Nonlinear finite elements/Homework 8/Solutions

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Problem 1: Plate Bending[edit]

Given:

Consider the flat plate shown in Figure 1.

Figure 1. Flat plate

Transverse normals are defined as the straight lines perpendicular to the midsurface.

If we assume that the plate is thin, the Kirchhoff-Love hypothesis holds, i.e.,

  • Transverse normals remain straight.
  • Transverse normals remain normal to the midsurface.
  • Transverse normals are inextensible.


Solutions[edit]

Part 1[edit]

The displacement field that satisfies the Kirchhoff-Love hypothesis is given by

where are the displacements of the midsurface.

Write the expanded form of the above equations after making the following substitutions.


After making the substitutions we get

The expanded form is

Part 2[edit]

Assume that the plate undergoes infinitesimal strains and rotations. Then the strain-displacement relations are

Show that, in terms of the midsurface displacements, the strain-displacement relations can be written in index notation as

where and .


The strain-displacement relations in index notation are

where . If the indices and run from 1 to 2, we can write the above strain-displacement relations as

The displacements are given by

If the index runs from 1 to 2, the displacements can be written as

The partial derivatives of the displacements are (since are functions only of and )

Therefore,

Part 3[edit]

The internal virtual work in the plate is given by (using index notation)

Note that if a component of the strain tensor is zero, then the variation of that strain component is also zero.

Show that the internal virtual work can be written as

where

Write down the expanded form of equation (3).


The strains are given by

Therefore, the variations of the strains are

The internal virtual work is

Therefore,

The expanded form of the above equation is

or,

Part 4[edit]

The equilibrium equations for the plate can be written as

where

Write the above equations in index notation.


The equilibrium equations are (written using indices ):

Using the summation convention, we get

The first two equations can then be written as

and the third equation can be written as

Therefore the equilibrium equations in index notation are

Part 5[edit]

Show that in index-free Gibbs notation the equilibrium equations for the plate can be written as

where

and , , and .


Equation (4) in index notation is

The above equations have been shown to be equivalent to the first two equilibrium equations in the previous part.

Let us now look at equation (5). Let

Then

For the next term, let

Then,

Hence

Therefore, equation (5) can be written as

The above equation has been shown to be equivalent to the third equilibrium equations in the previous part.

Part 6[edit]

Derive the symmetric weak forms of the equilibrium equations starting with equations (4) and (5). Use the coordinate-free Gibbs notation in your derivation.

For equation 4, we multiply the equation by a vector weighting function and integrate over the midsurface . Then,

Using the identity

we get

From the symmetry to the stress tensor . Hence,

Therefore

where is the normal to the boundary of the midsurface.

Hence the symmetric weak form of the first equation is

For equation 5, we multiply the equation by a scalar weighting function and integrate over the midsurface . Then,

Using the identity,

we get

Therefore,

In order to have natural BCs directly in terms of moments, we have to perform a similar set of operations the last term above. Thus,

Hence,

Similarly,

Therefore, the symmetric weak form of the second equation is

Using the distributive law for vectors

we can also write the symmetric weak form for the second equation as:

Part 7[edit]

Write down the natural boundary conditions in Gibbs notation.

From the expressions for the weak form, the weighted natural boundary conditions are

Therefore, the natural boundary conditions are:

Part 8[edit]

What are the essential boundary conditions that can be applied to the plate?

The essential boundary conditions that can be applied are

Problem 2: Transient Heat Conduction[edit]

Given:

The model problem shown in Figure 6.

Figure 6. Container with burning high energy material.

A 4340 steel casing is used to contain a solid high energy (HE) material. The HE material reacts on the inner (star-shaped) surface and generates gases. Since the gases are confined, the temperature of the gases increases according to the ideal gas law ().

The inner radius of the steel casing is 5 cm. The outer radius of the steel casing is 6 cm. The inner and outer radii of the star shaped region are 4 cm and 2 cm, respectively.

The high energy material has properties: Density = 1200 kg/m, Young's modulus = 1 GPa, Poisson's ratio = 0.49, Coefficient of thermal expansion = 1.210 /K, Thermal Conductivity = 5 W/m-K, Specific Heat Capacity = 1000 J/kg-K.

The burn-surface does not change position. The temperature is the same at all points on the burn surface. The temperature at the burn surface increases exponentially with time according to the relation

where is the initial temperature (300 K), is the temperature at time , K, and K/s.


