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1 ! Fundamentals of the Stiffness Method ! Member Local Stiffness Matrix ! Displacement and Force Transformation Matrices ! Member Global Stiffness Matrix ! Application of the Stiffness Method for Truss Analysis ! Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors ! Space-Truss Analysis TRUSSES ANALYSIS

# TRUSSES ANALYSIS · Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors! Space-Truss Analysis TRUSSES ANALYSIS. 2 2-Dimension Trusses. 3 Fundamentals of the Stiffness

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• 1

! Fundamentals of the Stiffness Method! Member Local Stiffness Matrix! Displacement and Force Transformation Matrices! Member Global Stiffness Matrix! Application of the Stiffness Method for Truss

Analysis! Trusses Having Inclined Supports, Thermal Changes

and Fabrication Errors! Space-Truss Analysis

TRUSSES ANALYSIS

• 2

2-Dimension Trusses

• 3

Fundamentals of the Stiffness Method

Node and Member Identification

Global and Member Coordinates

Degrees of Freedom

12

3 4

2

1

3

(x1, y1)

(x3, y3)

(x2, y2)

(x4, y4)

1

23

4

56

78

34

56

78

x

y

Known degrees of freedom D3, D4, D5, D6, D7 and D8 Unknown degrees of freedom D1 and D2

• 4

AE/LAE/L

x djAE/L x d´j

AE/L

x diAE/LAE/L x d´i

Member Local Stiffness Matrix

x´y´

i

j

q´i

q´j

AE/L

AE/L

jii dLAEd

LAEq ''' −=

AE/LAE/Ld´ i =

1

d´ j = 1

x di

x dj

jij dLAEd

LAEq ''' +−=

−=

j

i

j

i

dd

LAE

qq

''

1111

''

−=

1111

]'[L

AEk

[q´] = [k´][d´] ----------(1)

q´j

q´ix´

x´y´

• 5

x´y´

m

i

j

(xi,yi)

(xj,yj)x

y

Displacement and Force Transformation Matrices

θyθx

22 )()(cos

ijij

ijijxx

yyxx

xxL

xx

−+−

−=

−== θλ

22 )()(cos

ijij

ijijyy

yyxx

yyL

yy

−+−

−=

−== θλ

• 6

x

y

Global

m

i

jdjx

djy

dix

diy

djx

djyx´y´

m

i

j

Local

d´i

d´j

dix

diy

d´j

d´i

Displacement Transformation Matrices

yiyxixi ddd θθ coscos' +=

=

jy

jx

iy

ix

yx

yx

j

i

dddd

dd

λλλλ00

00''

θyθx

yjyxjxj ddd θθ coscos' +=

=

yx

yxTλλ

λλ00

00][

λx λy

[d´] = [T][d] ----------(2)

• 7

x

y

θyθx

m

i

j

x

y

Global

x´y´

m

i

j

Local

q´i

q´j

Force Transformation Matrices

θyθx

xiix qq θcos'=

yiiy qq θcos'=

xjjx qq θcos'=

yjjy qq θcos'=

=

j

i

y

x

y

x

jy

jx

iy

ix

qq

qqqq

''

00

00

λλ

λλ

=

y

x

y

x

TT

where

λλ

λλ

00

00

][

λx

λy

[q] = [T]T[q´] ----------(3)

qjx

qjy

qix

qiy

• 8

Member Global Stiffness Matrix

[ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF]

[ k ] [ k ] = [ T ]T[ k´ ][T]

[qF] = [ T ]T [q´F]

[q] = [T]T[q´] ----------(3)

Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result,

λyλx

λxλx

−λyλx

−λxλx

λyλy

λxλy

−λyλy

−λxλyλyλx

λxλx

−λyλx

−λxλx

λyλy

λxλy

−λyλy

−λxλy

V U VU

[ k ] = AELVUV

U

00λyλxλyλx00

-11

1-1

AEL[ k

] =

00

λy

λx

λy

λx

00

• 9

[Qa] = [K][D] + [QF]

[Qk] = [K11][Du] + [K12][Dk] + [QF]

Reaction Boundary Condition

Equilibrium Equation:

Partitioned Form:

Application of the Stiffness Method for Truss Analysis

[Du] = (([Qk] - [QF]) - [K12][Dk])[Ku]-1

Qk

Qu

Du

Dk=

K12

K22

K11

K21

+QFk

QFu

• 10

+

−=

jF

iF

j

i

j

i

qq

dd

LAE

qq

''

