[{"data":1,"prerenderedAt":1503},["ShallowReactive",2],{"public-task-104":3,"public-task-topics-en":30},{"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":29},104,140,"bolt-tightening-torque-wrench-en","Bolt Tightening Torque","en",103,"bolt-tightening-torque-with-wrench","Bolt Tightening Torque and Required Wrench Length — Problem","Calculate bolt tightening torque from force and wrench length, then determine the minimum wrench length required for a specified torque.","A practical statics problem involving the moment of a force during bolt tightening.","",5,[17,20,24,26],{"type":18,"html":19},"html","\u003Cp>A bolt is tightened using a wrench. A force of \u003Cstrong>F = 190 N\u003C\u002Fstrong> is applied to the end of the wrench \u003Cstrong>perpendicular\u003C\u002Fstrong> to its handle. The distance from the bolt axis to the point where the force is applied is \u003Cstrong>l = 190 mm\u003C\u002Fstrong>.\u003C\u002Fp>\u003Cp>\u003Cstrong>1.\u003C\u002Fstrong> Determine the bolt tightening torque in N·m.\u003C\u002Fp>",{"type":21,"html":22,"name":23},"number","1. Tightening torque M, N·m","moment",{"type":18,"html":25},"\u003Cp>\u003Cstrong>2.\u003C\u002Fstrong> Another bolt must be tightened to a torque of \u003Cstrong>112 N·m\u003C\u002Fstrong>. The maximum force that can be applied perpendicular to the wrench is \u003Cstrong>320 N\u003C\u002Fstrong>. Determine the minimum required wrench length in mm.\u003C\u002Fp>",{"type":21,"html":27,"name":28},"2. Minimum wrench length l, mm","length","2026-08-23 18:37:56",[31,45,715,1465,1494],{"id":32,"parent_id":33,"code":33,"slug":34,"name":35,"seo_title":36,"seo_description":37,"seo_text":38,"content":33,"locale":8,"uk_topic_id":39,"show_in_theory_list":40,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":34,"children":44},79,null,"school-knowledge-test","School Knowledge Test","School Physics and Math Assessment Test","Evaluate your foundation in physics and mathematics prior to studying mechanics.","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":46,"parent_id":33,"code":33,"slug":47,"name":48,"seo_title":49,"seo_description":50,"seo_text":51,"content":52,"locale":8,"uk_topic_id":53,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":47,"children":54},81,"theoretical-mechanics","Theoretical Mechanics","Theoretical Mechanics — Statics, Dynamics & Kinematics","Fundamental principles of rigid body mechanics for engineering students.","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,[55,275,470],{"id":56,"parent_id":46,"code":33,"slug":57,"name":58,"seo_title":59,"seo_description":60,"seo_text":61,"content":62,"locale":8,"uk_topic_id":63,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":57,"children":64},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,[65,138,196,235],{"id":66,"parent_id":56,"code":33,"slug":67,"name":68,"seo_title":69,"seo_description":70,"seo_text":71,"content":72,"locale":8,"uk_topic_id":73,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":67,"children":75},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,"draft",[76,86,97,107,118,128],{"id":77,"parent_id":66,"code":33,"slug":78,"name":79,"seo_title":80,"seo_description":81,"seo_text":82,"content":83,"locale":8,"uk_topic_id":84,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":78,"children":85},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":87,"parent_id":66,"code":33,"slug":88,"name":89,"seo_title":90,"seo_description":91,"seo_text":92,"content":93,"locale":8,"uk_topic_id":94,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":88,"children":96},82,"force-projections","Force Projections","Force Projections on Coordinate Axes — Statics","How to project force vectors onto Cartesian coordinate systems.","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>A \u003Cstrong>projection of a force onto an axis\u003C\u002Fstrong> is an algebraic scalar quantity representing the component of the force vector along a selected direction. Projections convert vector relations into scalar equations and are therefore fundamental to analytical solutions in statics.\u003C\u002Fp>\u003Ch2>Projection onto a coordinate axis\u003C\u002Fh2>{{chunk:statics-force-projection}}\u003Cp>If the force $\\vec F$ forms an angle $\\alpha$ with the positive $x$ direction, its projection is $F_x$. The projections $F_y$ and, in three-dimensional problems, $F_z$ are defined in the same way.\u003C\u002Fp>\u003Ch2>Signs of projections\u003C\u002Fh2>\u003Cp>A projection is positive when the corresponding force component points in the positive direction of the axis and negative when it points in the opposite direction. A force perpendicular to an axis has zero projection onto that axis.\u003C\u002Fp>\u003Ch2>Resolving a force in two dimensions\u003C\u002Fh2>\u003Cp>For a force in the $xy$ plane, $\\vec F=F_x\\vec i+F_y\\vec j$. If $\\alpha$ is measured from the positive $x$ axis, then $F_x=F\\cos\\alpha$ and $F_y=F\\sin\\alpha$, with signs determined by the actual component directions.\u003C\u002Fp>\u003Ch2>Checking the force magnitude\u003C\u002Fh2>\u003Cp>In a rectangular Cartesian coordinate system, the magnitude of a two-dimensional force satisfies $F=\\sqrt{F_x^2+F_y^2}$. This relation is useful for checking calculated components.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A force $F=10$ kN acts at $30^\\circ$ above the positive $x$ axis in the first quadrant. Then $F_x=10\\cos30^\\circ\\approx8.66$ kN and $F_y=10\\sin30^\\circ=5$ kN. Both projections are positive.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing an angle measured from the $x$ axis with one measured from the $y$ axis;\u003C\u002Fli>\u003Cli>using only the magnitude of a projection and ignoring its sign;\u003C\u002Fli>\u003Cli>using sine instead of cosine without checking which axis the angle is measured from;\u003C\u002Fli>\u003Cli>adding force magnitudes instead of algebraic force projections.\u003C\u002Fli>\u003C\u002Ful>\u003Cp>Equilibrium equations use the algebraic projections of all forces onto the selected coordinate axes.\u003C\u002Fp>",43,2,[],{"id":5,"parent_id":66,"code":33,"slug":98,"name":99,"seo_title":100,"seo_description":101,"seo_text":102,"content":103,"locale":8,"uk_topic_id":104,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":98,"children":106},"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":108,"parent_id":66,"code":33,"slug":109,"name":110,"seo_title":111,"seo_description":112,"seo_text":113,"content":114,"locale":8,"uk_topic_id":115,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":109,"children":117},183,"force-couples-couple-moment","Force Couples and Couple Moment","Force Couples and Couple Moment — Statics","Force couples in statics, the moment of a couple and its properties. Learn equivalent couples and their role in transforming force systems.","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> is a system of two parallel forces of equal magnitude, opposite direction, and different lines of action. The vector sum of the two forces is zero, but their moments produce a nonzero rotational effect.\u003C\u002Fp>\u003Ch2>Couple arm and couple moment\u003C\u002Fh2>\u003Cp>The \u003Cstrong>couple arm\u003C\u002Fstrong> $d$ is the shortest, perpendicular distance between the lines of action of the two forces.\u003C\u002Fp>{{chunk:statics-couple-moment}}\u003Cp>The SI unit of couple moment is N·m. A couple moment should not be confused with work or energy even though their SI dimensions have the same unit form.\u003C\u002Fp>\u003Ch2>Why the resultant force is zero\u003C\u002Fh2>\u003Cp>The two forces have equal magnitudes and opposite directions, so their vector sum is zero: $\\vec F+(-\\vec F)=0$. Their lines of action do not coincide, however, so the moments of the forces do not cancel. A couple therefore cannot be replaced by a single nonzero force.\u003C\u002Fp>\u003Ch2>Independence from the reference point\u003C\u002Fh2>\u003Cp>The sum of the moments of the two forces of a couple is the same about any reference point. The couple moment is therefore a free vector: in the rigid-body model, a couple may be moved to another location without changing its mechanical effect as long as the magnitude and direction of its moment remain unchanged.\u003C\u002Fp>\u003Ch2>Equivalent couples\u003C\u002Fh2>\u003Cp>Two couples are equivalent when their moment vectors are equal. Thus, a larger force acting with a smaller arm may be equivalent to a smaller force acting with a larger arm if the product $Fd$ remains the same.\u003C\u002Fp>\u003Ch2>Addition of couples\u003C\u002Fh2>\u003Cp>Several couples can be replaced by one couple whose moment is the vector sum of the original couple moments. In planar statics, this reduces to algebraic addition with the adopted moment signs.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Two forces of 300 N form a couple with an arm $d=0.20$ m. The couple moment is $M=300\\cdot0.20=60$ N·m. If the arm is increased to 0.30 m while the same moment is required, forces of only $F=60\u002F0.30=200$ N are needed.\u003C\u002Fp>\u003Ch2>Key points\u003C\u002Fh2>\u003Cul>\u003Cli>the vector sum of the forces in a couple is zero;\u003C\u002Fli>\u003Cli>a couple produces a pure rotational effect;\u003C\u002Fli>\u003Cli>the couple moment equals $Fd$ and is independent of the reference point;\u003C\u002Fli>\u003Cli>couples with equal moment vectors are equivalent;\u003C\u002Fli>\u003Cli>in planar problems, couple moments are added algebraically.\u003C\u002Fli>\u003C\u002Ful>",168,4,[],{"id":119,"parent_id":66,"code":33,"slug":120,"name":121,"seo_title":122,"seo_description":123,"seo_text":124,"content":125,"locale":8,"uk_topic_id":126,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":120,"children":127},184,"addition-forces-resultant-force-system","Addition of Forces. Resultant of a Force System","Addition of Forces and Resultant Force Systems","Addition of forces in statics and determination of the resultant force. Geometric and analytical methods for common force systems.","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,[],{"id":129,"parent_id":66,"code":33,"slug":130,"name":131,"seo_title":131,"seo_description":132,"seo_text":133,"content":134,"locale":8,"uk_topic_id":135,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":130,"children":137},185,"reduction-force-system-given-point","Reduction of a Force System to a Given Point","Reduce a general force system to a given point using the resultant force vector and resultant moment for statics and equilibrium analysis.","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":139,"parent_id":56,"code":33,"slug":140,"name":141,"seo_title":142,"seo_description":143,"seo_text":144,"content":145,"locale":8,"uk_topic_id":146,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":140,"children":147},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,[148,157,167,176,186],{"id":149,"parent_id":139,"code":33,"slug":150,"name":151,"seo_title":151,"seo_description":152,"seo_text":153,"content":154,"locale":8,"uk_topic_id":155,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":150,"children":156},186,"equilibrium-conditions-coplanar-force-systems","Equilibrium Conditions for Coplanar Force Systems","Equilibrium equations for coplanar force systems in statics. Use force components and moment sums to solve engineering mechanics problems.","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":158,"parent_id":139,"code":33,"slug":159,"name":160,"seo_title":161,"seo_description":162,"seo_text":163,"content":164,"locale":8,"uk_topic_id":165,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":159,"children":166},187,"constraints-constraint-reactions","Constraints and Constraint Reactions","Constraints and Constraint Reactions in Statics","Common constraints and reactions in statics: supports, hinges, cables and contacts. Learn to construct free-body diagrams for equilibrium.","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":168,"parent_id":139,"code":33,"slug":169,"name":170,"seo_title":170,"seo_description":171,"seo_text":172,"content":173,"locale":8,"uk_topic_id":174,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":169,"children":175},188,"equilibrium-rigid-body-two-dimensions","Equilibrium of a Rigid Body in Two Dimensions","Rigid-body equilibrium in two dimensions: free-body diagrams, support reactions and statics equations for determining unknown forces.","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":177,"parent_id":139,"code":33,"slug":178,"name":179,"seo_title":180,"seo_description":181,"seo_text":182,"content":183,"locale":8,"uk_topic_id":184,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":178,"children":185},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":187,"parent_id":139,"code":33,"slug":188,"name":189,"seo_title":190,"seo_description":191,"seo_text":192,"content":193,"locale":8,"uk_topic_id":194,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":188,"children":195},190,"distributed-loads-resultants","Distributed Loads and Their Resultants","Distributed Loads and Their Resultants — Statics","Distributed loads in statics: load intensity, equivalent resultant and its point of application for uniform and linearly varying loads.","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 finite region of a body. In planar beam problems it is commonly described by an intensity $q(x)$, which represents force per unit length. The SI unit of a line-load intensity is N\u002Fm.