Question

A cantilever beam of length \(L\) is fixed at the left end (\(x=0\)). It is subjected to a concentrated downward point load \(P\) and a concentrated clockwise moment \(M = \frac{PL}{2}\) at the midpoint (\(x = L/2\)). Which of the following descriptions correctly represents the Shear Force Diagram (SFD) for the beam?

• A. A rectangular block of constant positive height \(P\) from \(x=0\) to \(x=L/2\), and zero shear force from \(x=L/2\) to \(x=L\).
• B. A rectangular block of constant positive height \(P\) from \(x=0\) to \(x=L/2\), followed by another rectangular block of height \(P/2\) from \(x=L/2\) to \(x=L\).
• C. A triangular shape increasing linearly from \(x=0\) to \(x=L/2\).
• D. A rectangular block from \(x=0\) to \(x=L\), unaffected by the point load.

Hint:

Concentrated moments do not appear in the Shear Force Diagram (SFD); they only cause sudden jumps in the Bending Moment Diagram (BMD). Only vertical forces (point loads or distributed loads) affect the SFD.

Answer 1

The Correct Option is A

Step 1: Determine Support Reactions. At the fixed end (\(x=0\)), let the vertical reaction be \(R_y\). Summing vertical forces (\(\sum F_y = 0\)): \(R_y - P = 0 \implies R_y = P\) (upward).
Step 2: Sectional Analysis for Shear Force (\(V\)).

• Region \(0<x<L/2\): Take a section. Looking to the left, the only force is \(R_y\). So, \(V = +P\).
• At \(x = L/2\): The downward load \(P\) acts. The shear force drops by \(P\). New shear = \(P - P = 0\).
• Region \(L/2<x<L\): No additional vertical loads exist. The shear remains \(0\).

Step 3: Effect of the Moment. A concentrated moment \(M\) causes a jump in the Bending Moment Diagram (BMD) but has no effect on the SFD because it does not contribute to the summation of vertical forces.