slug":505,"name":506,"seo_title":507,"seo_description":508,"seo_text":509,"content":510,"locale":8,"uk_topic_id":511,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":505,"children":512},225,"spherical-motion-rigid-body","Spherical Motion of a Rigid Body","Spherical Motion of a Rigid Body — Angular Velocity & Instantaneous Axis","Learn spherical rigid-body motion with one fixed point: angular velocity vector, instantaneous axis of rotation, and point velocities in three-dimensional motion.","This topic introduces spherical motion of a rigid body with one fixed point, including the instantaneous axis of rotation and angular-velocity vector.","\u003Cp>\u003Cstrong>Spherical motion of a rigid body\u003C\u002Fstrong> is three-dimensional motion in which one point of the body remains fixed. Every other point moves on a spherical surface centered at that fixed point.\u003C\u002Fp>\u003Ch2>Geometry of spherical motion\u003C\u002Fh2>\u003Cp>Let point $O$ be fixed. The distance $OP$ to any body point $P$ is constant, so the trajectory of $P$ lies on a sphere of radius $OP$. Unlike rotation about a fixed axis, the direction of the instantaneous rotation axis generally changes with time.\u003C\u002Fp>\u003Ch2>Instantaneous angular velocity\u003C\u002Fh2>{{chunk:kinematics-spherical-motion-velocity}}\u003Cp>The vector $\\vec\\omega$ describes the instantaneous rotation of the body. The line through fixed point $O$ in the direction of $\\vec\\omega$ is the instantaneous axis of rotation.\u003C\u002Fp>\u003Ch2>Point velocities\u003C\u002Fh2>\u003Cp>The speed of point $P$ is $v_P=\\omega r_\\perp$, where $r_\\perp$ is the perpendicular distance from the point to the instantaneous axis. Points on the instantaneous axis therefore have zero velocity at that instant.\u003C\u002Fp>\u003Ch2>Changing instantaneous axis\u003C\u002Fh2>\u003Cp>The instantaneous axis is not generally one material line that remains fixed throughout the motion. Its position changes both within the body and in fixed space. This distinguishes general spherical motion from simple fixed-axis rotation.\u003C\u002Fp>\u003Ch2>Angular acceleration\u003C\u002Fh2>\u003Cp>Angular acceleration is $\\vec\\varepsilon=d\\vec\\omega\u002Fdt$ in the fixed reference frame. Because both magnitude and direction of $\\vec\\omega$ may change, $\\vec\\varepsilon$ and $\\vec\\omega$ are not generally parallel.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If at an instant $\\omega=6$ rad\u002Fs and a point is at a perpendicular distance of 0.15 m from the instantaneous axis, its speed is $v=6\\cdot0.15=0.9$ m\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>assuming the instantaneous axis remains fixed throughout spherical motion;\u003C\u002Fli>\u003Cli>using the full distance $OP$ instead of perpendicular distance to the instantaneous axis in $v=\\omega r_\\perp$;\u003C\u002Fli>\u003Cli>assuming $\\vec\\varepsilon$ is always parallel to $\\vec\\omega$;\u003C\u002Fli>\u003Cli>confusing spherical rigid-body motion with the motion of a single particle on a sphere.\u003C\u002Fli>\u003C\u002Ful>",210,[],{"id":514,"parent_id":495,"code":70,"slug":515,"name":516,"seo_title":517,"seo_description":518,"seo_text":519,"content":520,"locale":8,"uk_topic_id":521,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":515,"children":522},226,"general-motion-free-rigid-body","General Motion of a Free Rigid Body","General 3D Rigid-Body Motion — Velocity & Acceleration Formulas","Analyze general spatial motion of a free rigid body as translation plus rotation, with formulas for point velocities and accelerations in three dimensions.","This topic completes rigid-body kinematics with general spatial motion represented by translation of a reference point combined with rotation about that point.","\u003Cp>\u003Cstrong>General motion of a free rigid body\u003C\u002Fstrong> is the most general case of rigid-body kinematics, with no fixed point or fixed axis. It can be represented as translation of an arbitrarily selected reference point combined with spherical motion of the body relative to that point.\u003C\u002Fp>\u003Ch2>Decomposition of general motion\u003C\u002Fh2>\u003Cp>Body position is specified by the position of a selected reference point $A$ and the orientation of the body relative to the fixed coordinate system. The selected point affects the translational part of the description, but the instantaneous angular velocity $\\vec\\omega$ of the body does not depend on that choice.\u003C\u002Fp>\u003Ch2>Velocity of an arbitrary point\u003C\u002Fh2>{{chunk:kinematics-general-rigid-body-velocity}}\u003Cp>This relation is the three-dimensional generalization of the velocity relation for points of a plane rigid body.\u003C\u002Fp>\u003Ch2>Acceleration of an arbitrary point\u003C\u002Fh2>\u003Cp>For two points $A$ and $B$ of the same rigid body:\u003C\u002Fp>\u003Cp>$$\\vec a_B=\\vec a_A+\\vec\\varepsilon\\times\\vec r_{B\u002FA}+\\vec\\omega\\times(\\vec\\omega\\times\\vec r_{B\u002FA}).$$\u003C\u002Fp>\u003Cp>The second term is associated with angular acceleration and the third with instantaneous angular velocity.\u003C\u002Fp>\u003Ch2>Choosing the reference point\u003C\u002Fh2>\u003Cp>The reference point is chosen for convenience, often as the center of mass, a point with known motion, or a geometrically significant point of a mechanism. The physical motion of the body does not change with the choice.\u003C\u002Fp>\u003Ch2>Relation to special motions\u003C\u002Fh2>\u003Cp>If $\\vec\\omega=0$, general motion reduces to translation. If the selected point is fixed, it becomes spherical motion. Plane motion is a special case in which point trajectories lie in parallel planes and $\\vec\\omega$ has a fixed direction perpendicular to those planes.\u003C\u002Fp>\u003Ch2>Instantaneous screw motion\u003C\u002Fh2>\u003Cp>In the general spatial case, the rigid-body velocity field can be interpreted as an instantaneous screw motion: rotation about an instantaneous axis combined with translation along that axis. This generalizes the concept of an instantaneous rotation axis.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If reference point $A$ has velocity $\\vec v_A$ and the position of point $B$ relative to it is known, $B$’s velocity is obtained by adding $\\vec v_A$ to $\\vec\\omega\\times\\vec r_{B\u002FA}$. Even when $A$ is instantaneously at rest, other points may have nonzero velocities because of the rotational term.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating general motion as merely translation of the center of mass;\u003C\u002Fli>\u003Cli>omitting the rotational term $\\vec\\omega\\times\\vec r_{B\u002FA}$;\u003C\u002Fli>\u003Cli>assuming $\\vec\\omega$ depends on the selected reference point;\u003C\u002Fli>\u003Cli>using planar scalar direction rules instead of full three-dimensional vector analysis;\u003C\u002Fli>\u003Cli>confusing the instantaneous screw representation with the trajectory of a particular material point.\u003C\u002Fli>\u003C\u002Ful>",211,[],{"id":524,"parent_id":91,"code":70,"slug":525,"name":526,"seo_title":527,"seo_description":528,"seo_text":529,"content":530,"locale":8,"uk_topic_id":531,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":525,"children":532},138,"dynamics","Dynamics","Dynamics — Laws of Motion and Problems | Mechanics","Engineering dynamics: laws of motion, forces, work, energy and momentum. Core theory, formulas and practical theoretical mechanics problems.","Dynamics studies mechanical motion while accounting for forces and mass. This section covers fundamental laws of dynamics, equations of motion, work and power, kinetic energy, momentum and general dynamics theorems, with practical engineering problems.","\u003Cp>Dynamics relates the motion of material bodies to the forces acting on them. Kinematic quantities are combined with mass, forces and the fundamental laws of mechanics.\u003C\u002Fp>\u003Cp>The section is designed for problem solving: determining forces and motion parameters, applying equations of motion, and using energy, momentum and other general methods of dynamics.\u003C\u002Fp>",135,[533,583,653,723],{"id":534,"parent_id":524,"code":70,"slug":535,"name":536,"seo_title":537,"seo_description":538,"seo_text":539,"content":540,"locale":8,"uk_topic_id":541,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":535,"children":542},293,"particle-dynamics","Particle Dynamics","Particle Dynamics — Newton’s Laws, F = ma & Equations of Motion","Learn particle dynamics using Newton’s laws, F = ma, differential equations of motion, direct and inverse dynamics problems, and common force models.","This section covers Newton's laws, differential equations of particle motion, the fundamental dynamics problems, and motion under typical forces.","\u003Cp>\u003Cstrong>Particle dynamics\u003C\u002Fstrong> establishes the relationship between particle motion and the forces acting on it. Unlike kinematics, acceleration is now interpreted as a consequence of force interaction according to Newton's laws.\u003C\u002Fp>\u003Ch2>Newton's laws\u003C\u002Fh2>\u003Cp>The section is based on inertia, the relation between resultant force and acceleration, and the action-reaction principle. For a particle of constant mass, the equation of motion is written in vector form and projected onto convenient axes.\u003C\u002Fp>\u003Ch2>Differential equations of motion\u003C\u002Fh2>\u003Cp>When forces are known as functions of position, velocity, or time, Newton's second law produces differential equations. Their solution together with initial conditions determines the particle's motion.\u003C\u002Fp>\u003Ch2>Direct and inverse problems\u003C\u002Fh2>\u003Cp>In one class of problems, a prescribed motion is used to determine the required force or resultant. In the other, known forces and initial conditions are used to determine motion. Typical models include gravity, elastic forces, and resistance.\u003C\u002Fp>",289,[543,553,563,573],{"id":544,"parent_id":534,"code":70,"slug":545,"name":546,"seo_title":547,"seo_description":548,"seo_text":549,"content":550,"locale":8,"uk_topic_id":551,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":545,"children":552},248,"fundamental-laws-of-dynamics-newtons-laws","Fundamental Laws of Dynamics. Newton's Laws","Newton’s Laws — 1st, 2nd & 3rd Laws and F = ma","Learn Newton’s three laws of motion, inertial reference frames, mass and force, the equation F = ma, and the action–reaction principle in particle dynamics.","This topic introduces particle dynamics, inertial reference frames, mass, force, and Newton's three laws as the foundation of equations of motion.","\u003Cp>\u003Cstrong>Dynamics\u003C\u002Fstrong> studies the motion of bodies while accounting for the causes that produce that motion. Classical dynamics is based on Newton's laws, which relate force, mass, and acceleration.\u003C\u002Fp>\u003Ch2>Particle model and inertial frame\u003C\u002Fh2>\u003Cp>A particle is an idealized body whose dimensions can be neglected for the problem at hand. Newton's laws in their standard form apply in inertial reference frames, in which a free particle remains at rest or moves with constant velocity along a straight line.\u003C\u002Fp>\u003Ch2>Newton's first law\u003C\u002Fh2>\u003Cp>If the resultant force on a particle is zero, its velocity in an inertial frame remains constant: $\\sum\\vec F=0\\Rightarrow\\vec v=const$.\u003C\u002Fp>\u003Ch2>Newton's second law\u003C\u002Fh2>\u003Cp>For a particle of constant mass, the fundamental equation of dynamics is:\u003C\u002Fp>\u003Cp>$$m\\vec a=\\sum_i\\vec F_i.$$\u003C\u002Fp>\u003Cp>Acceleration is directed along the resultant force; its magnitude is proportional to force and inversely proportional to mass.\u003C\u002Fp>\u003Ch2>Newton's third law\u003C\u002Fh2>\u003Cp>The interaction forces of two bodies are equal in magnitude, opposite in direction, and act on different bodies: $\\vec F_{12}=-\\vec F_{21}$. They therefore must not be canceled on the free-body diagram of a single body.\u003C\u002Fp>\u003Ch2>Mass and force\u003C\u002Fh2>\u003Cp>Mass characterizes inertia. In SI, mass is measured in kilograms and force in newtons: $1\\,\\text{N}=1\\,\\text{kg}\\cdot\\text{m}\u002F\\text{s}^2$.\u003C\u002Fp>\u003Ch2>Procedure for applying Newton's second law\u003C\u002Fh2>\u003Col>\u003Cli>isolate the body or particle;\u003C\u002Fli>\u003Cli>show all external forces;\u003C\u002Fli>\u003Cli>choose coordinate axes;\u003C\u002Fli>\u003Cli>determine the acceleration;\u003C\u002Fli>\u003Cli>write $m\\vec a=\\sum\\vec F$ and project it onto the axes.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If a resultant force of 20 N acts along the $x$ axis on a 5 kg body, then $a_x=20\u002F5=4$ m\u002Fs².\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>mixing forces that act on different bodies;\u003C\u002Fli>\u003Cli>assuming that motion always requires a nonzero resultant force;\u003C\u002Fli>\u003Cli>confusing mass and weight;\u003C\u002Fli>\u003Cli>using the equilibrium equation $\\sum\\vec F=0$ for a particle with nonzero acceleration.\u003C\u002Fli>\u003C\u002Ful>",227,[],{"id":554,"parent_id":534,"code":70,"slug":555,"name":556,"seo_title":557,"seo_description":558,"seo_text":559,"content":560,"locale":8,"uk_topic_id":561,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":555,"children":562},249,"differential-equations-of-motion-of-a-particle","Differential Equations of Motion of a Particle","Particle Equations of Motion — F = ma in Cartesian & Path Coordinates","Write particle equations of motion from F = ma in vector, Cartesian, and tangential-normal coordinates, then apply initial conditions to solve dynamics problems.","This topic develops the fundamental equation of particle dynamics and its vector, Cartesian, and natural-coordinate representations.","\u003Cp>Differential equations of motion connect particle kinematics with the forces acting on the particle. For a particle of constant mass, they follow directly from Newton's second law.\u003C\u002Fp>\u003Ch2>Vector equation\u003C\u002Fh2>\u003Cp>The fundamental equation of motion is:\u003C\u002Fp>\u003Cp>$$m\\frac{d^2\\vec r}{dt^2}=\\sum_i\\vec F_i.$$\u003C\u002Fp>\u003Cp>If forces depend on position, velocity, or time, the right-hand side may be written as $\\vec F(\\vec r,\\vec v,t)$.\u003C\u002Fp>\u003Ch2>Cartesian equations\u003C\u002Fh2>\u003Cp>Projection onto fixed coordinate axes gives:\u003C\u002Fp>\u003Cp>$$m\\ddot x=\\sum F_x,\\qquad m\\ddot y=\\sum F_y,\\qquad m\\ddot z=\\sum F_z.$$\u003C\u002Fp>\u003Cp>This is a system of second-order differential equations for the particle coordinates.\u003C\u002Fp>\u003Ch2>Natural coordinates\u003C\u002Fh2>\u003Cp>For motion along a known path, projection onto the tangent and principal normal is often convenient:\u003C\u002Fp>\u003Cp>$$m\\frac{dv}{dt}=\\sum F_\\tau,\\qquad m\\frac{v^2}{\\rho}=\\sum F_n,$$\u003C\u002Fp>\u003Cp>where $\\rho$ is the radius of curvature of the path.\u003C\u002Fp>\u003Ch2>Initial conditions\u003C\u002Fh2>\u003Cp>To determine the motion uniquely, initial position and velocity are normally specified, for example $x(t_0)=x_0$ and $\\dot x(t_0)=v_{0x}$. The integration constants are found from these conditions.\u003C\u002Fp>\u003Ch2>Choosing coordinates\u003C\u002Fh2>\u003Cp>The coordinate system should match the geometry of motion and force directions. One axis is enough for rectilinear motion; natural coordinates are often efficient for motion along a curved path.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For vertical free fall without air resistance, with the $y$ axis directed upward, $m\\ddot y=-mg$, hence $\\ddot y=-g$. Integrating twice and applying the initial conditions gives the equation of motion.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing the sign of a force component with the force magnitude;\u003C\u002Fli>\u003Cli>omitting constraint reactions;\u003C\u002Fli>\u003Cli>integrating without applying initial conditions;\u003C\u002Fli>\u003Cli>treating $v^2\u002F\\rho$ as the total acceleration rather than its normal component.\u003C\u002Fli>\u003C\u002Ful>",228,[],{"id":564,"parent_id":534,"code":70,"slug":565,"name":566,"seo_title":567,"seo_description":568,"seo_text":569,"content":570,"locale":8,"uk_topic_id":571,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":565,"children":572},250,"two-fundamental-problems-of-particle-dynamics","Two Fundamental Problems of Particle Dynamics","Direct & Inverse Dynamics Problems — Find Motion or Forces","Learn the two fundamental particle-dynamics problems: determine forces from a prescribed motion or find the motion from known forces and initial conditions.","This topic explains the direct and inverse problems of particle dynamics and a systematic procedure for solving each type.","\u003Cp>Particle dynamics has two fundamental problem types. Both use the equation $m\\vec a=\\sum\\vec F$, but differ in which quantities are known and which must be determined.\u003C\u002Fp>\u003Ch2>First problem of dynamics\u003C\u002Fh2>\u003Cp>Given the particle's motion, its mass, and some of the forces, determine the unknown forces. Starting from $\\vec r(t)$, find velocity and acceleration and then use the equations of dynamics to determine the required forces or reactions.\u003C\u002Fp>\u003Ch2>Second problem of dynamics\u003C\u002Fh2>\u003Cp>Given the forces, mass, and initial conditions, determine the motion. Set up the differential equations, integrate them, and evaluate the integration constants from the initial conditions.\u003C\u002Fp>\u003Ch2>First problem: procedure\u003C\u002Fh2>\u003Col>\u003Cli>write the prescribed motion;\u003C\u002Fli>\u003Cli>differentiate the coordinates twice to obtain acceleration;\u003C\u002Fli>\u003Cli>draw the free-body diagram;\u003C\u002Fli>\u003Cli>write the components of $m\\vec a=\\sum\\vec F$;\u003C\u002Fli>\u003Cli>solve for the unknown forces.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Second problem: procedure\u003C\u002Fh2>\u003Col>\u003Cli>isolate the particle and identify the forces;\u003C\u002Fli>\u003Cli>choose coordinates;\u003C\u002Fli>\u003Cli>form the differential equations of motion;\u003C\u002Fli>\u003Cli>integrate them;\u003C\u002Fli>\u003Cli>apply the initial conditions;\u003C\u002Fli>\u003Cli>check the resulting motion and units.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Example of the first problem\u003C\u002Fh2>\u003Cp>If motion along an axis is prescribed by $x=2t^2$ m for a 3 kg particle, then $a_x=4$ m\u002Fs² and the required resultant force is $F_x=ma_x=12$ N.\u003C\u002Fp>\u003Ch2>Example of the second problem\u003C\u002Fh2>\u003Cp>If a constant force $F$ acts along an axis on a particle of mass $m$, then $\\ddot x=F\u002Fm$. With $x(0)=x_0$ and $\\dot x(0)=v_0$, integration gives $x=x_0+v_0t+Ft^2\u002F(2m)$.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>not distinguishing what is prescribed in the two problem types;\u003C\u002Fli>\u003Cli>omitting initial conditions in the second problem;\u003C\u002Fli>\u003Cli>trying to determine force from velocity instead of acceleration;\u003C\u002Fli>\u003Cli>omitting unknown constraint reactions from the equations of motion.\u003C\u002Fli>\u003C\u002Ful>",229,[],{"id":574,"parent_id":534,"code":70,"slug":575,"name":576,"seo_title":577,"seo_description":578,"seo_text":579,"content":580,"locale":8,"uk_topic_id":581,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":575,"children":582},251,"particle-motion-under-typical-forces","Particle Motion Under Typical Forces","Particle Motion Under Gravity, Springs, Friction & Drag","Model particle motion under common forces including gravity, linear springs, dry friction, and fluid resistance, and write the corresponding equations of motion.","This topic examines particle-motion models under common forces: gravity, elasticity, dry friction, and resistance of a medium.","\u003Cp>Dynamics problems repeatedly use several standard force models. Writing these forces correctly and choosing consistent directions makes the differential equation of motion much easier to formulate.\u003C\u002Fp>\u003Ch2>Gravity\u003C\u002Fh2>\u003Cp>Near Earth's surface, gravity is commonly treated as constant: $\\vec F_g=m\\vec g$. If air resistance is neglected, vertical motion has constant acceleration $g$ directed downward.\u003C\u002Fp>\u003Ch2>Elastic force\u003C\u002Fh2>\u003Cp>For a linear spring within Hooke's-law behavior, the restoring force is proportional to deformation and opposite to it:\u003C\u002Fp>\u003Cp>$$F_s=-kx.$$\u003C\u002Fp>\u003Cp>With no other variable forces, $m\\ddot x+kx=0$ describes free harmonic oscillation.\u003C\u002Fp>\u003Ch2>Dry sliding friction\u003C\u002Fh2>\u003Cp>In the simple Coulomb model, the sliding-friction magnitude is $F_f=\\mu N$, and its direction opposes relative sliding velocity. The sign of its component must be chosen from the actual direction of motion.\u003C\u002Fp>\u003Ch2>Resistance of a medium\u003C\u002Fh2>\u003Cp>At relatively low speeds, a linear model $\\vec R=-c\\vec v$ is often used. In other regimes, a quadratic model with resistance magnitude proportional to $v^2$ may be appropriate. The problem statement must specify or justify the model.\u003C\u002Fp>\u003Ch2>Motion on an inclined plane\u003C\u002Fh2>\u003Cp>With an axis along a plane inclined by $\\alpha$ to the horizontal, the gravity component along the plane has magnitude $mg\\sin\\alpha$, while the normal component is $mg\\cos\\alpha$. If no other normal forces act, $N=mg\\cos\\alpha$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A body slides down an incline with friction coefficient $\\mu$. Taking positive direction down the plane gives $ma=mg\\sin\\alpha-\\mu mg\\cos\\alpha$, hence $a=g(\\sin\\alpha-\\mu\\cos\\alpha)$.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>automatically directing friction opposite to the coordinate axis rather than opposite to sliding;\u003C\u002Fli>\u003Cli>writing spring force without its restoring direction;\u003C\u002Fli>\u003Cli>assuming $N=mg$ for every geometry;\u003C\u002Fli>\u003Cli>mixing linear and quadratic resistance models.\u003C\u002Fli>\u003C\u002Ful>",230,[],{"id":584,"parent_id":524,"code":70,"slug":585,"name":586,"seo_title":587,"seo_description":588,"seo_text":589,"content":590,"locale":8,"uk_topic_id":591,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":585,"children":592},294,"general-theorems-of-particle-dynamics","General Theorems of Particle Dynamics","Particle Dynamics Theorems — Momentum, Work & Energy Methods","Learn the main particle-dynamics theorems for linear and angular momentum, work, kinetic and potential energy, and conservation laws used to solve motion problems.","This section brings together momentum and energy methods for particle dynamics, including impulse, angular momentum, work, energy, and conservation laws.","\u003Cp>\u003Cstrong>General theorems of particle dynamics\u003C\u002Fstrong> provide alternatives to direct integration of the differential equations of motion. They relate changes in motion quantities to force impulse, force moments, and work.\u003C\u002Fp>\u003Ch2>Momentum methods\u003C\u002Fh2>\u003Cp>Linear momentum characterizes translational motion of a particle. The impulse-momentum theorem relates the change in momentum over a time interval to the impulse of the resultant force.\u003C\u002Fp>\u003Ch2>Angular momentum\u003C\u002Fh2>\u003Cp>Motion relative to a point or axis can be characterized by angular momentum. Its rate of change is determined by the moment of the acting forces about the same reference point or axis.\u003C\u002Fp>\u003Ch2>Work and energy\u003C\u002Fh2>\u003Cp>The work-energy theorem relates changes in speed to the work of forces and can often eliminate time from the analysis. For conservative forces, potential energy is introduced, leading under appropriate conditions to conservation of mechanical energy.\u003C\u002Fp>",290,[593,603,613,623,633,643],{"id":594,"parent_id":584,"code":70,"slug":595,"name":596,"seo_title":597,"seo_description":598,"seo_text":599,"content":600,"locale":8,"uk_topic_id":601,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":595,"children":602},252,"linear-momentum-of-a-particle-impulse-of-a-force","Linear Momentum of a Particle. Impulse of a Force","Linear Momentum & Impulse — p = mv and Impulse-Momentum Theorem","Learn linear momentum p = mv, impulse J = ∫Fdt, and the impulse-momentum theorem for calculating velocity changes in particle dynamics.","This topic introduces particle linear momentum, impulse of a force, and the integral form of the impulse-momentum theorem.","\u003Cp>\u003Cstrong>Linear momentum of a particle\u003C\u002Fstrong> is a vector measure of mechanical motion defined as the product of particle mass and velocity.\u003C\u002Fp>\u003Ch2>Linear momentum\u003C\u002Fh2>\u003Cp>For a particle of constant mass:\u003C\u002Fp>\u003Cp>$$\\vec p=m\\vec v.$$\u003C\u002Fp>\u003Cp>The vector $\\vec p$ has the same direction as velocity. Its SI unit is kg·m\u002Fs.\u003C\u002Fp>\u003Ch2>Impulse of a force\u003C\u002Fh2>\u003Cp>The impulse of a force from $t_1$ to $t_2$ is:\u003C\u002Fp>\u003Cp>$$\\vec J=\\int_{t_1}^{t_2}\\vec F\\,dt.$$\u003C\u002Fp>\u003Cp>For a constant force, $\\vec J=\\vec F\\Delta t$. Impulse measures the action of a force over a time interval.\u003C\u002Fp>\u003Ch2>Impulse-momentum theorem\u003C\u002Fh2>\u003Cp>Newton's second law for constant mass gives $d\\vec p\u002Fdt=\\sum\\vec F$. Integrating:\u003C\u002Fp>\u003Cp>$$\\vec p_2-\\vec p_1=\\int_{t_1}^{t_2}\\sum\\vec F\\,dt.$$\u003C\u002Fp>\u003Cp>Thus, the change in linear momentum equals the impulse of the resultant force.\u003C\u002Fp>\u003Ch2>Coordinate components\u003C\u002Fh2>\u003Cp>The vector theorem may be applied separately along coordinate directions, for example $mv_{2x}-mv_{1x}=\\int\\sum F_xdt$.\u003C\u002Fp>\u003Ch2>When the method is useful\u003C\u002Fh2>\u003Cp>The impulse approach is especially effective when velocities at two instants are needed but the detailed motion between them is not. It is also useful for large forces acting over short intervals.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 2 kg body initially moves at 3 m\u002Fs along the $x$ axis. A constant 8 N force acts in the same direction for 0.5 s. Its impulse is 4 N·s, so $2v_2-2\\cdot3=4$ and $v_2=5$ m\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing linear momentum $m\\vec v$ with kinetic energy;\u003C\u002Fli>\u003Cli>ignoring the vector nature of impulse;\u003C\u002Fli>\u003Cli>using $F\\Delta t$ for a variable force without justification;\u003C\u002Fli>\u003Cli>omitting forces from the total impulse.\u003C\u002Fli>\u003C\u002Ful>",231,[],{"id":604,"parent_id":584,"code":70,"slug":605,"name":606,"seo_title":607,"seo_description":608,"seo_text":609,"content":610,"locale":8,"uk_topic_id":611,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":605,"children":612},253,"angular-momentum-of-a-particle","Angular Momentum of a Particle","Angular Momentum of a Particle — Formula & Angular-Momentum Theorem","Calculate a particle’s angular momentum about a point or axis and use the angular-momentum theorem to relate its rate of change to applied moment.","This topic covers angular momentum of a particle and relates its rate of change to the moment of the applied force.","\u003Cp>\u003Cstrong>Angular momentum of a particle\u003C\u002Fstrong> characterizes the rotational aspect of particle motion relative to a selected point or axis.\u003C\u002Fp>\u003Ch2>Angular momentum about a point\u003C\u002Fh2>\u003Cp>For particle $M$ relative to point $O$:\u003C\u002Fp>\u003Cp>$$\\vec L_O=\\vec r\\times m\\vec v,$$\u003C\u002Fp>\u003Cp>where $\\vec r=\\overrightarrow{OM}$. The direction of $\\vec L_O$ follows from the right-hand rule.\u003C\u002Fp>\u003Ch2>Magnitude\u003C\u002Fh2>\u003Cp>The magnitude is $L_O=mvr\\sin\\theta=mv h$, where $h$ is the perpendicular distance from point $O$ to the velocity line.\u003C\u002Fp>\u003Ch2>Angular momentum about an axis\u003C\u002Fh2>\u003Cp>Angular momentum about an axis is the projection of $\\vec L_O$ onto that axis. For the $z$ axis, $L_z=(\\vec r\\times m\\vec v)\\cdot\\vec e_z$.\u003C\u002Fp>\u003Ch2>Angular-momentum theorem\u003C\u002Fh2>\u003Cp>For a fixed point $O$ in an inertial frame:\u003C\u002Fp>\u003Cp>$$\\frac{d\\vec L_O}{dt}=\\vec M_O,$$\u003C\u002Fp>\u003Cp>where $\\vec M_O=\\vec r\\times\\sum\\vec F$ is the moment of the resultant force about $O$.\u003C\u002Fp>\u003Ch2>Conservation\u003C\u002Fh2>\u003Cp>If the resultant external moment about the point is zero, then $\\vec L_O=const$. Likewise, if the sum of moments about a fixed axis is zero, the corresponding component of angular momentum is conserved.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 2 kg particle moves at 4 m\u002Fs perpendicular to a 0.5 m position vector. Its angular momentum magnitude is $L_O=mvr=2\\cdot4\\cdot0.5=4$ kg·m²\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing angular momentum with moment of force;\u003C\u002Fli>\u003Cli>reversing the cross-product order;\u003C\u002Fli>\u003Cli>using the full distance $r$ instead of the perpendicular lever arm $h$ when the vectors are not perpendicular;\u003C\u002Fli>\u003Cli>applying conservation without checking the external moment.\u003C\u002Fli>\u003C\u002Ful>",232,[],{"id":614,"parent_id":584,"code":70,"slug":615,"name":616,"seo_title":617,"seo_description":618,"seo_text":619,"content":620,"locale":8,"uk_topic_id":621,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":615,"children":622},254,"work-of-a-force-power","Work of a Force. Power","Work & Power in Dynamics — Formulas A = ∫F·dr and P = F·v","Calculate work and power in dynamics using A = ∫F·dr and P = F·v, including constant forces, common force models, signs, and rotational work.","This topic covers work done by a force along a displacement, mechanical power, the sign of work, and work of common forces.","\u003Cp>\u003Cstrong>Work of a force\u003C\u002Fstrong> measures the action of a force through a displacement, while \u003Cstrong>power\u003C\u002Fstrong> measures the rate at which work is done.\u003C\u002Fp>\u003Ch2>Elementary work\u003C\u002Fh2>\u003Cp>For an infinitesimal displacement $d\\vec r$:\u003C\u002Fp>\u003Cp>$$dA=\\vec F\\cdot d\\vec r=F\\,ds\\cos\\alpha,$$\u003C\u002Fp>\u003Cp>where $\\alpha$ is the angle between the force and displacement direction.\u003C\u002Fp>\u003Ch2>Work over a finite displacement\u003C\u002Fh2>\u003Cp>From point 1 to point 2:\u003C\u002Fp>\u003Cp>$$A_{1\\to2}=\\int_1^2\\vec F\\cdot d\\vec r.$$\u003C\u002Fp>\u003Cp>For a constant force over a straight displacement $s$, $A=Fs\\cos\\alpha$.\u003C\u002Fp>\u003Ch2>Sign of work\u003C\u002Fh2>\u003Cp>Work is positive for an acute angle between force and displacement, negative for an obtuse angle, and zero when the force is perpendicular to the instantaneous displacement.\u003C\u002Fp>\u003Ch2>Work of gravity\u003C\u002Fh2>\u003Cp>Near Earth's surface, the work of gravity depends only on the change in height: $A_g=mg(h_1-h_2)$. It is positive for downward motion and negative for upward motion.\u003C\u002Fp>\u003Ch2>Work of a spring force\u003C\u002Fh2>\u003Cp>For a spring with $F_x=-kx$:\u003C\u002Fp>\u003Cp>$$A_s=\\frac{kx_1^2}{2}-\\frac{kx_2^2}{2}.$$\u003C\u002Fp>\u003Ch2>Power\u003C\u002Fh2>\u003Cp>Instantaneous power is:\u003C\u002Fp>\u003Cp>$$P=\\frac{dA}{dt}=\\vec F\\cdot\\vec v.$$\u003C\u002Fp>\u003Cp>The SI unit is the watt: $1\\,\\text{W}=1\\,\\text{J}\u002F\\text{s}$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A constant 50 N force moves a point 3 m in the force direction. The work is 150 J. If this occurs over 5 s at a uniform average rate of doing work, the average power is 30 W.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using $Fs$ without accounting for the angle;\u003C\u002Fli>\u003Cli>confusing work and power;\u003C\u002Fli>\u003Cli>assuming work is always positive;\u003C\u002Fli>\u003Cli>using average power where instantaneous $\\vec F\\cdot\\vec v$ is required.\u003C\u002Fli>\u003C\u002Ful>",233,[],{"id":624,"parent_id":584,"code":70,"slug":625,"name":626,"seo_title":627,"seo_description":628,"seo_text":629,"content":630,"locale":8,"uk_topic_id":631,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":625,"children":632},255,"kinetic-energy-of-a-particle-work-energy-theorem","Kinetic Energy of a Particle. Work-Energy Theorem","Kinetic Energy of a Particle — mv²\u002F2 & Work-Energy Theorem","Calculate particle kinetic energy T = mv²\u002F2 and apply the work-energy theorem T2 − T1 = A to relate force work to changes in particle speed.","This topic explains particle kinetic energy and the work-energy theorem and shows how it is used to solve dynamics problems.","\u003Cp>\u003Cstrong>Kinetic energy\u003C\u002Fstrong> is a scalar measure of the mechanical motion of a particle. Unlike linear momentum, it does not depend on the direction of velocity.\u003C\u002Fp>\u003Ch2>Kinetic energy of a particle\u003C\u002Fh2>\u003Cp>For a particle of mass $m$ moving at speed $v$:\u003C\u002Fp>\u003Cp>$$T=\\frac{mv^2}{2}.$$\u003C\u002Fp>\u003Cp>Kinetic energy is nonnegative and is measured in joules in SI.\u003C\u002Fp>\u003Ch2>Differential form of the theorem\u003C\u002Fh2>\u003Cp>From $m\\vec a=\\sum\\vec F$ and $d\\vec r=\\vec vdt$:\u003C\u002Fp>\u003Cp>$$dT=\\sum\\vec F\\cdot d\\vec r=\\sum dA.$$\u003C\u002Fp>\u003Cp>Thus, the elementary change in kinetic energy equals the sum of the elementary works of the forces.\u003C\u002Fp>\u003Ch2>Integral form\u003C\u002Fh2>\u003Cp>Between positions 1 and 2:\u003C\u002Fp>\u003Cp>$$T_2-T_1=\\sum A_{1\\to2}.$$\u003C\u002Fp>\u003Cp>This is the work-energy theorem for a particle.\u003C\u002Fp>\u003Ch2>Advantages of the energy method\u003C\u002Fh2>\u003Cp>The method directly relates speed and position and often avoids solving for time. Reactions of ideal fixed constraints that are perpendicular to displacement do no work and therefore need not appear in the energy equation.\u003C\u002Fp>\u003Ch2>Sign of work and change of speed\u003C\u002Fh2>\u003Cp>Positive net work increases kinetic energy, while negative net work decreases it. However, because kinetic energy depends on $v^2$, the theorem determines speed rather than the direction of the velocity vector.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 2 kg body starts from rest and the total work of all forces over a displacement is 36 J. Then $mv^2\u002F2=36$, giving $v=6$ m\u002Fs.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>assigning kinetic energy a sign based on velocity direction;\u003C\u002Fli>\u003Cli>including the work of only one force instead of the total work;\u003C\u002Fli>\u003Cli>confusing the work-energy theorem with conservation of mechanical energy;\u003C\u002Fli>\u003Cli>using the energy equation to determine the velocity direction directly.\u003C\u002Fli>\u003C\u002Ful>",234,[],{"id":634,"parent_id":584,"code":70,"slug":635,"name":636,"seo_title":637,"seo_description":638,"seo_text":639,"content":640,"locale":8,"uk_topic_id":641,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":635,"children":642},256,"potential-force-field-potential-energy","Potential Force Field. Potential Energy","Potential Energy — Conservative Forces, Work & Force Relation","Learn conservative force fields and potential energy, the relation A = −ΔU, force from potential energy, and common gravity and spring examples.","This topic introduces conservative forces, potential force fields, and potential energy and relates force work to changes in potential energy.","\u003Cp>A \u003Cstrong>potential force field\u003C\u002Fstrong> is one in which the work done between two positions is independent of the path and depends only on the initial and final positions. Such forces are called conservative.\u003C\u002Fp>\u003Ch2>Potential energy\u003C\u002Fh2>\u003Cp>For a conservative force, potential energy $\\Pi$ is defined so that:\u003C\u002Fp>\u003Cp>$$A_{1\\to2}=\\Pi_1-\\Pi_2=-\\Delta\\Pi.$$\u003C\u002Fp>\u003Cp>The zero level of potential energy is arbitrary; only differences in potential energy have physical significance.\u003C\u002Fp>\u003Ch2>Force and potential energy\u003C\u002Fh2>\u003Cp>In three dimensions, a conservative force is related to potential energy by $\\vec F=-\\nabla\\Pi$. In one-dimensional motion this becomes $F_x=-d\\Pi\u002Fdx$.\u003C\u002Fp>\u003Ch2>Gravity near Earth's surface\u003C\u002Fh2>\u003Cp>With height $h$ measured upward, gravitational potential energy may be written $\\Pi_g=mgh+C$. Choosing zero potential at $h=0$ gives $\\Pi_g=mgh$.\u003C\u002Fp>\u003Ch2>Elastic force\u003C\u002Fh2>\u003Cp>For a linear spring with $F_x=-kx$, the elastic potential energy is:\u003C\u002Fp>\u003Cp>$$\\Pi_s=\\frac{kx^2}{2}+C.$$\u003C\u002Fp>\u003Cp>It is common to choose $\\Pi_s=0$ at $x=0$.\u003C\u002Fp>\u003Ch2>Properties of conservative forces\u003C\u002Fh2>\u003Cp>The work of a conservative force around any closed path is zero. Dry sliding friction and most resistance models are nonconservative because their work depends on the path traveled.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 3 kg body descends by 2 m. Its change in gravitational potential energy is $\\Delta\\Pi=-3g\\cdot2$, while the work of gravity is $A_g=6g\\approx58.9$ J for $g=9.81$ m\u002Fs².\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing work of a conservative force with change in potential energy; their signs are opposite;\u003C\u002Fli>\u003Cli>treating the absolute value of potential energy as unique without choosing a reference level;\u003C\u002Fli>\u003Cli>assigning potential energy to dry friction;\u003C\u002Fli>\u003Cli>omitting the minus sign in $\\vec F=-\\nabla\\Pi$.\u003C\u002Fli>\u003C\u002Ful>",235,[],{"id":644,"parent_id":584,"code":70,"slug":645,"name":646,"seo_title":647,"seo_description":648,"seo_text":649,"content":650,"locale":8,"uk_topic_id":651,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":645,"children":652},257,"conservation-of-mechanical-energy","Conservation of Mechanical Energy","Conservation of Mechanical Energy — K + U = Constant","Learn when mechanical energy is conserved, use K + U = constant for conservative systems, and distinguish conservation from work by nonconservative forces.","This topic explains the conditions for conservation of mechanical energy and the use of energy balance in dynamics problems.","\u003Cp>The \u003Cstrong>mechanical energy\u003C\u002Fstrong> of a system is the sum of its kinetic and potential energies: $E=T+\\Pi$. For a system acted on only by conservative forces, this sum remains constant.\u003C\u002Fp>\u003Ch2>Conservation law\u003C\u002Fh2>\u003Cp>If the work of all nonconservative forces is zero, then:\u003C\u002Fp>\u003Cp>$$T_1+\\Pi_1=T_2+\\Pi_2=const.$$\u003C\u002Fp>\u003Cp>Kinetic and potential energy may transform into each other while their sum remains unchanged.\u003C\u002Fp>\u003Ch2>Energy balance\u003C\u002Fh2>\u003Cp>More generally, the change in mechanical energy equals the work of nonconservative forces:\u003C\u002Fp>\u003Cp>$$E_2-E_1=A_{nc}.$$\u003C\u002Fp>\u003Cp>For example, negative work of dry friction reduces mechanical energy.\u003C\u002Fp>\u003Ch2>Gravitational system\u003C\u002Fh2>\u003Cp>For a particle moving in a uniform gravitational field without resistance, $mv^2\u002F2+mgh=const$. A decrease in height is accompanied by an increase in kinetic energy.\u003C\u002Fp>\u003Ch2>Elastic system\u003C\u002Fh2>\u003Cp>For a mass attached to an ideal spring with no losses, $mv^2\u002F2+kx^2\u002F2=const$. At extreme positions the speed may be zero while elastic potential energy is maximum.\u003C\u002Fp>\u003Ch2>Choice of potential-energy reference\u003C\u002Fh2>\u003Cp>The conservation law is independent of the chosen zero level of $\\Pi$, provided the same reference is used consistently in all states.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A body falls from rest through 5 m without resistance. Taking $\\Pi=0$ at the lower level gives $mgh=mv^2\u002F2$, so $v=\\sqrt{2gh}\\approx9.90$ m\u002Fs for $g=9.81$ m\u002Fs².\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>applying mechanical-energy conservation while omitting friction work;\u003C\u002Fli>\u003Cli>mixing different zero levels of potential energy;\u003C\u002Fli>\u003Cli>assuming kinetic and potential energy are separately constant;\u003C\u002Fli>\u003Cli>confusing conservation of mechanical energy with conservation of total energy of the physical system.\u003C\u002Fli>\u003C\u002Ful>",236,[],{"id":654,"parent_id":524,"code":70,"slug":655,"name":656,"seo_title":657,"seo_description":658,"seo_text":659,"content":660,"locale":8,"uk_topic_id":661,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":655,"children":662},295,"dynamics-of-mechanical-systems","Dynamics of Mechanical Systems","Dynamics of Mechanical Systems — Center of Mass, Momentum & Energy","Learn mechanical-system dynamics through center-of-mass motion, linear and angular momentum, kinetic energy, and the general theorems governing systems of particles.","This section covers the dynamics of systems of particles, including center-of-mass motion, momentum, angular momentum, and kinetic-energy theorems.","\u003Cp>\u003Cstrong>Dynamics of mechanical systems\u003C\u002Fstrong> extends particle dynamics to systems of interacting particles and rigid bodies. Instead of analyzing every particle separately, integral characteristics of the entire system are often used.\u003C\u002Fp>\u003Ch2>Center of mass\u003C\u002Fh2>\u003Cp>The center of mass is determined by the mass distribution. Its motion is governed by the resultant external force; internal forces do not change the motion of the center of mass of the complete system.\u003C\u002Fp>\u003Ch2>Momentum and angular momentum\u003C\u002Fh2>\u003Cp>Total linear momentum characterizes the translational aspect of system motion, while angular momentum characterizes rotational motion about a point or axis. Their change theorems connect these quantities to external forces and moments.\u003C\u002Fp>\u003Ch2>Kinetic energy of a system\u003C\u002Fh2>\u003Cp>The kinetic energy of a mechanical system is the sum of the kinetic energies of its particles. The work-energy theorem relates its change to force work and is a powerful method for analyzing complex systems.\u003C\u002Fp>",291,[663,673,683,693,703,713],{"id":664,"parent_id":654,"code":70,"slug":665,"name":666,"seo_title":667,"seo_description":668,"seo_text":669,"content":670,"locale":8,"uk_topic_id":671,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":665,"children":672},258,"dynamics-of-a-mechanical-system-center-of-mass","Dynamics of a Mechanical System. Center of Mass","Center of Mass of a Mechanical System — Formula & Dynamics","Learn how to calculate the center of mass of a mechanical system, distinguish internal and external forces, and relate center-of-mass motion to system dynamics.","This topic introduces mechanical systems, internal and external forces, total mass, and the center of mass as preparation for the general theorems of system dynamics.","\u003Cp>A \u003Cstrong>mechanical system\u003C\u002Fstrong> is a collection of particles or bodies whose motion is considered together. Moving from a single particle to a system leads to general dynamics theorems for the center of mass, momentum, angular momentum, and energy.\u003C\u002Fp>\u003Ch2>Mass of a system\u003C\u002Fh2>\u003Cp>For a system of $n$ particles, the total mass is:\u003C\u002Fp>\u003Cp>$$M=\\sum_{i=1}^{n}m_i.$$\u003C\u002Fp>\u003Cp>In classical mechanics of a closed system, this mass is treated as constant.\u003C\u002Fp>\u003Ch2>External and internal forces\u003C\u002Fh2>\u003Cp>\u003Cstrong>External forces\u003C\u002Fstrong> act on system particles from bodies outside the selected system. \u003Cstrong>Internal forces\u003C\u002Fstrong> are interactions between particles or bodies within the system. The classification depends on which bodies are included in the system.\u003C\u002Fp>\u003Ch2>Properties of internal forces\u003C\u002Fh2>\u003Cp>By Newton's third law, internal forces occur in equal and opposite pairs, so their resultant is zero. For central pair interactions, their total moment about any point is also zero.\u003C\u002Fp>\u003Ch2>Center of mass\u003C\u002Fh2>\u003Cp>The position of the center of mass $C$ is:\u003C\u002Fp>\u003Cp>$$\\vec r_C=\\frac{1}{M}\\sum_{i=1}^{n}m_i\\vec r_i.$$\u003C\u002Fp>\u003Cp>In Cartesian coordinates, $x_C=\\sum m_ix_i\u002FM$, $y_C=\\sum m_iy_i\u002FM$, and $z_C=\\sum m_iz_i\u002FM$.\u003C\u002Fp>\u003Ch2>Velocity and acceleration of the center of mass\u003C\u002Fh2>\u003Cp>For constant system mass, differentiation gives $\\vec v_C=(1\u002FM)\\sum m_i\\vec v_i$ and $\\vec a_C=(1\u002FM)\\sum m_i\\vec a_i$.\u003C\u002Fp>\u003Ch2>Continuous body\u003C\u002Fh2>\u003Cp>For a continuous mass distribution, sums are replaced by integrals: $\\vec r_C=(1\u002FM)\\int\\vec r\\,dm$. Symmetry often places the center of mass on an axis, plane, or center of symmetry.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Two particles of masses 2 kg and 3 kg lie on the $x$ axis at 0 m and 5 m. Then $x_C=(2\\cdot0+3\\cdot5)\u002F5=3$ m.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating the classification of forces as internal or external as independent of the selected system;\u003C\u002Fli>\u003Cli>using a simple average of coordinates when masses are unequal;\u003C\u002Fli>\u003Cli>confusing center of mass with the geometric center of a nonuniform body;\u003C\u002Fli>\u003Cli>concluding that internal forces do not affect the motion of individual parts of the system.\u003C\u002Fli>\u003C\u002Ful>",237,[],{"id":674,"parent_id":654,"code":70,"slug":675,"name":676,"seo_title":677,"seo_description":678,"seo_text":679,"content":680,"locale":8,"uk_topic_id":681,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":675,"children":682},259,"theorem-on-motion-of-center-of-mass","Theorem on the Motion of the Center of Mass","Center-of-Mass Motion Theorem — Equation M aC = ΣFext","Learn the center-of-mass motion theorem M aC = ΣFext, why internal forces cancel, and what zero resultant external force means for system motion.","This topic relates the acceleration of the center of mass to the resultant external force and develops important conservation consequences.","\u003Cp>The \u003Cstrong>theorem on the motion of the center of mass\u003C\u002Fstrong> describes the translational motion of a mechanical system by a single equation in which internal forces do not appear explicitly.\u003C\u002Fp>\u003Ch2>Derivation\u003C\u002Fh2>\u003Cp>For each particle, $m_i\\vec a_i=\\vec F_i^{e}+\\vec F_i^{i}$. Summing over the system cancels the internal forces, while $\\sum m_i\\vec a_i=M\\vec a_C$.\u003C\u002Fp>\u003Ch2>Fundamental equation\u003C\u002Fh2>\u003Cp>Therefore:\u003C\u002Fp>\u003Cp>$$M\\vec a_C=\\sum\\vec F^{e}=\\vec R^{e},$$\u003C\u002Fp>\u003Cp>where $\\vec R^{e}$ is the resultant of the external forces.\u003C\u002Fp>\u003Ch2>Coordinate components\u003C\u002Fh2>\u003Cp>In Cartesian coordinates:\u003C\u002Fp>\u003Cp>$$M\\ddot x_C=\\sum F_x^{e},\\qquad M\\ddot y_C=\\sum F_y^{e},\\qquad M\\ddot z_C=\\sum F_z^{e}.$$\u003C\u002Fp>\u003Cp>These equations have the same form as the equations of motion of a particle of mass $M$.\u003C\u002Fp>\u003Ch2>Zero external resultant\u003C\u002Fh2>\u003Cp>If $\\sum\\vec F^{e}=0$, then $\\vec a_C=0$ and $\\vec v_C=const$. The center of mass is either at rest or moves uniformly in a straight line.\u003C\u002Fp>\u003Ch2>Conservation along one coordinate\u003C\u002Fh2>\u003Cp>If the sum of external-force components along one axis, say $x$, is zero, then $v_{Cx}=const$. If additionally $v_{Cx}(0)=0$, the coordinate $x_C$ remains constant.\u003C\u002Fp>\u003Ch2>Internal motions\u003C\u002Fh2>\u003Cp>Internal forces can strongly change the relative positions of system parts, but by themselves cannot change the center-of-mass motion of an isolated system. Motion of one part is therefore accompanied by compensating motion of other parts.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A horizontal external resultant of 30 N acts on a system of mass 10 kg. Regardless of internal interactions, $a_C=30\u002F10=3$ m\u002Fs² in the force direction.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>including internal forces in the center-of-mass equation after summing the complete system;\u003C\u002Fli>\u003Cli>assuming $\\sum\\vec F^e=0$ means the center of mass must be stationary rather than have constant velocity;\u003C\u002Fli>\u003Cli>applying the theorem to only part of a system without reclassifying forces;\u003C\u002Fli>\u003Cli>identifying center-of-mass motion with the motion of every system particle.\u003C\u002Fli>\u003C\u002Ful>",238,[],{"id":684,"parent_id":654,"code":70,"slug":685,"name":686,"seo_title":687,"seo_description":688,"seo_text":689,"content":690,"locale":8,"uk_topic_id":691,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":685,"children":692},261,"impulse-momentum-theorem-mechanical-system","Impulse-Momentum Theorem for a Mechanical System","Impulse-Momentum Theorem for a System — Formula & Conservation","Apply the impulse-momentum theorem to a mechanical system using external-force impulse and learn when total linear momentum is conserved.","This topic develops the impulse-momentum theorem for a mechanical system and the conditions under which total linear momentum is conserved.","\u003Cp>The \u003Cstrong>linear momentum of a mechanical system\u003C\u002Fstrong> is the vector sum of the momenta of all its particles. The impulse-momentum theorem describes the translational aspect of system motion using only external forces.\u003C\u002Fp>\u003Ch2>Momentum of a system\u003C\u002Fh2>\u003Cp>For a system of $n$ particles:\u003C\u002Fp>\u003Cp>$$\\vec Q=\\sum_{i=1}^{n}m_i\\vec v_i.$$\u003C\u002Fp>\u003Cp>Since $\\vec v_C=(1\u002FM)\\sum m_i\\vec v_i$, an important relation follows:\u003C\u002Fp>\u003Cp>$$\\vec Q=M\\vec v_C.$$\u003C\u002Fp>\u003Ch2>Differential form\u003C\u002Fh2>\u003Cp>Summing the equations of motion of all particles and using cancellation of internal forces gives:\u003C\u002Fp>\u003Cp>$$\\frac{d\\vec Q}{dt}=\\sum\\vec F^{e}=\\vec R^{e}.$$\u003C\u002Fp>\u003Cp>The rate of change of system momentum equals the resultant external force.\u003C\u002Fp>\u003Ch2>Integral form\u003C\u002Fh2>\u003Cp>Over the interval from $t_1$ to $t_2$:\u003C\u002Fp>\u003Cp>$$\\vec Q_2-\\vec Q_1=\\int_{t_1}^{t_2}\\sum\\vec F^{e}\\,dt.$$\u003C\u002Fp>\u003Cp>Thus, the change in total momentum equals the total impulse of the external forces.\u003C\u002Fp>\u003Ch2>Conservation of momentum\u003C\u002Fh2>\u003Cp>If $\\sum\\vec F^{e}=0$, then $\\vec Q=const$. If only the resultant external-force component along one axis is zero, the corresponding momentum component is conserved.\u003C\u002Fp>\u003Ch2>Internal forces\u003C\u002Fh2>\u003Cp>Internal forces may change the velocities of individual parts but do not change total system momentum. This principle is useful in recoil, separation, and many impact problems when external impulse is negligible.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Two bodies of masses 2 kg and 3 kg move along one axis at 4 m\u002Fs and -1 m\u002Fs. The system momentum is $Q_x=2\\cdot4+3\\cdot(-1)=5$ kg·m\u002Fs. If external impulse along the axis is zero, this value remains constant.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>adding momentum magnitudes instead of vector components;\u003C\u002Fli>\u003Cli>including internal forces in the total external impulse;\u003C\u002Fli>\u003Cli>assuming conservation of $\\vec Q$ means every particle velocity remains unchanged;\u003C\u002Fli>\u003Cli>applying conservation without checking external impulse.\u003C\u002Fli>\u003C\u002Ful>",239,[],{"id":694,"parent_id":654,"code":70,"slug":695,"name":696,"seo_title":697,"seo_description":698,"seo_text":699,"content":700,"locale":8,"uk_topic_id":701,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":695,"children":702},262,"angular-momentum-theorem-mechanical-system","Angular-Momentum Theorem for a Mechanical System","Angular-Momentum Theorem for a System — Formula & Conservation","Calculate angular momentum of a mechanical system and use the angular-momentum theorem to relate its change to external moments and conservation conditions.","This topic develops the angular-momentum theorem for a mechanical system and explains conservation when the resultant external moment vanishes.","\u003Cp>The \u003Cstrong>angular momentum of a mechanical system\u003C\u002Fstrong> characterizes the rotational aspect of system motion relative to a selected point or axis.\u003C\u002Fp>\u003Ch2>Angular momentum about a point\u003C\u002Fh2>\u003Cp>About a fixed point $O$:\u003C\u002Fp>\u003Cp>$$\\vec K_O=\\sum_{i=1}^{n}\\vec r_i\\times m_i\\vec v_i.$$\u003C\u002Fp>\u003Cp>Angular momentum about an axis is the projection of this vector onto that axis.\u003C\u002Fp>\u003Ch2>Angular-momentum theorem\u003C\u002Fh2>\u003Cp>For a fixed point $O$ in an inertial frame:\u003C\u002Fp>\u003Cp>$$\\frac{d\\vec K_O}{dt}=\\sum\\vec M_O^{e}.$$\u003C\u002Fp>\u003Cp>The time derivative of system angular momentum equals the resultant moment of external forces about the same point.\u003C\u002Fp>\u003Ch2>Role of internal forces\u003C\u002Fh2>\u003Cp>For central pairwise internal forces, the moments of each interaction pair cancel. Thus only the total moment of external forces remains in the equation for the complete system.\u003C\u002Fp>\u003Ch2>Theorem about an axis\u003C\u002Fh2>\u003Cp>Projecting onto a fixed $z$ axis gives:\u003C\u002Fp>\u003Cp>$$\\frac{dK_z}{dt}=\\sum M_z^{e}.$$\u003C\u002Fp>\u003Ch2>Conservation\u003C\u002Fh2>\u003Cp>If $\\sum\\vec M_O^{e}=0$, then $\\vec K_O=const$. If only the resultant external moment about a particular axis is zero, only the corresponding angular-momentum component is conserved.\u003C\u002Fp>\u003Ch2>Rigid-body rotation\u003C\u002Fh2>\u003Cp>For a rigid body rotating about a fixed principal axis $z$, $K_z=I_z\\omega$. With constant $I_z$, the equation becomes $I_z\\dot\\omega=\\sum M_z^{e}$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If a rigid body has $I_z=2$ kg·m² and rotates at $\\omega=5$ rad\u002Fs, then $K_z=10$ kg·m²\u002Fs. With zero external moment about the axis, this value remains constant.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing angular momentum with moment of force;\u003C\u002Fli>\u003Cli>taking force moments about one point and angular momentum about another;\u003C\u002Fli>\u003Cli>using $K_z=I_z\\omega$ without checking the motion and axis conditions;\u003C\u002Fli>\u003Cli>claiming conservation of the full vector when only one external-moment component is zero.\u003C\u002Fli>\u003C\u002Ful>",240,[],{"id":704,"parent_id":654,"code":70,"slug":705,"name":706,"seo_title":707,"seo_description":708,"seo_text":709,"content":710,"locale":8,"uk_topic_id":711,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":705,"children":712},263,"kinetic-energy-of-a-mechanical-system","Kinetic Energy of a Mechanical System","Kinetic Energy of a Mechanical System — Rigid-Body Formulas","Calculate kinetic energy of particle systems and rigid bodies in translation, fixed-axis rotation, and plane motion using standard energy formulas.","This topic covers kinetic energy of a mechanical system and the standard expressions for the principal types of rigid-body motion.","\u003Cp>The \u003Cstrong>kinetic energy of a mechanical system\u003C\u002Fstrong> is the sum of the kinetic energies of all its particles. It is a scalar quantity, so the energies of individual parts are added algebraically.\u003C\u002Fp>\u003Ch2>General definition\u003C\u002Fh2>\u003Cp>For a system of $n$ particles:\u003C\u002Fp>\u003Cp>$$T=\\sum_{i=1}^{n}\\frac{m_iv_i^2}{2}.$$\u003C\u002Fp>\u003Cp>System kinetic energy is nonnegative and is measured in joules.\u003C\u002Fp>\u003Ch2>König's theorem\u003C\u002Fh2>\u003Cp>The kinetic energy of a system can be decomposed into the kinetic energy of translational motion of its center of mass and the kinetic energy of motion relative to the center of mass:\u003C\u002Fp>\u003Cp>$$T=\\frac{Mv_C^2}{2}+T_C.$$\u003C\u002Fp>\u003Cp>This decomposition is especially useful for rigid bodies.\u003C\u002Fp>\u003Ch2>Rigid-body translation\u003C\u002Fh2>\u003Cp>In pure translation, all points have the same velocity, so:\u003C\u002Fp>\u003Cp>$$T=\\frac{Mv_C^2}{2}.$$\u003C\u002Fp>\u003Ch2>Rotation about a fixed axis\u003C\u002Fh2>\u003Cp>If a rigid body rotates with angular velocity $\\omega$ about a fixed $z$ axis:\u003C\u002Fp>\u003Cp>$$T=\\frac{I_z\\omega^2}{2},$$\u003C\u002Fp>\u003Cp>where $I_z$ is the mass moment of inertia about the rotation axis.\u003C\u002Fp>\u003Ch2>Plane motion\u003C\u002Fh2>\u003Cp>For plane motion of a rigid body:\u003C\u002Fp>\u003Cp>$$T=\\frac{Mv_C^2}{2}+\\frac{I_C\\omega^2}{2},$$\u003C\u002Fp>\u003Cp>where $I_C$ is the mass moment of inertia about the axis through the center of mass perpendicular to the plane of motion.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 4 kg disk has center-of-mass speed 3 m\u002Fs, central mass moment of inertia 0.5 kg·m², and angular speed 4 rad\u002Fs. Then $T=4\\cdot3^2\u002F2+0.5\\cdot4^2\u002F2=18+4=22$ J.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using only translational energy for a body that is also rotating;\u003C\u002Fli>\u003Cli>using a mass moment of inertia about the wrong axis;\u003C\u002Fli>\u003Cli>adding velocities instead of kinetic energies;\u003C\u002Fli>\u003Cli>forgetting that relative kinetic energy is zero in pure translation.\u003C\u002Fli>\u003C\u002Ful>",241,[],{"id":714,"parent_id":654,"code":70,"slug":715,"name":716,"seo_title":717,"seo_description":718,"seo_text":719,"content":720,"locale":8,"uk_topic_id":721,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":715,"children":722},264,"work-energy-theorem-mechanical-system","Work-Energy Theorem for a Mechanical System","Work-Energy Theorem for a Mechanical System — Formula & Work","Apply T2 − T1 = A to mechanical systems using kinetic energy and the work of external and internal forces for translation, rotation, and plane motion.","This topic explains the work-energy theorem for a mechanical system and the roles of external and internal forces in changing kinetic energy.","\u003Cp>The \u003Cstrong>work-energy theorem for a mechanical system\u003C\u002Fstrong> relates the change in total kinetic energy to the work of forces acting on the particles of the system.\u003C\u002Fp>\u003Ch2>Differential form\u003C\u002Fh2>\u003Cp>For each particle, the elementary change in kinetic energy equals the elementary work of the applied forces. Summing over the system gives:\u003C\u002Fp>\u003Cp>$$dT=\\sum dA^{e}+\\sum dA^{i},$$\u003C\u002Fp>\u003Cp>where superscripts $e$ and $i$ denote external and internal forces.\u003C\u002Fp>\u003Ch2>Integral form\u003C\u002Fh2>\u003Cp>Between configurations 1 and 2:\u003C\u002Fp>\u003Cp>$$T_2-T_1=\\sum A_{1\\to2}^{e}+\\sum A_{1\\to2}^{i}.$$\u003C\u002Fp>\u003Cp>Unlike the momentum and angular-momentum theorems, the work of internal forces is not zero in a general mechanical system.\u003C\u002Fp>\u003Ch2>Internal forces in a rigid body\u003C\u002Fh2>\u003Cp>For an ideal rigid body, internal interaction forces do no net work because distances between body particles remain unchanged. In a deformable system, internal forces may perform work associated with changes in internal or potential energy.\u003C\u002Fp>\u003Ch2>Ideal constraints\u003C\u002Fh2>\u003Cp>Reactions of ideal fixed constraints often do no work when the point of application has no displacement in the reaction direction. This can eliminate unknown reactions from the energy equation.\u003C\u002Fp>\u003Ch2>Application to rigid bodies\u003C\u002Fh2>\u003Cp>First evaluate $T_1$ and $T_2$ using the expression appropriate to translation, rotation, or plane motion, then equate their difference to the total work of forces and couples.