\u003C\u002Fp>\u003Ch2>Equivalent concentrated force\u003C\u002Fh2>\u003Cp>A distributed load can be replaced by a resultant concentrated force when its magnitude, direction, and moment effect are preserved. For a load $q(x)$ acting from $a$ to $b$:\u003C\u002Fp>\u003Cp>$$R=\\int_a^b q(x)\\,dx.$$\u003C\u002Fp>\u003Cp>Geometrically, the magnitude of the resultant equals the area under the load-intensity diagram.\u003C\u002Fp>\u003Ch2>Uniformly distributed load\u003C\u002Fh2>{{chunk:statics-uniform-distributed-load-resultant}}\u003Cp>The load diagram is a rectangle, whose area centroid lies at its midpoint. The line of action of the equivalent force therefore passes through the midpoint of the loaded interval.\u003C\u002Fp>\u003Ch2>Location of the resultant for a varying load\u003C\u002Fh2>\u003Cp>The coordinate of the resultant line of action follows from moment equivalence:\u003C\u002Fp>\u003Cp>$$x_R=\\frac{\\int_a^b xq(x)\\,dx}{\\int_a^b q(x)\\,dx}.$$\u003C\u002Fp>\u003Cp>Thus, the resultant passes through the centroid of the area under the $q(x)$ diagram. For standard rectangular, triangular, and trapezoidal diagrams, familiar geometric centroid locations can be used.\u003C\u002Fp>\u003Ch2>Triangular load\u003C\u002Fh2>\u003Cp>If the intensity varies linearly from zero to a maximum value $q_{max}$ over a length $L$, the resultant magnitude equals the triangular area: $R=q_{max}L\u002F2$. Its line of action is located $L\u002F3$ from the side with maximum intensity, or $2L\u002F3$ from the zero-intensity side.\u003C\u002Fp>\u003Ch2>Trapezoidal load\u003C\u002Fh2>\u003Cp>A trapezoidal load diagram can conveniently be decomposed into a rectangular and a triangular part. Determine the resultant and location for each part, then combine the forces while preserving their moments.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>A uniformly distributed downward load $q=3$ kN\u002Fm acts over 4 m of a beam. Its equivalent force is $R=qL=3\\cdot4=12$ kN and acts at the midpoint of the loaded interval, 2 m from its beginning.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>confusing the load intensity $q$ in kN\u002Fm with a force in kN;\u003C\u002Fli>\u003Cli>placing the resultant at the midpoint for every load-diagram shape;\u003C\u002Fli>\u003Cli>ignoring the actual starting position of the loaded interval on the beam;\u003C\u002Fli>\u003Cli>calculating only the resultant magnitude without preserving its moment about a reference point.\u003C\u002Fli>\u003C\u002Ful>",175,[],{"id":197,"parent_id":56,"code":33,"slug":198,"name":199,"seo_title":200,"seo_description":201,"seo_text":202,"content":203,"locale":8,"uk_topic_id":204,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":198,"children":205},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,[206,216,226],{"id":207,"parent_id":197,"code":33,"slug":208,"name":209,"seo_title":210,"seo_description":211,"seo_text":212,"content":213,"locale":8,"uk_topic_id":214,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":208,"children":215},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":217,"parent_id":197,"code":33,"slug":218,"name":219,"seo_title":220,"seo_description":221,"seo_text":222,"content":223,"locale":8,"uk_topic_id":224,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":218,"children":225},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":227,"parent_id":197,"code":33,"slug":228,"name":229,"seo_title":229,"seo_description":230,"seo_text":231,"content":232,"locale":8,"uk_topic_id":233,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":228,"children":234},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":236,"parent_id":56,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":237,"children":244},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,[245,255,265],{"id":246,"parent_id":236,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":247,"children":254},194,"sliding-friction-coulombs-laws","Sliding Friction. Coulomb's Laws","Sliding Friction and Coulomb's Laws — Statics","Sliding friction in statics: friction force, coefficient of friction, limiting friction, angle and cone of friction, and Coulomb's laws.","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":256,"parent_id":236,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":257,"children":264},195,"equilibrium-with-friction","Equilibrium with Friction","Equilibrium with Friction — Engineering Statics","Equilibrium of bodies with dry friction: friction direction, limiting conditions and equilibrium equations for common engineering statics problems.","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":266,"parent_id":236,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":267,"children":274},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":276,"parent_id":46,"code":33,"slug":277,"name":278,"seo_title":279,"seo_description":280,"seo_text":281,"content":282,"locale":8,"uk_topic_id":283,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":277,"children":284},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,[285,333,381,416,443],{"id":286,"parent_id":276,"code":33,"slug":287,"name":288,"seo_title":288,"seo_description":289,"seo_text":290,"content":291,"locale":8,"uk_topic_id":292,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":287,"children":293},284,"particle-kinematics","Particle Kinematics","Methods of describing particle motion, velocity, acceleration, and special cases of 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,[294,304,314,324],{"id":295,"parent_id":286,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":296,"children":303},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":305,"parent_id":286,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":306,"children":313},213,"particle-velocity","Particle Velocity","Particle Velocity in Kinematics","Particle velocity: velocity vector, Cartesian components, magnitude and direction, and velocity in path coordinates.","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":315,"parent_id":286,"code":33,"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":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":316,"children":323},214,"particle-acceleration","Particle Acceleration","Particle Acceleration in Kinematics","Particle acceleration: acceleration vector and Cartesian components, tangential and normal components, and total acceleration.","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":325,"parent_id":286,"code":33,"slug":326,"name":327,"seo_title":327,"seo_description":328,"seo_text":329,"content":330,"locale":8,"uk_topic_id":331,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":326,"children":332},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":334,"parent_id":276,"code":33,"slug":335,"name":336,"seo_title":337,"seo_description":338,"seo_text":339,"content":340,"locale":8,"uk_topic_id":341,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":335,"children":342},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,[343,352,361,371],{"id":344,"parent_id":334,"code":33,"slug":345,"name":346,"seo_title":346,"seo_description":347,"seo_text":348,"content":349,"locale":8,"uk_topic_id":350,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":345,"children":351},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":353,"parent_id":334,"code":33,"slug":354,"name":355,"seo_title":355,"seo_description":356,"seo_text":357,"content":358,"locale":8,"uk_topic_id":359,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":354,"children":360},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":362,"parent_id":334,"code":33,"slug":363,"name":364,"seo_title":365,"seo_description":366,"seo_text":367,"content":368,"locale":8,"uk_topic_id":369,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":363,"children":370},218,"point-velocities-accelerations-fixed-axis-rotation","Velocities and Accelerations of Points in Fixed-Axis Rotation","Point Velocities and Accelerations in Fixed-Axis Rotation","Linear velocity, tangential, normal and total acceleration of rigid-body points during rotation about a fixed axis.","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":372,"parent_id":334,"code":33,"slug":373,"name":374,"seo_title":375,"seo_description":376,"seo_text":377,"content":378,"locale":8,"uk_topic_id":379,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":373,"children":380},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":382,"parent_id":276,"code":33,"slug":383,"name":384,"seo_title":384,"seo_description":385,"seo_text":386,"content":387,"locale":8,"uk_topic_id":388,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":383,"children":389},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,[390,398,407],{"id":391,"parent_id":382,"code":33,"slug":392,"name":384,"seo_title":384,"seo_description":393,"seo_text":394,"content":395,"locale":8,"uk_topic_id":396,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":392,"children":397},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":399,"parent_id":382,"code":33,"slug":400,"name":401,"seo_title":401,"seo_description":402,"seo_text":403,"content":404,"locale":8,"uk_topic_id":405,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":400,"children":406},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":408,"parent_id":382,"code":33,"slug":409,"name":410,"seo_title":410,"seo_description":411,"seo_text":412,"content":413,"locale":8,"uk_topic_id":414,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":409,"children":415},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":417,"parent_id":276,"code":33,"slug":418,"name":419,"seo_title":419,"seo_description":420,"seo_text":421,"content":422,"locale":8,"uk_topic_id":423,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":418,"children":424},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,[425,433],{"id":426,"parent_id":417,"code":33,"slug":427,"name":419,"seo_title":419,"seo_description":428,"seo_text":429,"content":430,"locale":8,"uk_topic_id":431,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":427,"children":432},223,"relative-motion-particle","Absolute, relative and transport motion of a particle and the velocity-addition theorem for moving reference frames.","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":434,"parent_id":417,"code":33,"slug":435,"name":436,"seo_title":437,"seo_description":438,"seo_text":439,"content":440,"locale":8,"uk_topic_id":441,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":435,"children":442},224,"acceleration-addition-coriolis-acceleration","Acceleration Addition. Coriolis Acceleration","Acceleration Addition and Coriolis Acceleration","Absolute, relative, transport and Coriolis acceleration of a particle in relative motion. Acceleration-addition theorem.","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":444,"parent_id":276,"code":33,"slug":445,"name":446,"seo_title":446,"seo_description":447,"seo_text":448,"content":449,"locale":8,"uk_topic_id":450,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":445,"children":451},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,[452,461],{"id":453,"parent_id":444,"code":33,"slug":454,"name":455,"seo_title":455,"seo_description":456,"seo_text":457,"content":458,"locale":8,"uk_topic_id":459,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":454,"children":460},225,"spherical-motion-rigid-body","Spherical Motion of a Rigid Body","Spherical motion of a rigid body with one fixed point: angular velocity, instantaneous axis of rotation and point velocities.","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":462,"parent_id":444,"code":33,"slug":463,"name":464,"seo_title":464,"seo_description":465,"seo_text":466,"content":467,"locale":8,"uk_topic_id":468,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":463,"children":469},226,"general-motion-free-rigid-body","General