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 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?
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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.
Updated On: Feb 14, 2026
- A rectangular block of constant positive height \(P\) from \(x=0\) to \(x=L/2\), and zero shear force from \(x=L/2\) to \(x=L\).
- 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 \(x=L\).
- A triangular shape increasing linearly from \(x=0\) to \(x=L/2\).
- A rectangular block from \(x=0\) to \(x=L\), unaffected by the point load.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We need to determine the Shear Force Diagram (SFD) for a cantilever beam with a point load and a couple (moment) applied at the midpoint. Step 2: Analyzing the Loads and Reactions:
We need to determine the Shear Force Diagram (SFD) for a cantilever beam with a point load and a couple (moment) applied at the midpoint. Step 2: Analyzing the Loads and Reactions:
- Support: Fixed at the left end (\(x=0\)).
- Loads at \(x = L/2\): Point load \(P\) (downward), Moment \(M\) (clockwise).
- Reaction at Support (\(x=0\)): From vertical force equilibrium \(\Sigma F_y = 0\): \[ R_A - P = 0 \Rightarrow R_A = P \, (\text{upwards}) \]
- Region \(0<x<L/2\): Looking to the left, the only force is the support reaction \(R_A = P\) (upwards). \[ V(x) = +P \] The diagram is a horizontal line (rectangle).
- At \(x = L/2\): There is a downward point load \(P\). The shear force drops by \(P\).
- Region \(L/2<x<L\): \[ V(x) = +P (\text{Reaction}) - P (\text{Load}) = 0 \] The shear force is zero in this section.
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