Example 8.3.2. Internal forces in a simply supported beam.
A beam of length \(L\) is supported by a pin at \(A\) and a roller at \(B\) and is subjected to a horizontal force \(F\) applied to point \(B\) and a uniformly distributed load over its entire length. The intensity of the distributed load is \(w\) with units of [force/length].
Find the internal forces at the midpoint of the beam.
Answer.
Solution.
At the midpoint of the beam,
\begin{align*}
A \amp = F\\
V \amp = 0 \\
M \amp = wL^2/8
\end{align*}

Find the external reactions.
Begin by drawing a freebody diagram of the entire beam, simplified by replacing the distributed load \(w\) with an equivalent concentrated load at the centroid of the rectangle.The magnitude of the equivalent load \(W\) is equal to the “area” under the rectangular loading curve.\begin{equation*} W = w(L) \end{equation*}Then apply and simplify the equations of equilibrium to find the external reactions at \(A\) and \(B\text{.}\)\begin{align*} \Sigma M_A \amp = 0\\ (wL)(\cancel{L}/2)+(B)\cancel{L} \amp = 0\\ B \amp = wL/2 \\ \\ \Sigma F_x \amp = 0 \\ A_x+F \amp = 0\\ A_x \amp = F\\ \\ \Sigma F_y \amp = 0\\ A_ywL +B_y \amp = 0\\ A_y \amp = wLwL/2\\ \amp = wL/2 \end{align*}

Cut the beam.Cut the beam at the point of interest and separate the beam into two sections. Notice that as the beam is cut in two, the distributed load \(w\) is cut as well. Each of these distributed load halves will support equivalent point loads of \(wL/2\) acting through the centroid of each cut half.

Add the internal forces.At each cut, a shear force, a normal force, and a bending moment will be exposed, and these need to be included on the freebody diagram.At this point, we don't know the actual directions of the internal forces, but we do know that they act in opposite directions. We will assume that they act in the positive sense as defined by the standard sign convention.Axial forces are positive in tension and act in opposite directions on the two halves of the cut beam.Positive shear forces act down when looking at the cut from the right, and up when looking at the cut from the left. An alternate definition of positive shears is that the positive shears cause clockwise rotation. This definition is useful if you are dealing with a vertical column instead of a horizontal beam.Bending moments are positive when the moment tends to bend the beam into a smiling Ushape. Negative moments bend the beam into a frowning shape.For vertical columns, positive bending moments bend a beam into a C shape and negative into a backward Cshape.The final freebody diagrams look like this.Horizontal beams should always have assumed internal loadings in these directions at the cut, indicating that you have assumed positive shear, positive normal force and positive bending moments at that point.

Solve for the internal forces.You may uses either FBD to find the internal forces using the techniques you have already learned. So, with a standard \(xy\) coordinate system, forces to the right or up are positive when summing forces and counterclockwise moments are positive when summing moments.Using the left freebody diagram and substituting in the reactions, we get:\begin{align*} \Sigma F_x \amp = 0\\ A_x+N \amp = 0\\ N \amp = A_x \\ \\ \Sigma F_y \amp = 0\\ A_ywL/2V \amp = 0\\ V \amp = wL/2  wL/2\\ V \amp = 0\\ \\ \Sigma M_\text{cut} \amp = 0\\ (wL/2)(L/4)(A_y)(L/2)+M \amp = 0\\ M \amp =  wL^2/8 + wL^2/4\\ M \amp = wL^2/8 \end{align*}Using the right side freebody diagram we get:\begin{align*} \Sigma F_x \amp = 0\\ N+F \amp = 0\\ N \amp = F\\ \\ \Sigma F_y \amp = 0\\ VwL/2+B_y \amp = 0\\ V \amp = wL/2  B_y\\ V \amp = wL/2  wL/2\\ V \amp = 0\\ \\ \Sigma M_\text{cut} \amp = 0\\ M(L/4)(wL/2)+(L/2)(B_y) \amp = 0\\ M \amp = w L^2/8 + wL^2/4\\ M \amp = w L^2/8 \end{align*}