Nonlinear Material – Elastic Perfectly Plastic

Model an alloy steel plate with a thickness of 0.25 inch.

Analytical Solution: If 15,000 lbs is applied at the right edge, then the nominal stress is . . .

𝜎𝑛𝑜𝑚 = 15,000 𝑙𝑏

(2 𝑖𝑛 − 1 𝑖𝑛) ∙ 0.25𝑖𝑛= 60,000 𝑝𝑠𝑖

From the chart, d/H = 0.5 and Kt = 2.15.

Thus, the maximum stress is 129,000 psi.

Alloy steel has a yield strength of 89,984 psi, so yielding is expected at the notch.

Choose the material as Alloy Steel:

Choose “Plasticity – von Mises” as the model type. This corresponds to an elastic – perfectly plastic

material. Notice that some materials have an “SS” after them. Click on this to see the nonlinear stress-

strain curve associated with this nonlinear behavior.

Boundary Conditions: Apply fixed displacements on the left edge and a uniform load of 15,000 lbs on

the right edge.

Click on the “2” under “Points” in the “Time curve” window to make a third row come up in the time

curve. For “Point 3” add a time of 2 and a “Y” value of “0”. This will make the load drop to zero at the

end of the run.

Study Properties: Right click on the study and select “Properties.” Input an “End time” of “2” and use a

“Direct Sparse” solver.

Click on Advanced. We won’t change any of these settings.

Mesh and Run: Apply mesh control on the hole to improve solution accuracy, or use h-adaptive on a

linear study to get an efficient mesh, and then copy it for use in the non-linear model.

The stresses at time = 1 are shown below. Notice that the stress does not exceed the yield strength for

an elastic – perfectly plastic material.

Probe the result and do a plot if stress versus time. See the residual stress that remains in the material.

Plot other stress, strain and displacement components at different times to understand what is going on

as the material yields and the plastic zone spreads away from the hole across the ligament.