MEH - Kuliah ke-3

8/9/2019 MEH - Kuliah ke-3

1/83

Gasal 2009 Truss Elements 1

Truss Structures

Kuliah ke-3

Oleh: Dr. Rudi W. Prastianto


2/83


Definitions

Truss structure a structurecomposed of bar elements that are

connected together by frictionless pins.Plane trussall bar elements lying ina common plane (2 dimensional plane)

Space trussall bar elements lying in

a 3 dimensional space


3/83

Bar Element

T

T

x1x1 f,d

u,x

x2x2 f,d

y

y

x

L

1

2


4/83


5/83


0xdudAE

xdd

ATA

xd

ud

E

x

=

=

Forceequilibrium

Hookes

law

Governing Eq.for linear-elasticbar

Strain-displ.relationship


6/83


Assumptions

The bar cannot resist shear forces.

That is:

Effects of transverse displacementsare ignored.

Hookes law applies.

That is:

0ff y2y1 =

xx E


7/83


Step 1 - Select the ElementType.

The bar element is

selected with theproperties as previouslydiscussed.


8/83


Step 2 - Select aDisplacement Function

Assume a displacement function

Assume a linear function.

Number of coefficients = numberof d-o-f

Write in matrix form:

u

xaau 21 +

[ ] = 21aa

x1u


9/83


xx

x

dLadLaaLu

adaau

12221

1121

)()(

)0()0(

+==+=

==+=

uExpress as function of andx1d x2d


10/83


[ ]

2x 1x1x

1x

2x

1x

1 2

2x

1 2

d d u x dL

d x xu 1

L L d

du N N

d

Where :

x xN 1 and N

L L

= + = =

= =


11/83


2

1

x

u

y

x

Lx1d

x2d

Displacement plotted along length of bar.


12/83


-Strain/Displacement andStress/Strain Relationships

( )2x 1x

1x

2x 1x

d d u x dL

d ddu

dx L

E

= + = = =


13/83


Element Stiffness Matrixand Equations

( )( ) =

=

Ld

d

AETf

LddAEAET

AT

x2x1x1

x1x2


14/83



( )

==L

ddAETf

Tf

x1x2x2

x2


15/83



[ ] =

=

1111

LAEk

d

d

11

11

L

AE

f

f

x2

x1

x2

x1


16/83


Step 5 - Assemble theElement Equations to Obtain

the Global Equations andIntroduce the B.C.

[ ] [ ]{ } { }

=

===

N

1e

)e(

N

1e

)e(

fF

kK


17/83


Step 6 - Solve for NodalDisplacements

[ ] { } { }!SolveThen

FdK

:Obtain =


18/83


Step 7 - Solve for ElementForces

Once displacements at each

node are known, then substituteback into element stiffness equations

to obtain element nodal forces.


19/83


Three Bar Assembly

1 21

2

3 x3 4

30 in 30 in30 in

90 in

3000 lb

Elements 1 & 2

E = 30 x 106 psi

A = 1 in2

Element 3

E = 15 x 106 psi

A = 2 in2


20/83

=

=

=

x4

x3

33

33

x4

x3

x3

x2

22

22

x3

x2

x2

x1

11

11

x2

x1

d

d

kk

kk

f

f

:3elementFor

d

d

kk

kk

f

f

:2elementFor

d

d

kk

kk

f

f

:1elementFor


21/83

( ) ( )( )30

10x301k

30L10x30E1A

LEAk

dd

kk

kk

ff

:2&1elementsFor

6

1

16

11

1

111

x2

x1

11

11

x2

x1

==

= =


22/83

[ ]

[ ] [ ])1()2(2

622

6)1(

kk

30L10x30E1A11

1110k

:2&1elementsFor

==

=


23/83

( ) ( )( )30

10x152k

30L10x15E2A

L

EAk

d

d

kk

kk

f

f

:3elementFor

6

3

1

6

11

3

333

x4

x3

33

33

x4

x3

==

=

=


24/83


[ ]

= 1111

10k

:3elementFor

6)3(


25/83


)3(x4x4

)3(x3

)2(x3x3

)2(

x2

)1(

x2x2

)1(x1x1

fF

ffF

ffF

fF

matrixforceGlobalAssemble

=++

=


26/83


[ ]

[ ]

=

=

11001210

0121

0011

10K

kk00

kkkk0

0kkkk

00kk

K

6

33

3322

2211

11

Global stiffness matrix

pp y ng oun ary


27/83


0d0d

s.'C.B

dd

d

d

11001210

0121

0011

10

FF

F

F

x4x1

x4

x3

x2

x1

6

x4

x3

x2

x1

=

=

pp y ng oun aryconditions


28/83


indind

Solution

d

d

xx

x

x

001.0002.0

:

11

1210

0

3000

32

3

26

==

=

Matrix partitioning


29/83


=

=

lb1000

lb0

lb3000

lb2000

F

F

F

F

0

001.0

002.0

0

1100

1210

0121

0011

10

F

F

F

F

x4

x3

x2

x1

6

x4

x3

x2

x1

Back substitution


30/83


Checking

F1x

+ F4x

= F2x

(equal in magnitude,

opposite in direction)

Equilibrium of the bar assemblageverified

T f ti f


31/83


Local coordinates convenientto represent individual element.

Global coordinates convenientto represent whole structure.

Developing a transformation

matrix global stiffness matrix fora bar element.

Transformation of aVector in 2 Dimensions

T f ti f V t


32/83

Transformation of a Vectorin 2 Dimensions

jijidyxyx

dddd

x

j

y

i

i

j

y

x

d

T f ti f V t


33/83

Transformation of a Vectorin 2 Dimensions

xj

y

i

i

j

y

x

ba ba

Relate i &j ??ji &

e a ons p: oca g o a un


34/83


jsinicosi

)j(sinb

icosa

sinbcosa

1i

cosia

iba

=

= Law ofCosines

i = unit vector

Magnitude of unitvector

a in direction,

b indirection.

ij

e a ons p: oca g o a unvector


35/83


jcosisinj

isinb

jcosa

jba

=

,


36/83

==+

+

y

x

y

x

yyx

xyx

yx

yx

yxyx

d

d

d

d

ddd-

ddd

dd

dd

dddd

CS

SC

cossin

sincos

ji)j

cosi

(sin)j

sini

(cos

jijid

Combine in eachsame unit vector

In matrixform


37/83


sinScosC

CS

SC

==

matrixionTranformat

Transformation matrix


38/83


Global Stiffness Matrix

{ } [ ] { }{ } [ ] { }dkf

:Want

dkf

d

d

11

11

L

AE

f

f

x2

x1

x2

x1

==

=

Localcoordinate system

o a ness a r x con


39/83


o a ness a r x con .)

{ } [ ] { }

{ } { }

=

=

=

y2

x2

y1

x1

y2

x2

y1

x1

d

d

d

d

d

f

f

f

f

f

dkf

In global coordinate:* 4 comp. of force* 4 comp. of displacement

)


40/83


{ } [ ]{ }dTd dd

d

d

SC00

00SC

d

d

sindcosddsindcosdd

*

y2

x2

y1x1

x2

x1

y2x2x2

y1x1x1

=

=

Transformation relationship

)


41/83


{ } [ ]{ }fTf ff

f

f

SC00

00SC

f

f

sinfcosff

sinfcosff

*

y2

x2

y1

x1

x2

x1

y2x2x2

y1x1x1

=

=

Similarly,


42/83


{ } [ ] { }{ } [ ] { }{ } [ ] [ ] { }{ } [ ] { }[ ] { } [ ] [ ]{ }dTkfT

fTf

dTkf

dTd

dkf

**

*

*

*

==

===

We mustinvert [T*]


43/83


{ } [ ] { }{ } [ ] { }fTf dTdd

d

d

d

CS00

SC00

00CS

00SC

d

d

d

d

y2

x2

y1

x1

y2

x2

y1

x1

==

=

Expand local d, f, and k


44/83


[ ]

=

CS00

SC0000CS

00SC

T


45/83


=

y2

x2

y1

x1

y2

x2

y1

x1

d

dd

d

0000

01010000

0101

LAE

f

f

f

f

Expand local [k] to 4x4 size:


46/83


[ ] { } [ ] [ ] { }{ } [ ] [ ] [ ] { }[ ] [ ]{ } [ ] [ ] [ ] { }[ ] [ ] [ ] [ ]TkTk dTkTf

TT

dTkTf

dTkfT

T

T

T1

1

===

==

Expanded form:

Remember:

f= kd


47/83


[ ]

=

22

22

2222

SCSSCSCSCCSC

SCSSCSCSCCSC

L

AEk

[k] in explicit form:


48/83


EXAMPLE

30o

x

y

x

L

A=2 in2

E=30 x 106 psi

L=60 in


49/83


[ ]

2

1S

2

3

C

30

SCSSCS

CSCCSC

SCSSCS

CSCCSC

L

AEk

o

22

22

22

22

===

=


50/83


[ ]

=

41

43

41

43

4

3

4

3

4

3

4

34

1

4

3

4

1

4

34

3

4

3

4

3

4

3

60)10x30)(2(k

6


51/83


[ ]

=25.0symmetric

433.075.025.0433.025.0

433.075.0433.075.0

10k 6


52/83

Stress Computation

=

x2

x1

x2

x1

d

d

11

11

L

AE

f

f

x

y

x

L

1

2x2f

x1

f

ress ompu a on con


53/83


ress ompu a on con .)