''

1111

''

jy

jx

iy

ix

yx

yx

dddd

λλλλ00

00

+

−=

jF

iF

jy

jx

iy

ix

yx

yx

j

i

qq

DDDD

LAE

qq

''

0000

1111

''

λλλλ

Member Forces

x

y

θyθx

x´y´

m

i

j

q´i

q´j

• 11

+

−−

−−=

jF

iF

jy

jx

iy

ix

yxyx

yxyx

j

i

qq

DDDD

LAE

qq

''

''

λλλλλλλλ

Dyi

Dxi

Dyj

Dxjq´j =

AEL −λx −λy λx λy qj´

F+

x

y

θyθx

x´y´

m

i

j

Member Forces

q´i

q´j

• 12

Member Forces

Dyi

Dxi

Dyj

Dxjqm =

AEL −λx −λy λx λy qj´

F+

x

y

θyθx

x´y´

m

i

j

Member Forces

qm

• 13

3 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

Example 1

For the truss shown, use the stiffness method to:(a) Determine the deflections of the loaded joint.(b) Determine the end forces of each member and reactions at supports.Assume EA to be the same for each member.

• 14

Ljyy

Lixx ijij

ij

)()( −+−

cosθx = λx cosθy = λy

1

2

3

434

56

78

1

2

(0,0)(-4,-3)

(-4,3)

(4,-3)

31

23 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

Member

#1

#2

λx λy

#3

λyλx

λxλxλyλy

λxλy

λyλx

λxλxλyλy

λxλy

−λyλx

−λxλx−λyλy

−λxλy

−λyλx

−λxλx−λyλy

−λxλy

Vi Uj VjUi

[ k ]m = AEL

ViUjVj

Ui

31

2

-4/5 = -0.8

-4/5 = -0.8

4/5 = 0.8

-3/5 = -0.6

3/5 = 0.6

-3/5 = -0.6

• 15

-0.48

1.08-0.48

1.92-0.48

0.64

0.36

-0.481

2

1 2

[K] = AE5

1

2

3

434

56

78

1

2

31

2

0.48

0.64

0.36

0.48-0.48

-0.64

-0.36

-0.48

-0.48

-0.64

-0.36

-0.48[ k ]1 =

2 3 41

AE5

23

4

1

[ k ]2 =

2 5 61

AE5

25

6

1

-0.48

0.64

0.36

-0.480.48

-0.64

-0.36

0.48

0.48

-0.64

-0.36

0.48

[ k ]3 =

2 7 81

AE5

27

8

1

-0.48

0.64

0.36

-0.480.48

-0.64

-0.36

0.48

0.48

-0.64

-0.36

0.48

31

2

34

56

78

Member

#1

#2

#3

λx

-0.8

- 0.8

0.8

λy

-0.6

0.6

-0.6

λx2 λx λy λy2

0.64 0.48 0.36

0.64 -0.48 0.36

0.64 -0.48 0.36

0.48

0.64

0.36

0.48

-0.48

0.64

0.36

-0.48

• 16

34

56

78

1

2

31

2

Global

Q1 = -50

Q2 = -80

D1

D2

D1

D2 =-250.65/AE

-481.77/AE

3 m

3 m

4 m 4 m

80 kN

50 kN

5 m

5 m5 m

1

2

1

-0.48

2

-0.48

1.08

1.92

= AE5

80 kN

50 kN

0

0+1

2

• 17

1

2

3

434

56

78

1

2

31

2

Local

= -97.9 kN (C)

[q´F]1 = AE5 0.8 0.6 -0.8 -0.6

D2=D1=

D4=D3=

-481.77/AE-250.65/AE

0.00.0

= +17.7 kN (T)

D2=D1=

D6=D5=

-481.77/AE-250.65/AE

0.00.0

= -17.7 kN (C)

80 kN

50 kN 36.87o

97.9 kN

17.7 kN

17.7 kN

#1 -0.8 -0.6#2 -0.8 0.6#3 0.8 -0.6

[q´F]2 = AE5 0.8 -0.6 -0.8 0.6

[q´F]3 = AE5 -0.8 +0.6 +0.8 -0.6

D2=D1=

D8=D7=

-481.77/AE-250.65/AE

0.00.0

Member

#1#2#3

λx λy

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 18

80 kN

50 kN 36.89o

97.9 kN

17.7 kN

17.7 kN

80 kN

50 kN

Member

#1#2#3

λx-0.8-0.80.8

λy-0.60.6-0.6

1

2

3

434

56

78

1

2

31

2

97.9(0.8)=78.32 kN

97.9(0.6)=58.74 kN

17.7(0.8)=14.16 kN

17.7(0.6)=10.62 kN

17.7(0.8)=14.16 kN

17.7(0.6)=10.62 kN

ΣFx ´ = 0:+ 17.7 + 17.7 +50cos 36.89 - 97.9cos73.78 - 80cos53.11 = 0, O.K

Check :

• 19

Example 2

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the deflections of the loaded joint.The support B settles downward 2.5 mm. Temperature in member BDincrease 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.

