LINEAR STRAIN TRIANGLE EQUATION

1. Chapter 8 – Linear-Strain Triangle Equations
Learning Objectives
• To develop the linear-strain triangular (LST)
element stiffness matrix.
• To describe how the LST stiffness matrix can be
determined.
• To compare the differences in results using the
CST and LST elements.
Development of the Linear-Strain Triangle Equations
Introduction
In this section we will develop a higher-order triangular element,
called the linear-strain triangle (LST).
This element has many advantages over the constant-strain
triangle (CST).
The LST element has six nodes and twelve displacement
degrees of freedom.
The displacement function for the triangle is quadratic.
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 1/31

2. The procedure to derive the LST element stiffness matrix and
element equations is identical to that used for the CST
element.
Step 1 - Discretize and Select Element Types
Consider the triangular element shown in the figure below:
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
Each node has two degrees of freedom: displacements in
the x and y directions.
We will let ui and vi represent the node i displacement
components in the x and y directions, respectively.
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 2/31

3. The nodal displacements for an LST element are:
 
1
2
3
4
5
6
d
d
d
d
d
d
d
 
 
 
 
  
 
 
 
 
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
The nodes are ordered
counterclockwise around the
element
1
1
2
2
3
3
4
4
5
5
6
6
u
v
u
v
u
v
u
v
u
v
u
v
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
Step 2 - Select Displacement Functions
Consider a straight-sided triangular element shown below:
1
3
2
y
x
6
4
5
The variation of the displacements over the element may be
expressed as:
2 2
1 2 3 4 5 6
( , )
u x y a a x a y a x a xy a y
     
2 2
7 8 9 10 11 12
( , )
v x y a a x a y a x a xy a y
     
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 3/31

4. The displacement compatibility among adjoining elements is
satisfied because the three nodes defining adjacent sides
define a unique a parabola.
The CST and LST triangles are variations of the Pascal
triangles as show below.
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
The general element displacement functions are:
 
1
2
2 2
2 2
11
12
ꞏ
1 0 0 0 0 0 0
ꞏ
0 0 0 0 0 0 1
a
a
x y x xy y
x y x xy y
a
a
 
 
 
 
 
   
 
   
 
 
 
    
*
M a
 
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 4/31

5. 2 2
1 1 1 1 1 1 1
2 2
2 2 2 2 2 2 2
2 2
6 6 6 6 6 6 6
2 2
1 1 1 1 1 1 1
2 2
5 5 5 5 5 5 5
2 2
6 6 6 6 6 6 6
1 0 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
u x y x x y y
u x y x x y y
u x y x x y y
v x y x x y y
v x y x x y y
v x y x x y y

 

 

 

 

 
  

  
  
 
 
 
 
  
            
            
1
2
6
7
11
12
a
a
a
a
a
a
  
  
  
  
  
 

 

 

 
 
 
 
 
 
 
   



To obtain the values for the a’s substitute the coordinates of the
nodal points into the above equations:
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Solving for the a’s and writing the results in matrix form gives
     
1
a x u


Development of the Linear-Strain Triangle Equations
where [x] is the 12 x 12 matrix on the right-hand-side of the
above equation.
2 2
1 1 1 1 1 1 1
2 2
2 2 2 2 2 2 2
2 2
6 6 6 6 6 6 6
2 2
7 1 1 1 1 1 1
2 2
11 5 5 5 5 5 5
2 2
12 6 6 6 6 6 6
1 0 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
a x y x x y y
a x y x x y y
a x y x x y y
a x y x x y y
a x y x x y y
a x y x x y y

 

 

 

 

 
  

 
 
 
 
 
 
  
            
            
1
1
2
6
1
5
6
u
u
u
v
v
v

  
  
  
  
  
 

 
 
 
 
 
 
 
 
 
 
 
   



CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 5/31

6. Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
The “best” way to invert [x] is to use a computer.
Note that only the 6 x 6 part of [x] really need be inverted.
    
