Documentation

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.

Import validation project into workspace

Geometry

Geometry of the bimetallic strip

The bimetallic strip has a length of l=10m

l=10m

, width of w=1m

w=1m

 and total height of h=0.1m

h=0.1m

 with each strip thickness of ta=tb=0.05m

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

Material:

Upper strip:

  • isotropic: Ea
    Ea

     

    = 200 GPa, ν

    ν

     

    = 0, ρ

    ρ

     

    = 7870 kg/m³, κ

    κ

     

    = 60 W/(mK), γa

    γa

     

    = 1e-5 1/K, Reference temperature = 300 K

Lower strip:

  • isotropic: Eb
    Eb

     

    = 200 GPa, ν

    ν

     

    = 0, ρ

    ρ

     

    = 7870 kg/m³, κ

    κ

     

    = 60 W/(mK), γb

    γb

     

    = 2e-5 1/K, Reference temperature = 300 K

Initial Conditions:

  • uniform: To
    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
    T

     

    = 400 K on face ABCD and EFGH

Contact:

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

Reference Solution

(1)

K1=4+6tatb+4(tatb)2+EaEb(tatb)3+EaEbtatb=16

K1=4+6tatb+4(tatb)2+EaEb(tatb)3+EaEbtatb=16

 

(2)

dx=6l(γbγa)(TTo)(ta+tb)(tb)2K1=0.0015 m

dx=6l(γbγa)(TTo)(ta+tb)(tb)2K1=0.0015 m

 

(3)

dz=3(l)2(γbγa)(TTo)(ta+tb)(tb)2K1=0.075 m

dz=3(l)2(γbγa)(TTo)(ta+tb)(tb)2K1=0.075 m

 

(4)

σ=(γbγa)(TTo)EaK1[3tatb+2EaEb(tatb)3]=50 MPa

σ=(γbγa)(TTo)EaK1[3tatb+2EaEb(tatb)3]=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

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
Contents

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