SS-V: 5030 Reactions at the Ends of Axially Loaded Plastic Bar

Test No. NVL04 Find reactions at the fixed ends and maximum displacement of a bar axially loaded beyond plasticity.

Definition

Bar dimensions are 10 x 10 x 200 mm. Distance between loaded point and left end A=50 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 material properties are:

Properties: Value
$K$ $K$: 530 MPa
$n$ $n$: 0.26
Poisson's Ratio: 0

The study was performed for the following load F values: 30000 N, 47000 N, 55000 N, and 60000 N. These loads cover the full range of elastic-plastic response of the bar.

Reference Solution

One-dimensional analytical reference solution is described here.

The length of the bar does not change under the load.(2)

\int_{0}^{A} ε 1 d x - \int_{0}^{L - A} ε 2 d x = 0

$\int_{0}^{A} ε 1 d x - \int_{0}^{L - A} ε 2 d x = 0$

or,(3)

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

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

Where,

$ε 1$ $ε 1$: Tensile strain at the left span of the bar.
$ε 2$ $ε 2$: Compressive strain at the right span of the bar,
$N$ $N$: Reaction force at left end of the bar.
$R = F - N$ $R = F - N$: Reaction force at the right end of the bar.
$A$ $A$: Bar cross-section area

From this equation you can find the reaction at the left end of the bar.(4)

N = F / (1 + {(a / b)}^{n})

$N = F / (1 + {(a / b)}^{n})$

and $R = F - N$ $R = F - N$ at the right end.

Results

Bar was modeled as a 3D solid with immovable ends. Axial force F could not be applied precisely at the solid bar axis, so four line spots were created at the bar sides and total load F was uniformly distributed over the spots (Figure 3).

The following table summarizes the reaction force results.

Force F [N]	SOL Reference, Reaction [N]	SimSolid, Reaction [N]	% Difference
30000	17128	18151	5.97%
47000	26834	27146	1.16%
55000	31401	31788	1.23%
60000	34256	34591	0.98%

Typical von Mises stress distributions are shown in Figure 4 and Figure 5. The distribution has high gradients at load application lines; yet the reactions values correlate to the 1D solution because the reactions are applied far from the active force.