N u
      
1
*
N M x


where
Development of the Linear-Strain Triangle Equations
    
*
M a
       
1
a x u

       
1
*
M x u

  
The general displacement expressions in terms of interpolation
functions and the nodal degrees of freedom are:
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
Step 3 - Define the Strain-Displacement and Stress-Strain
Relationships
Elemental Strains: The strains over a two-dimensional
element are:
{ }
x
y
xy

 

 
 
  
 
 
u
x
v
y
u v
y x
 

 

 

 
  

 
 
 

 
 
 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 6/31

7. Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
Step 3 - Define the Strain-Displacement and Stress-Strain
Relationships
Elemental Strains: The strains over a two-dimensional
element are:
 
1
2
11
12
0 1 0 2 0 0 0 0 0 0 0
ꞏ
0 0 0 0 0 0 0 1 0 0 2
ꞏ
0 0 1 0 2 0 1 0 2 0
a
a
x y
x y
x y x y
a
a

 
 
 
 
 
 
  
 
 
 
   
 
 
u v
Observe that the strains are linear over the triangular element;
therefore, the element is called a linear-strain triangle (LST).
The above equation may be written in matrix form as
    
'
M a
 
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
where [M ’] is based on derivatives of [M*].
 
0 1 0 2 0 0 0 0 0 0 0
' 0 0 0 0 0 0 0 1 0 0 2
0 0 1 0 2 0 1 0 2 0
x y
M x y
x y x y
 
 

 
 
 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 7/31

8. Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
If we substitute the values of a’s into the above equation gives:
    
B d

 
where [B] is a function of the nodal coordinates (x1, y1) through
(x6, y6).
    
1
'
B M x


    
'
M a
       
1
a x u


The stresses are given as:
[ ]
x x
y y
xy xy
D
 
 
 
   
   

   
   
   
For plane stress, [D] is:
 
2
1 0
[ ] 1 0
1
0 0 0.5 1
E
D




 
 
  

 

 
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
For plane strain, [D] is:
  
1 0
[ ] 1 0
1 1 2
0 0 0.5
E
D
 
 
 


 
 
 
 
 
 

 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 8/31

9. Step 4 - Derive the Element Stiffness Matrix and Equations
The stiffness matrix can be defined as:
[ ] [ ] [ ][ ]
T
V
k B D B dV
 
However, [B] is now a function of x and y; therefore, we must
integrate the above expression to develop the element
stiffness matrix.
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
The [B] matrix is:
 
1 2 3 4 5 6
1 2 3 4 5 6
1 1 2 2 3 3 4 4 5 5 6 6
0 0 0 0 0 0
1
0 0 0 0 0 0
2
B
A
     
     
           
 
 

 
 
 
The β’s and the γ’s are functions of x and y as well as the nodal
coordinates.
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 9/31

10. Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
The [B] matrix is:
The stiffness matrix is a 12 x 12 matrix and is very cumbersome
to compute in explicit form.
However, if the origin of the coordinates is the centroid of the
element, the integrations become more amenable.
Typically, the integrations are computed numerically.
 
1 2 3 4 5 6
1 2 3 4 5 6
1 1 2 2 3 3 4 4 5 5 6 6
0 0 0 0 0 0
1
0 0 0 0 0 0
2
B
A
     
     
           
 
 

 
 
 
The element body forces and surface forces should not be
automatically lumped at the nodes.
The following integration should be computed:
  [ ] { }
T
b
V
f N X dV
 
  [ ] { }
T
s
S
f N T dS
 
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 10/31

11. The element equations are:
     
1 11 12 13 1,12 1
1 21 22 23 2,12 1
2 31 32 33 3,12 2
6 12,1 12,2 12,3 12,12 6
12 1 12 12 12 1
x
y
x
y
f k k k k u
f k k k k v
f k k k k u
f k k k k v
     
     
     
   
 

   
 
   
 
   
 
   
 
 
 
  



      

Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
Steps 5, 6, and 7
Assembling the global stiffness matrix, determining the global
displacements, and calculating the stresses, are identical to
the procedures used for CST elements.
Derivation of the Linear-Strain Triangular Elemental
Stiffness Matrix and Equations
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 11/31

12. Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
Consider a straight-sided triangular element shown below:
The triangle has a base dimension of b and a height h, with mid-
side nodes.
We can calculate the coefficients a1 through a6 by evaluating
the displacement u at each node.
1 1
(0,0)
u u a
 
2
2 1 2 4
( ,0)
u u b a a b a b
   
2
3 1 3 6
(0, )
u u h a a h a h
   
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
2 2
1 2 3 4 5 6
( , )
u x y a a x a y a x a xy a y
     
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 12/31

13. 2 2
4 1 2 3 4 5 6
,
2 2 2 2 2 4 2
b h b h b bh h
u u a a a a a a
     
      
     
     
2
5 1 3 6
0,
2 2 2
h h h
u u a a a
   
   
   
   
2
6 1 2 4
,0
2 2 2
b b b
u u a a a
   
   
   
   
The triangle has a base dimension of b and a height h, with mid-
side nodes.
We can calculate the coefficients a1 through a6 by evaluating
the displacement u at each node.
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
Solving the above equations simultaneously for the a’s gives:
5 1 3
3
4 3
u u u
a
h
 

 
1 4 5 6
5
4 u u u u
a
bh
  

 
3 5 1
6 2
2 2
u u u
a
h
 

Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
1 1
a u
 6 1 2
2
4 3
u u u
a
b
 

 
2 6 1
4 2
2 2
u u u
a
b
 

CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 13/31

14. The u displacement equation is:
6 1 2 5 1 3
1
4 3 4 3
( , )
u u u u u u
u x y u x y
b h
   
   
  
   
   
   
2 6 1 1 4 5 6
2
2
2 2 4
u u u u u u u
x xy
b bh
   
    
 
   
   
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
 
3 5 1 2
2
2 2
u u u
y
h
 
 
  
 
   
2 6 1 1 4 5 6
2
2
2 2 4
v v v v v v v
x xy
b bh
   
    
 
   
   
6 1 2 5 1 3
1
4 3 4 3
( , )
v v v v v v
v x y v x y
b h
   
   
  
   
   
The v displacement equation can be determined in a manner
identical to that used for the u displacement:
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
 
3 5 1 2
2
2 2
v v v
y
h
 
 
  
 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 14/31

15. Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
The general form of the displacement expressions in terms of
the interpolation functions is given as 1
1
1 2 3 4 5 6
1 2 3 4 5 6
6
6
0 0 0 0 0 0
0 0 0 0 0 0
u
v
N N N N N N
u
N N N N N N
v
u
v
 
 
 
 
   

   
 
     
 
 
 

where the interpolation functions are:
2 2
1 2 2
3 3 2 4 2
1
x y x xy y
N
b h b bh h
     
2
3 2
2
y y
N
h h
  
2
2 2
2
x x
N
b b
  
4
4xy
N
bh

2
6 2
4 4 4
x xy x
N
b bh b
  
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
The general form of the displacement expressions in terms of
the interpolation functions is given as 1
1
1 2 3 4 5 6
1 2 3 4 5 6
6
6
0 0 0 0 0 0
0 0 0 0 0 0
u
v
N N N N N N
u
N N N N N N
v
u
u
 
 
 
 
   

   
 
     
 
 
 

where the interpolation functions are:
  
2
5 2
4 4 4
y xy y
N
h bh h
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 15/31

16. The element interpolation functions N have two basic shapes.
The behavior of the functions N1, N2, and N3 is similar except
referenced at different nodes.
The shape function N1 is shown below:
2
1
3
1
N1
y
x
6
4
5
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
The second type of interpolation function is valid for functions
N4, N5, and N6. The function N5 is shown below:
2
1
3
1
N5
y
x
6
4
5
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 16/31

17. The element strain is given as:
    
B d
 
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
where the [B] matrix is:
 
1 2 3 4 5 6
1 2 3 4 5 6
1 1 2 2 3 3 4 4 5 5 6 6
0 0 0 0 0 0
1
0 0 0 0 0 0
2
B
A
     
     
           
 
 

 
 
 

 
  

 
2 i
i
N
A
x


 
  

