Statics of Bending: Shear and Bending Moment DiagramsDavid RoylanceDepartment of Materials Science and EngineeringMassachusetts Institute of TechnologyCambridge, MA 02139November 15, 2000IntroductionBeams are long and slender structural elements, differing from truss elements in that they arecalled on to support transverse as well as axial loads. Their attachment points can also bemore complicated than those of truss elements: they may be bolted or welded together, so theattachments can transmit bending moments or transverse forces into the beam. Beams areamong 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 notbe rectangular, and often consists of a vertical web separating horizontal flanges at the top andbottom of the beam1.Figure 1: Beam nomenclature.As will be seen in Modules 13 and 14, the stresses and deflections induced in a beam underbending loads vary along the beam’s length and height. The first step in calculating these quan-tities and their spatial variation consists of constructing shear and bending moment diagrams,V (x)andM(x), which are the internal shearing forces and bending moments induced in thebeam, plotted along the beam’s length. The following sections will describe how these diagramsare made.1Figure 2: A cantilevered beam.Free-body diagramsAs a simple starting example, consider a beam clamped (“cantilevered”) at one end and sub-jected to a load P at the free end as shown in Fig. 2. A free body diagram of a section cuttransversely at position x shows that a shear force V and a moment M must exist on the cutsection to maintain equilibrium. We will show in Module 13 that these are the resultants of shearand normal stresses that are set up on internal planes by the bending loads. As usual, we willconsider section areas whose normals point in the +x direction to be positive; then shear forcespointing in the +y direction on +x faces will be considered positive. Moments whose vectordirection as given by the right-hand rule are in the +z direction (vector out of the plane of thepaper, or tending to cause counterclockwise rotation in the plane of the paper) will be positivewhen acting on +x faces. Another way to recognize positive bending moments is that they causethe bending shape to be concave upward. For this example beam, the statics equations give:XFy=0=V +P ⇒V = constant = −P (1)XM0=0=−M+Px ⇒ M = M(x)=Px (2)Note that the moment increases with distance from the loaded end, so the magnitude of themaximum value of M compared with V increases as the beam becomes longer. This is true ofmost beams, so shear effects are usually more important in beams with small length-to-heightratios.Figure 3: Shear and bending moment diagrams.1There is a standardized protocol for denoting structural steel beams; for instance W 8 × 40 indicates awide-flange beam with a nominal depth of 800and weighing 40 lb/ft of length2As stated earlier, the stresses and deflections will be shown to be functions of V and M,soitis important to be able to compute how these quantities vary along the beam’s length. Plots ofV (x)andM(x) are known as shear and bending moment diagrams, and it is necessary to obtainthem before the stresses can be determined. For the end-loaded cantilever, the diagrams shownin 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 choiceof origin is arbitrary. It is not always possible to guess the easiest way to proceed, so considerwhat would have happened if the origin were placed at the wall as in Fig. 4. Now when a freebody diagram is constructed, forces must be placed at the origin to replace the reactions thatwere imposed by the wall to keep the beam in equilibrium with the applied load. These reactionscan be determined from free-body diagrams of the beam as a whole (if the beam is staticallydeterminate), and must be found before the problem can proceed. For the beam of Fig. 4:XFy=0=−VR+P ⇒VR=PXMo=0=MR−PL ⇒ MR= PLThe shear and bending moment at x are thenV (x)=VR=P = constantM(x)=MR−VRx=PL−PxThis choice of origin produces some extra algebra, but the V (x)andM(x) diagrams shown inFig. 5 are the same as before (except for changes of sign): V is constant and equal to P ,andMvaries linearly from zero at the free end to PL at the wall.Distributed loadsTransverse loads may be applied to beams in a distributed rather than at-a-point manner asdepicted in Fig. 6, which might be visualized as sand piled on the beam. It is convenient todescribe these distributed loads in terms of force per unit length, so that q(x) dx would be theload applied to a small section of length dx by a distributed load q(x). The shear force V (x)setup in reaction to such a load isV (x)=−Zxx0q(ξ)dξ (3)3Figure 5: Alternative shear and bending moment diagrams for the cantilevered beam.Figure 6: A distributed load and a free-body section.where x0is the value of x at which q(x)begins,andξis a dummy length variable that looksbackward from x. Hence V (x) is the area under the q(x) diagram up to position x. The momentbalance is obtained considering the increment of load q(ξ) dξ applied to a small width dξ of beam,adistanceξfrom point x. The incremental moment of this load around point x is q(ξ) ξdξ,sothe moment M(x)isM=Zxx0q(ξ)ξdξ (4)This can be related to the centroid of the area under the q(x) curve up to x, whose distancefrom x is¯ξ =Rq(ξ) ξdξRq(ξ)dξHence Eqn. 4 can be writtenM = Q¯ξ (5)where Q =Rq(ξ) dξ is the area. Therefore, the distributed load q(x) is statically equivalent toa concentrated load of magnitude Q placed at the centroid of the area under the q(x) diagram.Example 1Consider a simply-supported beam carrying a triangular and a concentrated load as shown in Fig. 7. For4Figure 7: Distributed and concentrated loads.the purpose of determining the support reaction forces R1and R2, the distributed triangular load can bereplaced by its static equivalent. The magnitude of this equivalent force isQ =Z20(−600x) dx = −1200The equivalent force acts through the centroid of the triangular area, which is is 2/3 of the distance fromits narrow end (see Prob. 1). The reaction