CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
In Chapters 2 through 5, we considered only line elements.
Line elements are connected only at common nodes, forming
framed or articulated structures such as trusses, frames, and
grids.
Line elements have geometric properties such as crosssectional area and moment of inertia associated with their
cross sections.
Plane Stress and Plane Strain Equations
However, only one local coordinate along the length of the
element is required to describe a position along the element
(hence, they are called line elements).
Nodal compatibility is then enforced during the formulation of
the nodal equilibrium equations for a line element.
This chapter considers the two-dimensional finite element.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Two-dimensional (planar) elements are thin-plate elements
such that two coordinates define a position on the element
surface.
The elements are connected at common nodes and/or along
common edges to form continuous structures.
Plane Stress and Plane Strain Equations
Nodal compatibility is then enforced during the formulation of
the nodal equilibrium equations for two-dimensional
elements.
If proper displacement functions are chosen, compatibility
along common edges is also obtained.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
The two-dimensional element is extremely important for:
(1) Plane stress analysis, which includes problems such
as plates with holes, fillets, or other changes in
geometry that are loaded in their plane resulting in local
stress concentrations.
Plane Stress Problems
Plane Stress and Plane Strain Equations
The two-dimensional element is extremely important for:
(1) Plane stress analysis, which includes problems such
as plates with holes, fillets, or other changes in
geometry that are loaded in their plane resulting in local
stress concentrations.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
The two-dimensional element is extremely important for:
(2) Plane strain analysis, which includes problems such
as a long underground box culvert subjected to a
uniform load acting constantly over its length or a long
cylindrical control rod subjected to a load that remains
constant over the rod length (or depth).
Plane Strain Problems
Plane Stress and Plane Strain Equations
The two-dimensional element is extremely important for:
(2) Plane strain analysis, which includes problems such
as a long underground box culvert subjected to a
uniform load acting constantly over its length or a long
cylindrical control rod subjected to a load that remains
constant over the rod length (or depth).
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
We begin this chapter with the development of the stiffness
matrix for a basic two-dimensional or plane finite element,
called the constant-strain triangular element.
The constant-strain triangle (CST) stiffness matrix derivation is
the simplest among the available two-dimensional elements.
We will derive the CST stiffness matrix by using the principle
of minimum potential energy because the energy formulation
is the most feasible for the development of the equations for
both two- and three-dimensional finite elements.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
We will now follow the steps described in Chapter 1 to formulate
the governing equations for a plane stress/plane strain
triangular element.
First, we will describe the concepts of plane stress and plane
strain.
Then we will provide a brief description of the steps and basic
equations pertaining to a plane triangular element.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Plane Stress
Plane stress is defined to be a state of stress in which the
normal stress and the shear stresses directed
perpendicular to the plane are assumed to be zero.
That is, the normal stress z and the shear stresses xz and yz
are assumed to be zero.
Generally, members that are thin (those with a small z
dimension compared to the in-plane x and y dimensions) and
whose loads act only in the x-y plane can be considered to be
under plane stress.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Plane Strain
Plane strain is defined to be a state of strain in which the
strain normal to the x-y plane z and the shear strains xz
and yz are assumed to be zero.
The assumptions of plane strain are realistic for long bodies
(say, in the z direction) with constant cross-sectional area
subjected to loads that act only in the x and/or y directions and
do not vary in the z direction.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
The concept of two-dimensional state of stress and strain and
the stress/strain relationships for plane stress and plane strain
are necessary to understand fully the development and
applicability of the stiffness matrix for the plane stress/plane
strain triangular element.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
A two-dimensional state of stress is shown in the figure below.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
The infinitesimal element with sides dx and dy has normal
stresses x and y acting in the x and y directions (here on the
vertical and horizontal faces), respectively.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
The shear stress xy acts on the x edge (vertical face) in the y
direction. The shear stress yx acts on the y edge (horizontal
face) in the x direction.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
Since xy equals yx, three independent stress exist:
 
T
   x
 y  xy 
Recall, the relationships for principal stresses in twodimensions are:
2
x y
x y 
2
1 
 
   xy   max
2
2


2 
x y
2
 y 
2
  x
   xy   min
 2 
2
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
Also, p is the principal angle which defines the normal whose
direction is perpendicular to the plane on which the maximum
or minimum principle stress acts.
tan 2 p 
2 xy
x y
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
The general two-dimensional state of strain at a point is show
below.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
x 
u
x
y 
v
x
 xy 
u v

y x
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
x 
u
x
y 
v
x
 xy 
u v

y x
The strain may be written in matrix form as:
 
T
   x
y
 xy 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
For plane stress, the stresses z, xz, and yz are assumed to
be zero. The stress-strain relationship is:
 x 
E
 
 y  
2
  1  
 xy 
1 
  x 
0

 
0
 1
  y 
0 0 0.5 1      xy 
 x 
 x 
 
 
 y   D    y 
 
 
 xy 
 xy 
1 

0
E 

[D ] 
 1
0
2 

1 
0 0 0.5 1    
is called the stress-strain matrix (or the constitutive matrix),
E is the modulus of elasticity, and  is Poisson’s ratio.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
For plane strain, the strains z, xz, and yz are assumed to be
zero. The stress-strain relationship is:
 x 
1  
E
 
 
 y  

  1   1  2   0

 xy 
 x 
 x 
 
 
 y   D    y 
 
 
 xy 
 xy 

1 
0
0   x 
 
0   y 

0.5     xy 
1  
E
 
[D ] 
1   1  2   0


1 
0
0 
0 

0.5   
is called the stress-strain matrix (or the constitutive matrix),
E is the modulus of elasticity, and  is Poisson’s ratio.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Two-Dimensional State of Stress and Strain
The partial differential equations for plane stress are:
 2u  2u 1     2u  2v 





