Coupled temperature-displacement analysis: one-dimensional gap conductance and radiation

First we assume that the gap is open and that the gap thermal conductivity, $k_g$, increases linearly as the gap reduces, so

where $k_0$ and $k_u^{\prime }$ are nonnegative constants and $u_B$ is the displacement of point $B.$ Gap radiation is neglected in these calculations.

The thermal boundary conditions are that the temperatures at A and C, ${\theta }_A$ and ${\theta }_C$, are held constant. The mechanical boundary conditions are that points A and C are fixed. Since in this case the end of the rod never touches C, force equilibrium requires that

The equations given above define the problem. Their solution is readily developed as follows. Combining integrated force equilibrium with the linear elastic constitutive equation and the displacement relationship, $u\left(S\right)$, gives

Thermal equilibrium combined with Fourier's law and the gap heat flux equation then gives

With the assumed form of the gap thermal conductivity, $k_g$, and assuming $k_g>$ 0, this is

Substituting for $u_B$ then gives a quadratic equation for ${\theta }_B$:

The roots of this quadratic equation provide two solutions for ${\theta }_B.$ The solutions for $u_B$ are available once ${\theta }_B$ is determined. Only one of the two solutions gives a value of $u_B$ for which $k_g>$ 0; hence, this is the only physically acceptable solution.

As a numerical example the parameters are chosen in consistent units as $k_r=k_0=$ 1.0; $\alpha =$10⁻⁵, $L=$ 1.0; $k_u^{\prime }=$ 100; ${\theta }_A=$ 400°; and ${\theta }_C=$ 200°.

These values give ${\theta }_B=$ 285.4° or −4485°, so $u_B=$3.427 × 10⁻³ or −2.043 × 10⁻². The second solution must be rejected as it gives $k_g<$ 0. The first solution is valid so long as $d_0>u_B.$

Next we assume that the gap is open and that the gap radiation view factor, $F_g$, increases linearly as the gap reduces; so

where $F_0$ and $F_u^{\prime }$ are nonnegative constants and $u_B$ is the displacement of point $B.$ In this case gap conduction is neglected.

The thermal and mechanical boundary conditions are the same as the gap conduction problem considered above. Force equilibrium requires that

Following a procedure similar to that used in the gap conduction problem and combining integrated force equilibrium with the linear elastic constitutive equation and the displacement relationship, $u\left(S\right)$, gives

Combining thermal equilibrium with Fourier's law and the gap heat flux equation then gives

With the assumed form of the gap radiation, $q_g$, and assuming $F_g>$ 0, this is

Substituting for $u_B$ then gives the following equation for ${\theta }_B$:

${\theta }_B$ is obtained by solving the above equation numerically. The solutions for $u_B$ are available once ${\theta }_B$ is determined.

As a numerical example the parameters are chosen in consistent units as $F_0=$ 1.0; $\alpha =$10⁻⁵; $L=$ 1.0; $F_u^{\prime }=$ 50; $\sigma =$ 1.E−8; $ϵ=$1.0; ${\theta }_A=$ 400°; ${\theta }_C=$ 200°; and absolute zero $=$ –460.

These values give ${\theta }_B=$ 222.4°, so $u_B=$ 3.112 × 10⁻³. This solution is valid so long as $d_0>u_B.$ All other solutions must be rejected since they give $F_g<0$ or $d_0<u_B.$

In this case the rod is always in contact with C. Combining the integrated equilibrium equation with the mechanical constitutive model gives

Combining this with the temperature solution, $\theta$, gives

Integrating this along the rod using the displacement relationship, $u\left(S\right)$, gives the pressure as a function of the temperature at points A and B,

since $u_B=0.$

In this case the conductivity of the closed gap is proportional to the contact pressure:

where $k_a$ and $k_p^{\prime }$ are nonnegative constants. Since $k_r\left({\theta }_A-{\theta }_B\right)/L=k_g\left({\theta }_B-{\theta }_C\right)$, with this behavior for $k_g$ we have

Combining this with the equation for the pressure provides a quadratic equation for ${\theta }_B$:

The roots of this equation provide two solutions for ${\theta }_B$, and the corresponding values of p are then defined by $p=E\alpha \left({\theta }_A+{\theta }_B\right)/2.$ Only one solution gives a positive value for p; the other must be rejected because it is inconsistent with the assumption that the gap is closed.

As a numerical example the parameters are chosen in consistent units as $k_r=$ 10; $k_a=$ 2; $k_p^{\prime }=$ 0.2; $\alpha =$ 10⁻⁵; $E=$ 10⁵; $L=$ 1.0; ${\theta }_A=$ 200°; and ${\theta }_C=$ 100°.

These values give ${\theta }_B=$ 122.6° or −342.6°, so $p=$ 161.3 or −71.3. The second solution must be rejected as it gives $p<0.$