Motion of a Free Rigid Body","General three-dimensional rigid-body motion: translational and rotational components, point velocities and accelerations.","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":471,"parent_id":46,"code":33,"slug":472,"name":473,"seo_title":474,"seo_description":475,"seo_text":476,"content":477,"locale":8,"uk_topic_id":478,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":472,"children":479},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,[480,526,593,657],{"id":481,"parent_id":471,"code":33,"slug":482,"name":483,"seo_title":483,"seo_description":484,"seo_text":485,"content":486,"locale":8,"uk_topic_id":487,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":482,"children":488},293,"particle-dynamics","Particle Dynamics","Newton's laws, differential equations, fundamental dynamics problems, and particle motion under typical forces.","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,[489,499,508,517],{"id":490,"parent_id":481,"code":33,"slug":491,"name":492,"seo_title":493,"seo_description":494,"seo_text":495,"content":496,"locale":8,"uk_topic_id":497,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":491,"children":498},248,"fundamental-laws-of-dynamics-newtons-laws","Fundamental Laws of Dynamics. Newton's Laws","Fundamental Laws of Dynamics and Newton's Laws","Basic concepts of particle dynamics, inertial reference frames, and Newton's three laws of motion.","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":500,"parent_id":481,"code":33,"slug":501,"name":502,"seo_title":502,"seo_description":503,"seo_text":504,"content":505,"locale":8,"uk_topic_id":506,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":501,"children":507},249,"differential-equations-of-motion-of-a-particle","Differential Equations of Motion of a Particle","Particle equations of motion in vector, Cartesian, and natural-coordinate forms and their use in dynamics.","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":509,"parent_id":481,"code":33,"slug":510,"name":511,"seo_title":511,"seo_description":512,"seo_text":513,"content":514,"locale":8,"uk_topic_id":515,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":510,"children":516},250,"two-fundamental-problems-of-particle-dynamics","Two Fundamental Problems of Particle Dynamics","Direct and inverse dynamics problems: finding motion from forces and forces from a prescribed motion.","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":518,"parent_id":481,"code":33,"slug":519,"name":520,"seo_title":520,"seo_description":521,"seo_text":522,"content":523,"locale":8,"uk_topic_id":524,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":519,"children":525},251,"particle-motion-under-typical-forces","Particle Motion Under Typical Forces","Particle dynamics under gravity, elastic force, dry friction, and resistance of a medium.","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":527,"parent_id":471,"code":33,"slug":528,"name":529,"seo_title":529,"seo_description":530,"seo_text":531,"content":532,"locale":8,"uk_topic_id":533,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":528,"children":534},294,"general-theorems-of-particle-dynamics","General Theorems of Particle Dynamics","Momentum, angular momentum, work, kinetic and potential energy, and conservation laws for a particle.","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,[535,545,554,564,574,584],{"id":536,"parent_id":527,"code":33,"slug":537,"name":538,"seo_title":539,"seo_description":540,"seo_text":541,"content":542,"locale":8,"uk_topic_id":543,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":537,"children":544},252,"linear-momentum-of-a-particle-impulse-of-a-force","Linear Momentum of a Particle. Impulse of a Force","Linear Momentum of a Particle and Impulse of a Force","Linear momentum, impulse of a force, and the impulse-momentum theorem for a particle.","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":546,"parent_id":527,"code":33,"slug":547,"name":548,"seo_title":548,"seo_description":549,"seo_text":550,"content":551,"locale":8,"uk_topic_id":552,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":547,"children":553},253,"angular-momentum-of-a-particle","Angular Momentum of a Particle","Angular momentum of a particle about a point and an axis and the angular-momentum theorem.","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":555,"parent_id":527,"code":33,"slug":556,"name":557,"seo_title":558,"seo_description":559,"seo_text":560,"content":561,"locale":8,"uk_topic_id":562,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":556,"children":563},254,"work-of-a-force-power","Work of a Force. Power","Work of a Force and Power in Dynamics","Elementary and finite work of a force, work of common forces, and mechanical power.","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":565,"parent_id":527,"code":33,"slug":566,"name":567,"seo_title":568,"seo_description":569,"seo_text":570,"content":571,"locale":8,"uk_topic_id":572,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":566,"children":573},255,"kinetic-energy-of-a-particle-work-energy-theorem","Kinetic Energy of a Particle. Work-Energy Theorem","Kinetic Energy of a Particle and Work-Energy Theorem","Particle kinetic energy and the work-energy theorem relating its change to the work of forces.","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":575,"parent_id":527,"code":33,"slug":576,"name":577,"seo_title":578,"seo_description":579,"seo_text":580,"content":581,"locale":8,"uk_topic_id":582,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":576,"children":583},256,"potential-force-field-potential-energy","Potential Force Field. Potential Energy","Potential Force Field and Potential Energy","Conservative forces, potential force fields, potential energy, and the relation between work and potential-energy change.","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":585,"parent_id":527,"code":33,"slug":586,"name":587,"seo_title":587,"seo_description":588,"seo_text":589,"content":590,"locale":8,"uk_topic_id":591,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":586,"children":592},257,"conservation-of-mechanical-energy","Conservation of Mechanical Energy","Conservation of the sum of kinetic and potential energies in conservative mechanical systems.","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":594,"parent_id":471,"code":33,"slug":595,"name":596,"seo_title":596,"seo_description":597,"seo_text":598,"content":599,"locale":8,"uk_topic_id":600,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":595,"children":601},295,"dynamics-of-mechanical-systems","Dynamics of Mechanical Systems","Center of mass, momentum, angular momentum, kinetic energy, and general dynamics theorems for mechanical systems.","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,[602,612,621,630,639,648],{"id":603,"parent_id":594,"code":33,"slug":604,"name":605,"seo_title":606,"seo_description":607,"seo_text":608,"content":609,"locale":8,"uk_topic_id":610,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":604,"children":611},258,"dynamics-of-a-mechanical-system-center-of-mass","Dynamics of a Mechanical System. Center of Mass","Dynamics of a Mechanical System and Center of Mass","Mechanical systems, internal and external forces, system mass, and the center of mass.","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":613,"parent_id":594,"code":33,"slug":614,"name":615,"seo_title":615,"seo_description":616,"seo_text":617,"content":618,"locale":8,"uk_topic_id":619,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":614,"children":620},259,"theorem-on-motion-of-center-of-mass","Theorem on the Motion of the Center of Mass","Equation of motion of a mechanical system's center of mass and consequences for systems with zero resultant external force.","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":622,"parent_id":594,"code":33,"slug":623,"name":624,"seo_title":624,"seo_description":625,"seo_text":626,"content":627,"locale":8,"uk_topic_id":628,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":623,"children":629},261,"impulse-momentum-theorem-mechanical-system","Impulse-Momentum Theorem for a Mechanical System","Linear momentum of a mechanical system, impulse of external forces, and conservation of momentum.","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":631,"parent_id":594,"code":33,"slug":632,"name":633,"seo_title":633,"seo_description":634,"seo_text":635,"content":636,"locale":8,"uk_topic_id":637,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":632,"children":638},262,"angular-momentum-theorem-mechanical-system","Angular-Momentum Theorem for a Mechanical System","Angular momentum of a mechanical system about a point and axis, external moments, and conservation of angular momentum.","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":640,"parent_id":594,"code":33,"slug":641,"name":642,"seo_title":642,"seo_description":643,"seo_text":644,"content":645,"locale":8,"uk_topic_id":646,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":641,"children":647},263,"kinetic-energy-of-a-mechanical-system","Kinetic Energy of a Mechanical System","Kinetic energy of particle systems and rigid bodies in translation, fixed-axis rotation, and plane motion.","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":649,"parent_id":594,"code":33,"slug":650,"name":651,"seo_title":651,"seo_description":652,"seo_text":653,"content":654,"locale":8,"uk_topic_id":655,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":650,"children":656},264,"work-energy-theorem-mechanical-system","Work-Energy Theorem for a Mechanical System","Change in kinetic energy of a mechanical system and the work of external and internal forces.","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":658,"parent_id":471,"code":33,"slug":659,"name":660,"seo_title":661,"seo_description":662,"seo_text":663,"content":664,"locale":8,"uk_topic_id":665,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":659,"children":666},296,"analytical-mechanics-and-oscillations","Analytical Mechanics and Oscillations","Analytical Mechanics and Small Oscillations","D'Alembert's principle, virtual work, general equation of dynamics, Lagrange's equations, and 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,[667,677,687,697,706],{"id":668,"parent_id":658,"code":33,"slug":669,"name":670,"seo_title":671,"seo_description":672,"seo_text":673,"content":674,"locale":8,"uk_topic_id":675,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":669,"children":676},265,"dalemberts-principle-particle-system","D'Alembert's Principle for a Particle and a System","D'Alembert's Principle in Theoretical Mechanics","Inertia forces and D'Alembert's principle for a particle and a mechanical system.","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":678,"parent_id":658,"code":33,"slug":679,"name":680,"seo_title":681,"seo_description":682,"seo_text":683,"content":684,"locale":8,"uk_topic_id":685,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":679,"children":686},266,"virtual-displacements-principle-of-virtual-work","Virtual Displacements. Principle of Virtual Work","Virtual Displacements and Principle of Virtual Work","Ideal constraints, virtual displacements, virtual work, and the principle of virtual work for equilibrium.","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":688,"parent_id":658,"code":33,"slug":689,"name":690,"seo_title":691,"seo_description":692,"seo_text":693,"content":694,"locale":8,"uk_topic_id":695,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":689,"children":696},267,"general-equation-of-dynamics","General Equation of Dynamics","General Equation of Dynamics of a Mechanical System","Combination of D'Alembert's principle and virtual work for systems with ideal constraints.","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":698,"parent_id":658,"code":33,"slug":699,"name":700,"seo_title":700,"seo_description":701,"seo_text":702,"content":703,"locale":8,"uk_topic_id":704,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":699,"children":705},268,"lagranges-equations-of-the-second-kind","Lagrange's Equations of the Second Kind","Generalized coordinates, generalized forces, and Lagrange's equations for mechanical systems.","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":707,"parent_id":658,"code":33,"slug":708,"name":709,"seo_title":709,"seo_description":710,"seo_text":711,"content":712,"locale":8,"uk_topic_id":713,"show_in_theory_list":41,"is_published":40,"status":74,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":708,"children":714},269,"small-oscillations-one-degree-of-freedom-system","Small