Question 1.13: A copper rod with circular cross-sections, as shown in Figur...

If the copper rod is allowed to expand freely, it will elongate by

\begin{aligned} \Delta L_{ Cu } &=L_{ Cu } \alpha_{ Cu } \Delta T \\ &=(30+60) \times 18 \times 10^{-6} \times 40 \\ &=0.0648  cm =0.648  mm \end{aligned}

If steel frame is allowed to expand freely, it will elongate by

\begin{aligned} \Delta L_{ St } &=L_{ St } \alpha_{ St } \Delta T \\ &=(30+60+0.0025) \times 12 \times 10^{-6} \times 40 \\ &=0.0432  cm =0.432  mm \end{aligned}

It is clear that copper will try to elongate beyond steel which is not possible. So the copper rod will face compressive load, whereas steel frame will be in tension. Finally, the copper rod will contract and steel will elongate a little which will balance the load. Let the effective change in length for copper be \delta_{ Cu } \text { and for steel be } \delta_{ St } \text {, so }

\left(0.648-\delta_{ Cu }\right)=\left(0.432+\delta_{ St }+0.025\right)

This is the required compatibility equation,

In Figure 1.34, S–S is the initial position of the top of copper rod. S′–S′ is the initial position of the lower side of steel frame (note that the gap in between is 0.025 mm) and S′′–S′′ is the final position of both. Now,

\begin{aligned} \delta_{ Cu } &=\frac{P}{E_{ Cu }}\left[\frac{l_1}{A_1}+\frac{l_2}{A_2}\right] \\ &=\frac{P}{120 \times 10^3}\left[\frac{300}{\frac{\pi}{4}(20)^2}+\frac{600}{\frac{\pi}{4}(25)^2}\right] \\ &=P \times 1.814 \times 10^{-3}  mm \end{aligned}

Similarly,

\delta_{ St }=\frac{P}{2} \times \frac{1}{E_{ St }}\left[\frac{900+0.025}{6.25 \times 10^4}\right]

For explanation of P/2, refer Figure 1.31(b). Substituting the values, we get

\delta_{ St }=\frac{P}{2} \times \frac{1}{200 \times 10^3}\left[\frac{900.025}{6.25 \times 10^4}\right]=0.36 P \times 10^{-3}  mm

Putting these values in the compatibility equation, we get P = 8785.6 N.
So, force developed on copper rod is 8785.6 N.