Solution[edit]

The properties of normalized 4340 steel from Matweb are

Density 7850 kg/m 0.284 lbm/in
Young's Modulus 205 GPa 29.710 psi
Poisson's Ratio 0.29 0.29
CTE 12.610 /K 710/F
Thermal Conductivity 44.5 W/m-K 309 BTU-in/hr-ft-F
Specific Heat Capacity 475 J/kg-K 0.114 BTU/lb-F

Part 1[edit]

Assume that the initial temperature of the steel and the HE material is 300 K. Also assume that there is no heat flux on the external surface of the steel casing. Plot the temperature contours and stresses in the two materials at s and s.


The temperature contours for the two materials are shown in Figure 7. The low thermal conductivity of the HE material leads to a slow rate of diffusion of heat into the material. Therefore, the steel casing sees no change in temperature. However, the casing does deform due to the deformation of the HE material.

Figure 7(a). Temperature contours for no flux BC at t = 1 sec. Units are degrees K.
Figure 7(b). Temperature contours for no flux BC at t = 2 sec. Units are degrees K.

The stress contours for the two materials are shown in Figures 8, 9, 10, and 11. The principal stress is tensile in the steel casing and reaches a values of around 200 MPa (close to its yield point). In the HE material, is compressive in some regions and tensile in others and reaches values well beyond the ulimate tensile strength of the material inside some elements.

Figure 8(a). First Principal Stress contours for no flux BC at t = 1 sec. Units are Pa.
Figure 8(b). First Principal Stress contours for no flux BC at t = 2 sec. Units are Pa.
Figure 9(a). Second Principal Stress contours for no flux BC at t = 1 sec. Units are Pa.
Figure 9(b). Second Principal Stress contours for no flux BC at t = 2 sec. Units are Pa.
Figure 10(a). Third Principal Stress contours for no flux BC at t = 1 sec. Units are Pa.
Figure 10(b). Third Principal Stress contours for no flux BC at t = 2 sec. Units are Pa.
Figure 11(a). Shear Stress contours for no flux BC at t = 1 sec. Units are Pa.
Figure 11(b). Shear Stress contours for no flux BC at t = 2 sec. Units are Pa.

The ANSYS input file used for this problem is given below. The preprocessing stage is shown in the PDF file NFE_HW8Prob2Ans.pdf.

Figure 12. Preprocessing ANSYS input.

The solution stage is shown below:

!
! Solve
!
/solu
antype, trans
timint, off, struc
cnvtol, heat
cnvtol, f
autots, on
solcontrol, on
!
! Initial condition
!
ic, all, temp, 300
!
! Time stepping
!
time, 2
deltim, 0.1
outres, all, 10
solve
finish

Part 2[edit]

Assume that the initial temperature of the steel and the HE material is 300 K. Also assume that the external surface of the steel casing is maintained at 300 K. Plot the temperature contours and stresses in the two materials at s and s.

The temperature distribution from the previous problem suggests that the heat flux at the external surface is zero if the outside is maintained at 300 K (because of the low thermal conductivity of the HE material.) We don't expect to see any significant difference in the temperature and stress distributions.

Problem 3: Explicit Finite Elements[edit]

Given:

Figure 13(a) shows a beam that is built-in at one and loaded at the other. Figure 13(b) shows a plate that is built-in at one end and loaded at the other. The load that is applied is transient and has the form shown in Figure 13(c).

Figure 13(a) and (b). Built-in cantilevered beam and plate.
Figure 13(c). Load curve.

Assume that the beam and the plate are made of 4340 steel. The beam has dimensions 12 in. 0.25 in. 0.25 in. The plate has dimensions 12 in. 0.25 in. 6 in.

Simulate the problem for 10 secs. using LS-DYNA or some other explicit FE code. Use beam elements for the beam and shell elements for the plate. Use an element size of 2 in. for the beam and 2 in. 2 in. for the shell.


Solution[edit]

Part 1[edit]

Plot the displacement of the end of the beam as a function of time.

Figure 14 shows the displacement of the end of the beam as a function of time. Some numerical damping appears to occur if Hughes-Liu beam elements are used with 22 quadrature.

Figure 14. Displacement (in) of the end of the beam as a function of time (secs).

Part 2[edit]

Plot the axial stress at two points on the top of the beam as a function of time.

Figure 15 shows the axial stress at a Gauss point near the top of the beam for elements 2 and 5.