Oscillations of a One-Degree-of-Freedom System","Free and forced small oscillations of a one-degree-of-freedom mechanical system, natural frequency, 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":716,"parent_id":33,"code":717,"slug":718,"name":719,"seo_title":720,"seo_description":721,"seo_text":722,"content":723,"locale":8,"uk_topic_id":41,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":718,"children":724},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; stress and strain state; shear and torsion; bending and beam deflections; geometric properties of plane areas; 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>",[725,872,1004,1055,1095,1159,1222,1265,1296,1329,1372,1392],{"id":726,"parent_id":716,"code":727,"slug":728,"name":729,"seo_title":730,"seo_description":731,"seo_text":732,"content":733,"locale":8,"uk_topic_id":734,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":728,"children":735},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,[736,746,757,768,779,790,801,813,824,836,848,860],{"id":737,"parent_id":726,"code":738,"slug":739,"name":740,"seo_title":740,"seo_description":741,"seo_text":742,"content":743,"locale":8,"uk_topic_id":744,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":739,"children":745},151,"1.0.1","scope-and-objectives-of-strength-of-materials","Scope and Objectives of Strength of Materials","What Strength of Materials studies, the engineering questions it answers, and how real structures are converted into calculation models.","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":747,"parent_id":726,"code":748,"slug":749,"name":750,"seo_title":751,"seo_description":752,"seo_text":753,"content":754,"locale":8,"uk_topic_id":755,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":749,"children":756},152,"1.0.2","deformation-and-displacement","Deformation and Displacement","Deformation and Displacement in Strength of Materials","Deformation versus displacement, absolute and relative dimensional changes, small strains and their physical meaning.","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 a change in the position of a point or cross-section in space, whereas \u003Cstrong>deformation\u003C\u002Fstrong> describes changes in the relative positions of material points and therefore changes in shape and\u002For dimensions. A body may move as a whole without deforming.\u003C\u002Fp>\u003Ch2>Absolute and relative changes\u003C\u002Fh2>\u003Cp>For a bar of initial length $l$, the change $\\Delta l$ is an absolute elongation or shortening. The normal strain measures this change relative to the initial length: $\\varepsilon=\\Delta l\u002Fl$.\u003C\u002Fp>\u003Ch2>Small deformation\u003C\u002Fh2>\u003Cp>Many basic problems assume that strains and displacements are sufficiently small for equilibrium to be described using the original geometry. Large changes may require a geometrically nonlinear model.\u003C\u002Fp>\u003Ch2>Physical meaning\u003C\u002Fh2>\u003Cp>Deformation is a kinematic consequence of loading. A constitutive material model, such as Hooke’s law in the linear-elastic range, relates strain to stress.\u003C\u002Fp>",144,[],{"id":758,"parent_id":726,"code":759,"slug":760,"name":761,"seo_title":762,"seo_description":763,"seo_text":764,"content":765,"locale":8,"uk_topic_id":766,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":760,"children":767},153,"1.0.3","elastic-and-plastic-deformation","Elastic and Plastic Deformation","Elastic and Plastic Deformation of Materials","The difference between elastic and plastic deformation, unloading behavior, permanent strain, and limits of elastic models.","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":769,"parent_id":726,"code":770,"slug":771,"name":772,"seo_title":773,"seo_description":774,"seo_text":775,"content":776,"locale":8,"uk_topic_id":777,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":771,"children":778},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-en}}\u003Ch2>Simple and combined cases\u003C\u002Fh2>\u003Cp>Tension, compression, shear, torsion, and bending provide fundamental models for engineering analysis. In a real member, several internal force resultants may act simultaneously; this is treated as combined loading.\u003C\u002Fp>\u003Ch2>Why distinguish deformation modes?\u003C\u002Fh2>\u003Cp>Each mode has characteristic internal force resultants, stress distributions, and deformation equations. Correctly recognizing how a member works is therefore a first step in choosing an appropriate calculation model.\u003C\u002Fp>",146,[],{"id":780,"parent_id":726,"code":781,"slug":782,"name":783,"seo_title":784,"seo_description":785,"seo_text":786,"content":787,"locale":8,"uk_topic_id":788,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":782,"children":789},155,"1.0.5","strength-stiffness-and-stability","Strength, Stiffness and Stability","Strength, Stiffness and Stability of Structures","Three fundamental structural performance requirements: strength, stiffness, and stability, their differences and engineering meaning.","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":791,"parent_id":726,"code":792,"slug":793,"name":794,"seo_title":795,"seo_description":796,"seo_text":797,"content":798,"locale":8,"uk_topic_id":799,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":793,"children":800},156,"1.0.6","material-properties-and-models","Basic Material Properties and Models","Material Models: Homogeneity, Isotropy and Anisotropy","Homogeneous and heterogeneous, isotropic, anisotropic and orthotropic materials, plus linear and nonlinear material models.","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>Every calculation equation relies on a particular \u003Cstrong>material model\u003C\u002Fstrong>. Before using it, we must understand which properties are assumed to be the same at different points and directions and how the material responds to loading.\u003C\u002Fp>\u003Ch2>Homogeneity and heterogeneity\u003C\u002Fh2>\u003Cp>A homogeneous model assumes the same properties at all points of the considered volume. In a heterogeneous material, properties vary spatially. Real microstructures are heterogeneous, but at a suitable engineering scale a material can often be represented by effective homogeneous properties.\u003C\u002Fp>\u003Ch2>Isotropy and anisotropy\u003C\u002Fh2>\u003Cp>An \u003Cstrong>isotropic\u003C\u002Fstrong> material model has the same mechanical properties in every direction. In an \u003Cstrong>anisotropic\u003C\u002Fstrong> material, properties depend on direction. \u003Cstrong>Orthotropy\u003C\u002Fstrong> is an important special case with three mutually perpendicular material symmetry directions; it is commonly used for wood and laminated composites.\u003C\u002Fp>\u003Ch2>Linearity and nonlinearity\u003C\u002Fh2>\u003Cp>In a linear-elastic model, stress and strain are linearly related within the model’s range of applicability. Nonlinear response may result from material behavior, large deformation, or other physical effects.\u003C\u002Fp>\u003Ch2>Why the model matters\u003C\u002Fh2>\u003Cp>Equations and constants derived for isotropic materials cannot be applied to anisotropic materials without verification. The model must match the material, scale, loading direction, and required accuracy.\u003C\u002Fp>",148,[],{"id":802,"parent_id":726,"code":803,"slug":804,"name":805,"seo_title":806,"seo_description":807,"seo_text":808,"content":809,"locale":8,"uk_topic_id":810,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":811,"url_slug":804,"children":812},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":814,"parent_id":726,"code":815,"slug":816,"name":817,"seo_title":817,"seo_description":818,"seo_text":819,"content":820,"locale":8,"uk_topic_id":821,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":822,"url_slug":816,"children":823},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":825,"parent_id":726,"code":826,"slug":827,"name":828,"seo_title":829,"seo_description":830,"seo_text":831,"content":832,"locale":8,"uk_topic_id":833,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":834,"url_slug":827,"children":835},163,"1.0.9","loads-and-supports","Loads and Supports","Loads and Supports in Strength of Materials","Concentrated forces and moments, distributed and body loads, static and dynamic actions, supports and reaction forces.","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":837,"parent_id":726,"code":838,"slug":839,"name":840,"seo_title":841,"seo_description":842,"seo_text":843,"content":844,"locale":8,"uk_topic_id":845,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":846,"url_slug":839,"children":847},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":849,"parent_id":726,"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":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":858,"url_slug":851,"children":859},165,"1.0.11","external-and-internal-forces-method-of-sections","External and Internal Forces. Method of Sections","External and Internal Forces and the Method of Sections","The method of sections in Strength of Materials: cutting a body, isolating either part, and introducing internal resultants N, Q, M and 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":861,"parent_id":726,"code":862,"slug":863,"name":864,"seo_title":865,"seo_description":866,"seo_text":867,"content":868,"locale":8,"uk_topic_id":869,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":870,"url_slug":863,"children":871},166,"1.0.12","stress-and-strain-at-a-point","Stress and Strain at a Point","Stress and Strain at a Point: σ, τ, ε and γ","Introduction to normal and shear stresses and normal and shear strains as local measures of the mechanical state of a material.","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":873,"parent_id":716,"code":874,"slug":875,"name":876,"seo_title":877,"seo_description":878,"seo_text":879,"content":880,"locale":8,"uk_topic_id":95,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":875,"children":881},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>",[882,891,967,976,985,995],{"id":883,"parent_id":873,"code":884,"slug":885,"name":886,"seo_title":886,"seo_description":887,"seo_text":888,"content":889,"locale":8,"uk_topic_id":105,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":885,"children":890},47,"3","internal-forces-and-stresses","Internal Forces and Stresses in Cross-Sections","Method of sections, internal force diagrams (N, Q, M), and stress distributions.","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":892,"parent_id":873,"code":893,"slug":894,"name":895,"seo_title":896,"seo_description":897,"seo_text":898,"content":899,"locale":8,"uk_topic_id":116,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":894,"children":900},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>",[901,912,923,934,945,956],{"id":902,"parent_id":892,"code":903,"slug":904,"name":905,"seo_title":906,"seo_description":907,"seo_text":908,"content":909,"locale":8,"uk_topic_id":910,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":904,"children":911},110,"4.1","tensile-test-diagram-characteristic-points","Tensile Test Diagram and Characteristic Points","Tensile Test Diagram — Characteristic Points and Regions","Tensile stress–strain diagram: elastic and plastic regions, proportional, yield and ultimate strengths, necking, and specimen fracture.","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 diagram shows the relationship between stress and strain of a specimen during loading. Its shape is used to determine important elastic, ductility, and strength characteristics of a material.\u003C\u002Fp>\u003Ch2>Main regions\u003C\u002Fh2>\u003Cp>At the beginning of loading, deformation is predominantly elastic, and within the proportional region stress is approximately linearly related to strain.\u003C\u002Fp>{{chunk:uniaxial-hooke-law}}\u003Cp>After the material leaves the elastic range, irreversible plastic deformation develops. For ductile metals, the yield strength or offset yield strength is identified, while the maximum engineering stress is called the ultimate tensile strength.\u003C\u002Fp>\u003Ch2>After the maximum load\u003C\u002Fh2>\u003Cp>In a ductile specimen, localized reduction of cross-section called \u003Cstrong>necking\u003C\u002Fstrong> may occur. Further deformation concentrates in this region and eventually ends in fracture. The exact diagram shape and prominence of individual regions depend on the material and test conditions.