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## Q. 10.15

An object of weight W is dropped on to the middle of a simply supported beam form a height h. The beam has a rectangular cross-sectional area A. If $h \gg \delta_{s t} \text {, where } \delta_{s t}$ is the static deflection of beam due to W (i.e., $h \gg W L^3 / 48 E \bar{I}$ where L is the beam length) and that the mass of the object is large compared to the mass of the beam, obtain an expression for the maximum bending stress, $\sigma_{\max }$ , in the beam due to the falling weight.

## Verified Solution

Let us fix the datum at the final deflected position of the beam. Therefore,

Potential energy of the W = Final strain energy due to bending

Thus,

$W(h+\delta)=\frac{1}{2} k \delta^2$

where $k = 48EI/L^3$. So,

$k \delta^2-2 W \delta-2 W h=0 \Rightarrow \delta^2-2\left\lgroup \frac{W}{k} \right\rgroup \delta-2\left\lgroup\frac{W}{k} \right\rgroup h=0$

or                $\delta^2-2 \delta_{ st } \delta-2 \delta_{ st } h=0$

$\delta=\delta_{ st }+\sqrt{\delta_{ st }^2+2 \delta_{ st } h}=\delta_{ st }\left[1+\sqrt{1+\frac{2 h}{\delta_{ st }}}\right]$

The dynamic amplification factor (DAF) is given by

$DAF =1+\sqrt{1+\frac{2 h}{\delta_{ st }}}$

$=1+\sqrt{\frac{2 h}{\delta_{ st }}}\left\lgroup 1+\frac{\delta_{ st }}{2 h} \right\rgroup^{\frac{1}{2}} = \sqrt{\frac{2 h}{\delta_{ st }}} \quad \text { for } h \gg \delta_{ st }$

Now,

$\sigma_{\max }=\left(\sigma_{ st }\right)_{\max } \sqrt{\frac{2 h}{\delta_{ st }}}$

here,

$\left(\sigma_{ st }\right)_{\max }=\frac{6 M_{\max }}{b h^2}=\frac{3 W L}{2 A h^3}$

where $h$ is the depth and $b$ is the width of the section. Also,

$M_{\max }=W L^3 / 4 E A h^2$

and          $\delta_{ st }=\frac{W L^3}{48 E \bar{I}}=\frac{W L^3}{4 E b h^3}=\frac{W L^3}{4 E A h^2}$

Therefore,

$\sigma_{\max }=\sqrt{\frac{8 h E A h^2}{W L^3} \times \frac{9 W^2 L^2}{4 A^2 h^2}}=\sqrt{\frac{18 W E h}{A L}}$

The maximum bending stress on the beam is $\sqrt{18 W E h / A L}$