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
The bimetallic strip has a length of l=10m
, width of w=1m
and 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 |
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)
K1=4+6tatb+4(tatb)2+EaEb(tatb)3+EaEbtatb=16
(2)
dx=6l(γb−γa)(T−To)(ta+tb)(tb)2K1=0.0015 m
(3)
dz=3(l)2(γb−γa)(T−To)(ta+tb)(tb)2K1=0.075 m
(4)
σ=(γb−γa)(T−To)EaK1[3tatb+2−EaEb(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.