FEA Solution Procedure - West Virginia...

MAE 456 - Finite Element Analysis

FEA Solution Procedure

(demonstrated with a 1-D bar element problem)

Several slides from this set are adapted from B.S. Altan, Michigan Technological University


FEA Procedure for Static Analysis1. Prepare the FE model

a. discretize (mesh) the structureb. prescribe loadsc. prescribe supports

2. Perform calculationsa. generate stiffness matrix (k) for each elementb. combine elements (assemble global stiffness matrix K)c. assemble loads (into global load vector R)d. impose support conditionse. solve equations (KD=R) for displacements

3. View resultsa. displacements b. reaction forces at restraintsc. element strainsd. element stressese. element forces

2


The 1-D Bar ProblemGiven:

P

u1 u2 u3 u4 1

2 3 4

5

E1 = E5 = 200 GPa (1020 steel)

E2 = E3 = E4 = 70 GPa (1100 aluminum)

σy1 = σy5 = 295 MPa (1020 steel)

σy2 = σy3 = σy4 = 95 MPa (1100 aluminum)3


The 1-D Bar Problem

Find: a) displacements (u1, u2, u3, u4),b) strains (ε1, ε2, ε3, ε4, ε5),c) stresses (σ1, σ2, σ3, σ4, σ5),d) internal forces (Fi,1, Fi,2, Fi,3, Fi,4, Fi,5),e) safety factor (SF).

4

A1 = A4 = A5 = 100 mm2

A3 = 200 mm2 A2 = 200 mm2

L1 = L5 = 3 m L2 = L3 = L4 = 1 mP = 50,000 N

Node number


1. Prepare the FE Model• The structure is idealized in terms of nodes

and bar elements

P

1

2 3 4

5

P

u1 u2 u3 u4

21

3 45

5


2.a. Generate Elemental Equations• If we fix the left end of a bar (with constant cross

section) it’s end displacement is given by:

• If the left end is NOT fixed, the relationshipbetween force and displacement is given by:

221

121

FuuL

AE

FuuL

AE

F2

u1 u2

F1

6

AEPL

P


Generate Elemental Equations• These two equations can be conveniently expressed in

matrix form as:

• The different parts are known as:

– the elemental stiffness matrix

– the elemental displacement vector

– the elemental force vector

• This form allows us to easily combine the equations from all elements of a structure.

2

1

2

1

1111

FF

uu

LAE

1111

LAEk

2

1

uu

d

2

1

FF

f

7


Generate Elemental Equations

1,4

1,1

4

1

1

11

1111

FF

uu

LEA

1,4

1,1

4

16

67.667.667.667.6

10FF

uu

3,3

3,2

3

2

3

33

1111

FF

uu

LEA

Element numberNode number

2,2

2,1

2

1

2

22

1111

FF

uu

LEA

2,2

2,1

2

16

21212121

10FF

uu

3,3

3,2

3

26

14141414

10FF

uu

8


Generate Elemental Equations

5,4

5,1

4

1

5

55

1111

FF

uu

LEA

4,4

4,3

4

3

4

44

1111

FF

uu

LEA

5,4

5,1

4

16

67.667.667.667.6

10FF

uu

4,4

4,3

4

36

7777

10FF

uu

9


2.b. Combine Elements

Node 1: 1,1F2,1F5,1F

R

05,12,11,1 RFFF

Elements are combined by considering the equilibrium of forces at each node

10

Node 2: 2,2F 3,2F

03,22,2 FF


Combine ElementsNode 3: 3,3F 4,3F

04,33,3 FF

11

Node 4:1,4F4,4F5,4F

P

05,44,41,4 PFFF


Combine Elements

P

R

FFFFFFF

FFF

00

5,44,41,4

4,33,3

3,22,2

5,12,11,1

The equations for each node can then be combined into one vector equation.

