# Problem 724 | Propped beam with partially restrained wall support

**Problem 724**

The beam shown in Fig. P-724 is only partially restrained at the wall so that, after the uniformly distributed load is applied, the slope at the wall is $w_oL^3 / 48EI$ upward to the right. If the supports remain at the same level, determine $R$.

**Solution**

$\theta = \dfrac{w_oL^3}{48EI}$

$L \theta = \dfrac{w_oL^4}{48EI}$

$t_{A/B} = L \theta$

$t_{A/B} = \dfrac{w_oL^4}{48EI}$

$EI ~ t_{A/B} = \frac{1}{48}w_oL^4$

$\frac{1}{2}(L)(RL)(\frac{2}{3}L) - \frac{1}{3}(L)(\frac{1}{2}w_oL^2)(\frac{3}{4}L) = \frac{1}{48}w_oL^4$

$\frac{1}{3}RL^3 - \frac{1}{8}w_oL^4 = \frac{1}{48}w_oL^4$

$\frac{1}{3}RL^3 = \frac{7}{48}w_oL^4$

$R = \frac{7}{16}w_oL^4$ *answer*

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