\u003C\u002Fp>\u003Ch2>Work of a moment in rotation\u003C\u002Fh2>\u003Cp>For a moment $M_z$ acting during rotation about a fixed axis:\u003C\u002Fp>\u003Cp>$$A=\\int_{\\varphi_1}^{\\varphi_2}M_z\\,d\\varphi.$$\u003C\u002Fp>\u003Cp>For a constant moment, $A=M_z(\\varphi_2-\\varphi_1)$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A system starts from rest and the total work of its external and internal forces is 50 J. Its final kinetic energy is therefore 50 J. Determining individual velocities then depends on the type of motion and mass distribution.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>automatically neglecting work of all internal forces for every mechanical system;\u003C\u002Fli>\u003Cli>using the wrong kinetic-energy expression for the type of motion;\u003C\u002Fli>\u003Cli>including reactions that do no work or omitting reactions of moving constraints that do work;\u003C\u002Fli>\u003Cli>confusing the work of a moment with the moment itself.\u003C\u002Fli>\u003C\u002Ful>",242,[],{"id":724,"parent_id":524,"code":70,"slug":725,"name":726,"seo_title":727,"seo_description":728,"seo_text":729,"content":730,"locale":8,"uk_topic_id":731,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":725,"children":732},296,"analytical-mechanics-and-oscillations","Analytical Mechanics and Oscillations","Analytical Mechanics — D’Alembert, Virtual Work & Lagrange Equations","Learn analytical mechanics through D’Alembert’s principle, virtual displacements and work, Lagrange’s equations of the second kind, and an introduction to small oscillations.","This section combines foundations of analytical mechanics — D'Alembert's principle, virtual work, and Lagrange's equations — with an introduction to small oscillations.","\u003Cp>\u003Cstrong>Analytical mechanics and oscillations\u003C\u002Fstrong> brings together methods for describing constrained systems through virtual displacements and generalized coordinates. These methods are especially useful when direct Newtonian equations would introduce many unknown constraint reactions.\u003C\u002Fp>\u003Ch2>D'Alembert's principle\u003C\u002Fh2>\u003Cp>Introducing inertia forces allows equations of dynamics to be written in a form resembling equilibrium equations. This does not make the problem static; it is a mathematical reformulation that supports further analysis.\u003C\u002Fp>\u003Ch2>Virtual displacements\u003C\u002Fh2>\u003Cp>For ideal constraints, the total virtual work of constraint reactions is zero. Combined with D'Alembert's principle, this leads to the general equation of dynamics and allows ideal reactions to be eliminated.\u003C\u002Fp>\u003Ch2>Lagrange's equations\u003C\u002Fh2>\u003Cp>Using independent generalized coordinates leads to Lagrange's equations of the second kind, a systematic method for deriving equations of motion of constrained mechanical systems.\u003C\u002Fp>\u003Ch2>Small oscillations\u003C\u002Fh2>\u003Cp>Near a stable equilibrium, motion can often be linearized. For a one-degree-of-freedom system this produces the classical models of free, damped, and forced oscillations and introduces the concept of resonance.\u003C\u002Fp>",292,[733,743,753,763,773],{"id":734,"parent_id":724,"code":70,"slug":735,"name":736,"seo_title":737,"seo_description":738,"seo_text":739,"content":740,"locale":8,"uk_topic_id":741,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":735,"children":742},265,"dalemberts-principle-particle-system","D'Alembert's Principle for a Particle and a System","D’Alembert’s Principle — Inertia Force & Dynamic Equilibrium","Learn D’Alembert’s principle for a particle and mechanical system, define inertia force −ma, and rewrite equations of motion as formal equilibrium equations.","This topic introduces inertia forces and D'Alembert's principle, which rewrites dynamics equations in a formally static form.","\u003Cp>\u003Cstrong>D'Alembert's principle\u003C\u002Fstrong> rewrites equations of dynamics in a formally equilibrium-like form by adding inertia forces to the applied forces and constraint reactions.\u003C\u002Fp>\u003Ch2>Inertia force of a particle\u003C\u002Fh2>\u003Cp>For a particle of mass $m$ with acceleration $\\vec a$, define the inertia force:\u003C\u002Fp>\u003Cp>$$\\vec F^{in}=-m\\vec a.$$\u003C\u002Fp>\u003Cp>This is a computational construct, not an additional physical interaction with another body.\u003C\u002Fp>\u003Ch2>D'Alembert's principle for a particle\u003C\u002Fh2>\u003Cp>The equation $m\\vec a=\\sum\\vec F$ may be rewritten as:\u003C\u002Fp>\u003Cp>$$\\sum\\vec F+\\vec F^{in}=0.$$\u003C\u002Fp>\u003Cp>The resulting form resembles static equilibrium even though the particle may be accelerating.\u003C\u002Fp>\u003Ch2>Mechanical system\u003C\u002Fh2>\u003Cp>For every particle of a system, introduce $\\vec F_i^{in}=-m_i\\vec a_i$. The applied forces, constraint reactions, and inertia forces then form a formally balanced system in the sense of D'Alembert's principle.\u003C\u002Fp>\u003Ch2>Resultant inertia force\u003C\u002Fh2>\u003Cp>For a system of constant mass:\u003C\u002Fp>\u003Cp>$$\\vec R^{in}=\\sum\\vec F_i^{in}=-M\\vec a_C.$$\u003C\u002Fp>\u003Cp>Thus the resultant inertia force is determined by the acceleration of the center of mass.\u003C\u002Fp>\u003Ch2>Resultant moment of inertia forces\u003C\u002Fh2>\u003Cp>About a selected point $O$:\u003C\u002Fp>\u003Cp>$$\\vec M_O^{in}=\\sum\\vec r_i\\times\\vec F_i^{in}.$$\u003C\u002Fp>\u003Cp>Together with the resultant inertia force, it is useful in rigid-body dynamics and in determining support reactions.\u003C\u002Fp>\u003Ch2>Practical use\u003C\u002Fh2>\u003Cp>The method is convenient when constraint reactions must be found for a system whose motion is known. After introducing inertia forces, equilibrium-style equations may be used, provided all required inertial terms are included correctly.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A 5 kg body translates with acceleration 3 m\u002Fs² to the right. Its inertia force has magnitude 15 N and points to the left. In D'Alembert's equation it is included together with the real external forces.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>treating inertia force as an ordinary interaction force;\u003C\u002Fli>\u003Cli>directing $\\vec F^{in}$ along acceleration instead of opposite to it;\u003C\u002Fli>\u003Cli>omitting moments of inertia forces in rotational motion;\u003C\u002Fli>\u003Cli>using static equilibrium equations without all required inertial terms.\u003C\u002Fli>\u003C\u002Ful>",243,[],{"id":744,"parent_id":724,"code":70,"slug":745,"name":746,"seo_title":747,"seo_description":748,"seo_text":749,"content":750,"locale":8,"uk_topic_id":751,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":745,"children":752},266,"virtual-displacements-principle-of-virtual-work","Virtual Displacements. Principle of Virtual Work","Principle of Virtual Work — Virtual Displacements & Equilibrium","Learn virtual displacements, ideal constraints, and virtual work, then apply the principle of virtual work to equilibrium without solving constraint reactions directly.","This topic introduces virtual displacements, ideal constraints, and the principle of virtual work as a foundation of analytical mechanics.","\u003Cp>A \u003Cstrong>virtual displacement\u003C\u002Fstrong> is an infinitesimal imagined displacement compatible with the constraints at a given instant. The principle of virtual work provides equilibrium conditions without explicitly determining reactions of ideal constraints.\u003C\u002Fp>\u003Ch2>Virtual displacements\u003C\u002Fh2>\u003Cp>Virtual displacements of system particles are denoted $\\delta\\vec r_i$. They must satisfy the geometric restrictions imposed by the constraints but are not actual displacements occurring during a time interval $dt$.\u003C\u002Fp>\u003Ch2>Virtual work\u003C\u002Fh2>\u003Cp>The elementary virtual work of forces is:\u003C\u002Fp>\u003Cp>$$\\delta A=\\sum_i\\vec F_i\\cdot\\delta\\vec r_i.$$\u003C\u002Fp>\u003Cp>For applied couples, corresponding terms $M\\,\\delta\\varphi$ are added.\u003C\u002Fp>\u003Ch2>Ideal constraints\u003C\u002Fh2>\u003Cp>Constraints are ideal if the total virtual work of their reactions is zero for every admissible virtual displacement. Their reactions can then be omitted from the virtual-work equation.\u003C\u002Fp>\u003Ch2>Principle of virtual work\u003C\u002Fh2>\u003Cp>For equilibrium of a system with ideal constraints, under the usual conditions of stationary constraints, it is necessary and sufficient that:\u003C\u002Fp>\u003Cp>$$\\sum_i\\vec F_i^{a}\\cdot\\delta\\vec r_i=0$$\u003C\u002Fp>\u003Cp>for every admissible virtual displacement, where $\\vec F_i^{a}$ are the applied active forces.\u003C\u002Fp>\u003Ch2>Generalized coordinates\u003C\u002Fh2>\u003Cp>If the configuration is described by independent coordinates $q_j$, then $\\delta\\vec r_i=\\sum_j(\\partial\\vec r_i\u002F\\partial q_j)\\delta q_j$. Virtual work can be written $\\delta A=\\sum_jQ_j\\delta q_j$, where $Q_j$ are generalized forces.\u003C\u002Fp>\u003Ch2>Advantage of the method\u003C\u002Fh2>\u003Cp>The method is particularly effective for mechanisms with many constraint reactions: for ideal constraints, these reactions are automatically eliminated from the equation.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a lever that can rotate about a fixed axis, an admissible virtual displacement is described by a small rotation $\\delta\\varphi$. The condition $\\delta A=0$ reduces to zero algebraic sum of moments of the active forces about the axis.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing virtual displacement with actual displacement during $dt$;\u003C\u002Fli>\u003Cli>choosing a displacement incompatible with the constraints;\u003C\u002Fli>\u003Cli>discarding work of reactions of nonideal constraints;\u003C\u002Fli>\u003Cli>treating $\\delta$ as an ordinary time differential.\u003C\u002Fli>\u003C\u002Ful>",244,[],{"id":754,"parent_id":724,"code":70,"slug":755,"name":756,"seo_title":757,"seo_description":758,"seo_text":759,"content":760,"locale":8,"uk_topic_id":761,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":755,"children":762},267,"general-equation-of-dynamics","General Equation of Dynamics","General Equation of Dynamics — D’Alembert & Virtual Work","Derive equations of motion for systems with ideal constraints by combining D’Alembert’s principle with virtual work in the general equation of dynamics.","This topic develops the general equation of dynamics for systems with ideal constraints by combining D'Alembert's principle with the principle of virtual work.","\u003Cp>The \u003Cstrong>general equation of dynamics\u003C\u002Fstrong> combines D'Alembert's principle with the principle of virtual work. It provides equations of motion for systems with ideal constraints without explicitly introducing their reactions.\u003C\u002Fp>\u003Ch2>Starting idea\u003C\u002Fh2>\u003Cp>By D'Alembert's principle, inertia forces $\\vec F_i^{in}=-m_i\\vec a_i$ are added to active forces and constraint reactions. The principle of virtual work is then applied to the formally balanced system.\u003C\u002Fp>\u003Ch2>General equation\u003C\u002Fh2>\u003Cp>For a system with ideal constraints:\u003C\u002Fp>\u003Cp>$$\\sum_i(\\vec F_i^{a}-m_i\\vec a_i)\\cdot\\delta\\vec r_i=0.$$\u003C\u002Fp>\u003Cp>Reactions of ideal constraints do not appear because their total virtual work is zero.\u003C\u002Fp>\u003Ch2>Physical meaning\u003C\u002Fh2>\u003Cp>The equation does not mean that the system is in static equilibrium. The inertial terms represent actual accelerations, while virtual displacements are a mathematical device for eliminating constraint reactions.\u003C\u002Fp>\u003Ch2>Generalized coordinates\u003C\u002Fh2>\u003Cp>If a system has $s$ degrees of freedom described by $q_j$, admissible displacements can be expressed through independent variations $\\delta q_j$. Grouping coefficients of each $\\delta q_j$ yields $s$ independent equations of motion.\u003C\u002Fp>\u003Ch2>Advantages\u003C\u002Fh2>\u003Cp>The general equation is particularly useful for constrained systems in which direct application of Newton's laws introduces many unknown reactions.\u003C\u002Fp>\u003Ch2>Connection with Lagrange's equations\u003C\u002Fh2>\u003Cp>Passing to independent generalized coordinates and expressing the inertial terms through kinetic energy leads to Lagrange's equations of the second kind.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a one-degree-of-freedom system, all admissible $\\delta\\vec r_i$ are expressed through one $\\delta q$. Substitution gives $B(q,\\dot q,\\ddot q,t)\\delta q=0$. Since $\\delta q$ is arbitrary, $B=0$ is the equation of motion.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>retaining reactions of ideal constraints without using their zero virtual work;\u003C\u002Fli>\u003Cli>treating inertia forces as ordinary active forces;\u003C\u002Fli>\u003Cli>using dependent coordinate variations as if they were independent;\u003C\u002Fli>\u003Cli>confusing the general equation of dynamics with static equilibrium.\u003C\u002Fli>\u003C\u002Ful>",245,[],{"id":764,"parent_id":724,"code":70,"slug":765,"name":766,"seo_title":767,"seo_description":768,"seo_text":769,"content":770,"locale":8,"uk_topic_id":771,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":765,"children":772},268,"lagranges-equations-of-the-second-kind","Lagrange's Equations of the Second Kind","Lagrange’s Equations of the Second Kind — Formula & Coordinates","Learn Lagrange’s equations of the second kind using generalized coordinates, generalized forces, kinetic energy, and a systematic method for equations of motion.","This topic introduces generalized coordinates and forces and develops Lagrange's equations of the second kind as a systematic method for deriving equations of motion.","\u003Cp>\u003Cstrong>Lagrange's equations of the second kind\u003C\u002Fstrong> provide a systematic way to derive equations of motion using independent generalized coordinates without explicitly introducing reactions of ideal constraints.\u003C\u002Fp>\u003Ch2>Generalized coordinates\u003C\u002Fh2>\u003Cp>If a system has $s$ degrees of freedom, its configuration can be described by independent coordinates $q_1,\\ldots,q_s$. These may be linear displacements, angles, or other parameters that uniquely determine configuration.\u003C\u002Fp>\u003Ch2>Generalized velocities\u003C\u002Fh2>\u003Cp>The derivatives $\\dot q_j$ are generalized velocities. The kinetic energy is written as a function $T(q_j,\\dot q_j,t)$.\u003C\u002Fp>\u003Ch2>Generalized forces\u003C\u002Fh2>\u003Cp>The virtual work of active forces is written:\u003C\u002Fp>\u003Cp>$$\\delta A=\\sum_{j=1}^{s}Q_j\\delta q_j,$$\u003C\u002Fp>\u003Cp>where $Q_j$ is the generalized force corresponding to coordinate $q_j$.\u003C\u002Fp>\u003Ch2>Lagrange's equations\u003C\u002Fh2>\u003Cp>For a system with ideal constraints:\u003C\u002Fp>\u003Cp>$$\\frac{d}{dt}\\left(\\frac{\\partial T}{\\partial\\dot q_j}\\right)-\\frac{\\partial T}{\\partial q_j}=Q_j,\\qquad j=1,\\ldots,s.$$\u003C\u002Fp>\u003Cp>The number of independent equations equals the number of degrees of freedom.\u003C\u002Fp>\u003Ch2>Conservative forces\u003C\u002Fh2>\u003Cp>If forces have potential energy $\\Pi(q,t)$, their conservative generalized-force contribution is $Q_j=-\\partial\\Pi\u002F\\partial q_j$. With the Lagrangian $L=T-\\Pi$, the equations may be written $d(\\partial L\u002F\\partial\\dot q_j)\u002Fdt-\\partial L\u002F\\partial q_j=Q_j^{nc}$, where $Q_j^{nc}$ are nonconservative generalized forces.\u003C\u002Fp>\u003Ch2>Solution procedure\u003C\u002Fh2>\u003Col>\u003Cli>determine the number of degrees of freedom;\u003C\u002Fli>\u003Cli>choose independent $q_j$;\u003C\u002Fli>\u003Cli>express positions and velocities through $q_j,\\dot q_j$;\u003C\u002Fli>\u003Cli>calculate $T$;\u003C\u002Fli>\u003Cli>determine $Q_j$ or $\\Pi$;\u003C\u002Fli>\u003Cli>write one Lagrange equation for each coordinate.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a mass $m$ on a horizontal spring with coordinate $x$, $T=m\\dot x^2\u002F2$ and $\\Pi=kx^2\u002F2$. Lagrange's equation gives $m\\ddot x+kx=0$.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>choosing dependent coordinates as independent;\u003C\u002Fli>\u003Cli>assuming every generalized force has units of newtons — for an angular coordinate it has units of moment;\u003C\u002Fli>\u003Cli>omitting coordinate dependence of kinetic energy;\u003C\u002Fli>\u003Cli>counting the same conservative force both through $\\Pi$ and through $Q_j$.\u003C\u002Fli>\u003C\u002Ful>",246,[],{"id":774,"parent_id":724,"code":70,"slug":775,"name":776,"seo_title":777,"seo_description":778,"seo_text":779,"content":780,"locale":8,"uk_topic_id":781,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":775,"children":782},269,"small-oscillations-one-degree-of-freedom-system","Small Oscillations of a One-Degree-of-Freedom System","Single-DOF Oscillations — Natural Frequency, Forced Motion & Resonance","Learn free and forced small oscillations of a one-degree-of-freedom system, including the equation of motion, natural frequency, amplitude, damping, and resonance.","This topic covers linearized free and forced oscillations of a one-degree-of-freedom system, natural frequency, damping, and resonance.","\u003Cp>\u003Cstrong>Small oscillations\u003C\u002Fstrong> occur near a stable equilibrium when deviations are small enough for the equations of motion to be linearized. A one-degree-of-freedom system is described by one generalized coordinate.\u003C\u002Fp>\u003Ch2>Undamped free oscillations\u003C\u002Fh2>\u003Cp>The standard linear equation is:\u003C\u002Fp>\u003Cp>$$m\\ddot x+kx=0.$$\u003C\u002Fp>\u003Cp>The natural circular frequency is:\u003C\u002Fp>\u003Cp>$$\\omega_n=\\sqrt{\\frac{k}{m}},$$\u003C\u002Fp>\u003Cp>and the period is $T=2\\pi\u002F\\omega_n$.\u003C\u002Fp>\u003Ch2>Free-oscillation response\u003C\u002Fh2>\u003Cp>The solution may be written $x=C_1\\cos\\omega_nt+C_2\\sin\\omega_nt$ or $x=A\\cos(\\omega_nt+\\varphi)$. Amplitude and initial phase are determined from the initial conditions.\u003C\u002Fp>\u003Ch2>Viscous damping\u003C\u002Fh2>\u003Cp>With linear resistance $c\\dot x$:\u003C\u002Fp>\u003Cp>$$m\\ddot x+c\\dot x+kx=0.$$\u003C\u002Fp>\u003Cp>The type of motion depends on damping relative to its critical value. With light damping, the system undergoes decaying oscillations.\u003C\u002Fp>\u003Ch2>Forced oscillations\u003C\u002Fh2>\u003Cp>For harmonic excitation $F_0\\cos\\Omega t$:\u003C\u002Fp>\u003Cp>$$m\\ddot x+c\\dot x+kx=F_0\\cos\\Omega t.$$\u003C\u002Fp>\u003Cp>The steady-state response has excitation frequency $\\Omega$, while its amplitude depends on frequency ratio and damping.\u003C\u002Fp>\u003Ch2>Resonance\u003C\u002Fh2>\u003Cp>In an ideal undamped system, harmonic excitation at $\\Omega=\\omega_n$ produces resonant growth of amplitude. With damping, the amplitude remains finite and the frequency-response maximum shifts depending on damping.\u003C\u002Fp>\u003Ch2>Linearization near equilibrium\u003C\u002Fh2>\u003Cp>For a general system, the coordinate is measured from stable equilibrium and only first-order terms in the small deviation are retained. The result has the form $m_{eq}\\ddot q+c_{eq}\\dot q+k_{eq}q=Q(t)$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For $m=2$ kg and $k=50$ N\u002Fm without damping, $\\omega_n=\\sqrt{50\u002F2}=5$ rad\u002Fs and $T=2\\pi\u002F5\\approx1.26$ s.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing circular frequency in rad\u002Fs with ordinary frequency $f=\\omega\u002F(2\\pi)$ in hertz;\u003C\u002Fli>\u003Cli>using a linear small-oscillation model for large deviations without checking validity;\u003C\u002Fli>\u003Cli>ignoring damping when estimating resonant amplitude of a real system;\u003C\u002Fli>\u003Cli>using a physical mass directly instead of equivalent inertia for a compound mechanism without derivation.\u003C\u002Fli>\u003C\u002Ful>",247,[],{"id":784,"parent_id":70,"code":785,"slug":786,"name":787,"seo_title":788,"seo_description":789,"seo_text":790,"content":791,"locale":8,"uk_topic_id":78,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":786,"children":792},45,"1","strength-of-materials","Strength of Materials","Strength of Materials — Theory and Solved Problems","Learn Strength of Materials online: fundamentals, formulas, stress analysis, and structural design.","Strength of Materials is a core engineering discipline studying methods for calculating structural elements and machine parts for strength, stiffness, and stability. This section covers fundamental concepts of loads, internal forces, stresses, and strains. You will master classical hypotheses and assumptions, such as material continuity, isotropy, and Bernoulli's hypothesis of plane sections. Our online course includes detailed theoretical materials, graphical explanations, and step-by-step solutions for typical exam problems.","\u003Cp>\u003Cstrong>Strength of Materials\u003C\u002Fstrong> is an engineering discipline concerned with the strength, stiffness, and stability of structural members and machine components under load. Its central task is to connect external actions with internal forces, stresses, and strains so that components can be designed safely and efficiently.\u003C\u002Fp>\u003Ch2>What Strength of Materials studies\u003C\u002Fh2>\u003Cp>Real structural elements deform when loaded. Equilibrium equations and support reactions alone are therefore not enough: the internal force resultants must be determined, the stress and strain state evaluated, and the relevant allowable limits checked.\u003C\u002Fp>\u003Cp>The course considers bars, shafts, beams, thin-walled shells, and other common engineering members. Basic loading modes include tension and compression, shear, torsion, bending, and their combinations.\u003C\u002Fp>\u003Ch2>Three basic performance requirements\u003C\u002Fh2>\u003Cul>\u003Cli>\u003Cstrong>Strength\u003C\u002Fstrong> — the ability to carry load without fracture or unacceptable plastic deformation.\u003C\u002Fli>\u003Cli>\u003Cstrong>Stiffness\u003C\u002Fstrong> — the ability to keep displacements and deformations within specified limits.\u003C\u002Fli>\u003Cli>\u003Cstrong>Stability\u003C\u002Fstrong> — the ability to preserve the required equilibrium configuration and avoid loss of stability at a critical load.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Basic engineering calculation sequence\u003C\u002Fh2>\u003Col>\u003Cli>Define the structural model, geometry, material, supports, and loads.\u003C\u002Fli>\u003Cli>Determine reactions and internal force resultants, commonly using the method of sections.\u003C\u002Fli>\u003Cli>Calculate stresses and strains with the model appropriate to the loading mode.\u003C\u002Fli>\u003Cli>Identify the critical section or critical state.\u003C\u002Fli>\u003Cli>Check strength, stiffness, and, where required, stability.\u003C\u002Fli>\u003Cli>Select or adjust member dimensions based on the verification results.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Fundamental concepts\u003C\u002Fh2>\u003Cp>\u003Cstrong>Stress\u003C\u002Fstrong> describes the intensity of internal forces in a material, while \u003Cstrong>strain\u003C\u002Fstrong> describes changes in dimensions and shape. For simple axial loading, the average normal stress is:\u003C\u002Fp>\u003Cp>$$\\sigma=\\frac{N}{A},$$\u003C\u002Fp>\u003Cp>where N is the axial force and A is the cross-sectional area. In the linear-elastic range, normal stress and longitudinal strain are related by Hooke's law:\u003C\u002Fp>\u003Cp>$$\\sigma=E\\varepsilon.$$\u003C\u002Fp>\u003Cp>These elementary relations are a starting point. Bending, torsion, multiaxial stress states, stability, shell behavior, and contact problems require their corresponding specialized models.\u003C\u002Fp>\u003Ch2>Course structure\u003C\u002Fh2>\u003Cp>The material progresses from fundamental to more advanced models: tension and compression; shear and direct shear; torsion; stress and strain state; geometric properties of plane areas; bending and beam deflections; combined loading; energy methods; stability of compressed members; dynamic and cyclic loading; shell analysis; and contact stresses.\u003C\u002Fp>\u003Ch2>Limits of engineering models\u003C\u002Fh2>\u003Cp>Strength-of-materials formulas rely on assumptions about geometry, material behavior, deformation magnitude, and loading. Their applicability should be checked before use. This is especially important for plasticity, local stress concentrations, large deformations, contact problems, and loss of stability.\u003C\u002Fp>\u003Ch2>How to use this section\u003C\u002Fh2>\u003Cp>For each topic, first understand the physical model, sign convention, governing equations, units, and calculation procedure. Then consolidate the formulas through representative engineering examples and check each result for dimensional consistency, reasonable magnitude, and physical meaning.\u003C\u002Fp>",[793,941,1089,1109,1181,1221,1287,1352,1395,1427,1530,1550,1592,1666],{"id":794,"parent_id":784,"code":795,"slug":796,"name":797,"seo_title":798,"seo_description":799,"seo_text":800,"content":801,"locale":8,"uk_topic_id":802,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":796,"children":803},142,"1.0","basic-concepts-and-types-of-deformation","Basic Concepts and Types of Deformation","Basic Concepts and Types of Deformation | Strength of Materials","Introduction to strength of materials: deformation, displacement, strength, stiffness, stability, elastic and plastic deformation, tension, compression, shear, torsion and bending.","This introductory Strength of Materials topic explains deformation and displacement and distinguishes strength, stiffness, and stability as fundamental design requirements. It introduces elastic and plastic deformation and the main deformation modes of structural members: tension, compression, shear, torsion, and bending. These concepts provide the foundation for studying internal forces, stresses, strains, and engineering design checks.","\u003Cp>\u003Cstrong>Strength of Materials\u003C\u002Fstrong> studies the behavior of deformable structural members under load. Before studying tension, compression, torsion, or bending, it is useful to establish a common set of concepts: deformation, material and structural models, and the criteria used to assess structural performance.\u003C\u002Fp>\u003Cp>This introductory section covers the scope of Strength of Materials; deformation and displacement; elastic and plastic behavior; basic deformation modes; strength, stiffness and stability; isotropy, anisotropy and orthotropy; composite materials; and the principal assumptions and engineering idealizations used in calculation models.\u003C\u002Fp>\u003Ch2>Key concepts\u003C\u002Fh2>{{chunk:som-basic-concepts-en}}\u003Ch2>Map of basic deformation modes\u003C\u002Fh2>{{chunk:som-deformation-classification-en}}\u003Cp>The purpose of this section is not to introduce every specialized equation, but to establish the language and assumptions used throughout the rest of the course. Detailed equations and calculation methods are introduced in the corresponding loading and material-behavior topics.\u003C\u002Fp>",141,[804,815,826,837,848,859,870,882,893,905,917,929],{"id":805,"parent_id":794,"code":806,"slug":807,"name":808,"seo_title":809,"seo_description":810,"seo_text":811,"content":812,"locale":8,"uk_topic_id":813,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":807,"children":814},151,"1.0.1","scope-and-objectives-of-strength-of-materials","Scope and Objectives of Strength of Materials","What Is Strength of Materials? Scope, Objectives & Analysis Workflow","Learn what Strength of Materials studies and how engineers turn real structures into models to calculate internal forces, stresses, strains, displacements, and design checks.","This topic introduces the scope and objectives of Strength of Materials and the transition from a real structural component to an engineering calculation model, internal forces, stresses, strains, displacements, and design checks.","\u003Cp>\u003Cstrong>Strength of Materials\u003C\u002Fstrong> studies deformable structural members and machine components and provides engineering methods for analyzing them under load. Unlike rigid-body statics, it accounts for changes in shape and dimensions.\u003C\u002Fp>\u003Ch2>Main objective\u003C\u002Fh2>\u003Cp>The engineer determines internal force resultants, stresses, strains, and displacements and then checks whether the structure satisfies requirements for strength, stiffness, and stability.\u003C\u002Fp>\u003Ch2>From a real object to a model\u003C\u002Fh2>\u003Cp>A real structure is idealized into bars, beams, shafts, shells, and other calculation elements. Geometry, material properties, supports, and loads are specified. The validity of later equations depends on the quality of this model.\u003C\u002Fp>\u003Ch2>Typical analysis sequence\u003C\u002Fh2>\u003Col>\u003Cli>Construct the calculation model.\u003C\u002Fli>\u003Cli>Determine external loads and reactions.\u003C\u002Fli>\u003Cli>Find internal force resultants.\u003C\u002Fli>\u003Cli>Calculate stresses, strains, and displacements.\u003C\u002Fli>\u003Cli>Check strength, stiffness and, where required, stability.\u003C\u002Fli>\u003C\u002Fol>",143,[],{"id":816,"parent_id":794,"code":817,"slug":818,"name":819,"seo_title":820,"seo_description":821,"seo_text":822,"content":823,"locale":8,"uk_topic_id":824,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":818,"children":825},152,"1.0.2","deformation-and-displacement","Deformation and Displacement","Deformation vs Displacement — What’s the Difference?","Understand the difference between deformation and rigid-body displacement, absolute and relative dimensional change, strain, and the small-deformation assumption.","This topic distinguishes rigid-body displacement from deformation, introduces absolute and relative changes in dimensions, and explains the small-deformation assumption used in introductory Strength of Materials.","\u003Cp>\u003Cstrong>Displacement\u003C\u002Fstrong> describes the change in position of a point, whereas \u003Cstrong>deformation\u003C\u002Fstrong> describes a change in the body's shape or dimensions—that is, a change in the relative positions of its points.\u003C\u002Fp>\u003Ch2>Displacement of a point\u003C\u002Fh2>\u003Cp>If a point initially has position vector \\(\\vec r\\) and moves to \\(\\vec r'\\), its displacement vector is\u003C\u002Fp>\u003Cp>\\[\\vec u=\\vec r'-\\vec r.\\]\u003C\u002Fp>\u003Cp>In Cartesian coordinates,\u003C\u002Fp>\u003Cp>\\[\\vec u=u_x\\vec i+u_y\\vec j+u_z\\vec k.\\]\u003C\u002Fp>{{chunk:deformation-displacement-body-motion}}\u003Ch2>Motion without deformation\u003C\u002Fh2>\u003Cp>A rigid body may translate or rotate while the distance between every pair of its points remains unchanged. Therefore, motion of a body does not by itself imply deformation.\u003C\u002Fp>\u003Ch2>Deformation\u003C\u002Fh2>\u003Cp>In a deformable body, different points generally undergo different displacements. Spatial variation of the displacement field changes lengths, angles, and shape.\u003C\u002Fp>\u003Cp>For a small uniaxial elongation of a bar, the average normal strain is\u003C\u002Fp>\u003Cp>\\[\\varepsilon=\\frac{\\Delta L}{L}.\\]\u003C\u002Fp>\u003Cp>Strain is dimensionless, whereas displacement has units of length.\u003C\u002Fp>\u003Ch2>Small-deformation assumption\u003C\u002Fh2>\u003Cp>Elementary strength-of-materials models often assume displacements and rotations are small enough that equilibrium can be written using the initial geometry. Large displacement or rotation may require a geometrically nonlinear formulation.\u003C\u002Fp>\u003Ch2>Why the distinction matters\u003C\u002Fh2>\u003Cp>Serviceability and stiffness checks limit structural displacement and deformation. A beam may satisfy a stress-based strength requirement yet still be unacceptable because its deflection is excessive.\u003C\u002Fp>",144,[],{"id":827,"parent_id":794,"code":828,"slug":829,"name":830,"seo_title":831,"seo_description":832,"seo_text":833,"content":834,"locale":8,"uk_topic_id":835,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":829,"children":836},153,"1.0.3","elastic-and-plastic-deformation","Elastic and Plastic Deformation","Elastic vs Plastic Deformation — Key Differences Explained","Understand elastic and plastic deformation, unloading and permanent strain, time-dependent behavior, and when linear-elastic material models stop applying.","This topic explains reversible elastic deformation, irreversible plastic deformation, loading and unloading, and why linear-elastic equations have limited applicability.","\u003Cp>According to their behavior after unloading, deformations are commonly separated into \u003Cstrong>elastic\u003C\u002Fstrong> and \u003Cstrong>plastic\u003C\u002Fstrong> components. Real material response may also depend on time, loading rate, and temperature.\u003C\u002Fp>\u003Ch2>Elastic deformation\u003C\u002Fh2>\u003Cp>Elastic deformation disappears after the load is removed: within the adopted model, the body recovers its original shape and dimensions. At sufficiently low loads many engineering materials are approximated as linearly elastic.\u003C\u002Fp>\u003Ch2>Plastic deformation\u003C\u002Fh2>\u003Cp>Plastic deformation is irreversible. After unloading, permanent deformation remains. A real deformation process may contain both elastic and plastic components.\u003C\u002Fp>\u003Ch2>Time-dependent behavior\u003C\u002Fh2>\u003Cp>Some materials do not respond independently of loading duration. \u003Cstrong>Creep\u003C\u002Fstrong> is the development of deformation with time under sustained loading and can be especially important at elevated temperature. \u003Cstrong>Stress relaxation\u003C\u002Fstrong> is a decrease in stress with time while deformation is maintained. \u003Cstrong>Viscoelastic behavior\u003C\u002Fstrong> combines elastic response with time dependence.\u003C\u002Fp>\u003Ch2>Why the distinction matters\u003C\u002Fh2>\u003Cp>Linear-elastic equations cannot automatically be extended into ranges of significant plastic deformation or processes in which time and temperature are important. The calculation model must match the actual material response. Detailed stress-strain behavior is considered later with mechanical properties of materials.\u003C\u002Fp>",145,[],{"id":838,"parent_id":794,"code":839,"slug":840,"name":841,"seo_title":842,"seo_description":843,"seo_text":844,"content":845,"locale":8,"uk_topic_id":846,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":840,"children":847},154,"1.0.4","basic-modes-of-deformation","Basic Modes of Deformation","Basic Deformation Modes: Tension, Compression, Shear, Torsion and Bending","Overview of tension, compression, shear, torsion, bending, and combined loading of structural members.","This introductory topic organizes the principal deformation modes of structural members: tension, compression, shear, torsion, and bending, and introduces combined loading.","{{chunk:som-deformation-classification}}\u003Ch2>Rod deformation visualization\u003C\u002Fh2>{{chunk:threejs-rod-deformation-types}}\u003Ch2>Simple and combined cases\u003C\u002Fh2>\u003Cp>The division into tension, compression, shear, torsion, and bending provides the basic models used in engineering calculations. In a real member, several internal force resultants may act simultaneously; this is treated as combined loading.\u003C\u002Fp>\u003Ch2>Why classify deformation modes\u003C\u002Fh2>\u003Cp>Each case has characteristic internal force resultants, stress distributions, and deformation relations. Correctly recognizing how a member works is therefore the first step in selecting an appropriate calculation model.\u003C\u002Fp>\u003Cp>The following major sections of the course examine axial tension and compression, shear and torsion, bending, and their combinations in sequence.\u003C\u002Fp>",146,[],{"id":849,"parent_id":794,"code":850,"slug":851,"name":852,"seo_title":853,"seo_description":854,"seo_text":855,"content":856,"locale":8,"uk_topic_id":857,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":851,"children":858},155,"1.0.5","strength-stiffness-and-stability","Strength, Stiffness and Stability","Strength vs Stiffness vs Stability — What’s the Difference?","Understand the difference between strength, stiffness, and stability, why each requires a separate structural check, and which failure mode can govern design.","This topic distinguishes strength, stiffness, and stability and explains why a structure can be strong enough yet unacceptable because of excessive deformation or loss of equilibrium stability.","{{chunk:som-basic-concepts-en}}\u003Ch2>Three different checks\u003C\u002Fh2>\u003Cp>\u003Cstrong>Strength\u003C\u002Fstrong> concerns failure or an unacceptable material state. \u003Cstrong>Stiffness\u003C\u002Fstrong> limits deformation and displacement even when strength is adequate. \u003Cstrong>Stability\u003C\u002Fstrong> concerns the ability to preserve the intended equilibrium configuration.\u003C\u002Fp>\u003Ch2>Why strength alone is not enough\u003C\u002Fh2>\u003Cp>A beam may have acceptable stresses but excessive deflection. A slender compressed member may buckle at stresses below those associated with material failure. The governing performance criterion therefore depends on the structure and loading.\u003C\u002Fp>\u003Ch2>Engineering approach\u003C\u002Fh2>\u003Cp>Real designs often require several checks simultaneously. Later topics provide the equations used for these checks.\u003C\u002Fp>",147,[],{"id":860,"parent_id":794,"code":861,"slug":862,"name":863,"seo_title":864,"seo_description":865,"seo_text":866,"content":867,"locale":8,"uk_topic_id":868,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":862,"children":869},156,"1.0.6","material-properties-and-models","Basic Material Properties and Models","Material Models in Strength of Materials — Elastic & Plastic","Learn the main material models used in strength of materials: linear elastic, nonlinear elastic, elastic-plastic behavior, stiffness, and Poisson’s ratio.","This topic introduces key material idealizations in Strength of Materials: homogeneity, heterogeneity, isotropy, anisotropy and orthotropy, together with linear, nonlinear, elastic and elastoplastic behavior.","\u003Cp>In mechanics of deformable solids, a real material is represented by a \u003Cstrong>constitutive model\u003C\u002Fstrong>—an idealized law relating stress, strain, and, when required, loading history.\u003C\u002Fp>\u003Ch2>Basic properties\u003C\u002Fh2>\u003Cp>\u003Cstrong>Elasticity\u003C\u002Fstrong> is the ability to recover the original shape after unloading. \u003Cstrong>Plasticity\u003C\u002Fstrong> is the ability to develop permanent deformation without immediate fracture. \u003Cstrong>Stiffness\u003C\u002Fstrong> describes resistance to elastic deformation and is associated with elastic moduli.\u003C\u002Fp>\u003Ch2>Linear-elastic model\u003C\u002Fh2>\u003Cp>For a uniaxial state, the simplest model follows Hooke's law:\u003C\u002Fp>\u003Cp>\\[\\sigma=E\\varepsilon,\\]\u003C\u002Fp>\u003Cp>where \\(E\\) is Young's modulus. Complete unloading returns the material to its initial state.\u003C\u002Fp>{{chunk:material-model-stress-strain-response}}\u003Ch2>Nonlinear-elastic model\u003C\u002Fh2>\u003Cp>If the relation between \\(\\sigma\\) and \\(\\varepsilon\\) is nonlinear but complete unloading leaves no permanent strain, the response is nonlinear elastic.\u003C\u002Fp>\u003Ch2>Elastic-plastic model\u003C\u002Fh2>\u003Cp>In an elastic-plastic model, deformation is elastic before yielding. Once plastic deformation develops, part of the strain remains after unloading.\u003C\u002Fp>\u003Cp>In the ideal elastic-perfectly plastic model, continued plastic deformation after the yield stress \\(\\sigma_y\\) occurs without an increase in stress. Hardening models account for increasing resistance during plastic deformation.\u003C\u002Fp>\u003Ch2>Poisson's ratio\u003C\u002Fh2>\u003Cp>Under uniaxial tension or compression, longitudinal strain is accompanied by transverse strain. In linear elasticity, Poisson's ratio is\u003C\u002Fp>\u003Cp>\\[\\nu=-\\frac{\\varepsilon_{\\perp}}{\\varepsilon_{\\parallel}}.\\]\u003C\u002Fp>\u003Ch2>Why material models matter\u003C\u002Fh2>\u003Cp>The appropriate model depends on load level and the purpose of the analysis. Linear elasticity is often sufficient for small recoverable strains. Plasticity models are required when permanent deformation, post-yield response, or limit states are important.\u003C\u002Fp>",148,[],{"id":871,"parent_id":794,"code":872,"slug":873,"name":874,"seo_title":875,"seo_description":876,"seo_text":877,"content":878,"locale":8,"uk_topic_id":879,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":873,"children":881},157,"1.0.7","composite-materials","Composite Materials","Composite Materials in Strength of Materials","Matrix and reinforcement, fiber-reinforced, laminated and sandwich composites, directional properties, and characteristic failure mechanisms.","An introduction to composite materials for Strength of Materials: matrix and reinforcement, major composite architectures, anisotropy, directional stiffness and strength, and characteristic damage mechanisms.","\u003Cp>A \u003Cstrong>composite material\u003C\u002Fstrong> combines two or more constituents to obtain a useful set of properties. Many structural composites consist of a \u003Cstrong>matrix\u003C\u002Fstrong>, which binds the system and transfers load, and \u003Cstrong>reinforcement\u003C\u002Fstrong>, which strongly influences stiffness and strength.\u003C\u002Fp>\u003Ch2>Common types\u003C\u002Fh2>\u003Cul>\u003Cli>\u003Cstrong>Fiber-reinforced composites\u003C\u002Fstrong>, such as carbon- or glass-fiber reinforced polymers.\u003C\u002Fli>\u003Cli>\u003Cstrong>Laminates\u003C\u002Fstrong>, built from layers whose orientations are selected for the loading.\u003C\u002Fli>\u003Cli>\u003Cstrong>Sandwich structures\u003C\u002Fstrong>, combining strong thin faces with a lightweight thick core.\u003C\u002Fli>\u003Cli>\u003Cstrong>Particle-reinforced materials\u003C\u002Fstrong>, containing a distributed second phase.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Direction matters\u003C\u002Fh2>\u003Cp>In a unidirectional fiber composite, stiffness and strength along the fibers can differ greatly from properties transverse to them. Anisotropy and orthotropy are therefore central concepts in composite mechanics.\u003C\u002Fp>\u003Ch2>Characteristic damage\u003C\u002Fh2>\u003Cp>Composite failure may involve matrix cracking, fiber failure, delamination, or loss of bonding between constituents. Detailed composite analysis therefore requires models beyond elementary isotropic Strength of Materials.\u003C\u002Fp>\u003Cp>This introductory topic primarily establishes the limits of equations derived for homogeneous isotropic materials.\u003C\u002Fp>",149,7,[],{"id":883,"parent_id":794,"code":884,"slug":885,"name":886,"seo_title":886,"seo_description":887,"seo_text":888,"content":889,"locale":8,"uk_topic_id":890,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":891,"url_slug":885,"children":892},158,"1.0.8","assumptions-and-idealizations","Assumptions and Idealizations in Strength of Materials","Continuum assumption, small deformation, one-dimensional member models, Saint-Venant’s principle, and limits of simplified calculations.","This topic explains why Strength of Materials relies on idealizations and assumptions, including the continuum model, small deformation, beam and bar idealizations, Saint-Venant’s principle, and limits of applicability.","\u003Cp>Strength of Materials uses simplified models that make real structures accessible to engineering equations. Every simplification has a \u003Cstrong>range of applicability\u003C\u002Fstrong>.\u003C\u002Fp>\u003Ch2>Continuum assumption\u003C\u002Fh2>\u003Cp>Material is represented as a continuous medium even though its microscopic structure may be atomic, granular, fibrous, or otherwise heterogeneous. This allows stress and strain to be treated as spatial fields.\u003C\u002Fp>\u003Ch2>Small deformation and displacement\u003C\u002Fh2>\u003Cp>Classical linear problems assume sufficiently small changes in geometry. If displacement substantially changes the equilibrium geometry, a geometrically nonlinear formulation may be required.\u003C\u002Fp>\u003Ch2>Member idealization\u003C\u002Fh2>\u003Cp>When one dimension is much larger than the cross-sectional dimensions, an element can often be modeled as a bar, beam, or shaft. The full three-dimensional geometry is replaced by a longitudinal axis and cross-sectional properties.\u003C\u002Fp>\u003Ch2>Saint-Venant’s principle\u003C\u002Fh2>\u003Cp>At sufficient distance from a load application region, statically equivalent load distributions generally produce similar stress fields. This permits simplification of local load details, but does not remove local stress effects near the point of application.\u003C\u002Fp>\u003Ch2>Idealizations must be justified\u003C\u002Fh2>\u003Cp>Holes, stress concentrations, contact zones, large deformation, anisotropy, and complex geometry may require more detailed models.\u003C\u002Fp>",150,8,[],{"id":894,"parent_id":794,"code":895,"slug":896,"name":897,"seo_title":898,"seo_description":899,"seo_text":900,"content":901,"locale":8,"uk_topic_id":902,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":903,"url_slug":896,"children":904},163,"1.0.9","loads-and-supports","Loads and Supports","Loads & Supports — Reactions and Free-Body Diagrams Explained","Learn concentrated and distributed loads, support reactions, rollers, pins and fixed supports, and how to build a correct free-body diagram.","This introductory topic explains how external actions and restraints are represented in Strength of Materials, including concentrated forces, moments, distributed loads, body forces, time-dependent loading, and idealized supports.","\u003Cp>Structural analysis begins by defining the \u003Cstrong>external actions\u003C\u002Fstrong> and restraints. Loads and supports are transferred to a calculation model that preserves the mechanically important features of the real structure.\u003C\u002Fp>\u003Ch2>Main load types\u003C\u002Fh2>\u003Cul>\u003Cli>\u003Cstrong>Concentrated force\u003C\u002Fstrong> — an idealized force acting at a point or over a region whose dimensions are negligible at the scale of the model.\u003C\u002Fli>\u003Cli>\u003Cstrong>Concentrated moment\u003C\u002Fstrong> — an idealized couple producing a moment without a resultant force.\u003C\u002Fli>\u003Cli>\u003Cstrong>Distributed load\u003C\u002Fstrong> — an action distributed along a length, over a surface, or another geometric region and described by an intensity.\u003C\u002Fli>\u003Cli>\u003Cstrong>Body forces\u003C\u002Fstrong> — forces acting throughout the material volume, such as gravity or inertia forces.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Variation with time\u003C\u002Fh2>\u003Cp>Loads may be static or time-dependent. When inertia effects are negligible, a quasi-static model may be used. Impact, vibration, and other rapidly varying actions can require dynamic analysis.\u003C\u002Fp>\u003Ch2>Supports and reactions\u003C\u002Fh2>\u003Cp>A \u003Cstrong>support\u003C\u002Fstrong> models a connection between a structural element and another part of the structure or its foundation. It prevents selected translations or rotations. Each independent restrained motion is associated with a corresponding unknown reaction component.\u003C\u002Fp>\u003Cp>In a planar problem, a rigid body has three independent possible motions: translation along $x$, translation along $y$, and rotation in the plane. The support type determines which of these degrees of freedom remain possible.\u003C\u002Fp>\u003Ch3>Roller support\u003C\u002Fh3>\u003Cp>A \u003Cstrong>roller support\u003C\u002Fstrong> restrains translation in one direction while allowing motion along the supporting surface and allowing rotation. For an ideal smooth surface it produces \u003Cstrong>one reaction\u003C\u002Fstrong> normal to that surface.\u003C\u002Fp>\u003Cp>This model is commonly used for movable beam supports. An important purpose is to avoid unnecessary restraint, for example by permitting longitudinal movement caused by deformation or thermal expansion.\u003C\u002Fp>\u003Ch3>Pin support\u003C\u002Fh3>\u003Cp>A \u003Cstrong>pin support\u003C\u002Fstrong> restrains translation of the supported point in two independent planar directions but allows rotation about the pin. It therefore produces \u003Cstrong>two reaction components\u003C\u002Fstrong>, commonly written $R_x$ and $R_y$ or $A_x$ and $A_y$.\u003C\u002Fp>\u003Cp>An ideal pin does not transmit a reaction moment. The direction of the resultant reaction is not known in advance and follows from its components after solving the equilibrium equations.\u003C\u002Fp>\u003Ch3>Fixed support\u003C\u002Fh3>\u003Cp>A \u003Cstrong>fixed support\u003C\u002Fstrong> in a planar model restrains both translations and the rotation of the attached section. It therefore produces \u003Cstrong>three reaction quantities\u003C\u002Fstrong>: two force components $R_x$, $R_y$, and a reaction moment $M$.\u003C\u002Fp>\u003Cp>The fixed end of a cantilever beam is modeled this way when the connection to the foundation is sufficiently rigid relative to deformation of the beam.\u003C\u002Fp>\u003Ch3>Two-force link\u003C\u002Fh3>\u003Cp>An ideal straight link pinned at both ends and carrying no intermediate loads transmits a force \u003Cstrong>along its own axis\u003C\u002Fstrong>. The reaction direction is therefore known in advance, while its magnitude and actual sense are determined from equilibrium.\u003C\u002Fp>\u003Ch3>Flexible cable or rope\u003C\u002Fh3>\u003Cp>An ideal flexible cable can transmit only \u003Cstrong>tension\u003C\u002Fstrong> along its axis. It cannot provide a compressive reaction; if the geometry and loading would require compression, the cable becomes slack and that restraint is no longer active.\u003C\u002Fp>\u003Ch3>Contact with a smooth surface\u003C\u002Fh3>\u003Cp>Without friction, a smooth surface produces only a \u003Cstrong>normal contact reaction\u003C\u002Fstrong>. It prevents penetration into the surface but cannot transmit tangential force. A frictional contact model may additionally include a tangential component.\u003C\u002Fp>\u003Ch2>Supports in three dimensions\u003C\u002Fh2>\u003Cp>A rigid body in space has six independent possible motions: three translations and three rotations. A fully fixed spatial support can therefore develop three force components and three moment components. Spatial pins, guides, bearings, and other restraints remove only the degrees of freedom prohibited by their mechanical construction.\u003C\u002Fp>\u003Ch2>Real connection versus ideal support\u003C\u002Fh2>\u003Cp>The name of a real connection does not by itself determine its mathematical model. Bolted, welded, bearing, and other connections may behave differently depending on geometry and stiffness. The engineer must determine which motions are effectively restrained and which forces or moments the connection can transmit.\u003C\u002Fp>\u003Ch2>Why the support model matters\u003C\u002Fh2>\u003Cp>An incorrect load direction, application point, distribution, or support model changes reactions and internal force resultants. Excessive restraints may make the model statically indeterminate, while insufficient restraints may leave it kinematically unstable. Selecting supports is therefore part of the physical problem definition, not merely a graphical convention.\u003C\u002Fp>",159,9,[],{"id":906,"parent_id":794,"code":907,"slug":908,"name":909,"seo_title":910,"seo_description":911,"seo_text":912,"content":913,"locale":8,"uk_topic_id":914,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":915,"url_slug":908,"children":916},164,"1.0.10","structural-element-models","Structural Element Models","Structural Element Models: Bars, Beams, Shafts, Plates and Shells","Basic structural models in Strength of Materials: bars, beams, shafts, columns, plates, shells, solid bodies, axes and cross-sections.","This topic introduces the main geometric idealizations used in Strength of Materials: bars, beams, shafts, columns, plates, shells, and three-dimensional solid bodies, together with the concepts of axis and cross-section.","\u003Cp>A real component or structural part is replaced by a geometric model sufficient to describe the mechanics of interest. The chosen element type determines which dimensions, displacements, and internal force resultants are central to the analysis.\u003C\u002Fp>\u003Ch2>One-dimensional members\u003C\u002Fh2>\u003Cp>A \u003Cstrong>bar or member\u003C\u002Fstrong> has one dimension, its length, substantially greater than its cross-sectional dimensions. Its geometry is represented by a longitudinal axis and cross-sections.\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>Beam\u003C\u002Fstrong> — a member commonly characterized by bending.\u003C\u002Fli>\u003Cli>\u003Cstrong>Shaft\u003C\u002Fstrong> — a member commonly used to transmit torque.\u003C\u002Fli>\u003Cli>\u003Cstrong>Column\u003C\u002Fstrong> — a compressed slender member for which stability can govern design.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Plates and shells\u003C\u002Fh2>\u003Cp>A \u003Cstrong>plate\u003C\u002Fstrong> has a thickness small compared with its other two dimensions and is represented by a middle plane. A \u003Cstrong>shell\u003C\u002Fstrong> is also thin but has a curved middle surface.\u003C\u002Fp>\u003Ch2>Three-dimensional solids\u003C\u002Fh2>\u003Cp>When all three characteristic dimensions are comparable and lower-dimensional idealizations do not capture the required mechanics, a three-dimensional solid model is used.\u003C\u002Fp>\u003Ch2>Cross-section\u003C\u002Fh2>\u003Cp>For member models, cross-sectional shape and dimensions determine area and geometric properties that control stress, stiffness, and stability. Internal force resultants are introduced through a cross-section using the method of sections.\u003C\u002Fp>",160,10,[],{"id":918,"parent_id":794,"code":919,"slug":920,"name":921,"seo_title":922,"seo_description":923,"seo_text":924,"content":925,"locale":8,"uk_topic_id":926,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":927,"url_slug":920,"children":928},165,"1.0.11","external-and-internal-forces-method-of-sections","External and Internal Forces. Method of Sections","Method of Sections — How to Find Internal Forces N, Q, M & T","Learn the method of sections step by step: cut and isolate a member, apply equilibrium, and determine axial force N, shear Q, bending moment M, and torque T.","This topic explains the fundamental method of sections and how the action of the removed part is replaced by internal force and moment resultants, including axial force N, shear force Q, torque T, and bending moment M.","\u003Cp>\u003Cstrong>External forces\u003C\u002Fstrong> act on a body from other bodies or physical fields. Within a loaded body, its parts interact with one another; this interaction is represented by internal forces.\u003C\u002Fp>\u003Ch2>Method of sections\u003C\u002Fh2>\u003Cp>To expose the internal interaction, the body is conceptually cut at the location of interest. \u003Cstrong>Either side may then be removed\u003C\u002Fstrong> — left or right — and the action of the removed part on the retained part is replaced by internal force and moment resultants at the cut.\u003C\u002Fp>\u003Cp>Both choices describe the same internal interaction. Correct equilibrium equations therefore give consistent results whether the left or right portion is analyzed.\u003C\u002Fp>\u003Ch2>Internal resultants of a member\u003C\u002Fh2>\u003Cp>In a general spatial case, the resultant force and moment at a cross-section are resolved into components. Common notation includes \u003Cstrong>axial force $N$\u003C\u002Fstrong>, \u003Cstrong>shear forces $Q$\u003C\u002Fstrong>, \u003Cstrong>torque $T$\u003C\u002Fstrong>, and \u003Cstrong>bending moments $M$\u003C\u002Fstrong>. Which components are nonzero depends on the loading.\u003C\u002Fp>\u003Ch2>Using the method\u003C\u002Fh2>\u003Cp>After isolating a portion of the member, its internal resultants are found from equilibrium. These resultants are then used to determine stresses. For example, under centric axial tension or compression the primary internal resultant is $N$.\u003C\u002Fp>\u003Ch2>Physical meaning\u003C\u002Fh2>\u003Cp>An internal resultant represents the integrated effect of stresses over the entire cross-section. Stress describes how that interaction is distributed locally over the section.\u003C\u002Fp>",161,11,[],{"id":930,"parent_id":794,"code":931,"slug":932,"name":933,"seo_title":934,"seo_description":935,"seo_text":936,"content":937,"locale":8,"uk_topic_id":938,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":939,"url_slug":932,"children":940},166,"1.0.12","stress-and-strain-at-a-point","Stress and Strain at a Point","Stress and Strain Explained — σ, τ, ε and γ","Understand stress and strain at a point: normal stress, shear stress, normal strain, and shear strain, with clear definitions and the key physical differences.","This topic introduces local measures of stress and strain: normal stress, shear stress, normal strain, and shear strain, and distinguishes cross-sectional internal resultants from stress at a material point.","\u003Cp>Internal force resultants describe the total interaction between portions of a body across a section. To describe this interaction \u003Cstrong>locally\u003C\u002Fstrong> at a material point, stress is introduced. Local changes in geometry are described by strain.\u003C\u002Fp>\u003Ch2>Stress\u003C\u002Fh2>\u003Cp>The internal interaction acting on a small oriented area can be resolved into a component normal to the area and a component lying in its plane. These define \u003Cstrong>normal stress $\\sigma$\u003C\u002Fstrong> and \u003Cstrong>shear stress $\\tau$\u003C\u002Fstrong>.\u003C\u002Fp>\u003Cp>Stress is not a force. A force or internal resultant is an integrated quantity, while stress describes the intensity of distributed internal interaction. Its dimension is force per unit area.\u003C\u002Fp>\u003Ch2>Normal strain\u003C\u002Fh2>\u003Cp>\u003Cstrong>Normal strain $\\varepsilon$\u003C\u002Fstrong> describes the relative change in length of a material line element. For a one-dimensional small elongation, its average value is $\\varepsilon=\\Delta l\u002Fl$.\u003C\u002Fp>\u003Ch2>Shear strain\u003C\u002Fh2>\u003Cp>\u003Cstrong>Shear strain $\\gamma$\u003C\u002Fstrong> describes the change of an initially right angle between material directions. It is particularly important in shear and torsion.\u003C\u002Fp>\u003Ch2>State at a point\u003C\u002Fh2>\u003Cp>The values of $\\sigma$, $\\tau$, $\\varepsilon$, and $\\gamma$ depend on position and on the orientation of the plane or direction considered. A complete stress and strain state is multicomponent and is developed later in dedicated topics. At this introductory stage, the key distinction is between global internal resultants and local material measures.\u003C\u002Fp>",162,12,[],{"id":942,"parent_id":784,"code":943,"slug":944,"name":945,"seo_title":946,"seo_description":947,"seo_text":948,"content":949,"locale":8,"uk_topic_id":139,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":944,"children":950},46,"2","tension-and-compression","Tension and Compression","Tension and Compression — Stress & Strain Calculations","Comprehensive guide to axial loading: normal stress, axial strain, Hooke's law, and diagrams.","Tension and compression are basic forms of deformation where the only internal force factor in a structural member's cross-section is the axial force N. This section describes procedures for constructing normal force diagrams, calculating normal stresses, and linear strains. You will study Hooke's law under axial load, Young's modulus, and Poisson's ratio. Special attention is given to strength conditions and cross-sectional dimension design.","\u003Cp>\u003Cstrong>Tension and compression\u003C\u002Fstrong> are forms of axial deformation of a bar in which the external forces act along its longitudinal axis. In the simplest model of centric tension or compression, the cross-section carries one internal force resultant: the \u003Cstrong>axial force $N$\u003C\u002Fstrong>.