[{"data":1,"prerenderedAt":37},["ShallowReactive",2],{"topic-en-theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fdistributed-loads-resultants":3},{"topic":4,"trail":18,"children":32,"tasks":33,"alternates":34},{"id":5,"name":6,"locale":7,"path":8,"seo_title":9,"seo_description":10,"seo_text":11,"content_html":12,"content_chunks":13},190,"Distributed Loads and Their Resultants","en","theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fdistributed-loads-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>\u003Cp>For a uniformly distributed load of intensity $q$ acting over a length $L$, the magnitude of the equivalent concentrated force is:\u003C\u002Fp>\u003Cp>$$R=qL.$$\u003C\u002Fp>\u003Cp>The line of action of the resultant passes through the midpoint of the loaded interval.\u003C\u002Fp>\u003Cdiv data-formula-calculator-config=\"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\">\u003C\u002Fdiv>\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>",[14],{"id":15,"code":16,"type":17,"locale":7},155,"statics-uniform-distributed-load-resultant","formula",[19,23,27,31],{"id":20,"name":21,"path":22},81,"Theoretical Mechanics","theoretical-mechanics",{"id":24,"name":25,"path":26},136,"Statics","theoretical-mechanics\u002Fstatics",{"id":28,"name":29,"path":30},275,"Equilibrium of Coplanar Systems","theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems",{"id":5,"name":6,"path":8},[],[],{"en":35,"uk":36},"https:\u002F\u002Fmechclassroom.com\u002Fen\u002Ftopics\u002Ftheoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fdistributed-loads-resultants","https:\u002F\u002Fmechclassroom.com\u002Ftopics\u002Fteoretychna-mekhanika\u002Fstatyka\u002Frivnovaha-ploskykh-system\u002Frozpodilene-navantazhennia-ta-ioho-rivnodiina",1787712535394]