Nonlinear finite elements/Homework 11

Problem 1: Small Strain Elastic-Plastic Behavior edit

For small strains, the strain tensor is given by


In classical (small strain) rate-independent plasticity we start off with an additive decomposition of the strain tensor


Assuming linear elasticity, we have the following elastic stress-strain law


Let us assume that the   theory applies during plastic deformation of the material. Hence, the material obeys an associated flow rule


where   is the plastic flow rate,   is the yield function,   is the temperature, and   is an internal variable.

Answer the following questions. Show your derivations in a clear and step-by-step manner.

Part 1 edit

Let   be the equivalent plastic strain, defined as


Express the time derivative of   in terms of   and  . This is the evolution law for  .

Part 2 edit

For an adiabatic process, the rate of change of temperature can be written as


where   is the Taylor-Quinney coefficient,   is the density, and   is the specific heat. Express   in terms of   and  . This is the evolution law for  .

Part 3 edit

Write down the rate form of the elastic stress-strain law. Assume that deformations are small so that objectivity of the rates is not a concern.

Part 4 edit

The consistency condition during plastic flow requires that


Write down an expression for the time derivative of   using the chain rule.

Part 5 edit

Use the consistency condition and the expressions you have derived in the previous parts to derive an expression for   in terms of  ,  ,  ,  , and  .

Part 6 edit

The continuum elastic-plastic tangent modulus is defined by the following relation


Derive an expression for the elastic plastic tangent modulus using the results you have derived in the previous parts.

Part 7 edit

The   theory of plasticity also states that the material satisfies the von Mises yield condition


where   is the deviatoric part of the stress  . Derive an expression for   in terms of the normal to the yield surface


Part 8 edit

The yield stress   is given by the Johnson-Cook model


where   is the initial yield stress,   are constants,   is a reference temperature, and   is the melt temperature. Derive expressions for  , and   for the von Mises yield condition with the Johnson-Cook flow stress model.

Part 9 edit

Assume that the elastic response of the material is linear, i.e.,


Derive the expression for the elastic-plastic tangent modulus for a von Mises yield condition with Johnson-Cook flow stress for a linear elastic material using the expressions that you have derived in the previous parts.

Part 10 edit

Discretize the equations for   (equation 1),   (from part 1), and   (from part 2) using Forward Euler. Use the following notation in your discretized equations:


where   is the time step,   is the value of   at  ,   is the value of   at  .

Part 11 edit

The Kuhn-Tucker loading-unloading conditions are


Write down a discrete form of the Kuhn-Tucker conditions.

Part 12 edit

In the radial return algorithm, we define a trial elastic state as


where   are the stress and strain at   and   are the values at  . Show that, if the elastic response of the material is linear, equation (2) can be written as


Hint: Start by showing that


Part 13 edit

Starting from equation (3) show that


where   is the deviatoric part of   and   is the deviatoric part of  .

Part 14 edit

Show that


Hint: The stress   is given by


Express this equation in terms of   and  . Then use the discretized equation for   (part 10) and the relation for   for isotropic elasticity. Finally compute the deviatoric stress terms after showing that


Part 15 edit

The discretized form of the Kuhn-Tucker conditions in conjunction with the consistency condition gives us


Use this condition and the relations you have derived in the previous sections to arrive at a nonlinear equation in   that can be solved using Newton iterations.

Part 16 edit

Let the nonlinear equation be  . Recall that the Newton method requires that we iterate using the formula


where   is the Newton iteration number. Derive an expression for the derivative of   that is required in the above formula.

(You can use Computational Inelasticity by J.C. Simo and T.J.R. Hughes for pointers.)

Problem 2: Billet Upset Forging edit

Consider the isothermal upset forging of the cylindrical billet shown in Figure 2.

Figure 2. Upset forging of a cylindrical billet.

Assume that the dies are rigid. Also assume that sticking friction is in effect between the billet and the die faces when they are in contact.

The billet has an initial radius of 10 mm and its initial height is 30 mm. The shear modulus of the material is 384.6 MPa, the bulk modulus of the material is 833.3 MPa, the initial yield stress is 1 MPa and the linear hardening modulus is 3 MPa.

Model a quarter of the cylinder using symmetry boundary conditions.

Apply a compressive force of 1 kN to the die.

  1. Plot the final shape of the billet. Compare your results with those shown in Simo and Hughes (Fig. 9.8, p. 325).
  2. Plot a curve of the die force (kN) versus the die stroke (mm). Compare your results with those shown in Simo and Hughes. Do you observe any volumetric locking?

(Use an implicit software to solve these problems.)

Problem 3: Taylor Impact Tests edit

Consider the impact of a cylindrical Taylor impact specimen on a rigid target. The undeformed and deformed profiles of the specimen are shown in in Figure 3.

Figure 3. Taylor impact test of a cylindrical rod.

The initial length of the specimen is 30 mm. The initial diameter is 6 mm. The initial velocity is 188 m/s. The initial temperature is 718 K.

The material of the specimen is OFHC copper. The properties of the bar are (in SI units):

Density 8930.0
Thermal conductivity 386.0
Specific heat 414.0
Shear modulus 46.0e9
Bulk modulus 129.0e9
Coeff. Thermal Expansion 1.76e-5

The plastic deformation of the specimen is described by the Johnson-Cook model and   plasticity. The Johnson-Cook model parameters are (in SI units):

A 90.0e6
B 292.0e6
C 0.025
n 0.31
m 1.09
  • Use LS-DYNA to simulate the Taylor impact test. Assume that there is no friction between the anvil and the specimen. Plot the final deformed shape of the specimen and compare that with the experimentally determined shape given in the table below. What differences do you observe and why?
Point x (mm) y (mm)
1 0.000000 0.000000
2 5.436409 0.000000
3 4.711554 0.852540
4 4.611804 2.040725
5 4.581879 3.228910
6 4.615129 4.141866
7 4.585204 4.980980
8 4.448878 6.175878
9 4.312552 7.223092
10 4.073150 8.344149
11 3.870324 9.465205
12 3.597672 10.868203
13 3.388196 11.707317
14 3.218620 12.902215
15 3.152120 13.949429
16 2.982544 15.070486
17 2.952618 16.674871
18 0.000000 16.674871
19 0.000000 0.000000
You can generate a mesh in ANSYS and tranfer it to LS-DYNA if you want.
  • Show whether energy is conserved during your simulation.