\u003C\u002Fp>\u003Ch2>Internal force and stress\u003C\u002Fh2>\u003Cp>The axial force is determined by the method of sections from the equilibrium equations of a cut portion of the bar. Tension is commonly taken as positive and compression as negative. For a centrally loaded prismatic bar, sufficiently far from local disturbances, the normal stress is the axial force divided by the cross-sectional area.\u003C\u002Fp>{{chunk:axial-normal-stress}}\u003Ch2>Deformation\u003C\u002Fh2>\u003Cp>Under axial force, the length of the bar changes. It elongates in tension and shortens in compression. Axial strain characterizes the change in length relative to the initial length.\u003C\u002Fp>{{chunk:axial-bar-elongation}}\u003Cp>The relation between stress and strain in the linear-elastic range is described by Hooke's law.\u003C\u002Fp>{{chunk:uniaxial-hooke-law}}\u003Ch2>Strength and stiffness\u003C\u002Fh2>\u003Cp>Axial design is not limited to calculating stress. A member must satisfy strength requirements and, where relevant, stiffness requirements. The strength condition limits dangerous stresses, while the stiffness condition limits excessive deformation.\u003C\u002Fp>{{chunk:axial-strength-stiffness-check}}\u003Ch2>Topics covered in this section\u003C\u002Fh2>\u003Cp>The child topics address internal forces and normal stresses, mechanical properties of materials, axial deformations, strength and stiffness checks, thermal deformation, and statically indeterminate axial systems. This sequence moves from equilibrium and stress toward deformation and compatibility.\u003C\u002Fp>\u003Ch2>Basic calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine external loads and reactions.\u003C\u002Fli>\u003Cli>Use the method of sections to find the axial force $N$ in each segment.\u003C\u002Fli>\u003Cli>Calculate normal stresses.\u003C\u002Fli>\u003Cli>Determine strains and displacements when required.\u003C\u002Fli>\u003Cli>Check strength and stiffness conditions.\u003C\u002Fli>\u003C\u002Fol>\u003Cp>For stepped bars or systems made of different materials, perform the calculation segment by segment using the appropriate values of $N$, $A$, $E$, and $L$.\u003C\u002Fp>",[951,961,1047,1057,1067,1078],{"id":952,"parent_id":942,"code":953,"slug":954,"name":955,"seo_title":956,"seo_description":957,"seo_text":958,"content":959,"locale":8,"uk_topic_id":150,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":954,"children":960},47,"3","internal-forces-and-stresses","Internal Forces and Stresses in Cross-Sections","Method of Sections — Internal Forces N, Q, M, T & Stress","Learn how to use the method of sections to find internal forces N, Q, M and T, construct force diagrams, and relate axial force to normal stress.","Internal force components in any cross-section are determined using the method of sections. This section details sign conventions and algorithms for constructing axial force (N), shear force (Q), bending moment (M), and torque (T) diagrams. It also explores the physical nature of stress as internal force intensity, divided into normal and shear stresses. Worked examples illustrate the connection between external loads and internal material response.","\u003Cp>External loads produce internal forces within a bar that resist deformation. These internal actions are determined using the \u003Cstrong>method of sections\u003C\u002Fstrong>: imagine cutting the member at the required location and analyze the equilibrium of one of the resulting parts.\u003C\u002Fp>\u003Ch2>Internal force resultants\u003C\u002Fh2>\u003Cp>In the general case, a cross-section may carry an axial force $N$, shear forces $Q$, bending moments $M$, and a torque $T$. Under centric tension or compression, only the axial force $N$ acts.\u003C\u002Fp>{{chunk:section-method-axial-force}}\u003Ch2>Sign convention for axial force\u003C\u002Fh2>\u003Cp>The axial force $N$ is taken as positive in tension and negative in compression. Its value is determined separately for each segment from equilibrium equations.\u003C\u002Fp>\u003Ch2>Normal stress\u003C\u002Fh2>\u003Cp>The internal force represents the resultant action of the material over the section, whereas stress describes the intensity of that action. For centric tension or compression of a straight member with a uniform cross-section, normal stress is uniformly distributed sufficiently far from local disturbances.\u003C\u002Fp>{{chunk:axial-normal-stress}}\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Consider a bar with cross-sectional area $A=200\\ \\text{mm}^2$ carrying a tensile axial force $N=20\\ \\text{kN}$. Converting the force gives $N=20000\\ \\text{N}$. Therefore:\u003C\u002Fp>\u003Cp>$$\\sigma=\\frac{20000}{200}=100\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>The positive stress corresponds to tension.\u003C\u002Fp>\u003Ch2>Learning outcome\u003C\u002Fh2>\u003Cp>After studying this topic, you should be able to cut a bar conceptually, determine the axial force from equilibrium, construct an $N$ diagram, and calculate normal stress in a cross-section.\u003C\u002Fp>",[],{"id":962,"parent_id":942,"code":963,"slug":964,"name":965,"seo_title":966,"seo_description":967,"seo_text":968,"content":969,"locale":8,"uk_topic_id":161,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":964,"children":970},48,"4","mechanical-properties-of-materials","Mechanical Properties of Materials","Mechanical Properties of Materials — Tensile Testing","Stress-strain diagrams, yield strength, ultimate strength, Hooke's law, and elasticity.","Experimental testing of mechanical properties forms the foundation for proper material selection in engineering. This section analyzes the standard tensile stress-strain diagram for mild steel, highlighting proportionality, elasticity, yield, and ultimate strength limits. Differences between ductile and brittle materials, plastic deformation, strain hardening, and safety factors are thoroughly explained alongside allowable stress calculations.","\u003Cp>Mechanical properties describe how a material deforms and fails under load. For engineering calculations, particularly important properties include \u003Cstrong>elasticity, ductility, strength, and stiffness\u003C\u002Fstrong>. They are determined experimentally, for example by a standard tensile test.\u003C\u002Fp>\u003Ch2>Stress and strain in a tensile test\u003C\u002Fh2>\u003Cp>The test results are represented by a stress–strain diagram in coordinates of normal stress $\\sigma$ versus axial strain $\\varepsilon$. Axial strain is $\\varepsilon=\\Delta l\u002Fl_0$ and is dimensionless.\u003C\u002Fp>{{chunk:uniaxial-hooke-law}}\u003Cp>Young's modulus $E$ characterizes material stiffness in tension and compression: a larger $E$ produces a smaller elastic strain at the same stress.\u003C\u002Fp>\u003Ch2>Characteristic regions of the stress–strain curve\u003C\u002Fh2>\u003Cp>In the initial region, stress is approximately proportional to strain. Beyond the elastic range, irreversible plastic deformation may develop. For a ductile material, commonly used characteristics include:\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>proportional limit\u003C\u002Fstrong> — stress up to which the $\\sigma$–$\\varepsilon$ relation is approximately linear;\u003C\u002Fli>\u003Cli>\u003Cstrong>elastic limit\u003C\u002Fstrong> — a characteristic boundary below which unloading leaves no more than a specified small permanent strain;\u003C\u002Fli>\u003Cli>\u003Cstrong>yield strength\u003C\u002Fstrong> — stress associated with substantial plastic deformation or defined by an offset method;\u003C\u002Fli>\u003Cli>\u003Cstrong>ultimate tensile strength\u003C\u002Fstrong> — the maximum engineering stress on the tensile stress–strain curve.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Elastic and plastic deformation\u003C\u002Fh2>\u003Cp>\u003Cstrong>Elastic deformation\u003C\u002Fstrong> disappears after unloading. \u003Cstrong>Plastic deformation\u003C\u002Fstrong> does not disappear completely, leaving a permanent change in shape or dimensions. The ability to accumulate substantial plastic deformation before fracture is called ductility.\u003C\u002Fp>\u003Ch2>Ductile and brittle materials\u003C\u002Fh2>\u003Cp>Ductile materials generally exhibit noticeable permanent deformation before fracture. Brittle materials fracture with relatively little plastic deformation. This distinction influences the choice of design strength and safety factor.\u003C\u002Fp>\u003Ch2>Strain hardening\u003C\u002Fh2>\u003Cp>Plastic deformation can change material properties. During cold plastic deformation, \u003Cstrong>strain hardening\u003C\u002Fstrong> commonly increases resistance to further plastic flow while reducing the remaining ductility.\u003C\u002Fp>\u003Ch2>Allowable stress and factor of safety\u003C\u002Fh2>{{chunk:allowable-normal-stress}}\u003Cp>The factor of safety accounts for uncertainty in loads, material properties, the calculation model, and operating conditions. Its value is not universal and must follow the adopted design method or code.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose $E=200\\ \\text{GPa}$ and the elastic strain is $\\varepsilon=0.001$. Hooke's law gives:\u003C\u002Fp>\u003Cp>$$\\sigma=E\\varepsilon=200\\cdot10^9\\cdot0.001=200\\cdot10^6\\ \\text{Pa}=200\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>If the selected limiting material strength is $360\\ \\text{MPa}$ and $n=1.5$, the allowable stress is $[\\sigma]=360\u002F1.5=240\\ \\text{MPa}$. Thus $200\\ \\text{MPa}$ does not exceed $240\\ \\text{MPa}$.\u003C\u002Fp>\u003Ch2>Learning outcome\u003C\u002Fh2>\u003Cp>After studying this topic, you should be able to distinguish elastic and plastic deformation, interpret the main regions of a tensile stress–strain curve, explain Young's modulus and strength characteristics, apply Hooke's law within its valid range, and perform a basic allowable-stress check.\u003C\u002Fp>",[971,982,993,1004,1015,1025,1036],{"id":972,"parent_id":962,"code":973,"slug":974,"name":975,"seo_title":976,"seo_description":977,"seo_text":978,"content":979,"locale":8,"uk_topic_id":980,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":974,"children":981},110,"4.1","tensile-test-diagram-characteristic-points","Tensile Test Diagram and Characteristic Points","Tensile Test Diagram — Stress-Strain Curve and Key Points","Read a tensile test stress-strain diagram: elastic region, yield strength, ultimate tensile strength, necking, and fracture explained for engineering materials.","This topic explains the engineering tensile stress–strain diagram and the physical meaning of its characteristic regions and points. It covers proportional and elastic behavior, yield strength, ultimate tensile strength, plastic deformation, necking, and fracture, helping students interpret tensile-test results for engineering calculations.","\u003Cp>A tensile test records how a specimen responds to increasing axial tension. The resulting \u003Cstrong>engineering stress–strain diagram\u003C\u002Fstrong> is one of the most useful ways to see when a material behaves elastically, when permanent deformation begins, how much strengthening occurs during plastic deformation, and how failure develops.\u003C\u002Fp>\u003Cp>The familiar diagram with a pronounced yield region is characteristic of \u003Cstrong>mild low-carbon steel\u003C\u002Fstrong>. It should not be treated as a universal curve for every engineering material.\u003C\u002Fp>{{chunk:visual-brief-tensile-steel-interactive-110}}\u003Ch2>Characteristic regions of a mild-steel tensile diagram\u003C\u002Fh2>\u003Ch3>1. Linear elastic region\u003C\u002Fh3>\u003Cp>At the beginning of loading, stress is approximately proportional to strain:\u003C\u002Fp>\u003Cp>\\[\\sigma=E\\varepsilon.\\]\u003C\u002Fp>\u003Cp>The slope of this straight portion is Young's modulus \\(E\\). If the specimen is unloaded within the elastic range, it returns approximately to its original dimensions.\u003C\u002Fp>\u003Ch3>2. Proportional and elastic limits\u003C\u002Fh3>\u003Cp>The \u003Cstrong>proportional limit\u003C\u002Fstrong> marks the end of nearly linear stress–strain behavior. The \u003Cstrong>elastic limit\u003C\u002Fstrong> is the greatest stress for which deformation remains essentially recoverable after unloading. In real tests these two limits may be close and are not always identified as sharply separated points.\u003C\u002Fp>\u003Ch3>3. Yielding\u003C\u002Fh3>\u003Cp>Mild steel may show a distinct upper yield point followed by a lower yield level or a short \u003Cstrong>yield plateau\u003C\u002Fstrong>. During this stage, strain can increase substantially while engineering stress changes relatively little. This pronounced yield behavior is useful for understanding steel, but many other materials do not show it.\u003C\u002Fp>\u003Ch3>4. Strain hardening\u003C\u002Fh3>\u003Cp>After the yield region, continued plastic deformation requires increasing load. The material undergoes \u003Cstrong>strain hardening\u003C\u002Fstrong>, and engineering stress rises again.\u003C\u002Fp>\u003Ch3>5. Ultimate tensile strength\u003C\u002Fh3>\u003Cp>The maximum engineering stress on the tensile diagram is the \u003Cstrong>ultimate tensile strength\u003C\u002Fstrong> (UTS):\u003C\u002Fp>\u003Cp>\\[\\sigma_u=\\frac{F_{max}}{A_0}.\\]\u003C\u002Fp>\u003Cp>Here \\(F_{max}\\) is the maximum tensile force and \\(A_0\\) is the original cross-sectional area.\u003C\u002Fp>\u003Ch3>6. Necking and fracture\u003C\u002Fh3>\u003Cp>After the maximum load is reached, deformation becomes strongly localized and a \u003Cstrong>neck\u003C\u002Fstrong> forms. Because engineering stress is calculated using the original area \\(A_0\\), the engineering stress curve usually falls during necking until fracture. This descending branch should not be confused with the corresponding true-stress behavior.\u003C\u002Fp>\u003Ch2>How stress–strain diagrams differ between materials\u003C\u002Fh2>\u003Cp>The mild-steel curve is only one characteristic case. The shape of a tensile diagram depends on material class, composition, heat treatment, temperature, strain rate, specimen geometry, and test conditions.\u003C\u002Fp>{{chunk:visual-brief-tensile-material-curves-110}}\u003Ch3>Aluminum alloys: smooth yielding\u003C\u002Fh3>\u003Cp>Many aluminum alloys do \u003Cstrong>not\u003C\u002Fstrong> have a distinct upper\u002Flower yield point or a flat yield plateau. Their transition from elastic to plastic deformation is gradual. For this reason, engineering practice commonly specifies a \u003Cstrong>0.2% proof stress\u003C\u002Fstrong>: a line parallel to the initial elastic slope is offset by a strain of 0.002, and its intersection with the stress–strain curve defines a conventional yield strength.\u003C\u002Fp>\u003Ch3>Gray cast iron: brittle behavior\u003C\u002Fh3>\u003Cp>Gray cast iron in tension typically undergoes only limited plastic deformation before fracture. Its tensile diagram therefore lacks the long yield and strain-hardening regions associated with ductile mild steel, and a pronounced necking stage is generally absent. For brittle materials, fracture strength and the limited strain before failure are especially important.\u003C\u002Fp>\u003Ch3>Ductile polymers: large strains and strong nonlinearity\u003C\u002Fh3>\u003Cp>Ductile polymers can behave very differently from metals. They may have a much lower initial stiffness, pronounced nonlinear response, yielding followed by large plastic strain or drawing, and later strain hardening. Their characteristic strain scale can be far larger than that of structural metals, so directly overlaying polymer and metal curves on an unexplained common strain axis can be misleading.\u003C\u002Fp>\u003Cp>These diagrams should therefore be compared by their \u003Cstrong>qualitative features\u003C\u002Fstrong>—initial stiffness, presence or absence of a distinct yield point, plastic strain capacity, strain hardening, necking, and fracture—not by assuming that every material follows the same curve.\u003C\u002Fp>\u003Ch2>Engineering significance\u003C\u002Fh2>\u003Cp>From a tensile diagram, engineers can identify or estimate elastic stiffness, yield or proof strength, ultimate tensile strength, ductility, and the character of failure. These properties support material selection and strength calculations, but the values used in design should come from the relevant material standard or verified test data rather than from a schematic teaching diagram.\u003C\u002Fp>",96,[],{"id":983,"parent_id":962,"code":984,"slug":985,"name":986,"seo_title":987,"seo_description":988,"seo_text":989,"content":990,"locale":8,"uk_topic_id":991,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":985,"children":992},111,"4.2","elastic-properties-of-materials","Elastic Properties of Materials","Elastic Properties of Materials — E, G and Poisson's Ratio","Young's modulus E, shear modulus G, Poisson's ratio ν, and their physical meaning in linear-elastic deformation calculations.","This topic summarizes the main elastic constants of an isotropic material: Young's modulus E, shear modulus G, and Poisson's ratio ν. It explains their physical meaning, units, engineering use, and the relation between E, G, and ν for a linearly elastic isotropic material.","\u003Cp>Elastic properties describe a material's resistance to reversible deformation. For a linearly elastic isotropic material, the principal constants include Young's modulus $E$, shear modulus $G$, and Poisson's ratio $\\nu$.\u003C\u002Fp>{{chunk:uniaxial-hooke-law}}\u003Ch2>Poisson's ratio\u003C\u002Fh2>\u003Cp>Under uniaxial tension, longitudinal elongation is accompanied by transverse contraction. Poisson's ratio is defined as $\\nu=-\\varepsilon_{\\perp}\u002F\\varepsilon_{\\parallel}$, where $\\varepsilon_{\\perp}$ is transverse strain and $\\varepsilon_{\\parallel}$ is longitudinal strain.\u003C\u002Fp>\u003Ch2>Shear modulus\u003C\u002Fh2>\u003Cp>The shear modulus $G$ characterizes material stiffness in shear. For a linearly elastic isotropic material, the elastic constants are related by $G=E\u002F[2(1+\\nu)]$.\u003C\u002Fp>\u003Ch2>Engineering application\u003C\u002Fh2>\u003Cp>$E$ is used in tension, compression, and bending calculations; $G$ is used in shear and torsion; and $\\nu$ is needed to describe the coupling between longitudinal and transverse strains.\u003C\u002Fp>",97,[],{"id":994,"parent_id":962,"code":995,"slug":996,"name":997,"seo_title":998,"seo_description":999,"seo_text":1000,"content":1001,"locale":8,"uk_topic_id":1002,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":996,"children":1003},112,"4.3","ductility-and-brittleness-of-materials","Ductility and Brittleness of Materials","Ductility and Brittleness of Materials — Deformation and Fracture","Ductile and brittle behavior, permanent strain, elongation and reduction of area, and their significance for engineering material behavior.","This topic explains the difference between ductile and brittle material behavior. It covers permanent deformation, elongation and reduction of area after fracture, and the engineering importance of ductility for stress redistribution and warning before failure.","\u003Cp\r\n >\u003Cstrong>Ductility\u003C\u002Fstrong> is the ability of a material to undergo appreciable irreversible\r\n deformation before fracture. \u003Cstrong>Brittle behavior\u003C\u002Fstrong> is characterized by fracture with\r\n relatively little plastic deformation.\u003C\u002Fp\r\n>\u003Ch2>Measures of ductility\u003C\u002Fh2\r\n>\u003Cp\r\n >After a tensile test, ductility is commonly characterized by percentage elongation and percentage\r\n reduction of area. Under comparable test conditions, larger values indicate that the specimen\r\n accumulated more plastic deformation before fracture.\u003C\u002Fp\r\n>\u003Ch2>Why ductility matters\u003C\u002Fh2\r\n>\u003Cp\r\n >Plastic deformation can allow local redistribution of stress and may provide visible warning of\r\n overload before failure. Brittle fracture often develops with little preceding plastic\r\n deformation, so defects, stress concentrations, temperature, and loading conditions require\r\n particular attention.\u003C\u002Fp\r\n>\u003Ch2>Not an absolute classification\u003C\u002Fh2\r\n>\u003Cp\r\n >Material behavior depends not only on composition or material name. Temperature, strain rate,\r\n stress state, specimen geometry, and environment can change the observed mode of deformation and\r\n fracture.\u003C\u002Fp\r\n>\r\n",98,[],{"id":1005,"parent_id":962,"code":1006,"slug":1007,"name":1008,"seo_title":1009,"seo_description":1010,"seo_text":1011,"content":1012,"locale":8,"uk_topic_id":1013,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1007,"children":1014},113,"4.4","allowable-stresses-and-factor-of-safety","Allowable Stresses and Factor of Safety","Factor of Safety & Allowable Stress — Formula and Example","Learn how allowable stress and factor of safety are used in strength calculations, how limiting strength is selected, and see a simple worked example.","This topic explains the allowable-stress design approach. It covers selection of the limiting material characteristic, the physical meaning of the factor of safety, and verification of the strength condition. It also explains why the required margin depends on loads, material variability, model accuracy, manufacturing, service conditions, and reliability requirements.","\u003Cp>Real structures are not designed so that working stresses directly reach a limiting material characteristic. An \u003Cstrong>allowable stress\u003C\u002Fstrong> and a \u003Cstrong>factor of safety\u003C\u002Fstrong> provide a margin between normal operation and an unacceptable state.\u003C\u002Fp>{{chunk:allowable-normal-stress}}\u003Ch2>Selecting the limiting characteristic\u003C\u002Fh2>\u003Cp>The limiting characteristic depends on the material, loading type, and adopted calculation method. For ductile materials under static loading, yield strength is often a relevant reference, whereas brittle materials may require fracture-related strength characteristics.\u003C\u002Fp>\u003Ch2>What the safety margin accounts for\u003C\u002Fh2>\u003Cp>The factor of safety accounts for uncertainty in loads, scatter of material properties, simplifications of the mechanical model, geometric deviations, manufacturing conditions, and service environment. Its numerical value is determined by the applicable code or design methodology rather than by one universal rule.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>If $\\sigma_{lim}=360$ MPa and $n=1.5$, then $[\\sigma]=360\u002F1.5=240$ MPa. The calculated working stress under the adopted model should not exceed this value.\u003C\u002Fp>",99,[],{"id":1016,"parent_id":962,"code":70,"slug":1017,"name":1018,"seo_title":1019,"seo_description":1020,"seo_text":1021,"content":1022,"locale":8,"uk_topic_id":1023,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1017,"children":1024},388,"creep-of-materials","Creep of Materials","Creep of Materials: Stages, Creep Rate and Stress Relaxation","Creep of materials: the three stages of the creep curve, creep rate, effects of temperature and stress, long-term strength, and stress relaxation.","Creep of materials is time-dependent deformation under sustained loading. This topic covers the creep curve and its three stages, creep rate, the effects of temperature and stress, long-term strength, and stress relaxation.","\u003Cp\r\n >\u003Cstrong>Creep\u003C\u002Fstrong> is the time-dependent deformation of a material under a sustained load.\r\n Unlike instantaneous elastic deformation, creep strain develops gradually even when the load\r\n remains constant.\u003C\u002Fp\r\n>\u003Ch2>When creep becomes important\u003C\u002Fh2\r\n>\u003Cp\r\n >The intensity of creep depends on the material, temperature, stress level, and duration of\r\n loading. For metals, creep becomes particularly significant at elevated temperatures. For polymers\r\n and some other materials, time-dependent deformation may also be noticeable at ordinary\r\n temperatures.\u003C\u002Fp\r\n>\u003Ch2>Creep testing\u003C\u002Fh2\r\n>\u003Cp\r\n >In a typical creep test, a specimen is held for a long time under specified load and temperature,\r\n while strain $\\varepsilon$ is recorded as a function of time $t$. The resulting relationship is\r\n called the \u003Cstrong>creep curve\u003C\u002Fstrong>.\u003C\u002Fp\r\n>{{chunk:creep-strain-time-curve-en}}\u003Ch2>Stages of creep\u003C\u002Fh2\r\n>\u003Col\r\n >\u003Cli>\u003Cstrong>Primary creep.\u003C\u002Fstrong> The strain rate decreases with time.\u003C\u002Fli\r\n >\u003Cli\r\n >\u003Cstrong>Secondary, or steady-state, creep.\u003C\u002Fstrong> The strain rate is approximately\r\n constant.\u003C\u002Fli\r\n >\u003Cli\r\n >\u003Cstrong>Tertiary creep.\u003C\u002Fstrong> The strain rate increases due to accumulated damage; this\r\n stage ends in rupture.\u003C\u002Fli\r\n >\u003C\u002Fol\r\n>\u003Cp>The creep rate is characterized by the derivative $\\dot{\\varepsilon}=d\\varepsilon\u002Fdt$.\u003C\u002Fp\r\n>\u003Ch2>Effect of temperature and stress\u003C\u002Fh2\r\n>\u003Cp\r\n >As temperature and stress level increase, creep generally accelerates and the time to rupture\r\n decreases.\u003C\u002Fp\r\n>\u003Ch2>Long-term strength\u003C\u002Fh2\r\n>\u003Cp\r\n >\u003Cstrong>Long-term strength\u003C\u002Fstrong> characterizes a material’s ability to resist rupture for a\r\n specified time at a given temperature. Long-term design must also limit accumulated deformation\r\n and prevent failure during the required service life.\u003C\u002Fp\r\n>\u003Ch2>Stress relaxation\u003C\u002Fh2\r\n>\u003Cp\r\n >Creep is related to \u003Cstrong>stress relaxation\u003C\u002Fstrong>. If the total deformation of a component\r\n is kept constant, the stress in the material may decrease with time.\u003C\u002Fp\r\n>\u003Ch2>Engineering example\u003C\u002Fh2\r\n>\u003Cp\r\n >Superheated-steam pipelines at thermal power plants operate for years under the combined action\r\n of high temperature and internal pressure. The pressure produces mechanical stresses in the pipe\r\n wall, while the high temperature makes creep of the metal significant.\u003C\u002Fp\r\n>\u003Cp\r\n >Even if the pressure and temperature remain nearly constant during a given operating regime,\r\n deformation of the metal continues to accumulate gradually. At an early stage, the creep rate\r\n decreases; then, during a substantial part of the service life, it remains relatively constant.\r\n Over time, damage such as microvoids and microcracks may accumulate in the material. Tertiary\r\n creep then begins: deformation accelerates and can eventually lead to pipe rupture.\u003C\u002Fp\r\n>\u003Cp\r\n >For such equipment, checking only the instantaneous strength condition $\\sigma < [\\sigma]$ is\r\n therefore insufficient. The engineer must account for temperature, stress, and the duration of\r\n their action when assessing long-term strength and the remaining service life of the material. The\r\n practical conclusion is that a component may safely carry a given load today, but this does not\r\n mean it can safely carry the same load for tens of thousands of hours at high temperature.\u003C\u002Fp\r\n>\u003Ch2>What you should be able to do\u003C\u002Fh2\r\n>\u003Cp\r\n >After studying this topic, a student should be able to explain the physical meaning of creep,\r\n distinguish the three stages of the creep curve, determine creep rate, explain the effects of\r\n temperature and stress, distinguish creep from stress relaxation, and understand the need for\r\n long-term design calculations.\u003C\u002Fp\r\n>\r\n",387,[],{"id":1026,"parent_id":962,"code":1027,"slug":1028,"name":1029,"seo_title":1030,"seo_description":1031,"seo_text":1032,"content":1033,"locale":8,"uk_topic_id":1034,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1028,"children":1035},114,"4.5","strain-hardening-and-reloading","Strain Hardening and Reloading","Strain Hardening — Plastic Deformation, Unloading & Reloading","Learn how prior plastic deformation changes material response through strain hardening, residual strain, elastic unloading, reloading, strengthening, and reduced ductility.","This topic explains how previous plastic deformation changes subsequent material response. It covers strain hardening, strengthening, residual deformation, elastic unloading, reduced remaining ductility, and why loading history matters when evaluating the mechanical behavior of a component.","\u003Cp>After plastic deformation, a material may have different mechanical characteristics than it had initially. Therefore, subsequent loading may depend on the material's \u003Cstrong>deformation history\u003C\u002Fstrong>.\u003C\u002Fp>\u003Ch2>Strain hardening\u003C\u002Fh2>\u003Cp>During cold plastic deformation, many metals exhibit strain hardening: resistance to further plastic flow increases. At the same time, the remaining ductility usually decreases.\u003C\u002Fp>\u003Ch2>Unloading\u003C\u002Fh2>\u003Cp>During unloading, the elastic part of the deformation largely disappears, while the plastic part remains. The specimen therefore does not return completely to its original dimensions.\u003C\u002Fp>\u003Ch2>Reloading\u003C\u002Fh2>\u003Cp>On reloading, the response depends on the previous plastic deformation, loading direction, material, and thermal history. In elementary models, reloading after strain hardening illustrates how the stress level required for further plastic deformation can change.\u003C\u002Fp>\u003Ch2>Engineering significance\u003C\u002Fh2>\u003Cp>Strain hardening is deliberately used in cold-working processes, but it must also be considered when ductility and the ability of a component to tolerate subsequent overloads are important.\u003C\u002Fp>",100,[],{"id":1037,"parent_id":962,"code":1038,"slug":1039,"name":1040,"seo_title":1041,"seo_description":1042,"seo_text":1043,"content":1044,"locale":8,"uk_topic_id":1045,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1039,"children":1046},115,"4.6","external-factors-affecting-mechanical-properties","External Factors Affecting Mechanical Properties","What Affects Mechanical Properties? Temperature, Defects & Strain Rate","Learn how temperature, strain rate, stress concentrators, defects, surface condition, corrosion, and manufacturing processes change material mechanical properties.","This topic explains why mechanical properties measured under standard conditions are not universal constants for every service environment. It covers high and low temperature, strain rate, stress concentrations, defects, surface condition, corrosion, and manufacturing or heat treatment, including creep and brittle-fracture susceptibility.","\u003Cp>Mechanical properties measured in a standard test apply to specified conditions. In a real component, temperature, loading rate, surface condition, defects, environment, and manufacturing history can substantially change material behavior.\u003C\u002Fp>{{chunk:external-factors-mechanical-properties}}\u003Ch2>Temperature\u003C\u002Fh2>\u003Cp>Changes in temperature can alter elastic modulus, yield strength, strength, ductility, and fracture behavior. At elevated temperature under prolonged loading, \u003Cstrong>creep\u003C\u002Fstrong> may become important: deformation accumulates with time under a constant or nearly constant load. At low temperature, some structural materials become less ductile and more susceptible to brittle fracture.\u003C\u002Fp>\u003Ch2>Strain rate and loading type\u003C\u002Fh2>\u003Cp>Material response can depend on strain rate. Results from a slow static test therefore do not always describe impact behavior directly. Cyclic loading also requires separate analysis because fatigue failure can occur at stresses below the static ultimate strength.\u003C\u002Fp>\u003Ch2>Stress concentrations, defects, and size\u003C\u002Fh2>\u003Cp>Holes, abrupt section changes, notches, cracks, and other defects create local stress concentrations. Their effect depends on material, geometry, stress state, and loading type. Component size can also influence defect probability and the applicability of test data.\u003C\u002Fp>\u003Ch2>Surface and environment\u003C\u002Fh2>\u003Cp>Surface quality is particularly important under repeated loading. Scratches and corrosion damage can become crack-initiation sites. Aggressive environments may simultaneously reduce effective section and alter the material damage mechanism.\u003C\u002Fp>\u003Ch2>Manufacturing and heat treatment\u003C\u002Fh2>\u003Cp>Cold working, welding, machining, and heat treatment can change microstructure, hardness, strength, ductility, and residual stresses. Responsible calculations should therefore use properties representative of the actual state of the finished component.\u003C\u002Fp>\u003Ch2>Engineering conclusion\u003C\u002Fh2>\u003Cp>A handbook value should not automatically be transferred to every structure. Check service temperature, loading duration and rate, environment, surface condition, geometric stress concentrators, manufacturing history, and the factors required by the applicable design method.\u003C\u002Fp>",101,[],{"id":1048,"parent_id":942,"code":70,"slug":1049,"name":1050,"seo_title":1051,"seo_description":1052,"seo_text":1053,"content":1054,"locale":8,"uk_topic_id":1055,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1049,"children":1056},86,"thermal-deformations","Thermal Deformations and Stresses","Thermal Stress & Deformation — Formulas and Worked Example","Calculate free thermal expansion and thermal stress in restrained bars using αΔT and EαΔT, with compatibility equations and a worked example.","Temperature changes cause thermal expansion or contraction in structural materials. In statically indeterminate systems, constrained expansion induces severe thermal stresses. This section presents thermal stress evaluation formulas and code compliance procedures.","\u003Cp>A temperature change causes thermal expansion or contraction of a material. If a bar is free to change length, the temperature change produces deformation without mechanical stress. If movement is restrained by supports or other structural members, \u003Cstrong>thermal stresses\u003C\u002Fstrong> develop.\u003C\u002Fp>\u003Ch2>Free thermal strain\u003C\u002Fh2>{{chunk:thermal-strain-elongation}}\u003Cp>The coefficient $\\alpha$ depends on the material and temperature range. In elementary calculations it is commonly treated as constant over the specified temperature change.\u003C\u002Fp>\u003Ch2>Fully restrained bar\u003C\u002Fh2>\u003Cp>Consider a bar whose ends cannot move axially. During heating, it would tend to elongate, but the restraints prevent this motion. The support reactions create a mechanical strain opposite to the free thermal strain.\u003C\u002Fp>{{chunk:fully-restrained-thermal-stress}}\u003Ch2>Partial restraint\u003C\u002Fh2>\u003Cp>In the general case, displacement need not be zero. Thermal and mechanical components are then considered together. For a bar segment in a linear-elastic model:\u003C\u002Fp>\u003Cp>$$\\Delta l=\\frac{NL}{EA}+\\alpha\\Delta T\\,L.$$\u003C\u002Fp>\u003Cp>The signs of the terms follow the selected convention: tension and elongation are positive, while compression and shortening are negative.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A steel bar with $E=200\\ \\text{GPa}$ and $\\alpha=12\\cdot10^{-6}\\ \\text{°C}^{-1}$ is fully restrained and heated by $\\Delta T=40\\ \\text{°C}$. Assuming linear-elastic behavior:\u003C\u002Fp>\u003Cp>$$\\sigma_T=-200000\\cdot12\\cdot10^{-6}\\cdot40=-96\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>The resulting thermal stress is therefore $96\\ \\text{MPa}$ in compression.\u003C\u002Fp>\u003Ch2>Engineering significance\u003C\u002Fh2>\u003Cp>Thermal stresses are important in pipelines, rails, long metal structures, machine components, and assemblies made from materials with different thermal expansion coefficients. Expansion joints, compensators, and details that permit controlled movement are used to reduce them.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>\u003Cp>First determine the free thermal deformation, then identify the kinematic restraints. For a statically indeterminate system, write deformation-compatibility equations and solve them together with equilibrium equations to obtain reactions and internal forces.\u003C\u002Fp>",44,[],{"id":1058,"parent_id":942,"code":1059,"slug":1060,"name":1061,"seo_title":1062,"seo_description":1063,"seo_text":1064,"content":1065,"locale":8,"uk_topic_id":172,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1060,"children":1066},49,"5","statically-indeterminate-axial","Statically Indeterminate Problems under Axial Loading","Statically Indeterminate Axial Bars — Compatibility Method Step by Step","Solve statically indeterminate axial bars using equilibrium plus deformation compatibility, axial stiffness EA\u002FL, stepped members, temperature, and assembly effects.","When static equilibrium equations are insufficient to determine support reactions or internal forces, a structure is statically indeterminate. Solving these problems requires additional deformation compatibility equations based on geometric constraints. This section covers degrees of indeterminacy, calculation algorithms for stepped bars and pin-jointed truss systems under axial load, and thermal\u002Fassembly stress analysis.","\u003Cp>A system is \u003Cstrong>statically indeterminate\u003C\u002Fstrong> when the number of unknown reactions or internal forces exceeds the number of independent static-equilibrium equations. To determine all unknowns, equilibrium equations must be supplemented by \u003Cstrong>deformation-compatibility equations\u003C\u002Fstrong> describing the geometric constraints of the system.\u003C\u002Fp>\u003Ch2>Why equilibrium is not enough\u003C\u002Fh2>\u003Cp>In a statically determinate axial member, reactions can be obtained from equilibrium alone. Additional restraints or interacting members introduce redundant unknowns, so the force distribution also depends on member stiffnesses $EA$ and geometry.\u003C\u002Fp>\u003Ch2>Deformation equations\u003C\u002Fh2>{{chunk:axial-bar-elongation}}\u003Cp>For several segments, the total displacement is obtained by algebraically summing their elongations and shortenings. If temperature changes are present, thermal deformation is added to the mechanical deformation.\u003C\u002Fp>{{chunk:thermal-strain-elongation}}\u003Ch2>Compatibility condition\u003C\u002Fh2>\u003Cp>The compatibility condition follows from the actual geometry and restraints. If both ends of a bar are fixed and the distance between the supports does not change, the total change in length is zero: $\\sum\\Delta l_i=0$. In other systems, selected nodal displacements may be equal or related by a geometric constraint.\u003C\u002Fp>{{chunk:axial-static-indeterminacy-algorithm}}\u003Ch2>Simple example\u003C\u002Fh2>\u003Cp>Consider a uniform bar fixed between two immovable supports and loaded axially at an intermediate point. There are two support reactions but only one independent axial equilibrium equation, so the system is statically indeterminate to the first degree. The second equation follows from the requirement that the total change in distance between the supports is zero. The left and right segment deformations are expressed through their internal forces and axial stiffnesses $EA$.\u003C\u002Fp>\u003Ch2>Effect of stiffness\u003C\u002Fh2>\u003Cp>In statically indeterminate systems, forces are distributed among members according to their stiffness. For the same kinematic condition, a stiffer member generally carries a larger share of the load. Therefore, changing $A$, $E$, or $L$ can change reactions even if the external load remains unchanged.\u003C\u002Fp>\u003Ch2>Assembly and temperature effects\u003C\u002Fh2>\u003Cp>Manufacturing errors, initial gaps, forced assembly, or temperature changes may create internal forces even without an ordinary external mechanical load. These problems are solved by the same principle: equilibrium plus compatibility of total deformations.\u003C\u002Fp>\u003Ch2>Verification\u003C\u002Fh2>\u003Cp>After finding the reactions, verify equilibrium, the geometric compatibility condition, deformation signs, and units. Then calculate stresses in each segment and perform the required strength check.\u003C\u002Fp>",[],{"id":1068,"parent_id":942,"code":1069,"slug":1070,"name":1071,"seo_title":1072,"seo_description":1073,"seo_text":1074,"content":1075,"locale":8,"uk_topic_id":1076,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1070,"children":1077},116,"2.5","deformation-of-bars-in-tension-and-compression","Deformation of Bars in Tension and Compression","Axial Deformation of Bars — ΔL = NL\u002F(EA) Formula & Example","Calculate elongation or shortening of axially loaded bars with ΔL = NL\u002F(EA), including axial rigidity, stepped bars, sign convention, and a worked example.","This topic explains deformation of bars under centric tension and compression. It covers absolute change in length Δl, axial strain ε, the physical meaning of axial rigidity EA, the formula Δl = NL\u002F(EA), and summation of deformations for stepped bars with different N, E, A, and L.","\u003Cp>Under centric tension or compression, a bar changes its length. This change is described by the \u003Cstrong>absolute axial deformation $\\Delta l$\u003C\u002Fstrong> and the \u003Cstrong>axial strain $\\varepsilon$\u003C\u002Fstrong>.\u003C\u002Fp>\u003Ch2>Absolute and relative deformation\u003C\u002Fh2>\u003Cp>The absolute deformation is the difference between final and initial length: $\\Delta l=l-l_0$. It is positive for elongation and negative for shortening. Axial strain is the change in length divided by the initial length: $\\varepsilon=\\Delta l\u002Fl_0$.\u003C\u002Fp>\u003Ch2>Deformation of a prismatic bar\u003C\u002Fh2>{{chunk:axial-bar-elongation}}\u003Cp>The product $EA$ is the axial rigidity. For the same $N$ and $L$, increasing $E$ or $A$ reduces deformation, while increasing $L$ increases it.\u003C\u002Fp>\u003Ch2>Stepped bar\u003C\u002Fh2>\u003Cp>If axial force, cross-sectional area, or Young's modulus changes along the bar, divide it into segments. Determine $N_i$, $L_i$, $E_i$, and $A_i$ for each segment and algebraically sum the changes in length. Tensile segments contribute positive elongation and compressed segments negative shortening under the adopted convention.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A steel bar has $N=20\\ \\text{kN}$, $L=1000\\ \\text{mm}$, $A=200\\ \\text{mm}^2$, and $E=200000\\ \\text{MPa}$. Using N and mm consistently:\u003C\u002Fp>\u003Cp>$$\\Delta l=\\frac{20000\\cdot1000}{200000\\cdot200}=0.5\\ \\text{mm}.$$\u003C\u002Fp>\u003Cp>The bar therefore elongates by $0.5\\ \\text{mm}$ in tension.\u003C\u002Fp>\u003Ch2>Limits of applicability\u003C\u002Fh2>\u003Cp>The formula $\\Delta l=NL\u002F(EA)$ in this form assumes centric axial loading, linearly elastic behavior, and constant $N$, $E$, and $A$ within the considered segment. Variable quantities require segmentation or the corresponding integral form.\u003C\u002Fp>",102,[],{"id":1079,"parent_id":942,"code":1080,"slug":1081,"name":1082,"seo_title":1083,"seo_description":1084,"seo_text":1085,"content":1086,"locale":8,"uk_topic_id":1087,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1081,"children":1088},117,"2.6","strength-and-stiffness-design-in-tension-and-compression","Strength and Stiffness Design in Tension and Compression","Axial Strength & Stiffness Design — Formulas and Example","Check axially loaded bars for strength and stiffness, size the required cross-sectional area, determine allowable load, and follow a worked design example.","This topic systematizes the main engineering calculations for bars under centric tension and compression: strength verification, sizing of the required cross-sectional area, determination of allowable axial load, and stiffness verification. It uses |σmax| ≤ [σ], |Δl| ≤ [Δl], σ = N\u002FA, and Δl = NL\u002F(EA).","\u003Cp>An axially loaded bar must not only carry the applied forces safely but also remain within acceptable deformation limits. Therefore, calculations distinguish between \u003Cstrong>strength\u003C\u002Fstrong> and \u003Cstrong>stiffness\u003C\u002Fstrong> requirements.\u003C\u002Fp>{{chunk:axial-strength-stiffness-check}}\u003Ch2>Verification calculation\u003C\u002Fh2>\u003Cp>If geometry and loading are known, determine the axial force $N$, calculate $\\sigma=N\u002FA$, and compare the largest absolute stress with the allowable value. If the strength condition is satisfied, the section meets the adopted allowable-stress criterion.\u003C\u002Fp>\u003Ch2>Design calculation\u003C\u002Fh2>\u003Cp>If the load and allowable stress are known but the cross-section must be selected, estimate the required area from $A_{\\mathrm{req}}\\ge |N|\u002F[\\sigma]$. Then choose an actual standard or constructively acceptable section with an area not smaller than the calculated requirement and verify it again.\u003C\u002Fp>\u003Ch2>Allowable load\u003C\u002Fh2>\u003Cp>For a given section, the allowable axial force under the adopted condition can be estimated from $|N|\\le[\\sigma]A$. If the member has several segments, the governing external load is determined by the most critical segment together with the relation between its internal force and the applied load.\u003C\u002Fp>\u003Ch2>Stiffness check\u003C\u002Fh2>{{chunk:axial-bar-elongation}}\u003Cp>Even when stresses are acceptable, excessive elongation or shortening may interfere with service. The calculated displacement is therefore compared with an allowable value specified by service requirements or the applicable design method.\u003C\u002Fp>\u003Ch2>Area-sizing example\u003C\u002Fh2>\u003Cp>Let a bar carry $N=60\\ \\text{kN}$ in tension and let the allowable stress be $[\\sigma]=150\\ \\text{MPa}$. Using N and mm:\u003C\u002Fp>\u003Cp>$$A_{\\mathrm{req}}=\\frac{60000}{150}=400\\ \\text{mm}^2.$$\u003C\u002Fp>\u003Cp>An actual section with area not less than $400\\ \\text{mm}^2$ should be selected and then checked using its real properties and, where required, the stiffness condition.\u003C\u002Fp>",103,[],{"id":1090,"parent_id":784,"code":1091,"slug":1092,"name":1093,"seo_title":1094,"seo_description":1095,"seo_text":1096,"content":1097,"locale":8,"uk_topic_id":915,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1092,"children":1098},54,"10","shear-and-direct-shear","Shear and Direct Shear","Shear and Direct Shear — Stress, Strain and Connection Design","Pure shear, Hooke’s law in shear, average direct shear stress, single and double shear, and basic connection checks.","Shear and direct shear introduce shear stress, shear strain, and Hooke’s law in shear. The section covers pure shear, single and double direct shear, and basic checks for bolts, rivets, pins, and welds.","\u003Cp>\u003Cstrong>Shear and direct shear\u003C\u002Fstrong> cover material distortion caused by relative sliding of adjacent layers and the practical design of connectors that may be sheared by a transverse force. The principal quantities are shear stress \\(\\tau\\), shear strain \\(\\gamma\\), and shear modulus \\(G\\).\u003C\u002Fp>\n\u003Ch2>Pure shear\u003C\u002Fh2>\n\u003Cp>In pure shear, complementary shear stresses act on mutually perpendicular planes. Initially right angles change while the first-order linear dimensions of the element remain unchanged.\u003C\u002Fp>\n{{chunk:shear-hooke-law}}\n\u003Ch2>Direct shear\u003C\u002Fh2>\n\u003Cp>For a bolt, rivet, pin, or another short connector, the introductory design model assumes a uniform average shear stress over the shear plane:\u003C\u002Fp>\n\u003Cp>\\[\\tau_{avg}=\\frac{V}{A_s}.\\]\u003C\u002Fp>\n\u003Cp>For \\(n\\) identical active shear planes, the total resisting area is \\(nA_s\\). Single and double shear must therefore be distinguished.\u003C\u002Fp>\n\u003Ch2>Connection checks\u003C\u002Fh2>\n\u003Cul>\u003Cli>shear resistance of the bolt, rivet, or pin;\u003C\u002Fli>\u003Cli>bearing of the connected parts;\u003C\u002Fli>\u003Cli>resistance of the weakened net section;\u003C\u002Fli>\u003Cli>edge shear-out or tear-out;\u003C\u002Fli>\u003Cli>weld resistance where applicable.\u003C\u002Fli>\u003C\u002Ful>\n\u003Ch2>Limits of the model\u003C\u002Fh2>\n\u003Cp>The average-stress formula is a direct-shear model and does not describe the detailed nonuniform stress field. Transverse shear in beams belongs to the bending section.\u003C\u002Fp>\n\u003Ch2>Further study\u003C\u002Fh2>\n\u003Cp>The next topic treats pure shear, single and double shear, and the design of bolted, riveted, pinned, and welded joints. Torsion is organized as a separate section at the same hierarchy level.\u003C\u002Fp>",[1099],{"id":1100,"parent_id":1090,"code":1101,"slug":1102,"name":1103,"seo_title":1104,"seo_description":1105,"seo_text":1106,"content":1107,"locale":8,"uk_topic_id":927,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1102,"children":1108},55,"11","pure-shear-and-joints","Pure Shear and Design of Joints","Shear Stress in Bolts, Rivets & Welds — Joint Design","Calculate shear and bearing stress in bolts, rivets, pins, and welded joints using τ = F\u002FAs, shear planes, projected bearing area, and strength checks.","Pure shear is a stress state where only shear stresses act on two mutually perpendicular planes. This mathematical model underpins practical design for mechanical fasteners and joints. This page presents strength analysis for riveted, bolted, keyed, and welded connections under shear and bearing loads, with engineering formulas for calculating required rivet counts or weld lengths.","\u003Cp>\u003Cstrong>Shear\u003C\u002Fstrong> occurs when external forces tend to slide one part of a member relative to another along a separation plane. In simplified engineering calculations for bolts, rivets, pins, and some welded joints, shear stress is often assumed to be uniformly distributed over the calculated shear area.\u003C\u002Fp>\u003Ch2>Average shear stress\u003C\u002Fh2>\u003Cp>For direct shear, the average shear stress is:\u003C\u002Fp>\u003Cp>$$\\tau_{\\mathrm{avg}}=\\frac{F}{A_s},$$\u003C\u002Fp>\u003Cp>where $F$ is the force transmitted through the shear plane and $A_s$ is the total calculated shear area. When identical fasteners and shear planes work symmetrically, the total area includes their number.\u003C\u002Fp>\u003Ch2>Pure shear deformation\u003C\u002Fh2>{{chunk:shear-hooke-law}}\u003Cp>In a pure shear state, complementary shear stresses act on mutually perpendicular planes. These paired stresses satisfy moment equilibrium of a small material element.\u003C\u002Fp>\u003Ch2>Shear of fasteners\u003C\u002Fh2>\u003Cp>For a bolt or rivet, the number of shear planes must be identified correctly. A single-shear joint has one resisting cross-sectional area of the shank; a double-shear joint has two. In the average-stress model, the strength condition is $\\tau_{\\mathrm{avg}}\\le[\\tau]$.\u003C\u002Fp>\u003Ch2>Bearing stress\u003C\u002Fh2>\u003Cp>Contact between a bolt or rivet and the wall of a hole produces local contact stresses. In a simple design model they are represented by an average bearing stress. For a plate of thickness $t$ and a fastener of diameter $d$, the projected contact area is often taken as $A_b=dt$, giving $\\sigma_b=F\u002FA_b$.\u003C\u002Fp>\u003Ch2>Welded joints\u003C\u002Fh2>\u003Cp>In a simplified fillet-weld calculation, the load is related to the effective throat area of the weld. The actual stress distribution may be nonuniform, especially under eccentric loading, so code-based methods and appropriate coefficients are required for responsible design.\u003C\u002Fp>\u003Ch2>Verification procedure\u003C\u002Fh2>\u003Col>\u003Cli>Identify the load path through the joint.\u003C\u002Fli>\u003Cli>Determine the number of active fasteners and shear planes.\u003C\u002Fli>\u003Cli>Check fasteners in shear.\u003C\u002Fli>\u003Cli>Check bearing of the contacting surfaces.\u003C\u002Fli>\u003Cli>If required, check the net section weakened by holes or the effective weld section.\u003C\u002Fli>\u003C\u002Fol>",[],{"id":1110,"parent_id":784,"code":70,"slug":1111,"name":1112,"seo_title":1113,"seo_description":1114,"seo_text":1115,"content":1116,"locale":8,"uk_topic_id":1117,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1111,"children":1118},433,"torsion","Torsion","Torsion of Shafts — Shear Stress, Strength and Stiffness","Internal torque, torque diagrams, torsional shear stress, angle of twist, strength and stiffness of circular shafts, and advanced section types.","Learn shaft torsion through internal torque, torque diagrams, circular-shaft shear stress, angle of twist, power transmission, strength and stiffness checks.","\u003Cp>\u003Cstrong>Torsion\u003C\u002Fstrong> is loading by couples acting about a member's longitudinal axis. An internal torque \\(T\\) develops in each cross-section and the material is subjected primarily to shear deformation.\u003C\u002Fp>\n\u003Ch2>Internal torque\u003C\u002Fh2>\n\u003Cp>The torque at a section is found by cutting the shaft and applying rotational equilibrium to either segment. A torque diagram \\(T(x)\\) identifies critical regions and changes of sign.\u003C\u002Fp>\n\u003Ch2>Solid and hollow circular shafts\u003C\u002Fh2>\n\u003Cp>For a circular shaft, cross-sections remain plane and rotate as rigid discs. Shear stress varies linearly with radius:\u003C\u002Fp>\n\u003Cp>\\[\\tau(\\rho)=\\frac{T\\rho}{J},\\qquad \\tau_{max}=\\frac{T}{W_p}.\\]\u003C\u002Fp>\n\u003Cp>For a uniform segment, the angle of twist is\u003C\u002Fp>\n\u003Cp>\\[\\varphi=\\frac{TL}{GJ}.\\]\u003C\u002Fp>\n{{chunk:circular-shaft-torsion}}\n\u003Ch2>Strength and stiffness\u003C\u002Fh2>\n\u003Cp>Shaft dimensions must satisfy both \\(\\tau_{max}\\le\\tau_{allow}\\) and \\(|\\varphi|\\le\\varphi_{allow}\\). Adequate strength alone does not guarantee acceptable torsional stiffness.\u003C\u002Fp>\n\u003Ch2>Power transmission\u003C\u002Fh2>\n\u003Cp>For a rotating shaft, power, torque, and angular velocity are related by \\(P=T\\omega\\). This converts motor power and rotational speed into design torque.\u003C\u002Fp>\n\u003Ch2>Special cases\u003C\u002Fh2>\n\u003Cul>\u003Cli>statically indeterminate torque-loaded systems;\u003C\u002Fli>\u003Cli>stepped and composite shafts;\u003C\u002Fli>\u003Cli>stress concentrations at grooves, fillets, and holes;\u003C\u002Fli>\u003Cli>inelastic torsion and residual stresses;\u003C\u002Fli>\u003Cli>helical cylindrical springs.\u003C\u002Fli>\u003C\u002Ful>\n\u003Ch2>More complex cross-sections\u003C\u002Fh2>\n\u003Cp>The circular-shaft formula cannot be applied directly to rectangular, elliptical, or other shapes. See \u003Ca href=\"\u002Fen\u002Fstrength-of-materials\u002Ftorsion-of-solid-noncircular-shafts\">Torsion of Solid Noncircular Shafts\u003C\u002Fa> for an overview of warping and the torsion constant. Detailed thin-walled torsion is treated as a separate advanced section.\u003C\u002Fp>\n\u003Ch2>Calculation route\u003C\u002Fh2>\n\u003Cp>\u003Cstrong>Applied torques → reaction torques → torque diagram \\(T\\) → theory selected by section shape → shear stress → angle of twist → strength and stiffness checks.\u003C\u002Fstrong>\u003C\u002Fp>",432,[1119,1129,1139,1150,1160,1170],{"id":1120,"parent_id":1110,"code":70,"slug":1121,"name":1122,"seo_title":1123,"seo_description":1124,"seo_text":1125,"content":1126,"locale":8,"uk_topic_id":1127,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1121,"children":1128},435,"internal-torque-and-torque-diagrams","Internal Torque and Torque Diagrams","Internal Torque and Torque Diagrams for Shafts","Find internal torque by the method of sections, apply a consistent sign convention, and construct torque diagrams for shafts.","Internal torque follows from rotational equilibrium of a cut shaft segment. Learn torque sign conventions, diagram jumps, distributed torque and a systematic construction procedure.","\u003Cp>\u003Cstrong>Internal torque \\(T\\)\u003C\u002Fstrong> is the stress resultant developed at a shaft cross-section under torsional loading. It is found by the method of sections and rotational equilibrium of either cut segment.\u003C\u002Fp>\u003Ch2>Applied and internal torques\u003C\u002Fh2>\u003Cp>Applied torque may be produced by a motor, coupling, pulley, gear, or force couple. After an imaginary cut, the internal torque balances the algebraic sum of external torques acting on one side:\u003C\u002Fp>\u003Cp>\\[T(x)=-\\sum M_x.\\]\u003C\u002Fp>\u003Ch2>Sign convention\u003C\u002Fh2>\u003Cp>Choose a positive direction once and use it consistently. The right-hand rule about the longitudinal axis is convenient. Internal torques on opposite cut faces have opposite directions.\u003C\u002Fp>\u003Ch2>Torque diagram\u003C\u002Fh2>\u003Cp>On a segment without distributed torsional loading, \\(T\\) is constant. A concentrated applied torque produces a jump of equal magnitude. For distributed torque \\(m_t(x)\\):\u003C\u002Fp>\u003Cp>\\[\\frac{dT}{dx}=-m_t(x).\\]\u003C\u002Fp>\u003Ch2>Procedure\u003C\u002Fh2>\u003Col>\u003Cli>Show all applied and reaction torques.\u003C\u002Fli>\u003Cli>Find unknown reactions from \\(\\sum M_x=0\\).\u003C\u002Fli>\u003Cli>Divide the shaft into loading intervals.\u003C\u002Fli>\u003Cli>Cut each interval and calculate \\(T(x)\\).\u003C\u002Fli>\u003Cli>Draw the diagram and check jumps and overall equilibrium.\u003C\u002Fli>\u003C\u002Fol>",434,[],{"id":1130,"parent_id":1110,"code":1131,"slug":1132,"name":1133,"seo_title":1134,"seo_description":1135,"seo_text":1136,"content":1137,"locale":8,"uk_topic_id":939,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1132,"children":1138},56,"12","torsion-of-shafts","Torsion of Solid and Hollow Circular Shafts","Shaft Torsion — Shear Stress, Twist & Worked Example","Learn torsion of solid and hollow circular shafts: torque, polar moment, shear stress, angle of twist, strength checks, and a worked shaft example.","In circular and hollow shaft torsion, the plane sections assumption holds: cross-sections remain flat and rotate relative to one another. This section is devoted to calculating polar moments of inertia and section moduli, constructing torque and twist angle diagrams, and performing allowable shear stress strength and angular deformation stiffness calculations.","\u003Cp>\u003Cstrong>Torsion\u003C\u002Fstrong> is deformation of a member under moments acting about its longitudinal axis. For solid and hollow circular shafts, classical torsion theory assumes that cross-sections remain plane and rotate relative to one another.\u003C\u002Fp>\u003Ch2>Internal torque\u003C\u002Fh2>\u003Cp>The internal torque $T$ at a section is determined by the method of sections from moment equilibrium about the shaft axis. For a stepped shaft or a shaft carrying several applied torques, $T$ is determined separately for each segment and a torque diagram can be constructed.\u003C\u002Fp>\u003Ch2>Polar properties of the section\u003C\u002Fh2>{{chunk:circular-shaft-polar-properties}}\u003Cp>The polar second moment of area $J_p$ characterizes the geometric resistance of the section to torsion, while the polar section modulus $W_p$ is convenient for calculating the maximum shear stress.\u003C\u002Fp>\u003Ch2>Shear stresses\u003C\u002Fh2>{{chunk:circular-shaft-torsion}}\u003Cp>In a solid circular shaft, shear stress varies linearly with radius: $\\tau=0$ at the axis and reaches its maximum at the outer surface. In a hollow shaft, material near the axis is removed, so for a given amount of material a hollow section can use material more efficiently in torsion.