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I was reading about bi-metallic strips and came to know that on heating it forms an arc like shape. I also read a sentence that said the radius of such an arc can also be calculated which will be taught in the future grades. So I just wanted to know if there is any direct formula to find the radius of curvature?
Also, I think it can be found out using $A=l/r$ where $A$ is the angle subtended. But then again, how can the angle be found? So what is the formula for radius?

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The angle subtended is given by the arc length divided by the radius:

$$\phi = \frac{L_2}{R+t/2}=\frac{L_1}{R-t/2} $$

where $L_2$ is the length of the longer strip (at $R+t/2$) and $L_1$ the length of the shorter strip (at $R-t/2$), $t$ is the thickness of the strips. $R$ is the radius to the middle of the strips. Assumptions here are basically small bending and thin strips: $R\gg L_{1,2}\gg t$

Solving this equation for $R$ gives: $$R=\frac{t(L_1+L_2)}{2(L_2-L_1)}$$

The change in length is related to the thermal expansion coefficients $\alpha_{1,2}$ and the change in temperature $\Delta T$ of the materials:

$$L_{1,2}=L(1+\alpha_{1,2}\Delta T)$$

Plugging this into the equation for the radius gives: $$R=\frac{t}{\Delta\alpha\Delta T}$$

where $\Delta\alpha = \alpha_2-\alpha_1$ and a small term $\propto \Delta T$ has been neglected, basically assuming that the change in length due to temperature is small: $\Delta L \ll L$

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  • $\begingroup$ Thanks a lot. I had a doubt after seeing the video and you just clarified it. ^_^ $\endgroup$ – Lokesh Sangewar May 15 '17 at 9:20
  • $\begingroup$ @Lokesh What video? $\endgroup$ – user122395 May 15 '17 at 15:16
  • $\begingroup$ @paracetamol the one mentioned in steeven's answer. $\endgroup$ – Lokesh Sangewar May 16 '17 at 6:28
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The arc-like bending happens because the two metals expand differently when heated. They have different thermal expansion coefficients $\alpha$. The bending allows the outer metal strip to expand more than the inner, so that they can expand differently while staying stuck together.

The formnula for the radius of the curvature is:

$$R=\frac t{\Delta \alpha \Delta T}$$

where $t$ is the strip thickness (thickness of each strip if they are equally thick), $\Delta \alpha$ is the difference in thermal expansion coefficients, and $\Delta T$ is the temperature increase (the increase from the original temperature at which the metal strips were joined together).


Where does this formula come from?

Metal strips expand linearly when heated. The elongation per Kelvin is called $\alpha$, so the actual elongation is $\alpha \Delta T$ (how many times longer the strip becomes when the temperature rises $\Delta T$).

It takes force $F$ to expand such metal strip. It has some elasticity $E$ (Young's modulus), which resists the elongation. $E$ is how many Newton of force the metal strip holds back with per area, so the actual counteracting force is $EA$, which increases with the elongation (when pulling the strip to double the length, half of the elasticity is "already used" making the strip double as stiff) so we multiply them:

$$F=\alpha \Delta T EA\quad\Leftrightarrow \quad\sigma A=\alpha \Delta T EA\quad\Leftrightarrow \quad\sigma =\alpha \Delta T E$$

Such internal force corresponds to an internal stress $\sigma$, which is force per area $\sigma=F/A$. We want to express it with $\sigma$, because this bending metal strip can be viewed as a bending beam for which there are many well-known bending expressions, typically involving $\sigma$. The relevant bending expression for this simple situation is:

$$\frac{\sigma}{y}=\frac{E}{R}\quad\Leftrightarrow\quad \sigma=y\frac{E}{R}$$

where $\sigma$ is the stress and $y$ the depth to the neutral line (halfway into the thickness) when the beam is bent to a curvature radius of $R$. Now, here we are actually making the assumption that the metal strips together act as a uniform beam. This is not exactly the case, and there will be some non-uniform forces at the interface between them. But neglecting that gives much simplification and our final expression is still approximately valid. Also, $E$ is different for the materials, so the used $E$ here is an average (which luckily cancels out in the next step). We plug this into our expression:

$$y\frac{E}{R}=\alpha \Delta T E\\ R=\frac y{\alpha \Delta T}$$

We are almost there. The $\alpha$ here is for the entire strip, though, so it is in fact only the average of the $\alpha$ for each strip: $\alpha=\Delta \alpha/2$. This adds an unwanted factor $2$ to the expression - but at the same time, $y$ is the thickness to the centre which is half of the total thickness $y=t/2$. The two factors of $2$ cancel out, and we have our expression:

$$R=\frac{t/2}{\Delta \alpha/2\; \Delta T}=\frac{t}{\Delta \alpha \Delta T}$$

There are other ways to derive this, and it seems that this Youtube-video uses a more geometrical method, if you are unfamiliar with the bending beams.

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