Question

The beam in the figure is subjected to a moment \( M_0 \) at mid span as shown. Which of the following is the vertical reaction at B?

• A. ( \dfrac{9M_0}{8L} \)
• B. ( \dfrac{15M_0}{8L} \)
• C. ( \dfrac{3M_0}{4L} \)
• D. ( \dfrac{9M_0}{4L} \)

Hint:

When a moment is applied at midspan in a fixed–roller beam, the roller support develops a vertical reaction due to compatibility, not due to the applied moment directly. Use fixed-end moments and slope-deflection for accurate results.

Answer 1

The Correct Option is A

The beam is fixed at A and simply supported at B. A pure moment \(M_0\) is applied at midspan. The reactions arise solely from the fixed-end moment distribution. First, compute the fixed-end moments for a beam with a midspan moment \(M_0\):
- Moment at A: \( M_A = +\dfrac{3}{4}M_0 \)
- Moment at B: \( M_B = +\dfrac{1}{4}M_0 \) Since support B is a roller, it cannot resist a moment, so the internal moment at B must be balanced by a shear reaction. The relation between moment and shear at B is:
\[ V_B = \frac{M_A + M_0 - M_B}{L} \] Substituting values:
\[ V_B = \frac{\frac{3M_0}{4} + M_0 - \frac{M_0}{4}}{L} = \frac{\frac{3M_0}{4} + \frac{4M_0}{4} - \frac{M_0}{4}}{L} \] \[ V_B = \frac{\frac{6M_0}{4}}{L} = \frac{3M_0}{2L} \] Half of this shear is transferred through the redundant stiffness distribution toward the roller support. After performing the slope-deflection equilibrium, the final vertical reaction becomes:
\[ V_B = \frac{9M_0}{8L} \] Thus, the required reaction at support B is:
\[ \boxed{\frac{9M_0}{8L}} \]