Question

The figure shows a propped cantilever with uniform flexural rigidity \( EI \) (in N.m\(^2\)) and subjected to a moment \( M \) (in N.m). Consider forces and displacements in the upward direction as positive.
Find the upward reaction at the propped support B (in N) when this support settles by \( \Delta \) (in metres).
 

• A. \( \frac{3M}{2L} - \frac{6EI}{L^3} \)
• B. \( \frac{8M}{3L} - \frac{2EI}{3L^3} \)
• C. \( \frac{3M}{2L} - \frac{3EI}{L^3} \)
• D. \( \frac{M}{L} - \frac{8EI}{L^3} \)

Hint:

When analyzing reactions in a cantilever, break down the problem into moments and deflection components to accurately account for the effects of support movement and applied forces.

Answer 1

The Correct Option is C

Let’s start by considering the reaction at point \( B \) due to the applied moment.

1. Reaction due to Moment:
The moment \( M \) applied at the cantilever results in a reaction force at point \( B \). Since the flexural rigidity \( EI \) is constant, the moment at the support \( B \) is given by:
\[ R_m = \frac{3M}{2L} \]
2. Deflection at B:
The deflection at point \( B \) should be zero, thus:
\[ \frac{ML^2}{2EI} - R_m \times \frac{L^3}{3EI} = 0 \] Substituting \( R_m = \frac{3M}{2L} \) into this equation:
\[ \frac{ML^2}{2EI} - \frac{3M}{2L} \times \frac{L^3}{3EI} = 0 \] This simplifies to:
\[ R_m = \frac{3M}{2L} \]
3. Reaction at Propped End Due to Sinking of Support:
The reaction at the propped end due to the sinking of the support is given by:
\[ R_{\Delta} = \frac{3EI}{L^3} \times \Delta \]
4. Net Reaction at Propped End:
The net reaction at the propped end is the combination of the moment reaction and the sinking due to support movement. Therefore, the net reaction is:
\[ R_B = \frac{3M}{2L} - \frac{3EI}{L^3} \times \Delta \]
Thus, the net upward reaction at the propped support \( B \) is given by:
\[ \boxed{ \frac{3M}{2L} - \frac{3EI}{L^3} \times \Delta } \]