SS-V: 5040 Residual Deformations in an Axially Loaded Plastic Bar
Test No. VNL05 Find residual deformations in a bar axially loaded beyond plasticity.
Definition
Bar dimensions are 10 x 10 x 200 mm. Strain-stress curve of the bar material is
defined by the power law:(1)
σ=Kεn
Where,
- K
- Strength coefficient.
- n
- Must be in the range [0,1].
- n =0
- Material is perfectly plastic.
- n =1
- Material is elastic.
The left end of the bar is clamped and the right end is loaded with force F.
The material properties are:
Figure 2. Corresponded strain-stress curve
- Properties
- Value
- K
- 530 MPa
- n
- 0.26
- Poisson's Ratio
- 0
Note: Elasticity modulus defined by the first two points of the strain-stress
curve is E=2.67324e+10 Pa.
The study was performed for the following load F values: 20000 N, 25000 N, and 30000 N
Reference Solution
One-dimensional analytical reference solution is described here.
At strain
ε
and stress
σ
, the residual strain is:(2)
εr=ε−εe=(σK)1n−σE=(F(K∗A))1n−(F(E∗A))
Where,
- ε
- Total strain in the bar.
- εe
- Elastic component of the total strain.
- εr
- Residual strain.
- A
- Bar cross-section area.
Then residual displacement at the right end of the bar.(3)
Ur=L∫0(n√FK*A
Results
The bar was modeled as a 3D solid. The left end of the solid was fixed and the right
end loaded with axial force (Figure 3).
Figure 3.
The following table summarizes the residual deformations results.
Figure 4. von Mises stress distribution at load F=30000
N
| Force F [N] | SOL Reference, Residual Displacement [mm] | SimSolid, Residual Displacement [mm] | % Difference |
|---|---|---|---|
| 20000 | 3.22 | 3.43 | 6.78% |
| 25000 | 9.16 | 9.172 | 0.13% |
| 30000 | 19.077 | 19.14 | 0.33% |