Question 11.S-P.7: For the uniform beam and loading shown, determine the reacti...
For the uniform beam and loading shown, determine the reactions at the supports.
Castigliano’s Theorem. The beam is indeterminate to the first degree and we choose the reaction R_A as redundant. Using Castigliano’s theorem, we determine the deflection at A due to the combined action of R_A and the distributed load. Since EI is constant, we write
y_{A}=\int \frac{M}{E I}\left(\frac{\partial M}{\partial R_{A}}\right) d x=\frac{1}{E I} \int M \frac{\partial M}{\partial R_{A}} d x (1)
The integration will be performed separately for portions AB and BC of the beam. Finally, R_A is obtained by setting y_A equal to zero.
Free Body: Entire Beam. We express the reactions at B and C in terms of R_A and the distributed load
R_{B}=\frac{9}{4} w L-3 R_{A} R_{C}=2 R_{A}-\frac{3}{4} w L (2)
Portion AB of Beam. Using the free-body diagram shown, we find
M_{1}=R_{A} x-\frac{w x^{2}}{2} \frac{\partial M_{1}}{\partial R_{A}}=x
Substituting into Eq. (1) and integrating from A to B, we have
\frac{1}{E I} \int M_{1} \frac{\partial M}{\partial R_{A}} d x=\frac{1}{E I} \int_{0}^{L}\left(R_{A} x^{2}-\frac{w x^{3}}{2}\right) d x=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{3}-\frac{w L^{4}}{8}\right) (3)
Portion BC of Beam. We have
M_{2}=\left(2 R_{A}-\frac{3}{4} w L\right) v-\frac{w v^{2}}{2} \frac{\partial M_{2}}{\partial R_{A}}=2 v
Substituting into Eq. (1) and integrating from C, where v = 0 , to B, where v=\frac{1}{2} L, we have
\frac{1}{E I} \int M_{2} \frac{\partial M_{2}}{\partial R_{A}} d v=\frac{1}{E I} \int_{0}^{L / 2}\left(4 R_{A} v^{2}-\frac{3}{2} w L v^{2}-w v^{3}\right) d v
=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{w L^{4}}{16}-\frac{w L^{4}}{64}\right)=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{5 w L^{4}}{64}\right) (4)
Reaction at A. Adding the expressions obtained in (3) and (4), we determine y_{A} and set it equal to zero
y_{A}=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{3}-\frac{w L^{4}}{8}\right)+\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{5 w L^{4}}{64}\right)=0Solving for R_A, R_{A}=\frac{13}{32} w L R_{A}=\frac{13}{32} w L \uparrow
Reactions at B and C. Substituting for R_{A} into Eqs. (2), we obtain
R_{B}=\frac{33}{32} w L \uparrow R_{C}=\frac{w L}{16} \uparrow