2  y 2 xy 
x 2 y 2
 2v  2v 1     2v
 2u 





2  y 2 xy 
x 2 y 2
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Consider the problem of a thin plate subjected to a tensile load
as shown in the figure below:
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
Discretize the thin plate into a set of triangular elements. Each
element is define by nodes i, j, and m.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
We use triangular elements because boundaries of irregularly
shaped bodies can be closely approximated, and because the
expressions related to the triangular element are
comparatively simple.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
This discretization is called a coarse-mesh generation if few
large elements are used.
Each node has two degrees of freedom: displacements in the x
and y directions.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
We will let ui and vi represent the node i displacement
components in the x and y directions, respectively.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
The nodal displacements for an element with nodes i, j, and m
are:
 di 
d    d j 
d 
 m
where the nodes are ordered counterclockwise around the
element, and
u 
d i    i 
v i 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 1 - Discretize and Select Element Types
The nodal displacements for an element with nodes i, j, and m
are:
 ui 
v 
 i
u 
d    j 
v j 
um 
 
v m 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
u( x, y )
The general displacement function is:  i   

v ( x, y )
The functions u(x, y) and v(x, y) must be compatible with the
element type.
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
The general definitions of normal and shear strains are:
x 
u
x
y 
v
x
 xy 
u v

y x
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
For plane stress, the stresses z, xz, and yz are assumed to
be zero. The stress-strain relationship is:
 x 
E
 
 y  
2
  1  
 xy 
1 
  x 
0

 
0
 1
  y 
0 0 0.5 1      xy 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
For plane strain, the strains z, xz, and yz are assumed to be
zero. The stress-strain relationship is:
 x 
1  
E
 


 y  

  1   1  2   0

 xy 

1
0
  x 
 
 y 
0.5     xy 
0
0
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
Using the principle of minimum potential energy, we can derive
the element stiffness matrix.
f   [k ]d 
This approach is better than the direct methods used for onedimensional elements.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations and
Introduce Boundary Conditions
The final assembled or global equation written in matrix form is:
F   [K ]d 
where {F} is the equivalent global nodal loads obtained by
lumping distributed edge loads and element body forces at the
nodes and [K] is the global structure stiffness matrix.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 6 - Solve for the Nodal Displacements
Once the element equations are assembled and modified to
account for the boundary conditions, a set of simultaneous
algebraic equations that can be written in expanded matrix
form as:
F   [K ]d 
Step 7 - Solve for the Element Forces (Stresses)
For the structural stress-analysis problem, important secondary
quantities of strain and stress (or moment and shear force)
can be obtained in terms of the displacements determined in
Step 6.
Plane Stress and Plane Strain Equations
Derivation of the Constant-Strain Triangular Element
Stiffness Matrix and Equations
Step 1 - Discretize and Select Element Types
Consider the problem of a thin plate subjected to a tensile load
as shown in the figure below:
 ui 
v 
 i
u 
d    j 
v j 
um 
 
v m 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
We will select a linear displacement function for each triangular
element, defined as:
u
u( x, y )
Linear representation of u(x, y)
 i   

u
y
v ( x, y )
m
x
ui
uj
(xi, yi)
(xm, ym)
(xj, yj)
 a  a2 x  a3 y 
 1

a4  a5 x  a6 y 
A linear function ensures that the displacements along each
edge of the element and the nodes shared by adjacent
elements are equal.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
We will select a linear displacement function for each triangular
element, defined as:
 a1 
a 
 2
 a  a x  a3 y   1 x y 0 0 0  a3 
 i    1 2

 
a4  a5 x  a6 y  0 0 0 1 x y  a4 
a5 
 
a6 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
To obtain the values for the a’s substitute the coordinated of the
nodal points into the above equations:
ui  a1  a2 xi  a3 y i
v i  a4  a5 xi  a6 y i
u j  a1  a2 x j  a3 y j
v j  a4  a5 x j  a6 y j
um  a1  a2 xm  a3 y m
v m  a4  a5 xm  a6 y m
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
Solving for the a’s and writing the results in matrix forms gives:
 ui  1 xi
  
 u j   1 x j
u  1 x
m
 m 
y i   a1 
 
y j  a2 

y m  a3 

a   x  u
1
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The inverse of the [x] matrix is:
 i
1 
i
[x] 
2A 
  i
1
i
 j m 

 j m 
 j  m 
m
j
 i  x j y m  y j xm
i  y j  y m
 i  xm  x j
 j  xi y m  y i xm
 j  ym  yi
 j  xi  xm
 m  xi y j  y i x j
m  y i  y j
 m  x j  xi
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The inverse of the [x] matrix is:
 i
1 
i
[x] 
2A 
  i
1
 j m 

 j m 
 j  m 
1
2A  1
xi
xj
yi
yj
1
xm
ym
2 A  xi  y j  y m   x j  y m  y i   xm  y i  y j 
where A is the area of the triangle
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The values of a may be written matrix form as:
 i
 a1 
1 
 
a2  
 i
a  2 A  
 3
 i
 j  m   ui 
 
 j m   u j 
 j  m  um 
 i
a4 
1 
 
a5  
 i
2
A
a 
  i
 6
 j m   vi 
 
 j m   v j 
 j  m  v m 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
Expanding the above equations
u  1
x
 a1 
 
y  a2 
 
a3 
Substituting the values for a into the above equation gives:
 i  j  m   ui 
1
u  1 x y    i  j  m   u j 
2A
 
  i  j  m  um 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
We will now derive the u displacement function in terms of the
coordinates x and y.
 i ui   j u j   mum 


y    i u i   j u j   m um 
  i ui   j u j   mum 
Multiplying the matrices in the above equations gives:
1
u  1 x
2A
u ( x, y ) 

