Homework 10
Non-linear Problems
Holly Ibanez
MANE6960
Advanced Topics in Finite Elements
Summary
The finite element method can be applied to the solution of many non-linear problems encountered in
applications. For static non-linear problems, the FEM yields a system of non-linear algebraic equations.
The system is usually solved using iterative methods. However, convergence is not necessarily
guaranteed and in some cases it may be difficult to obtain. Transient problems can be even more
challenging as the solution of a non-linear system of algebraic equations is required at each time step.
In this report, the finite element method is applied to the solution of a few selected non-linear
problems.
Calculation and Discussion
A cylindrical sample (radius 0.0125 m, length 0.1 m) of HY-100 steel is pulled uniaxially using a tensile
testing machine. When the stress in the sample reaches about 790 MPa, it yields and begins to perform
plastically. To account for yielding, assume that the stress-strain curve in the plastic domain is linear
with a modulus 100 times smaller than the elastic modulus. The goal is to estimate the strain and stress
fields inside the bar as a function of the applied load. To create the model, material properties of HY100 steel were applied to the model. The effects of plasticity were also defined in the model. A
prescribed displacement was applied in the z-direction to the bottom edge of the system, and a tensile
load was applied to the top. Figure 1 shows the 2D geometry of the sample.
The governing equation for the system is as follows:
−∇ ∙ 𝜎 = 𝐹𝑣
The variational formulation of the problem is as follows:
(𝑢′ , 𝑣 ′ ) = (𝑓, 𝑣)
Hot water (T=343 K) enters a pipe (radius = 0.0125 m, length 0.1 m) whose wall is maintained at a
constant temperature (T=298 K). As the water flows inside the pipe, it cools. The system was created
with the same geometry shown in Figure 1. The geometry represents water flow through a pipe;
therefore, material properties of water are applied to the model. The conjugate heat transfer module
was used in COMSOL to consider both laminar flow and heat transfer of the system. The outer wall of
the cylinder was set to a no-slip condition. The inlet fluid velocity was set to 0.1 m/s.
The governing equation for this system is as follows:
𝜌𝐶𝑝 𝑢 ∙ ∇𝑇 = ∇ ∙ (𝑘∇𝑇)
The variational formulation for the problems is as follows:
(𝑢′ , 𝑣 ′ ) = (𝑓, 𝑣)
The final system represents the solidification of a molten metal in a mold. Molten copper rapidly fills an
iron mold of constant square cross section. The mold is initially at room temperature and copper is
initially at 1400 K. The cross section of the mold opening is 0.05 m by 0.05 m and the mold wall
thickness is 0.1 m. The copper and iron system was created in COMSOL as follows:
To account for the latent heat of solidification, assume that the specific heat is a function of
temperature such that within the melting range between delta Ts = 1360-1250 = 10, the effective
specific heat Cp,eff = L/Delta Ts, shown below:
The governing equation for this system is as follows:
𝜕𝑢 𝜕 2 𝑢 𝜕𝑢
=
+
=0
𝜕𝑡 𝜕𝑥 2 𝜕𝑦
The variational formulation for the problem is as follows:
(𝑢̇ (𝑡), 𝑣) + 𝐴(𝑢(𝑡), 𝑣) = (𝑓(𝑡), 𝑣)
Results and Conclusions
The following table shows the results the tensile test system. Normal, finer, and extremely fine meshes
were used to verify the temperature and stress results.
Mesh
Normal
Finer
Extremely Fine
Displacement (m)
1.6387 x 10-3
1.6387 x 10-3
1.6387 x 10-3
Von Mises Stress (N/m2)
8.1487 x 108
8.1487 x 108
8.1487 x 108
The figures below show graphic results of the displacement and stress of the extremely fine mesh.
The following table shows the COMSOL results of fluid flow in a pipe. Coarser, coarse, and normal
meshes were used to validate the velocity results of flow through the pipe. The water temperature does
not cool as expected, showing a possible error in the model creation. The temperature does change at
the outer edges of the pipe, showing that the temperature outside of the pipe does have a small cooling
effect on the water. The velocity, however, behaves as expected, with no water movement along the
walls of the pipe, and the greatest velocity at the center.
Mesh
Coarser
Coarse
Normal
Velocity (m/s)
0.1222
0.1226
0.1231
The figures below show the graphic temperature and velocity results computed by the normal mesh.
The following figures show the results of the solidification model. The time dependent solver evaluates
the model every 0.1 second for 10 seconds. The figures show the system at t=0, 5, and 10 seconds
respectively. The results of this study were verified using various mesh refinements.