Analysis of Bimetal Deformation

If two plates with different materials are bonded together, it may cause deformation by the temperature change. This deformation is caused by the different thermal expansion coefficients (CTE) of each material, and such materials are called bimetals.

In some devices, this property is actively utilized, such as bimetallic thermometers and thermostats. In the field of space applications, bimetals are used in thermal louvers for spacecraft. This device is desined to control the amount of radiative heat rejection, by rotating the louver blades with bimetal actuators. The advantages of this design are that the actuator can be activated automatically by the temperature of the actuator itself, and it does not require moving parts such as motors or bearings for rotation. On the other hand, the thermal louvers consist of the blades, actuators, and frames to hold them, which results in a certain amount of weight.

Thermal louvers are not commonly used in the recent space missions, but some applications can be found in exploration missions, where the spacecraft experiences significant thermal environment changes throughout the mission phases.

Table 1: Application Examples of Spacecraft Thermal Louver

Spacecraft Name	COSPAR ID	Reference
Mariner 4	1964-077A	[3]
Pioneer 6	1965-105A	[4]
Voyager 1, Voyager 2	1977-084A, 1977-076A	[5]
Ohzora	1984-015A	[6]
Sakigake	1985-001A	[6]
Suisei	1985-073A	[6]
Akebono	1989-016A	[6]
Magellan	1989-033B	[7]
HALCA	1997-005A	[6]
Landsat 7	1999-020A	[8]
Rosetta	2004-006A	[9]
Kaguya	2007-001A	[10]
Dawn	2007-043A	[11]
Parker Solar Probe	2018-065A	[12]
Psyche	2023-157A	[13]
Europa Clipper	2024-182A	[14]

Analytical Expression of Bending Radius

Under uniform temperature changes without external forces or moments, the deformation (bending radius $\rho$) of a bimetallic strip can be expressed analytically. As shown in Figure 1, the position of the neutral line where no strain occurs is set as the origin in the Z direction, and the position of the bottom surface of the object is set as $R$.

Figure 1: Deformation of Bi-metal Beam.

In this configuration, the axial force and moment can be expressed as follows:

$$\begin{align*} N_x &= \int^{R+h_1}_{R} \sigma_1 b\ dz + \int^{R+h_1+h_2}_{R+h_1} \sigma_2 b\ dz \\ &= \int^{R+h_1}_{R} E_1 \left( \frac{z}{\rho} - \alpha_1 \Delta T \right) b\ dz + \int^{R+h_1+h_2}_{R+h_1} E_2 \left( \frac{z}{\rho} - \alpha_2 \Delta T \right) b\ dz \\ &= E_1 b \left[ \frac{z^2}{2\rho} - \alpha_1 \Delta T z \right]^{R+h_1}_{R} + E_2 b \left[ \frac{z^2}{2\rho} - \alpha_2 \Delta T z \right]^{R+h_1+h_2}_{R+h_1} \\ &= E_1 b \left( \frac{2Rh_1 + h_1^2}{2\rho} - \alpha_1 \Delta T h_1 \right) + E_2 b \left( \frac{2(R+h_1)h_2 + h_2^2}{2\rho} - \alpha_2 \Delta T h_2 \right) \end{align*}$$ $$\begin{align*} M_y &= \int^{R+h_1}_{R} \sigma_1 zb\ dz + \int^{R+h_1+h_2}_{R+h_1} \sigma_2 zb\ dz \\ &= \int^{R+h_1}_{R} E_1 \left( \frac{z}{\rho} - \alpha_1 \Delta T \right) zb\ dz + \int^{R+h_1+h_2}_{R+h_1} E_2 \left( \frac{z}{\rho} - \alpha_2 \Delta T \right) zb\ dz \\ &= E_1 b \left[ \frac{z^3}{3\rho} - \frac{\alpha_1 \Delta T z^2}{2} \right]^{R+h_1}_{R} + E_2 b \left[ \frac{z^3}{3\rho} - \frac{\alpha_2 \Delta T z^2}{2} \right]^{R+h_1+h_2}_{R+h_1} \\ &= E_1 b \left( \frac{3R^2h_1 + 3Rh_1^2 + h_1^3}{3\rho} - \frac{\alpha_1 \Delta T (2Rh_1 + h_1^2)}{2} \right) \\ &+ E_2 b \left( \frac{3(R+h_1)^2h_2 + 3(R+h_1)h_2^2 + h_2^3}{3\rho} - \frac{\alpha_2 \Delta T (2(R+h_1)h_2 + h_2^2)}{2} \right) \end{align*}$$

The relationship between strain and bending radius is described by Eq. (1).

$$\begin{equation} \epsilon(z) = \frac{(\rho - z)d\theta - \rho d\theta}{\rho d\theta} = \frac{-z}{\rho} \end{equation}$$

Since we assume that there are no external forces, the total stress in the cross-section of the bimetallic strip (axial force) is zero. This leads to the relationship expressed in Eq. (2).

