Problem 1: Kinematics and Stress Rates[edit | edit source]
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
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
Also, the shape functions must satisfy the partition of unity condition
The required expressions are
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
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
The Jacobian determinant is
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.
For which values of and do we get invalid motions?
We get invalid motions when . Let us consider the case where
This is possible when
Therefore the values at which we get invalid motions are
Derive the expression for the Green (Lagrangian) strain tensor
for the element as a function of time.
The Green strain tensor is given by
Let us make the following substitutions
The Green strain is
Derive an expression for the velocity gradient as a function of
The velocity is the material time derivative of the motion.
Recall that the motion is given by
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
The product is
Note that the first matrix is symmetric while the second is skew-symmetric.
Therefore, the velocity gradient is
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:
Assume that and . Sketch the undeformed configuration and the deformed configuration at and . Draw both the deformed and undeformed configurations on the same plot
Recall that in the initial configuration
Also, the motion is
Plugging in the values of and , we get
The deformed and undeformed configurations are shown below.
Deformed and undeformed configurations.
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
The eigenvalues are (as expected)
The principal stretches are
The principal directions are (by inspection)
Now, the right stretch tensor is given by
Hence the right stretch is
At , we have
Therefore, the rotation is
At , we have
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
Using the values of and from the previous part, at ,
Therefore, the trace of the rate of defromation is
For the Jaumann rate
where the spin is
For the Truesdell rate
For , , , we have
Problem 2: Hyperelastic Pinched Cylinder Problem[edit | edit source]
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:
What do the authors mean by "enhanced assumed strain"?
See Wikipedia article on [w:Enhanced assumed strain|enhanced assumed strain]] .
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.
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 | edit source]
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:
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.
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.
Figure 6. ANSYS input listing for Problem 3.
Figure 7. ANSYS input listing for Problem 3.