Figure 15 (a). Axial stress (psi) at element 2 at the top of the beam as a function of time (secs).
Figure 15 (b). Axial stress (psi) at element 5 at the top of the beam as a function of time (secs).

The LS-Dyna keyword file for the beam is shown below.

*keyword
*title
HW8Prob3_1.k - cantilever beam
*control_termination
10
*database_binary_d3plot
0.1
*database_history_node
1, 2, 3, 4, 5, 6, 7
*database_nodout
0.1
*database_history_beam
1, 2, 3, 4, 5, 6
*database_elout
0.1
*part
steel beam
1,1,1
*section_beam
1, 1, 0.833, 2.0, 0.0
0.25, 0.25, 0.25, 0.25, 0.0, 0.0
*mat_elastic
1, 0.284, 29.7e6, 0.29
*node
1, 0, 0, 0, 7, 7
2, 2.0, 0, 0, 0, 0
3, 4.0, 0, 0, 0, 0
4, 6.0, 0, 0, 0, 0
5, 8.0, 0, 0, 0, 0
6, 10.0, 0, 0, 0, 0
7, 12.0, 0, 0, 0, 0
8, 5.0, 1.0, 0, 0, 0
*element_beam
1, 1, 1, 2, 8, 0, 0, 0, 0, 2
2, 1, 2, 3, 8, 0, 0, 0, 0, 2
3, 1, 3, 4, 8, 0, 0, 0, 0, 2
4, 1, 4, 5, 8, 0, 0, 0, 0, 2
5, 1, 5, 6, 8, 0, 0, 0, 0, 2
6, 1, 6, 7, 8, 0, 0, 0, 0, 2
*load_node_point
7, 2, 1, 1.0
*define_curve
1
0.0, 0.0
0.1, 200.0
0.2, 200.0
0.25, 0.0
10.0, 0.0
*end

Part 3[edit]

Plot the displacement of the end of the plate as a function of time.

Figure 16 shows the displacement of the end of the plate as a function of time. Hughes-Liu shell elements with 3 integration points are used.

Figure 16. Displacement of the end of the plate as a function of time.

Part 4[edit]

Plot the axial stress at two points on the top of the plate as a function of time.

Figure 17 shows the axial stress at a Gauss point near the top of the plate for elements 8 and 11.

Figure 17(a). Axial stress at element 8 at the top of the plate as a function of time.
Figure 17(b). Axial stress at element 11 at the top of the plate as a function of time.

The LS-Dyna keyword file for the beam is shown below.

*keyword
*title
HW8Prob3_2 - bending of shell element
*control_termination
10
*database_binary_d3plot
0.1
*database_history_node_set
2
*database_nodout
0.1
*database_history_shell_set
1
*database_elout
0.1
*part
steel shell
1,1,1
*section_shell
1, 1, 0.833, 3.0, 0.0
0.25, 0.25, 0.25, 0.25, 0.0, 0.0
*mat_elastic
1, 0.284, 29.7e6, 0.29
*node
1,0,0,0
2,2,0,0
3,4,0,0
4,6,0,0
5,8,0,0
6,10,0,0
7,12,0,0
8,0,2,0
9,2,2,0
10,4,2,0
11,6,2,0
12,8,2,0
13,10,2,0
14,12,2,0
15,0,4,0
16,2,4,0
17,4,4,0
18,6,4,0
19,8,4,0
20,10,4,0
21,12,4,0
22,0,6,0
23,2,6,0
24,4,6,0
25,6,6,0
26,8,6,0
27,10,6,0
28,12,6,0
*set_node_list
1
1, 8, 15, 22
*set_node_list
2
7, 14, 21, 28
*element_shell
1,1,1,2,9,8
2,1,2,3,10,9
3,1,3,4,11,10
4,1,4,5,12,11
5,1,5,6,13,12
6,1,6,7,14,13
7,1,8,9,16,15
8,1,9,10,17,16
9,1,10,11,18,17
10,1,11,12,19,18
11,1,12,13,20,19
12,1,13,14,21,20
13,1,15,16,23,22
14,1,16,17,24,23
15,1,17,18,25,24
16,1,18,19,26,25
17,1,19,20,27,26
18,1,20,21,28,27
*set_shell_list
1
8, 11
*boundary_spc_set
1, 0, 1, 1, 1, 1, 1, 1
*load_node_set
2, 3, 1, 1.0
*define_curve
1
0.0, 0.0
0.1, 200.0
0.2, 200.0
0.25, 0.0
10.0, 0.0
*end