Question 9.14: Determine the reaction at the supports for the prismatic bea...
Determine the reaction at the supports for the prismatic beam and loading shown (Fig. 9.71).
We consider the couple exerted at the fixed end A as redundant and replace the fixed end by a pin-and-bracket support. The couple M_{A} is now considered as an unknown load (Fig. 9.72a) and will be determined from the condition that the tangent to the beam at A must be horizontal. It follows that this tangent must pass through the support B and, thus, that the tangential deviation t_{B / A} of B with respect to A must be zero. The solution is carried out by computing separately the tangential deviation \left(t_{B / A}\right)_{w} caused by the uniformly distributed load w (Fig. 9.72b) and the tangential deviation \left(t_{B / A}\right)_{M} produced by the unknown couple M _{A} (Fig. 9.72c).
Considering first the free-body diagram of the beam under the known distributed load w (Fig. 9.73a), we determine the corresponding reactions at the supports A and B. We have
\left( R _{A}\right)_{1}=\left( R _{B}\right)_{1}=\frac{1}{2} w L \uparrow (9.64)
We can now draw the corresponding shear and (M/EI) diagrams (Figs. 9.73b and c). Observing that M/EI is represented by an arc of parabola, and recalling the formula, A=\frac{2}{3} b h, for the area under a parabola, we compute the first moment of this area about a vertical axis through B and write
\left(t_{B / A}\right)_{w}=A_{1}\left(\frac{L}{2}\right)=\left(\frac{2}{3} L \frac{w L^{2}}{8 EI}\right)\left(\frac{L}{2}\right)=\frac{w L^{4}}{24 E I} (9.65)
Considering next the free-body diagram of the beam when it is subjected to the unknown couple M _{A} (Fig. 9.74a), we determine the corresponding reactions at A and B:
\left( R _{A}\right)_{2}=\frac{M_{A}}{L} \uparrow \left( R _{B}\right)_{2}=\frac{M_{A}}{L} \downarrow (9.66)
Drawing the corresponding (M/EI) diagram (Fig. 9.74b), we apply again the second moment-area theorem and write
\left(t_{B / A}\right)_{M}=A_{2}\left(\frac{2 L}{3}\right)=\left(-\frac{1}{2} L \frac{M_{A}}{E I}\right)\left(\frac{2 L}{3}\right)=-\frac{M_{A} L^{2}}{3 E I} (9.67)
Combining the results obtained in (9.65) and (9.67), and expressing that the resulting tangential deviation t_{B / A} must be zero (Fig. 9.72), we have
t_{B / A}=\left(t_{B / A}\right)_{w}+\left(t_{B / A}\right)_{M}=0\frac{w L^{4}}{24 E I}-\frac{M_{A} L^{2}}{3 E I}=0
and, solving for M_{A},
M_{A}=+\frac{1}{8} w L^{2} M_{A}=\frac{1}{8} w L^{2}\circlearrowleft
Substituting for M_{A} into (9.66), and recalling (9.64), we obtain the values of R_{A} and R_{B}:
R_{A}=\left(R_{A}\right)_{1}+\left(R_{A}\right)_{2}=\frac{1}{2} w L+\frac{1}{8} w L=\frac{5}{8} w LR_{B}=\left(R_{B}\right)_{1}+\left(R_{B}\right)_{2}=\frac{1}{2} w L-\frac{1}{8} w L=\frac{3}{8} w L
