Nonlinear finite elements/Homework 11
Problem 1: Small Strain Elastic-Plastic Behavior[edit | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
The consistency condition during plastic flow requires that
Write down an expression for the time derivative of using the chain rule.
Part 5[edit | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
The Kuhn-Tucker loading-unloading conditions are
Write down a discrete form of the Kuhn-Tucker conditions.
Part 12[edit | edit source]
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 | edit source]
Starting from equation (3) show that
where is the deviatoric part of and is the deviatoric part of .
Part 14[edit | edit source]
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 | edit source]
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 | edit source]
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 | edit source]
Consider the isothermal upset forging of the cylindrical billet shown in Figure 2.
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.
- Plot the final shape of the billet. Compare your results with those shown in Simo and Hughes (Fig. 9.8, p. 325).
- 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 | edit source]
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.
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):
|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):
- 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.