Computer Aided Engineering- FE Lecture 4 Slides

7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides

1/23

Remember from last lecture, single element (2 nodes)matrix can be written as follows

Element e

1e 2e

u1e

u2e

F1e F2

e

ke

2

e

2

k ue

1

u-k

e2

k u

e

e1uk

e

eF=+

e1

F=

1 2

1

2

2

e

2

ke ue

1

u-k

e2

ke u

e

e1uk

e

eF=+

e1

F=

1 2

1

2

ke - ke

ke- ke

The stiffness matrix from:


2/23

Lets write the individual matrix for each element

1

2

3

4

P applied load

u

1

u

2

u

3

u

4

R1 unknown

reaction force

k2

k3

ele. 1

ele. 2

ele. 3

k1

5

R1

K1(u2 u1)

k4 ele. 4

K1(u2 u1)

K2(u3 u2)K2(u3 u2)

K3(u4 u3)

K3(u4 u3)

K4(u5 u4)

K4(u5 u4)

P

Node 1

Node 2

Node 3

Node 4

Node 5

ee

ee

kk

kk

11

11Element 1

Element (e) Nodes

1 1, 2

2 2, 3

3 3, 4

4 4, 5

00000

00000

00000

00000

00000

1 2 3 4 5

1

23

4

5

This is a plainmatrix


3/23

ee

ee

kk

kk

11

11Element 1

00000

00000

00000

000

000

1 2 3 4 5

1

2

3

4

5

1

2

3

4

P applied load

u

1

u

2

u3

u4

R1 unknown

reaction force

k2

k3

ele. 1

ele. 2

ele. 3

k1

5

R1

K1(u2 u1)

k4 ele. 4

K1(u2 u1)

K2(u3 u2)K2(u3 u2)

K3(u4 u3)

K3(u4 u3)

K4(u5 u4)

K4(u5 u4)

P

Node 1

Node 2

Node 3

Node 4

Node 5

Element 1 is elated to nodes 1 & 2So we can place this matrix in the

plain global matrix at theintersection of nodes 1 & 2

00000

00000

00000

00000

00000

1 2 3 4 5

1

2

3

4

5

ee

ee

kk

kk

11

11


4/23

ee

ee

kk

kk

22

22Element 2

00000

00000

000

00

000

1 2 3 4 5

1

2

3

4

5

1

2

3

4

P applied load

u

1

u

2

u3

u4

R1 unknown

reaction force

k2

k3

ele. 1

ele. 2

ele. 3

k1

5

R1

K1(u2 u1)

k4 ele. 4

K1(u2 u1)

K2(u3 u2)K2(u3 u2)

K3(u4 u3)

K3(u4 u3)

K4(u5 u4)

K4(u5 u4)

P

Node 1

Node 2

Node 3

Node 4

Node 5



00000

00000

00000

00000

00000

1 2 3 4 5

1

2

3

4

5

ee

ee

kk

kk

22

22


5/23

ee

ee

kk

kk

33

33Element 3

00000

000

00

00

000

1 2 3 4 5

1

2

3

4

5

1

2

3

4

P applied load

u

1

u

2

u3

u4

R1 unknown

reaction force

k2

k3

ele. 1

ele. 2

ele. 3

k1

5

R1

K1(u2 u1)

k4 ele. 4

K1(u2 u1)

K2(u3 u2)K2(u3 u2)

K3(u4 u3)

K3(u4 u3)

K4(u5 u4)

K4(u5 u4)

P

Node 1

Node 2

Node 3

Node 4

Node 5



00000

00000

00000

00000

00000

1 2 3 4 5

1

2

3

4

5

ee

ee

kk

kk

33

33


6/23

ee

ee

kk

kk

44

44Element 4

000

00

00

00

000

1 2 3 4 5

1

2

3

4

5

1

2

3

4

P applied load

u

1

u

2

u3

u4

R1 unknown

reaction force

k2

k3

ele. 1

ele. 2

ele. 3

k1

5

R1

K1(u2 u1)

k4 ele. 4

K1(u2 u1)

K2(u3 u2)K2(u3 u2)

K3(u4 u3)

K3(u4 u3)

K4(u5 u4)

K4(u5 u4)

P

Node 1

Node 2

Node 3

Node 4

Node 5



00000

00000

00000

00000

00000

1 2 3 4 5

1

2

3

4

5

ee

ee

kk

kk

44

44


7/23

The final global matrix of the system is obtained simply byassembling or adding the individual element matrices