1
 i   i x   i y  u i    j   j x   j y  u j
2A
  m   m x   m y um 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
We will now derive the v displacement function in terms of the
coordinates x and y.
 i v i   j v j   mv m 


y    i v i   j v j   mv m 
  i v i   j v j   mv m 
Multiplying the matrices in the above equations gives:
1
v   1 x
2A
v ( x, y ) 

1
 i   i x   i y  v i    j   j x   j y  v j
2A
  m   m x   m y v m 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The displacements can be written in a more convenience form
as:
u ( x, y )  N i u i  N j u j  N m u m
v ( x, y )  N i v i  N j v j  N m v m
where:
1
 i   i x   i y 
2A
1
Nj 
 j   j x   j y 
2A
1
Nm 
 m   m x   m y 
2A
Ni 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The elemental displacements can be summarized as:
u( x, y ) Ni ui  N j u j  Nmum 


v ( x, y ) Ni v i  N j v j  Nmv m 
 i   
In another form the above equations are:
N i
{ }  
0
0
Ni
{ }  [N ]{d }
Nj
0
0
Nj
Nm
0
 ui 
v 
 i
0  uj 
 
Nm   v j 
um 
 
v m 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
In another form the equations are: { }  [N ]{d }
N
0
Ni
N    0i

Nj
0
0
Nj
Nm
0
0 
Nm 
The linear triangular shape functions are illustrated below:
Nj
Ni
Nm
1
1
y
y
y
1
x
m
i
x
m
i
x
j
j
m
i
j
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
So that u and v will yield a constant value for rigid-body
displacement, Ni + Nj + Nm = 1 for all x and y locations on the
element.
The linear triangular shape functions are illustrated below:
Nj
Ni
Nm
1
1
y
y
y
1
x
m
i
j
x
m
i
j
x
m
i
j
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
So that u and v will yield a constant value for rigid-body
displacement, Ni + Nj + Nm = 1 for all x and y locations on the
element.
For example, assume all the triangle displaces as a rigid body in
the x direction: u = u0
u0 
u0  u0  N i  N j  N m 
0
 
Ni 0 N j 0 Nm 0  u0 
 Ni  N j  Nm  1
{ }  
 

 0 Ni 0 N j 0 Nm   0 
u0 
 
0
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
So that u and v will yield a constant value for rigid-body
displacement, Ni + Nj + Nm = 1 for all x and y locations on the
element.
For example, assume all the triangle displaces as a rigid body in
the y direction: v = v0
0
v 0  v 0  Ni  N j  Nm 
v 
0
 
N i 0 N j 0 N m 0   0 
 Ni  N j  Nm  1
{ }  
 

 0 Ni 0 N j 0 Nm  v 0 
0
 
v 0 
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The requirement of completeness for the constant-strain triangle
element used in a two-dimensional plane stress element is
illustrated in figure below.
The element must be able to translate uniformly in either the x
or y direction in the plane and to rotate without straining as
shown
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 2 - Select Displacement Functions
The reason that the element must be able to translate as a rigid
body and to rotate stress-free is illustrated in the example of a
cantilever beam modeled with plane stress elements.
By simple statics, the beam elements
beyond the loading are stress free.
Hence these elements must be free
to translate and rotate without
stretching or changing shape.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
Elemental Strains: The strains over a two-dimensional
element are:
 u 


  x   x 
   v 
{ }    y   

   y 
 xy   u v 
 

 y x 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
Substituting our approximation for the displacement gives:
u

 u,x 
 Ni ui  N j u j  Nm um 
x
x
u,x  Ni ,x ui  N j ,x u j  Nm,x um
where the comma indicates differentiation with respect to that
variable.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
The derivatives of the interpolation functions are:
Ni ,x 
N j ,x 

1 
 i   i x   i y   i
2 A x
2A
j
2A
Nm, x 
m
2A
Therefore:
u
1

  i ui   j u j   m um 
x 2 A
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
In a similar manner, the remaining strain terms are
approximated as:
v
1

  i v i   j v j   mv m 
y 2 A
u v
1


  i u i   i v i   j u j   j v j   m u m   mv m 
y x 2 A
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
We can write the strains in matrix form as:
 u 


 i 0  j 0
  x   x 
1 
   v 
{ }    y   
 
 0 i 0  j
   y  2 A    
j
i
j
 i
 xy   u v 
  
 y x 
{ }  [B ]{d }
{ }  Bi
Bj
 ui 
v 
m 0   i 
 u 
0 m   j 
v
 m  m   j 
um 
 
v m 
 di 
 
Bm   d j 
 
d m 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 3 - Define the Strain-Displacement and
Stress-Strain Relationships
Stress-Strain Relationship: The in-plane stress-strain
relationship is:
 x 
 x 
 
 
{ }  [D ][B ]{d }
 y   [D ]   y 
 
 
 xy 
 xy 
For plane stress [D] is:
1 

0
E 

[D ] 
 1
0
2 

1
0 0 0.5 1    
For plane strain [D] is:
[D ] 
1  
E
 
1   1  2   0


1
0



0.5   
0
0
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
The total potential energy is defined as the sum of the internal
strain energy U and the potential energy of the external forces
:
 U   
p
b
p
s
Where the strain energy is: U 
1
1
{ }T { }dV   { }T [D]{ }dV

