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    Bimetallic Strip Under Thermal Load


    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.

    Import validation project into workspace



    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

    Mesh used for the SimScale

    Simulation Setup


    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


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


    • T = 400 K on face ABCD and EFGH


    • 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.


    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%


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

    Last updated: January 29th, 2019

    What's Next

    part of: Buoyant flow