\u003C\u002Fp>\u003Ch2>Strength check\u003C\u002Fh2>\u003Cp>In a simple allowable-stress model, the strength condition is $|\\tau_{\\max}|\\le[\\tau]$. For a solid circular shaft, this relation can be used to select the required diameter from the known torque and allowable shear stress.\u003C\u002Fp>\u003Ch2>Stiffness check\u003C\u002Fh2>\u003Cp>Excessive twist may impair machine accuracy even when stresses are safe. Therefore, the total or specific angle of twist is compared with an allowable value. For a shaft consisting of several segments, the twists are added algebraically.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a solid shaft of diameter $d=40\\ \\text{mm}$ carrying $T=500\\ \\text{N·m}$, first convert the torque to N·mm: $T=500000\\ \\text{N·mm}$. The polar section modulus is:\u003C\u002Fp>\u003Cp>$$W_p=\\frac{\\pi40^3}{16}\\approx12566\\ \\text{mm}^3.$$\u003C\u002Fp>\u003Cp>Thus $\\tau_{\\max}=500000\u002F12566\\approx39.8\\ \\text{MPa}$.\u003C\u002Fp>\u003Ch2>Limits of applicability\u003C\u002Fh2>\u003Cp>These relations apply primarily to solid and hollow circular members in the linear-elastic range. Noncircular sections have a different stress and deformation distribution in torsion.\u003C\u002Fp>",[],{"id":1140,"parent_id":1110,"code":1141,"slug":1142,"name":1143,"seo_title":1144,"seo_description":1145,"seo_text":1146,"content":1147,"locale":8,"uk_topic_id":1148,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1142,"children":1149},118,"10.4","power-transmission-by-shafts","Power Transmission by Shafts","Power Transmission by Shafts — Torque, Power and Speed","Relationship between shaft power, torque, and rotational speed using P = Tω and T ≈ 9550P\u002Fn, followed by torsional strength and stiffness checks.","This topic explains the relationship between transmitted mechanical power, shaft torque, and rotational speed. It covers P = Tω and T ≈ 9550P\u002Fn, consistent units, conversion from drive parameters to design torque, and the subsequent shaft checks for torsional shear stress and angle of twist.","\u003Cp>In a mechanical drive, a shaft transmits energy between a motor, gearbox, coupling, and driven machine. For shaft strength calculations, power and rotational speed must first be converted into \u003Cstrong>torque\u003C\u002Fstrong>.\u003C\u002Fp>{{chunk:shaft-power-torque}}\u003Ch2>Physical meaning\u003C\u002Fh2>\u003Cp>For the same transmitted power, reducing rotational speed increases torque. This is why an idealized reduction gearbox that lowers output speed increases the torque available at its output shaft.\u003C\u002Fp>\u003Ch2>From drive parameters to shaft calculation\u003C\u002Fh2>\u003Cp>After determining $T$, use it in the torsion relations. For a circular shaft, determine the polar section properties, maximum torsional shear stress, and angle of twist.\u003C\u002Fp>{{chunk:circular-shaft-torsion}}\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A shaft transmits $P=15\\ \\text{kW}$ at $n=1500\\ \\text{rpm}$. Then:\u003C\u002Fp>\u003Cp>$$T\\approx\\frac{9550\\cdot15}{1500}=95.5\\ \\text{N·m}.$$\u003C\u002Fp>\u003Cp>This is the basic torque for the subsequent shaft calculation. A real drive may additionally require allowances for transmission efficiency, load nonuniformity, starting conditions, and dynamic effects.\u003C\u002Fp>\u003Ch2>Units and a common error\u003C\u002Fh2>\u003Cp>In the formula containing the coefficient 9550, power $P$ must be entered in kW and rotational speed $n$ in rpm; torque $T$ is obtained in N·m. When using $P=T\\omega$, all quantities must be expressed in a consistent SI unit system.\u003C\u002Fp>",104,[],{"id":1151,"parent_id":1110,"code":70,"slug":1152,"name":1153,"seo_title":1154,"seo_description":1155,"seo_text":1156,"content":1157,"locale":8,"uk_topic_id":1158,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1152,"children":1159},437,"strength-and-stiffness-design-of-shafts","Strength and Stiffness Design of Shafts","Shaft Design for Torsional Strength and Stiffness","Check circular shafts for torsional shear strength and angle-of-twist stiffness, combine stepped-shaft rotations, and select shaft size.","Torsional shaft design combines a maximum shear-stress check with an angle-of-twist check. Both must be satisfied after selecting a practical shaft size.","\u003Cp>Shaft design in torsion requires separate checks of \u003Cstrong>strength\u003C\u002Fstrong> from shear stress and \u003Cstrong>stiffness\u003C\u002Fstrong> from angle of twist.\u003C\u002Fp>\u003Ch2>Strength condition\u003C\u002Fh2>\u003Cp>For a solid or hollow circular shaft:\u003C\u002Fp>\u003Cp>\\[\\tau_{max}=\\frac{|T|}{W_p}\\le\\tau_{allow}.\\]\u003C\u002Fp>\u003Cp>The critical section has the largest ratio \\(|T|\u002FW_p\\), not necessarily the largest torque alone.\u003C\u002Fp>\u003Ch2>Stiffness condition\u003C\u002Fh2>\u003Cp>For a uniform segment:\u003C\u002Fp>\u003Cp>\\[\\varphi=\\frac{TL}{GJ}.\\]\u003C\u002Fp>\u003Cp>For a stepped shaft, segment twists are added algebraically:\u003C\u002Fp>\u003Cp>\\[\\varphi=\\sum_i\\frac{T_iL_i}{G_iJ_i}.\\]\u003C\u002Fp>\u003Cp>The specified limit may apply to total relative rotation or twist per unit length.\u003C\u002Fp>\u003Ch2>Design procedure\u003C\u002Fh2>\u003Col>\u003Cli>Construct the torque diagram.\u003C\u002Fli>\u003Cli>Calculate \\(J\\) and \\(W_p\\) for each segment.\u003C\u002Fli>\u003Cli>Find maximum shear stress and check strength.\u003C\u002Fli>\u003Cli>Calculate segment twists with their signs.\u003C\u002Fli>\u003Cli>Check stiffness.\u003C\u002Fli>\u003Cli>Select a practical standard size and repeat both checks.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Practical considerations\u003C\u002Fh2>\u003Cp>Keyways, shoulders, grooves, and holes introduce stress concentrations. Noncircular sections require the appropriate torsion constant instead of the circular polar second moment.\u003C\u002Fp>",436,[],{"id":1161,"parent_id":1110,"code":70,"slug":1162,"name":1163,"seo_title":1164,"seo_description":1165,"seo_text":1166,"content":1167,"locale":8,"uk_topic_id":1168,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1162,"children":1169},439,"statically-indeterminate-torsion","Statically Indeterminate Torsion","Statically Indeterminate Torsion of Shafts","Reaction torques, equilibrium, rotation compatibility, fixed-ended shafts, and composite coaxial shafts under torsional loading.","Statically indeterminate torsion is solved by combining torque equilibrium, torque–twist relations, and compatibility of shaft rotations.","\u003Cp>A torsional system is \u003Cstrong>statically indeterminate\u003C\u002Fstrong> when equilibrium alone cannot determine all reaction torques. Additional equations follow from compatibility of rotations.\u003C\u002Fp>\u003Ch2>Three equation groups\u003C\u002Fh2>\u003Col>\u003Cli>\u003Cstrong>Equilibrium:\u003C\u002Fstrong> \\(\\sum M_x=0\\).\u003C\u002Fli>\u003Cli>\u003Cstrong>Constitutive relation:\u003C\u002Fstrong> \\(d\\varphi\u002Fdx=T\u002F(GJ)\\).\u003C\u002Fli>\u003Cli>\u003Cstrong>Compatibility:\u003C\u002Fstrong> specified sections have equal rotation or zero relative rotation.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Shaft fixed at both ends\u003C\u002Fh2>\u003Cp>For an applied torque between two fixed supports, reaction torques satisfy equilibrium but require one additional condition:\u003C\u002Fp>\u003Cp>\\[\\varphi_{AB}=\\sum_i\\frac{T_iL_i}{G_iJ_i}=0.\\]\u003C\u002Fp>\u003Cp>After finding the reactions, construct the final torque diagram and check strength and stiffness.\u003C\u002Fp>\u003Ch2>Composite coaxial shafts\u003C\u002Fh2>\u003Cp>Rigidly connected coaxial shafts undergo the same angle of twist, while the applied torque is shared between components. The share depends on their torsional stiffnesses \\(GJ\u002FL\\).\u003C\u002Fp>\u003Ch2>Procedure\u003C\u002Fh2>\u003Col>\u003Cli>Introduce unknown reaction torques.\u003C\u002Fli>\u003Cli>Write equilibrium.\u003C\u002Fli>\u003Cli>Express internal torque on every segment.\u003C\u002Fli>\u003Cli>Write rotation compatibility.\u003C\u002Fli>\u003Cli>Solve reactions, construct the torque diagram, and complete design checks.\u003C\u002Fli>\u003C\u002Fol>",438,[],{"id":1171,"parent_id":1110,"code":1172,"slug":1173,"name":1174,"seo_title":1175,"seo_description":1176,"seo_text":1177,"content":1178,"locale":8,"uk_topic_id":1179,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1173,"children":1180},57,"13","design-of-helical-springs","Design of Helical Springs","Helical Spring Design — Stiffness, Deflection & Shear Stress","Calculate helical spring stiffness, axial deflection, and shear stress from wire diameter, coil diameter, active coils, shear modulus, and the Wahl factor.","Close-coiled helical springs are crucial machine components operating primarily under wire torsion driven by axial loading. This page provides engineering theory for spring coil design: calculating torsional torque, curvature correction factors, maximum shear stresses, total spring deflection, and determining active coil counts for required stiffness.","\u003Cp>A \u003Cstrong>cylindrical helical spring\u003C\u002Fstrong> converts an axial force primarily into torsion of the wire in each coil. This allows the spring to store elastic energy and produce relatively large axial displacements within compact dimensions.\u003C\u002Fp>\u003Ch2>Main geometric parameters\u003C\u002Fh2>\u003Cp>Design parameters include wire diameter $d$, mean coil diameter $D$, number of active coils $n$, and the spring index $C=D\u002Fd$. End coils may serve a structural function and are not always counted as active coils.\u003C\u002Fp>\u003Ch2>Torsion of the spring wire\u003C\u002Fh2>\u003Cp>An axial force $F$ produces a principal torque in the wire of approximately $T=FD\u002F2$. Therefore, the basic spring model is based on the torsion theory of a circular bar.\u003C\u002Fp>{{chunk:circular-shaft-polar-properties}}\u003Ch2>Deflection and stiffness\u003C\u002Fh2>{{chunk:helical-spring-stiffness}}\u003Cp>The formula shows the strong influence of wire diameter: stiffness is proportional to $d^4$. Increasing the mean coil diameter $D$ or the number of active coils $n$ reduces stiffness.\u003C\u002Fp>\u003Ch2>Shear stresses\u003C\u002Fh2>\u003Cp>The simplest torsion model gives a nominal shear stress due to torque. In a real helical spring, direct shear and wire curvature also affect the maximum stress. More accurate calculations therefore use correction factors related to the spring index, such as the Wahl factor in common engineering models.\u003C\u002Fp>\u003Ch2>Strength and stiffness\u003C\u002Fh2>\u003Cp>Spring design requires at least two checks: maximum shear stress must remain within the adopted allowable limit, and axial deformation must provide the required force–displacement characteristic.\u003C\u002Fp>\u003Ch2>Example of geometric influence\u003C\u002Fh2>\u003Cp>If $d$ is increased by a factor of 1.2 while $D$, $n$, and $G$ remain unchanged, the basic model predicts a stiffness increase by $1.2^4\\approx2.07$. Thus a relatively small increase in wire diameter can more than double spring stiffness.\u003C\u002Fp>\u003Ch2>Limits of the simple model\u003C\u002Fh2>\u003Cp>The basic formulas are most appropriate for close-coiled springs with small helix angle and elastic material behavior. Practical design may also require consideration of end coils, coil contact in compression, fatigue, buckling of long springs, manufacturing effects, and applicable standards.\u003C\u002Fp>",13,[],{"id":1182,"parent_id":784,"code":1183,"slug":1184,"name":1185,"seo_title":1186,"seo_description":1187,"seo_text":1188,"content":1189,"locale":8,"uk_topic_id":183,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1184,"children":1190},50,"6","stress-and-strain-state","Stress and Strain State","Stress State at a Point — Tensor, Principal Stresses & Strain","Understand plane and 3D stress states, stress-tensor components, principal planes and stresses, strain components, and the link to failure criteria.","The stress state at a point in a deformable body is fully characterized by the stress tensor across all possible planes passing through that point. We distinguish between uniaxial, biaxial (plane), and triaxial (3D) stress states. This page provides a theoretical analysis of stress components, coordinate transformation rules under rotation, and methods for identifying principal planes where shear stresses equal zero.","\u003Cp>The \u003Cstrong>stress and strain state at a point\u003C\u002Fstrong> describes the set of stresses and strains acting on planes of different orientations passing through the same material point. This framework is required whenever a simple uniaxial stress model is insufficient.\u003C\u002Fp>\u003Ch2>Stress components\u003C\u002Fh2>\u003Cp>On an arbitrary plane, the traction can be resolved into normal and shear components. In Cartesian coordinates, a three-dimensional stress state is described by normal components $\\sigma_x$, $\\sigma_y$, $\\sigma_z$ and shear components $\\tau_{xy}$, $\\tau_{yz}$, $\\tau_{zx}$. In classical continuum mechanics with moment equilibrium, the stress tensor is symmetric, so paired shear components are equal.\u003C\u002Fp>\u003Ch2>Plane and three-dimensional stress\u003C\u002Fh2>\u003Cp>For thin plates loaded in their own plane, a \u003Cstrong>plane stress\u003C\u002Fstrong> model is often appropriate, with $\\sigma_z$, $\\tau_{xz}$, and $\\tau_{yz}$ taken as approximately zero. A general triaxial state requires all three principal stresses.\u003C\u002Fp>\u003Ch2>Principal stresses\u003C\u002Fh2>\u003Cp>There are mutually perpendicular principal planes on which shear stresses vanish. The normal stresses acting on these planes are the principal stresses. They provide coordinate-independent characteristics that are convenient for strength assessment.\u003C\u002Fp>{{chunk:plane-stress-principal-stresses}}\u003Ch2>Strain state\u003C\u002Fh2>\u003Cp>Strain at a point is described by normal strains $\\varepsilon_x$, $\\varepsilon_y$, $\\varepsilon_z$ and engineering shear strains $\\gamma_{xy}$, $\\gamma_{yz}$, $\\gamma_{zx}$. For a linearly elastic isotropic material, stresses and strains are related by generalized Hooke's law.\u003C\u002Fp>{{chunk:generalized-hooke-3d}}\u003Ch2>Transition to strength assessment\u003C\u002Fh2>\u003Cp>Under a multiaxial stress state, a single relation such as $\\sigma=N\u002FA$ is not sufficient. For ductile materials, an equivalent stress based on the Tresca or von Mises criterion is commonly compared with the strength characteristic specified by the adopted design method.\u003C\u002Fp>\u003Ch2>Section structure\u003C\u002Fh2>\u003Cp>The child topics develop stress transformation and Mohr's circle, principal stresses, failure criteria, and generalized Hooke's law. Together they form the basis for analyzing combined loading.\u003C\u002Fp>",[1191,1201,1211],{"id":1192,"parent_id":1182,"code":1193,"slug":1194,"name":1195,"seo_title":1196,"seo_description":1197,"seo_text":1198,"content":1199,"locale":8,"uk_topic_id":880,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":1194,"children":1200},51,"7","stress-analysis-at-a-point","Stress Analysis at a Point (Principal Stresses)","Stress Analysis at a Point — Principal Stresses & Mohr's Circle","How to calculate principal stresses and use Mohr's Circle for 2D\u002F3D stress analysis.","Determining extreme values of normal and shear stresses is a critical step in structural strength evaluation. This section presents analytical formulas for principal stress calculations as well as Mohr's Circle—a graphical method for stress state transformation. Mohr's Circle provides visual insight into stress variation relative to plane inclination angles, simplifying peak shear stress determination.","\u003Cp>Stress components at a point depend on the orientation of the plane, while the physical stress state itself remains unchanged. Stress analysis determines the normal and shear stresses on rotated planes and identifies their extreme values.\u003C\u002Fp>\u003Ch2>Principal planes and principal stresses\u003C\u002Fh2>\u003Cp>\u003Cstrong>Principal planes\u003C\u002Fstrong> are planes on which shear stress is zero. The normal stresses acting on them are the principal stresses. For plane stress, they can be calculated directly from $\\sigma_x$, $\\sigma_y$, and $\\tau_{xy}$.\u003C\u002Fp>{{chunk:plane-stress-principal-stresses}}\u003Ch2>Stress transformation\u003C\u002Fh2>\u003Cp>When the coordinate axes are rotated through an angle $\\theta$, the components of plane stress transform according to trigonometric relations. With a commonly used sign convention:\u003C\u002Fp>\u003Cp>$$\\sigma_{x'}=\\frac{\\sigma_x+\\sigma_y}{2}+\\frac{\\sigma_x-\\sigma_y}{2}\\cos2\\theta+\\tau_{xy}\\sin2\\theta,$$\u003C\u002Fp>\u003Cp>$$\\tau_{x'y'}=-\\frac{\\sigma_x-\\sigma_y}{2}\\sin2\\theta+\\tau_{xy}\\cos2\\theta.$$\u003C\u002Fp>\u003Cp>The sign of the shear component must always be interpreted consistently with the selected convention.\u003C\u002Fp>\u003Ch2>Mohr's circle\u003C\u002Fh2>{{chunk:mohr-circle-plane-stress}}\u003Cp>Mohr's circle is a graphical representation of the same transformation equations. It provides a convenient way to identify principal stresses, maximum shear stresses, and the stresses acting on a plane of specified orientation.\u003C\u002Fp>\u003Ch2>Maximum shear stress\u003C\u002Fh2>\u003Cp>In the plane representation, the radius of Mohr's circle equals the maximum absolute shear stress among planes whose normals lie in the $x$–$y$ plane. For a complete three-dimensional assessment, all three principal stresses must be considered; the absolute maximum shear stress equals half the largest difference between them.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Let $\\sigma_x=80\\ \\text{MPa}$, $\\sigma_y=20\\ \\text{MPa}$, and $\\tau_{xy}=30\\ \\text{MPa}$. The circle center is $C=50\\ \\text{MPa}$ and its radius is:\u003C\u002Fp>\u003Cp>$$R=\\sqrt{30^2+30^2}\\approx42.43\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>Therefore, $\\sigma_1\\approx92.43\\ \\text{MPa}$, $\\sigma_2\\approx7.57\\ \\text{MPa}$, and the maximum in-plane shear stress is approximately $42.43\\ \\text{MPa}$.\u003C\u002Fp>\u003Ch2>Verification\u003C\u002Fh2>\u003Cp>The sum of the two principal stresses must equal $\\sigma_x+\\sigma_y$, and their average must equal the center coordinate of Mohr's circle. These relations provide useful checks on the calculation.\u003C\u002Fp>",[],{"id":1202,"parent_id":1182,"code":1203,"slug":1204,"name":1205,"seo_title":1206,"seo_description":1207,"seo_text":1208,"content":1209,"locale":8,"uk_topic_id":891,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":891,"url_slug":1204,"children":1210},52,"8","strength-theories","Strength Theories (Failure Criteria)","Failure Criteria — Tresca, von Mises & Mohr Explained","Compare failure criteria for multiaxial stress: Tresca maximum shear stress, von Mises distortion energy, Mohr-type criteria, and equivalent stress.","Strength theories (failure criteria) allow engineers to evaluate structural safety under complex multiaxial stress states by comparing them to simple uniaxial tensile testing. This section details maximum shear stress theory (Tresca criterion), distortion energy theory (von Mises criterion), and Mohr-Coulomb failure criterion for materials with asymmetric tensile and compressive strengths.","\u003Cp>Under a multiaxial stress state, a material is subjected simultaneously to several stress components. \u003Cstrong>Failure criteria\u003C\u002Fstrong> reduce such a state to an equivalent measure that can be compared with material strength obtained from simpler tests.\u003C\u002Fp>\u003Ch2>Why an equivalent stress is needed\u003C\u002Fh2>\u003Cp>A single principal stress does not always represent the severity of a complex stress state. For ductile metals, differences between principal stresses and the deviatoric part of the stress state are especially important. This motivates the widespread use of the Tresca and von Mises criteria.\u003C\u002Fp>{{chunk:tresca-von-mises-criteria}}\u003Ch2>Tresca criterion\u003C\u002Fh2>\u003Cp>The maximum-shear-stress criterion associates yielding with the largest difference between principal stresses. It is straightforward for hand calculations and is generally somewhat more conservative than von Mises for many loading states.\u003C\u002Fp>\u003Ch2>von Mises criterion\u003C\u002Fh2>\u003Cp>The distortion-energy criterion associates yielding with the energy of shape change. It is widely used for ductile isotropic metals in engineering calculations and numerical analysis.\u003C\u002Fp>\u003Ch2>Plane stress\u003C\u002Fh2>\u003Cp>When $\\sigma_z=0$, the von Mises equivalent stress can be written directly in terms of the plane-stress components:\u003C\u002Fp>\u003Cp>$$\\sigma_{\\mathrm{eq,VM}}=\\sqrt{\\sigma_x^2-\\sigma_x\\sigma_y+\\sigma_y^2+3\\tau_{xy}^2}.$$\u003C\u002Fp>\u003Cp>This form allows an assessment without first calculating the principal stresses.\u003C\u002Fp>\u003Ch2>Materials with different tensile and compressive strengths\u003C\u002Fh2>\u003Cp>For brittle materials, or materials whose tensile and compressive strengths differ substantially, other criteria may be more appropriate, including Mohr-type approaches. The selected relation and allowable material characteristics must follow the adopted code or calculation method.\u003C\u002Fp>\u003Ch2>Historical criteria\u003C\u002Fh2>\u003Cp>Educational courses also discuss the maximum-normal-stress and maximum-normal-strain criteria. They are useful for understanding the development of failure theories but are not universal criteria for modern design of ductile metals.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For principal stresses $\\sigma_1=100\\ \\text{MPa}$, $\\sigma_2=40\\ \\text{MPa}$, and $\\sigma_3=0$, Tresca gives $\\sigma_{\\mathrm{eq,T}}=100\\ \\text{MPa}$. Von Mises gives:\u003C\u002Fp>\u003Cp>$$\\sigma_{\\mathrm{eq,VM}}=\\sqrt{\\frac{60^2+40^2+100^2}{2}}\\approx87.2\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>The criterion must be selected according to the material, expected failure mechanism, and the rules of the specific design method.\u003C\u002Fp>",[],{"id":1212,"parent_id":1182,"code":1213,"slug":1214,"name":1215,"seo_title":1216,"seo_description":1217,"seo_text":1218,"content":1219,"locale":8,"uk_topic_id":903,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":903,"url_slug":1214,"children":1220},53,"9","generalized-hookes-law","Generalized Hooke's Law","Generalized Hooke’s Law — 3D Stress-Strain Relations","Learn generalized Hooke’s law for isotropic linear elasticity: 3D normal and shear stress-strain relations using Young’s modulus, Poisson’s ratio, and shear modulus.","Generalized Hooke's law defines linear relationships between components of the strain tensor and stress tensor for isotropic elastic bodies under triaxial stress states. This section covers linear strains along principal axes, shear strains, volumetric deformation, and bulk modulus. Practical engineering applications include stress analysis of thin-walled pressure vessels and pipes.","\u003Cp>\u003Cstrong>Generalized Hooke's law\u003C\u002Fstrong> relates stress and strain components in a linearly elastic isotropic material. Unlike the uniaxial relation $\\sigma=E\\varepsilon$, it accounts for simultaneous normal stresses in three directions as well as shear stresses.\u003C\u002Fp>\u003Ch2>Poisson effect\u003C\u002Fh2>\u003Cp>A normal stress acting in one direction produces not only longitudinal strain but also transverse strains. This coupling is described by Poisson's ratio $\\nu$.\u003C\u002Fp>{{chunk:generalized-hooke-3d}}\u003Ch2>Principal directions\u003C\u002Fh2>\u003Cp>If the coordinate axes are aligned with the principal directions, shear stresses on the principal planes are zero. The normal-strain relation can then be written in terms of $\\sigma_1$, $\\sigma_2$, and $\\sigma_3$:\u003C\u002Fp>\u003Cp>$$\\varepsilon_1=\\frac{1}{E}[\\sigma_1-\\nu(\\sigma_2+\\sigma_3)],$$\u003C\u002Fp>\u003Cp>with the other two equations obtained by cyclic permutation of the indices.\u003C\u002Fp>\u003Ch2>Volumetric strain\u003C\u002Fh2>{{chunk:volumetric-strain-bulk-modulus}}\u003Cp>The hydrostatic part of the stress state is associated with volume change, whereas the deviatoric part is associated with change of shape. This distinction is important for understanding energy-based failure criteria.\u003C\u002Fp>\u003Ch2>Plane stress\u003C\u002Fh2>\u003Cp>For a thin plate, $\\sigma_z$ is often taken as zero. However, $\\varepsilon_z$ is generally not zero: because of the Poisson effect, $\\sigma_x$ and $\\sigma_y$ produce strain through the thickness:\u003C\u002Fp>\u003Cp>$$\\varepsilon_z=-\\frac{\\nu}{E}(\\sigma_x+\\sigma_y).$$\u003C\u002Fp>\u003Cp>This is an important distinction between plane stress and plane strain.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Let $E=200\\ \\text{GPa}$, $\\nu=0.3$, $\\sigma_x=100\\ \\text{MPa}$, $\\sigma_y=40\\ \\text{MPa}$, and $\\sigma_z=0$. Then:\u003C\u002Fp>\u003Cp>$$\\varepsilon_x=\\frac{100-0.3\\cdot40}{200000}=0.00044,$$\u003C\u002Fp>\u003Cp>$$\\varepsilon_y=\\frac{40-0.3\\cdot100}{200000}=0.00005,$$\u003C\u002Fp>\u003Cp>$$\\varepsilon_z=-\\frac{0.3(100+40)}{200000}=-0.00021.$$\u003C\u002Fp>\u003Ch2>Limits of applicability\u003C\u002Fh2>\u003Cp>These relations assume small strains, a homogeneous isotropic material, and linear-elastic behavior. Anisotropic materials, plastic deformation, and large strains require different constitutive relations.\u003C\u002Fp>",[],{"id":1222,"parent_id":784,"code":1223,"slug":1224,"name":1225,"seo_title":1226,"seo_description":1227,"seo_text":1228,"content":1229,"locale":8,"uk_topic_id":1230,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1224,"children":1231},59,"15","geometric-properties-of-plane-areas","Geometric Properties of Plane Areas","Geometric Properties of Areas — Centroid, Moment of Inertia & Section Modulus","Calculate centroids, first and second moments of area, polar moment, section modulus, principal axes, and properties of composite cross-sections.","Cross-sectional geometric properties dictate a member's resistance to various deformation modes independent of material composition. This section explores first moments of area, centroid coordinates, planar and polar moments of inertia, parallel axis theorem (Steiner's theorem), and methods for locating principal axes of inertia for complex built-up rolled steel shapes (channels, I-beams, angles).","\u003Cp>\u003Cstrong>Geometric properties of a plane area\u003C\u002Fstrong> describe how area is distributed relative to selected axes and points. Unlike mechanical material properties, they depend only on the shape, dimensions, and orientation of the cross-section. These quantities are used in calculations of bending, torsion, stability, and combined loading.\u003C\u002Fp>\u003Ch2>Area and centroid\u003C\u002Fh2>\u003Cp>The area $A$ gives the overall size of a cross-section, while first moments of area are used to determine the position of its centroid. For composite sections, the centroid is obtained as an area-weighted average of the centroids of the simple component shapes.\u003C\u002Fp>{{chunk:area-static-moments-centroid}}\u003Ch2>Second moments of area\u003C\u002Fh2>\u003Cp>The second moments of area $I_x$ and $I_y$ characterize the distribution of area about the corresponding axes. Area elements farther from an axis contribute more strongly because the distance enters quadratically.\u003C\u002Fp>{{chunk:area-second-moments}}\u003Ch2>Axis transfer and rotation\u003C\u002Fh2>\u003Cp>When the required axis is parallel to a known centroidal axis, the parallel-axis theorem is used. When the orientation of the axes changes, transformation formulas are applied to determine principal centroidal axes and principal second moments of area.\u003C\u002Fp>{{chunk:parallel-axis-theorem}}\u003Ch2>Section moduli\u003C\u002Fh2>\u003Cp>In bending problems, the section modulus $W=I\u002Fy_{\\max}$ is used, where $y_{\\max}$ is the distance from the neutral axis to the corresponding extreme fiber. For an unsymmetrical cross-section, the section moduli for the upper and lower edges may differ.\u003C\u002Fp>\u003Ch2>Typical cross-section shapes\u003C\u002Fh2>{{chunk:threejs-typical-cross-sections}}\u003Ch2>Simple and composite sections\u003C\u002Fh2>{{chunk:rectangle-circle-area-properties}}\u003Cp>Complex profiles are divided into rectangles, triangles, circles, or other simple component areas. After the common centroid is found, the properties of each component are transferred to the common axes and summed.\u003C\u002Fp>{{chunk:composite-section-properties-algorithm}}\u003Ch2>Section structure\u003C\u002Fh2>\u003Cp>The child topics separately cover first moments and centroids, second moments of area, the parallel-axis theorem, axis rotation, principal properties, and section moduli. This provides the geometric basis for the subsequent study of stresses and deformations in bending.\u003C\u002Fp>",15,[1232,1243,1254,1265,1276],{"id":1233,"parent_id":1222,"code":1234,"slug":1235,"name":1236,"seo_title":1237,"seo_description":1238,"seo_text":1239,"content":1240,"locale":8,"uk_topic_id":1241,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1235,"children":1242},119,"15.1","centroid-and-first-moments-of-area","Centroid and First Moments of Area","How to Find a Section Centroid — First Moments of Area & Example","Find the centroid of simple and composite plane sections using first moments of area, symmetry, negative areas for holes, and a worked example.","This topic explains first moments of area and the determination of centroid coordinates for plane sections. It covers simple and composite shapes, algebraic summation of areas, treatment of holes as negative areas, and symmetry as a useful geometric check.","\u003Cp>The \u003Cstrong>centroid of an area\u003C\u002Fstrong> is the geometric point through which the centroidal axes of a plane section pass. Its location is determined using first moments of area.\u003C\u002Fp>{{chunk:area-static-moments-centroid}}\u003Ch2>Property of centroidal axes\u003C\u002Fh2>\u003Cp>The first moment of the complete area about an axis passing through its centroid is zero. If a shape has one axis of symmetry, its centroid lies on that axis; with two symmetry axes, the centroid lies at their intersection.\u003C\u002Fp>\u003Ch2>Composite section\u003C\u002Fh2>\u003Cp>Divide a composite shape into simple parts. Determine each area $A_i$ and the coordinates of its own centroid, then use the algebraic sums $\\sum A_i x_i$ and $\\sum A_i y_i$. Holes are treated as negative areas.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose two rectangles have areas $A_1=2000\\ \\text{mm}^2$ and $A_2=1000\\ \\text{mm}^2$, with centroid $y$-coordinates of $20$ and $80\\ \\text{mm}$. Then:\u003C\u002Fp>\u003Cp>$$y_c=\\frac{2000\\cdot20+1000\\cdot80}{3000}=40\\ \\text{mm}.$$\u003C\u002Fp>\u003Ch2>Common errors\u003C\u002Fh2>\u003Cp>Do not average component centroid coordinates without weighting them by area. All coordinates must be measured from the same reference axis, and the sign of a hole area must be handled consistently.\u003C\u002Fp>",105,[],{"id":1244,"parent_id":1222,"code":1245,"slug":1246,"name":1247,"seo_title":1248,"seo_description":1249,"seo_text":1250,"content":1251,"locale":8,"uk_topic_id":1252,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1246,"children":1253},120,"15.2","second-and-polar-moments-of-area","Second and Polar Moments of Area","Second and Polar Moments of Area — Ix, Iy and Jp","Definitions of Ix, Iy, Ixy and Jp, geometric meaning, units, and standard formulas for rectangular, circular, and annular sections.","This topic introduces the second moments of area Ix and Iy, product of inertia Ixy, and polar moment Jp as geometric measures of area distribution. It explains their integral definitions, units, geometric meaning, and standard formulas for rectangles, circles, and annuli.","\u003Cp>A \u003Cstrong>second moment of area\u003C\u002Fstrong> characterizes how cross-sectional area is distributed relative to a selected axis. It is a geometric property and must not be confused with a mass moment of inertia.\u003C\u002Fp>{{chunk:area-second-moments}}\u003Ch2>Geometric meaning\u003C\u002Fh2>\u003Cp>Area elements located farther from an axis contribute much more strongly because the distance enters quadratically. This is why sections that place material far from the centroidal axis can achieve high bending stiffness with relatively modest area.\u003C\u002Fp>\u003Ch2>Standard shapes\u003C\u002Fh2>{{chunk:rectangle-circle-area-properties}}\u003Ch2>Polar moment\u003C\u002Fh2>\u003Cp>For two mutually perpendicular axes $x$ and $y$ passing through the same point, $J_p=I_x+I_y$. For a circular shaft, this property enters directly into the classical torsion formulas.\u003C\u002Fp>\u003Ch2>Units\u003C\u002Fh2>\u003Cp>If geometric dimensions are given in millimetres, second moments of area are expressed in mm⁴. Because characteristic dimensions enter to the fourth power, relatively small dimensional changes can strongly affect $I$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a rectangle with $b=40\\ \\text{mm}$ and $h=80\\ \\text{mm}$, about its centroidal $x$-axis parallel to $b$:\u003C\u002Fp>\u003Cp>$$I_x=\\frac{40\\cdot80^3}{12}\\approx1.707\\cdot10^6\\ \\text{mm}^4.$$\u003C\u002Fp>",106,[],{"id":1255,"parent_id":1222,"code":1256,"slug":1257,"name":1258,"seo_title":1259,"seo_description":1260,"seo_text":1261,"content":1262,"locale":8,"uk_topic_id":1263,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1257,"children":1264},121,"15.3","parallel-axis-theorem-and-composite-sections","Parallel-Axis Theorem and Composite Sections","Parallel-Axis Theorem — Composite Sections Step by Step","Learn the parallel-axis theorem step by step for composite sections: find the centroid, transfer moments of inertia, handle holes, and avoid common mistakes.","This topic explains the parallel-axis theorem and its application to composite cross-sections. It covers decomposition into simple shapes, centroid determination, transfer of component second moments to common centroidal axes, algebraic summation, and treatment of holes.","\u003Cp>For a composite section, tabulated properties of individual simple shapes are not sufficient because their second moments must first be referred to a common axis. The \u003Cstrong>parallel-axis theorem\u003C\u002Fstrong> provides this transfer.\u003C\u002Fp>{{chunk:parallel-axis-theorem}}\u003Ch2>Why the centroid comes first\u003C\u002Fh2>\u003Cp>Centroidal second moments of a composite section are calculated about axes through the centroid of the complete area. Therefore, the first step is to determine that centroid.\u003C\u002Fp>{{chunk:area-static-moments-centroid}}\u003Ch2>Procedure\u003C\u002Fh2>{{chunk:composite-section-properties-algorithm}}\u003Ch2>Holes\u003C\u002Fh2>\u003Cp>A hole can be treated as a negative component: its area, first moments, and transferred second moment are subtracted from the corresponding sums for solid parts.\u003C\u002Fp>\u003Ch2>Example structure\u003C\u002Fh2>\u003Cp>For a T-section, represent the flange and web as two rectangles. First determine the centroid coordinate of the complete section. Then calculate the centroidal $I_x$ of each rectangle and add $A_i a_i^2$, where $a_i$ is the distance between the component centroid and the common centroidal axis.\u003C\u002Fp>\u003Ch2>Check\u003C\u002Fh2>\u003Cp>Verify that all distances are measured to the same reference axis, units are consistent, and the $Aa^2$ term has not been added to a second moment that is already taken about the required common axis.\u003C\u002Fp>",107,[],{"id":1266,"parent_id":1222,"code":1267,"slug":1268,"name":1269,"seo_title":1270,"seo_description":1271,"seo_text":1272,"content":1273,"locale":8,"uk_topic_id":1274,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1268,"children":1275},122,"15.4","product-of-inertia-and-axis-rotation","Product of Inertia and Axis Rotation","Product of Inertia Ixy — Axis Rotation & Principal Axes","Learn product of inertia Ixy and transform Ix, Iy, and Ixy under axis rotation, use the invariant Ix + Iy, and determine principal centroidal axes.","This topic explains the product of inertia Ixy and how second moments of area depend on coordinate-axis orientation. It presents the transformation equations for Ix, Iy, and Ixy, the invariant Ix + Iy, and the condition used to determine principal centroidal axes.","\u003Cp>The \u003Cstrong>product of inertia $I_{xy}$\u003C\u002Fstrong> characterizes the combined distribution of area relative to two mutually perpendicular axes. Unlike $I_x$ and $I_y$, it may be positive, negative, or zero.\u003C\u002Fp>{{chunk:area-second-moments}}\u003Ch2>Symmetry\u003C\u002Fh2>\u003Cp>If one centroidal axis is an axis of symmetry, the product of inertia with respect to that axis and the perpendicular centroidal axis is zero. However, $I_{xy}=0$ by itself does not necessarily imply geometric symmetry.\u003C\u002Fp>\u003Ch2>Axis rotation\u003C\u002Fh2>{{chunk:area-inertia-axis-rotation}}\u003Cp>Under rotation of axes, the sum $I_x+I_y$ remains unchanged. This invariant is useful for checking calculations.\u003C\u002Fp>\u003Ch2>Principal axes\u003C\u002Fh2>\u003Cp>Orientations for which $I_{xy}=0$ and $I_x$ and $I_y$ take extreme values are called principal axes of inertia. If they pass through the centroid, they are principal centroidal axes.\u003C\u002Fp>\u003Ch2>Engineering significance\u003C\u002Fh2>\u003Cp>For unsymmetrical sections, principal axes are required in unsymmetrical bending and other problems where the loading direction does not coincide with convenient geometric axes.\u003C\u002Fp>",108,[],{"id":1277,"parent_id":1222,"code":1278,"slug":1279,"name":1280,"seo_title":1281,"seo_description":1282,"seo_text":1283,"content":1284,"locale":8,"uk_topic_id":1285,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1279,"children":1286},123,"15.5","principal-axes-principal-moments-and-section-moduli","Principal Axes, Principal Moments, and Section Moduli","Principal Moments of Area & Section Modulus — Formulas for Bending","Learn principal centroidal axes and moments of area, calculate section modulus W = I\u002Fymax, and use these geometric properties in bending-stress calculations.","This topic explains principal centroidal axes and principal second moments of area, their determination for unsymmetrical sections, and their relationship to section modulus. It covers W = I\u002Fymax, separate section moduli for unequal extreme-fiber distances, and the use of these properties in bending calculations.","\u003Cp>\u003Cstrong>Principal centroidal axes\u003C\u002Fstrong> are mutually perpendicular axes through the centroid for which the product of inertia is zero and the second moments of area take extreme values.\u003C\u002Fp>{{chunk:area-inertia-axis-rotation}}\u003Ch2>Principal second moments\u003C\u002Fh2>\u003Cp>For known centroidal $I_x$, $I_y$, and $I_{xy}$, the principal values are:\u003C\u002Fp>\u003Cp>$$I_{1,2}=\\frac{I_x+I_y}{2}\\pm\\sqrt{\\left(\\frac{I_x-I_y}{2}\\right)^2+I_{xy}^2}.$$\u003C\u002Fp>\u003Cp>The larger value is commonly denoted $I_1$ and the smaller $I_2$. Their sum equals $I_x+I_y$.\u003C\u002Fp>\u003Ch2>Section modulus\u003C\u002Fh2>\u003Cp>For bending about a selected neutral axis, the geometric section modulus is:\u003C\u002Fp>\u003Cp>$$W=\\frac{I}{y_{\\max}}.$$\u003C\u002Fp>\u003Cp>It has dimensions of length cubed, for example mm³, and enters directly into the maximum normal-stress relation for simple bending: $|\\sigma_{\\max}|=|M|\u002FW$.\u003C\u002Fp>\u003Ch2>Unsymmetrical section\u003C\u002Fh2>\u003Cp>If the neutral axis does not divide the section depth symmetrically, the distances to the extreme fibers differ. Separate section moduli are then used: $W_+=I\u002Fy_+$ and $W_-=I\u002Fy_-$.\u003C\u002Fp>\u003Ch2>Standard shapes\u003C\u002Fh2>{{chunk:rectangle-circle-area-properties}}\u003Ch2>Engineering meaning\u003C\u002Fh2>\u003Cp>A larger second moment of area reduces beam curvature for a given bending moment, while a larger section modulus reduces the maximum normal stress. Efficient beam sections therefore place material appropriately relative to the neutral axis.\u003C\u002Fp>",109,[],{"id":1288,"parent_id":784,"code":1289,"slug":1290,"name":1291,"seo_title":1292,"seo_description":1293,"seo_text":1294,"content":1295,"locale":8,"uk_topic_id":1296,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1290,"children":1297},58,"14","bending","Bending","Beam Bending — Shear Force, Bending Moment, Stress & Deflection","Learn beam bending from Q–M diagrams to flexural and shear stress, neutral axis, flexural rigidity EI, slope, deflection, and strength and stiffness checks.","Bending is a deformation mode characterized by the curvature of a beam's longitudinal axis under transverse loading. This section covers pure bending, transverse bending, and direct bending in symmetric cross-sections. Learn the internal force generation theory for shear force Q and bending moment M, along with differential relationships between load intensity, shear force, and bending moment.","\u003Cp>\u003Cstrong>Bending\u003C\u002Fstrong> is a deformation mode in which the longitudinal axis of a member becomes curved under transverse loads or applied moments. A typical structural member working in bending is a beam.\u003C\u002Fp>\u003Ch2>Internal force resultants\u003C\u002Fh2>\u003Cp>In transverse bending, the beam sections carry a bending moment $M$ and a shear force $Q$. They are determined by the method of sections and represented by diagrams along the beam axis. Detailed construction of $Q$ and $M$ diagrams is covered in the corresponding child topic.\u003C\u002Fp>\u003Ch2>Normal stresses\u003C\u002Fh2>\u003Cp>The bending moment produces normal stresses: part of the cross-section is in tension and another part is in compression. Between them lies the neutral axis, where normal stress is zero in the classical simple-bending model.\u003C\u002Fp>{{chunk:bending-navier-stress}}\u003Ch2>Shear stresses\u003C\u002Fh2>\u003Cp>The shear force $Q$ produces shear stresses whose distribution depends on the cross-sectional shape. In classical beam theory, they are evaluated using the Zhuravsky formula.\u003C\u002Fp>{{chunk:beam-shear-zhuravsky}}\u003Ch2>Beam deformation\u003C\u002Fh2>\u003Cp>Under load, the beam axis becomes an elastic curve. The main kinematic quantities are transverse deflection $w$ and cross-section rotation $\\theta$. Resistance to curvature is characterized by the flexural rigidity $EI$.\u003C\u002Fp>{{chunk:beam-curvature-moment}}\u003Ch2>Strength and stiffness\u003C\u002Fh2>\u003Cp>Beam design normally includes a stress check at critical sections and a check of deflections or rotations. Excessive deflection may make a structure unserviceable even when the stresses remain within safe limits.\u003C\u002Fp>\u003Ch2>Section structure\u003C\u002Fh2>\u003Cp>The child topics successively cover internal forces and $Q$–$M$ diagrams, normal and shear stresses, deflections and rotations, the differential equation of the elastic curve, and the initial-parameter method. The geometric properties $I$ and $W$ are treated in a separate section on plane cross-sections.\u003C\u002Fp>",14,[1298,1308,1319],{"id":5,"parent_id":1288,"code":1299,"slug":1300,"name":1301,"seo_title":1302,"seo_description":1303,"seo_text":1304,"content":1305,"locale":8,"uk_topic_id":1306,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":1300,"children":1307},"16","internal-forces-in-bending","Internal Forces in Bending","Shear Force & Bending Moment Diagrams — Q and M Step by Step","Learn how to construct beam shear-force Q and bending-moment M diagrams using support reactions, the method of sections, sign conventions, and diagram checks.","Accurate determination of internal forces is the primary step in beam bending calculations. This section covers the method of sections for evaluating shear forces Q and bending moments M across beam segments. Detailed explanations include sign conventions, span segmentation rules, critical section identification, and differential consistency checks.","\u003Cp>In transverse beam bending, the main internal force resultants are the \u003Cstrong>shear force $Q$\u003C\u002Fstrong> and the \u003Cstrong>bending moment $M$\u003C\u002Fstrong>. They are determined by the method of sections, and their variation along the beam is represented graphically by diagrams.\u003C\u002Fp>\u003Ch2>Shear force and bending moment\u003C\u002Fh2>\u003Cp>The shear force $Q$ represents the resultant internal tangential action in a cross-section. The bending moment $M$ represents the resultant effect of internal normal forces that causes curvature of the beam axis.\u003C\u002Fp>\u003Cp>To determine $Q$ and $M$, make an imaginary cut at the required section and write equilibrium equations for one of the two resulting beam parts.\u003C\u002Fp>{{chunk:beam-qm-section-method}}\u003Ch2>Sign conventions\u003C\u002Fh2>\u003Cp>Use one consistent sign convention throughout the calculation. A positive bending moment is commonly associated with sagging, in which the lower fibers are in tension and the upper fibers in compression. The sign of $Q$ follows the selected convention for the cut portion. A negative calculated value means that the actual internal resultant acts opposite to the assumed positive direction.\u003C\u002Fp>\u003Ch2>$Q$ and $M$ diagrams\u003C\u002Fh2>\u003Cp>A diagram shows how an internal force resultant varies along the beam axis. Before constructing diagrams, divide the beam into segments at supports, concentrated forces and moments, and the start or end points of distributed loads.\u003C\u002Fp>{{chunk:beam-qm-differential-relations}}\u003Ch2>Characteristic diagram features\u003C\u002Fh2>\u003Cul>\u003Cli>a concentrated transverse force causes a jump in the $Q$ diagram;\u003C\u002Fli>\u003Cli>a concentrated applied moment causes a jump in the $M$ diagram;\u003C\u002Fli>\u003Cli>where distributed load is absent, $Q$ is constant and $M$ varies linearly;\u003C\u002Fli>\u003Cli>under a uniform distributed load, $Q$ varies linearly and $M$ is parabolic;\u003C\u002Fli>\u003Cli>where $Q$ crosses zero, $M$ has a local maximum or minimum.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Short example\u003C\u002Fh2>\u003Cp>Consider a cantilever beam of length $L$ carrying a concentrated force $F$ at its free end. No distributed load acts between the free end and the fixed support, so the shear force is constant in magnitude: $|Q|=F$. The bending moment varies linearly from zero at the free end to the maximum magnitude $|M|_{\\max}=FL$ at the fixed end. The signs depend on the adopted convention.\u003C\u002Fp>\u003Ch2>Calculation check\u003C\u002Fh2>\u003Cp>After constructing the diagrams, verify overall equilibrium, jumps at concentrated forces and moments, the expected curve shape on every segment, and consistency between the slope of the $M$ diagram and the sign of $Q$. These checks reveal many common calculation errors.\u003C\u002Fp>\u003Ch2>Learning outcome\u003C\u002Fh2>\u003Cp>After studying this topic, you should be able to determine support reactions, divide a beam into calculation segments, derive $Q(x)$ and $M(x)$, locate characteristic and extreme values, and construct shear-force and bending-moment diagrams.\u003C\u002Fp>",16,[],{"id":1309,"parent_id":1288,"code":1310,"slug":1311,"name":1312,"seo_title":1313,"seo_description":1314,"seo_text":1315,"content":1316,"locale":8,"uk_topic_id":1317,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1311,"children":1318},61,"17","stresses-in-bending","Stresses in Bending","Beam Bending Stress — Flexure Formula & Shear Stress","Calculate normal bending stress with σ = My\u002FI and shear stress with the Zhuravsky formula, including section modulus, neutral axis, and critical beam sections.","Transverse bending generates both normal flexural stresses and transverse shear stresses within beam cross-sections. This page presents the derivation of Navier's flexure formula for normal stress calculations using section modulus, alongside Jourawski's formula for shear stress distributions across rectangular, I-shaped, and circular profiles.","\u003Cp>In transverse bending, a beam cross-section may simultaneously carry \u003Cstrong>normal stresses\u003C\u002Fstrong> caused by the bending moment $M$ and \u003Cstrong>shear stresses\u003C\u002Fstrong> caused by the shear force $Q$. Their distributions across the section differ, so the critical points for $\\sigma$ and $\\tau$ do not necessarily coincide.\u003C\u002Fp>\u003Ch2>Normal stresses\u003C\u002Fh2>{{chunk:bending-navier-stress}}\u003Cp>In the classical simple-bending model, the neutral axis passes through the centroid and coincides with a principal centroidal axis of the section. Normal stress varies linearly: $\\sigma=0$ at the neutral axis and reaches its largest absolute values at the extreme fibers.\u003C\u002Fp>\u003Ch2>Section-modulus form\u003C\u002Fh2>\u003Cp>For a strength check, it is convenient to use $|\\sigma_{\\max}|=|M|\u002FW$. If the section is unsymmetrical about the neutral axis, the distances to the extreme fibers may differ, so the tensile and compressive sides should be checked separately.\u003C\u002Fp>\u003Ch2>Shear stresses\u003C\u002Fh2>{{chunk:beam-shear-zhuravsky}}\u003Cp>For a rectangular section, the shear-stress distribution over the depth is parabolic and its maximum at the neutral axis is $\\tau_{\\max}=3Q\u002F(2A)$. In an I-section, a large portion of the shear force is carried by the web because its local width is small while the first moment of the adjacent area can be substantial.\u003C\u002Fp>\u003Ch2>Circular section\u003C\u002Fh2>\u003Cp>For a solid circular section, shear stress is also zero at the external boundary and reaches its maximum near the neutral axis. For the classical solid circle, $\\tau_{\\max}=4Q\u002F(3A)$.\u003C\u002Fp>\u003Ch2>Strength assessment\u003C\u002Fh2>\u003Cp>Normal stresses are checked at sections with large $|M|$ and at extreme fibers. Shear stresses are checked at sections with large $|Q|$ and at characteristic points of the cross-section. If both $\\sigma$ and $\\tau$ are significant at the same point, a multiaxial failure criterion may be required.\u003C\u002Fp>\u003Ch2>Limits of the classical model\u003C\u002Fh2>\u003Cp>Navier's formula assumes linear-elastic behavior, small deformation, and the classical assumptions of beam theory. Near concentrated loads, supports, holes, and abrupt changes of section, the local stress distribution can differ substantially from the elementary model.\u003C\u002Fp>",17,[],{"id":1320,"parent_id":1288,"code":1321,"slug":1322,"name":1323,"seo_title":1324,"seo_description":1325,"seo_text":1326,"content":1327,"locale":8,"uk_topic_id":1328,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1322,"children":1329},62,"18","deflections-in-bending","Deflections in Bending","Beam Deflection & Slope — Flexural Rigidity EI and Methods","Learn beam deflection w and slope θ, flexural rigidity EI, boundary conditions, stiffness checks, and the main methods used to calculate beam displacements.","Under bending moments, the beam axis deforms into an elastic curve. This section introduces linear displacements (deflections w) and angular displacements (slope angles θ). Evaluating beam deflections is necessary to verify structural stiffness and ensure peak deflection does not exceed allowable serviceability limits set by building codes.","\u003Cp>Under bending moments, a beam deforms and its initially straight axis becomes an \u003Cstrong>elastic curve\u003C\u002Fstrong>. Stiffness calculations determine the transverse \u003Cstrong>deflection $w(x)$\u003C\u002Fstrong> and the \u003Cstrong>rotation $\\theta(x)$\u003C\u002Fstrong> of a cross-section.\u003C\u002Fp>\u003Ch2>Deflection and rotation\u003C\u002Fh2>\u003Cp>Deflection $w$ is the transverse displacement of a point on the beam axis. For small deformations, the rotation of a cross-section is approximately the derivative of deflection: $\\theta(x)\\approx w'(x)$.\u003C\u002Fp>\u003Ch2>Flexural rigidity\u003C\u002Fh2>{{chunk:beam-curvature-moment}}\u003Cp>The product $EI$ is called the \u003Cstrong>flexural rigidity\u003C\u002Fstrong>. Increasing $E$ or $I$ reduces curvature and, all else being equal, beam deflections. Therefore, the geometric distribution of material in the cross-section has a strong effect on stiffness.\u003C\u002Fp>{{chunk:threejs-i-beam-on-supports}}\u003Ch2>Boundary conditions\u003C\u002Fh2>{{chunk:beam-deflection-boundary-conditions}}\u003Cp>Boundary conditions determine the constants of integration and must represent the actual supports and connections. For a multi-segment beam, continuity of deflection and rotation is also imposed at locations without an internal hinge or discontinuity.\u003C\u002Fp>\u003Ch2>Methods for determining displacements\u003C\u002Fh2>\u003Cp>The child topics cover direct integration of the differential equation of the elastic curve and the initial-parameter method using Macaulay functions. In other parts of the course, beam displacements may also be found using energy methods.\u003C\u002Fp>\u003Ch2>Stiffness check\u003C\u002Fh2>\u003Cp>The calculated maximum deflection is compared with an allowable value prescribed by serviceability requirements or the applicable design procedure: $|w_{\\max}|\\le[w]$. Rotation may be limited in the same way when required.\u003C\u002Fp>\u003Ch2>Example of the influence of cross-section size\u003C\u002Fh2>\u003Cp>For a rectangular cross-section, $I=bh^3\u002F12$. If the depth $h$ is doubled while $b$, $E$, $L$, and the loading remain unchanged, $I$ increases by $2^3=8$ times. Within the same linear model, characteristic deflections, which are inversely proportional to $EI$, therefore decrease substantially.\u003C\u002Fp>",18,[1330,1341],{"id":1331,"parent_id":1320,"code":1332,"slug":1333,"name":1334,"seo_title":1335,"seo_description":1336,"seo_text":1337,"content":1338,"locale":8,"uk_topic_id":1339,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1339,"url_slug":1333,"children":1340},63,"19","differential-equation-elastic-curve","Differential Equation of the Elastic Curve","Beam Deflection Equation — EI·y″ = M(x) Step by Step","Calculate beam slope and deflection from the elastic-curve differential equation EI·y″ = M(x): integrate twice, apply boundary conditions, and track signs.","The analytical foundation for beam deformation evaluation is the approximate differential equation of the elastic curve: E*I*y' = -M(x). This section covers its derivation from geometric strain relations and Hooke's law, double integration techniques, boundary condition application at supports, and constructing analytical displacement functions.","\u003Cp>The differential equation of the elastic curve relates beam loading, through the bending-moment function $M(x)$, to geometric deformation expressed by deflection and rotation.\u003C\u002Fp>\u003Ch2>Curvature of the elastic curve\u003C\u002Fh2>{{chunk:beam-curvature-moment}}\u003Cp>The exact geometric curvature of a plane curve contains derivatives of deflection. For small rotations, when $|w'|\\ll1$, the denominator in the exact curvature expression is approximately one, so curvature is proportional to the second derivative of deflection.\u003C\u002Fp>\u003Ch2>Approximate differential equation\u003C\u002Fh2>\u003Cp>Depending on the adopted sign convention, the equation is written as $EIw''(x)=M(x)$ or $EIw''(x)=-M(x)$. Different sign conventions must not be mixed within one calculation.\u003C\u002Fp>\u003Ch2>Double integration\u003C\u002Fh2>{{chunk:beam-double-integration-algorithm}}\u003Cp>The first integration gives the rotation function, and the second gives the deflection function. For constant flexural rigidity $EI$, it may be taken outside the integral. If $E$ or $I$ changes along the beam, the variation must be handled segment by segment or directly in the differential equation.\u003C\u002Fp>\u003Ch2>Boundary conditions\u003C\u002Fh2>{{chunk:beam-deflection-boundary-conditions}}\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a cantilever of length $L$ carrying a force $F$ at the free end and having constant $EI$, integration with $w(0)=0$ and $w'(0)=0$ gives the familiar magnitudes at the free end:\u003C\u002Fp>\u003Cp>$$|\\theta(L)|=\\frac{FL^2}{2EI},\\qquad |w(L)|=\\frac{FL^3}{3EI}.$$\u003C\u002Fp>\u003Cp>The signs depend on the force direction and the selected positive direction for deflection.\u003C\u002Fp>\u003Ch2>Limits of applicability\u003C\u002Fh2>\u003Cp>The approximate equation assumes small deflections and rotations, linear-elastic material behavior, and applicability of the classical beam model. Large displacements require a geometrically nonlinear description.\u003C\u002Fp>",19,[],{"id":1342,"parent_id":1320,"code":1343,"slug":1344,"name":1345,"seo_title":1346,"seo_description":1347,"seo_text":1348,"content":1349,"locale":8,"uk_topic_id":1350,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1350,"url_slug":1344,"children":1351},64,"20","initial-parameter-method","Initial Parameter Method (Macaulay's Method)","Macaulay’s Method for Beam Deflection — Step-by-Step Procedure","Learn Macaulay’s method for beam slope and deflection: write one bending-moment equation, integrate it, apply boundary conditions, and avoid common errors.","When multiple loading segments exist, direct integration becomes cumbersome due to numerous integration constants. Macaulay's initial parameter method allows writing a single unified equation for slopes and deflections across the entire span. This page presents the governing equation, discontinuity function rules, and worked examples.","\u003Cp>The \u003Cstrong>initial-parameter method\u003C\u002Fstrong> describes beam rotation and deflection with unified expressions instead of introducing a separate pair of integration constants for every loading segment. Macaulay brackets provide a convenient notation for this purpose.\u003C\u002Fp>\u003Ch2>Macaulay brackets\u003C\u002Fh2>\u003Cp>The notation $\\langle x-a\\rangle^n$ means:\u003C\u002Fp>\u003Cp>$$\\langle x-a\\rangle^n=0\\quad\\text{for }x<a,$$\u003C\u002Fp>\u003Cp>$$\\langle x-a\\rangle^n=(x-a)^n\\quad\\text{for }x\\ge a.$$\u003C\u002Fp>\u003Cp>This allows a load that begins at coordinate $a$ to be included in one expression valid along the beam.\u003C\u002Fp>\u003Ch2>Relation to the beam equation\u003C\u002Fh2>{{chunk:beam-curvature-moment}}\u003Cp>The bending-moment function $M(x)$ is written using ordinary terms and Macaulay brackets and then integrated. Initial parameters, such as the deflection $w_0$ and rotation $\\theta_0$ at the chosen origin, play the role of integration constants.\u003C\u002Fp>\u003Ch2>Typical contributions\u003C\u002Fh2>\u003Cp>With signs defined by the adopted convention, a concentrated force $F$ at $x=a$ contributes a term proportional to $F\\langle x-a\\rangle^1$ to $M(x)$; a concentrated moment contributes $M_0\\langle x-a\\rangle^0$; and a uniform distributed load beginning at $a$ contributes a term proportional to $q\\langle x-a\\rangle^2\u002F2$. If a distributed load ends at another coordinate, a compensating term is introduced from that point.\u003C\u002Fp>\u003Ch2>Integration\u003C\u002Fh2>\u003Cp>Macaulay power brackets integrate like ordinary powers:\u003C\u002Fp>\u003Cp>$$\\int\\langle x-a\\rangle^n dx=\\frac{\\langle x-a\\rangle^{n+1}}{n+1}.$$\u003C\u002Fp>\u003Cp>After two integrations of the bending equation, unified expressions for $\\theta(x)$ and $w(x)$ are obtained.