2V
2V
The potential energy of the body force term is:
 b    { }T { X }dV
V
where {} is the general displacement function and {X} is the
body weight per unit volume.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
The total potential energy is defined as the sum of the internal
strain energy U and the potential energy of the external forces
:
 U   
p
b
p
s
Where the strain energy is: U 
1
1
{ }T { }dV   { }T [D]{ }dV

2V
2V
The potential energy of the concentrated forces is:
 p   {d }T {P }
where {P} are the concentrated forces and {d} are the nodal
displacements.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
The total potential energy is defined as the sum of the internal
strain energy U and the potential energy of the external forces
:
 U   
p
b
p
s
Where the strain energy is: U 
1
1
{ }T { }dV   { }T [D]{ }dV

2V
2V
The potential energy of the distributed loads is:
s    { }T {T }dS
S
where {} is the general displacement function and {T} are
the surface tractions.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
Then the total potential energy expression becomes:
1
T
T
 p   d  [B ]T [D][B ]d  dV   d  [N ]T { X }dV
2V
V
 d  P   d  [N ]T {T }dS
T
T
S
The nodal displacements {d} are independent of the general xy coordinates, therefore
1
T
T
 p  d   [B ]T [D][B ]dV d   d   [N ]T { X }dV
2
V
V
 d  P  d 
T
T
 [N ]
T
S
{T }dS
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
We can define the last three terms as:
f    [N ]T { X }dV  P   [N ]T {T }dS
V
S
Therefore:
1
T
T
d   [B]T [D][B]dV d   d  f 
2
V
p 
Minimization of p with respect to each nodal displacement
requires that:
 p
 d 
  [B ]T [D][B ]dV d   f   0
V
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
The above relationship requires:
 [B ]
T
[D][B ]dV d   f 
V
The stiffness matrix can be defined as:
[k ]   [B ]T [D][B ]dV
V
For an element of constant thickness, t, the above integral
becomes:
[k ]  t  [B ]T [D][B ] dx dy
A
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
The integrand in the above equation is not a function of x or y
(global coordinates); therefore, the integration reduces to:
[k ]  t [B ]T [D][B ] dx dy
A
[k ]  tA [B ]T [D][B ]
where A is the area of the triangular element.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
Expanding the stiffness relationship gives:
 [k ii ] [k ij ] [kim ] 


[k ]   [k ji ] [k jj ] [k jm ] 
[k mi ] [kmj ] [kmm ]
where each [kii] is a 2 x 2 matrix define as:
[kii ]  [Bi ]T [D ][Bi ] tA
[kim ]  [Bi ]T [D][Bm ] tA
[kij ]  [Bi ]T [D][B j ] tA
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 4 - Derive the Element Stiffness Matrix and Equations
Recall:
 i
1 
Bi   2A  0
  i
0
 i 
 i 
 m
1 
Bm   2A  0
  m
 j
1 
B j  
0
2A 
  j
0

j
 j 
0
 m 
 m 
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations to Obtain the
Global Equations and Introduce the Boundary
Conditions
The global stiffness matrix can be found by the direct stiffness
method.
N
[K ]   [ k ( e ) ]
e 1
The global equivalent nodal load vector is obtained by lumping
body forces and distributed loads at the appropriate nodes as
well as including any concentrated loads.
N
{F }   {f ( e ) }
e 1
36/69
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations to Obtain the
Global Equations and Introduce the Boundary
Conditions
The resulting global equations are: F   K d 
where {d} is the total structural displacement vector.
In the above formulation of the element stiffness matrix, the
matrix has been derived for a general orientation in global
coordinates.
Therefore, no transformation form local to global coordinates is
necessary.
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations to Obtain the
Global Equations and Introduce the Boundary
Conditions
However, for completeness, we will now describe the method to
use if the local axes for the constant-strain triangular element
are not parallel to the global axes for the whole structure.
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations to Obtain the
Global Equations and Introduce the Boundary
Conditions
To relate the local to global displacements, force, and stiffness
matrices we will use:
f   Tf
d   Td
k  T T k T
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 5 - Assemble the Element Equations to Obtain the
Global Equations and Introduce the Boundary
Conditions
The transformation matrix T for the triangular element is:
 C S 0 0 0 0
 S C 0 0 0 0 


 0 0 C S 0 0
T 

 0 0 S C 0 0 
 0 0 0 0 C S


 0 0 0 0 S C 
C  cos 
S  sin
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Formulation of the Plane Triangular Element Equations
Step 6 - Solve for the Nodal Displacements
Step 7 - Solve for Element Forces and Stress
Having solved for the nodal displacements, we can obtain
strains and stresses in x and y directions in the elements by
using:
{ }  [B ]{d }
{ }  [D ][B ]{d }
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Consider the structure shown in the figure below.
Let E = 30 x 106 psi,  = 0.25, and t = 1 in.
Assume the element nodal displacements have been
determined to be u1 = 0.0, v1 = 0.0025 in, u2 = 0.0012 in,
v2 = 0.0, u3 = 0.0, and v3 = 0.0025 in.
39/69
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Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
First, we calculate the element  ’s and  ’s as:
 i  y j  y m  0  1  1
 i  xm  x j  0  2  2
 j  y m  y i  0  ( 1)  2
 j  xi  xm  0  0  0
 m  y i  y j  1  0  1
 m  x j  xi  2  0  2
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Therefore, the [B] matrix is:
 i
1 
B   2 A  0
  i
0
j
0
m
i
i
0
j
j
0
j
m
 1 0 2 0 1 0 
0
  1  0 2 0 0 0 2 
m 

