Finite Element Methods Elastostatic Problems Finite Element Methods Two Dimensional Solid Instructor: Mohamed Abdou Mahran Kasem, Ph.D. Aerospace Engineering Department Cairo University.
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Finite Element Methods
Two Dimensional Solid
Instructor: Mohamed Abdou Mahran Kasem, Ph.D.
Aerospace Engineering Department
Cairo University
Plane stress
Plane stress is a state of stress in which the normal stress and the shear stresses directed perpendicular to the plane are assumed to be zero.
๐. ๐. ๐๐ง , ๐๐ฅ๐ง , ๐๐ฆ๐ง = 0
Plane strain
Plane strain is a state of strain in which the normal strain and the shear strains directed perpendicular to the plane are assumed to be zero.
๐. ๐. ๐๐ง, ๐พ๐ฅ๐ง , ๐พ๐ฆ๐ง = 0
Plane stress
As we mentioned before the governing equilibrium equation for elastic, static, linear analysis has the form
เถฑ๐ค ๐,๐ ๐๐๐ ๐ฮฉ = เถฑ๐ค๐๐๐ ๐ฮฉ + ๐ต. ๐. ๐๐๐ = ๐ถ๐๐๐๐๐๐๐ = ๐ถ๐๐๐๐๐ข๐,๐ + ๐ข๐,๐
2= ๐ถ๐๐๐๐๐ข ๐,๐
Plane stress
In this case the stress-strain relation is reduced to the form
๐11๐22๐12
=2๐ + ๐ ๐ 0
๐ 2๐ + ๐ 00 0 2๐
๐11๐22๐12
=๐ธ
1 โ ๐ฃ2
1 ๐ฃ 0๐ฃ 1 0
0 01 โ ๐ฃ
2
๐11๐22๐12
= ๐๐
ฮป and ฮผ are Lameยด constants. They are related to the well-known Youngโs Modulus (E) and Poissonโs ratio (ฯ ) by
the following relation
๐ =๐ธ ๐ฃ
1 + ๐ฃ 1 โ 2๐ฃ, ๐ =
๐ธ
2 1 + ๐ฃ
Plane stress
The strain-displacement relation takes the form
๐ฎ = ๐๐ โ ๐ = ๐๐ ๐ = ๐๐
Linear triangular element
Substitute by the BCโs
I
II
By solving the two-set of equations together, one can obtain the shape functions for linear
triangular element
Linear triangular element
By substitute in the weak form,
Where B is the strain displacement matrix and D in the material stiffness matrix
depends on the element either plane stress or strain.
เถฑ๐ค ๐,๐ ๐๐๐ ๐ฮฉ = เถฑ๐ค๐๐๐ ๐ฮฉ + ๐ต. ๐.
Linear triangular element
๐พ
=๐ธ๐ก
แป4๐ด(1 โ ๐2
๐ฝ๐2 โ
1
2(โ1 + ๐แป๐พ๐
2 1
2(1 + ๐แป๐ฝ๐๐พ๐ ๐ฝ๐๐ฝ๐ โ
1
2(โ1 + ๐แป๐พ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐ ๐ฝ๐๐ฝ๐ โ
1
2(โ1 + ๐แป๐พ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐
1
2(1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐
2 + ๐พ๐2 ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐ ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐
๐ฝ๐๐ฝ๐ โ1
2(โ1 + ๐แป๐พ๐๐พ๐ ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ ๐ฝ๐
2 โ1
2(โ1 + ๐แป๐พ๐
2 1
2(1 + ๐แป๐ฝ๐๐พ๐ ๐ฝ๐๐ฝ๐ โ
1
2(โ1 + ๐แป๐พ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐
โ1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐
1
2(1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐
2 + ๐พ๐2 ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐
๐ฝ๐๐ฝ๐ โ1
2(โ1 + ๐แป๐พ๐๐พ๐ ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ ๐ฝ๐๐ฝ๐ โ
1
2(โ1 + ๐แป๐พ๐๐พ๐ ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ ๐ฝ๐
2 โ1
2(โ1 + ๐แป๐พ๐
21
2(1 + ๐แป๐ฝ๐๐พ๐
โ1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐พ๐ + ๐๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐๐ฝ๐ + ๐พ๐๐พ๐
1
2(1 + ๐แป๐ฝ๐๐พ๐ โ
1
2(โ1 + ๐แป๐ฝ๐
2 + ๐พ๐2
Example
Evaluate the stiffness matrix for the element shown in Figure. The coordinates
are shown in units of inches. Assume plane stress conditions. Let ๐ธ = 30๐ฅ106psi,
๐ = 0.25, and thickness t = 1 in. Assume the element nodal displacements have been
determined to be ๐ข1 = 0, ๐ฃ1 = 0.0025 ๐๐, ๐ข2 = 0.0012 ๐๐, ๐ฃ2 = 0, ๐ข3 = 0, ๐ฃ3 = 0.0025 ๐๐
Determine the element stresses.
Example
For a thin plate subjected to the surface traction shown in Figure, determine the
nodal displacements and the element stresses.
The plate thickness t = 1 in., ๐ธ = 30๐ฅ106psi, and ๐ = 0.3.
Example
Comparing to analytical solution
- The analytical solution represents 1-D approximation, while the FE solution represents 2-
D approximation.
- We used a coarse mesh in the FE solution, which results in an inaccurate solution.
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