Question 9.CA.8: Determine the reactions at the supports for the prismatic be......

We consider the reaction at B as redundant and release the beam from the support. The reaction R _B is now considered as an unknown load (Fig. 9.22b) and will be determined from the condition that the deflection of the beam at B must be zero. The solution is carried out by considering separately the deflection \left(y_B\right)_w caused at B by the uniformly distributed load w (Fig. 9.22c) and the deflection \left(y_B\right)_R produced at the same point by the redundant reaction R_B (Fig. 9.22d).
From the table of Appendix F (cases 2 and 1),

\left(y_B\right)_w=-\frac{w L^4}{8 E I} \quad\left(y_B\right)_R=+\frac{R_B L^3}{3 E I}

Writing that the deflection at B is the sum of these two quantities and that it must be zero,

\begin{aligned} & y_B=\left(y_B\right)_w+\left(y_B\right)_R=0 \\ & y_B=-\frac{w L^4}{8 E I}+\frac{R_B L^3}{3 E I}=0 \end{aligned}

and, solving for R_B ,        R_B=\frac{3}{8} w L \quad R _B=\frac{3}{8} w L \uparrow

Drawing the free-body diagram of the beam (Fig. 9.22e) and writing the corresponding equilibrium equations,

+\uparrow \Sigma F_y=0: \quad R_A+R_B-w L=0           (1)

\begin{aligned} R_A=w L-R_B=w L-\frac{3}{8} w L & =\frac{5}{8} w L \\ R _A & =\frac{5}{8} w L \uparrow \end{aligned}

+⤻ \Sigma M_A=0: \quad M_A+R_B L-(w L)\left(\frac{1}{2} L\right)=0           (2)

\begin{aligned} M_A=\frac{1}{2} w L^2-R_B L=\frac{1}{2} w L^2-\frac{3}{8} w L^2 & =\frac{1}{8} w L^2 \\ M _A & =\frac{1}{8} w L^2 ⤻ \end{aligned}

Alternative Solution. We may 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. 9.22f ) and will be determined from the condition that the slope of the beam at A must be zero. The solution is carried out by considering separately the slope \left(\theta_A\right)_w caused at A by the uniformly distributed load w (Fig. 9.22g) and the slope \left(\theta_A\right)_M produced at the same point by the unknown couple M _A (Fig 9.22h).

Using the table of Appendix F (cases 6 and 7) and noting that A and B must be interchanged in case 7,

\left(\theta_A\right)_w=-\frac{w L^3}{24 E I} \quad\left(\theta_A\right)_M=\frac{M_A L}{3 E I}

Writing that the slope at A is the sum of these two quantities and that it must be zero gives

\begin{aligned} & \theta_A=\left(\theta_A\right)_w+\left(\theta_A\right)_M=0 \\ & \theta_A=-\frac{w L^3}{25 E I}+\frac{M_A L}{3 E I}=0 \end{aligned}

where M_A is

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

The values of R_A \text { and } R_B are found by using the equilibrium equations (1) and (2).