2(2) 
 m 
 2 1 0 2 2 1
 i  y j  y m  0  1  1
 i  xm  x j  0  2  2
 j  y m  y i  0  ( 1)  2
 j  xi  xm  0  0  0
 m  y i  y j  1  0  1
 m  x j  xi  2  0  2
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CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
For plane stress conditions, the [D] matrix is:
0.25
0 
 1
30  106 
[D ] 
0.25
1
0 
2 

1  (0.25)
 0
0
0.375 
Substitute the above expressions for [D] and [B] into the general
equations for the stiffness matrix:
[k ]  tA [B ]T [D ][B ]
Plane Stress and Plane Strain Equations
Plane Stress Example 1
[k ]  tA [B ]T [D][B ]
 1 0 2
 0 2 1

 1
0.25
0 
 1 0 2 0 1 0 
1 
(2)30  106  2 0
 1  0 2 0 0 0 2 
k
0.25
1
0


 2(2) 

4(0.9375)  2 0 2  
 0
 2 1 0 2 2 1
0
0.375 


 1 0 2 


 0 2 1
41/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Performing the matrix triple product gives:
1.25
2 1.5
0.5
0.25 
 2.5
1.25 4.375 1 0.75 0.25 3.625 




2

1
4
0

2
1
 lb
k  4  106 

1.5
1.5
0.75  in
 1.5 0.75 0
 0.5 0.25 2 1.5
2.5
1.25 


0.25 3.625 1 0.75 1.25 4.375 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The in-plane stress can be related to displacements by:
{ }  [D ][B ]{d }
0.0

0.0025 in 

 x 
0.25
0 
 1
 1 0 2 0 1 0  
6
0.0012 in 
1 
  30  10 


0.25
1
0
0 2 0 0 0 2 
 y  


 2(2) 
 0.0
  0.9375  0


0
0.375 

 2 1 0 2 2 1 0.0
 xy 



0.0025 in 
42/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The stresses are:
 x   19,200 psi 
  

 y    4,800 psi 
   15,000 psi 

 xy  
Recall, the relationships for principal stresses and principal
angle in two-dimensions are:
1 
2 
x y
2
x y
2
 y 
2
  x
   xy   max
2


2
p 
 2 xy 
1
tan1 

2
  x   y 
 y 
2
  x
   xy   min
2


2
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Therefore:
2
1 
19,200  4,800
2
 19,200  4,800 
 
   15,000   28,639 psi
2
2


2 
19,200  4,800
2
 19,200  4,800 
 
   15,000   4,639 psi
2
2


p 
1
 2( 15,000) 
o
tan1 
  32.3
2
19,200
4,800



2
43/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The general force vector is defined as:
f    [N ]T { X }dV  P   [N ]T {T }dS
V
S
Let’s consider the first term of the above equation.
fb    [N ]T { X }dV
V
 Xb 

 Yb 
X  
where Xb and Yb are the weight densities in the x and y
directions, respectively.
The force may reflect the effects of gravity, angular velocities, or
dynamic inertial forces.
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
For a given thickness, t, the body force term becomes:
fb    [N ]T { X }dV  t  [N ]T { X }dA
V
A
 Ni
 0

N
[N ]T   j
 0
Nm

 0
0 
Ni 

0 

Nj 
0 

Nm 
 Xb 

 Yb 
X  
44/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The integration of the {fb} is simplified if the origin of the
coordinate system is chosen at the centroid of the element, as
shown in the figure below.
With the origin placed at the centroid, we can use the definition
of a centroid.
 y dA  0
A
 x dA  0
A
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
Recall the interpolation functions for a plane stress/strain
triangle:
1
1
Nj 
Ni 
 i   i x   i y 
 j   j x   j y 
2A
2A
Nm 
1
 m   m x   m y 
2A
With the origin placed at the centroid, we can use the definition
of a centroid.
 y dA  0
A
 x dA  0
A
45/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
A
Therefore the terms in the integrand are:
bh
2
  y dA  0
  x dA  0
i
i
A
A
b 2h
h
bh
 i  x j y m  y j xm           0  
3
 2  3   3 
h
b 2h
bh
 j  xm y i  y m xi   0           
 3   2  3  3
 m  xi y j  y i x j              
 2  3   2  3  3
b
i   j  m 
h
b
h
bh
1
1
tA
 i dA  t  dA 
3
2A
3
A
A
2A
3
t
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
Therefore the terms in the integrand are:
fb    [N ]T { X }dV  t  [N ]T { X }dA
V
A
The body force at node i is given as:
fbi  
tA  X b 
 
3  Yb 
 fbix 
 Xb 
f 
Y 
biy 

 b
The general body force vector is:
 fbjx  tA  X b 
fb    f    Y 
 bjy  3  b 
fbmx 
 Xb 


 
 Yb 
fbmy 
46/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The third term in the general force vector is defined as:
fs    [N ]T {T }dS
S
Let’s consider the example of a uniform stress p acting between
nodes 1 and 3 on the edge of element 1 as shown in figure
below.
p 
p

0 
T    px   

y

Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The third term in the general force vector is defined as:
fs    [N ]T {T }dS
S
 N1
0

