RD-V: 0210 Yeoh Hyperelastic Material
A one element model in tension used as a test model and compare with analytical results.

The Yeoh model can be used to describe nearly impressible hyperelastic materials, like rubber. Set Yeoh parameter with LAW94 and compare one element tension Radioss results with analytical results.
Options and Keywords Used
/MAT/LAW94 (YEOH)
Input Files
Model Description

Units: mm, s, Mg, N, MPa
Properties: /PROP/SOLID with, Isolid=24, Ismstr=10, and Icpre=1
Simulation Iterations
- Yeoh theory:
The Yeoh energy density is:
W=3∑i=1[Ci0(ˉI1−3)i+1Di(J−1)2i]This example assumes an incompressible ( J=1 ) hyperelastic material, then the strain energy density of Yeoh simplified, as:
W=C10(ˉI1−3)1+C20(ˉI1−3)2+C30(ˉI1−3)3With
ˉI1=ˉλ21+ˉλ22+ˉλ23Where,- ˉλi
- Deviatoric stretch with ˉλi=J−13λi and λi=εi+1 .
- λ1,λ2,λ3
- The stretch in principal direction 1, 2, 3.
- εi
- Engineer strain in principal direction I.
It shows only three parameters C10,C20,C30 need to be defined for the (incompressible) Yeoh model.
Uniaxial test is used in this example, then:
λ1=λ and λ2=λ3=λ−12
The Cauchy stress of Yeoh model is computed as:
σi=λ1J∂W∂λ1In the uniaxial test, the Cauchy stress (true stress) in principal direction 1 is:
σ1=λ∂W∂ˉI1dˉI1dλThe engineer stress is:
σ1_eng=σ1(ε1_eng+1)In this example, three material parameters are defined below:C10=0.18C20=-2e-3C30=5e−5Figure 3. Engineer stress-strain curve - LAW94:
The hyperelastic model uses polynomial model and the Yeoh model is one of the polynomial model with only three parameters. In this example, viscous was not considered.
Results
In LAW94 Yeoh model, three parameters C10,C20,C30 defined same as analytical case.
D1 needs to be defined in the material card. If D1 is not defined in LAW94, then the default Poisson’s ratio 0.495 is used.
Then 1/D1=17.94
is printed in the Starter output file.
This can be checked as follows:
C10 . . . . . . . . . . . . . . . . . . .= 0.1800000000000
C01 . . . . . . . . . . . . . . . . . . .= 0.000000000000
C20 . . . . . . . . . . . . . . . . . . .=-2.0000000000000E-03
C11 . . . . . . . . . . . . . . . . . . .= 0.000000000000
C02 . . . . . . . . . . . . . . . . . . .= 0.000000000000
C30 . . . . . . . . . . . . . . . . . . .= 5.0000000000000E-05
C21 . . . . . . . . . . . . . . . . . . .= 0.000000000000
C12 . . . . . . . . . . . . . . . . . . .= 0.000000000000
C03 . . . . . . . . . . . . . . . . . . .= 0.000000000000
1/D1 . . . . . . . . . . . . . . . . . .= 17.94001716615
1/D2 . . . . . . . . . . . . . . . . . .= 0.000000000000
1/D3 . . . . . . . . . . . . . . . . . .= 0.000000000000


Since material incompressible is assumed in analytical calculation., Poisson’s ratio closer to 0.5 is better, but the computation time will increase. Just one element tension model from ν=0.495 to ν=0.49999995 run time is more than 20 times increased. In this example, use ν=0.4997 ; therefore, set D1=0.003334.