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Finite Element Method 

Chapter 8

Development of the Linear-Strain

Triangle Equations

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Stiffness Matrix of the Constant-Strain Triangular Element

Step 1: Discretize and Select Element Type

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Step 2: Select Displacement Functions

21211

210987

2

65

2

4321

),(

),(

 ya y xa xa ya xaa y xv

 ya y xa xa ya xaa y xu

T vuvuvuvuvuvud  665544332211}{  

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Step 2: Select Displacement Functions

12

2

1

22

22

1000000

0000001

}{

a

a

a

 y xy x y x

 y xy x y x

v

u

 

}{][}{   * a M  

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In Matrix Form

Solving for the a’ s

12

7

6

1

2666

2666

2111

2111

2666

2666

2111

2111

6

1

6

1

1000000

1000000

0000001

0000001

a

a

a

a

 y y x x y x

 y y x x y x

 y y x x y x

 y y x x y x

v

v

u

u

 

6

1

6

1

1

2666

2666

2111

2111

2

666

2

666

2111

2111

12

7

6

1

1000000

10000000000001

0000001

v

v

u

u

 y y x x y x

 y y x x y x y y x x y x

 y y x x y x

a

a

a

a

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}{][}{  1

d  X a 

* 1{ } [ ][ ] { }

[ ]{ }

 X d 

 N d 

   

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 0( , )

u

v

 N N N N N N u x y

 N N N N N N v x yu

v

 

 

   

6

1

6

1

{ }

i i

i

i i

i

 N u

 N v

  

 

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 x

v

 y

u

 y

v

 xu

 y x

 y

 x

 

 

 

 }{

Step 3: Define the Strain/Displacement and Stress/Strain Relationships

1

2

12

0 1 0 2 0 0 0 0 0 0 0

{ } 0 0 0 0 0 0 0 0 1 0 2

0 0 1 0 2 0 1 0 2 0

a x y

a x y

 x y x y a

 

 

 

12

2

1

22

22

1000000

0000001}{

a

a

a

 y xy x y x

 y xy x y x

v

u

 

Since

Then

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665544332211

654321

654321

000000

000000

2

1][

                  

      

            

 A B

  ' 1[ ]{ } B d 

 B M X 

 

where the b’s and  ’s are now functions of x and y as well as of the nodal coordinates 

1

{ } '

{ } [ ] { }

a

a X d 

 

The B matrix is illustrated for a specific linear-strain triangle in the next example 

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Stress Strain Relationship

 y x

 y

 x

 y x

 y

 x

 D

 

 

 

 

 

 

][

}{][][}{   d  B D 

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2

100

01

01

1][ 2

 

 

 

 

 E  D

2

2100

01

01

)21()1(][

 

  

  

  

 E  D

For Plane Strain Problems

For Plane Stress Problems

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Step 4 :Derive the Element Stiffness Matrix and Equations

),,,,,( mm j jii p p   vuvuvu    

 psb

 p

U 

U 

 

Total potential energy is defined as the sum of the internal

strain energy U and the potential energy of the external

forces Ω, that is:

For linear-elastic material, the internal strain energy is given by

V 

T  dV U    }{}{21   

V 

T dV  DU    }{][}{

2

1   

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The potential energy of the body forces:

V T 

b   dV  X }{}{ 

The potential energy of distributed loads or surface traction

S T s   dS T }{}{ 

}{}{   Pd   T  p  

The potential energy of concentrated loads

Step 4 :Derive the Element Stiffness Matrix and Equations

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Step 4 :Derive the Element Stiffness Matrix and Equations

V T 

V d  B D Bk    ][][][][

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The last three terms in equation represent the total load system or the

energy equivalent nodal forces on an element;

}{}{][}{][}{   PdS T  N dV  X  N  f 

S 

T 

V 

T 

Concentrated

nodal forces Body

forces Surface

Tractions 

}{}{}{][][][}{21  f d d V d  B D Bd 

  T 

V 

T T  p  

 

Step 4 :Derive the Element Stiffness Matrix and Equations

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V 

T V d  B D Bk    ][][][][

 A

T dydx B D Bt k    ][][][][

For an element with constant thickness t

Step 4 :Derive the Element Stiffness Matrix and Equations

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Step 5:  Assemble the Element Equations to Obtain the GlobalEquations and Introduce Boundary Conditions

 N 

e

ek K 

1

)( ][][

}{][}{   d K F   

 N 

e

e f F 

1

)( ][][

Step 6: Solve for the Nodal Displacements

Step 7: Solve for the Element Stresses

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Example: LST Stiffness Determination

Consider the following example.. The triangle is of base dimension b and

height h , with midside nodes.