[ ]

[ ] [ ]

=

x2

x1

x2

x1x2

x2

x1x2

x2

d

d11

L

E

d

d11

L

AE

A

1

A

f

d

d11L

AEf

A

f

ress ompu a on con


54/83


[ ]

{ } [ ] { }

[ ] [ ] { }[ ] { }dC

dTL

E

dTd

d

dLE

x

x

==

=

=

*

*

2

1

11

11


ress ompu a on con


55/83


[ ] { }[ ] [ ][ ] [ ]SCSCLEC

SC0000SC11

LEC

dC



56/83


Example

x

y

x

1

2

60o

A = 4 x 10-4 m2

E = 210 GPa

= 60od1x = 0.25 mmd1y = 0.0 mm

d2x = 0.50 mm

d2y

= 0.75 mm


57/83


Stress Computation

[ ]

{ }

=

=

m10x75.0

m10x50.0

m0.0

m10x25.0

d

d

d

d

d

2

3

2

1

2

3

2

1

m2

m/kN10x210C

3

3

3

y2

x2

y1

x1

6


58/83


Stress Computation

MPa32.81

mm/kN10x32.81

m10x75.0

m10x50.0

m0.0

m10x25.0

2

3

2

1

2

3

2

1

2

10x210

23

3

3

3

6

==


59/83


3-Bar Truss Example

45o

45o

10 ft

10

ft

3

41

2

3

21

Data for 3-Bar Truss


60/83


Data for 3 Bar TrussExample

Element Node i Node j L (ft) A (in2) C S C2 S2 CS1 1 2 10.00 2 90 0 1 0 1 0

2 1 3 10.00 2 45 0.7071 0.7071 0.5 0.5 0.5

3 1 4 14.14 2 0 1 0 1 0 0

E = 30 x 106 psi for all members


61/83


[ ]

=1010

0000

1010

0000

)12()10(

)2()10x30(

k

6)1(


62/83


[ ] ( ) ( ) ( )

=5.05.05.05.0

5.05.05.05.0

5.05.05.05.0

5.05.05.05.0

12210)2()10x30(k

6

)2(


63/83


[ ]

=

0000

0101

0000

0101

)12()10(

)2()10x30(

k

6)3(


64/83


[ ]

=

0000000000000000

00000000

000000000000kkkk

0000kkkk

0000kkkk

0000kkkk

K

)1(

44

)1(

43

)1(

42

)1(

41

)1(34

)1(33

)1(32

)1(31

)1(24

)1(23

)1(22

)1(21

)1(14

)1(13

)1(12

)1(11


65/83


+

00000000

00000000

00kk00kk

00kk00kk

00000000

00000000

00kk00kk00kk00kk

)2(44

)2(43

)2(42

)2(41

)2(34

)2(33

)2(32

)2(31

)2(24

)2(23

)2(22

)2(21

)2(

14

)2(

13

)2(

12

)2(

11


66/83


)3(44

)3(43

)3(42

)3(41

)3(34)3(33)3(32)3(31

)3(24

)3(23

)3(22

)3(21

)3(

14

)3(

13

)3(

12

)3(

11

kk0000kk

kk0000kk

00000000

0000000000000000

00000000

kk0000kkkk0000kk


67/83


[ ]

=

00000000

01000001

00354.0354.000354.0354.0

00354.0354.000354.0354.0

00001010

00000000

00354.0354.010354.1354.001354.0354.000354.0354.1

)500000(K


68/83


Point to Ponder

Why are rows and columns 3 & 8equal to zero?

x-displacement at node 2 is 3rd

d-o-fand y-displacement at node 4 is 8thd-o-f.

These displacements must be zerobecause of geometry (not B.C.)