\u003C\u002Fp>\u003Ch2>Boundary conditions\u003C\u002Fh2>{{chunk:beam-deflection-boundary-conditions}}\u003Ch2>Example of the elastic curve\u003C\u002Fh2>\u003Cp>The interactive model shows an overhanging beam subjected to two concentrated forces. Its elastic curve is calculated by integrating the bending-moment function expressed with Macaulay brackets and determining the constants from the zero-deflection conditions at the supports. The displayed deformation is magnified only for visibility.\u003C\u002Fp>{{chunk:threejs-beam-deflection-hybrid-primitives}}\u003Ch2>Procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the support reactions.\u003C\u002Fli>\u003Cli>Select the coordinate origin and one consistent sign convention.\u003C\u002Fli>\u003Cli>Write $M(x)$ using Macaulay brackets.\u003C\u002Fli>\u003Cli>Integrate $EIw''=\\pm M(x)$ twice.\u003C\u002Fli>\u003Cli>Determine the initial parameters or constants from boundary conditions.\u003C\u002Fli>\u003Cli>Evaluate the expressions at the required coordinates and verify the support conditions.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Advantages and limitations\u003C\u002Fh2>\u003Cp>The method is especially convenient for beams with several concentrated loads and distributed-load regions. Care is required when specifying the start and end coordinates of every load, the powers of the brackets, and the signs of all terms.\u003C\u002Fp>",20,[],{"id":1353,"parent_id":784,"code":1354,"slug":1355,"name":1356,"seo_title":1357,"seo_description":1358,"seo_text":1359,"content":1360,"locale":8,"uk_topic_id":1361,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":1355,"children":1362},65,"21","combined-loading","Combined Loading","Combined Loading — Axial, Bending & Torsion Stress Analysis","Analyze combined axial load, biaxial bending, eccentric loading, and torsion by superposition, then evaluate critical stresses with an appropriate failure criterion.","Combined loading refers to cases where multiple internal force components act simultaneously within a member's cross-section. This section outlines superposition principles for combining stresses and strains, covering unsymmetrical bending, bending with axial compression\u002Ftension, eccentric load application, and combined bending and torsion.","\u003Cp>\u003Cstrong>Combined loading\u003C\u002Fstrong> occurs when several internal force resultants act simultaneously in a member cross-section, such as axial force $N$, bending moments $M_x$ and $M_y$, torque $T$, or shear forces. Within the linear-elastic range, stresses caused by individual actions can often be calculated separately and then combined by superposition.\u003C\u002Fp>\u003Ch2>Principle of superposition\u003C\u002Fh2>\u003Cp>Superposition is applicable when material behavior is linear elastic, deformations are small, and changes in geometry do not significantly alter the loading scheme. For example, an axial force produces a uniform normal-stress component $N\u002FA$, while bending moments produce linearly varying normal stresses.\u003C\u002Fp>\u003Ch2>Biaxial and unsymmetrical bending\u003C\u002Fh2>{{chunk:unsymmetrical-bending-stress}}\u003Cp>Unsymmetrical bending is a typical combined-loading case: the bending moment is not aligned with one principal centroidal axis, so normal stress must include contributions from bending about both principal axes.\u003C\u002Fp>\u003Ch2>Eccentric axial loading\u003C\u002Fh2>{{chunk:eccentric-axial-stress}}\u003Cp>An eccentrically applied axial force is statically equivalent to a centric axial force plus one or two bending moments. The position of the load determines whether the whole section remains in compression or whether a tensile region appears.\u003C\u002Fp>\u003Ch2>Combined bending and torsion\u003C\u002Fh2>\u003Cp>In power-transmission shafts, bending creates normal stresses while torque creates shear stresses. Because these components act simultaneously at the same material point, strength is assessed using a multiaxial failure criterion.\u003C\u002Fp>{{chunk:bending-torsion-equivalent-stress}}\u003Ch2>General calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the internal force resultants and their critical combinations.\u003C\u002Fli>\u003Cli>Select principal centroidal axes and calculate the required section properties.\u003C\u002Fli>\u003Cli>Calculate normal and shear stress components produced by each internal action.\u003C\u002Fli>\u003Cli>Identify critical points of the section while accounting for stress signs and spatial distribution.\u003C\u002Fli>\u003Cli>For a multiaxial state, apply a failure criterion appropriate to the material.\u003C\u002Fli>\u003Cli>Where required, separately check stiffness, stability, or fatigue strength.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Limits of the approach\u003C\u002Fh2>\u003Cp>Simple superposition should not be applied automatically to plastic deformation, large displacements, significant geometric nonlinearity, or severe local stress concentrations. Such cases require an appropriately extended mechanical model.\u003C\u002Fp>",21,[1363,1374,1384],{"id":1364,"parent_id":1353,"code":1365,"slug":1366,"name":1367,"seo_title":1368,"seo_description":1369,"seo_text":1370,"content":1371,"locale":8,"uk_topic_id":1372,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":1366,"children":1373},66,"22","unsymmetrical-bending","Unsymmetrical Bending","Unsymmetrical Bending — Stress Formula & Neutral Axis","Calculate normal stress in unsymmetrical bending by resolving the bending moment about principal axes, then locate the neutral axis and critical section points.","Unsymmetrical bending occurs when the applied bending moment plane does not coincide with any principal axes of inertia. This page details moment vector decomposition along principal axes, combined normal stress formulas, neutral axis orientation equations, and strength verification procedures for extreme cross-sectional points.","\u003Cp>\u003Cstrong>Unsymmetrical bending\u003C\u002Fstrong> occurs when the plane of the resultant bending moment does not coincide with a principal centroidal plane of the cross-section. The moment is then resolved into two components about the principal axes, and the normal stress is obtained by superposing two simple-bending stress fields.\u003C\u002Fp>\u003Ch2>Principal axes as the calculation system\u003C\u002Fh2>\u003Cp>The most convenient coordinate system uses the principal centroidal axes $x$ and $y$, for which $I_{xy}=0$. Because $I_x$ and $I_y$ may differ substantially, the neutral axis is generally not perpendicular to the resultant bending-moment vector.\u003C\u002Fp>{{chunk:unsymmetrical-bending-stress}}\u003Ch2>Neutral axis\u003C\u002Fh2>\u003Cp>The neutral axis passes through the centroid because, in pure unsymmetrical bending, the resultant normal stress is zero along this line. Its position follows from $\\sigma(x,y)=0$.\u003C\u002Fp>\u003Cp>If $M_x\\ne0$ and $I_x$ and $I_y$ are known, the equation can be written as:\u003C\u002Fp>\u003Cp>$$y=\\frac{M_y I_x}{M_x I_y}x,$$\u003C\u002Fp>\u003Cp>with the signs of the moment components retained consistently.\u003C\u002Fp>\u003Ch2>Finding critical points\u003C\u002Fh2>\u003Cp>The largest absolute stress is not necessarily located at the point with the largest $|x|$ or $|y|$ alone. The critical points are those boundary points where the algebraic sum of the two bending contributions reaches its greatest positive or negative value. For polygonal sections, checking characteristic vertices is often convenient.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>For a rectangular section with known $I_x$ and $I_y$ under $M_x=4\\ \\text{kN·m}$ and $M_y=2\\ \\text{kN·m}$, evaluate the stress at each corner by substituting its coordinates into the unsymmetrical-bending formula. The largest positive value is checked against the allowable tensile stress, and the largest-magnitude negative value against the allowable compressive stress if these limits differ.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Locate the centroid and principal centroidal axes.\u003C\u002Fli>\u003Cli>Resolve the bending moment into $M_x$ and $M_y$.\u003C\u002Fli>\u003Cli>Determine $I_x$ and $I_y$.\u003C\u002Fli>\u003Cli>Write the neutral-axis equation.\u003C\u002Fli>\u003Cli>Calculate $\\sigma$ at characteristic extreme boundary points.\u003C\u002Fli>\u003Cli>Perform the strength check.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Limits\u003C\u002Fh2>\u003Cp>The formulas assume elastic bending, small deformation, and applicability of classical beam theory. For nonprincipal axes with $I_{xy}\\ne0$, independently adding terms of the form $M_x\u002FI_x$ and $M_y\u002FI_y$ without first transforming to principal axes is generally incorrect.\u003C\u002Fp>",22,[],{"id":1375,"parent_id":1353,"code":70,"slug":1376,"name":1377,"seo_title":1378,"seo_description":1379,"seo_text":1380,"content":1381,"locale":8,"uk_topic_id":1382,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1376,"children":1383},67,"eccentric-compression","Eccentric Compression","Eccentric Compression — Stress Formula & Section Core","Calculate stress under an eccentric compressive load using axial force plus bending, locate the neutral axis, and determine the section core for no-tension loading.","Eccentric compression occurs when an axial load is applied away from the centroid of a cross-section. This section presents normal stress distribution formulas combining axial forces and bending moments. It introduces the concept of the core of a section—the central region where load application guarantees no tensile stresses develop in brittle materials.","\u003Cp>\u003Cstrong>Eccentric compression\u003C\u002Fstrong> occurs when the line of action of a compressive force does not pass through the centroid of the cross-section. The force can then be replaced by a centric axial force together with one or two bending moments.\u003C\u002Fp>\u003Ch2>Eccentricity and force reduction\u003C\u002Fh2>\u003Cp>Eccentricity is the distance between the force line of action and the corresponding centroidal axis. If the force is offset from the centroid in both principal directions, biaxial bending accompanies axial compression.\u003C\u002Fp>{{chunk:eccentric-axial-stress}}\u003Ch2>Stress distribution\u003C\u002Fh2>\u003Cp>The $N\u002FA$ component is uniform across the section, while the bending components vary linearly with coordinates. For a small eccentricity, the entire section may remain in compression. As eccentricity increases, stress at one edge decreases to zero and then becomes tensile.\u003C\u002Fp>\u003Ch2>Section kern\u003C\u002Fh2>{{chunk:section-kern-basic}}\u003Cp>The kern concept is particularly important for materials and structural contacts that have little tensile capacity or where separation is undesirable. If the compressive force acts on the kern boundary, normal stress is zero at one extreme point.\u003C\u002Fp>\u003Ch2>Uniaxial eccentricity\u003C\u002Fh2>\u003Cp>If force $N$ is applied with eccentricity $e$ in only one principal plane, the bending moment is $M=Ne$. The extreme normal stresses can then be written as:\u003C\u002Fp>\u003Cp>$$\\sigma=\\frac{N}{A}\\pm\\frac{M}{W},$$\u003C\u002Fp>\u003Cp>where the actual sign at each edge depends on the selected convention and on the direction of eccentricity.\u003C\u002Fp>\u003Ch2>Rectangular-section example\u003C\u002Fh2>\u003Cp>For a rectangular section $b\\times h$ with eccentricity along the height, the condition for no tensile stress is $|e|\\le h\u002F6$. At $|e|=h\u002F6$, stress at one edge is zero. For larger eccentricity, the full-section linear model predicts a tensile region.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Locate the centroid and principal centroidal axes.\u003C\u002Fli>\u003Cli>Determine the force eccentricities.\u003C\u002Fli>\u003Cli>Calculate $N$, $M_x$, and $M_y$.\u003C\u002Fli>\u003Cli>Evaluate normal stress at characteristic extreme points.\u003C\u002Fli>\u003Cli>If tensile stress must be avoided, check whether the force acts inside the section kern.\u003C\u002Fli>\u003Cli>Perform the required strength check.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Contact limitation\u003C\u002Fh2>\u003Cp>If the member or foundation contact cannot transmit tension, once the resultant moves outside the kern the actual contact region may become only partial. In that case, the simple linear stress distribution over the full area no longer describes the contact completely and another contact model is required.\u003C\u002Fp>",37,[],{"id":1385,"parent_id":1353,"code":1386,"slug":1387,"name":1388,"seo_title":1389,"seo_description":1390,"seo_text":1391,"content":1392,"locale":8,"uk_topic_id":1393,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1387,"children":1394},68,"23","combined-bending-and-torsion","Combined Bending and Torsion","Combined Bending & Torsion — Shaft Stress and von Mises\u002FTresca","Design shafts under simultaneous bending and torsion: combine normal and shear stress, find the critical section, and check strength with von Mises or Tresca.","Simultaneous bending and torsion represents the primary loading state for power transmission shafts and gearboxes. This section covers combined moment diagrams, critical section identification, and equivalent bending moment equations evaluated via the maximum shear stress (Tresca) and distortion energy (von Mises) failure criteria.","\u003Cp>\u003Cstrong>Combined bending and torsion\u003C\u002Fstrong> is typical of shafts that transmit torque while also carrying transverse forces from gears, pulleys, chain drives, or other machine elements. At a critical section, bending produces normal stress and torsion produces shear stress.\u003C\u002Fp>\u003Ch2>Resultant bending moment\u003C\u002Fh2>\u003Cp>If a shaft bends in two mutually perpendicular planes, determine the components $M_x$ and $M_y$ and construct the corresponding bending-moment diagrams.\u003C\u002Fp>{{chunk:resultant-bending-moment}}\u003Ch2>Bending and torsional stresses\u003C\u002Fh2>\u003Cp>For a circular shaft, maximum bending normal stress occurs at the outer surface. Torsional shear stress is also maximum at the outer surface, so outer points of a critical section are typical candidates for multiaxial strength assessment.\u003C\u002Fp>{{chunk:bending-torsion-equivalent-stress}}\u003Ch2>Failure criteria\u003C\u002Fh2>{{chunk:tresca-von-mises-criteria}}\u003Cp>For ductile isotropic materials, the Tresca or von Mises criterion is commonly used. The criterion and allowable value must be consistent with the material properties and adopted design method.\u003C\u002Fp>\u003Ch2>Solid circular shaft\u003C\u002Fh2>\u003Cp>For diameter $d$:\u003C\u002Fp>\u003Cp>$$W=\\frac{\\pi d^3}{32},\\qquad W_p=\\frac{\\pi d^3}{16}=2W.$$\u003C\u002Fp>\u003Cp>Therefore, for a solid circular shaft the von Mises criterion can also be written in terms of moments:\u003C\u002Fp>\u003Cp>$$\\sigma_{\\mathrm{eq,VM}}=\\frac{32}{\\pi d^3}\\sqrt{M_b^2+\\frac{3}{4}T^2}.$$\u003C\u002Fp>\u003Cp>The Tresca equivalent stress becomes:\u003C\u002Fp>\u003Cp>$$\\sigma_{\\mathrm{eq,T}}=\\frac{32}{\\pi d^3}\\sqrt{M_b^2+T^2}.$$\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Let $M_b=600\\ \\text{N·m}$, $T=400\\ \\text{N·m}$, and $d=40\\ \\text{mm}$ for a solid shaft. Converting moments to N·mm:\u003C\u002Fp>\u003Cp>$$\\sigma_b=\\frac{32\\cdot600000}{\\pi40^3}\\approx95.5\\ \\text{MPa},$$\u003C\u002Fp>\u003Cp>$$\\tau_t=\\frac{16\\cdot400000}{\\pi40^3}\\approx31.8\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>According to von Mises:\u003C\u002Fp>\u003Cp>$$\\sigma_{\\mathrm{eq,VM}}\\approx\\sqrt{95.5^2+3\\cdot31.8^2}\\approx110.2\\ \\text{MPa}.$$\u003C\u002Fp>\u003Ch2>Shaft calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine forces from transmissions and support reactions.\u003C\u002Fli>\u003Cli>Construct $M_x$, $M_y$, and $T$ diagrams.\u003C\u002Fli>\u003Cli>Find $M_b$ and identify critical sections.\u003C\u002Fli>\u003Cli>Calculate $W$ and $W_p$.\u003C\u002Fli>\u003Cli>Determine $\\sigma_b$ and $\\tau_t$.\u003C\u002Fli>\u003Cli>Calculate equivalent stress using the selected failure criterion.\u003C\u002Fli>\u003Cli>Check strength and, where required, stiffness and fatigue.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Important practical note\u003C\u002Fh2>\u003Cp>Keyways, fillets, shoulders, fits, and other stress concentrators can significantly increase local stresses. Under cyclic shaft loading, a static equivalent-stress check is not sufficient by itself; fatigue must also be assessed.\u003C\u002Fp>",23,[],{"id":1396,"parent_id":784,"code":1397,"slug":1398,"name":1399,"seo_title":1400,"seo_description":1401,"seo_text":1402,"content":1403,"locale":8,"uk_topic_id":1404,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":891,"url_slug":1398,"children":1405},69,"24","energetic-methods","Energy Methods","Energy Methods for Deflection — Castigliano & Mohr Integral","Calculate structural displacements with strain energy, Castigliano’s theorem, the Mohr unit-load integral, and reciprocity methods for beams and frames.","Energy methods rely on energy conservation principles to calculate displacements in elastic structural systems. This section explores external work, strain energy accumulation, Clapeyron's theorem, Betti's reciprocal work theorem, Maxwell's reciprocal deflection theorem, and Castigliano's theorem for linear and angular deflection analysis.","\u003Cp>\u003Cstrong>Energy methods\u003C\u002Fstrong> determine displacements and deformations of elastic systems through external work and stored strain energy. Their main advantage is that a required displacement can often be found without constructing the complete deflected shape.\u003C\u002Fp>\u003Ch2>External work and strain energy\u003C\u002Fh2>\u003Cp>During gradual static loading, external forces perform work that is stored as strain energy in an ideal elastic system.\u003C\u002Fp>{{chunk:clapeyron-work-theorem}}\u003Cp>Strain energy can be expressed through internal force resultants and member stiffnesses, providing a direct connection between force analysis and displacement calculations.\u003C\u002Fp>{{chunk:strain-energy-basic-loadings}}\u003Ch2>Reciprocity of work and displacement\u003C\u002Fh2>{{chunk:betti-maxwell-reciprocity}}\u003Cp>Reciprocity is an important property of linear elastic systems and underlies several unit-load and energy methods.\u003C\u002Fp>\u003Ch2>Castigliano's theorem\u003C\u002Fh2>{{chunk:castigliano-second-theorem}}\u003Cp>By differentiating strain energy $U$ with respect to a required generalized force $P_i$ or moment $M_i$, the corresponding displacement $\\delta_i$ or rotation $\\varphi_i$ can be obtained.\u003C\u002Fp>\u003Ch2>Mohr integral\u003C\u002Fh2>{{chunk:mohr-integral-unit-load}}\u003Cp>The unit-load method is especially convenient for beams, frames, and member systems when one specific displacement or rotation is required.\u003C\u002Fp>\u003Ch2>Choosing a method\u003C\u002Fh2>\u003Cp>Castigliano's theorem is convenient when strain energy can be written easily as a function of loads. The Mohr integral naturally uses internal-force diagrams from the real and unit-load states. For piecewise-linear bending-moment diagrams, integration can sometimes be simplified by graphical multiplication methods such as the Vereshchagin rule.\u003C\u002Fp>\u003Ch2>Limits of applicability\u003C\u002Fh2>\u003Cp>The classical relations presented here assume small deformation and linearly elastic behavior. Nonlinear materials, large displacements, or loads depending on the deformed configuration require appropriate generalizations.\u003C\u002Fp>",24,[1406,1416],{"id":1407,"parent_id":1396,"code":1408,"slug":1409,"name":1410,"seo_title":1411,"seo_description":1412,"seo_text":1413,"content":1414,"locale":8,"uk_topic_id":15,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":15,"url_slug":1409,"children":1415},70,"25","work-of-external-forces-and-strain-energy","Work of External Forces and Strain Energy","Strain Energy — Formulas for Axial Load, Torsion & Bending","Learn strain-energy formulas for axial loading, shear, torsion, and bending, including energy density and the link to Castigliano’s theorem.","During elastic deformation, external loads perform work stored inside the body as strain energy U. This page provides formulas for strain energy under axial loading, pure shear, torsion, and bending. It also details strain energy density divided into volumetric change and shape distortion components.","\u003Cp>\u003Cstrong>Strain energy $U$\u003C\u002Fstrong> is the energy stored in an elastic body as a result of deformation. In a linearly elastic system under gradual static loading, it equals the work performed by the external forces.\u003C\u002Fp>{{chunk:clapeyron-work-theorem}}\u003Ch2>Energy expressed through internal force resultants\u003C\u002Fh2>{{chunk:strain-energy-basic-loadings}}\u003Cp>For a member subjected simultaneously to several deformation modes, the total energy within the applicable linear model is obtained by adding contributions from axial loading, bending, torsion, and, where necessary, transverse shear.\u003C\u002Fp>\u003Ch2>Axial loading\u003C\u002Fh2>\u003Cp>For a prismatic bar with constant $N$, $A$, and $E$:\u003C\u002Fp>\u003Cp>$$U_N=\\frac{N^2L}{2EA}.$$\u003C\u002Fp>\u003Cp>Since $\\Delta L=NL\u002F(EA)$, the same result can be written as $U_N=N\\Delta L\u002F2$.\u003C\u002Fp>\u003Ch2>Torsion\u003C\u002Fh2>\u003Cp>For a circular prismatic shaft with constant $T$, $G$, and $J_p$:\u003C\u002Fp>\u003Cp>$$U_T=\\frac{T^2L}{2GJ_p}=\\frac{1}{2}T\\varphi,$$\u003C\u002Fp>\u003Cp>where $\\varphi$ is the total angle of twist.\u003C\u002Fp>\u003Ch2>Bending\u003C\u002Fh2>\u003Cp>For a beam, the principal strain-energy contribution in the classical model is often associated with the bending moment:\u003C\u002Fp>\u003Cp>$$U_M=\\int_0^L\\frac{M^2(x)}{2EI}\\,dx.$$\u003C\u002Fp>\u003Cp>If transverse-shear deformation is significant, the corresponding shear contribution is added.\u003C\u002Fp>\u003Ch2>Strain-energy density\u003C\u002Fh2>\u003Cp>For a uniaxial linearly elastic state:\u003C\u002Fp>\u003Cp>$$u=\\frac{U}{V}=\\frac{1}{2}\\sigma\\varepsilon=\\frac{\\sigma^2}{2E}.$$\u003C\u002Fp>\u003Cp>For pure shear:\u003C\u002Fp>\u003Cp>$$u=\\frac{1}{2}\\tau\\gamma=\\frac{\\tau^2}{2G}.$$\u003C\u002Fp>\u003Ch2>Volumetric and distortional energy\u003C\u002Fh2>\u003Cp>For an isotropic linearly elastic material, the total strain-energy density can be separated into volumetric and deviatoric parts. The distortional-energy component is related to differences between principal stresses and forms the basis of the von Mises yield criterion.\u003C\u002Fp>{{chunk:tresca-von-mises-criteria}}\u003Ch2>Practical significance\u003C\u002Fh2>\u003Cp>The energy formulation is particularly useful for determining displacements by Castigliano's theorem or the unit-load method and for analyzing systems subjected to several simultaneous internal force resultants.\u003C\u002Fp>",[],{"id":1417,"parent_id":1396,"code":1418,"slug":1419,"name":1420,"seo_title":1421,"seo_description":1422,"seo_text":1423,"content":1424,"locale":8,"uk_topic_id":1425,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1425,"url_slug":1419,"children":1426},71,"26","castiglianos-theorem-and-mohrs-integral","Castigliano's Theorem and Mohr's Integral","Castigliano’s Theorem & Mohr’s Integral — Deflection Methods","Calculate beam and frame deflections with Castigliano’s theorem, the unit-load Mohr integral, and Vereshchagin’s diagram multiplication rule.","Mohr's integral evaluates structural displacements from arbitrary load configurations by applying dummy unit forces or unit moments. This section covers unit load method integration and Vereshchagin's visual diagram multiplication method for beams and frames using centroidal ordinates.","\u003Cp>\u003Cstrong>Castigliano's theorem and the Mohr integral\u003C\u002Fstrong> make it possible to determine individual linear and angular displacements of elastic member systems using strain energy or products of internal force resultants from two loading states.\u003C\u002Fp>\u003Ch2>Castigliano's theorem\u003C\u002Fh2>{{chunk:castigliano-second-theorem}}\u003Cp>For a beam in which bending strain energy dominates, $U=\\int M^2\u002F(2EI)\\,dx$. If $M$ depends on force $P$, differentiation gives:\u003C\u002Fp>\u003Cp>$$\\delta_P=\\frac{\\partial U}{\\partial P}=\\int\\frac{M}{EI}\\frac{\\partial M}{\\partial P}\\,dx.$$\u003C\u002Fp>\u003Cp>For a linear system, $\\partial M\u002F\\partial P$ corresponds to the bending-moment diagram produced by a unit force in the direction of $P$, showing the connection between Castigliano's theorem and the unit-load method.\u003C\u002Fp>\u003Ch2>Auxiliary force or moment\u003C\u002Fh2>\u003Cp>If no real force acts at the point of the required displacement, introduce an auxiliary parameter $X$ in the required direction. Write the internal force resultants as functions of $X$, evaluate $\\partial U\u002F\\partial X$, and set $X=0$ after differentiation.\u003C\u002Fp>\u003Ch2>Mohr integral\u003C\u002Fh2>{{chunk:mohr-integral-unit-load}}\u003Cp>For a linear displacement, apply a unit force in the direction of the required displacement. For a rotation, apply a unit moment. The sign of the result indicates whether the actual displacement agrees with the direction of the unit load.\u003C\u002Fp>\u003Ch2>Unit-load procedure\u003C\u002Fh2>\u003Col>\u003Cli>Analyze the system under the real loading and construct the required internal-force diagrams.\u003C\u002Fli>\u003Cli>Remove the real loading and apply a unit force or moment at the point and in the direction of the required displacement.\u003C\u002Fli>\u003Cli>Construct the unit-load diagrams.\u003C\u002Fli>\u003Cli>Write the Mohr integral over all segments and members.\u003C\u002Fli>\u003Cli>Use the actual $EA$, $EI$, $GJ_p$ and, where necessary, shear stiffness.\u003C\u002Fli>\u003Cli>Evaluate the integrals and sum all contributions.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Vereshchagin rule\u003C\u002Fh2>{{chunk:vereshchagin-diagram-multiplication}}\u003Cp>The Vereshchagin rule is especially efficient for beams and frames with constant-$EI$ segments and simple diagrams. It replaces part of the analytical integration with operations involving diagram areas and ordinates.\u003C\u002Fp>\u003Ch2>Reciprocity\u003C\u002Fh2>{{chunk:betti-maxwell-reciprocity}}\u003Cp>Reciprocity theorems provide a useful check on unit-load calculations and can sometimes suggest a simpler reciprocal loading state.\u003C\u002Fp>\u003Ch2>Common errors\u003C\u002Fh2>\u003Cp>Typical mistakes include choosing the wrong direction for the unit load, omitting members or segments, mixing diagram signs, using one $EI$ where stiffness changes, and mechanically applying the Vereshchagin rule to two nonlinear diagrams.\u003C\u002Fp>",26,[],{"id":1428,"parent_id":784,"code":1429,"slug":1430,"name":1431,"seo_title":1432,"seo_description":1433,"seo_text":1434,"content":1435,"locale":8,"uk_topic_id":1436,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":903,"url_slug":1430,"children":1437},72,"27","stability-of-compressed-members","Stability of Compressed Members","Column Buckling — Critical Load, Effective Length & Slenderness","Learn column buckling and stability: Euler critical load, effective length and end conditions, slenderness ratio, weak-axis buckling, and applicability limits.","Buckling is a sudden side-way deflection mode occurring in slender compression members when axial loads reach a critical threshold Pcr. This section covers stable, unstable, and neutral equilibrium states, factors influencing critical load capacity, and column stability verification principles.","\u003Cp>\u003Cstrong>Stability of compressed members\u003C\u002Fstrong> is their ability to preserve the initial equilibrium configuration under compression. A slender column may buckle laterally before its average compressive stress reaches the material strength limit.\u003C\u002Fp>\n\u003Ch2>Section roadmap\u003C\u002Fh2>\n\u003Cp>The section begins with the mechanism and modes of buckling. It then introduces radius of gyration, slenderness ratio and the weak buckling axis, followed by Euler's formula for long elastic columns.\u003C\u002Fp>\n\u003Cp>Separate topics cover critical-stress curves and inelastic buckling, column end conditions and effective length, practical design using a buckling reduction factor, and special cases such as imperfections, local buckling and non-prismatic columns.\u003C\u002Fp>\n\u003Ch2>Core equations\u003C\u002Fh2>\n\u003Cp>For an ideal elastic column,\u003C\u002Fp>\n\u003Cp>\\[P_{cr}=\\frac{\\pi^2EI_{min}}{(KL)^2}.\\]\u003C\u002Fp>\n\u003Cp>The slenderness ratio and Euler critical stress are\u003C\u002Fp>\n\u003Cp>\\[\\lambda=\\frac{KL}{r_{min}},\\qquad r_{min}=\\sqrt{\\frac{I_{min}}{A}},\\qquad \\sigma_{cr,E}=\\frac{\\pi^2E}{\\lambda^2}.\\]\u003C\u002Fp>\n\u003Cp>Real columns contain initial imperfections, eccentricities, residual stresses and non-ideal restraints. Euler's formula is therefore a fundamental theoretical model, while design resistance must be determined using the applicable structural design standard.\u003C\u002Fp>",27,[1438,1448,1458,1468,1479,1490,1500,1510,1520],{"id":1439,"parent_id":1428,"code":70,"slug":1440,"name":1441,"seo_title":1442,"seo_description":1443,"seo_text":1444,"content":1445,"locale":8,"uk_topic_id":1446,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1440,"children":1447},423,"mechanism-of-column-buckling","Mechanism of Column Buckling","Mechanism of Column Buckling and Loss of Stability","How column buckling develops: straight equilibrium, critical load, lateral deflection, second-order moment and loss of stability.","Column buckling is a change from straight equilibrium to a laterally deflected configuration under compression. The mechanism is governed by flexural stiffness and second-order effects.","\u003Cp>\u003Cstrong>Column buckling\u003C\u002Fstrong> is the transition of a compressed member from its initially straight equilibrium configuration to a laterally deflected configuration.\u003C\u002Fp>\n{{chunk:column-buckling-instability-transition}}\n\u003Ch2>How instability develops\u003C\u002Fh2>\n\u003Cp>At a small centric compressive load, the straight column is stable: after a small lateral disturbance it tends to return to its original position. As the load increases, flexural stiffness becomes less able to restore the straight configuration.\u003C\u002Fp>\n\u003Cp>At the critical load, an adjacent buckled equilibrium configuration becomes possible. A further load increase may produce a rapid growth of lateral displacement and bending moment.\u003C\u002Fp>\n\u003Ch2>Geometric mechanism\u003C\u002Fh2>\n\u003Cp>Once a lateral deflection \\(y\\) exists, the axial force \\(N\\) produces an additional moment approximately equal to\u003C\u002Fp>\u003Cp>\\[M=Ny.\\]\u003C\u002Fp>\n\u003Cp>The moment increases curvature, while the increased curvature produces a larger eccentricity. This feedback is the essential geometric mechanism of instability.\u003C\u002Fp>\n\u003Ch2>Critical load\u003C\u002Fh2>\n\u003Cp>The critical load is not simply a material failure load. It marks a stability limit of the structural model and depends on member length, flexural rigidity \\(EI\\), cross-sectional geometry, and end restraints.\u003C\u002Fp>\n\u003Ch2>Ideal and real columns\u003C\u002Fh2>\n\u003Cp>An ideal column remains straight up to its theoretical critical load. Real columns have initial crookedness, load eccentricity, residual stress and imperfect restraints, so lateral deflection generally develops progressively before the ideal bifurcation load is reached.\u003C\u002Fp>",417,[],{"id":1449,"parent_id":1428,"code":70,"slug":1450,"name":1451,"seo_title":1452,"seo_description":1453,"seo_text":1454,"content":1455,"locale":8,"uk_topic_id":1456,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1450,"children":1457},422,"column-buckling-modes","Column Buckling Modes","Column Buckling Modes: Flexural, Torsional and Higher Modes","First and higher Euler modes, flexural buckling, torsional buckling and flexural-torsional buckling of compression members.","Buckling modes describe the deformation pattern of a compression member. The governing mode is the one with the lowest critical load.","\u003Cp>A \u003Cstrong>buckling mode\u003C\u002Fstrong> describes the deformation pattern that becomes possible at a critical load. Several mathematical modes may exist, but the mode associated with the lowest critical load normally governs.\u003C\u002Fp>\n\u003Ch2>First Euler mode\u003C\u002Fh2>\n\u003Cp>For a pinned-pinned column, the first mode contains one half-wave:\u003C\u002Fp>\n\u003Cp>\\[y_1(x)=C\\sin\\frac{\\pi x}{L}.\\]\u003C\u002Fp>\n\u003Cp>It has no intermediate nodes and corresponds to \\(P_{cr,1}=\\pi^2EI\u002FL^2\\).\u003C\u002Fp>\n\u003Ch2>Higher modes\u003C\u002Fh2>\n\u003Cp>The higher ideal modes are\u003C\u002Fp>\n\u003Cp>\\[y_n(x)=C\\sin\\frac{n\\pi x}{L},\\qquad P_{cr,n}=n^2P_{cr,1}.\\]\u003C\u002Fp>\n\u003Cp>They contain two or more half-waves and require higher loads. They are important in eigenvalue analysis even though they usually do not govern a simple uniform column.\u003C\u002Fp>\n\u003Ch2>Flexural buckling\u003C\u002Fh2>\n\u003Cp>The member axis bends in a principal plane. With identical end conditions in both planes, buckling about the axis with the smaller flexural rigidity \\(EI\\) is critical.\u003C\u002Fp>\n\u003Ch2>Torsional and flexural-torsional buckling\u003C\u002Fh2>\n\u003Cp>Thin-walled open sections may buckle by twisting about the longitudinal axis. When twisting is coupled with lateral bending, the mode is called flexural-torsional buckling. It cannot be assessed by the elementary Euler formula using only one second moment of area.\u003C\u002Fp>",418,[],{"id":1459,"parent_id":1428,"code":70,"slug":1460,"name":1461,"seo_title":1462,"seo_description":1463,"seo_text":1464,"content":1465,"locale":8,"uk_topic_id":1466,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1460,"children":1467},424,"column-slenderness-weak-axis","Column Slenderness and Weak Buckling Axis","Column Slenderness Ratio and Weak Buckling Axis","Calculate radius of gyration, effective length and column slenderness ratio, and identify the weak axis that governs buckling.","Column slenderness is the ratio of effective length to radius of gyration. Both principal axes must be checked because end restraints may differ by plane.","\u003Cp>The \u003Cstrong>slenderness ratio\u003C\u002Fstrong> measures a compression member's tendency to buckle by combining its effective length with the geometric distribution of its cross-sectional area.\u003C\u002Fp>\n{{chunk:column-radius-slenderness}}\n\u003Ch2>Radius of gyration\u003C\u002Fh2>\n\u003Cp>For a selected centroidal axis,\u003C\u002Fp>\u003Cp>\\[r=\\sqrt{\\frac{I}{A}},\\]\u003C\u002Fp>\n\u003Cp>where \\(I\\) is the second moment of area and \\(A\\) is the cross-sectional area.\u003C\u002Fp>\n\u003Ch2>Slenderness in both planes\u003C\u002Fh2>\n\u003Cp>Calculate\u003C\u002Fp>\n\u003Cp>\\[\\lambda_x=\\frac{L_{eff,x}}{r_x},\\qquad \\lambda_y=\\frac{L_{eff,y}}{r_y}.\\]\u003C\u002Fp>\n\u003Cp>If the end restraints are identical, the smaller radius of gyration usually defines the weak buckling axis.\u003C\u002Fp>\n\u003Ch2>Governing axis\u003C\u002Fh2>\n\u003Cp>The governing direction is the one with the lower critical load. It is not always enough to select the smaller \\(I\\): different effective lengths or restraints in the two planes may change the result. Compare the complete slenderness ratios or critical loads.\u003C\u002Fp>\n\u003Ch2>Calculation sequence\u003C\u002Fh2>\n\u003Col>\u003Cli>Determine \\(A\\), \\(I_x\\), and \\(I_y\\).\u003C\u002Fli>\u003Cli>Calculate \\(r_x\\) and \\(r_y\\).\u003C\u002Fli>\u003Cli>Determine effective lengths in both planes.\u003C\u002Fli>\u003Cli>Calculate \\(\\lambda_x\\) and \\(\\lambda_y\\).\u003C\u002Fli>\u003Cli>Use the governing case in the stability check.\u003C\u002Fli>\u003C\u002Fol>",419,[],{"id":1469,"parent_id":1428,"code":1470,"slug":1471,"name":1472,"seo_title":1473,"seo_description":1474,"seo_text":1475,"content":1476,"locale":8,"uk_topic_id":1477,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1471,"children":1478},73,"28","euler-buckling-formula-critical-load","Euler Buckling Formula and Critical Buckling Load","Euler Buckling Formula & Critical Buckling Load | Pcr = π²EI\u002F(KL)²","Euler buckling formula for the critical buckling load of a slender column: Pcr = π²EI\u002F(KL)². Learn effective length, end conditions, assumptions and see a worked example.","Use the Euler buckling formula to calculate the critical buckling load of a slender column. This guide explains Pcr = π²EI\u002F(KL)², Euler critical load, effective length and end conditions, the weakest buckling axis, slenderness ratio, assumptions, units, and a worked numerical example.","\u003Cp>\u003Cstrong>The Euler buckling formula\u003C\u002Fstrong> gives the critical load of an ideal slender column that buckles while the material remains linearly elastic:\u003C\u002Fp>\n\u003Cp>\\[P_{cr}=\\frac{\\pi^2EI}{(KL)^2}.\\]\u003C\u002Fp>\n{{chunk:euler-column-critical-load}}\n\u003Ch2>Derivation for a pinned column\u003C\u002Fh2>\n\u003Cp>For a pinned-pinned column of length \\(L\\) under centric compression \\(P\\), a small lateral deflection \\(y(x)\\) produces the bending moment \\(M(x)=-Py(x)\\). The elastic-curve equation becomes\u003C\u002Fp>\n\u003Cp>\\[EI\\frac{d^2y}{dx^2}+Py=0.\\]\u003C\u002Fp>\n\u003Cp>With \\(k^2=P\u002F(EI)\\), the solution is\u003C\u002Fp>\n\u003Cp>\\[y(x)=C_1\\sin(kx)+C_2\\cos(kx).\\]\u003C\u002Fp>\n\u003Cp>The boundary conditions \\(y(0)=0\\) and \\(y(L)=0\\) require \\(C_2=0\\) and \\(\\sin(kL)=0\\). The first mode has \\(kL=\\pi\\), giving\u003C\u002Fp>\n\u003Cp>\\[P_{cr}=\\frac{\\pi^2EI}{L^2}.\\]\u003C\u002Fp>\n\u003Ch2>General form\u003C\u002Fh2>\n\u003Cp>Other end conditions are represented by the effective length \\(L_{eff}=KL\\):\u003C\u002Fp>\n\u003Cp>\\[P_{cr}=\\frac{\\pi^2EI_{min}}{L_{eff}^2}=\\frac{\\pi^2EI_{min}}{(KL)^2}.\\]\u003C\u002Fp>\n\u003Ch2>Physical meaning\u003C\u002Fh2>\n\u003Cp>The Euler load is proportional to Young's modulus and the relevant second moment of area, and inversely proportional to the square of effective length. It is a stability limit of an ideal model, not necessarily the failure load of a real column.\u003C\u002Fp>\n\u003Ch2>Worked example\u003C\u002Fh2>\n\u003Cp>For a pinned steel column with \\(E=200\\) GPa, \\(L=2\\) m and \\(I_{min}=2\\cdot10^{-6}\\) m⁴:\u003C\u002Fp>\n\u003Cp>\\[P_{cr}=\\frac{\\pi^2(200\\cdot10^9)(2\\cdot10^{-6})}{2^2}\\approx987\\ \\text{kN}.\\]\u003C\u002Fp>\n\u003Ch2>Assumptions\u003C\u002Fh2>\n\u003Cp>The classical model assumes elastic material, centric compression, an initially straight prismatic member, ideal end restraints and small deflections up to the critical state. Its applicability must be checked using slenderness and critical stress.\u003C\u002Fp>\n\u003Ch2>Euler buckling formula FAQ\u003C\u002Fh2>\n\u003Ch3>What is the Euler buckling formula?\u003C\u002Fh3>\u003Cp>For an ideal slender elastic column, \\(P_{cr}=\\pi^2EI\u002F(KL)^2\\).\u003C\u002Fp>\n\u003Ch3>Is the Euler load the actual failure load?\u003C\u002Fh3>\u003Cp>Not necessarily. Imperfections, residual stresses, local buckling and non-ideal restraints reduce the resistance of real columns.\u003C\u002Fp>",28,[],{"id":1480,"parent_id":1428,"code":1481,"slug":1482,"name":1483,"seo_title":1484,"seo_description":1485,"seo_text":1486,"content":1487,"locale":8,"uk_topic_id":1488,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1482,"children":1489},74,"29","limits-of-applicability-eulers-formula","Critical Buckling Stress, Euler Limits and Inelastic Column Formulas","Critical Buckling Stress, Euler Limits & Inelastic Buckling","Column buckling curves, Euler limiting slenderness, inelastic buckling, Johnson, Tetmajer, Rankine–Gordon and design-code approaches.","Euler's formula is valid only when critical stresses remain within a material's proportional limit. For intermediate and short columns buckling in the inelastic domain, empirical equations like the Yasinsky formula apply. This section details applicability boundaries and stress reduction factors.","\u003Cp>\u003Cstrong>Critical buckling stress\u003C\u002Fstrong> is the average compressive stress at which a column becomes unstable. Plotting it against slenderness produces a column-strength or buckling curve that distinguishes stocky, intermediate and slender members.\u003C\u002Fp>\n{{chunk:euler-applicability-limit}}\n\u003Ch2>Elastic Euler range\u003C\u002Fh2>\n\u003Cp>For a slender elastic column,\u003C\u002Fp>\n\u003Cp>\\[\\sigma_{cr,E}=\\frac{P_{cr}}{A}=\\frac{\\pi^2E}{\\lambda^2}.\\]\u003C\u002Fp>\n\u003Cp>Equating the Euler stress to the proportional limit \\(\\sigma_p\\) gives the limiting slenderness\u003C\u002Fp>\n\u003Cp>\\[\\lambda_{lim}=\\pi\\sqrt{\\frac{E}{\\sigma_p}}.\\]\u003C\u002Fp>\n\u003Cp>Euler's formula is applicable when the critical state remains within the elastic range.\u003C\u002Fp>\n\u003Ch2>Intermediate columns and inelastic buckling\u003C\u002Fh2>\n\u003Cp>Intermediate columns require empirical, semi-empirical, or code-based relations. The linear relation traditionally called the \u003Cstrong>Yasinsky formula\u003C\u002Fstrong> in Ukrainian and Eastern European mechanics courses is not commonly identified by that name in English-language engineering literature.\u003C\u002Fp>\n{{chunk:yasinsky-column-formula}}\n\u003Ch2>English terminology and alternative formulas\u003C\u002Fh2>\n\u003Cul>\n\u003Cli>\u003Cstrong>inelastic column buckling\u003C\u002Fstrong> is the usual general term for this range;\u003C\u002Fli>\n\u003Cli>\u003Cstrong>Tetmajer\u003C\u002Fstrong> or \u003Cstrong>Euler–Tetmajer formula\u003C\u002Fstrong> may denote historical empirical column curves used in Central European literature;\u003C\u002Fli>\n\u003Cli>\u003Cstrong>Johnson parabolic formula\u003C\u002Fstrong> is a common parabolic approximation for intermediate columns;\u003C\u002Fli>\n\u003Cli>\u003Cstrong>Rankine–Gordon formula\u003C\u002Fstrong> interpolates between crushing resistance and Euler buckling;\u003C\u002Fli>\n\u003Cli>\u003Cstrong>tangent-modulus theory\u003C\u002Fstrong> represents reduced material stiffness in the inelastic range;\u003C\u002Fli>\n\u003Cli>modern design standards use calibrated \u003Cstrong>column buckling curves\u003C\u002Fstrong> and reduction factors.\u003C\u002Fli>\n\u003C\u002Ful>\n\u003Ch2>Short columns\u003C\u002Fh2>\n\u003Cp>For low slenderness, overall buckling may no longer govern. Resistance can instead be controlled by yielding, crushing, local buckling, or another material or sectional limit state.\u003C\u002Fp>\n\u003Ch2>Model-selection procedure\u003C\u002Fh2>\n\u003Col>\u003Cli>Calculate slenderness and the Euler critical stress.\u003C\u002Fli>\u003Cli>Compare the result with the material proportional limit.\u003C\u002Fli>\u003Cli>Use Euler's formula for sufficiently slender elastic columns.\u003C\u002Fli>\u003Cli>Use the permitted inelastic or code-based relation for intermediate columns.\u003C\u002Fli>\u003Cli>Check strength and local stability for stocky members.\u003C\u002Fli>\u003C\u002Fol>",29,[],{"id":1491,"parent_id":1428,"code":70,"slug":1492,"name":1493,"seo_title":1494,"seo_description":1495,"seo_text":1496,"content":1497,"locale":8,"uk_topic_id":1498,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1492,"children":1499},425,"column-end-conditions-effective-length","Column End Conditions and Effective Length","Column End Conditions and Effective Length Factor K","How pinned, fixed and free end conditions affect column buckling, effective length KL and Euler critical load.","Column end restraints determine the buckled shape and effective-length factor K. Learn the standard K values and their influence on Euler critical load.","\u003Cp>Column end conditions strongly influence buckling resistance because they determine the buckled shape and the distance between inflection points. Their effect is represented by the \u003Cstrong>effective-length factor \\(K\\)\u003C\u002Fstrong>.\u003C\u002Fp>\n\u003Ch2>Effective length\u003C\u002Fh2>\n\u003Cp>\\[L_{eff}=KL,\\qquad P_{cr}=\\frac{\\pi^2EI_{min}}{(KL)^2}.\\]\u003C\u002Fp>\n\u003Cp>Because the Euler load varies with \\(1\u002FK^2\\), a change in end restraint can produce a large change in theoretical buckling resistance.\u003C\u002Fp>\n{{chunk:column-effective-length-factors}}\n\u003Ch2>Ideal end-condition values\u003C\u002Fh2>\n\u003Ctable>\u003Cthead>\u003Ctr>\u003Cth>End conditions\u003C\u002Fth>\u003Cth>\\(K\\)\u003C\u002Fth>\u003Cth>\\(L_{eff}\\)\u003C\u002Fth>\u003C\u002Ftr>\u003C\u002Fthead>\u003Ctbody>\n\u003Ctr>\u003Ctd>Pinned–pinned\u003C\u002Ftd>\u003Ctd>1.0\u003C\u002Ftd>\u003Ctd>\\(L\\)\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>Fixed–free\u003C\u002Ftd>\u003Ctd>2.0\u003C\u002Ftd>\u003Ctd>\\(2L\\)\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>Fixed–pinned\u003C\u002Ftd>\u003Ctd>approximately 0.7\u003C\u002Ftd>\u003Ctd>\\(0.7L\\)\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>Fixed–fixed\u003C\u002Ftd>\u003Ctd>0.5\u003C\u002Ftd>\u003Ctd>\\(0.5L\\)\u003C\u002Ftd>\u003C\u002Ftr>\n\u003C\u002Ftbody>\u003C\u002Ftable>\n\u003Ch2>Physical interpretation\u003C\u002Fh2>\n\u003Cp>A pin prevents lateral translation but permits rotation. A fixed end restrains both translation and rotation. A free end provides neither restraint. The more strongly rotations and translations are restrained, the shorter the effective buckling length.\u003C\u002Fp>\n\u003Ch2>Real structures\u003C\u002Fh2>\n\u003Cp>Actual joints have finite stiffness, and frames may sway laterally. The ideal tabulated factors should be used only when the assumed restraint model represents the structure. Frame effective length or global stability must be determined using the applicable design method.\u003C\u002Fp>",416,[],{"id":1501,"parent_id":1428,"code":70,"slug":1502,"name":1503,"seo_title":1504,"seo_description":1505,"seo_text":1506,"content":1507,"locale":8,"uk_topic_id":1508,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":1502,"children":1509},426,"column-buckling-reduction-factor","Column Stability Design Using Reduction Factor χ","Column Buckling Design Using Reduction Factor χ","Practical column stability verification using the buckling reduction factor χ, slenderness, design resistance and iterative section selection.","English-language structural design commonly denotes the buckling reduction factor by χ. It reduces the cross-sectional resistance to account for column instability and imperfections.","\u003Cp>Practical structural design commonly represents column instability by a \u003Cstrong>buckling reduction factor\u003C\u002Fstrong>. In English-language and Eurocode notation it is usually written as \\(\\chi\\), while some teaching and regional methods use \\(\\varphi\\).\u003C\u002Fp>\n\u003Ch2>General design form\u003C\u002Fh2>\n\u003Cp>A generic resistance check can be written as\u003C\u002Fp>\n\u003Cp>\\[N_{Ed}\\le \\chi A f_d,\\]\u003C\u002Fp>\n\u003Cp>where \\(N_{Ed}\\) is the design compression force, \\(A\\) is the relevant area, \\(f_d\\) is the design material strength, and \\(\\chi\\le1\\) is the buckling reduction factor.\u003C\u002Fp>\n\u003Ch2>What determines χ\u003C\u002Fh2>\n\u003Cp>The reduction factor decreases as nondimensional slenderness increases. The applicable relation may also depend on cross-section type, buckling axis, material, fabrication route, residual stresses, and the selected code buckling curve.\u003C\u002Fp>\n\u003Ch2>Calculation sequence\u003C\u002Fh2>\n\u003Col>\u003Cli>Determine the design axial force.\u003C\u002Fli>\u003Cli>Calculate area, second moments of area, and radii of gyration.\u003C\u002Fli>\u003Cli>Determine effective lengths and slenderness in both planes.\u003C\u002Fli>\u003Cli>Select the applicable buckling curve or reduction-factor relation.\u003C\u002Fli>\u003Cli>Calculate \\(\\chi\\) and verify the design resistance.\u003C\u002Fli>\u003Cli>Revise the section or restraint system if the check fails.\u003C\u002Fli>\u003C\u002Fol>\n\u003Ch2>Iterative section selection\u003C\u002Fh2>\n\u003Cp>Because the reduction factor depends on slenderness and slenderness depends on section size, column selection is usually iterative.\u003C\u002Fp>\n\u003Cp>\u003Cstrong>Important:\u003C\u002Fstrong> symbols, curves, safety factors, and resistance definitions must be taken from one consistent design standard. Values from different code systems are not interchangeable.\u003C\u002Fp>",420,[],{"id":1511,"parent_id":1428,"code":70,"slug":1512,"name":1513,"seo_title":1514,"seo_description":1515,"seo_text":1516,"content":1517,"locale":8,"uk_topic_id":1518,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":891,"url_slug":1512,"children":1519},427,"special-cases-column-stability","Special Cases in Column Stability","Special Cases in Column Stability and Buckling","Global and local buckling, imperfections, eccentric loading, stepped columns, variable stiffness, elastic restraints and flexural-torsional modes.","Real compression members require checks beyond ideal Euler buckling, including local buckling, imperfections, eccentricity, variable stiffness and flexible restraints.","\u003Cp>The elementary Euler model assumes an ideal straight prismatic column under centric compression. Real members may require additional stability considerations that reduce resistance or change the governing buckling mode.\u003C\u002Fp>\n\u003Ch2>Global and local buckling\u003C\u002Fh2>\n\u003Cp>Global buckling deflects the axis of the entire member. Local buckling deforms an individual plate element, such as a flange or web. Local buckling may occur first and reduce the effective cross-section.\u003C\u002Fp>\n\u003Ch2>Initial imperfections\u003C\u002Fh2>\n\u003Cp>Initial crookedness, dimensional tolerances, residual stresses, and joint imperfections produce lateral deflection before the Euler load is reached. This is one reason why design resistance is lower than the ideal elastic value.\u003C\u002Fp>\n\u003Ch2>Load eccentricity\u003C\u002Fh2>\n\u003Cp>An eccentric compression force produces bending from the beginning. Including lateral deflection gives a second-order moment\u003C\u002Fp>\u003Cp>\\[M=N(e+y).\\]\u003C\u002Fp>\n\u003Cp>The deformation increases the moment, and the increased moment further increases the deformation.\u003C\u002Fp>\n\u003Ch2>Stepped and non-prismatic columns\u003C\u002Fh2>\n\u003Cp>When \\(EI(x)\\) varies along the length, inserting a single second moment of area into the elementary Euler formula is generally inadequate. Critical loads may be obtained by a dedicated analytical, energy, or numerical model.\u003C\u002Fp>\n\u003Ch2>Intermediate and elastic restraints\u003C\u002Fh2>\n\u003Cp>Bracing can reduce the unrestrained buckling length, but its effectiveness depends on stiffness and strength. An elastic restraint is not automatically equivalent to an ideal fixed support.\u003C\u002Fp>\n\u003Ch2>Coupled modes and frames\u003C\u002Fh2>\n\u003Cp>Open thin-walled sections may buckle in torsional or flexural-torsional modes. In framed structures, joint translation and interaction between columns and the complete frame must also be considered.\u003C\u002Fp>",421,[],{"id":1521,"parent_id":1428,"code":70,"slug":1522,"name":1523,"seo_title":1524,"seo_description":1525,"seo_text":1526,"content":1527,"locale":8,"uk_topic_id":1528,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":903,"url_slug":1522,"children":1529},429,"code-based-design-steel-compression-members","Code-Based Design of Steel Compression Members","Steel Column Buckling Design to Eurocode 3: Reduction Factor χ","Eurocode 3 stability design of steel compression members: nondimensional slenderness, buckling curves, reduction factor χ and design buckling resistance.","Code-based column buckling design uses the Eurocode 3 reduction factor χ, nondimensional slenderness and buckling curves to determine the design resistance of steel compression members.","\u003Cp>Euler's formula explains ideal elastic buckling, but the design resistance of a real steel compression member is determined using code buckling curves. These curves account implicitly for initial crookedness, residual stresses, cross-section geometry, fabrication method and buckling axis.\u003C\u002Fp>\n\u003Ch2>Eurocode 3 design check\u003C\u002Fh2>\n\u003Cp>For a uniform member in compression, EN 1993-1-1 expresses the buckling resistance as\u003C\u002Fp>\n\u003Cp>\\[N_{b,Rd}=\\frac{\\chi A f_y}{\\gamma_{M1}},\\]\u003C\u002Fp>\n\u003Cp>for Class 1, 2 or 3 cross-sections. For a Class 4 cross-section, the effective area \\(A_{eff}\\) is used as required by the standard. The design condition is\u003C\u002Fp>\n\u003Cp>\\[N_{Ed}\\le N_{b,Rd}.\\]\u003C\u002Fp>\n\u003Ch2>Elastic critical load and nondimensional slenderness\u003C\u002Fh2>\n\u003Cp>For each possible buckling plane, determine the elastic critical load\u003C\u002Fp>\n\u003Cp>\\[N_{cr}=\\frac{\\pi^2EI}{L_{cr}^2}.\\]\u003C\u002Fp>\n\u003Cp>The nondimensional slenderness is then\u003C\u002Fp>\n\u003Cp>\\[\\bar\\lambda=\\sqrt{\\frac{A f_y}{N_{cr}}}\\]\u003C\u002Fp>\n\u003Cp>for Class 1–3 sections. Both principal axes and any relevant torsional or flexural-torsional mode must be considered.\u003C\u002Fp>\n\u003Ch2>Buckling reduction factor χ\u003C\u002Fh2>\n\u003Cp>Unlike the Ukrainian DBN notation \\(\\varphi\\), Eurocode 3 uses the symbol \\(\\chi\\). It is calculated from\u003C\u002Fp>\n\u003Cp>\\[\\chi=\\frac{1}{\\Phi+\\sqrt{\\Phi^2-\\bar\\lambda^2}}\\le1,\\]\u003C\u002Fp>\n\u003Cp>\\[\\Phi=\\frac12\\left[1+\\alpha(\\bar\\lambda-0.2)+\\bar\\lambda^2\\right],\\]\u003C\u002Fp>\n\u003Cp>where \\(\\alpha\\) is the imperfection factor associated with the selected buckling curve. The curve is chosen from the code tables according to section type, buckling axis, steel grade and fabrication route.\u003C\u002Fp>\n\u003Ch2>Eurocode calculation sequence\u003C\u002Fh2>\n\u003Col>\n\u003Cli>Determine \\(N_{Ed}\\), material strength and cross-section class.\u003C\u002Fli>\n\u003Cli>Calculate \\(A\\), \\(I_y\\), \\(I_z\\) and the relevant radii of gyration.\u003C\u002Fli>\n\u003Cli>Determine the buckling lengths or elastic critical loads for all relevant modes.\u003C\u002Fli>\n\u003Cli>Calculate \\(\\bar\\lambda\\) for each mode.\u003C\u002Fli>\n\u003Cli>Select the appropriate Eurocode buckling curve and imperfection factor \\(\\alpha\\).\u003C\u002Fli>\n\u003Cli>Calculate \\(\\chi\\) and \\(N_{b,Rd}\\).\u003C\u002Fli>\n\u003Cli>Verify \\(N_{Ed}\u002FN_{b,Rd}\\le1\\) for the governing mode.\u003C\u002Fli>\n\u003Cli>Check cross-section resistance, local buckling, member slenderness and any interaction with bending.\u003C\u002Fli>\n\u003C\u002Fol>\n\u003Ch2>Relation to the φ method\u003C\u002Fh2>\n\u003Cp>The DBN coefficient \\(\\varphi\\) and the Eurocode coefficient \\(\\chi\\) serve a similar purpose: both reduce the ideal cross-sectional compression resistance to account for member instability. Their numerical values must not be interchanged because the definitions of slenderness, resistance and safety factors belong to different code systems.\u003C\u002Fp>\n\u003Ch2>Scope and limitations\u003C\u002Fh2>\n\u003Cp>This procedure applies to the basic member-buckling check. Beam-columns, built-up members, frames with sway, non-uniform members and torsional or flexural-torsional buckling require the additional clauses of the applicable standard and its National Annex.\u003C\u002Fp>\n\u003Cp>\u003Ca href=\"https:\u002F\u002Feurocodes.jrc.ec.europa.eu\u002F\" target=\"_blank\" rel=\"noopener\">European Commission Joint Research Centre — Eurocodes\u003C\u002Fa>. See EN 1993-1-1, member buckling provisions and the applicable National Annex.\u003C\u002Fp>",428,[],{"id":1531,"parent_id":784,"code":70,"slug":1532,"name":1533,"seo_title":1534,"seo_description":1535,"seo_text":1536,"content":1537,"locale":8,"uk_topic_id":1538,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":915,"url_slug":1532,"children":1539},124,"shell-analysis","Shell Analysis","Shell Analysis — Membrane Theory and Thin-Walled Shells","Learn the fundamentals of thin-walled shell analysis and membrane theory: shell geometry, membrane forces, equilibrium, assumptions, and engineering applications.","This section introduces the strength analysis of thin-walled shells. It covers midsurface geometry, principal radii of curvature, membrane force resultants and stresses, Laplace equilibrium, cylindrical and spherical pressure vessels, strength assessment, and the limitations of membrane theory near edges and local disturbances.","\u003Cp>A \u003Cstrong>shell\u003C\u002Fstrong> is a thin-walled spatial structural element whose thickness $t$ is much smaller than the characteristic dimensions of its curved midsurface. Examples include tanks, large-diameter pipes, domes, pressure-vessel walls, and other curved thin structures.\u003C\u002Fp>\u003Ch2>Midsurface\u003C\u002Fh2>\u003Cp>The geometry of a thin shell is conveniently described by its midsurface, located approximately midway through the thickness. At each point, two principal directions of curvature can be identified with radii $R_1$ and $R_2$. For a shell of revolution, these commonly correspond to meridional and circumferential directions.\u003C\u002Fp>\u003Ch2>Internal force resultants\u003C\u002Fh2>\u003Cp>General shell theory may include membrane forces, transverse shear forces, bending moments, and twisting moments per unit length. In membrane theory, the dominant resultants lie in the tangent plane of the midsurface.\u003C\u002Fp>\u003Cp>For normal membrane resultants $N_1$ and $N_2$ and shell thickness $t$, the corresponding average stresses are:\u003C\u002Fp>\u003Cp>$$\\sigma_1=\\frac{N_1}{t},\\qquad \\sigma_2=\\frac{N_2}{t}.$$\u003C\u002Fp>\u003Ch2>Equilibrium of a curved element\u003C\u002Fh2>{{chunk:shell-laplace-equilibrium}}\u003Cp>Curvature allows in-plane membrane forces to balance loading normal to the surface. This is why shells can carry pressure efficiently with relatively small wall thickness.\u003C\u002Fp>\u003Ch2>Cylindrical and spherical shells\u003C\u002Fh2>{{chunk:thin-pressure-vessel-membrane-stress}}\u003Cp>These relations are basic examples of membrane action under internal pressure and illustrate how geometry affects stress distribution.\u003C\u002Fp>\u003Ch2>Strength assessment\u003C\u002Fh2>\u003Cp>After membrane stresses $\\sigma_1$ and $\\sigma_2$ are determined, the critical stress state is assessed. For ductile isotropic materials, an appropriate multiaxial failure criterion may be used; the selected criterion and allowable values depend on the material and design method.\u003C\u002Fp>\u003Ch2>When membrane theory is insufficient\u003C\u002Fh2>\u003Cp>Near rigid restraints, flanges, supports, openings, nozzles, joints, concentrated forces, and abrupt changes in thickness or curvature, edge and local effects can generate significant bending moments. Membrane theory should therefore not be used automatically for local verification in these regions.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>{{chunk:membrane-shell-calculation-algorithm}}\u003Cp>The child topic develops membrane theory in more detail, including assumptions, equilibrium relations, and typical pressure-shell calculations.\u003C\u002Fp>",87,[1540],{"id":1541,"parent_id":1531,"code":70,"slug":1542,"name":1543,"seo_title":1544,"seo_description":1545,"seo_text":1546,"content":1547,"locale":8,"uk_topic_id":1548,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":1542,"children":1549},125,"membrane-theory-of-shells","Membrane Theory","Membrane Theory of Shells — Equilibrium and Pressure Stresses","Membrane theory assumptions, Laplace equilibrium, meridional and hoop force resultants, and stresses in thin cylindrical and spherical shells.","This topic explains membrane theory for thin shells, where bending and twisting moments and transverse shear are neglected. It covers membrane force resultants, Laplace equilibrium, cylindrical and spherical pressure shells, stress calculations, assumptions, and regions where bending effects require a more advanced shell model.","\u003Cp>\u003Cstrong>Membrane theory of shells\u003C\u002Fstrong> is an approximate theory in which bending and twisting moments and transverse shear forces are neglected, while loading is carried primarily by membrane force resultants in the tangent plane of the midsurface.