Question

Two thin rods of same length but thermal coefficient of linear expansion $\alpha_1, \alpha_2$ are attached side by side to make a bimetallic strip. Assuming the thickness Of strip is d. Find radius of strip when heated for $\Delta T$ temperature Change.

Answer 1

R = \frac{d}{2(\alpha_2 - \alpha_1) \Delta T}

Answer 2

To determine the radius of curvature of the bimetallic strip, we consider the expansion of the two rods and their resulting arc lengths.

Let $L_0$ be the initial length of both rods.
Let $\alpha_1$ and $\alpha_2$ be the coefficients of linear expansion for the two rods. Assume $\alpha_2 > \alpha_1$, so the rod with $\alpha_2$ will be on the outer (convex) side of the bend.

The problem states that the "thickness of strip is d". This usually refers to the total thickness of the bimetallic strip. Since there are two rods attached side-by-side, we assume they have equal thickness. Therefore, the thickness of each individual rod is $t = d/2$.

When heated by a temperature change $\Delta T$, the length of the inner rod ($L_1$) and the outer rod ($L_2$) will be:
$L_1 = L_0 (1 + \alpha_1 \Delta T)$
$L_2 = L_0 (1 + \alpha_2 \Delta T)$

Let $R$ be the radius of curvature to the neutral axis of the inner rod. The neutral axis of a rod is at its center.
Since the thickness of each rod is $t = d/2$, the distance between the neutral axes of the two rods is $t = d/2$.
Therefore, the radius of curvature to the neutral axis of the outer rod will be $R + t = R + d/2$.

If the strip bends into an arc of a circle with angle $\theta$, the arc lengths are:
$L_1 = R \theta$
$L_2 = (R + d/2) \theta$

Now, we can find the ratio of the expanded lengths:
$\frac{L_2}{L_1} = \frac{(R + d/2) \theta}{R \theta} = 1 + \frac{d}{2R}$

Also, using the expansion formulas:
$\frac{L_2}{L_1} = \frac{L_0 (1 + \alpha_2 \Delta T)}{L_0 (1 + \alpha_1 \Delta T)} = \frac{1 + \alpha_2 \Delta T}{1 + \alpha_1 \Delta T}$

Equating the two expressions for $\frac{L_2}{L_1}$:
$1 + \frac{d}{2R} = \frac{1 + \alpha_2 \Delta T}{1 + \alpha_1 \Delta T}$

Rearranging to solve for $\frac{d}{2R}$:
$\frac{d}{2R} = \frac{1 + \alpha_2 \Delta T}{1 + \alpha_1 \Delta T} - 1$
$\frac{d}{2R} = \frac{(1 + \alpha_2 \Delta T) - (1 + \alpha_1 \Delta T)}{1 + \alpha_1 \Delta T}$
$\frac{d}{2R} = \frac{(\alpha_2 - \alpha_1) \Delta T}{1 + \alpha_1 \Delta T}$

Since $\alpha_1 \Delta T$ is typically very small (e.g., $10^{-5} \times 100 = 10^{-3}$), we can use the approximation $1 + \alpha_1 \Delta T \approx 1$.
So, the equation simplifies to:
$\frac{d}{2R} \approx (\alpha_2 - \alpha_1) \Delta T$

Solving for $R$:
$R = \frac{d}{2(\alpha_2 - \alpha_1) \Delta T}$

If $\alpha_1 > \alpha_2$, the strip would bend in the opposite direction, and the formula would involve $|\alpha_1 - \alpha_2|$.