Curvature of sheet due to thermal expansion I have a sheet of aluminum, $\rm 1.2m × 2.8m$ , which will be bonded to PTFE (also $\rm 1.2m × 2.8m$). The temperature of the laminate will then increase by $\rm ΔT$.
Material properties are:

*

*Coefficient of thermal expansion :

Aluminum : $\rm 24×10^-6/°C$
PTFE : $\rm 126×10^-6/°C$

*

*Young's Modulus :

Aluminum : $\rm 70,000\ MPa$
PTFE : $414\rm \ MPa$
Since the PTFE will expand more than the aluminum, the laminate will begin to curve as it is heated. Assuming it is flat at room temperature, how would I calculate the radius of curvature of the laminate once the material is heated up, as a function of thickness of each material ?
Thank you!
 A: Since your aspect ratio is >2, as a first pass, you could apply the bimetallic strip model:

Timoshenko published a solution for the corresponding curvature in "Analysis of Bi-Metal Thermostats," J. Opt. Soc. Am. 11, 233-255 (1925):
$$\kappa = \frac{1}{\rho}=\frac{6 E_1 E_2 (t_1 + t_2)t_1 t_2 \epsilon }{E_1^2 t_1^4 + 4 E_1 E_2 t_1^3 t_2 + 6 E_1 E_2 t_1^2 t_2^2 + 4 E_1 E_2 t_2^3 t_1 + E_2^2 t_2^4},$$
or
$$\kappa= \frac{\varepsilon}{t}\cdot\frac{6(1+m)^2 }{3(1+m)^2+(1+mn)\left(m^2+\frac{1}{mn}\right)},$$
or
$$\kappa= \frac{\varepsilon}{t}\cdot\frac{6 (m+2 + m^{-1})  }{n m^2 + 4 m + 6  + 4 m^{-1}  + n^{-1} m^{-2}},$$
where $\kappa$ is the curvature, $\rho$ is the radius of curvature, $t_1$ and $t_2$ are respectively the thicknesses of strips 1 and 2 (with $m=t_1/t_2$ and $t=t_1+t_2$), $E_1$ and $E_2$ are the corresponding Young moduli (with $n=E_1/E_2$), and $\varepsilon = (\alpha_1-\alpha_2)\Delta T$ is the thermal mismatch strain.
The derivation is as follows, adapted from the Wikipedia article on the bimetallic strip.

Let the layer on the concave side be layer 1 and on the convex side be layer 2. Layer 1 is in tension with a force outwards on each end of $F_1$, while layer 2 is compressed with a force inwards on each end of $F_2$. Because the system is in equilibrium, $F_1=F_2=F$.
At each end of layer 1, there is a bending moment $M_1$, and similarly for layer 2. If  $\rho$ is the radius of curvature, then $M_1=E_1I_1/\rho$ and $M_2=E_2I_2/\rho$, where $EI$ is the flexural rigidity, $E$ is Young's modulus and $I$ is the area moment of inertia or second moment of area. For a rectangular cross-section of  width $w$, $I_1=wt_1^3/12$ and $I_2=wt_2^3/12$.  The couple produced by the forces $F$ acting along the midlines of each layer and separated by $t_1/2+t_2/2=t/2$ is $Ft/2$, and because the strip is in equilibrium and there are no external applied torques, $Ft/2=M_1+M_2$. Hence,
$$\frac{Ft}{2}=\frac{w}{12\rho}(E_1t_1^3+E_2t_2^3).$$
We now consider the contact surface between the two layers. The length of this surface for layer 1 is $L_0\left(1 + \alpha_1\Delta T + \frac{F_1}{wt_1E_1} + \frac{t_1}{2\rho}\right)$ where $T_0$ is the temperature at which the strip is straight, $L_0$ is the length of the layer when the temperature $T=T_0$ (i.e., when it is straight and under no stress from layer 2), and $\alpha_1$ is the coefficient of thermal expansion (the fractional increase in length per unit increase in temperature). The second term here is the fractional change in length produced by the thermal expansion, and the third term is the strain induced by the stress $F_1/wt_1$ due to the force $F_1$ acting over the area of the end (positive because the force is tensile). The last term is the additional length of the contact surface relative to the midline of layer 1 (positive because the contact surface is the outer, convex surface). Similarly, the length of this surface for layer 2 is $L_0\left(1 + \alpha_2\Delta T - \frac{F_2}{wt_2E_2} - \frac{t_2}{2\rho}\right)$ (minus signs because the force is compressive and the contact is on the inner surface). Since the surfaces are bonded,
$$\alpha_1 \Delta T + \frac{F}{wt_1E_1} + \frac{t_1}{2\rho}=\alpha_2( T - T_0) - \frac{F}{wt_2E_2} - \frac{t_2}{2\rho}.$$
Rearranging to extract $\kappa=1/\rho$, collecting terms and eliminating $F$ using the equation above produces the equation above.

If $n$ is very small, as is the case here, then we can simplify the above equation to
$$\kappa\approx\frac{6\varepsilon(1+m)^2mn}{t}.$$
If the thicknesses are further similar, then
$$\kappa\approx\frac{24\varepsilon n}{t}.$$
Does this make sense?