A

BC

D

8 kN

4 kN

4 m

3 m+20

o C

∆B = 2.5 mm

• 20

λyλx

λxλxλyλy

λxλy

λyλx

λxλxλyλy

λxλy

−λyλx

−λxλx−λyλy

−λxλy

−λyλx

−λxλx−λyλy

−λxλy

∆B = 2.5 mm

A

BC

D

8 kN

4 kN

4 m

3 m+20

o C

Vi Uj VjUi

[ k ]m = AEL

ViUjVj

Ui

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

Member

#1

#2

λx λy

#3

-4/4 = -1

-4/5 = -0.8

0

0

-3/5 = -0.6

-3/3 = -1

Ljyy

Lixx ijij

ij

)()( −+−

cosθx = λx cosθy = λy

• 21

0.096

0.128

0.072

0.096

0

0

0.333

0

0

0.25

0

0

0

0.25

0

00

-0.25

0

0

0

-0.25

0

0[k]1 = 8x103

2 3 41

234

1

[k]3 = 8x103

2 7 81

278

1

0

0

0.333

00

0

-0.333

0

0

0

-0.333

0

Member

#1

#2

#3

λx

-1

- 0.8

0

λy

0

-0.6

-1

λx2/L λx λy/L λy2/L

0.25 0 0

0.128 0.096 0.072

0 0 0.333

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[k]2 = 8x103

2 5 61

256

1

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

1

23

4

56

78

[K] = 8x103

21

21

0.096

0.378

0.405

0.096

• 22

∆B = 2.5 mm

1.536 kN

1.152 kN

2+20oC1.536 kN

1.152 kN

2+20oC

1.92 kΝ =α(∆T1)AE = (12x10-6)(20)(8x103)

1.92 kN

1.536 kN

1.152 kN

1.536 kN

1.152 kN

12

34

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

∆B = 2.5 mm

+20o C

-1.536

1.536-1.152

1.152

+

1

2

6

5

q1q2

q1

q5

q2

q6

-1.536

-1.152+1

2

0

-2.5x10-35

6-0.096-0.128

-0.072

-0.096+ 8x1035 6

= 8x103

21

21

0.096

0.128

0.072

0.096 d1d2

d1

d5 = 0d2

d6 = -2.5x10-3

0.096

0.128

0.072

0.096

= 8x103

2 5 61

256

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

[q] = [k]m[d] + [qF]Member 2:

q1q2

-1.536

-1.152+1

2= 8x103

21

0.096

0.128

0.072

0.096 d1d2

1.92

1.44+

• 23

[Q] = [K][D] + [QF]

Global:

A

BC

D

8 kN

4 kN

4 m

3 m+20

o C

∆B = 2.5 mm

12

3 4

(0,0)

(-4,-3)

(-4,0)

(0,-3)

2

1

3

1

23

4

56

78

Q1 = -4Q2 = -8

= 8x103

21

21

0.096

0.378

0.405

0.096 D1D2

-1.536

-1.152+1.92

1.44+

D1D2

-0.8514x10-3 m

-2.356x10-3 m=

• 24

Member

#1

#2

#3

λx λy

Local12

3 4

2

1

3

1

23

4

56

78

[q´F]1 = 8x103 1.0 0.0 -1.0 0.04

= -1.70 kN (C)

D2=D1=

D4=D3=

0.00.0

-0.8514x10-3

-2.356x10-3

-1.92+[q´F]2 = 8x103 0.8 0.6 -0.8 -0.65

= -2.87 kN (C)

D2=D1=

D6=D5=

-0.00250.0

-0.8514x10-3

-2.356x10-3

= -6.28 kN (C)

