Question

sample problem 5.1
draw the shear and bending - moment diagrams for a
simply supported beam ab of span l subjected to a single
concentrated load p at it midpoint c.(fig. a)

Explanation:

Step1: Determine the support reactions

Since the beam is symmetrically loaded, the vertical reactions at \(A\) and \(B\), \(R_A = R_B=\frac{P}{2}\).

Step2: Shear force diagram

• For \(0 < x<\frac{L}{2}\): \(V = R_A=\frac{P}{2}\) (constant).
• At \(x = \frac{L}{2}\): There is a sudden change in shear force. The shear force drops from \(\frac{P}{2}\) to \(-\frac{P}{2}\) due to the concentrated load \(P\).
• For \(\frac{L}{2}

Step3: Bending - moment diagram

• For \(0\leq x\leq\frac{L}{2}\): \(M(x)=R_Ax=\frac{P}{2}x\). When \(x = 0\), \(M(0) = 0\); when \(x=\frac{L}{2}\), \(M(\frac{L}{2})=\frac{PL}{4}\).
• For \(\frac{L}{2}\leq x\leq L\): \(M(x)=R_Ax - P(x - \frac{L}{2})=\frac{P}{2}x-Px+\frac{PL}{2}=\frac{PL}{2}-\frac{P}{2}x\). When \(x = L\), \(M(L)=0\).

Answer:

The shear force diagram has a constant positive value of \(\frac{P}{2}\) from \(A\) to \(C\) and a constant negative value of \(-\frac{P}{2}\) from \(C\) to \(B\). The bending - moment diagram is a straight - line increasing from \(0\) at \(A\) to \(\frac{PL}{4}\) at \(C\) and then decreasing back to \(0\) at \(B\).