Validation Case: Bimetallic Strip Under Thermal Load

This validation case belongs to thermomechanics, with the case of a bimetallic strip under thermal load. The aim of this test case is to validate the following parameters:

• Thermomechanical solver
• Multiple materials and bonded contact

The simulation results of SimScale were compared to theoretical computations derived from [Roark’s]\(^1\).

Geometry

The geometry used for the case is as follows:

It represents a strip with length \(l\) of 10 \(m\), width \(w\) of 1 \(m\) and thickness \(t\) of 0.1 \(m\), composed of two strips each with thickness \(t_a , t_b\) of 0.05 \(m\). Nodes N1 and N3 are located at mid-thickness and node N2 is located at the bottom surface as shown in figure 1.

Analysis Type and Mesh

Tool Type: Code_Aster

Analysis Type: Thermomechanical steady state with static inertia effect

Mesh and Element Types:

The hexahedral mesh was computed locally and uploaded into the simulation project.

Simulation Setup

Material:

• Top Strip:
  • Elastic Modulus \(E_a = \) 200 \(GPa\)
  • Poison’s ratio \(\nu_a = \) 0
  • Density \( \rho_a = \) 7870 \( kg/m^3 \)
  • Thermal conductivity \( \kappa_a = \) 60 \( W/(mK) \)
  • Expansion ratio \( \gamma_a = \) 1e-5 \( 1/K \)
  • Reference temperature 300 \(K\)
  • Specific heat \(C_{pa} = \) 480 \( J/(kgK) \)
• Bottom Strip:
  • Elastic Modulus \(E_b = \) 200 \(GPa\)
  • Poison’s ratio \(\nu_b = \) 0
  • Density \( \rho_b = \) 7870 \( kg/m^3 \)
  • Thermal conductivity \( \kappa_b = \) 60 \( W/(mK) \)
  • Expansion ratio \( \gamma_b = \) 2e-5 \( 1/K \)
  • Reference temperature 300 \(K\)
  • Specific heat \(C_{pb} = \) 480 \( J/(kgK) \)

Boundary Conditions:

• Constraints:
  • N1 restrained with \( d_x = d_y = d_z = \) 0 \(m\)
  • N2 restrained with \( d_x = d_y = \) 0 \(m\)
  • N3 restrained with \( d_y = \) 0 \(m\)
  • Fixed temperature \(T = \) 400 \(K\) applied on top and bottom surfaces
• Contacts:
  • Bonded contact between the strips

Reference Solution

The reference solution is of the analytical type, as presented in [Roark’s]\(^1\). It is given in terms of the displacements of the free end of the strip and the stress at the bottom surface:

$$ d_x = l (T – T_0) \frac{\gamma_a + \gamma_b}{2} $$

$$ d_z = \frac{l^2}{2} \frac{ 6 (\gamma_b – \gamma_a) (T – T_0)(t_a + t_b) }{ t_b^2 K_1} $$

$$ \sigma_{bottom} = \frac{ (\gamma_b – \gamma_a)(T – T_0) E_b }{ K_1 } \Big[ 3 \frac{t_a}{t_b} + 2 – \frac{E_a}{E_b} \Big( \frac{t_a}{t_b} \Big)^3 \Big] $$

$$ K_1 = 4 + 6 \frac{t_a}{t_b} + 4 \Big( \frac{t_a}{t_b} \Big)^2 + \frac{E_a}{E_b} \Big( \frac{t_a}{t_b} \Big)^3 + \frac{E_b}{E_a} \frac{t_b}{t_a} $$

The computed solutions are:

\(d_x= 0.015\ m\)

\(d_z = 0.75\ m\)

\(\sigma_{bottom} = 50\ MPa\)

Result Comparison

A comparison of displacements at point N3 and stress \(\sigma_{XX}\) at point N2 with theoretical solution is presented below:

Illustration of the deformed shape and stress distribution on the bimetallic strip below:

Related Tutorials

Advanced Tutorial: Thermomechanical Analysis of an Engine Piston

Last updated: January 21st, 2021