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Chapter 10

Q. 10.10

A solid steel shaft of 6 mm diameter fits loosely inside a hollow steel shaft of inside diameter 6 mm and outside diameter of 9 mm as shown in Figure 10.26. A pin AA prevents relative rotation between the ends of the shafts at the left. Pin holes at the right are initially at right angles to each other as shown. The shafts are now twisted in the opposite directions until the pin holes at B line up and a pin at BB is inserted.
How much strain energy is stored in the system? Assume G = 82 GPa.

10.26

Step-by-Step

Verified Solution

Clearly, the hollow shaft and solid shaft will be subjected to equal and opposite torques, say, T. Also, if \theta_{ h } \text { and } \theta_{ s } be the angles of twist of hollow and solid shafts at right end, respectively; then from compatibility of deformation, we can write

\frac{T L}{G J_{ h }}+\frac{T L}{G J_{ s }}=\frac{\pi}{2}

or            \left\lgroup \frac{T L}{G} \right\rgroup\left[\frac{J_{ s }+J_{ h }}{J_{ s } J_{ h }}\right]=\frac{\pi}{2}

or            T=\left\lgroup\frac{\pi}{2} \right\rgroup\left\lgroup\frac{G}{L} \right\rgroup\left\lgroup\frac{J_{ s } J_{ h }}{J_{ s }+J_{ h }} \right\rgroup            (1)

Now, total strain energy of the entire system will be

U=U_{ h }+U_{ s }

U=\frac{T^2 L}{2 G J_{ h }}+\frac{T^2 L}{2 G J_{ s }}=\left\lgroup \frac{T^2 L}{2 G} \right\rgroup\left\lgroup \frac{J_{ h }+J_{ s }}{J_{ s } J_{ h }} \right\rgroup

Replacing T from Eq. (1),

U=\frac{1}{2}\left\lgroup \frac{\pi}{2} \right\rgroup^2\left\lgroup \frac{G}{L} \right\rgroup\left\lgroup \frac{J_{ h } J_{ s }}{J_{ h }+J_{ s }} \right\rgroup                  (2)

Now,

J_{ s }=\frac{\pi}{32} d_{ s }^4 \quad \text { and } \quad J_{ h }=\frac{\pi}{3}\left(d_0^4-d_i^4\right)

Therefore,

J_s=\frac{\pi}{32}(6)^4  mm ^4=127.235  mm ^4

and        J_{ h }=\frac{\pi}{32}\left(9^4-6^4\right)  mm ^4=516.89  mm ^4

From Eq. (2), putting the values we get

U=\left\lgroup \frac{\pi^2}{8} \right\rgroup\left\lgroup \frac{82 \times 10^9}{2.5} \right\rgroup \frac{(127.235)(516.89)\left(10^{-12}\right)}{(127.235+516.89)}  J

or              U = 4.13 J

The total strain energy of the system is 4.13 J.