Question

A long, thin metal bar of length l is clamped rigidly at its ends at temperature to. When the temperature is increased to t, the expanding bar will bow out, as shown below. If the bowing is not too large, a fair first approximation to the shape of the bar is two equal straight segments in the form of a wide V. What is the arch $\delta$ of the bow as a function of t? (This is the distance between the corner of the V and the straight line that represents the form of the bar at to.)

• A. $\delta = (l/2)\sqrt{2a(t-t_0)}$
• B. $\delta = (l/2)\sqrt{a(t-t_0)}$
• C. $\delta = (l)\sqrt{a(t-t_0)}$
• D. $\delta = (l)\sqrt{2a(t-t_0)}$

Answer 1

Answer: (A)

$\delta = (l/2)\sqrt{2a(t-t_0)}$

Answer 2

When the temperature of the metal bar increases by $\Delta t = t - t_0$, it tends to expand. If free to expand, its length would increase by $\Delta L = \alpha l \Delta t$, where $\alpha$ is the coefficient of linear thermal expansion. The total expanded length would be $l_{new} = l + \alpha l \Delta t$.

The problem approximates the bowed shape as two equal straight segments forming a V. Let the length of each segment be $L'$. The total length of the V-shape is $2L'$.

The total length of the bowed shape ($2L'$) must be equal to the expanded length of the bar ($l_{new}$). Thus, $2L' = l + \alpha l \Delta t$, which gives $L' = \frac{l}{2}(1 + \alpha \Delta t)$.

The V-shape, with arch $\delta$, forms two right-angled triangles. Each triangle has a base of $l/2$ (half the original length) and a height of $\delta$ (the arch). The hypotenuse of each triangle is $L'$. By the Pythagorean theorem, $L'^2 = (l/2)^2 + \delta^2$.

Substitute the expression for $L'$ into the Pythagorean equation: $\left(\frac{l}{2}(1 + \alpha \Delta t)\right)^2 = \left(\frac{l}{2}\right)^2 + \delta^2$ $\frac{l^2}{4}(1 + \alpha \Delta t)^2 = \frac{l^2}{4} + \delta^2$ $\delta^2 = \frac{l^2}{4}((1 + \alpha \Delta t)^2 - 1)$ $\delta^2 = \frac{l^2}{4}(1 + 2\alpha \Delta t + (\alpha \Delta t)^2 - 1)$ $\delta^2 = \frac{l^2}{4}(2\alpha \Delta t + (\alpha \Delta t)^2)$

Since the bowing is not too large, $\delta$ is small, which implies $\alpha \Delta t$ is small. We can neglect the term $(\alpha \Delta t)^2$ compared to $2\alpha \Delta t$. $\delta^2 \approx \frac{l^2}{4}(2\alpha \Delta t) = \frac{l^2}{2}\alpha \Delta t$ Taking the square root: $\delta \approx \sqrt{\frac{l^2}{2}\alpha \Delta t} = \frac{l}{\sqrt{2}}\sqrt{\alpha \Delta t} = l\sqrt{\frac{\alpha \Delta t}{2}}$ Replacing $\alpha$ with $a$ and $\Delta t$ with $(t-t_0)$: $\delta \approx l\sqrt{\frac{a(t-t_0)}{2}}$ This can be rewritten as: $\delta \approx \frac{l}{2}\sqrt{2}\sqrt{a(t-t_0)} = \frac{l}{2}\sqrt{2a(t-t_0)}$