# Problem 725 | Propped beam with partially restrained wall and settling simple support

**Problem 725**

If the support under the propped beam in Problem 724 settles an amount $\delta$, show that the propped reaction decreases by $3EI\delta / L^3$.

**Solution**

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

$EI ~ \theta = \frac{1}{48}w_oL^3$

$L(EI ~ \theta) = \frac{1}{48}w_oL^4$

$t_{A/B} = L ~ \theta - \delta$

$EI ~ t_{A/B} = L(EI ~ \theta) - EI ~ \delta$

$\frac{1}{3}RL^3 = \frac{7}{48}w_oL^4 - EI ~ \delta$

$R = \dfrac{7}{16}w_oL - \dfrac{3EI ~ \delta}{L^3}$

The quantity $\frac{7}{16} w_oL$ is the simple reaction when there is no settlement $\delta$ at the propped support, thus the reaction $R$ decreased by $3EI\delta / L^3$.

Tags:

Subscribe to MATHalino.com on