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Bimetallic Strip Under Thermal Load

Overview

The aim of this test case is to validate the following functions:

  • thermomechanical solver with different materials

The simulation results of SimScale were compared to the analytical results derived from [Roark]. The mesh used was created locally consisting of quadratic hexahedral elements and uploaded to the SimScale platform.

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Geometry

BimetallicStripUnderThermalLoad-geometry
BimetallicStripUnderThermalLoad-geometry

Geometry of the bimetallic strip

The bimetallic strip has a length of l=10m, width of w=1m and the total height of h=0.1m with each strip thickness of ta=tb=0.05m.

Analysis type and Domain

Tool Type : CalculiX

Analysis Type : Thermomechanical

Mesh and Element types :

Mesh type Number of nodes Number of 3D elements Element type
quadratic hexahedral 3652 600 3D isoparametric
BimetallicStripUnderThermalLoad-mesh
BimetallicStripUnderThermalLoad-mesh

Mesh used for the SimScale

Simulation Setup

Material:

Upper strip:

  • isotropic: Ea = 200 GPa, ν = 0, ρ = 7870 kg/m³, κ = 60 W/(mK), γa = 1e-5 1/K, Reference temperature = 300 K

Lower strip:

  • isotropic: Eb = 200 GPa, ν = 0, ρ = 7870 kg/m³, κ = 60 W/(mK), γb = 2e-5 1/K, Reference temperature = 300 K

Initial Conditions:

  • uniform: To = 300 K

Constraints:

  • Node N1 fixed in all directions
  • Node N2 fixed in x and y direction
  • Node N3 fixed in y direction

Temperature:

  • T = 400 K on face ABCD and EFGH

Contact:

  • Bonded contact with automatic ‘Position tolerance’ between two strips.

Reference Solution

(1)\[K_1 = 4 + 6 \frac {t_a}{t_b} + 4 \left(\frac {t_a}{t_b}\right)^2 + \frac {E_a}{E_b} \left(\frac {t_a}{t_b}\right)^3 + \frac {E_a}{E_b} \frac {t_a}{t_b} = 16\] (2)\[d_x = \frac {6 l (\gamma_b – \gamma_a) (T – T_o) (t_a + t_b)}{(t_b)^2 K_1} = 0.0015 \ m\] (3)\[d_z = \frac {3 (l)^2 (\gamma_b – \gamma_a) (T – T_o) (t_a + t_b)}{(t_b)^2 K_1} = 0.075 \ m\] (4)\[\sigma = \frac {(\gamma_b – \gamma_a) (T – T_o) E_a}{K_1} \left[3 \frac {t_a}{t_b} + 2 – \frac {E_a}{E_b} \left(\frac {t_a}{t_b}\right)^3\right] = 50 \ MPa\]

The equation (1)(2)(3) and (4) used to solve the problem is derived in [Roark]. Equations (2) and (3) are the displacements of the bimetallic strip in x and z direction respectively. Whereas, equation (4) is the normal stress in x direction at the bottom surface.

Results

Comparison of the x and z displacements computed on node N3 and σxx computed on node N2 with [Roark] formulations.

Comparison of the displacements and stress
Quantity [Roark] SimScale Error
dx (m) 0.0015 0.0015 0%
dz (m) 0.075 0.074975 0.03%
σ (Mpa) 50 48.79 2.42%

References

[Roark] (1234) (2011)”Roark’s Formulas For Stress And Strain, Eighth Edition”, W. C. Young, R. G. Budynas, A. M. Sadegh
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