\u003C\u002Fp>\u003Ch2>Engineering significance\u003C\u002Fh2>\u003Cp>The diagram is used to assess stiffness, the onset of plastic deformation, strength, and the ability of a material to deform before fracture. These characteristics support material selection and the determination of allowable stresses.\u003C\u002Fp>",96,[],{"id":913,"parent_id":892,"code":914,"slug":915,"name":916,"seo_title":917,"seo_description":918,"seo_text":919,"content":920,"locale":8,"uk_topic_id":921,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":915,"children":922},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":924,"parent_id":892,"code":925,"slug":926,"name":927,"seo_title":928,"seo_description":929,"seo_text":930,"content":931,"locale":8,"uk_topic_id":932,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":926,"children":933},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>\u003Cstrong>Ductility\u003C\u002Fstrong> is the ability of a material to undergo appreciable irreversible deformation before fracture. \u003Cstrong>Brittle behavior\u003C\u002Fstrong> is characterized by fracture with relatively little plastic deformation.\u003C\u002Fp>\u003Ch2>Measures of ductility\u003C\u002Fh2>\u003Cp>After a tensile test, ductility is commonly characterized by percentage elongation and percentage reduction of area. Under comparable test conditions, larger values indicate that the specimen accumulated more plastic deformation before fracture.\u003C\u002Fp>\u003Ch2>Why ductility matters\u003C\u002Fh2>\u003Cp>Plastic deformation can allow local redistribution of stress and may provide visible warning of overload before failure. Brittle fracture often develops with little preceding plastic deformation, so defects, stress concentrations, temperature, and loading conditions require particular attention.\u003C\u002Fp>\u003Ch2>Not an absolute classification\u003C\u002Fh2>\u003Cp>Material behavior depends not only on composition or material name. Temperature, strain rate, stress state, specimen geometry, and environment can change the observed mode of deformation and fracture.\u003C\u002Fp>",98,[],{"id":935,"parent_id":892,"code":936,"slug":937,"name":938,"seo_title":939,"seo_description":940,"seo_text":941,"content":942,"locale":8,"uk_topic_id":943,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":937,"children":944},113,"4.4","allowable-stresses-and-factor-of-safety","Allowable Stresses and Factor of Safety","Allowable Stresses and Factor of Safety in Strength Calculations","Allowable stress, selection of a limiting material strength, and the role of the factor of safety in engineering strength calculations.","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":946,"parent_id":892,"code":947,"slug":948,"name":949,"seo_title":950,"seo_description":951,"seo_text":952,"content":953,"locale":8,"uk_topic_id":954,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":948,"children":955},114,"4.5","strain-hardening-and-reloading","Strain Hardening and Reloading","Strain Hardening and Reloading of Materials","How prior plastic deformation affects subsequent loading: strain hardening, residual strain, elastic unloading, and changes in 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":957,"parent_id":892,"code":958,"slug":959,"name":960,"seo_title":961,"seo_description":962,"seo_text":963,"content":964,"locale":8,"uk_topic_id":965,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":959,"children":966},115,"4.6","external-factors-affecting-mechanical-properties","External Factors Affecting Mechanical Properties","External Factors Affecting Mechanical Properties of Materials","How temperature, strain rate, defects, surface condition, environment, and manufacturing processes affect 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":968,"parent_id":873,"code":33,"slug":969,"name":970,"seo_title":970,"seo_description":971,"seo_text":972,"content":973,"locale":8,"uk_topic_id":974,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":969,"children":975},86,"thermal-deformations","Thermal Deformations and Stresses","Calculating thermally induced stresses in constrained structural members.","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":977,"parent_id":873,"code":978,"slug":979,"name":980,"seo_title":980,"seo_description":981,"seo_text":982,"content":983,"locale":8,"uk_topic_id":15,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":979,"children":984},49,"5","statically-indeterminate-axial","Statically Indeterminate Problems under Axial Loading","Solving statically indeterminate axially loaded members using compatibility equations.","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":986,"parent_id":873,"code":987,"slug":988,"name":989,"seo_title":989,"seo_description":990,"seo_text":991,"content":992,"locale":8,"uk_topic_id":993,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":988,"children":994},116,"2.5","deformation-of-bars-in-tension-and-compression","Deformation of Bars in Tension and Compression","Axial and relative deformation of bars, Δl = NL\u002F(EA), stepped bars, and total elongation under centric axial loading.","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":996,"parent_id":873,"code":997,"slug":998,"name":999,"seo_title":999,"seo_description":1000,"seo_text":1001,"content":1002,"locale":8,"uk_topic_id":9,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":998,"children":1003},117,"2.6","strength-and-stiffness-design-in-tension-and-compression","Strength and Stiffness Design in Tension and Compression","Strength and stiffness checks for axially loaded bars, required cross-sectional area, and allowable axial load calculations.","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>",[],{"id":1005,"parent_id":716,"code":1006,"slug":1007,"name":1008,"seo_title":1009,"seo_description":1010,"seo_text":1011,"content":1012,"locale":8,"uk_topic_id":846,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":1007,"children":1013},54,"10","shear-and-torsion","Shear and Torsion","Shear and Torsion — Shafts and Joint Analysis","Torsional shear stress, torque diagrams, angle of twist, and shear deformation.","Shear and torsion deformations induce shear stresses across member cross-sections. This section covers pure shear fundamentals, the relationship between shear modulus G and Young's modulus E, and torsion theory for circular and hollow shafts. Key formulas include internal torque evaluation, maximum torsional shear stress via polar section modulus, and shaft strength and stiffness verification.","\u003Cp>\u003Cstrong>Shear and torsion\u003C\u002Fstrong> are deformation modes in which shear stresses play a central role. In simple shear, adjacent material layers tend to slide relative to one another; in torsion, cross-sections of a member rotate about its longitudinal axis.\u003C\u002Fp>\u003Ch2>Shear stress and shear strain\u003C\u002Fh2>\u003Cp>Shear stress $\\tau$ acts tangentially to a plane. Engineering shear strain $\\gamma$ measures the change of angle between initially perpendicular material lines. In the linear-elastic range, these quantities are related by Hooke's law in shear.\u003C\u002Fp>{{chunk:shear-hooke-law}}\u003Ch2>From shear to torsion\u003C\u002Fh2>\u003Cp>Torsion is produced by couples or torques acting about the longitudinal axis of a member. In circular shafts, torsion produces shear stresses that increase from zero at the axis to their maximum value at the outer surface.\u003C\u002Fp>{{chunk:circular-shaft-torsion}}\u003Ch2>Topics covered in this section\u003C\u002Fh2>\u003Cp>The child topics cover pure shear and practical joints, torsion of solid and hollow circular shafts, helical springs, and the relationship between torque, rotational speed, and transmitted power.\u003C\u002Fp>\u003Ch2>Strength and stiffness\u003C\u002Fh2>\u003Cp>Members under shear and torsion must be checked not only against allowable shear stress but also against deformation limits. For a shaft, the angle of twist may govern stiffness; for a spring, axial deflection and required spring rate are important.\u003C\u002Fp>\u003Ch2>Basic calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine external forces or applied torques.\u003C\u002Fli>\u003Cli>Find the internal force resultants in each calculation segment.\u003C\u002Fli>\u003Cli>Determine the required cross-sectional properties.\u003C\u002Fli>\u003Cli>Calculate shear stresses.\u003C\u002Fli>\u003Cli>Calculate deformations and verify strength and stiffness.\u003C\u002Fli>\u003C\u002Fol>",[1014,1024,1033,1044],{"id":1015,"parent_id":1005,"code":1016,"slug":1017,"name":1018,"seo_title":1019,"seo_description":1020,"seo_text":1021,"content":1022,"locale":8,"uk_topic_id":858,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":858,"url_slug":1017,"children":1023},55,"11","pure-shear-and-joints","Pure Shear and Design of Joints (Riveted and Welded)","Pure Shear and Design of Fastened Joints","Design of riveted, bolted, and welded connections subjected to shear and bearing.","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":1025,"parent_id":1005,"code":1026,"slug":1027,"name":1028,"seo_title":1028,"seo_description":1029,"seo_text":1030,"content":1031,"locale":8,"uk_topic_id":870,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":870,"url_slug":1027,"children":1032},56,"12","torsion-of-shafts","Torsion of Solid and Hollow Circular Shafts","Polar moment of inertia, torsional shear stress formulas, and shaft design.","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":1034,"parent_id":1005,"code":1035,"slug":1036,"name":1037,"seo_title":1038,"seo_description":1039,"seo_text":1040,"content":1041,"locale":8,"uk_topic_id":1042,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1042,"url_slug":1036,"children":1043},57,"13","design-of-helical-springs","Design of Helical Springs","Design of Helical Springs — Stress and Deflection","Formulas for calculating stresses and deflections in helical coil springs.","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":1045,"parent_id":1005,"code":1046,"slug":1047,"name":1048,"seo_title":1049,"seo_description":1050,"seo_text":1051,"content":1052,"locale":8,"uk_topic_id":4,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1053,"url_slug":1047,"children":1054},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>",14,[],{"id":1056,"parent_id":716,"code":1057,"slug":1058,"name":1059,"seo_title":1060,"seo_description":1061,"seo_text":1062,"content":1063,"locale":8,"uk_topic_id":136,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":1058,"children":1064},50,"6","stress-and-strain-state","Stress and Strain State","Stress and Strain State at a Point","3D and 2D stress state analysis, strain tensor, and principal stress planes.","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>",[1065,1075,1085],{"id":1066,"parent_id":1056,"code":1067,"slug":1068,"name":1069,"seo_title":1070,"seo_description":1071,"seo_text":1072,"content":1073,"locale":8,"uk_topic_id":811,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":811,"url_slug":1068,"children":1074},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":1076,"parent_id":1056,"code":1077,"slug":1078,"name":1079,"seo_title":1080,"seo_description":1081,"seo_text":1082,"content":1083,"locale":8,"uk_topic_id":822,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":822,"url_slug":1078,"children":1084},52,"8","strength-theories","Strength Theories (Failure Criteria)","Strength Theories (Failure Criteria) in Mechanics","Overview of failure criteria: Tresca, von Mises, Mohr-Coulomb theories.","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":1086,"parent_id":1056,"code":1087,"slug":1088,"name":1089,"seo_title":1090,"seo_description":1091,"seo_text":1092,"content":1093,"locale":8,"uk_topic_id":834,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":834,"url_slug":1088,"children":1094},53,"9","generalized-hookes-law","Generalized Hooke's Law","Generalized Hooke's Law — Stress-Strain Relations","Formulas for isotropic 3D stress-strain relations and Poisson's ratio effects.","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":1096,"parent_id":716,"code":1097,"slug":1098,"name":1099,"seo_title":1099,"seo_description":1100,"seo_text":1101,"content":1102,"locale":8,"uk_topic_id":1103,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":1098,"children":1104},59,"15","geometric-properties-of-plane-areas","Geometric Properties of Plane Areas","Centroids, moment of inertia, section modulus, and principal axes calculation.","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 cross-section\u003C\u002Fstrong> describe how area is distributed relative to selected axes and points. Unlike mechanical material properties, they depend only on the section shape, dimensions, and orientation. They are required in calculations of bending, torsion, stability, and combined loading.