Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 16 (Q.No. 25)
25.
A propped cantilever of span 'l' carries uniformly distributed load of 'w' per unit over its entire span. The value of prop react to keep the beam horizontal is
Discussion:
6 comments Page 1 of 1.
Nandu said:
5 years ago
Let the propped cantilever have a fixed end at A and propped end at B.
For a propped cantilever, deflection at the propped end (B) = 0.
A propped cantilever with a UDL is equivalent to a cantilever with a UDL(1) + a cantilever with an upward reaction at the propped end(2).
So at B, deflection due to case (1) = deflection due to case (2) ----------since total deflection = 0.
Therefore wL^4/8EI = (Rb)L^3/3EI ----------(3).
Where Rb is the reaction at B.
Also we have Ra+Rb = wL (equating forces) ------------(4).
From (3) and (4), Rb = 3wL/8.
For a propped cantilever, deflection at the propped end (B) = 0.
A propped cantilever with a UDL is equivalent to a cantilever with a UDL(1) + a cantilever with an upward reaction at the propped end(2).
So at B, deflection due to case (1) = deflection due to case (2) ----------since total deflection = 0.
Therefore wL^4/8EI = (Rb)L^3/3EI ----------(3).
Where Rb is the reaction at B.
Also we have Ra+Rb = wL (equating forces) ------------(4).
From (3) and (4), Rb = 3wL/8.
Nair BG said:
6 years ago
@Mudra ...
It's because on LHS it is UDL and on RHS it is Point load.
Total load by UDL = w x L.
It's because on LHS it is UDL and on RHS it is Point load.
Total load by UDL = w x L.
Mudra said:
6 years ago
@Chhaya : Why L^4 and not L^3. From your eqn I have understood, it's the equation for deflection.
Rahul kumar said:
6 years ago
Please give the clear explanation of the answer.
Chhaya said:
8 years ago
Here wL^4/8EI = WL^3/3EI.
W = 3wL/8.
W = 3wL/8.
Rakesh said:
9 years ago
Option B is the correct answer.
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