## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 10.20

For the beam shown in Figure 10.37, a 2 kg block is dropped from the position shown onto a 16-mmdiameter rod. Calculate (a) the maximum deflection of end A, (b) the maximum bending moment in the rod and (c) the maximum normal stress developed in the rod. Assume E = 200 GPa.

## Verified Solution

Let us draw the free-body diagram of the beam loading as shown in Figure 10.38, and first calculate the static parameters $\delta_{ A }\left(M_{\max }\right)_{ st } \text { and }\left(\sigma_{\max }\right)_{ st }$ as follows:

Now bending moment at any distance x from end C is

$M_x=\left\{\begin{array}{lr} -19.62 x N m ; & 0 \leq x \leq 0.6 m \\ -19.62 x+39.24 N m ; & x=0.6 m \\ 19.62 x-23.544 N m ; & 0.6 \leq x \leq 1.2 m \end{array}\right\}$

Therefore,

$\left|\left(M_x\right)_{\max }\right|=\left|M_{\max }\right|=(19.62) \times(0.6) N m =11.772 N m$

And maximum bending stress is

$\left(\sigma_{\max }\right)_{ st }=\frac{32 M_{\max }}{\pi d^3}=\frac{(32)(11.772)\left(10^3\right)}{\pi(16)^3}$

or          $\left(\sigma_{\max }\right)_{ st }=29.27 MPa$

Now, strain energy of beam is

$U=\left\lgroup \frac{2}{2 E I} \right\rgroup \int_0^{0.6} P^2 x^2 d x \text {; where } P=19.62 N$

$=\left\lgroup\frac{1}{E I}\right\rgroup \frac{P^2(0.6)^3}{3}$

Therefore, by Castigliano’s second theorem [refer Eq. (10.66)], we get deflection at point A where the mass strikes as

$\frac{\partial U}{\partial Q_i}=\Delta_i ; \quad 1 \leq i \leq n$            (10.66)

$\delta_{ A }=\frac{\partial U}{\partial P}=\left\lgroup \frac{2 P}{3 E I} \right\rgroup(0.6)^3=\frac{(2)(19.62)(0.6)^3}{(3)(200)\left(10^9\right) \frac{\pi}{64}(0.016)^4}$

or        $\delta_{ A }=4.39 mm$

Let dynamic amplification factor be μ, then

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

or          $\mu=1+\sqrt{1+\frac{80}{4.39}}=5.384$

And our static results are

$\left(\delta_{ A }\right)_{ st }=4.39 mm , \quad\left(M_{\max }\right)_{ st }=11.772 N m \quad \text { and } \quad\left(\sigma_{\max }\right)_{ st }=29.27 MPa$

The corresponding dynamic results are

\begin{aligned} \delta_{ A } &=\mu\left(\delta_{ A }\right)_{ st }=(5.384)(4.39) mm =23.64 mm \\ M_{\max } &=\mu\left(M_{\max }\right)_{ st }=(5.384)(11.772) N m =63.38 N m \\ \sigma_{\max } &=\mu\left(\sigma_{\max }\right)_{ st }=(5.384)(29.27) MPa =157.59 MPa \end{aligned}