\u003C\u002Fp>\u003Ch2>Area and centroid\u003C\u002Fh2>\u003Cp>The area $A$ measures the total size of the cross-section, while first moments of area are used to locate its centroid. For a composite section, the centroid is obtained as an area-weighted average of the centroids of its simple parts.\u003C\u002Fp>{{chunk:area-static-moments-centroid}}\u003Ch2>Second moments of area\u003C\u002Fh2>\u003Cp>The second moments $I_x$ and $I_y$ characterize the distribution of area relative to the corresponding axes. Area elements farther from an axis make a larger contribution because the distance enters quadratically.\u003C\u002Fp>{{chunk:area-second-moments}}\u003Ch2>Translation and rotation of axes\u003C\u002Fh2>\u003Cp>When the required axis is parallel to a known centroidal axis, the parallel-axis theorem is used. When the axis orientation changes, transformation formulas are applied to determine principal centroidal axes and principal second moments.\u003C\u002Fp>{{chunk:parallel-axis-theorem}}\u003Ch2>Section moduli\u003C\u002Fh2>\u003Cp>In bending calculations, the section modulus is $W=I\u002Fy_{\\max}$, where $y_{\\max}$ is the distance from the neutral axis to the relevant extreme fiber. For an unsymmetrical section, section moduli for the upper and lower extreme fibers may differ.\u003C\u002Fp>\u003Ch2>Simple and composite sections\u003C\u002Fh2>{{chunk:rectangle-circle-area-properties}}\u003Cp>Complex profiles are divided into rectangles, triangles, circles, or other simple parts. After locating the common centroid, the properties of each part are transferred to common axes and summed.\u003C\u002Fp>{{chunk:composite-section-properties-algorithm}}\u003Ch2>Section structure\u003C\u002Fh2>\u003Cp>The child topics separately address first moments and centroids, second moments of area, the parallel-axis theorem, rotation of axes, principal properties, and section moduli. This sequence provides the geometric foundation for subsequent stress and deformation calculations in bending.\u003C\u002Fp>",15,[1105,1116,1127,1137,1148],{"id":1106,"parent_id":1096,"code":1107,"slug":1108,"name":1109,"seo_title":1110,"seo_description":1111,"seo_text":1112,"content":1113,"locale":8,"uk_topic_id":1114,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":1108,"children":1115},119,"15.1","centroid-and-first-moments-of-area","Centroid and First Moments of Area","Centroid and First Moments of Area of Plane Sections","First moments of area Sx and Sy, centroid coordinates for simple and composite plane sections, symmetry rules, and treatment of holes.","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":1117,"parent_id":1096,"code":1118,"slug":1119,"name":1120,"seo_title":1121,"seo_description":1122,"seo_text":1123,"content":1124,"locale":8,"uk_topic_id":1125,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":1119,"children":1126},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":1128,"parent_id":1096,"code":1129,"slug":1130,"name":1131,"seo_title":1131,"seo_description":1132,"seo_text":1133,"content":1134,"locale":8,"uk_topic_id":1135,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":1130,"children":1136},121,"15.3","parallel-axis-theorem-and-composite-sections","Parallel-Axis Theorem and Composite Sections","Transfer second moments to parallel axes and calculate composite-section centroids and moments of inertia, including holes and simple components.","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":1138,"parent_id":1096,"code":1139,"slug":1140,"name":1141,"seo_title":1142,"seo_description":1143,"seo_text":1144,"content":1145,"locale":8,"uk_topic_id":1146,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":1140,"children":1147},122,"15.4","product-of-inertia-and-axis-rotation","Product of Inertia and Axis Rotation","Product of Inertia and Rotation of Area Axes","Product of inertia Ixy, transformation of Ix, Iy and Ixy under axis rotation, invariance of Ix + Iy, and the condition for principal 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":1149,"parent_id":1096,"code":1150,"slug":1151,"name":1152,"seo_title":1153,"seo_description":1154,"seo_text":1155,"content":1156,"locale":8,"uk_topic_id":1157,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":1151,"children":1158},123,"15.5","principal-axes-principal-moments-and-section-moduli","Principal Axes, Principal Moments, and Section Moduli","Principal Axes, Principal Moments and Section Moduli","Principal centroidal axes and moments of area, section modulus W = I\u002Fymax, and geometric properties used in bending 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":1160,"parent_id":716,"code":1161,"slug":1162,"name":1163,"seo_title":1164,"seo_description":1165,"seo_text":1166,"content":1167,"locale":8,"uk_topic_id":1053,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":1162,"children":1168},58,"14","bending","Bending","Bending of Beams — Direct and Transverse Bending","Beam bending theory, shear force and bending moment diagrams.","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>",[1169,1180,1191],{"id":1170,"parent_id":1160,"code":1171,"slug":1172,"name":1173,"seo_title":1174,"seo_description":1175,"seo_text":1176,"content":1177,"locale":8,"uk_topic_id":1178,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":1172,"children":1179},60,"16","internal-forces-in-bending","Internal Forces in Bending","Internal Forces in Bending — V and M Diagrams","Constructing shear force (V) and bending moment (M) diagrams for beams.","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":1181,"parent_id":1160,"code":1182,"slug":1183,"name":1184,"seo_title":1185,"seo_description":1186,"seo_text":1187,"content":1188,"locale":8,"uk_topic_id":1189,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":1183,"children":1190},61,"17","stresses-in-bending","Stresses in Bending","Stresses in Bending — Flexural and Shear Stresses","Flexure formula and Jourawski shear stress equation for beam design.","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":1192,"parent_id":1160,"code":1193,"slug":1194,"name":1195,"seo_title":1196,"seo_description":1197,"seo_text":1198,"content":1199,"locale":8,"uk_topic_id":1200,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":1194,"children":1201},62,"18","deflections-in-bending","Deflections in Bending","Deflections in Bending — Beam Slope and Deflection","Methods for calculating beam deflections and slope angles under load.","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>. Beam stiffness is assessed using 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 cross-section rotation is approximately the first 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 \u003Cstrong>flexural rigidity\u003C\u002Fstrong>. Increasing $E$ or $I$ reduces curvature and, all else being equal, reduces beam deflections. This is why the geometric distribution of material within the cross-section strongly influences stiffness.\u003C\u002Fp>\u003Ch2>Boundary conditions\u003C\u002Fh2>{{chunk:beam-deflection-boundary-conditions}}\u003Cp>Boundary conditions determine integration constants and must correspond to the actual supports and connections. For a multi-segment beam, continuity of deflection and rotation is also imposed at ordinary internal boundaries without hinges or discontinuities.\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 sections of the course, beam displacements may also be evaluated by energy methods.\u003C\u002Fp>\u003Ch2>Stiffness check\u003C\u002Fh2>\u003Cp>The calculated maximum deflection is compared with the allowable value specified by serviceability requirements or the adopted design method: $|w_{\\max}|\\le[w]$. Where necessary, rotations are limited in a similar way.\u003C\u002Fp>\u003Ch2>Influence of section depth\u003C\u002Fh2>\u003Cp>For a rectangular section, $I=bh^3\u002F12$. If $h$ is doubled while $b$, $E$, $L$, and loading remain unchanged, $I$ increases by $2^3=8$. Within the same linear model, characteristic deflections, which are inversely proportional to $EI$, decrease substantially.\u003C\u002Fp>",18,[1202,1212],{"id":1203,"parent_id":1192,"code":1204,"slug":1205,"name":1206,"seo_title":1206,"seo_description":1207,"seo_text":1208,"content":1209,"locale":8,"uk_topic_id":1210,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1210,"url_slug":1205,"children":1211},63,"19","differential-equation-elastic-curve","Differential Equation of the Elastic Curve","Derivation and integration of the elastic curve differential equation.","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":1213,"parent_id":1192,"code":1214,"slug":1215,"name":1216,"seo_title":1216,"seo_description":1217,"seo_text":1218,"content":1219,"locale":8,"uk_topic_id":1220,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1220,"url_slug":1215,"children":1221},64,"20","initial-parameter-method","Initial Parameter Method (Macaulay's Method)","Using Macaulay's method to write unified beam deflection equations.","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>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":1223,"parent_id":716,"code":1224,"slug":1225,"name":1226,"seo_title":1227,"seo_description":1228,"seo_text":1229,"content":1230,"locale":8,"uk_topic_id":1231,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":136,"url_slug":1225,"children":1232},65,"21","combined-loading","Combined Loading","Combined Loading Analysis — Bending, Torsion & Axial","Analyzing structural members subjected to multiple simultaneous loads.","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,[1233,1244,1254],{"id":1234,"parent_id":1223,"code":1235,"slug":1236,"name":1237,"seo_title":1238,"seo_description":1239,"seo_text":1240,"content":1241,"locale":8,"uk_topic_id":1242,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":1236,"children":1243},66,"22","unsymmetrical-bending","Unsymmetrical Bending","Unsymmetrical Bending — Neutral Axis & Stresses","Calculating stresses in unsymmetrical bending and locating the neutral axis.","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":1245,"parent_id":1223,"code":33,"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":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":41,"url_slug":1246,"children":1253},67,"eccentric-compression","Eccentric Compression","Eccentric Compression — Secant Formula & Section Core","Stress distribution under eccentric axial loads and section core determination.","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":1255,"parent_id":1223,"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":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":95,"url_slug":1257,"children":1264},68,"23","combined-bending-and-torsion","Combined Bending and Torsion","Combined Bending and Torsion in Shaft Design","Equivalent bending moment equations and shaft strength verification.","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":1266,"parent_id":716,"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":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":811,"url_slug":1268,"children":1275},69,"24","energetic-methods","Energy Methods","Energy Methods in Structural Mechanics","Strain energy principles, Castigliano's theorem, and virtual work method.","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,[1276,1286],{"id":1277,"parent_id":1266,"code":1278,"slug":1279,"name":1280,"seo_title":1280,"seo_description":1281,"seo_text":1282,"content":1283,"locale":8,"uk_topic_id":1284,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1284,"url_slug":1279,"children":1285},70,"25","work-of-external-forces-and-strain-energy","Work of External Forces and Strain Energy","Formulas for elastic strain energy in axial, torsional, and flexural loading.","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>",25,[],{"id":1287,"parent_id":1266,"code":1288,"slug":1289,"name":1290,"seo_title":1290,"seo_description":1291,"seo_text":1292,"content":1293,"locale":8,"uk_topic_id":1294,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1294,"url_slug":1289,"children":1295},71,"26","castiglianos-theorem-and-mohrs-integral","Castigliano's Theorem and Mohr's Integral","Calculating structural deflections using Mohr's integral and Vereshchagin's 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":1297,"parent_id":716,"code":1298,"slug":1299,"name":1300,"seo_title":1301,"seo_description":1302,"seo_text":1303,"content":1304,"locale":8,"uk_topic_id":1305,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":822,"url_slug":1299,"children":1306},72,"27","stability-of-compressed-members","Stability of Compressed Members","Stability of Compressed Members — Buckling","Buckling analysis of columns, critical load, and elastic stability.","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 a compressed member\u003C\u002Fstrong> is its ability to preserve the initial equilibrium configuration under a compressive force. A long, slender column may lose stability by lateral buckling before the average compressive stress reaches the material strength limit.