\u003C\u002Fp>\u003Ch2>Main assumptions\u003C\u002Fh2>\u003Cul>\u003Cli>shell thickness $t$ is small compared with the characteristic radii of curvature;\u003C\u002Fli>\u003Cli>through-thickness stresses can be represented by average membrane values;\u003C\u002Fli>\u003Cli>the analyzed region is sufficiently far from local edge disturbances;\u003C\u002Fli>\u003Cli>geometry and loading permit equilibrium without significant bending moments.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Membrane force resultants\u003C\u002Fh2>\u003Cp>In the principal midsurface directions, normal force resultants $N_1$ and $N_2$ have units of force per unit length, such as N\u002Fmm. For thickness $t$:\u003C\u002Fp>\u003Cp>$$\\sigma_1=\\frac{N_1}{t},\\qquad \\sigma_2=\\frac{N_2}{t}.$$\u003C\u002Fp>\u003Ch2>Laplace equilibrium equation\u003C\u002Fh2>{{chunk:shell-laplace-equilibrium}}\u003Cp>The normal equilibrium equation alone is generally insufficient to determine two unknown membrane resultants $N_1$ and $N_2$. A second relation follows from equilibrium of a cut-off portion of the shell, symmetry, or other membrane-equilibrium equations.\u003C\u002Fp>\u003Ch2>Cylindrical shell under internal pressure\u003C\u002Fh2>\u003Cp>For a cylinder, one principal radius of curvature equals $r$, while curvature along the generator is zero, so the corresponding radius is formally infinite. Normal equilibrium gives the hoop resultant $N_\\theta=pr$. The longitudinal resultant $N_z$ in a closed cylinder follows from equilibrium of the end portion.\u003C\u002Fp>{{chunk:thin-pressure-vessel-membrane-stress}}\u003Ch2>Derivation of the longitudinal cylinder resultant\u003C\u002Fh2>\u003Cp>Pressure $p$ on the end cap produces a resultant $p\\pi r^2$. It is balanced by longitudinal membrane force around the circumference:\u003C\u002Fp>\u003Cp>$$2\\pi rN_z=p\\pi r^2,$$\u003C\u002Fp>\u003Cp>hence:\u003C\u002Fp>\u003Cp>$$N_z=\\frac{pr}{2}.$$\u003C\u002Fp>\u003Ch2>Spherical shell\u003C\u002Fh2>\u003Cp>For a sphere, $R_1=R_2=r$ and symmetry gives $N_1=N_2=N$. From the Laplace equation:\u003C\u002Fp>\u003Cp>$$\\frac{N}{r}+\\frac{N}{r}=p,$$\u003C\u002Fp>\u003Cp>so $N=pr\u002F2$ and $\\sigma=pr\u002F(2t)$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A thin-walled closed cylinder has internal radius $r=500\\ \\text{mm}$, thickness $t=10\\ \\text{mm}$, and internal gauge pressure $p=2\\ \\text{MPa}$. The membrane stresses are:\u003C\u002Fp>\u003Cp>$$\\sigma_\\theta=\\frac{2\\cdot500}{10}=100\\ \\text{MPa},$$\u003C\u002Fp>\u003Cp>$$\\sigma_z=\\frac{2\\cdot500}{2\\cdot10}=50\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>Thus, for this idealized model, the hoop stress $\\sigma_\\theta$ is more critical than the longitudinal stress $\\sigma_z$.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>{{chunk:membrane-shell-calculation-algorithm}}\u003Ch2>Limitations\u003C\u002Fh2>\u003Cp>The membrane state is commonly disturbed near rigid edges, supports, flanges, nozzles, openings, joints, concentrated loads, and abrupt changes in curvature or thickness. Bending and edge stresses then require bending shell theory or numerical analysis.\u003C\u002Fp>",88,[],{"id":9,"parent_id":784,"code":1551,"slug":1552,"name":1553,"seo_title":1554,"seo_description":1555,"seo_text":1556,"content":1557,"locale":8,"uk_topic_id":1558,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":927,"url_slug":1552,"children":1559},"30","dynamic-and-cyclic-loading","Dynamic and Cyclic Loading","Dynamic and Cyclic Loading — Fatigue & Impact","Structural analysis under dynamic impact, inertia forces, and cyclic fatigue.","Engineering structures frequently undergo time-dependent or high-velocity loads causing dynamic effects. This page covers dynamic load categories: accelerated motion, impact loading, and cyclic variable stresses, explaining dynamic magnification factors and structural material responses.","\u003Cp>\u003Cstrong>Dynamic loading\u003C\u002Fstrong> changes rapidly enough, or is accompanied by accelerations large enough, that inertia forces significantly affect internal forces, stresses, and displacements. \u003Cstrong>Cyclic loading\u003C\u002Fstrong> repeats many times and may cause fatigue failure even when stresses remain below the static strength limit.\u003C\u002Fp>\u003Ch2>Dynamics versus statics\u003C\u002Fh2>\u003Cp>In a static calculation, accelerations of the structural mass are neglected. Dynamic analysis must account for inertia, the time variation of loading, velocity or acceleration, and, where required, the vibration properties of the system.\u003C\u002Fp>{{chunk:d-alembert-inertia-force}}\u003Ch2>Impact loading\u003C\u002Fh2>\u003Cp>During impact, kinetic and potential energy of a moving body is transferred over a short time into structural strain energy, while part of the energy may be dissipated by plasticity, friction, contact effects, and other mechanisms. In an idealized elastic model, the maximum response can often be estimated by an energy method using the dynamic factor $K_d$.\u003C\u002Fp>{{chunk:impact-dynamic-factor}}\u003Ch2>Cyclic loading and fatigue\u003C\u002Fh2>\u003Cp>Under repeated cycles, not only the maximum stress $\\sigma_{max}$ but also stress amplitude $\\sigma_a$, mean stress $\\sigma_m$, stress ratio $R$, and number of cycles $N$ are important.\u003C\u002Fp>{{chunk:fatigue-cycle-parameters}}{{chunk:fatigue-sn-curve-endurance}}\u003Ch2>Influence of the real component\u003C\u002Fh2>\u003Cp>Fatigue strength measured on a laboratory specimen cannot be transferred directly to a real component without considering its geometry, surface condition, size, environment, and stress concentrations.\u003C\u002Fp>{{chunk:fatigue-strength-factors}}\u003Ch2>Selecting a calculation model\u003C\u002Fh2>\u003Col>\u003Cli>Identify the loading type: accelerated motion, sudden application, impact, or repeated cycles.\u003C\u002Fli>\u003Cli>For motion with known accelerations, include inertia forces.\u003C\u002Fli>\u003Cli>For impact, evaluate the energy balance and dynamic factor $K_d$ within the adopted model.\u003C\u002Fli>\u003Cli>For cyclic loading, determine $\\sigma_a$, $\\sigma_m$, $R$, and the required life $N$.\u003C\u002Fli>\u003Cli>Account for stress concentrators, surface condition, size, and other real-component factors.\u003C\u002Fli>\u003Cli>Check whether the conditions exceed the limits of the simplified linear-elastic model.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Section structure\u003C\u002Fh2>\u003Cp>The child topics separately address inertia forces, impact calculations, and material fatigue. Vibration and resonance problems require a dedicated dynamic model and cannot be reduced to a static dynamic-factor calculation alone.\u003C\u002Fp>",30,[1560,1570,1581],{"id":1561,"parent_id":9,"code":70,"slug":1562,"name":1563,"seo_title":1564,"seo_description":1565,"seo_text":1566,"content":1567,"locale":8,"uk_topic_id":1568,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":77,"url_slug":1562,"children":1569},76,"inertia-forces","Inertia Forces","Inertia Forces — D’Alembert’s Principle, ma & Dynamic Stress","Use D’Alembert’s principle to calculate inertia forces and dynamic stress in accelerating bars and rotating components, including distributed inertia loading.","Analyzing machine components moving with acceleration (hoist cables, flywheel rims, rotating shafts) requires incorporating inertia forces. This section applies D'Alembert's principle to convert dynamic equilibrium into static equivalent systems for stress analysis.","\u003Cp>When a structural member moves with acceleration, its mass produces an \u003Cstrong>inertia effect\u003C\u002Fstrong> that must be considered when determining internal forces and stresses. D'Alembert's principle provides a convenient way to formulate such problems.\u003C\u002Fp>{{chunk:d-alembert-inertia-force}}\u003Ch2>Translational motion\u003C\u002Fh2>\u003Cp>For a body of mass $m$ with translational acceleration $a$, the magnitude of the inertia force is $ma$ and its direction in the D'Alembert calculation scheme is opposite to the acceleration. If mass is distributed along a member, the inertia loading may also be distributed.\u003C\u002Fp>\u003Cp>For a bar with constant mass per unit length $m_l$ and uniform acceleration $a$ along its length, the magnitude of the distributed inertia load is:\u003C\u002Fp>\u003Cp>$$q_i=m_l a.$$\u003C\u002Fp>\u003Ch2>Stress in an accelerating bar\u003C\u002Fh2>\u003Cp>If a straight uniform bar of length $L$ and area $A$ accelerates along its own axis, the axial force at a section must accelerate the portion of mass on one side of that section. Therefore $N$ varies along the length and normal stress is $\\sigma=N\u002FA$.\u003C\u002Fp>\u003Cp>For example, if a bar of density $\\rho$ is pulled at one end and the entire bar has axial acceleration $a$, then for a section at distance $x$ from the free end:\u003C\u002Fp>\u003Cp>$$N(x)=\\rho Aax,\\qquad \\sigma(x)=\\rho ax.$$\u003C\u002Fp>\u003Cp>The largest stress occurs near the end through which the accelerating force is transmitted.\u003C\u002Fp>\u003Ch2>Rotational motion\u003C\u002Fh2>\u003Cp>For angular acceleration $\\varepsilon$, an inertia moment $J\\varepsilon$ is introduced. During steady rotation with angular velocity $\\omega$, material points have centripetal acceleration $a_n=\\omega^2r$, so rotating components develop stresses associated with distributed mass forces.\u003C\u002Fp>\u003Ch2>Example: thin ring\u003C\u002Fh2>\u003Cp>For an idealized thin ring of material density $\\rho$, radius $r$, and angular velocity $\\omega$, the circumferential stress in a simple membrane model is:\u003C\u002Fp>\u003Cp>$$\\sigma_\\theta=\\rho\\omega^2r^2.$$\u003C\u002Fp>\u003Cp>Thick disks, complex rims, and nonuniform rotating components require more detailed stress models.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the law of motion and accelerations of the masses.\u003C\u002Fli>\u003Cli>Determine concentrated or distributed inertia forces.\u003C\u002Fli>\u003Cli>Add them to the calculation scheme according to D'Alembert's principle.\u003C\u002Fli>\u003Cli>Determine internal force resultants.\u003C\u002Fli>\u003Cli>Calculate stresses and displacements using the usual strength-of-materials relations within the adopted model.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Limitation\u003C\u002Fh2>\u003Cp>If acceleration changes rapidly, vibration develops, or deformation waves propagate through the member, a quasi-static inertia-force representation may be insufficient. A full dynamic analysis is then required.\u003C\u002Fp>",38,[],{"id":1571,"parent_id":9,"code":1572,"slug":1573,"name":1574,"seo_title":1575,"seo_description":1576,"seo_text":1577,"content":1578,"locale":8,"uk_topic_id":1579,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1579,"url_slug":1573,"children":1580},77,"31","impact-loading-calculations","Impact Loading Calculations","Impact Loading — Dynamic Factor, Stress & Deflection Formula","Calculate impact response with energy balance and the dynamic factor Kd: falling weights, suddenly applied loads, maximum stress, deflection, and model limits.","Under impact loading, kinetic energy from falling masses transforms instantaneously into strain energy. This section presents procedures for calculating dynamic impact factors under vertical and horizontal collisions, alongside dynamic stress and deflection evaluation formulas.","\u003Cp>\u003Cstrong>Impact loading\u003C\u002Fstrong> occurs when a body with nonzero velocity contacts a structure and transfers energy to it over a short time. Maximum forces, stresses, and displacements during impact may be several times larger than the static values produced by the same weight.\u003C\u002Fp>\u003Ch2>Energy model\u003C\u002Fh2>\u003Cp>In the simplest elastic model, the loss of potential energy of a falling weight is converted into strain energy of the structure. If a weight $P$ falls through height $h$ and the maximum additional deformation after contact is $\\delta$, the work of the weight through the distance $h+\\delta$ is equated to the maximum elastic strain energy.\u003C\u002Fp>\u003Cp>For a linear system whose static displacement under $P$ is $\\delta_{st}$:\u003C\u002Fp>\u003Cp>$$P(h+\\delta)=\\frac{1}{2}\\frac{P}{\\delta_{st}}\\delta^2.$$\u003C\u002Fp>\u003Cp>Solving this quadratic relation gives the dynamic factor $K_d$.\u003C\u002Fp>{{chunk:impact-dynamic-factor}}\u003Ch2>Static displacement\u003C\u002Fh2>\u003Cp>$\\delta_{st}$ is the displacement of the impact point in the load direction that would be produced by statically applying force $P$. It is calculated by ordinary strength-of-materials methods for an axial bar, beam, spring, or another linearly elastic system.\u003C\u002Fp>\u003Ch2>Suddenly applied load\u003C\u002Fh2>\u003Cp>If force $P$ is applied instantaneously without a drop height and without initial velocity, the idealized model has $h=0$ and $K_d=2$. Thus the maximum elastic displacement and stress are twice their corresponding static values.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose a weight falls through $h=20\\ \\text{mm}$ and the static displacement under its weight is $\\delta_{st}=0.5\\ \\text{mm}$. Then:\u003C\u002Fp>\u003Cp>$$K_d=1+\\sqrt{1+\\frac{2\\cdot20}{0.5}}=1+\\sqrt{81}=10.$$\u003C\u002Fp>\u003Cp>If the static stress from the weight is $\\sigma_{st}=25\\ \\text{MPa}$ and the system remains linear, the estimated maximum impact stress is $\\sigma_{dyn}=250\\ \\text{MPa}$.\u003C\u002Fp>\u003Ch2>Horizontal impact and specified velocity\u003C\u002Fh2>\u003Cp>If the initial energy is specified as kinetic energy $mv^2\u002F2$ rather than by a drop height, write the energy balance directly using that kinetic energy. The resulting response depends on impactor mass $m$, velocity $v$, structural compliance, and the adopted contact model.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the weight or initial kinetic energy of the impactor.\u003C\u002Fli>\u003Cli>Calculate the static displacement $\\delta_{st}$ of the impact point under the corresponding static force.\u003C\u002Fli>\u003Cli>Write the energy balance for the adopted model.\u003C\u002Fli>\u003Cli>Determine the maximum displacement $\\delta_{max}$ or dynamic factor $K_d$.\u003C\u002Fli>\u003Cli>Use linear proportionality to determine maximum internal forces and stresses.\u003C\u002Fli>\u003Cli>Verify that the stresses remain within the range where the elastic model is acceptable.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Limitations\u003C\u002Fh2>\u003Cp>Real impacts may involve plastic deformation, local contact crushing, rebound, damping, and stress-wave propagation. In such cases, the simple energy formula may overestimate or underestimate the actual maximum response and a more detailed dynamic model is required.\u003C\u002Fp>",31,[],{"id":1582,"parent_id":9,"code":1583,"slug":1584,"name":1585,"seo_title":1586,"seo_description":1587,"seo_text":1588,"content":1589,"locale":8,"uk_topic_id":1590,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1590,"url_slug":1584,"children":1591},78,"32","material-fatigue-and-endurance-limit","Material Fatigue and Endurance Limit","Fatigue Strength — S–N Curve, Endurance Limit & Goodman","Learn fatigue analysis with stress amplitude and mean stress, Wöhler S–N curves, endurance limit, stress concentration, Goodman relation, and fatigue-life checks.","Repeated cyclic stress variations lead to structural fatigue failure at stress levels far below ultimate tensile strength. This page covers cyclic stress parameters, endurance limits, Wöhler S-N curves, stress concentration effects (notches, fillets), and size factors.","\u003Cp>\u003Cstrong>Material fatigue\u003C\u002Fstrong> is the process of damage accumulation under repeatedly varying stresses, which may end in crack initiation, crack propagation, and final fracture. Fatigue failure can occur at maximum stresses below the static ultimate strength.\u003C\u002Fp>\u003Ch2>Stress-cycle parameters\u003C\u002Fh2>{{chunk:fatigue-cycle-parameters}}\u003Cp>The amplitude $\\sigma_a$ characterizes the cyclic part of loading, while mean stress $\\sigma_m$ shifts the entire cycle toward tension or compression. Therefore, two cycles with the same $\\sigma_a$ but different $\\sigma_m$ may have different fatigue severity.\u003C\u002Fp>\u003Ch2>S–N curve\u003C\u002Fh2>{{chunk:fatigue-sn-curve-endurance}}\u003Cp>To obtain an S–N curve, groups of specimens are tested at different cyclic stress levels and the number of cycles $N$ to failure is recorded. The number of cycles is commonly plotted on a logarithmic scale.\u003C\u002Fp>\u003Ch2>Endurance limit\u003C\u002Fh2>\u003Cp>For materials with a distinct endurance limit under a specified stress cycle, a stress level can be identified below which a laboratory specimen survives the large reference number of cycles adopted by the test method without fatigue failure. For many nonferrous alloys and other materials without a clear horizontal asymptote, fatigue strength is instead specified at a particular number of cycles $N$.\u003C\u002Fp>\u003Ch2>Stress concentration\u003C\u002Fh2>\u003Cp>Holes, threads, keyways, grooves, and abrupt section transitions create local stress maxima and promote fatigue-crack initiation. In fatigue design, the theoretical geometric stress-concentration factor is not always identical to the effective fatigue-strength reduction factor because material notch sensitivity must also be considered.\u003C\u002Fp>{{chunk:fatigue-strength-factors}}\u003Ch2>Mean-stress effect\u003C\u002Fh2>\u003Cp>When $\\sigma_m$ is nonzero, allowable amplitude is assessed using experimental or code-based diagrams. Common educational approximations include the Goodman and Soderberg lines and the Gerber parabola. The selected relation must be appropriate for the material and adopted method.\u003C\u002Fp>\u003Cp>For example, a linear Goodman relation for tensile mean stress is often written as:\u003C\u002Fp>\u003Cp>$$\\frac{\\sigma_a}{\\sigma_{-1}}+\\frac{\\sigma_m}{\\sigma_u}\\le\\frac{1}{n},$$\u003C\u002Fp>\u003Cp>where $\\sigma_{-1}$ is a reference endurance limit for fully reversed loading under the specified conditions, $\\sigma_u$ is ultimate strength, and $n$ is the selected safety factor. For a real component, the reference endurance characteristic is corrected according to the adopted design method.\u003C\u002Fp>\u003Ch2>Typical fatigue-failure development\u003C\u002Fh2>\u003Col>\u003Cli>Localization of cyclic plasticity or damage in a critical region.\u003C\u002Fli>\u003Cli>Initiation of a small crack, often at the surface or a stress concentrator.\u003C\u002Fli>\u003Cli>Progressive crack growth over many cycles.\u003C\u002Fli>\u003Cli>Final rapid fracture of the remaining section when it can no longer carry the load.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Verification procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine $\\sigma_{max}$ and $\\sigma_{min}$ at the critical point.\u003C\u002Fli>\u003Cli>Calculate $\\sigma_a$, $\\sigma_m$, and $R$.\u003C\u002Fli>\u003Cli>Determine the reference fatigue characteristic for the required life $N$ and stress cycle.\u003C\u002Fli>\u003Cli>Account for stress concentration, surface condition, size, temperature, and environment.\u003C\u002Fli>\u003Cli>Account for mean stress using the selected criterion.\u003C\u002Fli>\u003Cli>Determine the fatigue safety factor or allowable life.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Limitations\u003C\u002Fh2>\u003Cp>Fatigue assessment depends strongly on experimental data and the adopted standard. Variable-amplitude loading, multiaxial fatigue, low-cycle fatigue, and crack-growth analysis require specialized models beyond the basic S–N approach.\u003C\u002Fp>",32,[],{"id":1593,"parent_id":784,"code":1594,"slug":1594,"name":1595,"seo_title":1596,"seo_description":1597,"seo_text":1598,"content":1599,"locale":8,"uk_topic_id":1600,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":939,"url_slug":1594,"children":1601},126,"contact-stresses","Contact Stresses","Contact Stress & Hertz Theory — Formulas and Applications","Learn contact stress and Hertz theory for spheres, cylinders, and rollers: pressure distributions, contact geometry, subsurface stresses, and contact strength.","This section introduces contact stresses in engineering components. It covers contact geometry, reduced elastic and curvature parameters, Hertz pressure distributions, point and line contact, subsurface stress states, and the factors governing contact strength and fatigue damage.","\u003Cp>\u003Cstrong>Contact stresses\u003C\u002Fstrong> arise in a localized region through which two bodies transmit a normal force $F$. Because the contact area can be small, local stresses may be much higher than nominal stresses elsewhere in the components.\u003C\u002Fp>\u003Ch2>Contact geometry\u003C\u002Fh2>\u003Cp>Before loading, smooth curved surfaces may touch at a point or along a line. Elastic deformation under $F$ creates a finite contact patch or strip whose shape depends on the local surface curvatures $R_1$ and $R_2$.\u003C\u002Fp>\u003Ch2>Hertz theory\u003C\u002Fh2>\u003Cp>Classical Hertz theory describes local elastic contact of smooth bodies for small deformation and a contact region that is small compared with the characteristic body dimensions. The material properties of the two bodies are combined through the reduced modulus $E^*$, while geometry is represented through the reduced radius $R^*$.\u003C\u002Fp>{{chunk:hertz-reduced-elastic-modulus}}{{chunk:hertz-reduced-radius}}\u003Ch2>Contact pressure\u003C\u002Fh2>\u003Cp>Pressure $p$ in a Hertz contact region is nonuniform: it reaches the maximum value $p_0$ near the center and decreases to zero at the idealized contact boundary.\u003C\u002Fp>{{chunk:hertz-pressure-distribution}}\u003Ch2>Point and line contact\u003C\u002Fh2>\u003Cp>Axisymmetric spherical contact produces a circular patch of radius $a$, while long parallel cylinders produce a narrow strip of half-width $b$. Contact dimensions increase as load and material compliance increase.\u003C\u002Fp>\u003Ch2>Subsurface stresses\u003C\u002Fh2>\u003Cp>The contact pressure produces a three-dimensional stress state beneath the surface. Maximum shear or equivalent stresses may occur below the surface rather than at it, which is important for rolling-contact fatigue and pitting.\u003C\u002Fp>\u003Ch2>Contact strength\u003C\u002Fh2>\u003Cp>Machine-component design requires more than the maximum Hertz pressure $p_0$ alone. Repeated loading, material hardness and microstructure, roughness, lubrication, misalignment, residual stresses, and other service factors can strongly affect damage.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>{{chunk:hertz-contact-calculation-algorithm}}\u003Cp>The child topics develop contact geometry, Hertz theory, sphere and cylinder contact, subsurface stress state, and contact strength in greater detail.\u003C\u002Fp>",89,[1602,1613,1623,1634,1645,1656],{"id":1603,"parent_id":1593,"code":1604,"slug":1605,"name":1606,"seo_title":1607,"seo_description":1608,"seo_text":1609,"content":1610,"locale":8,"uk_topic_id":1611,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":915,"url_slug":1605,"children":1612},127,"contact-geometry","contact-types-and-geometry","Contact of Two Bodies: Types and Geometry","Point, Line & Surface Contact — Contact Geometry Explained","Understand point, line, and surface contact between bodies, contact-zone geometry, local curvature and reduced radius, and when Hertz contact theory applies.","This topic explains point, line, and surface contact between two bodies and the geometric parameters that control the contact region. It covers local principal curvatures, reduced radius, elastic deformation of the initial contact, and selection of Hertz or more general contact models.","\u003Cp>\u003Cstrong>Contact geometry\u003C\u002Fstrong> determines the shape of the contact region and strongly influences local pressure $p$. It is important to distinguish the initial geometric contact of unloaded bodies from the finite contact region created by elastic deformation.\u003C\u002Fp>\u003Ch2>Initial point contact\u003C\u002Fh2>\u003Cp>Two curved surfaces, such as two spheres or a sphere and a plane, may ideally touch at one point. Under a normal force $F$, that point expands into a small contact patch. It is circular for an axisymmetric case and generally elliptical when the principal curvatures differ in two directions.\u003C\u002Fp>\u003Ch2>Initial line contact\u003C\u002Fh2>\u003Cp>Two parallel cylinders or a cylinder and a plane ideally touch along a line. Deformation produces a narrow contact strip of finite half-width $b$.\u003C\u002Fp>\u003Ch2>Surface contact\u003C\u002Fh2>\u003Cp>If unloaded bodies already have a finite nominal contact area, the problem may not correspond to the classical Hertz model of initially point or line contact. Pressure then depends on geometry, compliance, restraint, and actual surface conformity.\u003C\u002Fp>\u003Ch2>Curvature\u003C\u002Fh2>\u003Cp>The local shape of a smooth surface is characterized by principal radii of curvature. In simple axisymmetric problems, two surfaces can be combined into one reduced radius $R^*$.\u003C\u002Fp>{{chunk:hertz-reduced-radius}}\u003Cp>For convex-concave contact, curvature signs depend on the adopted convention. Very conformal surfaces may produce a large $R^*$ and broad contact area, so the small-contact assumption of Hertz theory requires careful checking.\u003C\u002Fp>\u003Ch2>Material properties\u003C\u002Fh2>\u003Cp>The size of the elastic contact region depends not only on geometry but also on the combined compliance of both bodies, represented by $E^*$.\u003C\u002Fp>{{chunk:hertz-reduced-elastic-modulus}}\u003Ch2>Choosing a model\u003C\u002Fh2>\u003Col>\u003Cli>Identify the initial contact type: point, line, or finite area.\u003C\u002Fli>\u003Cli>Determine local principal radii of curvature $R_1$ and $R_2$.\u003C\u002Fli>\u003Cli>Estimate whether the contact region is small compared with body dimensions and radii.\u003C\u002Fli>\u003Cli>For a small elastic patch or strip, consider Hertz theory.\u003C\u002Fli>\u003Cli>For conformal contact, sharp edges, large contact regions, or complex geometry, use a more detailed contact model.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Examples\u003C\u002Fh2>\u003Cp>A sphere on a plane is initially point contact; a long roller on a plane is initially line contact; a flat bearing plate on a foundation is surface contact. The same normal force $F$ can therefore produce fundamentally different pressure distributions.\u003C\u002Fp>",90,[],{"id":1614,"parent_id":1593,"code":1615,"slug":1615,"name":1616,"seo_title":1617,"seo_description":1618,"seo_text":1619,"content":1620,"locale":8,"uk_topic_id":1621,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1350,"url_slug":1615,"children":1622},128,"hertz-contact-theory","Hertz Theory: Main Assumptions and Formulas","Hertz Contact Theory — Formulas, Assumptions & Contact Stress","Learn Hertz contact theory with formulas for reduced modulus, contact radius, maximum pressure, sphere contact, assumptions, and limits of the Hertz model.","This topic presents Hertz theory as the basic model of local elastic contact. It covers assumptions, reduced elastic modulus and curvature, nonuniform contact pressure, sphere and cylinder contact formulas, the calculation sequence, and the limits of the classical Hertz model.","\u003Cp>\u003Cstrong>Hertz contact theory\u003C\u002Fstrong> describes the local stresses and elastic deformations that arise when two curved elastic bodies are pressed together over a small contact region.\u003C\u002Fp>\n\u003Cp>Before loading, the bodies may touch at a point or along a line. Elastic deformation under the applied normal force creates a finite contact area with a nonuniform pressure distribution.\u003C\u002Fp>\n{{chunk:hertz-contact-patch-material-response}}\n\u003Ch2>Main assumptions\u003C\u002Fh2>\n\u003Cul>\u003Cli>The materials are homogeneous, isotropic, and linearly elastic.\u003C\u002Fli>\u003Cli>Deformations are small.\u003C\u002Fli>\u003Cli>The dimensions of the contact region are small compared with the characteristic radii of curvature of the bodies.\u003C\u002Fli>\u003Cli>The contacting surfaces are smooth in the local contact region.\u003C\u002Fli>\u003Cli>Classical Hertz theory neglects tangential traction and treats the load transfer as frictionless normal contact.\u003C\u002Fli>\u003C\u002Ful>\n\u003Ch2>Reduced elastic properties\u003C\u002Fh2>\n\u003Cp>For two bodies with Young's moduli \\(E_1,E_2\\) and Poisson's ratios \\(\\nu_1,\\nu_2\\), the reduced modulus is defined by\u003C\u002Fp>\n\u003Cp>\\[\\frac{1}{E^*}=\\frac{1-\\nu_1^2}{E_1}+\\frac{1-\\nu_2^2}{E_2}.\\]\u003C\u002Fp>\n\u003Cp>For spherical surfaces in the simplest external-contact case, the reduced radius is\u003C\u002Fp>\n\u003Cp>\\[\\frac{1}{R^*}=\\frac{1}{R_1}+\\frac{1}{R_2}.\\]\u003C\u002Fp>\n\u003Ch2>Circular Hertz contact\u003C\u002Fh2>\n\u003Cp>For axisymmetric contact between spherical surfaces, or a sphere and an elastic half-space, the radius of the circular contact area is\u003C\u002Fp>\n\u003Cp>\\[a=\\left(\\frac{3FR^*}{4E^*}\\right)^{1\u002F3}.\\]\u003C\u002Fp>\n\u003Cp>The maximum contact pressure at the center is\u003C\u002Fp>\n\u003Cp>\\[p_0=\\frac{3F}{2\\pi a^2}.\\]\u003C\u002Fp>\n\u003Cp>The pressure distribution over the contact radius is\u003C\u002Fp>\n\u003Cp>\\[p(r)=p_0\\sqrt{1-\\frac{r^2}{a^2}},\\qquad 0\\le r\\le a.\\]\u003C\u002Fp>\n{{chunk:hertz-pressure-distribution-interactive}}\n\u003Cp>The pressure is therefore highest at the center and decreases smoothly to zero at the edge of the contact area.\u003C\u002Fp>\n\u003Ch2>What controls Hertz contact stresses?\u003C\u002Fh2>\n\u003Cp>The contact dimensions and pressure level depend on the normal load, local surface curvature, and elastic properties of both materials. Increasing load enlarges the contact region while also changing the maximum pressure.\u003C\u002Fp>\n\u003Ch2>Applications and limits\u003C\u002Fh2>\n\u003Cp>Hertz theory is widely used as a first model for rolling-element bearings, gears, wheel-rail contact, rollers, cams, and other components with localized elastic contact.\u003C\u002Fp>\n\u003Cp>The classical solution does not account for plastic indentation, strong frictional effects, adhesion, surface roughness, or large deformation. Such cases require more advanced contact models.\u003C\u002Fp>",91,[],{"id":1624,"parent_id":1593,"code":1625,"slug":1626,"name":1627,"seo_title":1628,"seo_description":1629,"seo_text":1630,"content":1631,"locale":8,"uk_topic_id":1632,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1558,"url_slug":1626,"children":1633},129,"sphere-contact","contact-of-spheres-and-spherical-surfaces","Contact of Spheres and Spherical Surfaces","Hertz Contact of Spheres — Contact Radius, Pressure & Example","Calculate Hertz contact between spheres or a sphere and plane: reduced properties, contact radius, maximum pressure, worked example, and applicability limits.","This topic develops Hertz contact for spheres and spherical surfaces. It covers reduced elastic modulus and curvature, circular contact-patch radius, maximum pressure, load scaling, a sphere-on-plane example, elliptical-contact limitations, and engineering applications.","\u003Cp>\u003Cstrong>Contact of spherical surfaces\u003C\u002Fstrong> is the classical example of initially point contact. Under a normal force $F$, elastic deformation creates a finite contact patch. For two spheres with an axisymmetric local geometry, the patch is circular.\u003C\u002Fp>\u003Ch2>Reduced properties\u003C\u002Fh2>{{chunk:hertz-reduced-elastic-modulus}}{{chunk:hertz-reduced-radius}}\u003Cp>For a sphere on a plane, one radius is infinite, so in the simple convex case $R^*=R$ of the sphere.\u003C\u002Fp>\u003Ch2>Contact radius and maximum pressure\u003C\u002Fh2>{{chunk:hertz-sphere-contact}}\u003Cp>Increasing $F$ expands the patch as $F^{1\u002F3}$. The maximum pressure $p_0$ also increases as $F^{1\u002F3}$ because contact area does not grow rapidly enough to offset the increasing force completely.\u003C\u002Fp>\u003Ch2>Example: steel sphere on steel plane\u003C\u002Fh2>\u003Cp>Let $F=1000\\ \\text{N}$, $R=10\\ \\text{mm}$, $E_1=E_2=210000\\ \\text{MPa}$, and $\\nu_1=\\nu_2=0.30$. Then:\u003C\u002Fp>\u003Cp>$$E^*=\\left[2\\frac{1-0.3^2}{210000}\\right]^{-1}\\approx115385\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>For the plane, $R^*=10\\ \\text{mm}$. The contact radius is:\u003C\u002Fp>\u003Cp>$$a=\\left(\\frac{3\\cdot1000\\cdot10}{4\\cdot115385}\\right)^{1\u002F3}\\approx0.402\\ \\text{mm}.$$\u003C\u002Fp>\u003Cp>The maximum pressure is:\u003C\u002Fp>\u003Cp>$$p_0=\\frac{3\\cdot1000}{2\\pi\\cdot0.402^2}\\approx2.96\\cdot10^3\\ \\text{MPa}.$$\u003C\u002Fp>\u003Cp>This high local value shows why contact strength cannot be assessed from average stress over the entire component. Before using the result in design, verify that the purely elastic contact assumptions remain acceptable.\u003C\u002Fp>\u003Ch2>Elliptical contact\u003C\u002Fh2>\u003Cp>If the principal curvatures differ in two perpendicular directions, the contact patch is generally elliptical. Its semi-axes require the full Hertz solution based on the principal curvatures; the circular-patch formula for $a$ is then not directly applicable.\u003C\u002Fp>\u003Ch2>Applications\u003C\u002Fh2>\u003Cp>Spherical contact is a basic model for ball bearings, spherical supports, ball mechanisms, and localized interaction of rounded components.\u003C\u002Fp>\u003Ch2>Procedure\u003C\u002Fh2>{{chunk:hertz-contact-calculation-algorithm}}",92,[],{"id":1635,"parent_id":1593,"code":1636,"slug":1637,"name":1638,"seo_title":1639,"seo_description":1640,"seo_text":1641,"content":1642,"locale":8,"uk_topic_id":1643,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":98,"url_slug":1637,"children":1644},130,"cylinder-roller-contact","contact-of-cylinders-and-rollers","Contact of Cylinders and Rollers","Hertzian Contact Between Cylinders — Line Contact","Hertzian line contact between cylinders and rollers: calculate contact width and maximum pressure using reduced radius, elastic modulus, and load per unit length.","This topic explains the Hertz line-contact model for cylinders and rollers. It covers load per unit length, reduced elastic modulus and curvature, contact-strip half-width, maximum pressure, a numerical example, edge effects, and applications to rollers and rolling bearings.","\u003Cp>Contact between two cylinders with parallel axes is the classical Hertzian case of \u003Cstrong>line contact\u003C\u002Fstrong>. Before loading, the ideal surfaces touch along a line; after elastic deformation, the load is transmitted through a narrow rectangular contact strip.\u003C\u002Fp>\n{{chunk:hertz-cylinder-line-contact-deformation}}\n\u003Ch2>Reduced properties\u003C\u002Fh2>\n\u003Cp>For external contact between two convex cylinders, the reduced radius is\u003C\u002Fp>\n\u003Cp>\\[\\frac{1}{R^*}=\\frac{1}{R_1}+\\frac{1}{R_2}.\\]\u003C\u002Fp>\n\u003Cp>The reduced elastic modulus is\u003C\u002Fp>\n\u003Cp>\\[\\frac{1}{E^*}=\\frac{1-\\nu_1^2}{E_1}+\\frac{1-\\nu_2^2}{E_2}.\\]\u003C\u002Fp>\n\u003Ch2>Contact-strip width\u003C\u002Fh2>\n\u003Cp>Let \\(w=F\u002FL\\) be the normal load per unit contact length. For classical Hertzian line contact, the contact half-width is\u003C\u002Fp>\n\u003Cp>\\[b=\\sqrt{\\frac{4wR^*}{\\pi E^*}}.\\]\u003C\u002Fp>\n\u003Cp>The total strip width is \\(2b\\).\u003C\u002Fp>\n\u003Ch2>Pressure distribution\u003C\u002Fh2>\n\u003Cp>Across the strip, the contact pressure follows a semi-elliptical distribution:\u003C\u002Fp>\n\u003Cp>\\[p(x)=p_0\\sqrt{1-\\frac{x^2}{b^2}},\\qquad |x|\\le b,\\]\u003C\u002Fp>\n\u003Cp>with the maximum pressure at the center:\u003C\u002Fp>\n\u003Cp>\\[p_0=\\frac{2w}{\\pi b}=\\sqrt{\\frac{wE^*}{\\pi R^*}}.\\]\u003C\u002Fp>\n{{chunk:hertz-pressure-distribution-interactive}}\n\u003Ch2>Physical interpretation\u003C\u002Fh2>\n\u003Cp>Increasing \\(w\\) makes the contact strip wider and increases the maximum pressure. A larger reduced radius also broadens the strip and reduces pressure concentration. Increasing \\(E^*\\) makes the contact narrower for the same geometry and load.\u003C\u002Fp>\n\u003Ch2>Engineering examples\u003C\u002Fh2>\n\u003Cp>The model is used for bearing rollers, cylindrical rollers and supports, and as a simplified two-dimensional model of wheel-rail contact. For rollers of finite length, edge effects can significantly modify the stress field near the ends.\u003C\u002Fp>",93,[],{"id":1646,"parent_id":1593,"code":1647,"slug":1648,"name":1649,"seo_title":1650,"seo_description":1651,"seo_text":1652,"content":1653,"locale":8,"uk_topic_id":1654,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1182,"url_slug":1648,"children":1655},131,"subsurface-contact-stress","subsurface-contact-stress-state","Stress State Beneath the Contact Surface","Subsurface Hertz Contact Stress — Shear Stress & Critical Depth","Understand the stress state beneath a Hertz contact surface: principal and shear stresses, critical subsurface regions, friction effects, and fatigue relevance.","This topic explains the three-dimensional stress state beneath a contact surface. It covers the localization and depth variation of normal, principal, shear, and equivalent stresses, the role of friction, and why subsurface stresses are important for rolling-contact fatigue and contact-strength assessment.","\u003Cp>\u003Cstrong>Beneath a contact surface\u003C\u002Fstrong>, a complex three-dimensional stress state develops. Surface contact pressure $p(x,y)$ is only a boundary condition; several normal and shear stress components arise inside the material.\u003C\u002Fp>\u003Ch2>Localization of the field\u003C\u002Fh2>\u003Cp>The highest contact-related stresses are concentrated in a volume whose characteristic dimensions are of the same order as the contact-patch radius $a$ or contact-strip half-width $b$. Stresses decrease rapidly with distance from the contact.\u003C\u002Fp>\u003Ch2>Normal stresses\u003C\u002Fh2>\u003Cp>Within the contact area, compressive normal pressure $p(x,y)$ acts on the surface. Beneath it, three-dimensional elastic interaction also creates normal stresses in other directions, so the state cannot be represented by a single value $-p$.\u003C\u002Fp>\u003Ch2>Shear and principal stresses\u003C\u002Fh2>\u003Cp>Even in frictionless normal contact, differences between principal normal stresses produce nonzero maximum shear stress:\u003C\u002Fp>\u003Cp>$$\\tau_{max}=\\frac{\\sigma_1-\\sigma_3}{2},$$\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>$\\tau_{max}$\u003C\u002Fstrong> — maximum shear stress at the considered point;\u003C\u002Fli>\u003Cli>\u003Cstrong>$\\sigma_1$\u003C\u002Fstrong> — algebraically largest principal stress;\u003C\u002Fli>\u003Cli>\u003Cstrong>$\\sigma_3$\u003C\u002Fstrong> — algebraically smallest principal stress.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>In classical Hertz problems, $\\tau_{max}$ often occurs at a finite depth rather than directly at the surface. Its exact magnitude and position depend on contact type and Poisson's ratio $\\nu$, so a universal numerical value should not be used without specifying the problem.\u003C\u002Fp>\u003Ch2>Why subsurface stress matters\u003C\u002Fh2>\u003Cp>During repeated rolling, the subsurface region experiences a changing multiaxial stress state many times. This can promote fatigue-crack initiation below the surface and subsequent pitting or spalling.\u003C\u002Fp>\u003Ch2>Effect of friction\u003C\u002Fh2>\u003Cp>With tangential force or sliding, surface shear tractions are added to the normal Hertz problem. They change the principal-stress field and may move the critical region closer to the surface. A frictionless model should therefore not be applied automatically to contacts with substantial traction or sliding.\u003C\u002Fp>\u003Ch2>Assessment criteria\u003C\u002Fh2>\u003Cp>Depending on material and damage mechanism, engineers may examine maximum contact pressure $p_0$, principal stresses, $\\tau_{max}$, equivalent stress, or specialized contact-fatigue criteria. For ductile isotropic materials, local equivalent stress may be assessed with Tresca or von Mises criteria, but rolling-contact life requires a separate fatigue model.\u003C\u002Fp>{{chunk:tresca-von-mises-criteria}}\u003Ch2>Practical procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the contact-pressure distribution $p(x,y)$.\u003C\u002Fli>\u003Cli>Use the contact solution to obtain stress components in the subsurface region.\u003C\u002Fli>\u003Cli>Calculate principal, shear, or equivalent stresses.\u003C\u002Fli>\u003Cli>Locate the critical point.\u003C\u002Fli>\u003Cli>Relate the result to the expected damage mechanism and the appropriate strength or life criterion.\u003C\u002Fli>\u003C\u002Fol>",94,[],{"id":1657,"parent_id":1593,"code":1658,"slug":1659,"name":1660,"seo_title":1661,"seo_description":1662,"seo_text":1663,"content":1664,"locale":8,"uk_topic_id":4,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":5,"url_slug":1659,"children":1665},132,"contact-strength-machine-elements","contact-strength-of-machine-elements","Contact Strength of Machine Elements","Contact Strength — Hertz Stress, Pitting & Rolling Fatigue","Assess contact strength in bearings, gears, rollers, and wheels using Hertz pressure, pitting and rolling-contact fatigue, surface condition, lubrication, and alignment.","This topic applies contact-stress theory to machine elements. It covers allowable contact stress concepts, Hertz pressure, pitting and rolling-contact fatigue, material hardness, surface condition, lubrication, misalignment, and engineering checks for bearings, gears, rollers, and wheels.","\u003Cp>\u003Cstrong>Contact strength\u003C\u002Fstrong> is the ability of working surfaces to transmit localized contact loads without unacceptable plastic deformation, fatigue pitting, cracking, or another specified damage mode.\u003C\u002Fp>\u003Ch2>Contact-stress check\u003C\u002Fh2>\u003Cp>In a simple verification model, the calculated maximum contact pressure $p_0$ is compared with an allowable value:\u003C\u002Fp>\u003Cp>$$p_0\\le[p_H],$$\u003C\u002Fp>\u003Cp>or an equivalent standard-specific form such as:\u003C\u002Fp>\u003Cp>$$\\sigma_H\\le[\\sigma_H].$$\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>$p_0$\u003C\u002Fstrong> — calculated maximum contact pressure;\u003C\u002Fli>\u003Cli>\u003Cstrong>$[p_H]$\u003C\u002Fstrong> — allowable contact pressure according to the adopted method;\u003C\u002Fli>\u003Cli>\u003Cstrong>$\\sigma_H$\u003C\u002Fstrong> — calculated contact stress in a standard-specific formulation;\u003C\u002Fli>\u003Cli>\u003Cstrong>$[\\sigma_H]$\u003C\u002Fstrong> — allowable contact stress in the same formulation.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>The definitions of calculated and allowable contact stress depend on the component type and design standard; there is no single universal allowable value for all contacts.\u003C\u002Fp>\u003Ch2>Pitting and contact fatigue\u003C\u002Fh2>\u003Cp>In rolling bearings, gears, and roller pairs, contact loading repeats many times. The cyclic subsurface stress state can initiate cracks and cause local surface material removal known as pitting or spalling. Therefore, a static check based only on $p_0$ does not replace a life calculation.\u003C\u002Fp>\u003Ch2>Factors affecting contact strength\u003C\u002Fh2>\u003Cul>\u003Cli>\u003Cstrong>Material and hardness.\u003C\u002Fstrong> Heat treatment and surface hardening can strongly change resistance to contact damage.\u003C\u002Fli>\u003Cli>\u003Cstrong>Geometry.\u003C\u002Fstrong> Smaller radii of curvature generally increase local pressure under otherwise comparable conditions.\u003C\u002Fli>\u003Cli>\u003Cstrong>Roughness.\u003C\u002Fstrong> Real contact occurs through asperities that create local pressure peaks.\u003C\u002Fli>\u003Cli>\u003Cstrong>Lubrication.\u003C\u002Fstrong> A lubricant film can separate surfaces and alter friction, temperature, and damage mechanisms.\u003C\u002Fli>\u003Cli>\u003Cstrong>Misalignment and edge contact.\u003C\u002Fstrong> Nonuniform load distribution can sharply increase local pressure.\u003C\u002Fli>\u003Cli>\u003Cstrong>Cyclic loading.\u003C\u002Fstrong> Cycle count and load spectrum govern contact life.\u003C\u002Fli>\u003Cli>\u003Cstrong>Temperature and environment.\u003C\u002Fstrong> They affect material properties, lubrication, and surface processes.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Connection with Hertz analysis\u003C\u002Fh2>\u003Cp>Hertz analysis provides the elastic contact-region dimensions $a$ or $b$ and the maximum pressure $p_0$ that form inputs to subsequent strength assessment.\u003C\u002Fp>{{chunk:hertz-contact-calculation-algorithm}}\u003Ch2>Rolling bearings\u003C\u002Fh2>\u003Cp>In rolling bearings, balls or rollers repeatedly load the raceways. Practical bearing-life calculations use specialized standardized load-rating and life relations rather than only comparing $p_0$ with one allowable stress.\u003C\u002Fp>\u003Ch2>Gears\u003C\u002Fh2>\u003Cp>Contact between tooth flanks can locally be approximated as contact of curved bodies. Real gear calculations additionally account for mesh geometry, face-load distribution, dynamics, manufacturing accuracy, lubrication, and factors defined by the selected standard.\u003C\u002Fp>\u003Ch2>Rollers and wheels\u003C\u002Fh2>\u003Cp>For rollers, Hertz line contact and edge effects are important. For wheel-rail contact, normal contact is combined with traction forces, sliding, and a complex cyclic loading history.\u003C\u002Fp>\u003Ch2>Engineering verification procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine contact force $F$ and geometry.\u003C\u002Fli>\u003Cli>Check applicability of the elastic Hertz model.\u003C\u002Fli>\u003Cli>Calculate the contact region and $p_0$.\u003C\u002Fli>\u003Cli>Account for actual load distribution, misalignment, and edge effects.\u003C\u002Fli>\u003Cli>Identify the expected damage mechanism: plasticity, pitting, wear, scuffing, or another mode.\u003C\u002Fli>\u003Cli>Apply a criterion or standard intended for the specific component and operating regime.\u003C\u002Fli>\u003Cli>For cyclic contact, verify the required service life separately.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Important limitation\u003C\u002Fh2>\u003Cp>A high Hertz pressure $p_0$ alone is not a complete failure criterion. Contact strength combines the local stress state, properties of the surface layer, cyclic loading, and tribological conditions.\u003C\u002Fp>",[],{"id":1667,"parent_id":784,"code":70,"slug":1668,"name":1669,"seo_title":1670,"seo_description":1671,"seo_text":1672,"content":1673,"locale":8,"uk_topic_id":1674,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":1675,"url_slug":1668,"children":1676},431,"torsion-of-solid-noncircular-shafts","Torsion of Solid Noncircular Shafts","Torsion of Solid Noncircular Shafts — Formulas and Coefficients","Saint-Venant torsion of solid noncircular shafts: warping, torsion constant, rectangular-section formulas, and alpha-beta coefficient table.","Torsion of solid noncircular shafts, including rectangular-section strength and stiffness formulas, torsion constant, warping, and tabulated coefficients.","\u003Cp>\u003Cstrong>Torsion of solid noncircular shafts\u003C\u002Fstrong> differs fundamentally from torsion of circular shafts. Cross-sections generally do not remain plane: they warp, and shear stress is distributed nonuniformly over the area.\u003C\u002Fp>\n\u003Ch2>Why the circular-shaft formula does not apply\u003C\u002Fh2>\n\u003Cp>For a circular section, the geometry is compatible with circular shear paths and \\(\\tau=T\\rho\u002FJ\\) can be used. For a rectangular, elliptical, or other solid noncircular section, the polar second moment of area alone defines neither torsional stiffness nor stress distribution.\u003C\u002Fp>\n\u003Ch2>Saint-Venant torsion\u003C\u002Fh2>\n\u003Cp>For a prismatic member, the twist rate is\u003C\u002Fp>\n\u003Cp>\\[\\frac{d\\varphi}{dx}=\\frac{T}{GJ_t},\\]\u003C\u002Fp>\n\u003Cp>where \\(T\\) is torque, \\(G\\) is shear modulus, and \\(J_t\\) is the torsion constant. It depends on section shape and equals the polar second moment only for circular shafts. The shear-stress field follows from the Saint-Venant torsion problem, for example through Prandtl's stress function.\u003C\u002Fp>\n\u003Ch2>Solid rectangular section\u003C\u002Fh2>\n\u003Cp>Let \\(a\\) be the longer side and \\(b\\) the shorter side, so that \\(a\\ge b\\). Engineering calculations commonly use\u003C\u002Fp>\n\u003Cp>\\[\\tau_{\\max}=\\frac{T}{\\alpha a b^2},\\qquad J_t=\\beta a b^3,\\qquad \\varphi=\\frac{TL}{GJ_t}=\\frac{TL}{\\beta G a b^3}.\\]\u003C\u002Fp>\n\u003Cp>Maximum shear stress occurs at the midpoint of the long sides. It is zero at the corners and at the center. The coefficients \\(\\alpha\\) and \\(\\beta\\) depend on the aspect ratio:\u003C\u002Fp>\n\u003Ctable>\n\u003Cthead>\u003Ctr>\u003Cth>\\(a\u002Fb\\)\u003C\u002Fth>\u003Cth>\\(\\alpha\\)\u003C\u002Fth>\u003Cth>\\(\\beta\\)\u003C\u002Fth>\u003C\u002Ftr>\u003C\u002Fthead>\n\u003Ctbody>\n\u003Ctr>\u003Ctd>1.0\u003C\u002Ftd>\u003Ctd>0.208\u003C\u002Ftd>\u003Ctd>0.141\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>1.5\u003C\u002Ftd>\u003Ctd>0.231\u003C\u002Ftd>\u003Ctd>0.196\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>2.0\u003C\u002Ftd>\u003Ctd>0.246\u003C\u002Ftd>\u003Ctd>0.229\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>2.5\u003C\u002Ftd>\u003Ctd>0.258\u003C\u002Ftd>\u003Ctd>0.249\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>3.0\u003C\u002Ftd>\u003Ctd>0.267\u003C\u002Ftd>\u003Ctd>0.263\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>4.0\u003C\u002Ftd>\u003Ctd>0.282\u003C\u002Ftd>\u003Ctd>0.281\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>6.0\u003C\u002Ftd>\u003Ctd>0.299\u003C\u002Ftd>\u003Ctd>0.299\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>8.0\u003C\u002Ftd>\u003Ctd>0.307\u003C\u002Ftd>\u003Ctd>0.307\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>10.0\u003C\u002Ftd>\u003Ctd>0.313\u003C\u002Ftd>\u003Ctd>0.313\u003C\u002Ftd>\u003C\u002Ftr>\n\u003Ctr>\u003Ctd>\\(\\infty\\)\u003C\u002Ftd>\u003Ctd>0.333\u003C\u002Ftd>\u003Ctd>0.333\u003C\u002Ftd>\u003C\u002Ftr>\n\u003C\u002Ftbody>\u003C\u002Ftable>\n\u003Cp>Linear interpolation may be used for intermediate \\(a\u002Fb\\). For a very narrow rectangle, \\(a\u002Fb\\gg1\\), \\(\\alpha\\approx\\beta\\approx1\u002F3\\), hence \\(J_t\\approx ab^3\u002F3\\) and \\(\\tau_{\\max}\\approx3T\u002F(ab^2)\\).\u003C\u002Fp>\n\u003Ch3>Approximation for the torsion constant\u003C\u002Fh3>\n\u003Cp>When tabulated data are unavailable, the following approximation is convenient for \\(a\\ge b\\):\u003C\u002Fp>\n\u003Cp>\\[J_t\\approx\\frac{ab^3}{3}\\left[1-0.63\\frac{b}{a}+0.052\\left(\\frac{b}{a}\\right)^5\\right].\\]\u003C\u002Fp>\n\u003Cp>It closely reproduces the tabulated \\(\\beta\\) values; for a square it gives \\(J_t\\approx0.141a^4\\). The torsion constant must not be replaced by the rectangular section's polar second moment \\(I_p=I_x+I_y\\).\u003C\u002Fp>\n\u003Ch3>Strength and stiffness checks\u003C\u002Fh3>\n\u003Cp>The two criteria are checked separately:\u003C\u002Fp>\n\u003Cp>\\[\\tau_{\\max}\\le\\tau_{allow},\\qquad \\varphi=\\frac{TL}{GJ_t}\\le\\varphi_{allow}.\\]\u003C\u002Fp>\n\u003Cp>If twist per unit length is limited, check \\(\\theta=T\u002F(GJ_t)\\le\\theta_{allow}\\). These relations apply away from torque application zones, abrupt section changes, and restraints. Fillets, keyways, and other discontinuities require a suitable stress-concentration factor.\u003C\u002Fp>\n\u003Ch2>Other solid noncircular sections\u003C\u002Fh2>\n\u003Cp>Elliptical, triangular, and other solid shapes use their own analytical solutions, tabulated coefficients, or numerical analysis. The general route remains the same: determine \\(J_t\\), calculate twist, and check maximum shear stress.\u003C\u002Fp>\n\u003Ch2>Scope\u003C\u002Fh2>\n\u003Cp>This section covers only \u003Cstrong>solid\u003C\u002Fstrong> noncircular sections. Open and closed thin-walled members require different models and belong to a separate advanced section.\u003C\u002Fp>\n\u003Ch2>Calculation route\u003C\u002Fh2>\n\u003Col>\u003Cli>Define the shape and dimensions of the solid section.\u003C\u002Fli>\u003Cli>Obtain \\(J_t\\) and the maximum-stress coefficient from a formula or table.\u003C\u002Fli>\u003Cli>Calculate maximum shear stress.\u003C\u002Fli>\u003Cli>Calculate total twist or twist per unit length.\u003C\u002Fli>\u003Cli>Check strength, stiffness, and stress concentration where relevant.\u003C\u002Fli>\u003C\u002Fol>",430,998,[],{"id":1678,"parent_id":70,"code":70,"slug":1679,"name":1680,"seo_title":1681,"seo_description":1682,"seo_text":1683,"content":1684,"locale":8,"uk_topic_id":1685,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1679,"children":1686},83,"structural-mechanics","Structural Mechanics","Structural Mechanics — Beams, Trusses, Frames & Analysis Methods","Learn structural mechanics for beams, trusses, frames, and arches: reactions, internal forces, displacements, force and displacement methods, and matrix stiffness.","Structural mechanics covers the analysis of load-bearing systems including statically determinate beams, trusses, frames and arches, structural displacements, statically indeterminate systems using force and displacement methods, and the fundamentals of matrix stiffness analysis.","\u003Cp>\u003Cstrong>Structural mechanics\u003C\u002Fstrong> develops methods for analyzing load-bearing structural systems under external loads and imposed actions. Its principal models include beams, trusses, frames, arches, and more general framed structures. The emphasis is on the response of the structure as a system: reactions, internal forces, displacements, and static or kinematic determinacy.\u003C\u002Fp>\u003Ch2>Statically determinate structures\u003C\u002Fh2>\u003Cp>For a statically determinate structure, all reactions and internal forces can be obtained from equilibrium equations. Beams, planar trusses, frames, and three-hinged arches provide the fundamental models used before more advanced systems are considered.\u003C\u002Fp>\u003Ch2>Structural displacements\u003C\u002Fh2>\u003Cp>Displacements and rotations are required for stiffness assessment and for the analysis of indeterminate structures. Energy methods, Mohr integrals, and unit-load methods provide systematic tools for determining these quantities.\u003C\u002Fp>\u003Ch2>Statically indeterminate structures\u003C\u002Fh2>\u003Cp>When equilibrium alone is insufficient, compatibility conditions must be introduced. Classical approaches include the force method and displacement method. For larger systems, these ideas lead naturally to matrix stiffness analysis, which forms the basis of framed finite-element models.\u003C\u002Fp>",41,[1687,1727,1777,1787,1797,1837,1917,1967,2017,2067],{"id":1688,"parent_id":1678,"code":70,"slug":1689,"name":1690,"seo_title":1691,"seo_description":1692,"seo_text":1693,"content":1694,"locale":8,"uk_topic_id":1695,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1689,"children":1696},304,"fundamentals-of-structural-mechanics","Fundamentals of Structural Mechanics","Fundamentals of Structural Mechanics | Structural Models","Structural mechanics fundamentals: structural idealization, constraints, degrees of freedom, kinematic analysis, stability, and static determinacy.","An introduction to structural mechanics covering structural idealization, calculation models, constraints and degrees of freedom, kinematic analysis, geometric stability, and classification of structures by static determinacy.","\u003Cp>\u003Cstrong>Fundamentals of structural mechanics\u003C\u002Fstrong> establish the rules for converting a real structure into a mechanical model suitable for analysis. Before reactions, internal forces, or displacements can be determined, the analyst must identify the members and constraints, the motions they restrain, and whether the adopted structural model is geometrically stable.\u003C\u002Fp>\u003Ch2>Structural model\u003C\u002Fh2>\u003Cp>A real structure is idealized as a system of members, joints, and supports. Rigid and pinned connections, support types, and load models are selected according to the structural behavior relevant to the problem.\u003C\u002Fp>\u003Ch2>Kinematic analysis\u003C\u002Fh2>\u003Cp>Degrees of freedom and the arrangement of constraints are examined to distinguish a geometrically stable structure from a mechanism or an instantaneously unstable configuration. Merely counting constraints is not sufficient; their geometric arrangement also matters.\u003C\u002Fp>\u003Ch2>Static determinacy\u003C\u002Fh2>\u003Cp>If all reactions and internal force effects can be found from equilibrium alone, the structure is statically determinate. Redundant constraints create static indeterminacy and require additional equations based on structural deformation and compatibility.\u003C\u002Fp>",297,[1697,1707,1717],{"id":1698,"parent_id":1688,"code":70,"slug":1699,"name":1700,"seo_title":1701,"seo_description":1702,"seo_text":1703,"content":1704,"locale":8,"uk_topic_id":1705,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1699,"children":1706},337,"structural-models-and-idealization","Structural Models and Idealization","Structural Idealization — Members, Joints, Supports & Loads","Learn how to convert a real structure into an analytical model by idealizing members, joints, supports, and loads while keeping the assumptions relevant to structural response.","This topic explains structural idealization in structural mechanics, including analytical models of members, joints, supports and loads and the limitations of common modeling assumptions.","\u003Cp>A \u003Cstrong>structural model\u003C\u002Fstrong> is an idealized mechanical representation that retains the properties needed to determine reactions, internal forces, and displacements while neglecting secondary detail. A correct calculation therefore depends not only on the equations but also on whether the adopted model represents the intended structural behavior.\u003C\u002Fp>\u003Ch2>Member idealization\u003C\u002Fh2>\u003Cp>Beams, columns, braces, and other slender components are represented by their centroidal or reference axes, while geometry and stiffness properties are assigned separately. Connections may be modeled as rigid, pinned, or elastic according to the relative motions they permit.\u003C\u002Fp>\u003Ch2>Support idealization\u003C\u002Fh2>\u003Cp>A support is replaced by constraints that suppress selected displacement components. Each independent restrained component produces a corresponding reaction. In a planar model, a roller commonly gives one reaction, a pin two force reactions, and a fixed support two force reactions plus a moment.\u003C\u002Fp>\u003Ch2>Loads and imposed actions\u003C\u002Fh2>\u003Cp>Real actions are represented by concentrated forces and moments, distributed loads, temperature changes, prescribed support movements, and similar idealized inputs. The chosen representation should preserve the mechanical effect relevant to the analysis.