Plane Trusses - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...

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PLANE TRUSS ELEMENTS MNTamin, CSMLab SME 3033 FINITE ELEMENT METHOD Plane Trusses (Initial notes are designed by Dr. Nazri Kamsah)

Transcript of Plane Trusses - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...

PLANE TRUSS ELEMENTS

MNTamin, CSMLab

SME 3033 – FINITE ELEMENT METHOD

Plane Trusses (Initial notes are designed by Dr. Nazri Kamsah)

PLANE TRUSS ELEMENTS

MNTamin, CSMLab

SME 3033 – FINITE ELEMENT METHOD 4-3 Plane Trusses

A typical two-dimensional plane truss is shown. It comprises of two-force

members, connected by frictionless joints. All loads and reaction forces are

applied at the joints only.

Note: There are two displacement components at a given node j, denoted by Q2j-1 and Q2j.

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4-4 Local & Global Coordinate Systems

Local and global coordinate systems are

shown at left.

In local coordinate (x’), every node has one

degree of freedom, while in global coordinate

(x, y), every node has two degrees of

freedom.

The nodal displacements, in the local

coordinate system is,

In the global coordinate system,

Local and global coordinate systems.

' ' '

1 2

T

q q q

1 2 3 4

Tq q q q q

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4-5 Relation Between Coordinate Systems

Consider a deformed truss member as shown. We can now establish a

relationship between {q’} and {q} as follows:

sincos

sincos

43

'

2

21

'

1

qqq

qqq

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4-6 Direction Cosines

cosl and sinm

The relation between {q’} and {q} can now be written as

To eliminate the terms from previous equations, we define direction

cosines, such that

which can be written in matrix form as

)()(

)()(

43

'

2

21

'

1

mqlqq

mqlqq

4

3

2

1

'

2

'

1

00

00

q

q

q

q

ml

ml

q

q

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

where [L] is a rectangular matrix called the transformation matrix,

given by,

Or, in a condensed matrix form

qLq '

ml

mlL

00

00

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4-7 Formula for Evaluating l and m

Using a trigonometry relation, we observe

Note: Coordinates (xi, yi) are based on local coordinate system.

2 1

2 1

2 2

2 1 2 1

cos

sin

( ) ( )

e

e

e

x xl

l

y ym

l

l x x y y

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD 4-8 Element Stiffness Matrix

A truss element is a one-dimensional (bar) element, when it is viewed in

local coordinate system.

Thus, element stiffness matrix for a truss element in local coordinate,

The internal strain energy in the truss element, in local coordinate system is,

Substituting

' 1 1

1 1truss bare

AEk k

l

''''

2

1qkqU truss

T

e

qLq 'we get,

qLkLqU

qLkqLU

truss

TT

e

truss

T

e

''

''

2

1

2

1

We need the expression

for [k] when viewed in

global coordinate…

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

In the global coordinates system,

Since internal strain energy is independent of coordinate system, Ue = U’e.

Therefore,

Simplifying,

qkqU truss

T

e2

1

'

0

0 1 1 0 0

0 1 1 0 0

0

T

truss truss

trusse

k L k L

l

m l mAEk

l l ml

m

2 2

2 2

2 2

2 2

trusse

l lm l lm

lm m lm mAEk k

l l lm l lm

lm m lm m

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD Exercise 4-1

For each element of a plane truss structure shown,

a) Determine the transformation matrix, [L]e;

b) Write the element stiffness matrix, [k]e;

c) Assemble the global stiffness matrix, [K].

P = 50 kN

0.8 m

0.6 m

A

B

C

Use: E = 180 GPa; d = 15 mm for all members.

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4-9 System of Linear Equations

The system of linear equations for a single plane truss element in local

coordinate system can be expressed as

Where {q} is nodal displacement vector and {f} is nodal force vector, in the global

coordinate direction. Substituting, we get

Note: To assemble the global stiffness matrix, a local-global nodal connectivity

will be required.

fqk

2 21 1

2 22 2

2 23 3

2 24 4

e

q fl lm l lm

q flm m lm mAE

q fl l lm l lm

q flm m lm m

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

Exercise 4-2

Reconsider Exercise 4-1. a) Assemble global system of linear equations for

the structure; b) Apply the boundary conditions, and c) Write the reduced

system of linear equations.

P = 50 kN

0.8 m

0.6 m

A

B

C

Use: E = 180 GPa; d = 15 mm for all members.

