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.
𝑖. 𝑒. 𝜎𝑧 , 𝜏𝑥𝑧 , 𝜏𝑦𝑧 = 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.
𝑖. 𝑒. 𝜀𝑧, 𝛾𝑥𝑧 , 𝛾𝑦𝑧 = 0
Plane stress
As we mentioned before the governing equilibrium equation for elastic, static, linear analysis has the form
න𝑤 𝑖,𝑗 𝜎𝑖𝑗 𝑑Ω = න𝑤𝑖𝑓𝑖 𝑑Ω + 𝐵. 𝑇. 𝜎𝑖𝑗 = 𝐶𝑖𝑗𝑘𝑙𝜀𝑘𝑙 = 𝐶𝑖𝑗𝑘𝑙𝑢𝑘,𝑙 + 𝑢𝑙,𝑘
2= 𝐶𝑖𝑗𝑘𝑙𝑢 𝑖,𝑗
Plane stress
In this case the stress-strain relation is reduced to the form
𝜎11𝜎22𝜎12
=2𝜇 + 𝜆 𝜆 0
𝜆 2𝜇 + 𝜆 00 0 2𝜇
𝜀11𝜀22𝜀12
=𝐸
1 − 𝑣2
1 𝑣 0𝑣 1 0
0 01 − 𝑣
2
𝜀11𝜀22𝜀12
= 𝐃𝛆
λ and μ are Lame´ constants. They are related to the well-known Young’s Modulus (E) and Poisson’s ratio (υ) by
the following relation
𝜆 =𝐸 𝑣
1 + 𝑣 1 − 2𝑣, 𝜇 =
𝐸
2 1 + 𝑣
Linear triangular element
Substitute by the BC’s
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.
න𝑤 𝑖,𝑗 𝜎𝑖𝑗 𝑑Ω = න𝑤𝑖𝑓𝑖 𝑑Ω + 𝐵. 𝑇.
Linear triangular element
𝐾
=𝐸𝑡
ሻ4𝐴(1 − 𝜈2
𝛽𝑖2 −
1
2(−1 + 𝜈ሻ𝛾𝑖
2 1
2(1 + 𝜈ሻ𝛽𝑖𝛾𝑖 𝛽𝑖𝛽𝑗 −
1
2(−1 + 𝜈ሻ𝛾𝑖𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑗𝛾𝑖 + 𝜈𝛽𝑖𝛾𝑗 𝛽𝑖𝛽𝑚 −
1
2(−1 + 𝜈ሻ𝛾𝑖𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑚𝛾𝑖 + 𝜈𝛽𝑖𝛾𝑚
1
2(1 + 𝜈ሻ𝛽𝑖𝛾𝑖 −
1
2(−1 + 𝜈ሻ𝛽𝑖
2 + 𝛾𝑖2 𝜈𝛽𝑗𝛾𝑖 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛽𝑗 + 𝛾𝑖𝛾𝑗 𝜈𝛽𝑚𝛾𝑖 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛽𝑚 + 𝛾𝑖𝛾𝑚
𝛽𝑖𝛽𝑗 −1
2(−1 + 𝜈ሻ𝛾𝑖𝛾𝑗 𝜈𝛽𝑗𝛾𝑖 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛾𝑗 𝛽𝑗
2 −1
2(−1 + 𝜈ሻ𝛾𝑗
2 1
2(1 + 𝜈ሻ𝛽𝑗𝛾𝑗 𝛽𝑗𝛽𝑚 −
1
2(−1 + 𝜈ሻ𝛾𝑗𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑚𝛾𝑗 + 𝜈𝛽𝑗𝛾𝑚
−1
2(−1 + 𝜈ሻ𝛽𝑗𝛾𝑖 + 𝜈𝛽𝑖𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛽𝑗 + 𝛾𝑖𝛾𝑗
1
2(1 + 𝜈ሻ𝛽𝑗𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑗
2 + 𝛾𝑗2 𝜈𝛽𝑚𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑗𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑗𝛽𝑚 + 𝛾𝑗𝛾𝑚
𝛽𝑖𝛽𝑚 −1
2(−1 + 𝜈ሻ𝛾𝑖𝛾𝑚 𝜈𝛽𝑚𝛾𝑖 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛾𝑚 𝛽𝑗𝛽𝑚 −
1
2(−1 + 𝜈ሻ𝛾𝑗𝛾𝑚 𝜈𝛽𝑚𝛾𝑗 −
1
2(−1 + 𝜈ሻ𝛽𝑗𝛾𝑚 𝛽𝑚
2 −1
2(−1 + 𝜈ሻ𝛾𝑚
21
2(1 + 𝜈ሻ𝛽𝑚𝛾𝑚
−1
2(−1 + 𝜈ሻ𝛽𝑚𝛾𝑖 + 𝜈𝛽𝑖𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑖𝛽𝑚 + 𝛾𝑖𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑚𝛾𝑗 + 𝜈𝛽𝑗𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑗𝛽𝑚 + 𝛾𝑗𝛾𝑚
1
2(1 + 𝜈ሻ𝛽𝑚𝛾𝑚 −
1
2(−1 + 𝜈ሻ𝛽𝑚
2 + 𝛾𝑚2
Example
Evaluate the stiffness matrix for the element shown in Figure. The coordinates
are shown in units of inches. Assume plane stress conditions. Let 𝐸 = 30𝑥106psi,
𝜐 = 0.25, and thickness t = 1 in. Assume the element nodal displacements have been
determined to be 𝑢1 = 0, 𝑣1 = 0.0025 𝑖𝑛, 𝑢2 = 0.0012 𝑖𝑛, 𝑣2 = 0, 𝑢3 = 0, 𝑣3 = 0.0025 𝑖𝑛
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., 𝐸 = 30𝑥106psi, and 𝜈 = 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.