12


Combine Elements

P

R

F

F

FFF

FFF

F

F

00

000

0

0

0

000

0

5,4

5,1

4,4

4,33,3

3,22,2

2,1

1,4

1,1

Grouping forces from each element into individual vectors gives:

13


Combine Elements

4

3

2

1

6

1,4

1,1

67.60067.60000000067.60067.6

1000

uuuu

F

FElement 1:

Element 2:

4

3

2

1

62,2

2,1

00000000002121002121

10

00

uuuu

FF

14


Combine Elements

4

3

2

1

6

3,3

3,2

00000141400141400000

10

0

0

uuuu

FF

Element 3:

Element 4:

4

3

2

1

6

4,4

4,3

77007700

00000000

1000

uuuu

FF

15


Combine Elements

4

3

2

1

6

5,4

5,1

67.60067.60000000067.60067.6

1000

uuuu

F

FElement 5:

16


Combine Elements

P

R

uuuu

00

67.6767.67067.667.67714140

01414212167.667.602167.62167.6

10

1

4

3

2

1

6

17

P

R

uuuu

00

67.60067.60000000067.60067.6

77007700

00000000

00000141400141400000

00000000002121002121

67.60067.60000000067.60067.6

10

1

4

3

2

1

6


Combine ElementsThe resulting overall matrix equation is:

P

R

uuuu

00

207013721140

01435211302134

10

1

4

3

2

1

6

18

“global stiffness matrix”

“global displacement vector”“global applied force vector”

KD=R


2.c. Assemble LoadsInserting the load value(s):

6

1

4

3

2

1

6

1005.000

207013721140

01435211302134

10

R

uuuu

19


2.d. Support Conditions• We have one support condition: u1= 0.• Support conditions can be solved two ways:

– Penalty method• Multiply equation ui = i by a large number and add to row i.• 1000x106 u1 = 0

P

u1 u2 u3 u4 1

2 3 4

5

Stiff spring

6

1

4

3

2

1

6

1005.000

207013721140

014352113021341000

10

R

uuuu

• This is equivalent to adding a stiff spring

20


Support Conditions– Matrix partitioning method

• Partition global equation into 2 separate matrix equations

• Solve for D2 first, then solve for R1:– Step 1:

– Step 2:

6

1

4

3

26

1005.000

0

207013721140

01435211302134

10

R

uuu

2

1

2

1

2221

1211

RR

DD

KKKK

1112121

1212222

DKDKRDKRDK

21


Support Conditions– Step 1: Solve for u2, u3, u4

05.000

207072114

01435

4

3

2

uuu

– Step 2: Solve for R1

4

3

26

1 1302110uuu

R

22


2.e. Solving for Displacements• The global matrices (with applied boundary

conditions) are solved for displacements (D):D=K-1R

• In practice, the computer does not actually calculate K-1 , but solves for D directly, using techniques such as Gaussian Elimination.

23


3.a. Solving for Strain• Once we know the displacements, we can

calculate the strain for each element.• In a one dimensional problem, the strain is

given by:

• For the constant cross-section bar, dxdu

Luu 12

24


3.b. Stress and Internal Force• Below the yield stress, for uniaxial stress,

the stress is given by:

• The force in each bar can be calculated by:

or by:

E

AF

LuuAEF )( 12

25


3.c. Safety Factor• For a statically loaded member, the Safety

Factor (SF) with respect to yielding failure can be computed as:

where sy is the material yield stress, andmax is the maximum stress.

• The SF for the entire structure is equal to the minimum of the SF of each member.

26


Recommended Safety Factors• SF = 1.25 to 1.5 for exceptionally reliable materials used under

controllable conditions, subject to loads that can be determined with certainty – where low weight is important.

• SF = 1.5 to 2 for well-known materials, under reasonably constant environmental conditions, subject to loads that can be determined readily.

• SF = 2 to 2.5 for average materials operated in ordinary environments and subject to loads that can be determined.

• SF = 2.5 to 3 for less tried materials or for brittle materials under average conditions of environment and load.

• SF = 3 to 4 for untried materials used under average conditions of environment and load.

• SF = 3 to 4 also for better-known materials that are to be used in uncertain environments or subject to uncertain loads.

• Adjust SF for repeated loads, impact forces, and brittle materials.27

Adapted from Juvinall & Marshek, Fundamentals of Component Design, 4th Ed.


What about Planar (2-D) Problems?• The equation for a bar element with an arbitrary

orientation in planar space is obtained by transforming the local element coordinate system to the global coordinate system.

28


What about Planar (2-D) Problems?• Mathematically this is done by multiplying

the elemental stiffness equation by a rotation matrix:

29


What about Spatial (3D) Problems?• Transformation matrices are again used to

transform the equations in the element coordinate system to the global coordinate system. (This time using the direction cosines.)

30

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