Boundary conditions:

Q1 = Q2 = Q4 = 0

(homogeneous type)

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD 4-10 Stress Calculations

Normal stress in a plane truss element, in local coordinate system is,

In the global coordinate system, since

Expanding the [B] and [L] matrices,

Or

'qBE

qLq '

E B L q

1

2

3

4

0 011 1

0 0

q

ql mE

ql ml

q

4

3

2

1

q

q

q

q

mlmll

E

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

Exercise 4-3

Reconsider Exercise 4-2. a) Determine the unknown nodal displacements

at B and C; b) Compute the stresses in the member AC and BC.

P = 50 kN

0.8 m

0.6 m

A

B

C

Use: E = 180 GPa; d = 15 mm for all members.

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

Homework 4-1

For the truss structure shown, rod (1)

has modulus E = 29 x 103 ksi and a

yield point Y = 36 ksi. Rod (2) has

modulus E = 10 x 103 ksi and a yield

point Y = 60 ksi. If the horizontal

force P = 10 kips, determine the

cross-sectional areas of the two rods

so that the yield stress in each rod is

not exceeded.

Note: This is a trial and error

problem, so you may use Nastran

simulation to help you solve the

problem. Work in group.

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD Example 4-1

For a plane truss shown, determine: a) displacements at nodes 2 and 3; b)

stresses in each element; c) reaction forces at the supports.

Use: Ee = 29.5 x 106 psi and Ae = 1 in2 for all elements.

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD Solution

1. Establish nodal coordinate data and element connectivity

information

Node x y

1 0 0

2 40 0

3 40 30

4 0 30

Element connectivity

Element Node 1 Node 2

1 1 2

2 3 2

3 1 3

4 4 3

Nodal coordinate

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SME 3033 – FINITE ELEMENT METHOD

2. Compute the direction cosines

Element le l m

1 40 1 0

2 30 0 -1

3 50 0.8 0.6

4 40 1 0

3. Write the stiffness matrix for each element

0000

0101

0000

0101

40

1105.29 6)1(

k

Element 1

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SME 3033 – FINITE ELEMENT METHOD

For elements 2, 3, and 4

1010

0000

1010

0000

30

1105.29 6)2(

k

36.048.036.048.0

48.064.048.064.0

36.048.036.048.0

48.064.048.064.0

50

1105.29 6)3(

k

0000

0101

0000

0101

40

1105.29 6)4(

k

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

4. Assemble the global stiffness matrix [K] for the entire truss structure

6

22.68 5.76 15.0 0 7.68 5.76 0 0

5.76 4.32 0 0 5.76 4.32 0 0

15.0 0 15.0 0 0 0 0 0

0 0 0 20.0 0 20.0 0 029.5 10

7.68 5.76 0 0 22.68 5.76 15.0 0600

5.76 4.32 0 20.0 5.76 24.32 0 0

0 0 0 0 15.0 0 15.0 0

0 0 0 0 0 0 0 0

K

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

5. Assemble the global system of linear equations.

1

2

3

6

Q22.68 5.76 15.0 0 7.68 5.76 0 0

Q5.76 4.32 0 0 5.76 4.32 0 0

Q15.0 0 15.0 0 0 0 0 0

Q0 0 0 20.0 0 20.0 0 029.5 10

7.68 5.76 0 0 22.68 5.76 15.0 0600

5.76 4.32 0 20.0 5.76 24.32 0 0

0 0 0 0 15.0 0 15.0 0

0 0 0 0 0 0 0 0

4

5

6

7

8

0

0

20000

0

Q 0

Q 25000

Q 0

Q 0

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

6. Impose boundary conditions & write the reduced system of

linear equations

25000

0

20000

32.2476.50

76.568.220

0015

600

105.29

6

5

36

Q

Q

Q

7. Solve the reduced equations using the Gaussian elimination method,

we get

in

1025.22

1065.5

1012.27

3

3

3

6

5

3

Q

Q

Q

We have; Q1 = Q2 = Q4 = Q7 = Q8 = 0.

Using elimination method, the above equations reduced to

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

8. Compute the stress in each element

Element 1

psi20000

0

1012.27

0

0

010140

105.293

6

1

Element 2

psi21880

0

1012.27

1025.22

1065.5

101030

105.293

3

3

6

2

psi4167

psi5208

4

3

In the same manner,

PLANE TRUSS ELEMENTS

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SME 3033 – FINITE ELEMENT METHOD

9. Determine support reactions

lb

0

4167

21879

3126

15833

8

7

4

2

1

R

R

R

R

R

Using the 1st, 2nd, 4th, 7th, and the 8th equations, we get