Stiffness 3

8/2/2019 Stiffness 3

1/121

Lecture No. : 3


2/121

Remember :F1

1 2

k1 k2

F3

3

k3

F2

d1 d2 d3

k11F1

F2 = k21

F = K D

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3


3/121

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2d3

First column in Stiffness matrix

d1 =1 d2 =0 d3 =0

1 2

k1 k2

3

k3

d1=1

k11

k21

k31

=

F1

F2

F3

Remember :


4/121

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3

Second column in Stiffness matrix

d1 =0 d2 =1 d3 =0

k12

k22

k32

=

F1

F2

F3

1 2

k1 k2

3

k3

d2=1

Remember :


5/121

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3

Third column in Stiffness matrix

d1 =0 d2 =0 d3 =1

k13

k23

k33

=

F1

F2

F3

1 2

k1 k2

3

k3

d3=1

Remember :


6/121


7/121


8/121

a

b

D

q

D1= D cos q

D1 D2

D2= D sin q


9/121

D

P

P

AE

LPD=

AE

L

P D=

Relation between force andDisplacement in truss element


10/121

Degrees of freedom (DOF)

d1

d2

d1


11/121

AB

C 4 m

3

Example 1:

Construct the stiffness matrix for the shown

truss where EA is constant for all members


12/121

Modeling

A

B

C

4

3F1d1

F2 d2


13/121

F = K D

k11F1

F2=

k21

k12

k22

d1

d2


14/121


d1 =1 d2 =0k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3F1d

1

F2

d2

k11F1

F2= k21

k12

k22

1

0

k11 F1

F2

=k21


15/121


d1 =1 d2 =0k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3d1

1


16/121

A

B

C

4

3

1A

1

q

A cos qEL

AE

LAE

L

A cos qEL

AB

C 4 m3


17/121

A

B

C

4

3

1A

1

q

A x 0.8E

5

AE

4AE

4

A x 0.8E

5

AB

C 4 m3


18/121

4

3q

0.16 EA

0.25 EA 0.25 EA

0.16 EA

AE

4AE

4

A x 0.8E

5

A x 0.8E5


19/121

A

B

C

4

30.25 EA

0.16 EA

q

0.16 EA

0.25 EA 0.25 EA

0.16 EA


20/121

A

B

C

4

30.25 EA

0.16 EA

q 0.16 EA cos q

0.16 EA sin q

cos q=0.8sin q=0.6


21/121

A

B

C

4

30.25 EA

0.16 EA cos q

0.16 EA sin q

0.16 EA x 0.8

0.16 EA x 0.6

0.128 EA

0.096 EA


22/121

A

B

C

4

30.25 EA

0.128 EA

0.096 EA

0.378 EA


23/121

A

B

C

4

3

0.096 EA

0.378 EA

A

B

C

F1

F2

F1 = 0.378 EAF2 = 0.096 EA


24/121

A

B

C

4

3

0.096 EA

0.378 EA

F1 = 0.378 EAF2 = 0.096 EA

k11 F1

F2

=k21

k11 =k21

0.378 EA

0.096 EA


25/121

second column in Stiffness matrix

d1 =0 d2 =1k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3F1d

1

F2

d2

k11F1

F2

= k21

k12

k22

0

1

k12 F1

F2

=k22


26/121


d1 =0 d2 =1k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3

d2

1


27/121

A

B

C

4

3

1

A

1

a

A cos aEL

0

A cos aEL

0

cos a = 0.6L = 5


28/121

A

B

C

4

3

1

A

1

a

A x 0.6E5

0

A x 0.6E

5

0


29/121

4

3

0.12 EA

0 0

0.12 EA


30/121

A

B

C

4

30

0.12 EA

q

0.12 EA

0 0

0.12 EA


31/121

A

B

C

4

30

0.12 EA

q 0.12 EA cos q

0.12 EA sin q

cos q=0.8sin q=0.6


32/121

A

B

C

4

30

0.12 EA cos q

0.12 EA sin q

0.12 EA x 0.8

0.12 EA x 0.6

0.096 EA

0.072 EA

cos q=0.8sin q=0.6


33/121

A

B

C

4

3

0.096 EA

0.072 EA


34/121

A

B

C

4

3

0.072 EA

A

B

C

F1

F2

F1 = 0.096 EAF2 = 0.072 EA

0.096 EA


35/121

A

B

C

4

3 k12 F1F2=k22k12 =k22

0.096 EA

0.072 EA

F1 = 0.096 EAF2 = 0.072 EA

0.072 EA

0.096 EA

S


36/121

Summary

k11F1

F2=

k21

F = K D

k12

k22

d1

d2

k11 =k21

0.378 EA

0.096 EA

k12 =k22

0.096 EA

0.072 EA


37/121

k11 =k21

0.378 EA

0.096 EA

k12 =k22

0.096 EA

0.072 EA

K =

0.378 EA

0.096 EA

0.096 EA

0.072 EA

K =0.378

0.096

0.096

0.072

EA

E l 2


38/121

AB

C 4 m

3

Example 2:

Calculate the joint displacements for the

shown truss due to the shown force whereEA = 105 kN for all members

12 kN


39/121

k11F1

F2=

k21

F = K D

k12

k22

d1

d2

From the previous example

K =0.378

0.096

0.096

0.072

EA

F th i l


40/121

From the previous example

K =

0.378

0.096

0.096

0.072

EA

EA = 105 kN

K = 0.378

0.096

0.096

0.072

105 = 378

96

96

72

100


41/121

AB

C 4 m

312 kN

A

B

C

F1

F2

F1

F2=

0

-12

Force vector


42/121

=378

96

96

72

100d1

d2

0

-12

d1

d2= 100

378

96

96

72

0

-12

1

-1

d1

d2 =105

4

-5.33

-5.33

21

0

-12

1

=

0.0006

-0.0025=

0. 6

-2.5m mm

E l 3


43/121

AB

C 4 m

3

Example 3:

Calculate the joint displacements for the

shown truss due to the shown force whereE = 2000 kN/cm2, ABA= 100 cm

2, ACA= 200 cm2

12 kN

M d li


44/121

Modeling

A

B

C

4

3F1d1

F2 d2

EABA= 2x105 kN

EACA= 4x105

kN

E = 2000 kN/cm2

, ABA= 100 cm2

, ACA= 200 cm2


45/121

k11F1

F2=

k21

F = K D

k12

k22

d1

d2

First column in Stiffness matrix k k d


46/121


d1 =1 d2 =0k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3F1d1

F2 d2

k11F1

F2

=k21

k12

k22

1

0

k11 F1

F2

=k21

First column in Stiffness matrix k k d


47/121


d1 =1 d2 =0k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3d1

1

1


48/121

A

B

C

4

3

1A

1

q

A cos qEL

AE

LAE

L

A cos qEL

EABA= 2x105 kN

EACA= 4x105 kN

LBA= 400 cmLCA= 500 cm

cos q=0.8

1


49/121

A

B

C

4

3

1A

1

q

4x105x0.8

500

2x105

4002x105

400

4x105x0.8

500


50/121

4

3q

640

500 500

640


51/121

A

B

C

4

3500

640

q


52/121

A

B

C

4

3500

640

q 640 cos q

640 sin q

cos q=0.8sin q=0.6


53/121

A

B

C

4

3500

512

384

1012


54/121

A

B

C

4

3

384

1012

A

B

C

F1

F2

F1 = 1012F2 = 384


55/121

A

B

C

4

3

384

1012

F1 = 1012F2 = 384

k11 F1

F2

=k21

k11 =k21

1012

384

second column in Stiffness matrix k11F k12 d


56/121


d1 =0 d2 =1k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3F1d1

F2 d2

k11F1

F2

=k21

k12

k22

0

1

k12 F1

F2

=k22

second column in Stiffness matrix k11F k12 d1


57/121


d1 =0 d2 =1k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3

d2

1


58/121

A

B

C

4

3

1

A

1

a

A cos aEL

0

A cos aEL

0

EABA= 2x105 kN

EACA= 4x105 kN

LBA= 400 cmLCA= 500 cm

cos a=0.6


59/121

A

B

C

4

3

1

A

1

a

4x105x0.65

0 0

4x105x0.6

5


60/121

4

3

480

0 0

480


61/121

A

B

C

4

30

480

q


62/121

A

B

C

4

30

480

q 480 cos q

480 sin q


63/121

A

B

C

4

3

384

288

F 38


64/121

A

B

C

4

3

288

A

B

C

F1

F2

F1 = 384F2 = 288

384

F 384


65/121

A

B

C

4

3 k12 F1F2=k22k

12 =k22

384

288

F1 = 384F2 = 288

288

384

Summary


66/121

K =1012

384

384

288

k11 =

k21

1012

384

k12 =k22

384

288

F1

F2=

0

-12

=d1

d2

0

-12

1012

384

384

288

Summary

d0 1012 384


67/121

=d1

d2

0

-12

d1

d2 =103

2

-2.67

-2.67

7.03

0

-12

1

=

0.0320

-0.0843 =

0.320

-0.843cm mm

1012

384

384

288

d1

d2=

0

-12

-1

1012

384

384

288

Calculation of internal force in truss member


68/121

Calculation of internal force in truss member

a

b

q

DbxDby

DaxDa

y

a

b

q

DbxDby

DaxDay

-

-


69/121

a

b

q

Dbx

DbyDax

Day-

-D= ( )cos q + ( )sin q

bxD

ax- D

byD

ay-

AEL

N D=( )cos q+ ( )sin qbx Dax- Dby Day-

AE

L

N= [ ]