$$\begin{equation} R (E_1 h_1 + E_2 h_2) + E_2 h_1 h_2 + \frac{1}{2} (E_1 h_1^2 + E_2 h_2^2)= \rho \Delta T (E_1 \alpha_1 h_1 + E_2 \alpha_2 h_2) \end{equation}$$

Similarly, as the moment is also zero over the cross-section, the following relationship can be obtained.

$$\begin{align} &R^2 (E_1 h_1 + E_2 h_2) + R (E_1 h_1^2 + 2 E_2 h_1 h_2 + E_2 h_2^2) + \frac{1}{3} E_1 h_1^3 + E_2 h_1^2 h_2 + E_2 h_1 h_2^2 + \frac{1}{3} E_2 h_2^3 \notag \\ &= E_1 \alpha_1 \Delta T \rho R h_1 + E_2 \alpha_2 \Delta T \rho (R + h_1)h_2 + \frac{1}{2} E_1 \alpha_1 \Delta T \rho h_1^2 + \frac{1}{2} E_2 \alpha_2 \Delta T \rho h_2^2 \end{align}$$

Using the axial force relation Eq. (2), we can reduce the order of $R$ in the moment relation Eq. (3).

$$\begin{align} & \frac{1}{2} R (E_1 h_1^2 + 2 E_2 h_1 h_2 + E_2 h_2^2) + \frac{1}{3} E_1 h_1^3 + E_2 h_1^2 h_2 + E_2 h_1 h_2^2 + \frac{1}{3} E_2 h_2^3 \notag \\ &= \rho E_2 \alpha_2 \Delta T h_1 h_2 + \frac{1}{2} \rho E_1 \alpha_1 \Delta T h_1^2 + \frac{1}{2} \rho E_2 \alpha_2 \Delta T h_2^2 \end{align}$$

Deleting $R$ from the following two equations, we can obtain the bending radius $\rho$.

$$\begin{align*} &R (E_1 h_1 + E_2 h_2) (E_1 {h_1}^2 + 2 E_2 h_1 h_2 + E_2 {h_2}^2) \notag \\ &+ \frac{2}{3} (E_1 h_1 + E_2 h_2) ( E_1 {h_1}^3 + 3 E_2 {h_1}^2 h_2 + 3 E_2 h_1 {h_2}^2 + E_2 {h_2}^3 ) \\ &~~= \rho \Delta T (E_1 h_1 + E_2 h_2) (2 E_2 \alpha_2 h_1 h_2 + E_1 \alpha_1 {h_1}^2 + E_2 \alpha_2 {h_2}^2) \end{align*}$$ $$\begin{align*} &R (E_1 h_1 + E_2 h_2) (E_1 {h_1}^2 + 2 E_2 h_1 h_2 + E_2 {h_2}^2) + \frac{1}{2} (E_1 {h_1}^2 + 2 E_2 h_1 h_2 +E_2 {h_2}^2)^2 \\ &~~= \rho \Delta T (E_1 \alpha_1 h_1 + E_2 \alpha_2 h_2)(E_1 {h_1}^2 + 2 E_2 h_1 h_2 + {E_2}^2) \end{align*}$$

Some intermediate steps are shown below.

$$\begin{align*} &\rho \Delta T (E_1 h_1 + E_2 h_2) (2 E_2 \alpha_2 h_1 h_2 + E_1 \alpha_1 {h_1}^2 + E_2 \alpha_2 {h_2}^2) \notag \\ &- \frac{2}{3} (E_1 h_1 + E_2 h_2) ( E_1 {h_1}^3 + 3 E_2 {h_1}^2 h_2 + 3 E_2 h_1 {h_2}^2 + E_2 {h_2}^3 ) \\ &= \rho \Delta T (E_1 \alpha_1 h_1 + E_2 \alpha_2 h_2)(E_1 {h_1}^2 + 2 E_2 h_1 h_2 + {E_2}^2) - \frac{1}{2} (E_1 {h_1}^2 + 2 E_2 h_1 h_2 +E_2 {h_2}^2)^2 \\ \end{align*}$$ $$\begin{align*} &\rho \Delta T E_1 E_2 (\alpha_2 {h_1}^2 h_2 - \alpha_1 {h_1}^2 h_2 + \alpha_2 h_1 {h_2}^2 - \alpha_1 h_1 {h_2}^2) \notag \\ &= \frac{1}{6} ({E_1}^2 {h_1}^4 + 4 E_1 E_2 {h_1}^3 h_2 + 6 E_1 E_2 {h_1}^2 {h_2}^2 + 4 E_1 E_2 h_1 {h_2}^3 + {E_2}^2 {h_2}^4 ) \end{align*}$$

Finally, the bending radius $\rho$ can be expressed as follows.

$$\begin{equation} \frac{1}{\rho} = \frac{6(\alpha_2 - \alpha_1) \Delta T E_1 E_2 h_1 h_2 (h_1 + h_2)}{{E_1}^2 {h_2}^4 + 2 E_1 E_2 h_1 h_2 (2 {h_1}^2 + 3 h_1 h_2 + 2{h_2}^2) + {E_2}^2 {h_2}^4} \end{equation}$$

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