000

00

00

00

000

1 2 3 4 5

1

2

3

4

5ee

ee

kk

kk

44

44

ee

ee

kk

kk

33

33

ee

ee

kkkk

22

22

ee

ee

kk

kk

11

11

The overlappingK will be addedin the system

global matrix

The assembly of the above matrix gives the global matrix of the system

+

+

+

44

4433

3322

2211

11

000

00

00

00

000

kk

kkkk

kkkk

kkkk

kkThis method givesthe same matrixwhich we havedeveloped fromindividual equilibrium

equation for eachnode


8/23

To calculate the stiffness of any element

k = stiffness

E = Modules of elasticity

A = cross sectional area ( width x thickness )

l= length of the element

1

2

3

4

A

B

D

C

F

a

a

a

b c bwall


9/23

For the suspended plate which is fixed to the roof as shown in Fig. belowa. Model the plate in suitable FE form.b. Calculate the stiffness of each element, ke

c. Write the element nodes table.d. Set the global stiffness matrix of the model.e. Solve for the elongation in each element

E = Modulus of Elasticity = 100 GPa

All dimensions in mmt = thickness of the plate = 1mm

a = 20 mm

a a a a a

200

100

10 kN

Example 1


10/23

Model the plate in suitable FE form.

Example 1

1

2

3

Element 1 Element 2 Element 3

10kN

Element(e) Nodes

1 1, 2

2 1, 2

3

2, 34

1, 2

To calculate the stiffness

Element 4

k = stiffness

E = Modules of elasticity

A = cross sectional area ( width x thickness )

l= length of the element

Since elements 1, 2, & 3 have the same area. Module ofelasticity & length, hence their stiffness are equalK1 = K2 = K3

E = 100 GPa

t = thickness of the plate = 1mm

a = 20 mm

a a a a a

200

100

10 kN


11/23

Individual matrix of each element

ee

ee

kk

kk

11

11

ee

ee

kk

kk

22

22

ee

ee

kk

kk

33

33

Element 1 Element 2 Element 3 Element(e) Nodes

1 1, 2

2 1, 2

3

2, 34

1, 2

000

000

000

1 2 3

1

2

3

0

0

ee

ee

kk

kk

11

11

ee

ee

kk

kk

22

22

ee

ee

kk

kk

33

33

ee

ee

kk

kk

44

44

Element 4

ee

ee

kk

kk

44

44

+++

++

44

44321321

321321

0

0

kk

kkkkkkkk

kkkkkk


12/23

=

550

5116

066

107G

K

+++

++

44

44321321

321321

0

0

kkkkkkkkkk

kkkkkk

Substitute the above k values into the developed global matrix gives


13/23

Remember from last lecture

So we can write the matrix as

}F{}u{]K[ GG =

=

10000

0

0

550

5116

066

10

3

2

1

7

u

u

u

Zero because its a fixed point

=

10000

0

0

550

5116

066

10

3

2

1

7

u

u

u

In any matrix if there zero then the columnand row of the zero will be cancelled

=

10000

0

55

51110

3

27

u

u

U1=0 because its a fixed poin


14/23

=

10000

0

55

511

103

27

u

u We get from this matrix

two equations

Two equations with two unknowns can be easily solved and get

U2=0.167mm

U3=0.367mm

U1=0

Multiplying the matrices (refer to matrix lecture slides) gives


15/23

Example 2:

The beam shown in the figure below is

cantilevered at one end and is pulled by a force of

100 N at the other end. Calculate the totalelongation of the beam.

38

E1 E2 100 N

m0275.0

m0425.0

2

1

=

=

h

h

h1h2

0.1m 0.1m

E= 200000 Pa


16/23

Ex 2: Solution

Use 2 uniform cross section elements.

39

E1 E2 100 N

m0275.0

m0425.0

2

1

=

=

h

h

h1h2


17/23

E1 E2 100 N

100 N

k1 k2Node 1 Node 2 Node 3

R1

Two Elements Stiffness Matrix


18/23

K=

By using the method to assemble the global matrix fromindividual elements matrix. The resulted global stiffness

matrix for the system is:


19/23

Ex 2: Solution

Since this is a uniaxial 1-D problem, the elementscan be simplified as 2 springs in series.

42

100 N

k1 k2Node 1 Node 2 Node 3

R1

with element stiffness,ki = E * Ai / L

Where Ai=2h (cross section area is square)

k1 = [(200000*0.04252)/0.1] = 3612.5 N/mk2 = [(200000*0.0275

2)/0.1] = 1512.5 N/m

E1 E2 100 N

m0275.0

m0425.0

=

=

h

h


20/23

Ex 1: Solution

43

K= =

Global Stiffness Matrix**:


21/23

44

Ex 2: Solution

Arranging the problem in matrix form (K*x = F) yield;

A known solution, u1 = 0. Thus the problem can be

reduced to (by eliminating first column & first row):

=

=


22/23

45

Option 1 solution: Using the inverse matrix method, x = K * F ;-1

Ex 1: Solution

= =

=

-1

Refer to previous matrix

lecture slides


23/23

Option 2 solution: Use two equations solution

=

Two equations with two unknowns can be easily solved and get

U2=0.027

U3=0.093

U1=0

Computer Aided Engineering- FE Lecture 4 Slides

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