Section 8.5 Section Cut Method
In this section we’ll extend the method of Section 8.3 where we found the shear force and bending moment at a specific point to make shear and bending moment diagrams. The procedure is similar except that the cut is taken at a variable position designated by \(x\) instead of at a specified point. The analysis produces equations for shear and bending moments as functions of \(x\text{.}\) Shear and bending moment diagrams are plots of these equations, and the internal forces at any particular point can be found by substituting the point’s location into the equations.
As an example, we will use a cantilevered beam fixed to a wall on its left end and subject to a vertical force \(P\) on its right end as an example. Global equilibrium requires that the reactions at the fixed support at \(A\) are a vertical force \(A_y = P\text{,}\) and a counter-clockwise moment \(M_A = P L\text{.}\)
By taking a cut at a distance \(x\) from the left we can draw two free-body diagrams with lengths \(x\) and \((L-x)\text{.}\) This beam has one loading segment, because no matter where \(x\) is chosen, the free-body diagrams shown in Figure 8.5.1 (b) and (c) are correct. The internal loadings are named \(V(x)\) and \(M(x)\) to indicate that they are functions of \(x\text{.}\)
To find the shear and bending moment functions, we apply the equilibrium to one of the free-body diagrams. Either side will work, so we’ll select the right-hand portion as it doesn’t require us to find the reactions at \(A\text{.}\) Letting \(L\) be the length of the beam and \((L-x)\) the length of the right portion, we find
\begin{align*}
\Sigma F_y \amp = 0 \amp \Sigma M_\text{cut} \amp = 0\\
V(x) \amp = P \amp M(x) \amp = - P(L-x)
\end{align*}
The plots of the equations for \(V(x)\) and \(M(x)\) are shown below in Figure 8.5.2. These equations indicate that the shear force \(V(x)\) is constant \(P\) over the length of the beam and the moment \(M(x)\) is a linear function of the position of the cut, \(x\) starting at \(-PL\) at \(x=0\) and linearly increasing to zero at \(x=L\text{.}\) Note that the graphs are only valid from \(0 \le x \le L\text{,}\) so the curves outside this range is show as dotted lines. These two graphs are usually drawn stacked beneath the diagram of the beam and loading.
The previous example was simple because only one FBD was necessary for any point on the beam, but many beams are more complex. Beams with multiple loads must be divided into loading segments between the points where loads are applied or where distributed loads begin or end.
Consider the simply supported beam \(AD\) with a uniformly distributed load \(w\) over the first segment from \(A\) to \(B\text{,}\) and two vertical loads \(B\) and \(C\text{.}\)
This beam has three loading segments so you must draw three free-body diagrams and analyze each segment independently. For each, make an imaginary cut through the segment, then draw a new free-body diagram of the portion to the left (or right) of the cut. Always assume that the exposed internal shear force and internal bending moment act in the positive direction according to the sign convention.
After the equilibrium equations are applied to each segment, the resulting equations \(V(x)\) and \(M(x)\) from each segment are joined to plot the shear and moment diagrams. These diagrams help us visualize the values of \(V\) and \(M\) throughout the beam.