\u003C\u002Fp>\u003Ch2>Critical load\u003C\u002Fh2>\u003Cp>The \u003Cstrong>critical load $N_{cr}$\u003C\u002Fstrong> is the compressive force at which the straight equilibrium configuration becomes unstable and a laterally deflected configuration becomes possible. For an ideal slender elastic column, the critical load is given by Euler's formula.\u003C\u002Fp>{{chunk:euler-column-critical-load}}\u003Ch2>Effect of end conditions\u003C\u002Fh2>\u003Cp>Buckling resistance depends not only on the material and cross-section but also on the end restraints. Their influence is represented by the effective length $l_{eff}=\\mu L$.\u003C\u002Fp>{{chunk:column-effective-length-factors}}\u003Ch2>Slenderness\u003C\u002Fh2>\u003Cp>The dimensionless slenderness ratio $\\lambda$ compares the effective column length with the geometric stiffness of the cross-section. Buckling occurs about the weaker axis, so the minimum radius of gyration $i_{min}$ is decisive.\u003C\u002Fp>{{chunk:column-radius-slenderness}}\u003Ch2>Euler's formula is not universal\u003C\u002Fh2>\u003Cp>Euler's formula describes elastic buckling of sufficiently slender columns. If the calculated critical stress $\\sigma_{cr,E}$ lies beyond the proportional range of the material, the assumptions of the Euler model are no longer satisfied. Intermediate-slenderness members require empirical or code-based relations.\u003C\u002Fp>{{chunk:euler-applicability-limit}}\u003Ch2>Practical calculation procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine the actual member length $L$ and end-restraint scheme.\u003C\u002Fli>\u003Cli>Calculate area $A$, centroidal second moments of area, and $I_{min}$.\u003C\u002Fli>\u003Cli>Calculate $i_{min}$, effective length $l_{eff}$, and slenderness $\\lambda$.\u003C\u002Fli>\u003Cli>Determine which critical-state model is applicable for the resulting slenderness and material.\u003C\u002Fli>\u003Cli>Calculate $N_{cr}$ or the allowable compressive force according to the adopted method.\u003C\u002Fli>\u003Cli>Check material strength separately when it may govern.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Real columns\u003C\u002Fh2>\u003Cp>Real members have initial crookedness, load eccentricity, residual stresses, and imperfect restraints. Design practice therefore uses code factors and buckling curves, while Euler's formula remains the fundamental idealized model.\u003C\u002Fp>",27,[1307,1318],{"id":1308,"parent_id":1297,"code":1309,"slug":1310,"name":1311,"seo_title":1312,"seo_description":1313,"seo_text":1314,"content":1315,"locale":8,"uk_topic_id":1316,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1316,"url_slug":1310,"children":1317},73,"28","eulers-formula-critical-load","Euler's Formula for Critical Load","Buckling Load Formula for Columns | Euler Critical Load","Euler buckling load formula for slender columns: Pcr = π²EI\u002F(KL)², effective length factors, slenderness, units, assumptions, and a worked example.","Use Euler's buckling load formula to calculate the elastic critical load of a slender column. The page explains Pcr = π²EI\u002F(KL)², effective length and end conditions, the weakest buckling axis, slenderness ratio, units, assumptions, and a worked numerical example.","\u003Cp>\u003Cstrong>Euler's buckling load formula\u003C\u002Fstrong> gives the elastic critical load of an idealized slender column. It is one of the standard formulas used to estimate the load at which a straight compressed member becomes unstable by buckling.\u003C\u002Fp>\u003Ch2>Column buckling load formula\u003C\u002Fh2>{{chunk:euler-column-critical-load}}\u003Cp>Using the effective-length notation $l_{eff}=KL$, the same relation is commonly written as:\u003C\u002Fp>\u003Cp>$$P_{cr}=\\frac{\\pi^2EI_{min}}{(KL)^2}.$$\u003C\u002Fp>\u003Cul>\u003Cli>\u003Cstrong>$P_{cr}$ or $N_{cr}$\u003C\u002Fstrong> — Euler critical buckling load;\u003C\u002Fli>\u003Cli>\u003Cstrong>$E$\u003C\u002Fstrong> — Young's modulus;\u003C\u002Fli>\u003Cli>\u003Cstrong>$I_{min}$\u003C\u002Fstrong> — minimum second moment of area of the cross-section;\u003C\u002Fli>\u003Cli>\u003Cstrong>$L$\u003C\u002Fstrong> — actual column length;\u003C\u002Fli>\u003Cli>\u003Cstrong>$K$\u003C\u002Fstrong> — effective-length factor determined by the end conditions;\u003C\u002Fli>\u003Cli>\u003Cstrong>$KL=l_{eff}$\u003C\u002Fstrong> — effective buckling length.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>How to use Euler's formula\u003C\u002Fh2>\u003Col>\u003Cli>Determine the column end conditions and the effective-length factor $K$.\u003C\u002Fli>\u003Cli>Find the minimum second moment of area $I_{min}$, because buckling normally occurs about the weaker principal axis.\u003C\u002Fli>\u003Cli>Use consistent units for $E$, $I$, and $L$.\u003C\u002Fli>\u003Cli>Calculate $P_{cr}=\\pi^2EI_{min}\u002F(KL)^2$.\u003C\u002Fli>\u003Cli>Check that the member is sufficiently slender for elastic Euler buckling to be applicable.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Effective length and end conditions\u003C\u002Fh2>{{chunk:column-effective-length-factors}}\u003Cp>Because the critical load varies with $1\u002FK^2$, end restraint has a strong effect on buckling resistance. For the same $E$, $I$, and $L$, a fixed-free column with $K=2$ has only one quarter of the Euler load of a pinned-pinned column with $K=1$.\u003C\u002Fp>\u003Ch2>Worked example\u003C\u002Fh2>\u003Cp>Consider a pinned-pinned steel column with $E=200\\ \\text{GPa}$, $L=2\\ \\text{m}$, and $I_{min}=2\\cdot10^{-6}\\ \\text{m}^4$. For pinned ends, $K=1$:\u003C\u002Fp>\u003Cp>$$P_{cr}=\\frac{\\pi^2(200\\cdot10^9)(2\\cdot10^{-6})}{(1\\cdot2)^2}\\approx9.87\\cdot10^5\\ \\text{N}=987\\ \\text{kN}.$$\u003C\u002Fp>\u003Cp>If the same member were fixed-free, $K=2$ and the critical load would be four times smaller, $\\approx247\\ \\text{kN}$.\u003C\u002Fp>\u003Ch2>Radius of gyration and slenderness\u003C\u002Fh2>{{chunk:column-radius-slenderness}}\u003Cp>Substituting $I_{min}=Ai_{min}^2$ and $l_{eff}=\\lambda i_{min}$ gives the Euler critical stress:\u003C\u002Fp>\u003Cp>$$\\sigma_{cr,E}=\\frac{P_{cr}}{A}=\\frac{\\pi^2E}{\\lambda^2}.$$\u003C\u002Fp>\u003Cp>This form shows directly why long, slender columns buckle at lower average compressive stress.\u003C\u002Fp>\u003Ch2>Weakest buckling axis\u003C\u002Fh2>\u003Cp>If $I_x\\ne I_y$, calculate or compare buckling resistance in both principal planes. With the same end conditions, the governing mode is normally associated with the smaller second moment of area and smaller radius of gyration.\u003C\u002Fp>\u003Ch2>When Euler's buckling formula applies\u003C\u002Fh2>\u003Cp>The classical formula assumes a sufficiently slender member, linearly elastic material behavior, centric compression, an initially straight column, idealized end restraints, and small imperfections. Short and intermediate columns require other relations or code-based buckling curves.\u003C\u002Fp>{{chunk:euler-applicability-limit}}",28,[],{"id":1319,"parent_id":1297,"code":1320,"slug":1321,"name":1322,"seo_title":1323,"seo_description":1324,"seo_text":1325,"content":1326,"locale":8,"uk_topic_id":1327,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1327,"url_slug":1321,"children":1328},74,"29","limits-of-applicability-eulers-formula","Limits of Applicability of Euler's Formula (Yasinsky Formula)","Limits of Euler's Formula & Engesser\u002FYasinsky Equations","Inelastic buckling analysis using empirical formulas for medium slenderness.","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>Euler's formula describes buckling only when the column material remains within the linearly elastic range up to the critical state. Therefore, before using it, the \u003Cstrong>column slenderness $\\lambda$\u003C\u002Fstrong> and corresponding critical stress $\\sigma_{cr,E}$ must be checked.\u003C\u002Fp>{{chunk:euler-applicability-limit}}\u003Ch2>Limiting slenderness\u003C\u002Fh2>\u003Cp>If $\\lambda\\ge\\lambda_{lim}$, Euler's critical stress does not exceed the proportional limit $\\sigma_p$ and the classical elastic model may be applicable. If $\\lambda$ is smaller, Euler's formula predicts stresses for which the assumption of linear elasticity is no longer satisfied.\u003C\u002Fp>\u003Ch2>Intermediate-slenderness columns\u003C\u002Fh2>\u003Cp>For the intermediate range, empirical relations or code-based buckling curves are used. In traditional strength-of-materials courses, one such approximation is the Yasinsky formula.\u003C\u002Fp>{{chunk:yasinsky-column-formula}}\u003Ch2>Yasinsky coefficients\u003C\u002Fh2>\u003Cp>The coefficients $a$ and $b$ must be taken for the specific material from the table or method on which the calculation is based. Coefficients for one material must not be transferred to another, and the formula must not be used outside its specified slenderness range.\u003C\u002Fp>\u003Ch2>Low slenderness\u003C\u002Fh2>\u003Cp>For short, stocky compression members, overall buckling may cease to be the governing limit state. Load capacity may instead be controlled by material strength, local effects, or other mechanisms included in the adopted model.\u003C\u002Fp>\u003Ch2>Practical code-based form\u003C\u002Fh2>\u003Cp>In applied design methods, the stability of compression members is often represented by a reduction factor $\\varphi$ or an equivalent buckling resistance that depends on slenderness, material, cross-section type, imperfections, and other parameters. The actual relation must be taken from the applicable standard rather than automatically identified with the Yasinsky formula.\u003C\u002Fp>\u003Ch2>Model-selection procedure\u003C\u002Fh2>\u003Col>\u003Cli>Determine $i_{min}$ and slenderness $\\lambda$.\u003C\u002Fli>\u003Cli>Use the material properties to determine the elastic range and $\\lambda_{lim}$.\u003C\u002Fli>\u003Cli>For sufficiently large $\\lambda$, use Euler's formula.\u003C\u002Fli>\u003Cli>For intermediate $\\lambda$, use the permitted empirical or code-based relation.\u003C\u002Fli>\u003Cli>For low $\\lambda$, check material strength and other relevant limit states.\u003C\u002Fli>\u003Cli>For design calculations, apply the safety factors and rules required by the selected standard.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Example of the applicability boundary\u003C\u002Fh2>\u003Cp>If $E=200\\ \\text{GPa}$ and the proportional limit is $\\sigma_p=200\\ \\text{MPa}$:\u003C\u002Fp>\u003Cp>$$\\lambda_{lim}=\\pi\\sqrt{\\frac{200000}{200}}\\approx99.3.$$\u003C\u002Fp>\u003Cp>Thus, in this idealized estimate, Euler's formula satisfies $\\sigma_{cr,E}\\le\\sigma_p$ approximately for $\\lambda\\ge99.3$.