Example 4:


70/121

AB

C 4 m

3

Example 4:

Calculate the internal forces for the shown

truss due to the shown force whereE = 2000 kN/cm2, ABA= 100 cm

2, ACA= 200 cm2

12 kN

EABA= 2x105 kN EACA= 4x10

5 kN


71/121

EABA= 2x10 kN EACA= 4x10 kN

D= ( )cos q + ( )sin qbx Dax- Dby Day-

d1

d2=

0.0320

-0.0843 cm

D = ( )cos 0+ ( )sin 0Ax DBx- DAy DBy-BA = 0.032 cm

+ ( )sin 37= ( )cos 37Ax DCx- DAy DCy-CA = -0.025 cmNBA = 2 x 10

5 x 0.032 / 400 = 16 kN

AE

L

ND

=

NCA = 4 x 105 x -0.025 / 500 = - 20 kN

From example 3:

Tension

Compression

Check of resultsAB


72/121

Check of results

C 4 m

312 kN

For equilibrium of

joint A :

FCA

FBA qA

12 kNS Fy = 0S Fx = 0FCA sin q - 12 = 0 FCA = 12 / sin q = 20 kN

FCA cos q - FBA = 0 FBA = 20 x cos q = 16 kN

For large trusses


73/121

For large trusses

a

b

q

F1d1

F2 d2

F3d3

F4 d4


74/121

a

b

q

F1d1=1

F2

F3

F4

d = cos q


75/121

a

b

q

F1d1=1

F2

F3

F4

d = cos q


76/121

a

b

q

F1d1=1

F2

F3

F4

EA/L cos q

EA/L cos q


77/121

a

b

q

F1d1=1

F2

F3

F4

EA/L cos q

EA/L cos q

= EA/L cos2q

= EA/L cos qsin q

= - EA/L cos2q

= - EA/L cos qsin q


78/121

a

b

q

F1d1=1

F2

F3

F4

= EA/L cos2q

= EA/L cos qsin q

= - EA/L cos2q

= - EA/L cos qsin q


79/121

a

b

q

F1

F2

F3

F4

d = sin qd2=1


80/121

a

b

q

F1

F2

F3

F4

d = sin qd2=1


81/121

a

b

q

F1

F2

F3

F4

EA/L sin q

EA/L sin q

d2=1


82/121

a

b

qF1

F2

F3

F4

EA/L sin q

EA/L sin q

= EA/L sin2q

= EA/L sin qcos q

= - EA/L sin qcos q

= - EA/L sin2

q


83/121

a

b

qF1

F2

F3

F4

= EA/L sin2q

= EA/L sin qcos q

= - EA/L sin qcos q

= - EA/L sin2

q

d2=1

Example 5:


84/121

A B

O

5

p

Calculate the internal forces for the shown

truss due to the shown forces whereEA = 4x105 kN for all members

200 kN

100 kN

C D E

150 13560 45

Modeling


85/121

A B

O

5

C D E

150 13560 45

F1d1

d2F2k11F1

F2

=k21

k12

k22

d1

d2

Modeling



86/121

a

b

q

F1d1=1

F2

F3

F4

= EA/L cos2q= EA/L cos qsin q

= - EA/L cos2

q

= - EA/L cos qsin q

k11=

k21

d1 =1 d2 =0

S EA/L cos2qS EA/L cos qsin q

Second column in Stiffness matrix


87/121

a

b

qF1

F2

F3

F4

= EA/L sin2q

= EA/L sin qcos q

= - EA/L sin qcos q

= - EA/L sin2q

d2=1

d1 =0 d2 =1

k12=

k22 S EA/L sin2qS EA/L cos qsin q

A B C D E

150135

60 45

L= 500 / sin


88/121

O

5 m150

135

12719851039S

282.9282.9282.9.707.70756670745OE

300519.6173.2.866.569357760OD

080001080050090OC

-282.9282.9282.9.707-.707566707135OB

-173100300.5-.8664001000150OA

member L sin qcos qEAL

EA/Lcos2q

EA/Lsin2q

EA/Lsinqcosq

L= 500 / sin qEA = 4x105 kN

EA/L2

EA/L EA/Lsin


89/121

12719851039S

cos2q sin2q sinqcosq

k11=

k21

S EA/L cos2qS EA/L cos qsin q

k12=

k22 S EA/L sin2qS EA/L cos qsin q

k11

K= k21

k12

k22

1039

= 127

127

1985

Force vector


90/121

A B

O

5200 kN

100 kN

C D E

150 13560 45

F1

F2=

100

-200=F

F = K D


91/121

F = K Dk11F1

F2=

k21

k12

k22

d1

d2

100

-200

=

1039

127

127

1985

d1

d2

d1

d2

1039

127

127

1985=

-1100

-200

=

0. 109

-0. 108

Internal forces in truss elements


92/121

( )cos q+ ( )sin qbx Dax- Dby Day-AE

LN=

a

b

q

Dbx

Dby

DaxDay

[ ]

AFor member OA


93/121

A

O

150= 400

EA

L

= 0.5sin q= -.866cos q= 150oq

( )cos q+ ( )sin qAx Dox- DAy Doy-AE

L

NoA= [ ]

Dox = 0.109Doy= -0.108

- 0.109 cos 150+ 0.108sin 150400NoA= [ ]59.4 kNNoA= Tension

150

330

Another way for member AO


94/121

A

O

330

= 400EA

L

= -0.5sin q=.866cos q= 330oq

( )cos q+ ( )sin qox DAx- Doy DAy-AE

L

NAo= [ ]

Dox = 0.109Doy= -0.108

0.109 cos 330 - 0.108sin 330400NAo= [ ]59.4 kNNAo= Tension

A B C D E

150135

60 45


95/121

O

5 m150

-0.4.707.70756645OE

27.8660.569360OD

86.41080090OC

86.8.707-.707566135OB

59.40.5-.866400150OA

member sin qcos qEAL

N

- 0.019 cos q + 0.0108sin qAEL

Noi = [ ]

Example 6:


96/121

C

B

D

4 m

4

Calculate the joint displacements for theshown truss due to the shown forces where

