zint1001 - engineering computational tools - part c - engineering applications

8/4/2019 ZINT1001 - Engineering Computational Tools - Part C - Engineering Applications

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Engineering Applications

By

Dr Murat Tahtalı



Application to Statics

• If you recall from Statics, solving a trussproblem is a quite onerous task.

• It involves the repetitive solution member

forces and reactions.• In fact, the problem is merely the solution of

N-equations for N-unknowns.

• The brand of application to solve this typeproblems, and more, is called Finite ElementAnalysis (FEA).

UNSW@ADFASEIT & PEMS, S2, 2010

Dr Murat Tahtalı & Dr Isaac Towers 2



Simple FEA using Matlab

• A truss system is comprised of repeated truss-elements.

• Each element can be represented in matrix

form.• The whole truss system can be represented in

matrix forms.

• The solution of N-equation for N-unknownscan be easily written in matrix form.

• Matlab is well suited to solve matrix problem.





FEA steps

• Idealise the system

• Discretise the system

• Obtain element matrices

• Assemble element matrices• Apply boundary conditions (BCs)

• Solve for unknowns

• Post-processing – Display results

– Do further calculations

– ...





A simple linear spring system

k 2 k 1

F A C B

Given: k1=100 N/mm, k2 =50 N/mm, F=1000 N

• Preliminary observations:• It is a natural discrete system,

• Natural idealisation is supplied,

• System is linear and one-dimensional



Step 1: Idealisation

• Inherently given by the system as:

A CB

Element [1] Element [2]



Step2:Discretisation

• Also inherently supplied:

1 2 [1] f (2)[1] f (1)

[1]u(1) [1]u(2)

[2] f (2)[2] f (1)

[2]u(1) [2]u(2)

1 2

[1]k=k 1[2]k=k 2



Elemental Stiffness Matrices

Element [1]

Element [2]

uk f

]1[

1

]1[

11

11

uk f ]2[2

]2[

1111



Condition of nodal force equilibrium:

2

1

1

1

1]1[1

1

1][1 uk uk f f F i

i

3

2

2

21

1

1

1]2[2]1[2

1

2][2

)( uk uk k uk

f f f F i

i

3

2

2

2

3]1[3

3

3][3

uk uk f f F i

i

3

2

1

22

2211

11

3

2

1

0

)(

0

u

u

u

k k

k k k k

k k

F

F

F

U K F



Expanded form of the elemental

stiffness matrices

3

2

1

11

11

)2(]1[

)1(]1[

]1[]1[

000

0

0

0 u

u

u

k k

k k

f

f

uk f

3

2

1

22

22

)2(]2[

)1(]2[

]2[]2[

0

0

0000

u

u

u

k k

k k

f

f uk f

uk uk f f F

]2[]1[]2[]1[

uk k F

K

]2[]1[



Step 5: Introduce Boundary Conditions

• Node <1>: u<1>=0 (displacement specified)

Node <2>: F<2>=0 (force specified)

Node <3>: F<3>=1000 N (force specified)

The system of equations is now:

mm

u

umm

N

N

F

3

2

1 0

50500

50150100

0100100

1000

0

mm

u

u

mm

N

N

3

2

5050

50150

1000

0

Or, after deleting relevant rows and columns:



Step 6: Solve for unknown

displacements

• Solution of the reduced set of equations

yields:

u<2>=10 mm and u<3>=30 mm



Step 6: Solve for unknown forces

• F <1>=-1000 N

as it should be in equilibrium



Illustrative step-by-step Example



k1 k3

x

<1>

e2

<3>

e1 e3 e4

k2

<2> <4>

k4



Interconnectivity Table

Element # Local node # System node #

1 (1) <1>

(2) <2>

2 (1) <1>

(2) <3>

3 (1) <2>

(2) <3>

4 (1) <3>

(2) <4>



Element 1

2

1

21

11

111

k k

k k k

4

3

2

1

0000

0000

00

00

11

111

4321

k k

k k

k

1 (1) <1>

(2) <2>



Element 2

2 (1) <1>

(2) <3>

3

1

31

22

222

k k

k k k

4

3

2

1

0000

00

0000

00

22

222

4321

k k

k k

k



Element 3

3 (1) <2>

(2) <3>

3

2

32

33

333

k k

k k k

4

3

2

1

0000

00

00

0000

33

33

3

4321

k k

k k k



Element 4

4 (1) <3>

(2) <4>

4

3

43

44

444

k k

k k k

4

3

2

1

00

00

0000

0000

44

44

4

4321

k k

k k

k



Global Stiffness Matrix

44

443232

3311

2121

4

1

00

0

0

k k

k k k k k k

k k k k

k k k k

k K e

e



System of equations

4444343242141

3434333232131

2424323222121

1414313212111

F xK xK xK xK

F xK xK xK xK

F xK xK xK xK

F xK xK xK xK

One way to include prescribed nodal variables while retaining itsoriginal n n arrangement is to modify the matrices [K ] and {F } asfollows.