N
[N ]T   2
0
 N3

0
0
N1 

0

N2 
0

N3 
evaluated at x = a
p 
p

0 
T    px   

y

47/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
Therefore, the traction force vector is:
fs    [N ]T {T }dS
S
 N1
0

t L
N
fs      2
0
0 0
 N3

0
0
 N1p 
 0 

N1



L
N2 p 
0  p
 dy
   dy dz  t  
0 
N2   0 
0
 N3 p 
0



N3  x = a
 0 x=a
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The interpolation function for i = 1 is:
Ni 
1
 i   i x   i y 
2A
For convenience, let’s choose the coordinate system shown in
the figure below.
 i  x j y m  y j xm
with i = 1, j = 2, and m = 3
1  x 2 y 3  y 2 x3
  0  0    0  a   0
48/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The interpolation function for i = 1 is:
Ni 
1
 i   i x   i y 
2A
For convenience, let’s choose the coordinate system shown in
the figure below.
Similarly, we can find:
1  0
N1 
1  a
ay
2A
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The remaining interpolation function, N2 and N2 are:
N1 
ay
2A
N2 
L(a  x )
2A
N3 
Lx  ay
2A
Evaluating Ni along the 1-3 edge of the element (x = a) gives:
N1 
ay
2A
N2  0
N3 
a L  y 
2A
49/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
Substituting the interpolation function in the traction force vector
expression gives:
 y 
 N1p 
 0 
 0 




L
L


0
atp


N
p
fs    [N ]T {T }dS  t   02  dy  2 A   0  dy
0
0
S


L  y 
 N3 p 




 0 
 0 
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
Therefore, the traction force vector is:
 fs1x 
L2 
 1
f 
 
0 
 s1y 
 0
 


f


0 
0
pLt
atp
fs    [N ]T {T }dS  fs 2 x     
 
4 A  0
2 0 
S
 s2y 
L2 
fs 3 x 
 1
 
 
 
0 
 0
fs 3 y 
A
aL
2
50/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
The figure below shows the results of the surface load
equivalent nodal for both elements 1 and 2:
From Element 1
From Element 2
Plane Stress and Plane Strain Equations
Treatment of Body and Surface Forces
For the CS triangle, a distributed load on the element edge can
be treated as concentrated loads acting at the nodes
associated with the loaded edge.
However, for higher-order elements, like the linear strain triangle
(discussed in Chapter 8), load replacement should be made
by using the principle of minimum potential energy.
For higher-order elements, load replacement by potential
energy is not equivalent to the apparent statically equivalent
one.
51/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Explicit Expression for the Constant-Strain Triangle
Stiffness Matrix
Usually the stiffness matrix is computed internally by computer
programs, but since we are not computers, we need to
explicitly evaluate the stiffness matrix.
For a constant-strain triangular element, considering the plane
strain case, recall that: [k ]  tA[B ]T [D ][B ]
where [D] for plane strain is:
1  
E
 
[D ] 
1   1  2   0



0 

0.5   
0
1
0
Plane Stress and Plane Strain Equations
Explicit Expression for the Constant-Strain Triangle
Stiffness Matrix
Substituting the appropriate definition into the above triple
product gives:
[k ]  tA[B ]T [D ][B ]
 i
0

j
tE
[k ] 

4 A(1   )(1  2 ) 0

 m

0
0
i
0
j
0
m
i 
 i 

1  
j 


1
 j  
 0
0
m  

m 
0
0
j
0
m
0
  i
0

0.5      i
i
i
0
j
j
0
j
m
0

m 
 m 
52/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Explicit Expression for the Constant-Strain Triangle
Stiffness Matrix
Substituting the appropriate definition into the above triple
product gives:
The stiffness matrix is a function of the global coordinates x and
y, the material properties, and the thickness and area of the
element.
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Consider the thin plate subjected to the surface traction shown
in the figure below.
Assume plane stress conditions. Let E = 30 x 106 psi,  = 0.30,
and t = 1 in.
Determine the nodal displacements and the element stresses.
53/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Discretization
Let’s discretize the plate into two elements as shown below:
20 in.
10 in.
This level of discretization will probably not yield practical results
for displacement and stresses; however, it is useful example
for a longhand solution.
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Discretization
For element 2, The tensile traction forces can be converted into
nodal forces as follows:
fs 2
fs 3 x 
 1 5,000 lb 
 1
f 

0 
0  
0
 s3y 

 
  
 fs1x  pLt 0  1,000 psi (1 in )10 in 0  
0


 






0
2 0 
2
 fs1y 

0  
fs 4 x 
 1 5,000 lb 
 1
 

 
  
0

0 
0  
fs 4 y 
54/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Discretization
For element 2, The tensile traction forces can be converted into
nodal forces as follows:
20 in.
5 kips
F   A  1,000 psi 10 in 1in 
10 in.
 10,000 lb
5 kips
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
The governing global matrix equations are: {F }  [K ]{d }
Expanding the above matrices gives:
 F1x   R1x 
 0 
 d1x 
F  
 0 



d
 1y   R1y 
 
 1y 
F2 x   R2 x 
 0 
d 2 x 
  
 



F2 y   R2 y 
 0 
d 2 y 
 
  [ K ]    [K ]  
F3 x  5,000 lb 
d 3 x 
d 3 x 
F3 y   0

d 3 y 
d 3 y 
  

 
 
F4 x  5,000 lb 
d 4 x 
d 4 x 
F4 y   0

 
 
 
d 4 y 
d 4 y 
55/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Assemblage of the Stiffness Matrix
The global stiffness matrix is assembled by superposition of the
individual element stiffness matrices.
The element stiffness matrix is: [k ]  tA[B ]T [D ][B ]
 i
1 
B   2 A  0
  i
0
j
0
m
i
i
0
j
j
0
j
1  
E
 
[D ] 
1   1  2   0

m

1
0
0

m 
 m 

0 

0.5   
0
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
For element 1: the coordinates are xi = 0, yi = 0, xj = 20, yj = 10,
xm = 0, and ym = 10. The area of the triangle is:
A
bh (20)(10)