Assembling the Global


69/83


Assembling the GlobalStiffness Matrix - [K]

If there are 2 degrees of freedom and element eiconnects nodes i & j then the Global [K] matrix is

assembled as follows:


70/83


Position in local [k] adds to Position in Global [K]

Upper Left Quadrant:

row m 2i-1 if m=1

2i if m=2

column n 2i-1 if n=1

2i if n=2

Upper Right Quadrant:

row m 2i-1 if m=1

2i if m=2

columnn

2j-1 ifn=3

2j if n=4


71/83



Lower Left Quadrant:

row m 2j-1 if m=3

2j if m=4


2i if n=2

Lower Right Quadrant:

row m 2j-1 if m=3

2j if m=4

columnn

2j-1 ifn=3

2j if n=4


72/83



row m column n row column

1 1 2i-1 2i-1

1 2 2i-1 2i1 3 2i-1 2j-1

1 4 2i-1 2j

2 1 2i 2i-1

2 2 2i 2i

2 3 2i 2j-1

2 4 2i 2j

3 1 2j-1 2i-1

3 2 2j-1 2i

3 3 2j-1 2j-1

3 4 2j-1 2j

4 1 2j 2i-14 2 2j 2i

4 3 2j 2j-1

4 4 2j 2j

Suppose i=1 and j= 3


73/83


Suppose i=1 and j= 3



1 1 1 1

1 2 1 21 3 1 5

1 4 1 6

2 1 2 1

2 2 2 2

2 3 2 52 4 2 6

3 1 5 1

3 2 5 2

3 3 5 5

3 4 5 64 1 6 1

4 2 6 2

4 3 6 5

4 4 6 6


74/83


0000000000000000

00kk00kk

00kk00kk

00000000

0000000000kk00kk

00kk00kk

)2(44

)2(43

)2(42

)2(41

)2(34

)2(33

)2(32

)2(31

)2(24

)2(23

)2(22

)2(21

)2(14

)2(13

)2(12

)2(11

If there are 3 d o f per node (3D truss):


75/83


If there are 3 d-o-f per node (3D truss):


Upper Left Quadrant:

row m 3i-2 if m=13i-1 if m =2

3i if m =3


3i-1 if n =2

3i if n =3

Upper Right Quadrant:

row m 3i-2 if m=4

3i-1 if m =5

3i if m =6column n 3j-2 if n=1

3j-1 if n =2

3j if n =3



76/83





row m 3j-2 if m=43j-1 if m =5

3j if m =6


3i-1 if n =2

3i if n =3


row m 3j-2 if m=4

3j-1 if m =5

3j if m =6column n 3j-2 if n=4

3j-1 if n =5

3j if n =6



77/83



1 1 3i-2 3i-2

1 2 3i-2 3i-1

1 3 3i-2 3i1 4 3i-2 3j-2

1 5 3i-2 3j-1

1 6 3i-2 3j

2 1 3i-1 3i-2

2 2 3i-1 3i-1

2 3 3i-1 3i

2 4 3i-1 3j-2

2 5 3i-1 3j-1

2 6 3i-1 3j

3 1 3i 3i-2

3 2 3i 3i-13 3 3i 3i

3 4 3i 3j-2

3 5 3i 3j-1

3 6 3i 3j



78/83



4 1 3j-2 3i-2

4 2 3j-2 3i-1

4 3 3j-2 3i4 4 3j-2 3j-2

4 5 3j-2 3j-1

4 6 3j-2 3j

5 1 3j-1 3i-2

5 2 3j-1 3i-1

5 3 3j-1 3i

5 4 3j-1 3j-2

5 5 3j-1 3j-1

5 6 3j-1 3j

6 1 3j 3i-2

6 2 3j 3i-16 3 3j 3i

6 4 3j 3j-2

6 5 3j 3j-1

6 6 3j 3j


79/83


[ ]

=

00000000

01000001

00354.0354.000354.0354.0

00354.0354.000354.0354.0

00001010

00000000

00354.0354.010354.1354.001354.0354.000354.0354.1

)500000(K


80/83


{ } { }

=

=

0

0

0

0

0

0d

d

d

F

F

F

F

F

F

10000

0

F

y1

x1

y4x4

y3

x3

y2

x2


81/83


=

=

in10x59.1

in10x414.0

d

d

dd

354.1354.0354.0354.1)500000(

100000

2

2

y1

x1

y1

x1

10x4140 2


82/83


[ ]

[ ]

=

0

0

10x59.1

10x414.0

0101120

10x30

0

0

10x59.1

10x414.0

2

2

2

2

2

2

2

2

120

10x30

0

0

10x59.1

10x414.0

1010120

10x30

2

2

6)3(

2

2

6)2(

26)1(


83/83

psi1035

psi1471

psi3965

)3(

)2(

)1(

==

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MEH - Kuliah ke-3