Question:

The beam PQRS is subjected to a vertical point load of \(10\) kN at point S as shown in the figure. The magnitude of fixed end moment at P is _________ kN‑m.


 

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An internal hinge (zero moment) splits the beam; find reactions on one side, then treat the adjacent span as a cantilever.
Updated On: Apr 25, 2025
  • \(50\)
  • \(10\)
  • \(30\)
  • \(40\)
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The Correct Option is B

Solution and Explanation

Step 1: Internal hinge at Q. Since Q is an internal hinge, \(M_Q=0\). The portion Q–R–S behaves as a simply supported span Q–R of length 2 m with an overhang RS of 1 m carrying the 10 kN at S.
Step 2: Reactions on Q–R–S.
Taking moments about Q:
\[ -R_R\cdot2 + 10\,(2+1)=0 \;\Longrightarrow\;R_R=\frac{10\cdot3}{2}=15{ \,kN}. \] Vertical equilibrium gives:
\[ R_Q + R_R - 10 = 0 \;\Longrightarrow\;R_Q = 10 - 15 = -5{ \,kN} \] (downward 5 kN at Q).
Step 3: Cantilever PQ.
Span PQ is a cantilever of length 2 m with an end shear of \(|R_Q|=5\) kN at Q. The fixed-end moment at P is:
\[ M_P = 5{ \,kN}\times2{ \,m} = 10{ \,kN·m}. \]
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