Finite Element Method

Chapter 8

Development of the Linear-Strain

Triangle Equations

Stiffness Matrix of the Constant-Strain Triangular Element

Step 1: Discretize and Select Element Type

Step 2: Select Displacement Functions

21211

210987

265

24321

),(

),(

yayxaxayaxaayxv

yayxaxayaxaayxu

Tvuvuvuvuvuvud 665544332211}{

Step 2: Select Displacement Functions

12

2

1

22

22

1000000

0000001}{

a

a

a

yxyxyx

yxyxyx

v

u

}{][}{ * aM

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

yyxxyx

yyxxyx

yyxxyx

yyxxyx

v

v

u

u

6

1

6

1

1

2666

2666

2111

2111

2666

2666

2111

2111

12

7

6

1

1000000

1000000

0000001

0000001

v

v

u

u

yyxxyx

yyxxyx

yyxxyx

yyxxyx

a

a

a

a

}{][}{ 1 dXa

* 1{ } [ ][ ] { }

[ ]{ }

M 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 Nu x y

N N N N N Nv x y

u

v

6

1

6

1

{ }

i i

i

i i

i

N u

N v

x

v

y

u

y

v

x

u

yx

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

ax y

ax y

x y x ya

12

2

1

22

22

1000000

0000001}{

a

a

a

yxyxyx

yxyxyx

v

u

Since

Then

665544332211

654321

654321

000000

000000

2

1][

AB

' 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

{ } '

{ } [ ] { }

M a

a X d

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

Stress Strain Relationship

yx

y

x

yx

y

x

D

][

}{][][}{ dBD

2

100

01

01

1][

2

ED

2

2100

01

01

)21()1(][

ED

For Plane Strain Problems

For Plane Stress Problems

Step 4 :Derive the Element Stiffness Matrix and Equations

),,,,,( mmjjiipp 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 dVU }{}{21

V

T dVDU }{][}{21

The potential energy of the body forces:

V

Tb dVX}{}{

The potential energy of distributed loads or surface traction

S

Ts dST}{}{

}{}{ Pd Tp

The potential energy of concentrated loads

Step 4 :Derive the Element Stiffness Matrix and Equations

Step 4 :Derive the Element Stiffness Matrix and Equations

V

T VdBDBk ][][][][

The last three terms in equation represent the total load system or the energy equivalent nodal forces on an element;

}{}{][}{][}{ PdSTNdVXNf

S

T

V

T

Concentrated nodal forces

Body forces Surface Tractions

}{}{}{][][][}{21 fddVdBDBd T

V

TTp

Step 4 :Derive the Element Stiffness Matrix and Equations

V

T VdBDBk ][][][][

A

T dydxBDBtk ][][][][

For an element with constant thickness t

Step 4 :Derive the Element Stiffness Matrix and Equations

However, instead of constant stresses in each element, we now

have a linear variation of the stresses in each element.

Common practice was to use the centroidal element stresses.

Current practice is to use the average of the nodal element stresses.

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

N

e

ekK1

)( ][][

}{][}{ dKF

N

e

efF1

)( ][][

Step 6: Solve for the Nodal Displacements

Step 7: Solve for the Element Stresses

Example: LST Stiffness Determination

Consider the following example.. The triangle is of base dimension b and height h, with midside nodes.

Example: LST Stiffness Determination

2 2

1 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 a1 through a6 by evaluating the displacement u at each of the six known coordinates of each node as follows:

Example: LST Stiffness Determination

Solving the previous equations simultaneously for the ai , we 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

Example: LST Stiffness Determination

Similarly, solving for a7 through a12 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 ui 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:

Example: LST Stiffness Determination

These shape functions are then given by:

Example: LST Stiffness Determination

6

1

6

1

{ }

i i

i

i i

i

N uu

vN v

x

v

y

u

y

v

x

u

yx

y

x

}{

[ ]{ }B d

Since:

665544332211

654321

654321

000000

000000

2

1][

AB

Example1

Performing the differentiations indicated on u and v, we obtain

2 2

1 2 2

11 2

3 3 2 4 21

3 4 4 42 3 4

For E

x y x xy yN

b h b bh h

N x y hxA bh h y

x b b

xampl

bh

e

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 h

b b

y

hxy h y

b

byb x

h

byb x

h

byb x x

h

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 for the 12 x 12 stiffness matrix, being extremely cumbersome to obtain, is not given here.

A

T dydxBDBtk ][][][][

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

Comparison of Elements

For 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 simple CST 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

Comparison of Elements

Consider the cantilever beam subjected to a parabolic load.E=30x106 psi and =0.25

Comparison of Elements

Comparison of Elements

Comparison of Elements

In conclusion, The LST model might be preferred over the CST model for plane stress

applications when relatively small numbers of nodes are used.

However, the use of triangular elements of higher order, such as the LST, is not visibly advantageous when large numbers of nodes are used, particularly when the cost of formation of the element stiffnesses, equation bandwidth, and overall complexities involved in the computer modeling are considered.

Summary of equations using LST elements:

}{][}{ dkf

1) For each element, we find

1a) Element stiffness matrix:

A

T dydxBDBtk ][][][][

1 b) Element nodal force vector

}{}{][}{][}{ PdSTNdVXNf

S

T

V

T

Summary of equations using CST elements:

2) Assemble

N

e

ekK1

)( ][][

N

e

efF1

)( ][][

}{][}{ dKF

3) Solve for global nodal displacements

4) Find element strains and stresses

}{][}{ dB

}{][][}{ dBD

HW: 8.3, 8.4 and 8.5