A copper rod with circular cross-sections, as shown in Figure 1.33(a), rests inside a frame made of steel (with Young’s modulus, E = 200 GPa and coefficient of linear thermal expansion, α = 12 = 10^−6 /°C). E for copper is 120 GPa and a for copper is 18 × 10^−6 /°C. At 0°C, there exists a small gap

Question

A copper rod with circular cross-sections, as shown in Figure 1.33(a), rests inside a frame made of steel (with Young’s modulus, E = 200 GPa and coefficient of linear thermal expansion, [latex] \alpha=12 imes 10^{-6} /{ }^{\circ} C [/latex]). E for copper is 120 GPa and a for copper is [latex] 18 imes 10^{-6} /{ }^{\circ} C [/latex]. At 0°C, there exists a small gap of 0.025 mm between the upper end of the rod and the steel frame. Find out the compressive force developed in copper rod if the temperature is raised to 40°C. Assume that cross-sectional area of frame = 6.25 cm².

Accepted Answer

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

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

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

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

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 [latex] \delta_{ Cu } ext { and for steel be } \delta_{ St } ext {, so } [/latex]

[latex] \left(0.648-\delta_{ Cu } ight)=\left(0.432+\delta_{ St }+0.025 ight) [/latex]

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,

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

Similarly,

[latex] \delta_{ St }=\frac{P}{2} imes \frac{1}{E_{ St }}\left[\frac{900+0.025}{6.25 imes 10^4} ight] [/latex]

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

[latex] \delta_{ St }=\frac{P}{2} imes \frac{1}{200 imes 10^3}\left[\frac{900.025}{6.25 imes 10^4} ight]=0.36 P imes 10^{-3} mm [/latex]

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

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

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