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 = K ε^{n}

$σ = K ε^{n}$

Where,

$K$ $K$: Strength coefficient.
$n$ $n$: Must be in the range [0,1].

$n$ $n$ =0

Material is perfectly plastic.

$n$ $n$ =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:

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

One-dimensional analytical reference solution is described here.

At strain

ε

$ε$ and stress

σ

$σ$ , the residual strain is:(2)

ε r = ε - ε e = (\frac{σ}{K}) \frac{1}{n} - \frac{σ}{E} = (\frac{F}{(K * A)}) \frac{1}{n} - (\frac{F}{(E * A)})

$ε r = ε - ε e = (\frac{σ}{K}) \frac{1}{n} - \frac{σ}{E} = (\frac{F}{(K * A)}) \frac{1}{n} - (\frac{F}{(E * A)})$

Where,

Then residual displacement at the right end of the bar.(3)

U r = \int_{0}^{L} (\sqrt[n]{\frac{F}{K * A}} - F / (E * A)) d x

$U r = \int_{0}^{L} (\sqrt[n]{\frac{F}{K * A}} - F / (E * A)) d x$

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).

The following table summarizes the residual deformations results.

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%