Question 9.CA.14: Determine the reaction at the supports for the prismatic bea......

Consider the couple exerted at the fixed end A as redundant and replace the fixed end by a pin-and-bracket support. Couple M _A is now considered to be an unknown load (Fig. 9.40b) and will be determined from the condition that the tangent to the beam at A must be horizontal. Thus, this tangent must pass through the support B, and 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. 9.40c) and the tangential deviation \left(t_{B / A}\right)_M produced by the unknown couple M _A (Fig. 9.40d).
Using the free-body diagram of the beam under the known distributed load w (Fig. 9.40e), determine the corresponding reactions at the supports A and B.

\left( R _A\right)_1=\left( R _B\right)_1=\frac{1}{2} w L \uparrow             (1)

Now draw the corresponding shear and M/EI diagrams (Fig. 9.40e). 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, the first moment of this area about a vertical axis through B is

\left(t_{B / A}\right)_w=A_1\left(\frac{L}{2}\right)=\left(\frac{2}{3} L \frac{w L^2}{8 E I}\right)\left(\frac{L}{2}\right)=\frac{w L^4}{24 E I}           (2)

Using the free-body diagram of the beam when it is subjected to the unknown couple M _A (Fig. 9.40f), the corresponding reactions at A and B are

\left( R _A\right)_2=\frac{M_A}{L} \uparrow \quad\left( R _B\right)_2=\frac{M_A}{L} \downarrow             (3)

Drawing the corresponding M/EI diagram (Fig. 9.40f), the second moment-area theorem is applied to obtain

\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}         (4)

Combining the results obtained in Eqs. (2) and (4) and expressing that the resulting tangential deviation t_{B / A} must be zero (Fig. 9.40b–d),

\begin{gathered} 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 \end{gathered}

and solving for M_A ,

M_A=+\frac{1}{8} w L^2 \quad M _A=\frac{1}{8} w L^2 ⤻

Substituting for M_A into Eq. (3), and recalling Eq. (1), the values of R_A and R_B  are