Project 1

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 Figure 1, L is the length, or span; b is the width, and hℎ is the height (also called the depth). The cross-sectional shape need not be rectangular, and often consists of a vertical web separating horizontal fllanges at the top and bottom of the beam (There is a standardized protocol for denoting structural steel beams; for instance W 8 × 40 indicates a wide-ffllange beam with a nominal depth of 8′′ and weighing 40 lb/ft of length)

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.

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 Figure 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:

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.

As stated earlier, the stresses and defflections 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 Figure 3 are obvious from Eqns. 4.1.1 and 4.1.2.

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 Figure 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 Figure 4:

∑Fy=0=−VR+P⇒VR=P∑

The shear and bending moment at x are then

V(x)=VR=P=constantt

This choice of origin produces some extra algebra, but the V(x)) and M(x) diagrams shown in Figure 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 PL at the wall.

Distributed loads

Transverse loads may be applied to beams in a distributed rather than at-a-point manner as depicted in Figure 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) would be the load applied to a small section of length dx by a distributed load q. The shear force V(x) set up in reaction to such a load is

where x0 is the value of x at which \q(x)\) begins, and ξ 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(ξ)dξ applied to a small width dξ of beam, a distance ξ from point x. The incremental moment of this load around point x� is q(ξ)ξdξ, so the moment M(x)) is

This can be related to the centroid of the area under the q(x) curve up to x, whose distance from x is

ξ¯=∫q(ξ)ξdξ∫q(ξ)dξ

Hence Equation 4.1.4 can be written

where Q=∫q(ξ)dξ 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.