Nonlinear finite elements/Homework 10/Solutions

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Problem 1: Kinematics and Stress Rates[edit]

Given:

Figure 1 shows a linear three-noded triangular element in the reference configuration.

Figure 1. Three-noded triangular element.

The motion of the nodes is given by:

The configuration () of the element at time is given by

Solutions[edit]

Part 1[edit]

Write down expressions for , , and in terms of the initial configuration () ?

In the initial configuration . Therefore,

Therefore, the initial configuration is given by

Substituting the values of and , we get

Hence

Also, the shape functions must satisfy the partition of unity condition

Therefore,

The required expressions are

Part 2[edit]

Derive expressions for the deformation gradient and the Jacobian determinant for the element as functions of time.

The deformation gradient is given by

Before computing the derivatives, let us express in terms of . Recall

Therefore,

Substituting in the expressions for and , we get

In the above expression, the parent coordinates are no longer useful. Therefore, we write

where . Taking derivatives, we get

Therefore,

The Jacobian determinant is

or

Part 3[edit]

What are the values of and for which the motion is isochoric?

For isochoric motion, . Therefore,

One possibility is

This is a pure rotation.

Another possibility is that

This is a combination of shear and rotation where the volume remains constant.

Part 4[edit]

For which values of and do we get invalid motions?

We get invalid motions when . Let us consider the case where . Then

This is possible when

If then

That is,

Therefore the values at which we get invalid motions are

Part 5[edit]

Derive the expression for the Green (Lagrangian) strain tensor

for the element as a function of time.

The Green strain tensor is given by

Recall

Let us make the following substitutions

Then

Therefore,

Hence,

The Green strain is

Part 6[edit]

Derive an expression for the velocity gradient as a function of

time.

The velocity is the material time derivative of the motion.

Recall that the motion is given by

Therefore,

We could compute the velocity gradient using

after expressing in terms of . However, that makes the expression quite complicated. Instead, we will use the relation

The time derivative of the deformation gradient is

The inverse of the deformation gradient is

Using the substitutions

we get

and

The product is

Note that the first matrix is symmetric while the second is skew-symmetric.

Therefore, the velocity gradient is

Part 7[edit]

Compute the rate of deformation tensor and the spin tensor.

The rate of deformation is the symmetric part of the velocity gradient:

The rate of deformation is the skew-symmetric part of the velocity gradient:

Part 8[edit]

Assume that  and . Sketch the undeformed configuration and the deformed configuration at  and .  Draw both the deformed and undeformed configurations on the same plot

and label.

Recall that in the initial configuration

Also, the motion is

Plugging in the values of and , we get

At ,

At ,

The deformed and undeformed configurations are shown below.

Deformed and undeformed configurations.

Part 9[edit]

Compute the polar decomposition of the deformation gradient with the above values of and ,

The deformation gradient is

The right Cauchy-Green deformation tensor is

The eigenvalue problem is

This problem has a solution if

i.e.,

The eigenvalues are (as expected)

The principal stretches are

The principal directions are (by inspection)

Now, the right stretch tensor is given by

Therefore,

Hence the right stretch is

At , we have

Now

and

Therefore, the rotation is

At , we have

Part 10[edit]

Assume an isotropic, hypoelastic constitutive equation for the material of the element. Compute the material time derivative of the Cauchy stress at using (a) the Jaumann rate and (b) the Truesdell rate.

A hypoelastic material behaves according to the relation

For an isotropic material

Therefore,

Recall that

Using the values of and from the previous part, at ,

Therefore, the trace of the rate of defromation is

Therefore,

For the Jaumann rate

where the spin is

Therefore,

or,

For the Truesdell rate

Therefore,

Recall,

For , , , we have

Therefore,

and

Hence,


Problem 2: Hyperelastic Pinched Cylinder Problem[edit]

Read the following paper on shells:

Buchter, N., Ramm, E., and Roehl, D., 1994, "Three-dimensional extension of non-linear shell formulation based on the enhanced assumed strain concept," Int. J. Numer. Meth. Engng., 37, pp. 2551-2568.

Answer the following questions:


Solution[edit]

Part 1[edit]

What do the authors mean by "enhanced assumed strain"?

See Wikipedia article on [w:Enhanced assumed strain|enhanced assumed strain]] .

Part 2[edit]

Example 8.2 (and Figures 3 and 4 and Table III) in the paper discusses the simulation of a hyperelastic cylinder. Perform the same simulation using ANSYS for a shell thickness of 0.2 cm. Use shell elements and the Neo-Hookean hyperelastic material model that ANSYS provides.

The following material properties are used:

kN/cm, , kN/cm, and cm/kN. Symmetry is used and only half of the model is meshed. At the support, the model is constraint in all directions. The load of 36 kN is applied on the top of the cylinder (Fig~2. ANSYS input listing is shown Fig~6 of Appendix.

Figure 2. Meshed model

Part 3[edit]

Compare the total load needed to achieve an edge displacement of 16 cm with the results given in Table III. Comment on your results.

The plot of force vs. edge displacement (vertical) and the deformed model are shown in Fig 3. From the plot, one sees that a load of 35.1 kN is required to deform the edge by 16 cm. This is less than 1% difference compared to the result given in the paper using 7-parameter shell theory.

Figure 3. Left: Force vs. edge displacement. Right: Deformed model.

Problem 3: Elastic-Plastic Punch Indentation[edit]

Read the following paper on elastic-viscoplastic FEA:

Rouainia, M. and Peric, D., 1998, "A computational model for elasto-viscoplastic solids at finite strain with reference to thin shell applications," Int. J. Numer. Meth. Engng., 42, pp. 289-311.

Answer the following questions:

Solution[edit]

Part 1[edit]

Example 5.4 of the paper shows a simulation of the deformation of a thin sheet by a square punch. Perform the same simulation for 6061-T6 aluminum. Assume linear isotropic hardening and no rate dependence.

The following data are used for the thin plate: GPa, , Mpa, Mpa, see Fig 4 for the stress-strain data. Symmetry is used and only half of the model is meshed (Fig 4). At the support, the model is constrained in all directions. A plastic finite strained shell (SHELL43) is chosen for this simulation.

The following material properties are used for the punch and die: GPa, . The value of Young's modulus is arbitrary chosen so long as it is high enough to remain rigid during the simulation.

The meshed model is illustrated in Fig 4. A load of 10 kN is applied on the top of the punch. Load steps are split into two steps. First load step the punch is moved close to the plate to activate the contact elements (CONTAC49). The second load step, 30 kN is applied on top of the punch. ANSYS input listing is shown Fig 7 of the Appendix.

Figure 4. Left: Bi-linear stress-strain data. Right: Meshed model.

Part 2[edit]

Draw a plot of the punch force vs. punch travel and compare your result with the results shown in Figure 13 of the paper (qualitative comparison only).

The punch force vs. punch travel and the plot of the deformation at the final load step is shown in Fig 5. The curve displays similar pattern as those shown in the paper.

Figure 5. Left: Punch force vs. punch travel. Right: Sketch of deformation at final load step.

Appendix[edit]

Figure 6. ANSYS input listing for Problem 3.
Figure 7. ANSYS input listing for Problem 3.