the axial stiffness = 400 kN/cm for all members

100 kN

A

50 kN

Modeling


97/121

d1d2

C

B

D

A

d3d4

d5

D C


98/121

-0.50.50.5.707-.707400566135BD

0.50.50.5.707.70740056645AC

010-10400400270DA

0010-1400400180CD

0101040040090BC

001014004000AB

member L sin qcos qEAL

cos2q sin2q sinqcosq

BA

F = K D


99/121

k11F1

F2

=

k21

F3 k31

k12

k22

k32

k13

k23

k33

F4

F5

k41

k51

k42

k52

k43

k53

k14

k24

k34

k15

k25

k35

k44

k54

k45

k55

d1

d2

d3

d4

d5


100/121

a

b

q

F1

d1=1

F2

F3

F4


= - EA/L cos2q

= - EA/L cos qsin q


101/121

a

b

q F1

F2

F3

F4

= EA/L sin2q

= EA/L sin qcos q

= - EA/L sin qcos q

= - EA/L sin2q

d2=1

d1d2

d3d4First column in Stiffness matrix


102/121

C

B

D

A d5

-0.50.50.5BD

0.50.50.5AC

010DA

001CD010BC

001AB

member cos2q sin2q sinqcosq

d1 =1k11

k21

k31

k41

k51

[S cos2q]DA,DB,DC

=EA

L

[S cosq sinq]DA,DB,DC[-cos2q]DC[-cosq sinq]DC[-cos2q]DB

=

1.5

-.5

-1

0

-.5

EA

L

d1d2

d3d4Second column in Stiffness matrix


103/121

C

B

D

A d5

-0.50.50.5BD

0.50.50.5AC

010DA

001CD010BC

001AB


d2 =1k12

k22

k32

k42

k52

[S sin2q]DA,DB,DC=

EA

L

[S cosq sinq]DA,DB,DC

[-sin2q]DC

[-cosq sinq]DC =

-.5

1.5

0

0

0.5

EA

L

[-cosq sinq]DB

d1d2

d3d4Third column in Stiffness matrix


104/121

C

B

D

A d5

-0.50.50.5BD

0.50.50.5AC

010DA

001CD010BC

001AB


d3 =1k13

k23

k33

k43

k53

[S cos2q]CA,CB,CD=EA

L

[S cosq sinq]CA,CB,CD

[-cos2q]DC[-cosq sinq]DC

[-cos2q]CB

=

-1

0

1.5

0.5

0

EA

L

d1d2

d3d4Fourth column in Stiffness matrix


105/121

C

B

D

A d5

-0.50.50.5BD

0.50.50.5AC

010DA

001CD010BC

001AB


d4 =1k14

k24

k34

k44

k54

[S sin2q]CA,CB,CD=

EA

L[S cosq sinq]CA,CB,CD[-sin2q]CD

[-cosq sinq]CD

=

0

0

0.5

1.5

0

EA

L

[-cosq sinq]CB

d1d2

d3d4Fifth column in Stiffness matrix


106/121

C

B

D

A d5

-0.50.50.5BD

0.50.50.5AC

010DA

001CD010BC

001AB


d5 =1k15

k25

k35

k45

k55 [S cos2q]BA,BC,BD

=EA

L

[-cos2q]BD[-cosq sinq]BD

=

-.5

0.5

0

0

1.5

EA

L[-cos2q]BC[-cosq sinq]BC

k11 1.5 k12 -.5 k13 -1 k14 0 k15 -.5


107/121

k21

k31

k41

k51

=

-.5

-1

0

-.5

EA

L

k22

k32

k42

k52

=

1.5

0

0

0.5

EA

L

k23

k33

k43

k53

=

0

1.5

0.5

0

EA

L

k24

k34

k44

k54

=

0

0.5

1.5

0

EA

L

k25

k35

k45

k55

=

0.5

0

0

1.5

EA

L

k11 1.5 k12 -.5 k13 -1 k14 0 k15 -.5


108/121

k21

k31

k41

k51

=

-.5

-1

0

-.5

EA

L

k22

k32

k42

k52

=

1.5

0

0

0.5

EA

L

k23

k33

k43

k53

=

0

1.5

0.5

0

EA

L

k24

k34

k44

k54

=

0

0.5

1.5

0

EA

L

k25

k35

k45

k55

=

0.5

0

0

1.5

EA

L

1.5

=

-.5

K -1

-.5

1.5

0

-1

0

1.5

0

-.5

0

0.5

0.5

0

0

0

0.5

-.5

0.5

0

1.5

0

0

1.5

EA

L

d1d2

d3d4Force vector


109/121

C

B

D

A d5

C

B

D

100 kN

A

50 kN

F1

F2

=F3

F4

F5

50

-100

0

0

0

F = K D


110/121

=

d1

d2

d3

d4

d5

50

-100

0

0

0

1.5

-.5

-1

-.5

1.5

0

-1

0

1.5

0

-.5

0

0.5

0.5

0

0

0

0.5

-.5

0.5

0

1.5

0

0

1.5

EA

L

400

D = K-1 F


111/121

=

d1

d2

d3

d4

d5

50

-100

0

0

0

1.5

-.5

-1

-.5

1.5

0

-1

0

1.5

0

-.5

0

0.5

0.5

0

0

0

0.5

-.5

0.5

0

1.5

0

0

1.5

-1

1

400

D = K-1 F


112/121

=

d1

d2

d3

d4

d5

50

-100

0

0

0

2

.5

1.5

.5

.875

.375

1.5

.375

1.875

-.5

.5

-.125

-.125

-.625

.375

-.5

-.125

-.625

.5

-.125

.375

.875

-.125

-.125

.875

1

400

D = K-1 F


113/121

=

d1

d2

d3

d4

d5

0.1250

-0.15625

-0.09375

-0.03125

-0.09375


114/121

Summary

ModelingF d


115/121

A

B

C 4

3F1d1

F2 d2

F = K Dk11F1

F2=

k21

k12

k22

d1

d2

First column in Stiffness matrixk11F1

F2

=k21

k12

k22

d1

d2

Summary


116/121

d1 =1 d2 =0

A

B

C

4

3d1

1


k11F1

F2

=k21

k12

k22

d1

d2

Summary


117/121


d1 =0

d2 =1

F2 k21 k22 2

A

B

C

d2

1


118/121

a

b

q

F1d

1=1

F2

F3

F4


= - EA/L cos2q

= - EA/L cos qsin q


119/121

a

b

q F1

F2

F3

F4

= EA/L sin2q

= EA/L sinqcos

q

= - EA/L sin qcos q

= - EA/L sin2q

d2=1

For internal forces in truss elementsSummary


120/121

( )cos q+ ( )sin qbx Dax- Dby Day-AE

L

N=

a

b

q

Dbx

Dby

Dax

Day

[ ]

For internal forces in truss elements


121/121

Questions

Documents

Stiffness 3