 
2 i
i
N
A
y

Therefore, since
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
2 2
1 2 2
3 3 2 4 2
1
x y x xy y
N
b h b bh h
     
2
2 2
2
x x
N
b b
  
2
3 2
2
y y
N
h h
  

 
  

 
2 i
i
N
A
x

1
4
3 4
hx
h y
b
    
2
4hx
h
b
   
3 0
 
2
bh
A 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 17/31

18. 2
6 2
4 4 4
x xy x
N
b bh b
  
  
2
5 2
4 4 4
y xy y
N
h bh h
Therefore, since
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations

 
  

 
2 i
i
N
A
x

4
4xy
N
bh
 4 4y
 
5 4y
  
6
8
4 4
hx
h y
b
   
2
bh
A 
1
4
3 4
by
b x
h
    
3
4by
b
h
   

 
  

 
2 i
i
N
A
y

Therefore, since
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
2 2
1 2 2
3 3 2 4 2
1
x y x xy y
N
b h b bh h
     
2
2 2
2
x x
N
b b
  
2
3 2
2
y y
N
h h
  
2 0
 
2
bh
A 
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 18/31

19. 4 4x
 
5
8
4 4
by
b x
h
   
2
6 2
4 4 4
x xy x
N
b bh b
  
  
2
5 2
4 4 4
y xy y
N
h bh h
Therefore, since
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
4
4xy
N
bh


 
  

 
2 i
i
N
A
y

6 4x
  
2
bh
A 
The stiffness matrix for a constant thickness element can be
obtained by substituting the β’s and the γ’s into the [B] and
then substituting [B] into the following expression and
evaluating the integral numerically.
[ ] [ ] [ ][ ]
T
V
k B D B dV
 
Example LST Stiffness Determination
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 19/31

20. For a given number of nodes, a better representation of true
stress and displacement is generally obtained using LST
elements than is obtained using the same number of nodes
a finer subdivision of CST elements.
For example, a single LST element gives better results than four
CST elements.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Consider the following cantilever beam with E = 30 x 106 psi,
 = 0.25, and t = 1 in.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 20/31

21. Table 1 lists the series of tests run to compare results using the
CST and LST elements.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 1. Comparison of CST and LST results
Series of Test
Runs
Number of
Nodes
Degrees of
Freedom, nd
Number of
Elements
A-1 4 x 16 85 160 128 CST
A-2 8 x 32 297 576 512 CST
B-1 2 x 8 85 160 32 LST
B-2 4 x 16 297 576 128 LST
Table 2 shows comparisons of free-end (tip) deflection and
stress for each element type used to model the cantilever
beam.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 2. Comparison of CST and LST results
Runs nd Bandwidth, nd
Tip Deflection
(in)
x
(ksi)
Location
(x, y)
A-1 160 14 -0.29555 67.236 (2.250,11.250)
A-2 576 22 -0.33850 81.302 (1.125,11.630)
B-1 160 18 -0.33470 58.885 (4.500,10.500)
B-2 576 22 -0.35159 69.956 (2.250,11.250)
Exact Solution -0.36133 80.000 (0,12)
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 21/31

22. The larger the number of degrees of freedom for a given type
of triangular element, the closer the solution converges to the
exact one (compare run A-1 to run A-2, and B-1 to B-2).
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 2. Comparison of CST and LST results
Runs nd Bandwidth, nd
Tip Deflection
(in)
x
(ksi)
Location
(x, y)
A-1 160 14 -0.29555 67.236 (2.250,11.250)
A-2 576 22 -0.33850 81.302 (1.125,11.630)
B-1 160 18 -0.33470 58.885 (4.500,10.500)
B-2 576 22 -0.35159 69.956 (2.250,11.250)
Exact Solution -0.36133 80.000 (0,12)
For a given number of nodes, the LST analysis yields
somewhat better results than the CST analysis (compare run
A-1 to run B-1).
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 2. Comparison of CST and LST results
Runs nd Bandwidth, nd
Tip Deflection
(in)
x
(ksi)
Location
(x, y)
A-1 160 14 -0.29555 67.236 (2.250,11.250)
A-2 576 22 -0.33850 81.302 (1.125,11.630)
B-1 160 18 -0.33470 58.885 (4.500,10.500)
B-2 576 22 -0.35159 69.956 (2.250,11.250)
Exact Solution -0.36133 80.000 (0,12)
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 22/31

