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A brass rod, 0.60 m long, and a steel rod, 0.40 m long, both of cross-section 2.0 cm², lie end to…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A brass rod, $0.60\ \text{m}$ long, and a steel rod, $0.40\ \text{m}$ long, both of cross-section $2.0\ \text{cm}^2$ , lie end to end between two rigid walls, as shown in the figure. At $20\ ^\circ\text{C}$ they just fill the space, with no force between them. Brass has $\alpha = 1.9 \times 10^{-5}\ \text{K}^{-1}$ and Young's modulus $1.0 \times 10^{11}\ \text{Pa}$ ; steel has $\alpha = 1.2 \times 10^{-5}\ \text{K}^{-1}$ and Young's modulus $2.0 \times 10^{11}\ \text{Pa}$ . The rods lie inside a smooth, close-fitting guide tube, which keeps them straight (they cannot bend or buckle) without holding them along their length. They are heated to $70\ ^\circ\text{C}$ .

Two rigid walls with two rods end to end between them: a brass rod 0.60 m long on the left and a steel rod 0.40 m long on the right, both of cross-section 2.0 square cm, just touching each other and the walls.

Find the force with which the rods press on each other and on the walls, and the stress in the rods.
The joint between the rods moves. By how much, and towards which wall?
Suppose instead that a gap of $0.30\ \text{mm}$ had been left between the steel rod and its wall at $20\ ^\circ\text{C}$ . At what temperature would the gap close, and what would the force be at $70\ ^\circ\text{C}$ ?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Let the rods expand freely, then push them back to the space between the walls. The same force acts along both.

Step 1. Free expansion. Over $50\ \text{K}$ , brass grows by $\alpha L\Delta T = 0.570\ \text{mm}$ and steel by $0.240\ \text{mm}$ , together $0.810\ \text{mm}$ .

Step 2. Squeezing back (a). A force $F$ shortens a rod by $\dfrac{FL}{AY}$ . For brass, $\dfrac{L}{AY} = 3.0 \times 10^{-8}\ \text{m/N}$ ; for steel, $1.0 \times 10^{-8}\ \text{m/N}$ . The walls do not move, so

\displaystyle F\left(\frac{L_1}{AY_1} + \frac{L_2}{AY_2}\right) = 8.100 \times 10^{-4}\ \text{m} \quad\Rightarrow\quad F = \frac{8.100 \times 10^{-4}}{4.0 \times 10^{-8}}\ \text{N} = 20250\ \text{N}

The stress is $\dfrac{F}{A} = 1.013 \times 10^{8}\ \text{Pa}$ , the same in both rods.

Step 3. The joint (b). The brass expands freely by $0.570\ \text{mm}$ but is squeezed by $F \times 3.0 \times 10^{-8} = 0.6075\ \text{mm}$ . It ends $0.0375\ \text{mm}$ shorter than it started, so the joint moves $0.0375\ \text{mm}$ towards the brass rod's wall. The steel rod grows by the same amount.

Step 4. With a gap (c). Together the rods grow by $1.620 \times 10^{-5}\ \text{m}$ per kelvin, so a gap of $0.30\ \text{mm}$ closes after $\dfrac{3.0 \times 10^{-4}}{1.620 \times 10^{-5}}\ \text{K}$ , at $38.5\ ^\circ\text{C}$ . At $70\ ^\circ\text{C}$ the walls only take back the expansion beyond the gap:

\displaystyle F = \frac{8.100 \times 10^{-4} - 3.0 \times 10^{-4}}{4.0 \times 10^{-8}}\ \text{N} = 12750\ \text{N}

Answer: (a) $20250\ \text{N}$ ; $1.01 \times 10^{8}\ \text{Pa}$ . (b) $0.0375\ \text{mm}$ , towards the brass rod's wall. (c) At $38.5\ ^\circ\text{C}$ ; $12750\ \text{N}$ .

Check. The steel rod's change in length is $0.240\ \text{mm} - F \times 1.0 \times 10^{-8}\ \text{m/N} = 0.0375\ \text{mm}$ , the exact opposite of the brass rod's, so the total length is unchanged, as the walls demand.

Watch out. Brass expands more, so it is natural to expect the joint to move towards the steel. But brass is also softer, so the shared force squeezes it more. Here the squeeze wins, and the brass ends up a little shorter.

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