\u003C\u002Fp>",29,[],{"id":1330,"parent_id":716,"code":1331,"slug":1332,"name":1333,"seo_title":1334,"seo_description":1335,"seo_text":1336,"content":1337,"locale":8,"uk_topic_id":1338,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":834,"url_slug":1332,"children":1339},75,"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,[1340,1350,1361],{"id":1341,"parent_id":1330,"code":33,"slug":1342,"name":1343,"seo_title":1344,"seo_description":1345,"seo_text":1346,"content":1347,"locale":8,"uk_topic_id":1348,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":1342,"children":1349},76,"inertia-forces","Inertia Forces","Inertia Forces — D'Alembert's Principle in Mechanics","Calculating dynamic stresses caused by accelerated structural motion.","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":1351,"parent_id":1330,"code":1352,"slug":1353,"name":1354,"seo_title":1355,"seo_description":1356,"seo_text":1357,"content":1358,"locale":8,"uk_topic_id":1359,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1359,"url_slug":1353,"children":1360},77,"31","impact-loading-calculations","Impact Loading Calculations","Impact Loading Calculations — Dynamic Factor","Evaluating stress and deformation under impact loads using energy balance.","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":1362,"parent_id":1330,"code":1363,"slug":1364,"name":1365,"seo_title":1366,"seo_description":1367,"seo_text":1368,"content":1369,"locale":8,"uk_topic_id":1370,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1370,"url_slug":1364,"children":1371},78,"32","material-fatigue-and-endurance-limit","Material Fatigue and Endurance Limit","Material Fatigue and Endurance Limit — S-N Curve","Fatigue life prediction, stress concentration factors, and endurance limit.","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":1373,"parent_id":716,"code":33,"slug":1374,"name":1375,"seo_title":1376,"seo_description":1377,"seo_text":1378,"content":1379,"locale":8,"uk_topic_id":1380,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":846,"url_slug":1374,"children":1381},124,"shell-analysis","Shell Analysis","Shell Analysis — Membrane Forces and Thin-Walled Structures","Thin-walled shell analysis: midsurface geometry, membrane force resultants, stresses, Laplace equilibrium, pressure vessels, and strength checks.","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,[1382],{"id":1383,"parent_id":1373,"code":33,"slug":1384,"name":1385,"seo_title":1386,"seo_description":1387,"seo_text":1388,"content":1389,"locale":8,"uk_topic_id":1390,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":1384,"children":1391},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":1393,"parent_id":716,"code":1394,"slug":1394,"name":1395,"seo_title":1396,"seo_description":1397,"seo_text":1398,"content":1399,"locale":8,"uk_topic_id":1400,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":858,"url_slug":1394,"children":1401},126,"contact-stresses","Contact Stresses","Contact Stresses — Fundamentals of Hertz Contact","Contact stresses in engineering: contact geometry, Hertz theory, pressure distributions, sphere and cylinder contact, subsurface stresses, and 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,[1402,1413,1423,1433,1444,1455],{"id":1403,"parent_id":1393,"code":1404,"slug":1405,"name":1406,"seo_title":1407,"seo_description":1408,"seo_text":1409,"content":1410,"locale":8,"uk_topic_id":1411,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":846,"url_slug":1405,"children":1412},127,"contact-geometry","contact-types-and-geometry","Contact of Two Bodies: Types and Geometry","Types and Geometry of Contact Between Two Bodies","Point, line, and surface contact: contact-zone geometry, surface curvature, reduced radius, and choosing an appropriate contact model.","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":1414,"parent_id":1393,"code":1415,"slug":1415,"name":1416,"seo_title":1417,"seo_description":1418,"seo_text":1419,"content":1420,"locale":8,"uk_topic_id":1421,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1220,"url_slug":1415,"children":1422},128,"hertz-contact-theory","Hertz Theory: Main Assumptions and Formulas","Hertz Theory for Contact Stresses","Hertz elastic contact theory: assumptions, reduced modulus and curvature, contact pressure, sphere and cylinder formulas, and applicability limits.","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 theory\u003C\u002Fstrong> describes local elastic contact of smooth curved bodies. It relates normal force $F$, local geometry, and elastic material properties to the size of the contact region and the pressure distribution.\u003C\u002Fp>\u003Ch2>Main assumptions\u003C\u002Fh2>\u003Cul>\u003Cli>materials are homogeneous, isotropic, and linearly elastic;\u003C\u002Fli>\u003Cli>deformations are small;\u003C\u002Fli>\u003Cli>the contact region is small compared with characteristic body dimensions and radii of curvature;\u003C\u002Fli>\u003Cli>surfaces are locally smooth;\u003C\u002Fli>\u003Cli>the basic normal-contact problem does not include significant tangential forces;\u003C\u002Fli>\u003Cli>plastic deformation is not large enough to invalidate the elastic model.\u003C\u002Fli>\u003C\u002Ful>\u003Ch2>Reduced modulus\u003C\u002Fh2>{{chunk:hertz-reduced-elastic-modulus}}\u003Cp>The reduced modulus $E^*$ accounts for elastic deformation of both contacting bodies.\u003C\u002Fp>\u003Ch2>Reduced geometry\u003C\u002Fh2>{{chunk:hertz-reduced-radius}}\u003Cp>General elliptical contact requires the principal curvatures of both surfaces in two directions. A single $R^*$ is directly convenient for axisymmetric spherical contact and plane line-contact models.\u003C\u002Fp>\u003Ch2>Pressure distribution\u003C\u002Fh2>{{chunk:hertz-pressure-distribution}}\u003Cp>Hertz pressure $p$ has a smooth maximum $p_0$ at the center and decreases to zero at the contact boundary.\u003C\u002Fp>\u003Ch2>Spherical contact\u003C\u002Fh2>{{chunk:hertz-sphere-contact}}\u003Cp>For fixed $E^*$ and $R^*$, the contact radius scales as $a\\propto F^{1\u002F3}$ and the maximum pressure also scales as $p_0\\propto F^{1\u002F3}$.\u003C\u002Fp>\u003Ch2>Line contact\u003C\u002Fh2>{{chunk:hertz-cylinder-line-contact}}\u003Cp>For long cylinders, contact-strip half-width $b$ varies with the square root of load per unit length $F'$. End effects are omitted from the simple two-dimensional model.\u003C\u002Fp>\u003Ch2>Calculation procedure\u003C\u002Fh2>{{chunk:hertz-contact-calculation-algorithm}}\u003Ch2>Applicability limits\u003C\u002Fh2>\u003Cp>The classical model requires refinement for significant plasticity, large contact regions, substantial friction or sliding, rough surfaces, conformal geometry, thin coatings, edge contact, and strongly heterogeneous materials.\u003C\u002Fp>",91,[],{"id":1424,"parent_id":1393,"code":1425,"slug":1426,"name":1427,"seo_title":1427,"seo_description":1428,"seo_text":1429,"content":1430,"locale":8,"uk_topic_id":1431,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1338,"url_slug":1426,"children":1432},129,"sphere-contact","contact-of-spheres-and-spherical-surfaces","Contact of Spheres and Spherical Surfaces","Elastic contact of spheres and spherical surfaces: contact radius, maximum Hertz pressure, reduced properties, examples, 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":1434,"parent_id":1393,"code":1435,"slug":1436,"name":1437,"seo_title":1438,"seo_description":1439,"seo_text":1440,"content":1441,"locale":8,"uk_topic_id":1442,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":53,"url_slug":1436,"children":1443},130,"cylinder-roller-contact","contact-of-cylinders-and-rollers","Contact of Cylinders and Rollers","Contact of Cylinders and Rollers — Hertz Line Contact","Hertz line contact of cylinders and rollers: contact-strip width, maximum pressure, reduced properties, edge effects, and engineering applications.","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>\u003Cstrong>Contact of cylinders and rollers\u003C\u002Fstrong> is a typical model of initially line contact. For long parallel cylinders under a normal force $F$, the initial contact line expands into a narrow strip.\u003C\u002Fp>\u003Ch2>Load per unit length\u003C\u002Fh2>\u003Cp>If total normal force $F$ is transmitted approximately uniformly over effective contact length $L$, define:\u003C\u002Fp>\u003Cp>$$F'=\\frac{F}{L},$$\u003C\u002Fp>\u003Cp>where $F'$ has units of force per unit length.\u003C\u002Fp>\u003Ch2>Reduced properties\u003C\u002Fh2>{{chunk:hertz-reduced-elastic-modulus}}{{chunk:hertz-reduced-radius}}\u003Cp>For a cylinder on a plane, $R^*=R$ in the transverse plane. For two convex cylinders, their curvatures add; for a convex-concave pair, curvature signs follow the adopted convention.\u003C\u002Fp>\u003Ch2>Contact width and pressure\u003C\u002Fh2>{{chunk:hertz-cylinder-line-contact}}\u003Cp>The full contact-strip width is $2b$. Pressure $p(x)$ is maximum at the centerline, where $p(0)=p_0$, and decreases to zero at $x=\\pm b$.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Let a steel roller of radius $R=20\\ \\text{mm}$ contact a steel plane over $L=50\\ \\text{mm}$ under $F=5000\\ \\text{N}$. For identical steels with $E_1=E_2=210000\\ \\text{MPa}$ and $\\nu_1=\\nu_2=0.30$, $E^*\\approx115385\\ \\text{MPa}$ and $F'=100\\ \\text{N\u002Fmm}$.\u003C\u002Fp>\u003Cp>Then:\u003C\u002Fp>\u003Cp>$$b=\\sqrt{\\frac{4\\cdot100\\cdot20}{\\pi\\cdot115385}}\\approx0.149\\ \\text{mm}.$$\u003C\u002Fp>\u003Cp>The maximum pressure is:\u003C\u002Fp>\u003Cp>$$p_0=\\frac{2\\cdot100}{\\pi\\cdot0.149}\\approx427\\ \\text{MPa}.$$\u003C\u002Fp>\u003Ch2>Edge effects\u003C\u002Fh2>\u003Cp>A real roller has finite length $L$. Misalignment, sharp edges, and nonuniform load distribution can raise pressure near the ends, an effect absent from the ideal two-dimensional model. Crowning or other profile modifications may be used to reduce edge concentration.\u003C\u002Fp>\u003Ch2>Applications\u003C\u002Fh2>\u003Cp>The line-contact model is used for roller bearings, rollers, wheels on rails, and other components where $L$ is much greater than the elastic strip width $2b$.\u003C\u002Fp>\u003Ch2>Procedure\u003C\u002Fh2>{{chunk:hertz-contact-calculation-algorithm}}",93,[],{"id":1445,"parent_id":1393,"code":1446,"slug":1447,"name":1448,"seo_title":1449,"seo_description":1450,"seo_text":1451,"content":1452,"locale":8,"uk_topic_id":1453,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1056,"url_slug":1447,"children":1454},131,"subsurface-contact-stress","subsurface-contact-stress-state","Stress State Beneath the Contact Surface","Stress State Beneath a Contact Surface","Subsurface contact stresses: normal, principal, and shear components, depth variation, friction effects, and identifying critical regions for strength.","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":1456,"parent_id":1393,"code":1457,"slug":1458,"name":1459,"seo_title":1459,"seo_description":1460,"seo_text":1461,"content":1462,"locale":8,"uk_topic_id":1463,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":1170,"url_slug":1458,"children":1464},132,"contact-strength-machine-elements","contact-strength-of-machine-elements","Contact Strength of Machine Elements","Contact-strength assessment for bearings, rollers, gears, and wheels: Hertz pressure, pitting, fatigue life, lubrication, roughness, and edge effects.","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>",95,[],{"id":1466,"parent_id":33,"code":33,"slug":1467,"name":1468,"seo_title":1469,"seo_description":1470,"seo_text":1471,"content":33,"locale":8,"uk_topic_id":1472,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":105,"url_slug":1467,"children":1473},83,"structural-mechanics","Structural Mechanics","Structural Mechanics — Analysis of Framed Structures","Analysis of trusses, beams, and frames for structural engineering.","Structural mechanics develops methods for analyzing structural frameworks (trusses, beams, arches, frames) under static and dynamic loading. This section introduces force and displacement methods for solving statically indeterminate building structures.",41,[1474,1484],{"id":1475,"parent_id":1466,"code":33,"slug":1476,"name":1477,"seo_title":1478,"seo_description":1479,"seo_text":1480,"content":1481,"locale":8,"uk_topic_id":1482,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":40,"url_slug":1476,"children":1483},84,"trusses","Trusses","Truss Analysis | Method of Joints and Sections","Learn plane truss analysis: support reactions, zero-force members, method of joints, method of sections, tension and compression signs, and equilibrium checks.","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>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":1485,"parent_id":1466,"code":33,"slug":1486,"name":1487,"seo_title":1488,"seo_description":1489,"seo_text":1490,"content":1491,"locale":8,"uk_topic_id":1492,"show_in_theory_list":41,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":15,"url_slug":1486,"children":1493},85,"frames","Frames","Frame Analysis in Structural Mechanics | N, V, M Diagrams","Learn structural frame analysis: support reactions, member free-body diagrams, axial force, shear force and bending moment, sign conventions, and a step-by-step workflow.","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":1495,"parent_id":33,"code":33,"slug":1496,"name":1497,"seo_title":1498,"seo_description":1499,"seo_text":1500,"content":33,"locale":8,"uk_topic_id":1501,"show_in_theory_list":40,"is_published":41,"status":42,"visibility":43,"canonical":33,"noindex":40,"sort_order":116,"url_slug":1496,"children":1502},80,"comprehensive-problems","Comprehensive Problems","Comprehensive Engineering Mechanics Problems","Collection of advanced multi-topic problems in mechanics and strength of materials.","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,[],1787715902300]