\u003C\u002Fp>\u003Ch2>Limits of a model\u003C\u002Fh2>\u003Cp>The same physical structure may require different analytical models for different questions. A connection may be treated as pinned in a global truss model while its local stiffness is studied separately. Modeling assumptions should therefore be stated explicitly.\u003C\u002Fp>",311,[],{"id":1708,"parent_id":1688,"code":70,"slug":1709,"name":1710,"seo_title":1711,"seo_description":1712,"seo_text":1713,"content":1714,"locale":8,"uk_topic_id":1715,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1709,"children":1716},338,"kinematic-analysis-geometric-stability","Kinematic Analysis and Geometric Stability","Kinematic Analysis — Degrees of Freedom, Constraints & Stability","Learn how to assess geometric stability of framed structures using degrees of freedom, constraints, mechanisms, instantaneous instability, and kinematic analysis.","This topic covers degrees of freedom and constraints, geometric stability of framed systems, mechanisms and instantaneously unstable configurations, and basic kinematic analysis.","\u003Cp>\u003Cstrong>Kinematic analysis\u003C\u002Fstrong> determines whether a framed system can preserve its geometric configuration without deformation of its members. A structural model needs not only enough constraints but also a suitable geometric arrangement of those constraints.\u003C\u002Fp>\u003Ch2>Degrees of freedom and constraints\u003C\u002Fh2>\u003Cp>A free rigid body in a plane has three independent motions: two translations and one rotation. Constraints remove selected degrees of freedom. For structures made of several members or rigid parts, both external supports and internal connections must be considered.\u003C\u002Fp>\u003Ch2>Geometric stability\u003C\u002Fh2>\u003Cp>A geometrically stable structure cannot change its configuration without deformation of its members. A triangle is the simplest stable planar pin-jointed form, which is why triangulation plays a fundamental role in truss stability.\u003C\u002Fp>\u003Ch2>Mechanisms and instantaneous instability\u003C\u002Fh2>\u003Cp>A system that can undergo finite motion without member deformation is a mechanism. An instantaneously unstable configuration is a special case in which an infinitesimal motion is possible because of a singular geometric arrangement of constraints.\u003C\u002Fp>\u003Ch2>Why counting is not enough\u003C\u002Fh2>\u003Cp>A model can contain the nominally required number of constraints and still be unstable if some constraints are dependent or improperly arranged. Numerical counting must therefore be supplemented by geometric inspection.\u003C\u002Fp>",312,[],{"id":1718,"parent_id":1688,"code":70,"slug":1719,"name":1720,"seo_title":1721,"seo_description":1722,"seo_text":1723,"content":1724,"locale":8,"uk_topic_id":1725,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1719,"children":1726},339,"static-determinacy-framed-structures","Static Determinacy of Framed Structures","Static Determinacy — How to Identify Indeterminate Structures","Learn how to distinguish statically determinate and indeterminate framed structures, count redundant constraints, and separate external from internal indeterminacy.","This topic explains static determinacy and indeterminacy of framed structures, distinguishes external and internal indeterminacy, and introduces redundant reactions and constraints.","\u003Cp>\u003Cstrong>Static determinacy\u003C\u002Fstrong> indicates whether equilibrium equations are sufficient to determine all reactions and internal force effects. This classification should be applied to a geometrically stable structure; an unstable mechanism should not be labeled statically determinate merely because the number of unknowns matches the number of equations.\u003C\u002Fp>\u003Ch2>Statically determinate structures\u003C\u002Fh2>\u003Cp>In a statically determinate structure, force unknowns can be obtained from equilibrium. A planar rigid body provides three independent equilibrium equations, while internal hinges and separation into structural parts can provide additional independent equilibrium conditions in compound structures.\u003C\u002Fp>\u003Ch2>Static indeterminacy\u003C\u002Fh2>\u003Cp>If the number of independent force unknowns exceeds the available independent static equations, the structure is statically indeterminate. The excess gives the degree of static indeterminacy when the structural topology has been counted correctly.\u003C\u002Fp>\u003Ch2>External and internal indeterminacy\u003C\u002Fh2>\u003Cp>External indeterminacy is associated with redundant support restraints, whereas internal indeterminacy results from redundant connections between structural parts. Both types may occur in the same structure.\u003C\u002Fp>\u003Ch2>Mechanical consequence\u003C\u002Fh2>\u003Cp>Equilibrium alone cannot solve an indeterminate structure. Member deformation and displacement compatibility must also be considered, leading to the force method, displacement method, and matrix stiffness method.\u003C\u002Fp>",313,[],{"id":1728,"parent_id":1678,"code":70,"slug":1729,"name":1730,"seo_title":1731,"seo_description":1732,"seo_text":1733,"content":1734,"locale":8,"uk_topic_id":1735,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1729,"children":1736},305,"statically-determinate-beams","Statically Determinate Beams","Statically Determinate Beams | Reactions, Shear and Moment","Analysis of statically determinate beams: support reactions, internal forces, shear and bending-moment diagrams, and Gerber beams.","This section covers statically determinate beams: support reactions, shear forces and bending moments, internal-force diagrams, differential load-shear-moment relations, and hinged multispan beam systems.","\u003Cp>\u003Cstrong>Statically determinate beams\u003C\u002Fstrong> are fundamental structural systems whose support reactions and internal force effects can be determined from equilibrium. Because basic reactions and the section method are also covered in mechanics of materials, this section emphasizes structural-system analysis and verification rather than repeating the entire introductory treatment.\u003C\u002Fp>\u003Ch2>From loads to internal forces\u003C\u002Fh2>\u003Cp>The analysis begins with support reactions. The beam is then divided into characteristic intervals and section equilibrium is used to determine shear force $V$ or $Q$ and bending moment $M$.\u003C\u002Fp>\u003Ch2>Diagrams as analytical tools\u003C\u002Fh2>\u003Cp>Shear-force and bending-moment diagrams describe the variation of internal actions along the structure. In structural mechanics, their shapes are checked against load distribution, discontinuities, differential relations, and boundary conditions.\u003C\u002Fp>\u003Ch2>Compound beams\u003C\u002Fh2>\u003Cp>Internal hinges can be used to form statically determinate multispan systems. Their analysis requires an understanding of how forces are transferred between structural parts and in what sequence those parts should be solved.\u003C\u002Fp>",298,[1737,1747,1757,1767],{"id":1738,"parent_id":1728,"code":70,"slug":1739,"name":1740,"seo_title":1741,"seo_description":1742,"seo_text":1743,"content":1744,"locale":8,"uk_topic_id":1745,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1739,"children":1746},340,"support-reactions-internal-forces-beams","Support Reactions and Internal Forces in Beams","Beam Support Reactions & Internal Forces — Section Method","Calculate support reactions in statically determinate beams, then use the section method to find shear force V\u002FQ and bending moment M at any section.","This topic covers beam idealization, support reactions, and the use of sections to determine shear force and bending moment in statically determinate beams.","\u003Cp>\u003Cstrong>Support reactions and internal forces\u003C\u002Fstrong> are treated here as a concise review of tools needed throughout structural mechanics. The introductory theory of supports, external and internal forces, and the section method also belongs to mechanics of materials.\u003C\u002Fp>\u003Ch2>Support reactions\u003C\u002Fh2>\u003Cp>After replacing supports by their reaction components, equilibrium equations are written for the entire beam. A suitable moment center often eliminates several unknowns at once. Calculated reactions should be checked with an independent equilibrium equation.\u003C\u002Fp>\u003Ch2>Internal force resultants\u003C\u002Fh2>\u003Cp>A general section of a planar beam may carry axial force $N$, shear force $V$ or $Q$, and bending moment $M$. In common transverse-bending problems the primary quantities are shear and bending moment.\u003C\u002Fp>\u003Ch2>Section method\u003C\u002Fh2>\u003Cp>The beam is cut at the required location and equilibrium is written for one of the two parts. Choosing the side with fewer external actions usually simplifies the calculation. The sign convention for internal forces must be applied consistently.\u003C\u002Fp>\u003Ch2>Verification\u003C\u002Fh2>\u003Cp>An error in the reactions propagates through all subsequent diagrams. Before constructing $V(x)$ and $M(x)$, check global equilibrium, dimensions, and whether the resulting reaction directions are mechanically plausible.\u003C\u002Fp>",314,[],{"id":1748,"parent_id":1728,"code":70,"slug":1749,"name":1750,"seo_title":1751,"seo_description":1752,"seo_text":1753,"content":1754,"locale":8,"uk_topic_id":1755,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1749,"children":1756},341,"shear-force-bending-moment-diagrams","Shear-Force and Bending-Moment Diagrams","Shear Force & Bending Moment Diagrams — How to Draw V and M","Learn how to construct shear-force V\u002FQ and bending-moment M diagrams for statically determinate beams, handle jumps, and check key sections.","This topic covers V\u002FQ and M diagrams for statically determinate beams, characteristic intervals, jumps due to concentrated actions, and checks at key sections.","\u003Cp>\u003Cstrong>Shear-force and bending-moment diagrams\u003C\u002Fstrong> are used in structural mechanics not only as calculation outputs but also as tools for checking and interpreting beam behavior. The basic section method is supplemented here by structural features that a correct diagram must satisfy.\u003C\u002Fp>\u003Ch2>Characteristic points\u003C\u002Fh2>\u003Cp>Intervals are separated by concentrated forces or moments, the start and end of distributed loads, supports, and internal hinges. Within each interval, the mathematical form of $V(x)$ and $M(x)$ follows from the load distribution.\u003C\u002Fp>\u003Ch2>Jumps and continuity\u003C\u002Fh2>\u003Cp>A concentrated transverse force causes a jump in the shear diagram. A concentrated couple causes a jump in the bending-moment diagram. Without the corresponding concentrated action, the relevant function remains continuous across the point.\u003C\u002Fp>\u003Ch2>Moment extrema\u003C\u002Fh2>\u003Cp>On a smooth interval, a local extremum of $M$ occurs where $V=0$. Zeros of the shear diagram are therefore important control points for the bending-moment diagram.\u003C\u002Fp>\u003Ch2>Diagram checks\u003C\u002Fh2>\u003Cp>Use equilibrium, boundary conditions, and the relationships among load, shear, and moment. At an unloaded internal hinge the bending moment is zero. On an interval with no distributed load, shear is constant and bending moment varies linearly.\u003C\u002Fp>\u003Ch2>Sign convention\u003C\u002Fh2>\u003Cp>Textbooks may use different sign conventions. What matters is consistent use of the selected convention and agreement between the diagram and equilibrium of the isolated structural part.\u003C\u002Fp>",315,[],{"id":1758,"parent_id":1728,"code":70,"slug":1759,"name":1760,"seo_title":1761,"seo_description":1762,"seo_text":1763,"content":1764,"locale":8,"uk_topic_id":1765,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1759,"children":1766},342,"differential-relations-load-shear-moment","Differential Relations between Load, Shear, and Moment","Load, Shear & Moment Relations — dV\u002Fdx and dM\u002Fdx Formulas","Learn the differential relations among distributed load q, shear force V, and bending moment M, their geometric meaning, and how to use them to check beam diagrams.","This topic explains the differential relationships among distributed load q, shear force V or Q, and bending moment M and their geometric interpretation for beam diagrams.","\u003Cp>\u003Cstrong>Differential relations\u003C\u002Fstrong> connect the distributed load intensity $q(x)$, shear force $V(x)$ or $Q(x)$, and bending moment $M(x)$. They make it possible to predict diagram shapes and verify results without repeating a full section calculation at every point.\u003C\u002Fp>\u003Ch2>Basic relations\u003C\u002Fh2>\u003Cp>For one common sign convention:\u003C\u002Fp>\u003Cp>$$\\frac{dV}{dx}=-q(x),\\qquad \\frac{dM}{dx}=V(x).$$\u003C\u002Fp>\u003Cp>If a different sign convention is adopted for load or shear, the sign of the first equation may change; the underlying geometric relationships remain the same.\u003C\u002Fp>\u003Ch2>Geometric interpretation\u003C\u002Fh2>\u003Cp>Shear gives the slope of the bending-moment diagram, while distributed load controls the rate of change of shear. If $q=0$, shear is constant and moment is linear. If $q$ is constant, shear is linear and moment is quadratic.\u003C\u002Fp>\u003Ch2>Integral form\u003C\u002Fh2>\u003Cp>Over an interval from $x_1$ to $x_2$:\u003C\u002Fp>\u003Cp>$$V(x_2)-V(x_1)=-\\int_{x_1}^{x_2}q(x)\\,dx,$$\u003C\u002Fp>\u003Cp>$$M(x_2)-M(x_1)=\\int_{x_1}^{x_2}V(x)\\,dx.$$\u003C\u002Fp>\u003Cp>Thus changes in shear and moment can be read from signed areas of the preceding diagrams.\u003C\u002Fp>\u003Ch2>Practical verification\u003C\u002Fh2>\u003Cp>Zeros of shear correspond to stationary points of moment on smooth intervals. The sign of shear controls whether moment rises or falls, and the sign of load controls how shear changes. These checks quickly reveal inconsistent diagrams.\u003C\u002Fp>",316,[],{"id":1768,"parent_id":1728,"code":70,"slug":1769,"name":1770,"seo_title":1771,"seo_description":1772,"seo_text":1773,"content":1774,"locale":8,"uk_topic_id":1775,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1769,"children":1776},343,"hinged-multispan-beams","Hinged Multispan Beams","Gerber Beams — Hinged Multispan Beam Analysis Step by Step","Analyze statically determinate Gerber beams using the structural hierarchy, sequential equilibrium, support reactions, and shear-force and bending-moment diagrams.","This topic covers hinged multispan or Gerber beams, static determinacy, structural hierarchy diagrams, sequential support-reaction calculations, and internal-force diagrams.","\u003Cp>A \u003Cstrong>hinged multispan beam\u003C\u002Fstrong> consists of several beam segments connected by internal hinges. A suitable arrangement of hinges produces a geometrically stable, statically determinate structure spanning several bays.\u003C\u002Fp>\u003Ch2>Internal hinge\u003C\u002Fh2>\u003Cp>An internal hinge transfers forces between adjacent parts but does not transfer bending moment. Therefore, unless an external couple is applied directly at the hinge, $M=0$ there. This condition allows the compound structure to be separated into individual free bodies.\u003C\u002Fp>\u003Ch2>Structural hierarchy\u003C\u002Fh2>\u003Cp>It is useful to identify which beam segments are supporting components and which are suspended from them. This relationship is often represented by a hierarchy or storey diagram. Analysis begins with the suspended components, whose reactions are then applied as loads to the supporting components.\u003C\u002Fp>\u003Ch2>Analysis sequence\u003C\u002Fh2>\u003Cp>Equilibrium is applied to each component to determine reactions, after which shear and bending-moment diagrams are constructed. Interaction forces at an internal hinge are equal in magnitude and opposite in direction on the two connected parts.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>In addition to equilibrium of individual components, verify equilibrium of the complete beam. The bending-moment diagram must pass through zero at unloaded internal hinges, and force transfer between structural levels must satisfy action and reaction.\u003C\u002Fp>",317,[],{"id":1778,"parent_id":1678,"code":70,"slug":1779,"name":1780,"seo_title":1781,"seo_description":1782,"seo_text":1783,"content":1784,"locale":8,"uk_topic_id":1785,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1779,"children":1786},84,"statically-determinate-trusses","Statically Determinate Trusses","Truss Analysis — Method of Joints & Sections Step by Step","Analyze statically determinate plane trusses: find support reactions and zero-force members, then calculate member forces with the methods of joints and sections.","Truss analysis determines support reactions and axial member forces in a pin-jointed structure. This introduction covers ideal plane-truss assumptions, static determinacy, zero-force members, the method of joints, the method of sections, tension and compression signs, and practical equilibrium checks.","\u003Cp>\u003Cstrong>Truss analysis\u003C\u002Fstrong> determines the support reactions and axial forces in the members of a pin-jointed structural system. Under the classical idealization, straight truss members carry axial tension or compression rather than bending and shear.\u003C\u002Fp>\u003Ch2>Ideal truss model\u003C\u002Fh2>\u003Cp>Classical static analysis of a plane truss is based on several assumptions:\u003C\u002Fp>\u003Cul>\u003Cli>joints are ideal frictionless pins;\u003C\u002Fli>\u003Cli>external loads and support reactions act at the joints;\u003C\u002Fli>\u003Cli>member centerlines meet at the joint centers;\u003C\u002Fli>\u003Cli>member self-weight is neglected or converted into equivalent joint loads;\u003C\u002Fli>\u003Cli>each member behaves as a two-force member and carries only an axial force $N$.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>With these assumptions, ideal truss members have no bending moment or shear force. Real joint rigidity, loads applied between joints, and eccentric connections can introduce secondary bending effects.\u003C\u002Fp>\u003Ch2>Spatial structure of a truss\u003C\u002Fh2>\u003Cp>The interactive 3D model below helps visualize joints, longitudinal and diagonal members, spatial bracing, supports, and joint loads. Rotate the model to see how triangulation works beyond a single plane.\u003C\u002Fp>{{chunk:threejs-test-truss}}\u003Ch2>Static determinacy of a plane truss\u003C\u002Fh2>\u003Cp>For a simple plane truss with $m$ members, $j$ joints, and $r$ external reaction components, the necessary counting condition for static determinacy is:\u003C\u002Fp>\u003Cp>$$m+r=2j.$$\u003C\u002Fp>\u003Cp>For the common case of three independent support reactions, this becomes $m=2j-3$. The count alone does not guarantee geometric stability: an unfavorable member arrangement can still form a mechanism.\u003C\u002Fp>\u003Ch2>Truss analysis procedure\u003C\u002Fh2>\u003Col>\u003Cli>\u003Cstrong>Check the structural model.\u003C\u002Fstrong> Identify joints, members, supports, geometry, and applied loads.\u003C\u002Fli>\u003Cli>\u003Cstrong>Calculate support reactions.\u003C\u002Fstrong> Treat the entire truss as a rigid body and apply $\\sum F_x=0$, $\\sum F_y=0$, and $\\sum M=0$.\u003C\u002Fli>\u003Cli>\u003Cstrong>Identify zero-force members.\u003C\u002Fstrong> Recognizing them early can simplify the calculation considerably.\u003C\u002Fli>\u003Cli>\u003Cstrong>Select an analysis method.\u003C\u002Fstrong> Use the method of joints when forces in many members are required; use the method of sections when only a few selected member forces are needed.\u003C\u002Fli>\u003Cli>\u003Cstrong>Use one sign convention.\u003C\u002Fstrong> A convenient approach is to assume each unknown member force is tensile; a negative result then indicates compression.\u003C\u002Fli>\u003Cli>\u003Cstrong>Check equilibrium.\u003C\u002Fstrong> The final member forces must satisfy joint equilibrium and global equilibrium.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Method of joints\u003C\u002Fh2>\u003Cp>In the \u003Cstrong>method of joints\u003C\u002Fstrong>, individual pin joints are isolated successively. All forces acting at a joint are concurrent, so a plane joint provides two independent equilibrium equations:\u003C\u002Fp>\u003Cp>$$\\sum F_x=0,\\qquad \\sum F_y=0.$$\u003C\u002Fp>\u003Cp>After the support reactions are known, begin with a joint containing no more than two unknown member forces. Solve those forces, then move to an adjacent joint where the number of remaining unknowns has been reduced.\u003C\u002Fp>\u003Cp>It is convenient to draw an unknown member force pointing away from the isolated joint, initially assuming tension. If the calculated value is positive, the member is in tension under that convention; if it is negative, the actual member force is compression.\u003C\u002Fp>\u003Ch2>Method of sections\u003C\u002Fh2>\u003Cp>The \u003Cstrong>method of sections\u003C\u002Fstrong> determines selected member forces without solving the complete truss joint by joint. Pass an imaginary cut through the members of interest, isolate one side of the cut, and apply:\u003C\u002Fp>\u003Cp>$$\\sum F_x=0,\\qquad \\sum F_y=0,\\qquad \\sum M=0.$$\u003C\u002Fp>\u003Cp>For a statically determinate plane truss, a useful section normally cuts no more than three members with unknown forces. If the lines of action of two cut-member forces intersect, taking moments about their intersection eliminates both and can give the third member force directly.\u003C\u002Fp>\u003Ch2>Zero-force members\u003C\u002Fh2>\u003Cp>Several common zero-force cases can be recognized without numerical calculation:\u003C\u002Fp>\u003Cul>\u003Cli>if an unloaded joint connects only two non-collinear members, both member forces are zero;\u003C\u002Fli>\u003Cli>if an unloaded joint connects three members and two are collinear, the non-collinear member has zero force.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>A zero-force member is not necessarily unnecessary. It may stabilize the geometry, become active under another load case, or provide construction and bracing functions.\u003C\u002Fp>\u003Ch2>Method of joints or method of sections?\u003C\u002Fh2>\u003Cp>Use the method of joints when the goal is to determine forces throughout most of the truss. The method of sections is usually faster when only one or several particular member forces are required, especially for members far from the supports. In practical calculations, the two methods are often combined.\u003C\u002Fp>\u003Ch2>Common mistakes in truss analysis\u003C\u002Fh2>\u003Cul>\u003Cli>applying distributed load directly to an ideal truss member instead of converting it to joint loads;\u003C\u002Fli>\u003Cli>starting the method of joints at a joint with more than two unknown member forces;\u003C\u002Fli>\u003Cli>confusing a negative calculated force with an error rather than interpreting it as compression under the assumed tension-positive convention;\u003C\u002Fli>\u003Cli>using incorrect direction cosines for inclined members;\u003C\u002Fli>\u003Cli>cutting too many unknown members in the method of sections;\u003C\u002Fli>\u003Cli>treating $m+r=2j$ as sufficient proof that the truss is geometrically stable.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Checking the result\u003C\u002Fh2>\u003Cp>After solving the truss, verify equilibrium at one or more joints that were not used to obtain the corresponding member forces. The complete structure must satisfy global equilibrium, and every isolated joint must satisfy horizontal and vertical force balance. These checks are effective for finding sign, angle, and support-reaction errors.\u003C\u002Fp>",42,[],{"id":1788,"parent_id":1678,"code":70,"slug":1789,"name":1790,"seo_title":1791,"seo_description":1792,"seo_text":1793,"content":1794,"locale":8,"uk_topic_id":1795,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1789,"children":1796},85,"statically-determinate-frames","Statically Determinate Frames","Frame Analysis — Reactions and N, V, M Diagrams Step by Step","Analyze statically determinate plane frames: calculate support reactions and member internal forces, handle rigid joints and hinges, and construct N, V, M diagrams.","Frame analysis determines support reactions and internal axial force, shear force, and bending moment in structural frames. This introduction focuses on statically determinate plane frames, member-by-member equilibrium, rigid-joint behavior, N-V-M diagrams, essential checks, and how the same ideas extend to indeterminate frames.","\u003Cp>\u003Cstrong>Frame analysis\u003C\u002Fstrong> determines the reactions and internal forces that develop in a structural frame under applied loads. Unlike a truss idealization, a frame member generally carries not only axial force but also shear force and bending moment. This page gives a practical introduction to the analysis of plane frames, with emphasis on statically determinate systems.\u003C\u002Fp>\u003Ch2>What is a structural frame?\u003C\u002Fh2>\u003Cp>A frame is an assemblage of members connected at joints. A \u003Cstrong>rigid joint\u003C\u002Fstrong> can transmit force and moment between connected members, so bending is normally an essential part of frame behavior. Internal hinges, when present, release bending moment locally and can divide a frame into parts that are convenient for equilibrium analysis.\u003C\u002Fp>\u003Cp>In a plane-frame model, external loads and the main structural response lie in one plane. The usual internal force resultants in a member are:\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>$N$\u003C\u002Fstrong> — axial (normal) force;\u003C\u002Fli>\u003Cli>\u003Cstrong>$V$\u003C\u002Fstrong> or \u003Cstrong>$Q$\u003C\u002Fstrong> — shear force;\u003C\u002Fli>\u003Cli>\u003Cstrong>$M$\u003C\u002Fstrong> — bending moment.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Statically determinate frame analysis\u003C\u002Fh2>\u003Cp>For a statically determinate plane frame, reactions and internal forces can be obtained from equilibrium alone. For the whole frame, use:\u003C\u002Fp>\u003Cp>$$\\sum F_x=0,\\qquad \\sum F_y=0,\\qquad \\sum M=0.$$\u003C\u002Fp>\u003Cp>The useful feature of frame analysis is that equilibrium can then be applied again to individual members or isolated parts of the structure. Forces acting at a common joint must be equal and opposite on the connected member free-body diagrams.\u003C\u002Fp>\u003Ch2>Step-by-step frame analysis\u003C\u002Fh2>\u003Col>\u003Cli>\u003Cstrong>Idealize the structure.\u003C\u002Fstrong> Identify members, rigid joints, internal hinges, supports, dimensions, and applied loads.\u003C\u002Fli>\u003Cli>\u003Cstrong>Draw the free-body diagram of the entire frame.\u003C\u002Fstrong> Replace supports by their reaction components.\u003C\u002Fli>\u003Cli>\u003Cstrong>Calculate support reactions.\u003C\u002Fstrong> Apply global equilibrium and use internal hinges or other releases when they provide additional useful equilibrium conditions.\u003C\u002Fli>\u003Cli>\u003Cstrong>Separate the frame into members or convenient parts.\u003C\u002Fstrong> Draw a free-body diagram for each part and show the joint forces and moments acting on it.\u003C\u002Fli>\u003Cli>\u003Cstrong>Determine $N$, $V$, and $M$.\u003C\u002Fstrong> Use sections or member equilibrium. Keep one sign convention throughout the calculation.\u003C\u002Fli>\u003Cli>\u003Cstrong>Construct the internal-force diagrams.\u003C\u002Fstrong> Plot axial-force, shear-force, and bending-moment distributions along the members.\u003C\u002Fli>\u003Cli>\u003Cstrong>Check equilibrium and compatibility at joints.\u003C\u002Fstrong> Member-end actions at a joint must balance, and an ideal internal hinge must have zero bending moment.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Internal-force diagrams in frames\u003C\u002Fh2>\u003Cp>The diagrams are drawn along the local axis of each member. This is particularly important at corners: the global horizontal and vertical directions change their role relative to a member's local axial and transverse directions.\u003C\u002Fp>\u003Cp>For a member without a distributed axial load, $N$ is constant between concentrated axial actions. The shear-force diagram changes according to transverse loading, while the bending-moment diagram is related to shear by the usual beam relations. Concentrated forces cause jumps in the corresponding force diagrams; an applied concentrated couple causes a jump in the bending-moment diagram.\u003C\u002Fp>\u003Ch2>Rigid joints and internal hinges\u003C\u002Fh2>\u003Cp>A rigid joint does \u003Cem>not\u003C\u002Fem> imply that the bending moment is zero. It transfers end forces and moments between members. By contrast, an ideal internal hinge cannot transmit bending moment, so $M=0$ at the hinge. The hinge may still transmit force components, and those forces appear with opposite directions on the two separated free bodies.\u003C\u002Fp>\u003Ch2>Common mistakes in analysis of frames\u003C\u002Fh2>\u003Cul>\u003Cli>treating every joint as a pin and therefore incorrectly setting member-end moments to zero;\u003C\u002Fli>\u003Cli>using only the free-body diagram of the complete frame when member equilibrium is also required;\u003C\u002Fli>\u003Cli>mixing global $x$-$y$ directions with a member's local axial and transverse directions;\u003C\u002Fli>\u003Cli>changing the sign convention for $N$, $V$, or $M$ midway through the solution;\u003C\u002Fli>\u003Cli>forgetting that an internal hinge gives a zero-moment condition but can transmit forces;\u003C\u002Fli>\u003Cli>drawing an internal-force diagram that does not satisfy the calculated member-end actions.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Determinate and indeterminate frames\u003C\u002Fh2>\u003Cp>Equilibrium equations are sufficient only for a statically determinate frame. A \u003Cstrong>statically indeterminate frame\u003C\u002Fstrong> has additional unknown reactions or internal actions, so deformation compatibility and member stiffness must also be considered. Common approaches include force methods, displacement methods, slope-deflection and matrix stiffness methods. Those methods are beyond the scope of this introductory page; the equilibrium and member free-body concepts developed here remain their foundation.\u003C\u002Fp>\u003Ch2>Frame deflections\u003C\u002Fh2>\u003Cp>When a displacement or rotation is required, energy methods can be applied after the internal-force state has been established. For example, the unit-load method evaluates the contribution of the real and auxiliary internal-force diagrams:\u003C\u002Fp>{{chunk:mohr-integral-unit-load}}\u003Ch2>Practical checks\u003C\u002Fh2>\u003Cp>A completed structural frame analysis should satisfy global equilibrium, equilibrium of every isolated member or subassembly, action-reaction at connected member ends, zero moment at ideal hinges, and consistency between loads and the shapes of the $N$, $V$, and $M$ diagrams. These checks often reveal sign errors before a numerical result is used for design.\u003C\u002Fp>",36,[],{"id":1798,"parent_id":1678,"code":70,"slug":1799,"name":1800,"seo_title":1801,"seo_description":1802,"seo_text":1803,"content":1804,"locale":8,"uk_topic_id":1805,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1799,"children":1806},306,"arches","Arches","Statically Determinate Arches | Three-Hinged Arches","Analysis of statically determinate arches: three-hinged systems, reactions, horizontal thrust, internal forces, and the funicular axis.","This section covers statically determinate arch systems, especially three-hinged arches: support reactions and horizontal thrust, internal forces, the influence of arch geometry, and the concept of a funicular or rational arch axis.","\u003Cp>An \u003Cstrong>arch\u003C\u002Fstrong> is a curved structural member in which vertical loading commonly produces both vertical support reactions and horizontal thrust. Because of its geometry, an arch can transmit a substantial part of the load through axial compression and may develop much smaller bending moments than a beam spanning the same distance.\u003C\u002Fp>\u003Ch2>Statically determinate arches\u003C\u002Fh2>\u003Cp>The classical example is the three-hinged arch, with hinges at both supports and a third hinge at the crown or another point on the arch axis. The internal hinge cannot transmit bending moment, so the condition $M=0$ provides an additional equilibrium relation for determining the horizontal thrust.\u003C\u002Fp>\u003Ch2>Internal force effects\u003C\u002Fh2>\u003Cp>An arch section generally carries axial force $N$, shear force $V$ or $Q$, and bending moment $M$. Unlike a straight beam, the local section axes rotate along the curved member, so global force components must be resolved into tangential and normal directions.\u003C\u002Fp>\u003Ch2>Geometry and structural action\u003C\u002Fh2>\u003Cp>Bending moment depends on both loading and arch geometry. For a specified load pattern, a rational or funicular axis can be selected so that bending moments vanish or become small and the arch acts predominantly in axial compression.\u003C\u002Fp>",299,[1807,1817,1827],{"id":1808,"parent_id":1798,"code":70,"slug":1809,"name":1810,"seo_title":1811,"seo_description":1812,"seo_text":1813,"content":1814,"locale":8,"uk_topic_id":1815,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1809,"children":1816},344,"three-hinged-arches","Three-Hinged Arches","Three-Hinged Arch — Support Reactions & Horizontal Thrust","Learn how to analyze a three-hinged arch, calculate support reactions and horizontal thrust, and use the zero-moment condition at the crown hinge.","This topic explains the structural model of a three-hinged arch, calculation of vertical reactions and horizontal thrust, and the role of the internal hinge in static determinacy.","\u003Cp>A \u003Cstrong>three-hinged arch\u003C\u002Fstrong> has hinges at both supports and one internal hinge, commonly at the crown. With a proper geometry this system is statically determinate: all support reactions can be obtained from equilibrium without using member stiffness.\u003C\u002Fp>\u003Ch2>Reactions and horizontal thrust\u003C\u002Fh2>\u003Cp>For an arch whose supports are at the same elevation under vertical loading, the vertical reactions can be obtained in the same way as for the corresponding simply supported beam. The horizontal reactions form the thrust $H$, which is found from the zero-moment condition at the internal hinge.\u003C\u002Fp>\u003Ch2>Crown-hinge condition\u003C\u002Fh2>\u003Cp>If the internal hinge has coordinates $(x_c,y_c)$ relative to a support, the moment of all external actions on either isolated half about that hinge must vanish. The equilibrium equation contains the contribution $H y_c$ and can therefore be solved for the horizontal thrust.\u003C\u002Fp>\u003Ch2>Corresponding-beam relation\u003C\u002Fh2>\u003Cp>For equal support elevations and vertical loading, it is useful to compare the arch with a simply supported beam of the same span. At a section $(x,y)$,\u003C\u002Fp>\u003Cp>$$M(x)=M_0(x)-H\\,y(x),$$\u003C\u002Fp>\u003Cp>where $M_0(x)$ is the bending moment in the corresponding beam. This relation shows directly how horizontal thrust reduces bending in the arch.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>After calculating the reactions, verify global equilibrium and $M=0$ at every ideal hinge. For a symmetric arch under symmetric vertical loading, the vertical reactions should also be symmetric.\u003C\u002Fp>",318,[],{"id":1818,"parent_id":1798,"code":70,"slug":1819,"name":1820,"seo_title":1821,"seo_description":1822,"seo_text":1823,"content":1824,"locale":8,"uk_topic_id":1825,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1819,"children":1826},345,"internal-forces-in-arches","Internal Forces in Arches","Internal Forces in Arches — How to Find N, V & M","Calculate axial force N, shear force V, and bending moment M in statically determinate arch sections using local axes and horizontal thrust.","This topic covers N, V\u002FQ, and M in arch sections, transformation from global force components to local section directions, and the influence of horizontal thrust.","\u003Cp>A general section of a planar arch carries three main internal force effects: \u003Cstrong>axial force $N$, shear force $V$ or $Q$, and bending moment $M$\u003C\u002Fstrong>. The distinctive feature of an arch is that the local section directions rotate continuously along its curved axis.\u003C\u002Fp>\u003Ch2>Local coordinate system\u003C\u002Fh2>\u003Cp>At a point on the arch axis, define a tangent direction $t$ and a normal direction $n$. If the tangent forms an angle $\\varphi$ with the global $x$ axis, the resultant force on an isolated part is resolved along $t$ and $n$ to obtain axial and shear components according to the selected sign convention.\u003C\u002Fp>\u003Ch2>Bending moment\u003C\u002Fh2>\u003Cp>The bending moment at a section follows from moment equilibrium of either part of the arch. For a three-hinged arch with supports at the same elevation and vertical loading, a convenient relation is\u003C\u002Fp>\u003Cp>$$M(x)=M_0(x)-H\\,y(x),$$\u003C\u002Fp>\u003Cp>where $M_0$ is the moment in the corresponding beam, $H$ is horizontal thrust, and $y$ is the arch ordinate.\u003C\u002Fp>\u003Ch2>Axial force and shear\u003C\u002Fh2>\u003Cp>Once the global horizontal and vertical components of the section resultant are known, they are projected onto the tangent and normal. Consequently, the same global force contributes differently to $N$ and $V$ at different locations along the arch.\u003C\u002Fp>\u003Ch2>Structural interpretation\u003C\u002Fh2>\u003Cp>A well-shaped arch carries a large share of its load through axial compression. As the arch axis approaches the funicular shape for a given loading, bending moment and usually bending deformation decrease.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>Bending moment must be zero at ideal hinges. Symmetric geometry and loading should produce the corresponding symmetry or antisymmetry in the internal-force distributions. At every section, the calculated actions must equilibrate the isolated portion of the arch.\u003C\u002Fp>",319,[],{"id":1828,"parent_id":1798,"code":70,"slug":1829,"name":1830,"seo_title":1831,"seo_description":1832,"seo_text":1833,"content":1834,"locale":8,"uk_topic_id":1835,"show_in_theory_list":78,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1829,"children":1836},346,"funicular-arch","Funicular Arch","Funicular Arch: Shape, Equation, Thrust Line & Examples","Learn what a funicular arch is, how its shape follows the thrust line, why bending moment becomes zero, and when the ideal arch is parabolic or catenary.","A funicular arch is shaped so that, for a specified loading, the internal thrust follows the arch axis and bending is eliminated. This guide explains the funicular shape, thrust line, zero-moment equation, parabolic arches under uniform vertical load, the distinction between funicular and catenary arches, and the effect of changing the load pattern.","\u003Cp\r\n >A \u003Cstrong>funicular arch\u003C\u002Fstrong> is an arch whose axis is shaped to carry a specified loading by\r\n axial compression without bending. For that particular load case, the line of thrust coincides\r\n with the arch axis and the bending moment is zero along the idealized arch. The same idea is also\r\n called a \u003Cstrong>rational arch axis\u003C\u002Fstrong>.\u003C\u002Fp\r\n>\u003Ch2>What is a funicular arch?\u003C\u002Fh2\r\n>\u003Cp\r\n >An arch develops bending whenever the internal thrust does not pass through its structural axis.\r\n A funicular form removes this eccentricity for a chosen loading. As a result, the idealized arch\r\n carries the load through compression rather than a combination of compression and bending.\u003C\u002Fp\r\n>\u003Cp\r\n >The word \u003Cem>funicular\u003C\u002Fem> emphasizes that the required geometry depends on the loading. There\r\n is no single funicular shape that is optimal for every possible load pattern.\u003C\u002Fp\r\n>\u003Ch2>Funicular arch and the thrust line\u003C\u002Fh2\r\n>\u003Cp\r\n >For a three-hinged arch with supports at the same elevation under vertical loading, the bending\r\n moment at a section can be written as\u003C\u002Fp\r\n>\u003Cp>$$M(x)=M_0(x)-H\\,y(x),$$\u003C\u002Fp\r\n>\u003Cp\r\n >where $M_0(x)$ is the bending moment in the corresponding simply supported beam, $H$ is the\r\n horizontal thrust, and $y(x)$ is the ordinate of the arch axis above the support line.\u003C\u002Fp\r\n>\u003Cp>For a perfectly funicular axis,\u003C\u002Fp>\u003Cp>$$M(x)=0,$$\u003C\u002Fp>\u003Cp>so that\u003C\u002Fp\r\n>\u003Cp>$$y(x)=\\frac{M_0(x)}{H}.$$\u003C\u002Fp\r\n>\u003Cp\r\n >This result gives a useful interpretation:\r\n \u003Cstrong\r\n >the funicular arch shape is proportional to the bending-moment diagram of the corresponding\r\n beam for the same loading\u003C\u002Fstrong\r\n >.\u003C\u002Fp\r\n>\u003Cp>In the comparison below, all four systems have the same span $L=12\\,\\text{m}$ and rise $f=3\\,\\text{m}$. Only the loading changes. The axis ordinates are calculated from $y(x)=M_0(x)\u002FH$, with the horizontal thrust selected so that each form reaches the specified rise.\u003C\u002Fp>\n{{chunk:funicular-shapes-by-loading-en}}\n\u003Cp>A single point load produces a triangular $M_0$ diagram, so the funicular axis degenerates into two straight members. Two point loads produce three straight segments. A vertical load uniform over the horizontal projection produces a parabola. Adding a point load at midspan leaves two parabolic branches but creates a kink at the load point, with a slope jump $\\Delta y'=P\u002FH$.\u003C\u002Fp>\n\u003Cp>\u003Cstrong>Effect of self-weight.\u003C\u002Fstrong> The forms shown above correspond to the specified external loads. In masonry and reinforced-concrete arches, self-weight is often a major part of the total loading, so a strictly funicular axis must be determined for the external load plus self-weight. For an arch of constant thickness, self-weight per unit of horizontal projection is generally not uniform; it increases on steeper portions approximately in proportion to $\\sqrt{1+(y')^2}$. The initial parabola therefore requires a small correction. In practice, the geometry is refined iteratively: assume an initial axis, calculate its self-weight, rebuild the thrust line, and repeat.\u003C\u002Fp>\u003Ch2>Funicular arch shape\u003C\u002Fh2\r\n>\u003Cp\r\n >The ideal shape follows directly from the load distribution. Different loads produce different\r\n beam moment diagrams and therefore different funicular curves. A geometry that gives zero bending\r\n for one loading will generally develop bending when the loading changes.\u003C\u002Fp\r\n>\u003Ch2>Parabolic funicular arch\u003C\u002Fh2\r\n>\u003Cp\r\n >For a vertical load uniformly distributed over the \u003Cstrong>horizontal projection\u003C\u002Fstrong> of a\r\n span $L$, the corresponding funicular curve is a parabola. If the rise at midspan is $f$, the axis\r\n can be written as\u003C\u002Fp\r\n>\u003Cp>$$y(x)=\\frac{4f}{L^2}x(L-x).$$\u003C\u002Fp\r\n>\u003Cp>For a three-hinged arch under this loading, the horizontal thrust is\u003C\u002Fp>\u003Cp>$H=\\frac{qL^2}{8f},$\u003C\u002Fp>\u003Cp>where $q$ is the uniform vertical load per unit of horizontal length.\u003C\u002Fp>\u003Ch2>Funicular arch vs catenary arch\u003C\u002Fh2\r\n>\u003Cp\r\n >A funicular arch is \u003Cstrong>not necessarily a catenary\u003C\u002Fstrong>. “Funicular” describes the\r\n relationship between shape and loading; “catenary” names a particular mathematical curve. An\r\n inverted catenary is funicular for the loading associated with a freely hanging uniform cable\r\n under its own weight. By contrast, a uniform vertical load specified per unit of horizontal\r\n projection produces a parabolic funicular form.\u003C\u002Fp\r\n>\u003Cp\r\n >Therefore, statements such as “the ideal arch is always a catenary” are incomplete unless the\r\n load distribution is also specified.\u003C\u002Fp\r\n>\u003Ch2>Funicular arch vs an ordinary arch\u003C\u002Fh2\r\n>\u003Cp\r\n >An ordinary arch may have a circular, parabolic, catenary, pointed, or other axis. Its geometric\r\n name alone does not tell us whether it is funicular. The arch is funicular only when its axis\r\n matches the thrust line for the load case being considered.\u003C\u002Fp\r\n>\u003Cp\r\n >If the thrust line departs from the axis, an eccentricity develops and the arch must resist\r\n bending in addition to axial force.\u003C\u002Fp\r\n>\u003Ch2>Why does the funicular shape depend on loading?\u003C\u002Fh2\r\n>\u003Cp\r\n >The zero-moment condition contains $M_0(x)$, which is determined by the applied loads. Changing\r\n the magnitude, position, or distribution of the load changes $M_0(x)$ and therefore changes the\r\n required funicular axis.\u003C\u002Fp\r\n>\u003Cp\r\n >This is especially important for real structures. An arch may be nearly funicular under permanent\r\n loads but experience bending under asymmetric live load, wind, temperature effects, support\r\n movement, or other load cases.\u003C\u002Fp\r\n>\u003Ch2>Worked example: horizontal thrust\u003C\u002Fh2\r\n>\u003Cp\r\n >Consider a three-hinged parabolic arch with span $L=20\\,\\text{m}$, rise $f=5\\,\\text{m}$, and a\r\n uniformly distributed vertical load $q=10\\,\\text{kN\u002Fm}$ over the horizontal projection.\u003C\u002Fp\r\n>\u003Cp>Using\u003C\u002Fp>\u003Cp>$$H=\\frac{qL^2}{8f},$$\u003C\u002Fp>\u003Cp>we obtain\u003C\u002Fp\r\n>\u003Cp>$$H=\\frac{10\\times20^2}{8\\times5}=100\\,\\text{kN}.$$\u003C\u002Fp\r\n>\u003Cp\r\n >For this ideal load case, the parabolic axis is funicular, so the bending moment is zero\r\n throughout the idealized three-hinged arch while the supports must resist a horizontal thrust of\r\n $100\\,\\text{kN}$.\u003C\u002Fp\r\n>\u003Ch2>Engineering significance and limitations\u003C\u002Fh2\r\n>\u003Cp\r\n >Funicular action can make an arch structurally efficient because it reduces bending and makes\r\n greater use of compression. This is particularly advantageous for materials and structural systems\r\n that perform well in compression.\u003C\u002Fp\r\n>\u003Cp\r\n >However, reducing bending in the arch does not eliminate structural demands. Horizontal thrust\r\n must be transferred into abutments, foundations, or ties. In addition, a shape optimized for one\r\n load case may not remain funicular under other actions, so practical arch design requires checking\r\n all relevant load combinations and stability effects.\u003C\u002Fp\r\n>\u003Ch2>Frequently asked questions\u003C\u002Fh2>\u003Ch3>What is a funicular arch?\u003C\u002Fh3\r\n>\u003Cp\r\n >It is an arch whose axis coincides with the thrust line for a specified loading, producing axial\r\n compression without bending in the idealized model.\u003C\u002Fp\r\n>\u003Ch3>Why is the bending moment zero in a funicular arch?\u003C\u002Fh3\r\n>\u003Cp\r\n >Because the compressive resultant passes through the arch axis. With no eccentricity between the\r\n thrust line and the axis, the load does not create a bending moment about the axis.\u003C\u002Fp\r\n>\u003Ch3>Is a funicular arch always a catenary?\u003C\u002Fh3\r\n>\u003Cp\r\n >No. The funicular curve depends on how the load is distributed. A catenary is one particular\r\n funicular form; a uniformly distributed vertical load over the horizontal projection gives a\r\n parabola.\u003C\u002Fp\r\n>\u003Ch3>What is the ideal shape of an arch?\u003C\u002Fh3\r\n>\u003Cp\r\n >There is no universal ideal shape independent of loading. For minimum bending, the ideal axis\r\n follows the thrust line for the governing load case or is selected to perform acceptably across\r\n the relevant combination of load cases.\u003C\u002Fp\r\n>\r\n",320,[],{"id":1838,"parent_id":1678,"code":70,"slug":1839,"name":1840,"seo_title":1841,"seo_description":1842,"seo_text":1843,"content":1844,"locale":8,"uk_topic_id":1845,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1839,"children":1846},364,"influence-lines","Influence Lines","Influence Lines in Structural Analysis — Moving Loads Guide","Learn influence lines for reactions, shear, bending moments, and truss forces under moving loads, including construction methods and the Müller-Breslau principle.","This section covers influence lines for statically determinate structures: moving loads, influence lines for reactions, shear forces and bending moments, static and kinematic construction methods, the Müller-Breslau principle, loading of influence lines, hinged multispan beams, and trusses.","\u003Cp>An \u003Cstrong>influence line\u003C\u002Fstrong> shows how a selected reaction, internal force, or other structural response changes as a unit moving load occupies successive positions on the structure. Influence lines are fundamental for bridges, crane girders, and other systems subjected to loads whose position is not fixed.\u003C\u002Fp>\u003Ch2>Influence line versus internal-force diagram\u003C\u002Fh2>\u003Cp>A shear or bending-moment diagram gives an internal force at different sections for a \u003Cem>fixed\u003C\u002Fem> load system. An influence line instead refers to one selected response quantity and gives its value as a function of the \u003Cem>position of a moving load\u003C\u002Fem>.\u003C\u002Fp>\u003Ch2>Construction\u003C\u002Fh2>\u003Cp>For statically determinate structures, influence-line ordinates can be obtained directly from equilibrium as a unit load moves across the structure. A kinematic approach relates the influence-line shape to the displacement mechanism created by releasing the restraint associated with the response quantity.\u003C\u002Fp>\u003Ch2>Use with real loads\u003C\u002Fh2>\u003Cp>Once the influence line is known, the effect of concentrated loads is obtained by summing each load multiplied by the corresponding ordinate. Distributed loads are handled by integration, or by signed influence-line areas when the load intensity is constant.\u003C\u002Fp>\u003Ch2>Critical load position\u003C\u002Fh2>\u003Cp>The sign and shape of an influence line indicate where moving concentrated or distributed loads should be placed to maximize or minimize the selected structural response.\u003C\u002Fp>",363,[1847,1857,1867,1877,1887,1897,1907],{"id":1848,"parent_id":1838,"code":70,"slug":1849,"name":1850,"seo_title":1851,"seo_description":1852,"seo_text":1853,"content":1854,"locale":8,"uk_topic_id":1855,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1849,"children":1856},372,"influence-lines-moving-loads","Influence Lines and Moving Loads","Influence Lines and Moving Loads | Fundamentals","Definition of an influence line, its difference from an internal-force diagram, and the response to a moving unit load.","This topic introduces the influence line as a graph of a selected response quantity as a unit load moves across a structure and explains how it differs from a diagram produced by a fixed load system.","\u003Cp>An \u003Cstrong>influence line\u003C\u002Fstrong> for a response $S$ is the graph of $S(x)$ as a unit load moves along a specified path on the structure. The response may be a support reaction, shear force, bending moment, or axial force in a particular member.\u003C\u002Fp>\u003Ch2>What changes during construction\u003C\u002Fh2>\u003Cp>The reaction, section, or member for which the influence line is constructed remains fixed. Only the coordinate of the moving load changes. This is the essential distinction between an influence line and an ordinary internal-force diagram.\u003C\u002Fp>\u003Ch2>Unit load\u003C\u002Fh2>\u003Cp>In the classical construction, a dimensionless unit force is moved across the structure. The ordinate at each load position is numerically equal, including sign, to the selected response caused by that unit force.\u003C\u002Fp>\u003Ch2>Superposition\u003C\u002Fh2>\u003Cp>For a linear structural system, once the influence line is known, a set of concentrated loads gives\u003C\u002Fp>\u003Cp>$$S=\\sum_i P_i y_i,$$\u003C\u002Fp>\u003Cp>where $y_i$ is the influence-line ordinate beneath $P_i$. For a distributed load,\u003C\u002Fp>\u003Cp>$$S=\\int q(x)y(x)\\,dx.$$\u003C\u002Fp>\u003Ch2>Sign of an ordinate\u003C\u002Fh2>\u003Cp>A positive ordinate means that a unit load at that position produces a positive value of the selected response under the adopted sign convention; a negative ordinate produces the opposite response.\u003C\u002Fp>",365,[],{"id":1858,"parent_id":1838,"code":70,"slug":1859,"name":1860,"seo_title":1861,"seo_description":1862,"seo_text":1863,"content":1864,"locale":8,"uk_topic_id":1865,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1859,"children":1866},373,"influence-lines-beam-support-reactions","Influence Lines for Beam Support Reactions","Beam Support-Reaction Influence Lines — Step-by-Step Construction","Learn how to construct influence lines for vertical support reactions of statically determinate beams under a moving unit load and determine characteristic ordinates.","This topic covers analytical construction of support-reaction influence lines for beams under a moving unit load, characteristic ordinates, and the use of equilibrium equations.","\u003Cp>For a simply supported beam, support-reaction influence lines follow directly from equilibrium as a unit load moves across the span. Let the span be $L$ and let the unit load be at distance $x$ from the left support $A$.\u003C\u002Fp>\u003Ch2>Left support reaction\u003C\u002Fh2>\u003Cp>Taking moments about the right support gives\u003C\u002Fp>\u003Cp>$$R_A(x)=\\frac{L-x}{L}.$$\u003C\u002Fp>\u003Cp>The influence line for $R_A$ is therefore a straight line from ordinate $1$ at $A$ to $0$ at the right support $B$.\u003C\u002Fp>\u003Ch2>Right support reaction\u003C\u002Fh2>\u003Cp>Similarly,\u003C\u002Fp>\u003Cp>$$R_B(x)=\\frac{x}{L}.$$\u003C\u002Fp>\u003Cp>Its influence line rises linearly from $0$ at $A$ to $1$ at $B$.\u003C\u002Fp>\u003Ch2>Equilibrium check\u003C\u002Fh2>\u003Cp>At every position of the unit vertical load,\u003C\u002Fp>\u003Cp>$$R_A(x)+R_B(x)=1,$$\u003C\u002Fp>\u003Cp>which follows immediately from vertical force equilibrium.\u003C\u002Fp>\u003Ch2>Real moving loads\u003C\u002Fh2>\u003Cp>For concentrated loads $P_i$, a reaction is $R=\\sum P_i y_i$, where $y_i$ are the corresponding influence-line ordinates. For a uniform distributed load, the reaction equals the load intensity multiplied by the signed area of the loaded portion of the influence line.\u003C\u002Fp>",366,[],{"id":1868,"parent_id":1838,"code":70,"slug":1869,"name":1870,"seo_title":1871,"seo_description":1872,"seo_text":1873,"content":1874,"locale":8,"uk_topic_id":1875,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1869,"children":1876},374,"influence-lines-shear-bending-moment","Influence Lines for Shear and Bending Moment","Influence Lines for Shear & Bending Moment — Beam Section V and M","Learn how to construct influence lines for shear V and bending moment M at a specified beam section, determine key ordinates, and handle the shear discontinuity.","This topic covers influence lines for shear and bending moment at a beam section, characteristic ordinates, the discontinuity of the shear influence line, and equilibrium-based construction.","\u003Cp>Influence lines for \u003Cstrong>shear force $V$ and bending moment $M$\u003C\u002Fstrong> are constructed for one fixed beam section. Let section $C$ be located a distance $a$ from the left support and $b=L-a$ from the right support.\u003C\u002Fp>\u003Ch2>Bending-moment influence line\u003C\u002Fh2>\u003Cp>As a unit load moves across a simply supported beam, the bending moment at $C$ varies linearly on either side of the section. The ordinate directly beneath $C$ is\u003C\u002Fp>\u003Cp>$$y_C=\\frac{ab}{L}.$$\u003C\u002Fp>\u003Cp>The ordinates at both supports are zero, giving a triangular influence line for $M_C$.\u003C\u002Fp>\u003Ch2>Shear influence line\u003C\u002Fh2>\u003Cp>The influence line for $V_C$ has a discontinuity at section $C$. As the unit load crosses the section, the shear value changes by one unit, producing two linear branches separated by a unit jump.\u003C\u002Fp>\u003Ch2>Origin of the jump\u003C\u002Fh2>\u003Cp>Shear at a section depends on which side of the cut contains the moving force. At the instant the unit load passes through $C$, it transfers from one isolated free body to the other, which produces the unit discontinuity.\u003C\u002Fp>\u003Ch2>Use with moving loads\u003C\u002Fh2>\u003Cp>To calculate $M_C$ or $V_C$ from a real moving load system, multiply each load by the corresponding influence-line ordinate and sum the products. Large loads placed near large positive moment ordinates tend to produce large positive bending moments at the section.\u003C\u002Fp>",367,[],{"id":1878,"parent_id":1838,"code":70,"slug":1879,"name":1880,"seo_title":1881,"seo_description":1882,"seo_text":1883,"content":1884,"locale":8,"uk_topic_id":1885,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1879,"children":1886},375,"kinematic-method-muller-breslau-principle","Kinematic Method and Müller-Breslau Principle","Müller-Breslau Principle | Kinematic Method","Kinematic construction of influence lines and the Müller-Breslau principle for reactions and internal force effects.","This topic explains the kinematic approach to influence lines by releasing the corresponding restraint, imposing a unit generalized displacement, and using the Müller-Breslau principle to determine the influence-line shape.","\u003Cp>The \u003Cstrong>kinematic method for influence lines\u003C\u002Fstrong> is based on the Müller-Breslau principle. It determines the influence-line shape without moving a unit force through many positions and is particularly useful for qualitative analysis of more complex structures.\u003C\u002Fp>\u003Ch2>Müller-Breslau principle\u003C\u002Fh2>\u003Cp>To obtain the influence line for a reaction or internal force effect, release the restraint associated with that response and impose a small positive generalized displacement in its positive direction. The resulting compatible displacement shape of the released system is proportional to the influence line.\u003C\u002Fp>\u003Ch2>Support reaction\u003C\u002Fh2>\u003Cp>For a vertical support reaction, remove the corresponding vertical restraint and impose a unit vertical displacement in the positive reaction direction. The kinematically admissible displaced shape indicates the sign and geometry of the influence line.\u003C\u002Fp>\u003Ch2>Bending moment and shear\u003C\u002Fh2>\u003Cp>For bending moment at a section, introduce a hinge at that section, thereby releasing moment transfer, and impose a positive relative rotation of the two parts. For shear, introduce the corresponding relative transverse displacement across the cut.\u003C\u002Fp>\u003Ch2>Statically determinate structures\u003C\u002Fh2>\u003Cp>Releasing one restraint in a determinate structure creates a mechanism composed of rigid parts, so its displacement shape is often piecewise linear. The method gives shape and sign; the scale follows from the imposed unit generalized displacement.\u003C\u002Fp>",368,[],{"id":1888,"parent_id":1838,"code":70,"slug":1889,"name":1890,"seo_title":1891,"seo_description":1892,"seo_text":1893,"content":1894,"locale":8,"uk_topic_id":1895,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":172,"url_slug":1889,"children":1896},376,"loading-influence-lines","Loading Influence Lines","Loading Influence Lines | Concentrated and Distributed Loads","Using influence lines with concentrated loads, moving load systems, and distributed loads to calculate structural response.","This topic covers loading influence lines with concentrated forces, groups of moving forces, and distributed loads, using ordinates and signed areas and identifying unfavorable load positions.","\u003Cp>Once an influence line has been constructed, it can be used to evaluate the response to \u003Cstrong>real moving loads\u003C\u002Fstrong>. A single influence line can therefore be reused for many load positions and combinations.