1. Plane Stress Analysis in Finite
Element Methods
BY
Venkata Hemanth Gubbala
13/ME/86

2. What is Finite Element Method?
• FEM is a mathematical technique used to predict the response of structures
and materials to environmental factors.
• Finite Element Analysis uses FEM, as a powerful engineering tool , to
numerically simulate the real world without the need to test prototype in a
lab.
• FEM cuts a structure into several elements (pieces of structure).
• Then reconnects elements at nodes as if nodes were pins that hold elements
together.
• This process results in a set of simultaneous algebraic equations.

3. Plane Stress
• Plane stress is defined to be a state of stress in which the normal stress, z , and
the shear stresses, xz and yz ,directed perpendicular to the x-y plane are
assumed to be zero.
• Assume that the negligible principal stress is oriented in the z-direction. To
reduce the 3D stress matrix to 2D plane stress matrix, remove all components
with z subscripts to get,
x xy
yx y

4. Element Characteristics
• Displacement Functions: Figure shows the typical triangular element
considered, with nodes i, j, m numbered in anticlockwise direction. The
displacements of a node have two components ai =
• And the six element displacements are listed as a vector ae=
• The displacements within an element have
to be uniquely defined by these six values.
u = 1+2x+3y
v= 4+5x+6y
ui
vi
ai
aj
am

5. • The six constants  can be evaluated easily by solving two sets of three simultaneous
equations which will arise if the nodal coordinates are inserted and the displacements
equated to appropriate nodal displacements.
ui = 1+2xi+3yi
uj = 1+2xj+3yj
um= 1+2xm+3ym
• We can easily solve for 1,2,3 in terms of nodal displacements ui, uj, um and obtain finally,
u = 1/2 [(ai + bix + ciy)ui+(aj + bjx + cjy)uj+(am + bmx + cmy)um]
2=det =2.(area of triangle ijm)
• Hence in standard form
u= = Nae = [INi , INj , INm ]ae
• I is a two by two identity matrix, and Ni = ai + bix + ciy/2
1 xi yi
1 xj yj
1 xm ym
u
v

6. If coordinates are taken from the centroid of the element then
xi + xj + xm = yi + yj + ym =0 and ai = 2/3 = aj = am
Substituting the equations we get
=Bae = [Bi , Bj , Bm]
With a typical matrix Bi given by
Bi=SNi= = 1/2
It will be noted that in this case the B matrix is independent of the position within the element , and
hence the strains are constant through it .
ai
aj
am
Ni/x 0
0 Ni/y
Ni/y Ni/x
bi , 0
0 , ci
ci , bi
STRAIN (TOTAL): The total strain at any point within the element can be defined by
its three components which contribute to internal work . Thus
 = = = Su
x
y
xy
/x 0
0 /y
/y /x
u
v

7. Elasticity Matrix :
The matrix D of the equation
= =D --0
Can be explicitly stated for any material .To consider the special cases in two dimensions it is
convenient to start from the form = D-1  + 0 and impose the conditions of plane stress or plane strain.
Plane Stress-Isotropic Material : For plane stress in an isotropic material we have by definition
x = x/E - vy/E + x0
y = -vx/E + y/E + y0
xy= 2(1+v)xy/E + xy0
Solving the above for stresses , we obtain the matrix D as D = E/1-v2
and the initial strain as 0= , in which E is the Elastic Modulus and v is Poisson’s ratio.
x
y
xy
x
y
xy
1 v 0
v 1 0
0 0 (1-v)/2
x0
x0
xy0

8. The Stiffness Matrix: The stiffness matrix of the element ijm is defined from the general
relationship with the coefficients
Ke
ij =  Bt
i DBj t dxdy
Where t is the thickness of the element and the integration is taken over the area of the
triangle . If the thickness of the element is assumed to be constant, an assumption
convergent to the as the size of elements decreases then neither of the matrices contains x
or y. Ke
ij = Bt
i DBj t
where  is the area of the triangle . This form is now sufficiently explicit for computation with the
actual matrix operations being left.
Nodal Forces due to Initial Strain: Nodal forces due to initial strain is directly given by,
(fi)e
0 = -Bt
i D0 t,
These ‘initial strain’ forces contribute to the nodes of an element in an unequal manner and require
precise evaluation. Similar expressions are derived for initial stress forces .

9. Evaluation of Stresses
• The derived formula enable the full stiffness matrix of the structure to be assembled, and a solution
for displacement to be obtained .
• The stresses are by basic assumption constant within the element .It is usual to assign these to the
centroid of the element
• An alternative consists of obtaining stress values at the nodes by averaging the values in the adjacent
elements.

10. Finite Element Analysis of a Flat
Rectangular Plate with a Hole

11. • Software used: Ansys - 12.0 for win -10
• Material properties:
 Young’s Modulus(EX) = 2.105 kg/cm2
 Poisson’s ratio(PRXY) = 0.3
• Geometry:

12. • Element: Plane82 , 2-D, 8 node
• Boundary Conditions: All the DOF has been constrained at the outer edge because
outer edge is fixed
• Load: There is a load of 20N/mm distributed on the right hand side of the plate

13. Loads Applied

14. Solution: The stress at the top of the hole in the plate was found to be
approximately 3.8 MPa.