**Problem 722**

For the beam shown in Fig. P-722, compute the reaction R at the propped end and the moment at the wall. Check your results by letting b = L and comparing with the results in Problem 707.

**Solution**

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$(Area_{AB}) \cdot \bar X_A = 0$

$Mb(L - \frac{1}{2}b) - \frac{1}{2}L(RL)(\frac{2}{3}L) = 0$

$MbL - \frac{1}{2}Mb^2 - \frac{1}{3}RL^3 = 0$

$\frac{1}{3}RL^3 = MbL - \frac{1}{2}Mb^2$

$R = \dfrac{3Mb}{L^2} - \dfrac{3Mb^2}{2L^3}$

$R = \dfrac{3Mb}{2L^3}(2L - b)$ *answer*

$M_{wall} = M - RL$

$M_{wall} = M - \dfrac{3Mb}{2L^3}(2L - b)L$

$M_{wall} = M - \dfrac{3Mb}{2L^2}(2L - b)$ *answer*

When $b = L$

$R = \dfrac{3ML}{2L^3}(2L - L) = \dfrac{3M}{2L}$

$M_{wall} = M - \dfrac{3ML}{2L^2}(2L - L) = M - \frac{3}{2}M = -\frac{3}{2}M$

For comparison, see Problem 707, a similar situation but with moment load at the simple support.