23. Although the CST element is rather poor in modeling bending,
we observe that the element can be used to model a beam in
bending if sufficient number of elements is used.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 2. Comparison of CST and LST results
Runs nd Bandwidth, nd
Tip Deflection
(in)
x
(ksi)
Location
(x, y)
A-1 160 14 -0.29555 67.236 (2.250,11.250)
A-2 576 22 -0.33850 81.302 (1.125,11.630)
B-1 160 18 -0.33470 58.885 (4.500,10.500)
B-2 576 22 -0.35159 69.956 (2.250,11.250)
Exact Solution -0.36133 80.000 (0,12)
In general, both the LST and CST analyses yield sufficient
results for most plane stress/strain problems provided a
sufficient number of elements are used.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Table 2. Comparison of CST and LST results
Runs nd Bandwidth, nd
Tip Deflection
(in)
x
(ksi)
Location
(x, y)
A-1 160 14 -0.29555 67.236 (2.250,11.250)
A-2 576 22 -0.33850 81.302 (1.125,11.630)
B-1 160 18 -0.33470 58.885 (4.500,10.500)
B-2 576 22 -0.35159 69.956 (2.250,11.250)
Exact Solution -0.36133 80.000 (0,12)
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 23/31

24. That the LST model might be preferred over the CST model
for plane stress applications when a relatively small number
of nodes is used.
The use of triangular elements of higher order, such as the
LST, is not visibly more advantageous when large numbers
of nodes are used, particularly when the cost of the
formation of element stiffnesses, equation bandwidth, and
overall complexities involved in the computer modeling are
considered.
The results of Table 2 indicate:
Comparison of Elements
Development of the Linear-Strain Triangle Equations
Most commercial programs incorporate the use of CST and/or
LST elements for plane stress/strain problems although these
elements are used primarily as transition elements (usually
during mesh generation).
Also, recall that finite element displacements will always be less
than the exact ones, because finite element models are
always predicted to be stiffer than the actual structures when
using the displacement formulation of the finite element
method.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 24/31

25. A comparison of CST and LST models of a plate subjected to
parabolically distributed edge loads is shown below.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
The LST model converges to the exact solution for horizontal
displacement at point A faster than does the CST model.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 25/31

26. However, the CST model is quite acceptable even for modest
numbers of degrees of freedom.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
For example, a CST model with 100 nodes (200 degrees of
freedom) often yields nearly as accurate a solution as does
an LST model with the same number of degrees of freedom.
Comparison of Elements
Development of the Linear-Strain Triangle Equations
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 26/31

27. Comparing CST and LST triangles – 2 elements
20 in.
10 in.
Development of the Linear-Strain Triangle Equations
CST LST
Comparing CST and LST triangles – 2 elements
Development of the Linear-Strain Triangle Equations
CST LST
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 27/31

28. Comparing CST and LST triangles – 4 elements
Development of the Linear-Strain Triangle Equations
LST
CST
Comparing CST and LST triangles – 4 elements
Development of the Linear-Strain Triangle Equations
LST
CST
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 28/31

29. Plane Stress and Plane Strain Equations
Comparing CST and LST triangles – 8 elements
CST LST
Plane Stress and Plane Strain Equations
Comparing CST and LST triangles – 8 elements
CST LST
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 29/31

30. Plane Stress and Plane Strain Equations
Comparing CST and LST triangles – 16 elements
CST LST
Plane Stress and Plane Strain Equations
Comparing CST and LST triangles – 16 elements
CST LST
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 30/31

31. Problems
16. Work problems 8.3, 8.6, and 8.7 in your textbook.
17. Rework the plane stress problem given on page 364 and
6.13 in your textbook using Camp's LST code.
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?
Development of the Linear-Strain Triangle Equations
End of Chapter 8
CIVL 7/8117 Chapter 8 - Linear-Strain Triangle Equations 31/31