 100 in.2
2
2
 i  y j  y m  10  10  0
 i  xm  x j  0  20  20
 j  y m  y1  10  0  10
 j  xi  xm  0  0  0
 m  y i  y j  0  10  10
 m  xi  x j  20  0  20
56/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Therefore, the [B] matrix is:
0
 0
1 
[B ] 
0
20
200 
 20 0
10
0
10
0
0
0
0
10
20
0 
20  1
 in
10 
 i  y j  y m  10  10  0
 i  xm  x j  0  20  20
 j  y m  y1  10  0  10
 j  xi  xm  0  0  0
 m  y i  y j  0  10  10
 m  xi  x j  20  0  20
Plane Stress and Plane Strain Equations
Plane Stress Example 1
For plane stress conditions, the [D] matrix is:
1 

0
6
E 
 30  10

[D ] 
1
0


0.91
1  2 
0 0 0.5 1    
0 
 1 0.3
0.3 1
0  psi


 0
0 0.35 
Substitute the above expressions for [D] and [B] into the general
equations for the stiffness matrix:
[k ]  tA [B ]T [D ][B ]
57/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
0
20 
 0
Therefore:
 0
20 0 

  1 0.3
0 
0
0 
30(106 )  10
T
[B ] [D ] 
0  lb 3

 0.3 1
 in
200(0.91)  0
0
10 
 0
0 0.35 
 10 0
20 


20 10 
 0
 0
 6

6
 10
30(10
)
[B ]T [D] 

200(0.91)  0
 10

 6
0
20
3
0
3
20
7 
0 

0  lb
3

3.5  in
7 

3.5 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
[k ]  tA [B ]T [D][B ]
 0
 6

(0.15)(106 )  10
1(100)

0.91
 0
 10

 6
0
20
3
0
3
20
7 
0 

0
 0
0 
1 


0
20

3.5  200 
 20 0
7 

3.5 
10
0
10
0
0
0
10
0
20
0 
20 

10 
58/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Simplifying the above expression gives:
u1
v1
u3
v3
u2
 140

0

0
75,000 
[k (1) ] 

0.91  70
 140

 70
0
400
60
0
60
400
0
60
100
0
100
70
0
0
35
70
35
140
60
100
70
240
130
60
v2
70 
400 

60  lb
 in
35 
130 

435 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Rearranging the rows and columns gives:
u1
v1
u2
v2
 140

0

75,000  140
[k (1) ] 

0.91  70

0

 70
0
400
60
400
60
0
140
60
240
130
100
70
70
400
130
435
60
35
u3
0
60
100
60
100
0
v3
70 
0 

70  lb
 in
35 
0

35 
59/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
For element 2: the coordinates are xi = 0, yi = 0, xj = 20, yj = 0,
xm = 20, and ym = 10. The area of the triangle is:
A
bh (20)(10)

 100 in.2
2
2
 i  y j  y m  0  10  10
 i  xm  x j  20  20  0
 j  y m  y1  10  0  10
 j  xi  xm  0  20  20
m  y i  y j  0  0  0
 m  xi  x j  20  0  20
Plane Stress and Plane Strain Equations
Plane Stress Problem 2
Therefore, the [B] matrix is:
10
0
 10 0
1 
[B ] 
0
0
0
20
200 
 0
10 20 10
0
0
20
0
20
0

1
 in

 i  y j  y m  0  10  10
 i  xm  x j  20  20  0
 j  y m  y1  10  0  10
 j  xi  xm  0  20  20
m  y i  y j  0  0  0
 m  xi  x j  20  0  20
60/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
For plane stress conditions, the [D] matrix is:
1 

0
6
E 
 30  10

[D ] 
1
0


0.91
1  2 
0 0 0.5 1    
0 
 1 0.3
0.3 1
0  psi


 0
0 0.35 
Substitute the above expressions for [D] and [B] into the general
equations for the stiffness matrix:
[k ]  tA [B ]T [D][B ]
Plane Stress and Plane Strain Equations
Plane Stress Example 1
0 
 10 0
 0
10 
0

  1 0.3
0 
6
10
0
20



30(10
)
T
[B ] [D ] 
0 

 0.3 1

20 10  
200(0.91)  0
 0
0 0.35 

 0
0
20 


20
0 
 0
Therefore:
 10
 0

6
 10
30(10
)
[B ]T [D] 

200(0.91)  0
 6

 6
3
0
3
20
0
20
0 
3.5 

7 

3.5 
7 

0 
61/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
[k ]  tA [B ]T [D][B ]
 10
 0

(0.15)(106 )  10
1(100)

0.91
 0
 6

 6
3
0
3
20
0
20
0 
3.5 

10
0
 10 0
7 
1 


0
0
0
20

3.5  200 
10 20 10
 0
7 

0 
0
0
20
0 
20 

0 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Simplifying the above expression gives:
u1
 100
 0

75,000  100
[k (2) ] 

0.91  60
 0

 60
v1
0
35
70
35
70
0
u4
v4
u3
v3
100 60
60 
0
35
70
70
0 

240 130 140 60 

130 435
400 
70
140 70
140
0 

400
60
0
400 
62/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Rearranging the rows and columns gives:
u1
 100
 0

 0
75,000
[k (2) ] 

0.91  60
 100

 60
v1
u3
0
35
70
0
70
35
0
70
140
0
140
70
v3
u4
v4
60 100 60 
35 
0
70

140 70 
0

400 
400
60
60
240 130 

400 130 435 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
In expanded form, element 1 is:
u1
v1
u2
v2
u3
v3
0
28 14
0
14
 28
 0
80
12 80 12
0

