Nonlinear Material Elastic Perfectly Plastic Material Response

Transcription

1 Nonlinear Material Elastic Perfectly Plastic Material Response Model an alloy steel plate with a thickness of 0.25 inch. Analytical Solution: If 15,000 lbs is applied at the right edge when the left edge is fixed, the nominal stress is... σ nom = 15,000 lb = 60,000 psi (2 in 1 in) 0.25in From the chart, d/h = 0.5 and Kt = Thus, the maximum stress is... σ max = K t σ nom = ,000 psi = 129,000 psi Alloy steel has a yield strength of 89,984 psi, so yielding is expected at the notch.

2 Elastic Finite Element Model. Fix the left edge of the model and apply a uniform tensile load of 15,000 lbs to the right side of the model. Performing an elastic analysis of this body shows a peak von Mises stress of 130,221 psi. This is very close to the predicted value of σ max = 129,000 psi computed above. We know that the stress cannot exceed the yield strength. So, it is clear that a static elastic analysis does not account for yielding. In fact, the yield stress listed in the contour plot above is for reference purposes only.

3 stress Elastic Perfectly Plastic Model. An elastic perfectly plastic material model does not account for strain hardening of the material. When using this model, the stress increases linearly until the yield strength is reached, and then the material offers no further resistance to deformation, as shown by the stressstrain plot below. Choose the material as Alloy Steel: strain 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 stressstrain curve associated with this nonlinear behavior; we could define a material that has a custom stress strain response by entering our own data. Alternative, we could consider kinematic hardening by entering a tangent modulus in the table shown above (we would actually need to copy this material definition and modify the copied version).

4 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 loading history will cause the load to increase to 15,000 lbs at time=1 and then drop back down to zero at time=2. Removing the load in this way will allow us to check to see whether or not residual stresses exist.

5 Study Properties: Right click on the study and select Properties. Input an End time of 2 the large displacement and large strain options. Since we are using a uniform normal force, let s click the (Applicable only for normal uniform pressure and normal force) option. The entries available in this window allow us to control the minimum and maximum time steps (or load steps in this case) during the solution process. These values influence solution accuracy. Also, choosing smaller time steps would provide us with the system response at a smaller increment in loading.

6 Click on Advanced. We won t change any of these settings. These settings allow you to apply control over the nonlinear solution process.

7 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.

8 Residual Stresses. At time = 2, the loading has been reduced back to zero. Notice that we have considerable residual stresses (deformation-induced stresses that exist in the absence of external loading). Plotting σ x (or SX) allows us to see which residual stresses are tensile and which are compressive. Note that residual compressive stresses (minimum = -36,745 psi) are balanced by residual tensile stresses (maximum = 9,856 psi). The sum of forces in the x-direction due to integrating these stresses across the ligament between the hole and the edge must equal zero. Compressive stresses at notches like this hole can have beneficial effects for fatigue loading situations since the tendency for cracking is mitigated by the compressive stress field.