Question

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: Calculate support reactions

Since the beam is symmetric and the load \(P\) is at mid - point.
By symmetry, \(R_{A}=R_{B}=\frac{P}{2}\)

Step2: Shear force diagram (SFD)

• For \(0\leq x<\frac{L}{2}\): \(V(x)=R_{A}=\frac{P}{2}\)
• For \(\frac{L}{2}

At \(x = \frac{L}{2}\), there is a sudden change in shear force from \(\frac{P}{2}\) to \(-\frac{P}{2}\) (magnitude change of \(P\))

Step3: Bending - moment diagram (BMD)

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

Answer:

• Shear Force Diagram: A constant positive shear force \(\frac{P}{2}\) in the left half (\(0\leq x<\frac{L}{2}\)) and a constant negative shear force \(-\frac{P}{2}\) in the right half (\(\frac{L}{2}
• Bending - Moment Diagram: A linear increase from \(0\) to \(\frac{PL}{4}\) in the left half (\(0\leq x\leq\frac{L}{2}\)) given by \(M(x)=\frac{P}{2}x\) and a linear decrease from \(\frac{PL}{4}\) to \(0\) in the right half (\(\frac{L}{2}\leq x\leq L\)) given by \(M(x)=\frac{P}{2}(L - x)\)