 28 12 48 26 20 14

12
7
375,000  14 80 26 87
[k (1) ] 
0.91  0
20
0
12 20 12

14
0
7
7
 14 0
 0
0
0
0
0
0

0
0
0
0
0
 0
u4
v4
0
0 
0
0 
0
0 
0
0
0
0
0
0
0
0
0
0









63/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
In expanded form, element 2 is:
u1
v1
0
 20
 0
7

0
 0
 0
0
375,000 
[k (2) ] 
14
0.91  0

 12 0
 20 14

 12 7
u2
v2
u3
v3
u4
v4
0
0
0
12
20
12 
0
0
14
0
14
7 
0
0
0
0
0
0 
0
0
0
0
0
0
0
0
0
0
0
0
0
0



28 14 
28
0

0
80
12 80 
28 12
48 26 

14 80 26 87 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Using the superposition, the total global stiffness matrix is:
u1
v1
u2
v2
u3
v3
u4
v4
28 14
26 20 12 
0
0
 48
 0
87
12 80 26
0
14
7 


0
0 
 28 12 48 26 20 14


80 26 87
12
0
0
7
375,000  14

k
[ ]
0.91  0
48
0
26 20 12
28 14 


14
0
87
12 80 
7
 26 0
 20 14
0
0
48 26 
28 12


0
0
14 80 26 87 
 12 7
64/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Applying the boundary conditions: d1x  d1y  d 2 x  d 2 y  0
0
28 14
0
26 20 12   d01x 
 R1x 
 48
 
 R

 0
87
12 80 26
0
14
7   d01y 
1y




 R2 x 
0
0   d02 x 
 28 12 48 26 20 14



 
0
0  d02 y 
 R2 y  375,000  14 80 26 87 12 7
 



 d 3 x 
5,000
0.91
0

26

20
12
48
0

28
14
lb




 0

14
7
0
87
12 80  d 3 y 
 26 0


 
 20 14
0
0
28 12
48 26  d 4 xx 
 500 lb 


 0


0
0
14 80 26 87  
 12 7
d 4 yy 
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The governing equations are:
0
28 14  d3 x 
5,000 lb 
 48
 
 0

 0
87 12 80  d 3 y 

 375,000 



 
0.91  28 12
48 26  d 4 x 
5,000 lb 


 0

 14 80 26 87  d 4 y 
Solving the equations gives:
d 3 x 
d 
 3y 
6
   10
d
4
x
 
d 4 y 


609.6 


 4.2

 in
663.7 
 104.1
65/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The exact solution for the displacement at the free end of the
one-dimensional bar subjected to a tensile force is:

PL (10,000)20

 670  10 6 in
AE 10(30  106 )
The two-element FEM solution is:
d 3 x 
d 
 3y 
6
   10
d
 4x 
d 4 y 


609.6 


 4.2

 in
663.7


 104.1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The in-plane stress can be related to displacements by:
{ }  [D ][B ]{d }
1 
 i
0
E


{ } 
 1
0
0
2 A(1   2 ) 
0 0 0.5 1       i
0
j
0
m
i
i
0
j
j
0
j
m
 d ix 
d 
iy 
0 
d

 jx 
m   
d jy
 m   
d mx 


d my 
66/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Element 1: { }  [D][B ]{d }
 x 
 
 y  
 
 xy 
6

1
0.3


0.3
1
0
0
30(10 )(10 ) 
6
0.96(200)
 0
0  0

  20
0.35 
0
0
10
0
10
20
0
0
0
0
0
10
20
 0.0 
 0.0 

0 
609.6 

20 


4.2 

10 

0.0 


 0.0 
 x  1,005 psi 
  

 y    301 psi 
   2.4 psi 

 xy  
Plane Stress and Plane Strain Equations
Plane Stress Example 1
Element 2: { }  [D][B ]{d }
 x 
 
 y  
 
 xy 
6

1
0.3


0.3
1
0
0
30(10 )(10 ) 
6
0.96(200)
 x   995 psi 
  

 y    1.2 psi 
   2.4 psi 

 xy  
  10
0  0

  0
0.35 
0
0
10
0
0
0
0
20
0
10
20
10
20
 0.0 
 0.0 

0 
663.7 

20  
 104.1


0 

609.6 


 4.2 
67/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Plane Stress Example 1
The principal stresses and principal angle in element 1 are:
2
1005  301
 1005  301 
 
 (2.4)2  1005 psi
1 

2
2


2
2 
1005  301
 1005  301 
2
 
  (2.4)  301 psi
2
2


1

2(2.4)

o
 p  tan 1 
0
2
1005
301



Plane Stress and Plane Strain Equations
Plane Stress Example 1
The principal stresses and principal angle in element 2 are:
2
1 
995  1.2
 995  1.2 
2
 
  ( 2.4)  995 psi
2
2


2 
995  1.2
 995  1.2 
2
 
  ( 2.4)  1.1 psi
2
2


2
1
 2( 2.4) 
 p  tan 1 
 0o

2
 995  1.2 
68/69
CIVL 7/8117
Chapter 6 - Plane Stress/Plane Strain Stiffness Equations - Part 1
Plane Stress and Plane Strain Equations
Problems
13. Do problems 6.6a, 6.6c, 6.7, 6.9a-c, 6.11, and 6.13 on
pages 377 - 383 in your textbook “A First Course in the
Finite Element Method” by D. Logan.
14. Rework the plane stress problem given on page 356 in your
textbook “A First Course in the Finite Element Method” by
D. Logan using WinFElt to do analysis.
Start with the simple two element model. Continuously
refine your discretization by a factor of two each time until
your FEM solution is in agreement with the exact solution
for both displacements and stress.
How many elements did you need?
End of Chapter 6
69/69