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Example: LST Stiffness Determination

2 21 2 3 4 5 6( , )u x y a a x a y a x a x y a y

Using the first six equations we calculate the coefficients a 1 through a 6  by

evaluating the displacement u at each of the six known coordinates of each node

as follows:

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Example: LST Stiffness Determination

Solving the previous equations simultaneously for the a i  , w e obtain

Substituting into the following equation

2 2

1 2 3 4 5 6( , )u x y a a x a y a x a x y a y

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Example: LST Stiffness Determination

Similarly, solving for a 7  through a 1 2 bye valuating the displacement v at

each of the six nodes, we obtain

where the shape functions are obtained by collecting coefficients that

multiply each u i  term in previous equation.

For instance, collecting all terms that multiply by u1, we obtain N1.

We can express the general displacement expressions in terms of the shape

functions as:

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Example: LST Stiffness Determination

These shape functions are then given by:

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Example: LST Stiffness Determination

6

1

6

1

{ }

i i

i

i i

i

 N u

uv

 N v

   

   

 x

v

 y

u

 yv

 x

u

 y x

 y

 x

 

 

 

 }{

  [ ]{ } B d    

Since:

665544332211

654321

654321

000000000000

2

1][

                  

                  

 A B

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Example1

Performing the differentiations indicated on u and v, we obtain

2 2

1   2

1   2

1

2

1

3 3 2 4 21

3 4 4 4

2 3 4

 x y x x y y N 

b h b bh h

 N x y h

as an Exa

 x

 A bh

mple

h y x b b bh b  

  

   

1 2

3 4

5 6

1 2

3 1

5 6

4 43 4

0 4

84 4 4

43 4 0

44

84 4 4

hx hxh y hb b

 y

hx y h y

b

byb x

h

byb x

h

byb x xh

   

   

   

 

 

 

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Example: LST Stiffness Determination

These ’s and ’s are specific to the element in this example,

using calculus to set up the appropriate integration. The explicit expression forthe 12 x 12 stiffness matrix, being extremely cumbersome to obtain, is not given

here.

 A

T dydx B D Bt k    ][][][][

We can use numerical Integration to evaluate this integration as in Chapter 10

1 1 2 2 3 3 4 4 5 5 6 6

1 1 2 2 3 3 4 4 5 5 6 6

1 1 1 1 6 6 6 6

1

2

1

2

1

2

 x

 y

 x y

u u u u u u A

v v v v v v A

u v u v

 

 

 

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Comparison of ElementsFor a given number of nodes, a better representation of true stress and

displacement is Generally obtained using the LST element than is obtained

with the same number of nodes using a much finer subdivision into simpleCST elements.

For example, using one LST yields better results than using four CST

elements with the same number of nodes and hence the same number of

degrees of freedom

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Comparison of ElementsConsider the cantilever beam subjected to a parabolic load.

E=30x10 6  psi and  =0.25 

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Comparison of Elements

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Comparison of Elements

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Comparison of Elements

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Summary of equations using LST elements:

}{][}{   d k  f   

1) For each element, we find

1a) Element sti ffness matrix:

 A

T dydx B D Bt k    ][][][][

1 b) Element nodal force vector 

}{}{][}{][}{   PdS T  N dV  X  N  f 

S 

T 

V 

T 

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Summary of equations using CST elements:

2) Assemble

 N 

e

ek K 

1

)( ][][

 N 

e

e f F 

1

)( ][][

}{][}{   d K F    3) Solve for global nodal displacements

4) Find element strains and stresses

}{][}{   d  B 

}{][][}{   d  B D 

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HW:

8.3, 8.4 and 8.5