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Chapter 10

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.

10.37

Step-by-Step

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}

10.38