\u003C\u002Fp>\u003Ch2>Concentrated loads\u003C\u002Fh2>\u003Cp>For loads $P_i$ located over influence-line ordinates $y_i$,\u003C\u002Fp>\u003Cp>$$S=\\sum_i P_i y_i.$$\u003C\u002Fp>\u003Cp>The ordinates are algebraic. Loads acting over a negative portion reduce a positive response or produce a negative response.\u003C\u002Fp>\u003Ch2>Distributed load\u003C\u002Fh2>\u003Cp>For an intensity $q(x)$,\u003C\u002Fp>\u003Cp>$$S=\\int q(x)y(x)\\,dx.$$\u003C\u002Fp>\u003Cp>If $q$ is constant over the loaded interval, $S=qA$, where $A$ is the signed area of the loaded part of the influence line.\u003C\u002Fp>\u003Ch2>Critical load position\u003C\u002Fh2>\u003Cp>To maximize a positive response, a freely placeable distributed load is applied over positive influence-line regions; to maximize a negative response, it is applied over negative regions. For a moving train of concentrated loads, the critical position depends on both load magnitudes and influence-line geometry.\u003C\u002Fp>\u003Ch2>Loads with fixed spacing\u003C\u002Fh2>\u003Cp>When a group of loads moves as one system, their mutual spacing remains fixed. Evaluate characteristic positions, especially when one of the loads crosses a peak or a point where the influence-line slope changes.\u003C\u002Fp>\u003Ch2>Sign check\u003C\u002Fh2>\u003Cp>Before summing numerically, identify which loads act on positive and negative regions. This gives an immediate expectation for the sign of the result and helps detect ordinate-selection errors.\u003C\u002Fp>",369,[],{"id":1898,"parent_id":1838,"code":70,"slug":1899,"name":1900,"seo_title":1901,"seo_description":1902,"seo_text":1903,"content":1904,"locale":8,"uk_topic_id":1905,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":183,"url_slug":1899,"children":1906},377,"influence-lines-hinged-multispan-beams","Influence Lines in Hinged Multispan Beams","Influence Lines for Gerber Beams — Hinged Multispan Systems","Learn how to construct influence lines for reactions and internal forces in Gerber beams using structural hierarchy and moving-load transfer through internal hinges.","This topic covers influence lines in Gerber and other hinged multispan beam systems, transfer of moving-load effects through the structural hierarchy, and construction of characteristic influence-line segments.","\u003Cp>In \u003Cstrong>hinged multispan beams\u003C\u002Fstrong>, a moving load may act on one component while the response of interest occurs in another. Influence-line construction therefore relies on the structural hierarchy and on force transfer through internal hinges.\u003C\u002Fp>\u003Ch2>Structural hierarchy\u003C\u002Fh2>\u003Cp>First identify supporting and suspended beam components. When the unit load is on a suspended member, that member's reactions are transmitted to lower supporting members as concentrated forces. The effect of the moving load therefore propagates through the structural hierarchy.\u003C\u002Fp>\u003Ch2>Piecewise construction\u003C\u002Fh2>\u003Cp>Within each component of a statically determinate system, influence-line ordinates follow equilibrium relations and are often linear. Values at supports, hinges, and ends of suspended members define the characteristic segments of the complete influence line.\u003C\u002Fp>\u003Ch2>Transfer through a hinge\u003C\u002Fh2>\u003Cp>An internal hinge does not transmit bending moment but does transmit force. A load on one beam segment can therefore influence reactions and internal forces in another through the hinge force. The spans cannot always be analyzed as independent beams.\u003C\u002Fp>\u003Ch2>Verification\u003C\u002Fh2>\u003Cp>Place the unit load at several characteristic positions and calculate the selected response independently from equilibrium. These values should match the ordinates of the constructed influence line.\u003C\u002Fp>",370,[],{"id":1908,"parent_id":1838,"code":70,"slug":1909,"name":1910,"seo_title":1911,"seo_description":1912,"seo_text":1913,"content":1914,"locale":8,"uk_topic_id":1915,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":1909,"children":1916},378,"influence-lines-truss-member-forces","Influence Lines for Truss Member Forces","Influence Lines for Truss Member Forces — Sections & Joints","Learn how to construct influence lines for axial forces in statically determinate truss members under moving joint loads using section and joint equilibrium methods.","This topic covers influence lines for axial member forces in trusses, transfer of a moving load through panel joints, and use of section and joint equilibrium methods.","\u003Cp>In trusses, influence lines are used to determine \u003Cstrong>axial member forces caused by moving loads\u003C\u002Fstrong>. In the classical truss model, loads enter the truss through joints, so a load moving between adjacent panel points is transferred to those joints through the deck system or an equivalent linear distribution.\u003C\u002Fp>\u003Ch2>Joint-load transfer\u003C\u002Fh2>\u003Cp>When a unit load lies between two adjacent panel points, its effect on the truss is replaced by two joint loads whose magnitudes vary linearly with load position. The influence line for a member force is therefore normally linear between panel points.\u003C\u002Fp>\u003Ch2>Method of sections\u003C\u002Fh2>\u003Cp>Pass a section through the member of interest and write equilibrium for one part of the truss. Consider the unit load separately in the characteristic regions of its path. A moment equation can often isolate one cut-member force directly.\u003C\u002Fp>\u003Ch2>Method of joints\u003C\u002Fh2>\u003Cp>If influence lines are needed for several members meeting near one joint, determine the joint loads for characteristic unit-load positions and apply joint equilibrium successively.\u003C\u002Fp>\u003Ch2>Force sign\u003C\u002Fh2>\u003Cp>With tension taken as positive, positive influence-line ordinates represent tension and negative ordinates compression. If the influence line changes sign, the same member may change from tension to compression as the load moves.\u003C\u002Fp>\u003Ch2>Application\u003C\u002Fh2>\u003Cp>After the influence line is known, the extreme force from a train of moving loads is found by placing the loads on regions that are most favorable in sign and magnitude. This is particularly important in bridge trusses and similar structures subjected to traffic loads.\u003C\u002Fp>",371,[],{"id":1918,"parent_id":1678,"code":70,"slug":1919,"name":1920,"seo_title":1921,"seo_description":1922,"seo_text":1923,"content":1924,"locale":8,"uk_topic_id":1925,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":880,"url_slug":1919,"children":1926},307,"displacements-in-framed-structures","Displacements in Framed Structures","Structural Displacements | Unit-Load and Mohr Integral Methods","Displacements in beams, frames, and trusses using strain energy, the unit-load method, Mohr integrals, and graphical integration methods.","Energy methods for linear and angular displacements in framed structures: external and internal work, strain energy, Mohr integrals, the unit-load method, and graphical techniques for evaluating displacement integrals.","\u003Cp>\u003Cstrong>Displacements in framed structures\u003C\u002Fstrong> are translations of points and rotations of sections caused by member deformation. Their calculation is required for serviceability checks, compatibility equations in indeterminate structures, and interpretation of the deformed shape.\u003C\u002Fp>\u003Ch2>Types of displacement\u003C\u002Fh2>\u003Cp>A linear displacement is defined for a specified point and direction; a rotation is defined for a specified section and positive sense. The sign of the calculated result is interpreted relative to that chosen direction.\u003C\u002Fp>\u003Ch2>Energy approach\u003C\u002Fh2>\u003Cp>For a linear-elastic structure, external work is related to strain energy. This relation allows displacement calculations to be expressed through internal axial force, bending moment, shear, torsion, and the corresponding member stiffnesses.\u003C\u002Fp>\u003Ch2>Unit-load method\u003C\u002Fh2>{{chunk:mohr-integral-unit-load}}\u003Cp>In beams and frames, bending deformation often provides the dominant contribution. In ideal pin-jointed trusses, axial member deformation is normally the principal contribution.\u003C\u002Fp>\u003Ch2>Graphical evaluation\u003C\u002Fh2>{{chunk:vereshchagin-diagram-multiplication}}\u003Ch2>Other sources of displacement\u003C\u002Fh2>\u003Cp>Displacements can also result from temperature change, temperature gradients, prescribed support movements, and settlements. In statically indeterminate structures these imposed deformations may also generate additional internal forces.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>Verify dimensions, signs, the direction of the unit action, the consistency of real and unit-load diagrams, and the stiffness used on each segment. A negative result means that the actual displacement is opposite to the assumed unit-load direction.\u003C\u002Fp>",300,[1927,1937,1947,1957],{"id":1928,"parent_id":1918,"code":70,"slug":1929,"name":1930,"seo_title":1931,"seo_description":1932,"seo_text":1933,"content":1934,"locale":8,"uk_topic_id":1935,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1929,"children":1936},347,"work-and-strain-energy","Work and Strain Energy","Work & Strain Energy — Formulas for Beams, Frames & Trusses","Learn external work and strain energy in framed structures, including axial and bending energy formulas used as the basis for structural displacement methods.","This topic introduces external and internal work and strain energy in framed structures and provides the energy foundation for displacement methods.","\u003Cp>\u003Cstrong>Work and strain energy\u003C\u002Fstrong> provide the energy basis for many structural-analysis methods. In an elastic structure, work done by external actions is stored as strain energy in the members.\u003C\u002Fp>\u003Ch2>Work of a gradually applied force\u003C\u002Fh2>\u003Cp>If a force $P$ increases from zero to its final value and the structure is linear elastic, with final displacement $\\delta$ in the force direction,\u003C\u002Fp>\u003Cp>$$A=\\frac12 P\\delta.$$\u003C\u002Fp>\u003Cp>The factor $1\u002F2$ appears because force and displacement increase proportionally during gradual loading.\u003C\u002Fp>\u003Ch2>Member strain energy\u003C\u002Fh2>\u003Cp>For an elastic member, strain energy can be expressed through internal force effects. Typical contributions are\u003C\u002Fp>\u003Cp>$$U_N=\\int\\frac{N^2}{2EA}\\,dx,\\qquad U_M=\\int\\frac{M^2}{2EI}\\,dx.$$\u003C\u002Fp>\u003Cp>Torsional and shear strain-energy terms can be added when they are included in the structural model.\u003C\u002Fp>\u003Ch2>Energy balance\u003C\u002Fh2>\u003Cp>For quasistatic elastic loading without energy loss, external work equals the stored strain energy. This provides a way to relate loads and displacements without directly integrating the differential equation of the elastic curve.\u003C\u002Fp>\u003Ch2>Virtual and reciprocal work\u003C\u002Fh2>\u003Cp>For linear-elastic systems, reciprocal-work relations provide the basis of the unit-load method: a real internal-force state is combined with an auxiliary unit state, and products of corresponding internal forces are integrated along the members.\u003C\u002Fp>\u003Ch2>Applications\u003C\u002Fh2>\u003Cp>Energy relations are used to calculate deflections and rotations, coefficients of force-method compatibility equations, and independent checks of structural calculations.\u003C\u002Fp>",321,[],{"id":1938,"parent_id":1918,"code":70,"slug":1939,"name":1940,"seo_title":1941,"seo_description":1942,"seo_text":1943,"content":1944,"locale":8,"uk_topic_id":1945,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1939,"children":1946},348,"mohr-integral-unit-load-method","Mohr Integral and Unit-Load Method","Mohr Integral & Unit-Load Method — Structural Deflection Formula","Calculate beam, frame, and truss displacements with the Mohr integral and unit-load method: build the auxiliary unit state, integrate internal-force products, and interpret the sign.","This topic explains the unit-load method and Mohr integral for structural displacements, including construction of the auxiliary unit state and integration of real and unit internal-force functions.","\u003Cp>The \u003Cstrong>Mohr integral and unit-load method\u003C\u002Fstrong> determine a translation or rotation by combining internal force effects from two structural states: the real loaded state and an auxiliary unit state.\u003C\u002Fp>{{chunk:mohr-integral-unit-load}}\u003Ch2>Procedure\u003C\u002Fh2>\u003Col>\u003Cli>Specify the point, direction, and type of required displacement.\u003C\u002Fli>\u003Cli>Determine internal forces caused by the real loading.\u003C\u002Fli>\u003Cli>Create the unit state: apply a unit force for a translation or a unit moment for a rotation.\u003C\u002Fli>\u003Cli>Construct the corresponding unit-state internal-force diagrams.\u003C\u002Fli>\u003Cli>Multiply corresponding real and unit internal-force functions and divide by the relevant stiffness on each segment.\u003C\u002Fli>\u003Cli>Integrate and sum the contributions of all members.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Beams and frames\u003C\u002Fh2>\u003Cp>For slender beams and frames, the bending term $\\int M\\bar M\u002F(EI)\\,dx$ is often sufficient. Axial, shear, and torsional terms must be added when their deformations are significant.\u003C\u002Fp>\u003Ch2>Trusses\u003C\u002Fh2>\u003Cp>For an ideal truss with constant $N_i$, $\\bar N_i$, $E_i$, and $A_i$ in each member, the integral reduces to\u003C\u002Fp>\u003Cp>$$\\delta=\\sum_i\\frac{N_i\\bar N_iL_i}{E_iA_i}.$$\u003C\u002Fp>\u003Ch2>Sign\u003C\u002Fh2>\u003Cp>A positive result means displacement in the direction of the unit force or moment. A negative result means the actual displacement is opposite to the assumed unit-state direction.\u003C\u002Fp>",322,[],{"id":1948,"parent_id":1918,"code":70,"slug":1949,"name":1950,"seo_title":1951,"seo_description":1952,"seo_text":1953,"content":1954,"locale":8,"uk_topic_id":1955,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1949,"children":1956},349,"vereshchagin-rule","Vereshchagin Rule","Vereshchagin Rule for Structural Displacements","Graphical multiplication of diagrams using the Vereshchagin rule to evaluate displacement integrals in framed structures.","This topic explains graphical multiplication of internal-force diagrams using the Vereshchagin rule, its applicability conditions, and its use in structural displacement calculations.","\u003Cp>The \u003Cstrong>Vereshchagin rule\u003C\u002Fstrong> is a graphical procedure for evaluating the Mohr integral on segments of constant bending stiffness. It replaces direct integration with the area of one diagram and an ordinate of the other.\u003C\u002Fp>{{chunk:vereshchagin-diagram-multiplication}}\u003Ch2>Geometric interpretation\u003C\u002Fh2>\u003Cp>If one moment function is linear over a segment, the integral of the product can be evaluated using the first moment of area of the other diagram. The ordinate of the linear diagram must be taken directly beneath the centroid of the selected diagram area, not simply at the midpoint of the member segment.\u003C\u002Fp>\u003Ch2>Signs\u003C\u002Fh2>\u003Cp>Areas and ordinates are algebraic. Diagrams having the same sign over a contribution produce a positive term; opposite signs produce a negative term.\u003C\u002Fp>\u003Ch2>Subdivision\u003C\u002Fh2>\u003Cp>Trapezoidal, broken, or sign-changing diagrams can be decomposed into rectangles, triangles, or other simple shapes with known areas and centroid locations. Their contributions are calculated separately and summed.\u003C\u002Fp>\u003Ch2>Limitation\u003C\u002Fh2>\u003Cp>The simple rule must not be applied mechanically when both diagrams are nonlinear over the same segment. Use further subdivision, choose the diagram that is linear where possible, or evaluate the integral directly.\u003C\u002Fp>",323,[],{"id":1958,"parent_id":1918,"code":70,"slug":1959,"name":1960,"seo_title":1961,"seo_description":1962,"seo_text":1963,"content":1964,"locale":8,"uk_topic_id":1965,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":1959,"children":1966},350,"temperature-support-settlement-displacements","Displacements due to Temperature and Support Settlement","Temperature and Support-Settlement Displacements","Structural displacements caused by temperature changes, imposed strains, and support settlements or translations.","This topic covers imposed actions in framed structures, including thermal elongation and curvature, support settlement and translation, and their contribution to structural displacements.","\u003Cp>Structural displacements can occur without applied mechanical forces because of \u003Cstrong>temperature change, a temperature gradient, or prescribed support movement\u003C\u002Fstrong>. In a statically determinate structure these actions can often produce free deformation without additional stress resultants; in an indeterminate structure compatibility may generate reactions and internal forces.\u003C\u002Fp>\u003Ch2>Uniform temperature change\u003C\u002Fh2>\u003Cp>For a free straight member of length $L$ subjected to a uniform temperature change $\\Delta T$,\u003C\u002Fp>\u003Cp>$$\\Delta L=\\alpha\\Delta T\\,L,$$\u003C\u002Fp>\u003Cp>where $\\alpha$ is the coefficient of linear thermal expansion.\u003C\u002Fp>\u003Ch2>Temperature gradient\u003C\u002Fh2>\u003Cp>If temperature varies across the section depth, one side tends to expand more than the other and a thermal curvature develops. For a linear temperature difference between extreme fibers, the curvature is proportional to $\\alpha\\Delta T_h\u002Fh$, where $h$ is the distance between those fibers.\u003C\u002Fp>\u003Ch2>Support settlement and prescribed movement\u003C\u002Fh2>\u003Cp>Support settlement is a prescribed kinematic action. A determinate structure may accommodate it without additional reactions when its kinematics permit. In an indeterminate structure, prescribed support movements enter the compatibility conditions and can generate additional internal forces.\u003C\u002Fp>\u003Ch2>Superposition\u003C\u002Fh2>\u003Cp>In a linear analysis, displacements caused by mechanical loading, temperature, and prescribed support movements can be evaluated separately and then combined algebraically.\u003C\u002Fp>\u003Ch2>Signs and units\u003C\u002Fh2>\u003Cp>Define the sign of $\\Delta T$, the direction of support movement, and the positive direction of the required displacement explicitly. Thermal elongation has units of length, whereas thermal curvature has units of inverse length.\u003C\u002Fp>",324,[],{"id":1968,"parent_id":1678,"code":70,"slug":1969,"name":1970,"seo_title":1971,"seo_description":1972,"seo_text":1973,"content":1974,"locale":8,"uk_topic_id":1975,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":891,"url_slug":1969,"children":1976},308,"force-method","Force Method","Force Method in Structural Mechanics | Compatibility Equations","Force method for statically indeterminate structures: degree of indeterminacy, primary structure, compatibility equations, flexibility coefficients, and checks.","The force method analyzes statically indeterminate framed structures by releasing redundants and imposing deformation compatibility. Topics include the primary system, redundant forces, canonical compatibility equations, flexibility coefficients, load terms, and verification.","\u003Cp>The \u003Cstrong>force method\u003C\u002Fstrong> is a classical procedure for analyzing linear-elastic statically indeterminate structures. Its unknowns are redundant reactions or internal force resultants that cannot be obtained from equilibrium alone.\u003C\u002Fp>\u003Ch2>Basic idea\u003C\u002Fh2>\u003Cp>Remove as many restraints as the degree of static indeterminacy. Replace the actions of those released restraints by unknown generalized forces $X_1,X_2,\\ldots,X_n$. The resulting primary structure must be statically determinate and geometrically stable.\u003C\u002Fp>\u003Ch2>Compatibility\u003C\u002Fh2>\u003Cp>Displacements of the primary structure in the released directions, caused by external loading and the unknown redundants, must reproduce the deformation constraints of the original structure. When the corresponding original displacement is zero,\u003C\u002Fp>\u003Cp>$$\\sum_{j=1}^{n}\\delta_{ij}X_j+\\Delta_{iP}=0.$$\u003C\u002Fp>\u003Cp>Here $\\delta_{ij}$ is the displacement in direction $i$ caused by a unit value of redundant $X_j$, and $\\Delta_{iP}$ is the displacement caused by the prescribed loading.\u003C\u002Fp>\u003Ch2>Coefficients\u003C\u002Fh2>\u003Cp>Flexibility coefficients and load terms are evaluated by displacement methods such as the Mohr integral or graphical diagram multiplication. For a linear-elastic reciprocal system, $\\delta_{ij}=\\delta_{ji}$.\u003C\u002Fp>\u003Ch2>Analysis sequence\u003C\u002Fh2>\u003Col>\u003Cli>Determine the degree of static indeterminacy.\u003C\u002Fli>\u003Cli>Select redundants and a primary structure.\u003C\u002Fli>\u003Cli>Construct the load state and unit states.\u003C\u002Fli>\u003Cli>Evaluate the compatibility coefficients.\u003C\u002Fli>\u003Cli>Solve for the redundants $X_j$.\u003C\u002Fli>\u003Cli>Recover reactions and internal forces by superposition.\u003C\u002Fli>\u003Cli>Check equilibrium and deformation compatibility.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Imposed deformations\u003C\u002Fh2>\u003Cp>Temperature changes, support settlements, and prescribed support movements enter the compatibility equations as additional kinematic terms, so the force method is not limited to mechanical force loading.\u003C\u002Fp>",301,[1977,1987,1997,2007],{"id":1978,"parent_id":1968,"code":70,"slug":1979,"name":1980,"seo_title":1981,"seo_description":1982,"seo_text":1983,"content":1984,"locale":8,"uk_topic_id":1985,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":1979,"children":1986},351,"static-indeterminacy-primary-structure","Degree of Static Indeterminacy and Primary Structure","Static Indeterminacy — Degree & Primary Structure in Force Method","Learn how to determine the degree of static indeterminacy, select redundant forces or reactions, and construct a stable statically determinate primary structure for the force method.","This topic explains how to count static indeterminacy, choose redundant forces or reactions, and construct a statically determinate primary structure for the force method.","\u003Cp>The \u003Cstrong>degree of static indeterminacy\u003C\u002Fstrong> is the number of independent force unknowns that cannot be determined from equilibrium alone. In the force method it also gives the number of redundants $X_i$ and the number of compatibility equations.\u003C\u002Fp>\u003Ch2>External and internal indeterminacy\u003C\u002Fh2>\u003Cp>Redundancy may arise from support reactions or from internal force connections within the structure. Counting support reactions alone is therefore sufficient only for simple cases; frames, closed loops, and combined systems may contain internal redundancies.\u003C\u002Fp>\u003Ch2>Selecting redundants\u003C\u002Fh2>\u003Cp>Choose reactions or internal force resultants whose release produces a convenient statically determinate structure. Different valid sets of redundants can produce different primary structures, but the final response of the original structure must be the same.\u003C\u002Fp>\u003Ch2>Requirements for the primary structure\u003C\u002Fh2>\u003Cp>After releasing the redundants, the primary structure must be \u003Cstrong>statically determinate and geometrically stable\u003C\u002Fstrong>. A release that creates a mechanism or leaves hidden static indeterminacy is not valid.\u003C\u002Fp>\u003Ch2>Replacing released restraints\u003C\u002Fh2>\u003Cp>Each released restraint is replaced by its corresponding unknown generalized force or moment $X_i$. The primary structure is then analyzed for the prescribed loading and separately for a unit value of each redundant.\u003C\u002Fp>\u003Ch2>Efficient choice\u003C\u002Fh2>\u003Cp>A good primary structure produces simple internal-force diagrams, few integration segments, and makes use of symmetry when available. The choice affects computational effort, not the physical solution.\u003C\u002Fp>",325,[],{"id":1988,"parent_id":1968,"code":70,"slug":1989,"name":1990,"seo_title":1991,"seo_description":1992,"seo_text":1993,"content":1994,"locale":8,"uk_topic_id":1995,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":1989,"children":1996},352,"canonical-equations-force-method","Canonical Equations of the Force Method","Force Method Canonical Equations — Flexibility Coefficients & Compatibility","Learn the force-method canonical equations, flexibility coefficients δij, load displacement terms ΔiP, and compatibility conditions used to solve statically indeterminate structures.","This topic covers the canonical equations of the force method, unit flexibility coefficients, load terms, evaluation through displacement methods, and deformation compatibility conditions.","\u003Cp>The \u003Cstrong>canonical equations of the force method\u003C\u002Fstrong> express displacement compatibility in the directions of the released redundant restraints. A structure with degree of static indeterminacy $n$ produces $n$ linear equations.\u003C\u002Fp>\u003Ch2>General form\u003C\u002Fh2>\u003Cp>If each released direction has zero displacement in the original structure,\u003C\u002Fp>\u003Cp>$$\\delta_{i1}X_1+\\delta_{i2}X_2+\\cdots+\\delta_{in}X_n+\\Delta_{iP}=0,\\qquad i=1,\\ldots,n.$$\u003C\u002Fp>\u003Cp>In matrix form,\u003C\u002Fp>\u003Cp>$$[\\delta]\\{X\\}=-\\{\\Delta_P\\}.$$\u003C\u002Fp>\u003Ch2>Meaning of the coefficients\u003C\u002Fh2>\u003Cp>$\\delta_{ij}$ is the displacement in redundant direction $i$ caused by a unit value of redundant $X_j$ in the primary structure. $\\Delta_{iP}$ is the displacement in the same direction caused by the prescribed loading.\u003C\u002Fp>\u003Ch2>Evaluation\u003C\u002Fh2>\u003Cp>For bending-dominated structures with constant or piecewise-constant stiffness,\u003C\u002Fp>\u003Cp>$$\\delta_{ij}=\\sum\\int\\frac{M_iM_j}{EI}\\,dx,$$\u003C\u002Fp>\u003Cp>and\u003C\u002Fp>\u003Cp>$$\\Delta_{iP}=\\sum\\int\\frac{M_iM_P}{EI}\\,dx.$$\u003C\u002Fp>\u003Cp>Axial, torsional, and shear contributions are added when required by the model.\u003C\u002Fp>\u003Ch2>Matrix properties\u003C\u002Fh2>\u003Cp>For a linear-elastic conservative reciprocal system, $\\delta_{ij}=\\delta_{ji}$. Diagonal coefficients $\\delta_{ii}$ are positive for independent nonzero unit states.\u003C\u002Fp>\u003Ch2>Recovery\u003C\u002Fh2>\u003Cp>After solving for the redundants, final internal forces follow by superposition, for example $M=M_P+\\sum X_jM_j$. Check both equilibrium and the prescribed compatibility conditions.\u003C\u002Fp>",326,[],{"id":1998,"parent_id":1968,"code":70,"slug":1999,"name":2000,"seo_title":2001,"seo_description":2002,"seo_text":2003,"content":2004,"locale":8,"uk_topic_id":2005,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":1999,"children":2006},353,"beams-frames-force-method","Beams and Frames by the Force Method","Force Method for Beams & Frames — Step-by-Step Analysis","Analyze statically indeterminate beams and frames with the force method: choose a primary structure, build load and unit diagrams, solve compatibility equations, and recover final forces.","This topic applies the force method to beams and frames: selecting a primary structure, constructing load and unit diagrams, solving compatibility equations, and recovering final internal-force diagrams.","\u003Cp>Analysis of \u003Cstrong>beams and frames by the force method\u003C\u002Fstrong> reduces an indeterminate problem to a series of analyses of one statically determinate primary structure. The main practical choices are the redundants and an efficient evaluation of compatibility displacements.\u003C\u002Fp>\u003Ch2>Primary structure\u003C\u002Fh2>\u003Cp>For a continuous beam, redundants may be support reactions or moments at intermediate supports. For frames, one may release support restraints, introduce hinges, or cut a closed loop and replace the released force connections by unknowns $X_i$.\u003C\u002Fp>\u003Ch2>Load state\u003C\u002Fh2>\u003Cp>Apply the prescribed external loading to the primary structure and construct the load diagrams $M_P$ and, where required, $N_P$ and $V_P$. Use one consistent sign convention in every state.\u003C\u002Fp>\u003Ch2>Unit states\u003C\u002Fh2>\u003Cp>For each redundant, set $X_i=1$ while the other redundants are zero and construct the unit diagram $M_i$. These unit states are used to evaluate $\\delta_{ij}$ and $\\Delta_{iP}$.\u003C\u002Fp>\u003Ch2>Superposition\u003C\u002Fh2>\u003Cp>After solving the compatibility equations, final bending moments are\u003C\u002Fp>\u003Cp>$$M=M_P+X_1M_1+\\cdots+X_nM_n,$$\u003C\u002Fp>\u003Cp>with analogous expressions for other internal force effects included in the model.\u003C\u002Fp>\u003Ch2>Symmetric structures\u003C\u002Fh2>\u003Cp>For symmetric frames, separating symmetric and antisymmetric loading or unit states can make some coefficients vanish and substantially reduce the equation system.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>Final reactions must satisfy global equilibrium, and internal-force diagrams must satisfy hinge and free-end conditions. The essential deformation check is that displacements in the restored restraint directions equal their prescribed values.\u003C\u002Fp>",327,[],{"id":2008,"parent_id":1968,"code":70,"slug":2009,"name":2010,"seo_title":2011,"seo_description":2012,"seo_text":2013,"content":2014,"locale":8,"uk_topic_id":2015,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":2009,"children":2016},354,"temperature-support-settlement-force-method","Temperature and Support Settlement in the Force Method","Temperature & Support Settlement in the Force Method — Compatibility","Learn how thermal strains and prescribed support settlements enter force-method compatibility equations and generate redundant reactions and internal forces.","This topic explains how thermal effects and prescribed support displacements enter force-method compatibility conditions and how the resulting redundant forces are determined.","\u003Cp>In a statically indeterminate structure, \u003Cstrong>thermal deformation and support settlement\u003C\u002Fstrong> cannot generally occur freely because redundant restraints limit deformation and therefore generate additional reactions and internal forces. In the force method these effects enter the compatibility equations directly.\u003C\u002Fp>\u003Ch2>Compatibility with imposed actions\u003C\u002Fh2>\u003Cp>A general equation in redundant direction $i$ may be written as\u003C\u002Fp>\u003Cp>$$\\sum_{j=1}^{n}\\delta_{ij}X_j+\\Delta_{iP}+\\Delta_{iT}=\\Delta_i^{\\mathrm{given}},$$\u003C\u002Fp>\u003Cp>where $\\Delta_{iT}$ is the free displacement of the primary structure caused by temperature and $\\Delta_i^{\\mathrm{given}}$ is a prescribed displacement such as support settlement. Exact signs depend on the adopted positive directions.\u003C\u002Fp>\u003Ch2>Uniform temperature change\u003C\u002Fh2>\u003Cp>A free member has thermal elongation $\\alpha\\Delta T L$. If redundant restraints prevent that elongation, the unknown forces $X_j$ must generate compensating elastic deformation.\u003C\u002Fp>\u003Ch2>Temperature gradient\u003C\u002Fh2>\u003Cp>Nonuniform temperature through the section depth creates free curvature. In frames and continuous beams, restraint of this curvature can generate bending moments even when no external mechanical forces are applied.\u003C\u002Fp>\u003Ch2>Support settlement\u003C\u002Fh2>\u003Cp>If the original structure has a prescribed nonzero displacement in a released direction, the corresponding compatibility equation must reproduce that displacement. Settlement should therefore not automatically be modeled as an additional external force.\u003C\u002Fp>\u003Ch2>Superposition and verification\u003C\u002Fh2>\u003Cp>For linear material behavior, mechanical, thermal, and prescribed-movement states may be evaluated separately and superposed. Check both equilibrium and that the restored structure reproduces the specified support and restraint displacements.\u003C\u002Fp>",328,[],{"id":2018,"parent_id":1678,"code":70,"slug":2019,"name":2020,"seo_title":2021,"seo_description":2022,"seo_text":2023,"content":2024,"locale":8,"uk_topic_id":2025,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":903,"url_slug":2019,"children":2026},309,"displacement-method","Displacement Method","Displacement Method — Compatibility Equations for Beams & Frames","Analyze continuous beams and frames with the displacement method using kinematic indeterminacy, joint rotations and translations, a primary system, and canonical equations.","The displacement method uses independent joint translations and rotations as the primary unknowns. This section covers kinematic indeterminacy, the primary system, reactions due to unit displacements, and canonical equilibrium equations.","\u003Cp>The \u003Cstrong>displacement method\u003C\u002Fstrong> analyzes linear-elastic statically indeterminate framed structures using independent joint translations and rotations as the primary unknowns.\u003C\u002Fp>\u003Ch2>Basic idea\u003C\u002Fh2>\u003Cp>Unlike the force method, which uses redundant force quantities, the displacement method temporarily restrains all independent joint movements by additional constraints. The resulting structure is the primary system of the displacement method.\u003C\u002Fp>\u003Ch2>Primary unknowns\u003C\u002Fh2>\u003Cp>The unknowns $Z_1,Z_2,\\ldots,Z_n$ represent independent rotations of rigid joints and independent translations of joints or joint groups. Their number is the degree of kinematic indeterminacy of the adopted structural model.\u003C\u002Fp>\u003Ch2>Canonical equations\u003C\u002Fh2>\u003Cp>The artificial restraints do not exist in the real structure, so their total reactions must vanish after superposition:\u003C\u002Fp>\u003Cp>$$\\sum_{j=1}^{n}r_{ij}Z_j+R_{iP}=0,\\qquad i=1,\\ldots,n.$$\u003C\u002Fp>\u003Cp>Here $r_{ij}$ is the reaction in artificial restraint $i$ caused by unit displacement $Z_j=1$, and $R_{iP}$ is the reaction in the same restraint caused by the prescribed loading while all primary displacements are restrained.\u003C\u002Fp>\u003Ch2>Analysis states\u003C\u002Fh2>\u003Cp>The primary system is analyzed under the prescribed loading and under each unit displacement separately. Once $Z_j$ are known, member end moments, shears, axial forces, and support reactions are recovered by superposition.\u003C\u002Fp>\u003Ch2>Applications\u003C\u002Fh2>\u003Cp>The method is especially effective for continuous beams and frames when the number of independent joint displacements is smaller than the number of force redundants. Its stiffness-based logic leads directly to matrix structural analysis.\u003C\u002Fp>",302,[2027,2037,2047,2057],{"id":2028,"parent_id":2018,"code":70,"slug":2029,"name":2030,"seo_title":2031,"seo_description":2032,"seo_text":2033,"content":2034,"locale":8,"uk_topic_id":2035,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":2029,"children":2036},355,"kinematic-indeterminacy-primary-system","Kinematic Indeterminacy and Primary System","Kinematic Indeterminacy — Primary System in Displacement Method","Learn how to identify independent joint rotations and translations, determine kinematic indeterminacy, and construct the primary system for displacement-method analysis.","This topic explains the degree of kinematic indeterminacy, selection of independent joint rotations and translations, and introduction of restraints in the displacement-method primary system.","\u003Cp>\u003Cstrong>Kinematic indeterminacy\u003C\u002Fstrong> is the number of independent joint displacements required to describe the deformed configuration within the adopted structural model. In the classical displacement method these displacements are the primary unknowns.\u003C\u002Fp>\u003Ch2>Rotational unknowns\u003C\u002Fh2>\u003Cp>Each rigid joint that can rotate independently may contribute a rotational coordinate $Z_i$. Member ends rigidly connected at one joint share the same joint rotation.\u003C\u002Fp>\u003Ch2>Translational unknowns\u003C\u002Fh2>\u003Cp>Translations occur when a joint or group of joints can move without violating the adopted kinematic constraints. A common example in frames is a horizontal storey or sway displacement.\u003C\u002Fp>\u003Ch2>Primary system\u003C\u002Fh2>\u003Cp>To form the primary system, all independent unknown displacements are temporarily restrained by artificial constraints. A rotational restraint suppresses a joint rotation; a translational restraint suppresses the corresponding linear displacement.\u003C\u002Fp>\u003Ch2>Unit displacements\u003C\u002Fh2>\u003Cp>To obtain equation coefficients, impose $Z_j=1$ at one restraint while all other primary displacements remain zero. The resulting reactions in the artificial restraints form the stiffness coefficients $r_{ij}$.\u003C\u002Fp>\u003Ch2>Dependence on modeling assumptions\u003C\u002Fh2>\u003Cp>The degree of kinematic indeterminacy depends on the model. For example, neglecting axial deformation may constrain translations of several joints to be equal and reduce the number of independent coordinates.\u003C\u002Fp>",329,[],{"id":2038,"parent_id":2018,"code":70,"slug":2039,"name":2040,"seo_title":2041,"seo_description":2042,"seo_text":2043,"content":2044,"locale":8,"uk_topic_id":2045,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":2039,"children":2046},356,"canonical-equations-displacement-method","Canonical Equations of the Displacement Method","Displacement Method Canonical Equations — Stiffness Coefficients","Learn the canonical equations of the displacement method, stiffness coefficients rij, reactions from unit displacements, load terms, and equilibrium conditions.","This topic covers the physical meaning of displacement-method canonical equations, stiffness coefficients, reactions due to unit generalized displacements, and fixed-end load terms.","\u003Cp>The \u003Cstrong>canonical equations of the displacement method\u003C\u002Fstrong> express the condition that the artificial restraints introduced in the primary system carry no final reaction after the structure acquires its actual joint displacements.\u003C\u002Fp>\u003Ch2>General form\u003C\u002Fh2>\u003Cp>For $n$ independent displacements,\u003C\u002Fp>\u003Cp>$$r_{i1}Z_1+r_{i2}Z_2+\\cdots+r_{in}Z_n+R_{iP}=0,\\qquad i=1,\\ldots,n.$$\u003C\u002Fp>\u003Cp>In matrix form,\u003C\u002Fp>\u003Cp>$$[r]\\{Z\\}=-\\{R_P\\}.$$\u003C\u002Fp>\u003Ch2>Stiffness coefficients\u003C\u002Fh2>\u003Cp>$r_{ij}$ is the reaction in artificial restraint $i$ caused by unit displacement $Z_j=1$ while all other primary displacements are zero. The diagonal coefficient $r_{ii}$ measures the structural resistance associated with coordinate $Z_i$.\u003C\u002Fp>\u003Ch2>Load terms\u003C\u002Fh2>\u003Cp>$R_{iP}$ is the reaction in restraint $i$ caused by the prescribed external loading when all primary displacements are blocked. It is obtained from member end forces and equilibrium of the corresponding joint or structural part.\u003C\u002Fp>\u003Ch2>Reciprocity\u003C\u002Fh2>\u003Cp>For a linear-elastic conservative system, the coefficient matrix is symmetric: $r_{ij}=r_{ji}$. This provides a useful check on the unit states and equation assembly.\u003C\u002Fp>\u003Ch2>Force recovery\u003C\u002Fh2>\u003Cp>After solving for $Z_j$, any internal-force quantity can be recovered by superposition, for example\u003C\u002Fp>\u003Cp>$$M=M_P+Z_1M_1+\\cdots+Z_nM_n.$$\u003C\u002Fp>\u003Cp>Joint equilibrium, support reactions, and boundary conditions should then be verified.\u003C\u002Fp>",330,[],{"id":2048,"parent_id":2018,"code":70,"slug":2049,"name":2050,"seo_title":2051,"seo_description":2052,"seo_text":2053,"content":2054,"locale":8,"uk_topic_id":2055,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":2049,"children":2056},357,"continuous-beams-displacement-method","Continuous Beams by the Displacement Method","Continuous Beams by Displacement Method — Rotations & Moment Diagrams","Analyze statically indeterminate continuous beams by the displacement method using joint rotations, canonical equations, member-end moments, and the final bending-moment diagram.","This topic applies the displacement method to continuous beams: selecting generalized displacements, forming equations, determining member-end moments, and constructing the final bending-moment diagram.","\u003Cp>For a \u003Cstrong>continuous beam\u003C\u002Fstrong>, the displacement method is especially convenient because, in the absence of support settlement, the primary unknowns are often only rotations of the rigid intermediate joints. Ordinary support translations are prescribed or restrained.\u003C\u002Fp>\u003Ch2>Primary system\u003C\u002Fh2>\u003Cp>Artificial rotational restraints are introduced at the rigid intermediate joints. Each span of the primary system is then treated as a beam element with restrained or prescribed end displacements.\u003C\u002Fp>\u003Ch2>Load state\u003C\u002Fh2>\u003Cp>External loading acting while the joint rotations are restrained produces member end moments. These moments generate the load reactions of the artificial rotational restraints.\u003C\u002Fp>\u003Ch2>Unit rotations\u003C\u002Fh2>\u003Cp>Each independent joint is given a unit rotation in turn. Adjacent spans resist that rotation according to their flexural stiffness $EI$ and length. Reactions produced by the unit states form the coefficients of the canonical equations.\u003C\u002Fp>\u003Ch2>Joint equilibrium\u003C\u002Fh2>\u003Cp>After the actual rotations have been found, the sum of member end moments meeting at a joint, together with any applied joint moment, must satisfy moment equilibrium. This is the physical basis of the corresponding canonical equation.\u003C\u002Fp>\u003Ch2>Final moment diagram\u003C\u002Fh2>\u003Cp>Member end moments and $M(x)$ in each span are recovered by superposing the load state and unit-displacement states. Span equilibrium then gives shears and support reactions.\u003C\u002Fp>\u003Ch2>Support settlement\u003C\u002Fh2>\u003Cp>Prescribed vertical support movements produce additional member end moments and enter the equations as known kinematic effects.\u003C\u002Fp>",331,[],{"id":2058,"parent_id":2018,"code":70,"slug":2059,"name":2060,"seo_title":2061,"seo_description":2062,"seo_text":2063,"content":2064,"locale":8,"uk_topic_id":2065,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":2059,"children":2066},358,"frames-displacement-method","Frames by the Displacement Method","Frame Analysis by the Displacement Method — Slope-Deflection","Learn frame analysis by the displacement method: choose joint degrees of freedom, write member end moments, enforce equilibrium, and recover internal forces.","This topic covers frame analysis by the displacement method, including joint rotations, lateral translations, canonical equations, and recovery of member-end actions.","\u003Cp>In \u003Cstrong>frames\u003C\u002Fstrong>, the displacement method uses rigid-joint rotations and, for sway frames, independent translations of joints or storeys as unknowns. A kinematic analysis is therefore required before the equilibrium equations are assembled.\u003C\u002Fp>\u003Ch2>Non-sway frames\u003C\u002Fh2>\u003Cp>If geometry, supports, or bracing prevent joint translation, the unknowns are primarily joint rotations. Each rigid joint receives a moment-equilibrium equation after the end moments of all connected members have been expressed in terms of those rotations.\u003C\u002Fp>\u003Ch2>Sway frames\u003C\u002Fh2>\u003Cp>In a sway frame, one or more translational coordinates are added. A unit horizontal translation bends the columns and creates reactions in the temporary translational restraint. The canonical equation for that coordinate expresses equilibrium of the corresponding generalized horizontal force.\u003C\u002Fp>\u003Ch2>Member end actions\u003C\u002Fh2>\u003Cp>Member end moments depend on external loading, rotations at both ends, and relative transverse translation of the member ends. In classical hand analysis these relations are represented by tabulated primary-system reactions or slope-deflection-type equations.\u003C\u002Fp>\u003Ch2>Equation assembly\u003C\u002Fh2>\u003Cp>For each rotational coordinate, write moment equilibrium at the associated joint. For each translational coordinate, write force or generalized-reaction equilibrium in the displacement direction. Together these conditions form the system for $Z_i$.\u003C\u002Fp>\u003Ch2>Recovering results\u003C\u002Fh2>\u003Cp>After solving for joint displacements, calculate member end moments, then shears and axial forces, and finally support reactions. The completed solution must satisfy equilibrium of every joint and of the frame as a whole.\u003C\u002Fp>\u003Ch2>Connection to the matrix method\u003C\u002Fh2>\u003Cp>The classical displacement method already contains the central stiffness-method idea: joint displacements are unknowns and joint forces are linear functions of them. Matrix structural analysis systematizes this procedure for large numbers of elements and degrees of freedom.\u003C\u002Fp>",332,[],{"id":2068,"parent_id":1678,"code":70,"slug":2069,"name":2070,"seo_title":2071,"seo_description":2072,"seo_text":2073,"content":2074,"locale":8,"uk_topic_id":2075,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":915,"url_slug":2069,"children":2076},310,"matrix-structural-analysis","Matrix Structural Analysis","Matrix Structural Analysis | Matrix Stiffness Method","Matrix stiffness method fundamentals: local and global stiffness matrices, assembly, boundary conditions, nodal displacements, and member forces.","An introduction to matrix analysis of framed structures: local stiffness matrices, coordinate transformations, assembly of the global equation system, boundary conditions, solution for nodal displacements, and recovery of member forces.","\u003Cp>\u003Cstrong>Matrix structural analysis\u003C\u002Fstrong> systematizes the displacement method and provides a general algorithm for framed structures with many elements and degrees of freedom. Its principal form is the matrix stiffness method.\u003C\u002Fp>\u003Ch2>Governing equation\u003C\u002Fh2>\u003Cp>After numbering the nodal degrees of freedom, structural equilibrium is written as\u003C\u002Fp>\u003Cp>$$[K]\\{d\\}=\\{F\\},$$\u003C\u002Fp>\u003Cp>where $[K]$ is the global stiffness matrix, $\\{d\\}$ is the nodal displacement vector, and $\\{F\\}$ is the equivalent nodal load vector.\u003C\u002Fp>\u003Ch2>Element approach\u003C\u002Fh2>\u003Cp>The structure is divided into member elements. Each element is first described by local degrees of freedom and a local stiffness matrix $[k]$, relating its local nodal displacements to member-end forces.\u003C\u002Fp>\u003Ch2>Coordinate transformation\u003C\u002Fh2>\u003Cp>Because members may have different orientations, their local matrices are transformed to a common global coordinate system. Contributions from all elements are then assembled into the global matrix $[K]$.\u003C\u002Fp>\u003Ch2>Boundary conditions\u003C\u002Fh2>\u003Cp>Supports prescribe selected nodal displacements, usually zero. After these conditions are imposed, the reduced system is solved for unknown displacements and support reactions are recovered.\u003C\u002Fp>\u003Ch2>Member-force recovery\u003C\u002Fh2>\u003Cp>The global end displacements of each element are transformed back to local coordinates. Local stiffness relations then provide member-end forces and moments, from which internal-force diagrams can be constructed.\u003C\u002Fp>\u003Ch2>Connection to finite elements\u003C\u002Fh2>\u003Cp>For framed structures, the matrix stiffness method is a natural foundation of the finite element method: a model is composed of elements with their own matrices and degrees of freedom, and the global problem is obtained by assembly.\u003C\u002Fp>",303,[2077,2087,2097,2107],{"id":2078,"parent_id":2068,"code":70,"slug":2079,"name":2080,"seo_title":2081,"seo_description":2082,"seo_text":2083,"content":2084,"locale":8,"uk_topic_id":2085,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":78,"url_slug":2079,"children":2086},359,"local-stiffness-matrix-structural-element","Local Stiffness Matrix of a Structural Element","Local Stiffness Matrix — Structural Element Forces & Displacements","Learn how a structural element’s local stiffness matrix relates nodal displacements to end forces, including axial, beam-bending, and planar frame elements.","This topic introduces the local stiffness matrix of a structural element, its nodal degrees of freedom, and the relationship between local nodal displacement and force vectors.","\u003Cp>A \u003Cstrong>local stiffness matrix\u003C\u002Fstrong> describes the mechanical behavior of one member element in coordinates aligned with its own axis. It provides a linear relation between nodal displacements and member-end forces.\u003C\u002Fp>\u003Ch2>Basic relation\u003C\u002Fh2>\u003Cp>For an element without initial strains or member loading,\u003C\u002Fp>\u003Cp>$$\\{q\\}=[k]\\{u\\},$$\u003C\u002Fp>\u003Cp>where $\\{u\\}$ is the local nodal-displacement vector, $\\{q\\}$ is the local member-end force vector, and $[k]$ is the local stiffness matrix.\u003C\u002Fp>\u003Ch2>Truss member\u003C\u002Fh2>\u003Cp>For a two-node bar carrying axial deformation only, the local degrees of freedom are the axial end displacements and\u003C\u002Fp>\u003Cp>$$[k]=\\frac{EA}{L}\\begin{bmatrix}1&-1\\\\-1&1\\end{bmatrix}.$$\u003C\u002Fp>\u003Cp>The factor $EA\u002FL$ represents the axial stiffness associated with relative displacement of the member ends.\u003C\u002Fp>\u003Ch2>Beam element\u003C\u002Fh2>\u003Cp>For a planar bending element, typical local coordinates are transverse displacement and rotation at each end. Its stiffness coefficients contain combinations of $EI\u002FL^3$, $EI\u002FL^2$, and $EI\u002FL$, representing flexural resistance.\u003C\u002Fp>\u003Ch2>Plane-frame element\u003C\u002Fh2>\u003Cp>A plane-frame element combines axial and bending behavior. Each node commonly has three local degrees of freedom: axial translation, transverse translation, and rotation, giving a full $6\\times6$ local matrix.\u003C\u002Fp>\u003Ch2>Properties\u003C\u002Fh2>\u003Cp>For a linear-elastic conservative element, $[k]$ is symmetric. Before sufficient boundary conditions are imposed it may be singular because rigid-body motions cause no deformation and therefore no internal force.\u003C\u002Fp>",333,[],{"id":2088,"parent_id":2068,"code":70,"slug":2089,"name":2090,"seo_title":2091,"seo_description":2092,"seo_text":2093,"content":2094,"locale":8,"uk_topic_id":2095,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":139,"url_slug":2089,"children":2096},360,"coordinate-transformation-global-stiffness","Coordinate Transformation and Global Element Stiffness","Local-to-Global Stiffness Transformation — Matrix Formula TᵀkT","Learn how to transform a structural element from local to global coordinates using direction cosines and obtain its global stiffness matrix with TᵀkT.","This topic explains coordinate transformation matrices, element orientation, and transformation of local stiffness relationships into the global coordinate system of the structure.","\u003Cp>Each structural element is conveniently described in a \u003Cstrong>local coordinate system\u003C\u002Fstrong>, while the equations of the complete structure must use one common global system. A coordinate-transformation matrix connects the two descriptions.\u003C\u002Fp>\u003Ch2>Local and global displacements\u003C\u002Fh2>\u003Cp>Let $\\{d_e\\}$ be the element nodal-displacement vector in global coordinates and $\\{u_e\\}$ the corresponding local vector. With a consistent definition of $[T]$:\u003C\u002Fp>\u003Cp>$$\\{u_e\\}=[T]\\{d_e\\}.$$\u003C\u002Fp>\u003Cp>For planar structures, $[T]$ contains the direction cosines $c=\\cos\\alpha$ and $s=\\sin\\alpha$ of the member axis.\u003C\u002Fp>\u003Ch2>Element stiffness in global coordinates\u003C\u002Fh2>\u003Cp>Combining the displacement transformation with the consistent force transformation gives\u003C\u002Fp>\u003Cp>$$[k_e^{(g)}]=[T]^T[k_e][T].$$\u003C\u002Fp>\u003Cp>The matrix $[k_e^{(g)}]$ represents the same physical element, but its rows and columns correspond to global nodal displacement components.\u003C\u002Fp>\u003Ch2>Why transformation is necessary\u003C\u002Fh2>\u003Cp>Two identical members with the same $E$, $A$, $I$, and $L$ but different orientations have the same local matrix and different global matrices. The transformation accounts for their orientation within the structure.\u003C\u002Fp>\u003Ch2>Force recovery\u003C\u002Fh2>\u003Cp>After solving the global problem, the end displacements of a particular member are extracted from the global vector and transformed to local coordinates. Axial force, shear, and end moments can then be evaluated in the member’s natural coordinate system.\u003C\u002Fp>\u003Ch2>Checks\u003C\u002Fh2>\u003Cp>A coordinate transformation must not alter the physical strain energy. For an orthogonal rotational transformation it also preserves stiffness symmetry and the consistency of nodal-force work.\u003C\u002Fp>",334,[],{"id":2098,"parent_id":2068,"code":70,"slug":2099,"name":2100,"seo_title":2101,"seo_description":2102,"seo_text":2103,"content":2104,"locale":8,"uk_topic_id":2105,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":150,"url_slug":2099,"children":2106},361,"assembly-global-equation-system","Assembly of the Global Equation System","Global Stiffness Matrix Assembly — Build the System Kd = F","Learn how to assemble element stiffness matrices into the global matrix K, map degrees of freedom, form the nodal load vector F, and obtain the structural system Kd = F.","This topic covers degree-of-freedom numbering, assembly of element contributions, construction of the global stiffness matrix, and formation of the global nodal load vector.","\u003Cp>\u003Cstrong>Assembly\u003C\u002Fstrong> combines the stiffness contributions of individual elements into the global equation system of the complete structure. It is based on compatibility of nodal displacements and equilibrium of nodal forces.\u003C\u002Fp>\u003Ch2>Global numbering\u003C\u002Fh2>\u003Cp>Each independent structural degree of freedom is assigned a global number. Every element then has a connectivity map relating its element coordinates to these global degree-of-freedom numbers.\u003C\u002Fp>\u003Ch2>Adding element contributions\u003C\u002Fh2>\u003Cp>Each global coefficient $K_{IJ}$ is the sum of contributions from all elements containing the corresponding global degrees of freedom $I$ and $J$. Schematically,\u003C\u002Fp>\u003Cp>$$[K]=\\sum_e[A_e]^T[k_e^{(g)}][A_e],$$\u003C\u002Fp>\u003Cp>where $[A_e]$ maps element degrees of freedom to global coordinates.\u003C\u002Fp>\u003Ch2>Load vector\u003C\u002Fh2>\u003Cp>Concentrated nodal forces and moments are added directly to $\\{F\\}$. Distributed and other member loads are represented by equivalent nodal forces according to the adopted element formulation.\u003C\u002Fp>\u003Ch2>Global system\u003C\u002Fh2>\u003Cp>Assembly produces\u003C\u002Fp>\u003Cp>$$[K]\\{d\\}=\\{F\\}.$$\u003C\u002Fp>\u003Cp>Before supports are imposed, the matrix may be singular because rigid-body motion is possible. Correct boundary conditions remove the corresponding kinematic modes in a stable structure.\u003C\u002Fp>\u003Ch2>Matrix structure\u003C\u002Fh2>\u003Cp>The global matrix is normally sparse because an element directly couples only the degrees of freedom of its own nodes. For a linear-elastic conservative model, $[K]$ is symmetric.\u003C\u002Fp>\u003Ch2>Model checks\u003C\u002Fh2>\u003Cp>Unexpected singularity after applying supports often indicates a mechanism, insufficient restraint, connectivity error, incorrect member-end release, or duplicated independent degrees of freedom.\u003C\u002Fp>",335,[],{"id":2108,"parent_id":2068,"code":70,"slug":2109,"name":2110,"seo_title":2111,"seo_description":2112,"seo_text":2113,"content":2114,"locale":8,"uk_topic_id":2115,"show_in_theory_list":78,"is_published":77,"status":88,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":2109,"children":2116},362,"boundary-conditions-displacements-force-recovery","Boundary Conditions, Displacements, and Force Recovery","Matrix Stiffness Method — Boundary Conditions & Force Recovery","Learn how to impose support boundary conditions, solve for nodal displacements, calculate reactions, and recover local member-end forces in matrix structural analysis.","This topic completes the matrix stiffness workflow: imposing boundary conditions, solving the global equations, calculating reactions, and recovering local member-end forces.","\u003Cp>After the global system has been assembled, \u003Cstrong>boundary conditions\u003C\u002Fstrong> are imposed, unknown nodal displacements are solved, and support reactions and member forces are recovered.\u003C\u002Fp>\u003Ch2>Free and constrained degrees of freedom\u003C\u002Fh2>\u003Cp>Partition the displacement vector into unknown free coordinates $\\{d_f\\}$ and prescribed coordinates $\\{d_c\\}$. For a fixed support component, the corresponding $d_c$ is zero; a prescribed support movement may give a nonzero value.\u003C\u002Fp>\u003Ch2>Partitioned equations\u003C\u002Fh2>\u003Cp>After reordering coordinates,\u003C\u002Fp>\u003Cp>$$\\begin{bmatrix}K_{ff}&K_{fc}\\\\K_{cf}&K_{cc}\\end{bmatrix}\\begin{Bmatrix}d_f\\\\d_c\\end{Bmatrix}=\\begin{Bmatrix}F_f\\\\F_c\\end{Bmatrix}.$$\u003C\u002Fp>\u003Cp>The free-coordinate equations give\u003C\u002Fp>\u003Cp>$$K_{ff}d_f=F_f-K_{fc}d_c.$$\u003C\u002Fp>\u003Ch2>Support reactions\u003C\u002Fh2>\u003Cp>Once all displacements are known, reactions at constrained coordinates are recovered from the complete equilibrium equation. Unknown reactions should not be treated as prescribed external loads before the displacement solution.\u003C\u002Fp>\u003Ch2>Element displacements\u003C\u002Fh2>\u003Cp>For each member, its components are extracted from the global displacement vector to form $\\{d_e\\}$ and transformed to local coordinates:\u003C\u002Fp>\u003Cp>$$\\{u_e\\}=[T]\\{d_e\\}.$$\u003C\u002Fp>\u003Ch2>Member-end forces\u003C\u002Fh2>\u003Cp>Equivalent nodal actions and initial strains must be included consistently with the adopted element formulation. For the simplest unloaded element,\u003C\u002Fp>\u003Cp>$$\\{q_e\\}=[k_e]\\{u_e\\}.$$\u003C\u002Fp>\u003Cp>The components of $\\{q_e\\}$ provide local axial forces, shears, and end moments.\u003C\u002Fp>\u003Ch2>Verification\u003C\u002Fh2>\u003Cp>Check global equilibrium of applied loads and reactions, joint equilibrium, satisfaction of prescribed displacements, matrix symmetry where applicable, and consistency of local member-force sign conventions.\u003C\u002Fp>",336,[],{"id":2118,"parent_id":70,"code":70,"slug":2119,"name":2120,"seo_title":2121,"seo_description":2122,"seo_text":2123,"content":70,"locale":8,"uk_topic_id":2124,"show_in_theory_list":77,"is_published":78,"status":79,"visibility":80,"canonical":70,"noindex":77,"sort_order":161,"url_slug":2119,"children":2125},80,"comprehensive-problems","Comprehensive Problems","Engineering Mechanics Practice Problems — Multi-Topic Problems","Practice advanced engineering mechanics and strength-of-materials problems combining stress analysis, buckling, fatigue, and multiple calculation methods.","This section gathers advanced multi-topic problems combining combined stress analysis with buckling or fatigue life evaluations. Solving comprehensive problems builds deep engineering intuition and aids in course project and exam preparation.",39,[],1789532558119]