Statics of Bending Shear and Bending Moment Diagrams David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge MA 02139 November 15 2000 Introduction Beams are long and slender structural elements differing from truss elements in that they are called on to support transverse as well as axial loads Their attachment points can also be more complicated than those of truss elements they may be bolted or welded together so the attachments can transmit bending moments or transverse forces into the beam Beams are among the most common of all structural elements being the supporting frames of airplanes buildings cars people and much else The nomenclature of beams is rather standard as shown in Fig 1 L is the length or span b is the width and h is the height also called the depth The cross sectional shape need not be rectangular and often consists of a vertical web separating horizontal flanges at the top and bottom of the beam1 Figure 1 Beam nomenclature As will be seen in Modules 13 and 14 the stresses and deflections induced in a beam under bending loads vary along the beam s length and height The first step in calculating these quantities and their spatial variation consists of constructing shear and bending moment diagrams V x and M x which are the internal shearing forces and bending moments induced in the beam plotted along the beam s length The following sections will describe how these diagrams are made 1 Figure 2 A cantilevered beam Free body diagrams As a simple starting example consider a beam clamped cantilevered at one end and subjected to a load P at the free end as shown in Fig 2 A free body diagram of a section cut transversely at position x shows that a shear force V and a moment M must exist on the cut section to maintain equilibrium We will show in Module 13 that these are the resultants of shear and normal stresses that are set up on internal planes by the bending loads As usual we will consider section areas whose normals point in the x direction to be positive then shear forces pointing in the y direction on x faces will be considered positive Moments whose vector direction as given by the right hand rule are in the z direction vector out of the plane of the paper or tending to cause counterclockwise rotation in the plane of the paper will be positive when acting on x faces Another way to recognize positive bending moments is that they cause the bending shape to be concave upward For this example beam the statics equations give X X Fy 0 V P V constant P 1 M0 0 M P x M M x P x 2 Note that the moment increases with distance from the loaded end so the magnitude of the maximum value of M compared with V increases as the beam becomes longer This is true of most beams so shear effects are usually more important in beams with small length to height ratios Figure 3 Shear and bending moment diagrams 1 There is a standardized protocol for denoting structural steel beams for instance W 8 40 indicates a wide flange beam with a nominal depth of 800 and weighing 40 lb ft of length 2 As stated earlier the stresses and deflections will be shown to be functions of V and M so it is important to be able to compute how these quantities vary along the beam s length Plots of V x and M x are known as shear and bending moment diagrams and it is necessary to obtain them before the stresses can be determined For the end loaded cantilever the diagrams shown in Fig 3 are obvious from Eqns 1 and 2 Figure 4 Wall reactions for the cantilevered beam It was easiest to analyze the cantilevered beam by beginning at the free end but the choice of origin is arbitrary It is not always possible to guess the easiest way to proceed so consider what would have happened if the origin were placed at the wall as in Fig 4 Now when a free body diagram is constructed forces must be placed at the origin to replace the reactions that were imposed by the wall to keep the beam in equilibrium with the applied load These reactions can be determined from free body diagrams of the beam as a whole if the beam is statically determinate and must be found before the problem can proceed For the beam of Fig 4 X X Fy 0 VR P VR P Mo 0 MR P L MR P L The shear and bending moment at x are then V x VR P constant M x MR VR x P L P x This choice of origin produces some extra algebra but the V x and M x diagrams shown in Fig 5 are the same as before except for changes of sign V is constant and equal to P and M varies linearly from zero at the free end to P L at the wall Distributed loads Transverse loads may be applied to beams in a distributed rather than at a point manner as depicted in Fig 6 which might be visualized as sand piled on the beam It is convenient to describe these distributed loads in terms of force per unit length so that q x dx would be the load applied to a small section of length dx by a distributed load q x The shear force V x set up in reaction to such a load is V x Z 3 x x0 q d 3 Figure 5 Alternative shear and bending moment diagrams for the cantilevered beam Figure 6 A distributed load and a free body section where x0 is the value of x at which q x begins and is a dummy length variable that looks backward from x Hence V x is the area under the q x diagram up to position x The moment balance is obtained considering the increment of load q d applied to a small width d of beam a distance from point x The incremental moment of this load around point x is q d so the moment M x is Z M x x0 q d 4 This can be related to the centroid of the area under the q x curve up to x whose distance from x is R q d R q d Hence Eqn 4 can be written R M Q 5 where Q q d is the area Therefore the distributed load q x is statically equivalent to a concentrated load of magnitude Q placed at the centroid of the area under the q x diagram Example 1 Consider a simply supported beam carrying a triangular and a concentrated load as shown in Fig 7 For 4 Figure 7 Distributed and concentrated loads the purpose of determining the support reaction forces R1 and R2 the distributed triangular load can be replaced by its static equivalent The magnitude of this equivalent force is Z Q 0 2 …