Introduce the known displacements

4444343242141

33

2424323222121

11

F xK xK xK xK

x

F xK xK xK xK

x



Replace the known displacements in

each equation

4444343242141

33

2424323222121

11

F xK K xK K

x

F xK K xK K

x



Rearrange the equations so that allunknowns are on the LHS and all

the constants are on the RHS

3431414444242

33

3231212424222

11

K K F xK xK

x

K K F xK xK

x



Re-write in matrix form

1 0 0 0

0 0

0 0 1 0

0 0

22 24

42 44

1

2

3

4

1

2 21 1 23 3

3

4 41 1 43 3

K K

K K

x

x

x

x

R K K

R K K

• If i is the subscript of a prescribed nodal variable, the i -th rowand the i -th column of [K ] are set equal to zero and K

ii is set to

unity.• The term F

i of the column vector {F } is replaced by the known

value of x i ,, which is i .• Each of the n-1 remaining terms of {F } is modified by

subtracting from it the value of the prescribed nodal variablemultiplied by the appropriate column term from the original[K ] matrix. This procedure is repeated for each prescribed x i until all of them have been included.



2D Truss Element

y

x

F 4 ,

L

A = Cross-sectionalE = Young’s Modulus

L 2

Node 1

Node 2

(a)

F 3 ,

(x 2 , y 2 )

(x1 , y1 ) F1 ,

F 2 ,

L 2=sin( )

(b)



K K K K

K K K K

K K K K K K K K

u

v

uv

F

F

F F

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

1

1

2

2

1

2

3

4

1

2

3

4

1

1

2

2

u

v

u

v



2D Truss Element

y

x

F 4 ,

L

A = Cross-sectionalE = Young’s Modulus

L 2

Node 1

Node 2

(a)

F 3 ,

(x 2 , y 2 )

(x1 , y1 ) F1 ,

F 2 ,

L 2=sin( )

(b)nn

nn

v

u

2

12



2D Truss

Equilibrium Equations

K K K K

K K K K

K K K K K K K K

u

v

uv

F

F

F F

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

1

1

2

2

1

2

3

4

1

2

3

4

1

1

2

2

u

v

u

v



y

x

F 4

F 3

F1

F 2 F

ode 1

ode 2

sinsin

sinsin

sin

0sin

0

22

2

22

2

L

E AK

L

E AF

L L

E AF

F F

F y



y

x

F 4

F 3

F1

F 2 F

ode 1

ode 2

cossin

cossin

cos

0cos

0

32

3

23

3

L

E AK

L

E AF

L L

E AF

F F

F x



Elemental stiffness matrix

22

22

22

22

sinsincossinsincos

sincoscossincoscos

sinsincossinsincos

sincoscossincoscos

L

E AK

L x x y y ( ) ( )1 2

2

1 2

2

cos x x L

2 1 sin y y L

2 1



Example 3-1:

Numerical application

L

L

1.4142 L

30

100kN

Fv2

Fu2

Fv3

Fu3

u1

v1

2

2 1

3

1

(a) (b)



Element # Local node

#

System

node #

Direction Displacement

#

1

(1) <2>x 3

y 4

(2) <1>x 1

y 2

2

(1) <1>x 1

y 2

(2) <3>x 5

6

Table 3-4 Correspondence between local and global numbering

schemes

L

L

1.4142 L

30

100kN

Fv2

Fu2

Fv3

Fu3

u1

v1

2

2 1

3

1

(a) (b)

nn

nn

v

u

2

12



2

1

4

3

0000

0101

0000

01 01

/ 10300

2143

61 m N k


#

System

node #


#

1

(1) <2>x 3

y 4

(2) <1>x 1

y 2

2(1) <1>

x 1

y 2

(2) <3>x 5

y 6

6

5

4

3

2

1

000000

000000

000000

0003000300

000000

0003000300

654321

/ 106

1

m N k




#

System

node #


#

1

(1) <2>x 3

y 4

(2) <1>x 1

y 2

2(1) <1>

x 1

y 2

(2) <3>x 5

y 6

6

5

2

1

1111

1111

1111

11 11

/ 10294.98

6521

62 m N k

6

5

4

3

21

47.4947.490047.4947.49

47.4947.490047.4947.49

000000

000000

47.4947.490047.4947.4947.4947.490047.4947.49

65432 1

/ 106

2

m N k



50



3

3

2

2

6.86

50

0

0

00

1

1

v

u

v

u

R

R

R R

v

u

2221

1211

KKKK

6.86

50

1

1

11v

uK

6.86

501

11

1

1K

v

u

1

1

47.4947.49

47.4947.49

00

0300

3

3

2

2

v

u

R

R

R

R

v

u

v

u



1

2

3

4

5

3m

4m

4kN

3kN 6kN

2 / 300 mmkN E

Member A [mm2] L[m]

1 30 3

2 27 4

3 30 3

4 17 5

5 17 5

21 23 4



Global Node # x-coordinate [m] y-coordinate [m] Displacement #

1 0 01

2

2 4 03

4

3 0 35

6

4 4 3

7

8

1

2

3

4

5

1 2

1

2

1

2

1

2

1

2

1 2

3 4

x

yGlobal Coordinate

System (GCS)

element

element node

ystem node

nn

nn

v

u

2

12

Element # Local node # System node # Direction Displacement #



y p

1

(1) <1>

x 1

y 2

(2) <3>

x 5

y 6

2

(1) <3>

x 5

y 6

(2) <4>

x 7

y 8

3

(1) <2>

x 3

y 4

(2) <4>

x 7

y 8

4

(1) <2>

x 3

y 4

(2) <3> x 5

y 6

5

(1) <1>

x 1

y 2

(2) <4>

x 7

y 8

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