D2=D1=

D8=D7=

0.00.0

-0.8514x10-3

-2.356x10-3[q´F]3 = 8x103 0.0 1.0 0.0 -1.0

3

2+20oC

1.92 kN

1.92 kN

-1 0

- 0.8 -0.6

0 -1

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 25

4 kN

8 kN

6.28 kN

1.70 kN

2.87 kN

12

3 4

2

1

3

1

23

4

56

78

Member

#1

#2

#3

cosθx

-1

- 0.8

0

cosθy

0

-0.6

-1

[q´]m

-1.70

-2.87

-6.28

4 kN

8 kN

2

1

3

1.70 kN

6.28 kN

2.87(0.8) = 2.30 kN

2.87(0.6) = 1.72 kN

• 26

∆ AD = +

3 mm

AB

C

3 m

D

8 kN

4 kN

4 m 4 m∆ = - 4 m

m

Example 3

For the truss shown, use the stiffness method to:(a) Determine the end forces of each member and reactions at supports.(b) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

• 27

∆ AD = +

3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = - 4 mm

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

λyλx

λxλxλyλy

λxλy

λyλx

λxλxλyλy

λxλy

−λyλx

−λxλx−λyλy

−λxλy

−λyλx

−λxλx−λyλy

−λxλy

Vi Uj VjUi

[ k ]m = AEL

ViUjVj

Ui

Member

#1

#2

λx λy

#3

#4

#5

-4/5 =-0.8

0

4/5 = 0.8

4/4 = 1

4/4 = 1

-3/5 = -0.6

-3/3 = -1

-3/5 = -0.6

0

0

Ljyy

Lixx ijij

ij

)()( −+−

cosθx = λx cosθy = λy

• 28

[k]1 = 8x103

2 3 41

234

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

0.096

0.128

0.072

0.096

[k]2 = 8x103

2 5 61

256

1

0

0

0.333

00

0

-0.333

0

0

0

0.333

0

0

0

-0.333

0

[k]3 = 8x103

2 7 81

278

1

-0.096

0.128

0.072

-0.0960.096

-0.128

-0.072

0.096

0.096

-0.128

-0.072

0.096-0.096

0.128

0.072

-0.096

Member

#1

#2

λy

#3

λx-0.8

0

0.8

-0.6

-1

-0.6

λx2/L λy2/Lλxλy/L

0.128 0.0720.096

0 0.3330

0.128 0.072-0.096

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

5 m

3 m 5 m

4 m 4 m

• 29

21

4

3

5

1

2

34

56

78

1

2 3 4

(0,0)

(-4,-3) (4,-3)(0,-3)

5 m

3 m 5 m

4 m 4 m

Member

#4

#5

λyλx1

1

0

0

λx2/L λy2/Lλxλy/L

0

0.25

0

00

-0.25

0

0

0

0.25

0

0

0

-0.25

0

0

0

0.25

0

00

-0.25

0

0

0

0.25

0

0

0

-0.25

0

0[k]5= 8x103

6 7 85

678

5

0.25 00

0.25 00

Global Stiffness Matrix

[K] = 8x103

2 5 71

257

1

34 6

8

[k]4= 8x103

4 5 63

456

3

• 30

Global Stiffness Matrix

0.2560.0 0.477

0.0 0.00.0

0.0 0.0-0.128 0.096

-0.1280.096

0.50-0.25

-0.250.378

[K] = 8x103

2 5 71

257

1

[k]1 = 8x103

2 3 41

234

1

0.096

0.128

0.072

0.096

-0.096

-0.128

-0.072

-0.096-0.096

-0.128

-0.072

-0.096

0.096

0.128

0.072

0.096

[k]2 = 8x103

2 5 61

256

1

0

0

0.333

00

0

-0.333

0

0

0

0.333

0

0

0

-0.333

0

[k]3 = 8x103

2 7 81

278

1

-0.096

0.128

0.072

-0.0960.096

-0.128

-0.072

0.096

0.096

-0.128

-0.072

0.096-0.096

0.128

0.072

-0.096

0

0.250

-0.25

0

00

0

0

0.25

0

-0.250

0

0

0[k]4= 8x103

4 5 63

456

3

0

0.250

-0.25

0

00

0

0

0.25

0

-0.250

0

0

0[k]5= 8x103

6 7 85

678

5

• 31

∆AE/L = 10.67 kN

∆ AD = +

3 mm1

3.84 kN

2.88 kN

3.84 kN

2.88 kN

∆ AD = +

3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = - 4 mm

1

2

21

4

3

534

56

78

Global Fixed end forces

0.00.0

257

1∆AE/L = 4.8 kN

4.8 kN

∆AE/L = 10.67 kN

10.67 kN∆ = -4 m

m

2 Fixed End3.84 kN

2.88 kN-3.84-2.88 + 10.67 = 7.79

• 32

∆ AD = +

3 mm

AB

C

D

8 kN

4 kN

3 m

4 m 4 m

∆ = -4 mm

Global:

1

2

21

4

3

534

56

78

[Q] = [K][D] + [QF]

Q1 = 4

Q5 = 0Q2 = -8

Q7 = 0

= 8x103

2 5 71

257

1

-0.250.500.00.0

0.378-0.250.096

-0.128

-0.1280.00.0

0.256

0.0960.0

0.4770.0 D1

D5

D2

D7

-3.84

0.07.79

0.0

+

D1

D5D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

8 kN

4 kN

• 33

D1

D5D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

-4.8+D2

D1

00

= -1.54 kN (C)

1

2

21

4

3

534

56

78

Member forces

= -3.17 kN (C)

10.67+

[q´F]1 = 8x103 0.8 0.6 -0.8 -0.65

[q´F]2 = 8x103 0.0 1.0 0.0 -1.03

D2

D1

0D5

Member

#1

#2

λiyλix

-0.8 -0.6

0 -1

1

4.8 kN

= + 3 m

m

10.67 kN

10.67 kN

2

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 34

D1

D5D2

D7

6.4426x10-3 m

2.6144x10-3 m-5.1902x10-3 m

5.2288x10-3 m

=

1

2

21

4

3

534

56

78

00

0D5

[q´F]4 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

0D5

0D7

[q´F]5 = 8x103 -1.0 0.0 1.0 0.04

= 5.23 kN (T)

D2

D1

0D7

[q´F]3 = 8x103 -0.8 0.6 0.8 -0.65

= -6.54 kN (C)

Member λx λy#3#4

#5

0.8 -0.61 0

1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 35

1

2

21

4

3

534

56

78

-0.8

0

0.81

1

-0.6

-1

-0.60

0

Member

#1

#2

λx

#3#4

#5

λy [q´]

4 kN8 kN

3.17 kN

3.17 kN

6.54 kN

6.54 kN

1.54 kN

1.54 kN

5.23 kN5.23 kN

4 kN8 kN

0.92 kN

4 kN

3.17 kN 3.92 kN

-1.54

-3.17

-6.545.23

5.23

• 36

Special Trusses (Inclined roller supports)

• 37

λix = cos θiλiy = sin θi

λjx = cos θjλjy = sin θj

[ q* ] = [ T ]T[ q´ ] 1

2

3*

4*5

6

78

3

2

4

5

1

x *

y *

θi

x

y

θj

1

i

j

q´i

q´j

q*3

q1

q*4

q2

q´iq´j

[T]T

00λiyλixλjyλjx00

[ T ] = [[ T ]T]T =

=00

λiy

λix

λjy

λjx

00

1

i

j1

2

3*

4*

Transformation Matrices

• 38

[ k ] = [ T ]T[ k´ ][T]

λjyλjx

λjxλjx

−λiyλjx

−λixλjx

λjyλjy

λjxλjy

−λiyλjy

−λixλjyλiyλix

λixλix

−λjyλix

−λjxλix

λiyλiy

λixλiy

−λjyλiy

−λjxλiy

Vi Uj VjUi

[ k ]m = AEL

ViUjVj

Ui

00λiyλixλjyλjx00

-11

1-1

AEL[ k

]m =

00

λiy

λix

λjy

λjx

00

• 39

Example 5

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

3 m

4 m

45o

30 kN

• 40

3 m

4 m45o

2

1

3

1

2

5

63*

4*

Member 1:

15

63*

4*

[q*]θi = 0,λix = cos 0 = 1,λiy = sin 0 = 0

θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

[q*] = [T*]T[q´] + [T*]T[q´F]

q5

q3*

q6

q4*

q´iq´j

[T*]T

=0001

-0.7070.707

00

q´i q´ji j

1

• 41

15

63*

4*

[q*]θi = 0 ;λix = cos 0 = 1,λiy = sin 0 = 0

θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

0-0.707

00.707

00

10

AE4

1-1

-11[k*]1 =

0001

-0.7070.707

00

Member 1:

[k*]1 = AE63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

q´i q´ji j

1

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[ ]TL

AETk T

−=

1111

][*][

• 42

Member 2:q´i

q´j

i

j

22

1

2

3*4*

θi = -90o = 270o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

90o+45o =135o

90o

θj = -135o = 215o ,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

0-0.707

0-0.707

-10

00

AE3

1-1

-11[k*]2 =

00-10

-0.707-0.707

00

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[ ]TL

AETk T

−=

1111

][*][

• 43

[ ]TL

AETk T

−=

1111

][][

36.87o

Member 3:

θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

00.6

00.8

0.60

0.80

AE5

1-1

-11[k]3 =

00

0.60.8

0.60.800

3

1

2

5

63

q´i

q´j

i

j36.87o

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m45o

2

1

3

1

2

5

63*

4*

• 44

3 m

4 m45o

2

1

3

1

2

5

63*

4*

[k*]1 = AE63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

5

6 4*

Global Stiffness:

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1[K] = AE 2

3*

11

00.0960.128

2

-0.23570.40530.096

0.2917-0.2357

03*

• 45

Global :

[Q] = [K][D] + [QF]

Q1 = 30

Q3*= 0

Q2 = 0D1

D3*D2 = AE 2

3*

11

00.0960.128

2

-0.23570.40530.096

0.2917-0.2357

03*

D1

D3*D2 =

352.5

-127.3-157.5AE

1

3 m

4 m45o

30 kN

3 m

4 m45o

2

1

3

1

2

5

63*

4*

5

6 4*

• 46

[q´F]1 = AE00

0D3*

-1 0 0.707 -0.7074

= -22.50 kN, (C)

D2

D1

0D3*

[q´F]2 = AE 0 1 -0.707 -0.7073

= -22.50 kN, (C)

00

D2

D1[q´F]3 = AE -0.8 -0.6 0.8 0.6

5 = 37.50 kN, (T)

D1

D3*D2 =

352.5

-127.3-157.5AE

1

Member Forces :

Member

#1

λiyλix λjx λjy

#2

#3

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

3 m

4 m45o

2

1

3

1

2

5

63*

4*

5

6 4*

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 47

36.87o 45o45o

3 m

4 m45o

2

1

3

1

2

5

63*

4*

Member

Member Force (kN)

[q´]2[q´]1 [q´]3

-22.50 -22.50 37.50

22.50 kN

22.50 kN

37.50 kN

22.50 kN

7.50 kN31.82 kN

Reactions :

3 m

4 m45o

30 kN

• 48

Example 6

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.

3 m

4 m

45o

30 kN

4 m

• 49

15

63*

4*

[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0

θij = -45oλix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

[q*] = [T*]T[q´] + [T*]T[q´F]

q5

q3*

q6

q4*

q´iq´j

[T*]T

=0001

-0.7070.707

00

q´i q´ji j

1

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78Member 1:

• 50

3 m

4 m 4 m

15

63*

4*

[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0

θij = -45o = 315o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707

45o

0-0.707

00.707

00

10

AE4

1-1

-11[k*]1 =

0001

-0.7070.707

00

[k*]1 = AE63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

q´i q´ji j

1

2

1

3

4

5

12

5

6 3*4*

78Member 1:

[ ]TL

AETk T

−=

1111

][*][

• 51

q´i

q´j

i

j

22

1

2

3*4*

θi = -90o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1

90o

θj = -135o = 215o,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707

0-0.707

0-0.707

-10

00

AE3

1-1

-11[k*]2 =

00-10

-0.707-0.707

00

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78Member 2:

[ ]TL

AETk T

−=

1111

][*][

90o+45o =135o

• 52

36.87o

Member 3:

θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6

00.6

00.8

0.60

0.80

AE5

1-1

-11[k]3 =

00

0.60.8

0.60.800

3

1

2

5

63

q´i

q´j

i

j36.87o

[k]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

[ ]TL

AETk T

−=

1111

][*][

• 53

3 m

4 m 4 m

θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0

00

01

00

10

AE4

1-1

-11[k]1 =

0001

0100

q´i q´ji j

1

2

1

3

4

5

12

5

6 3*4*

78Member 4:

47

8

[q]1

2

[k]4 = AE278

17

00.25

0-0.25

8

0000

1

0-0.25

00.25

2

0000

[ ]TL

AETk T

−=

1111

][*][

• 54

Member 5:

θi = - 8.13o;λix = cos (- 8.13o) = 0.9899,λiy = sin(- 8.13o) = -0.1414

00.6

00.8

-0.14140

0.98990

AE5

1-1

-11

5

q´i

q´j

i

j36.87o

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

53*4*

7

8

[k*]5 =00

-0.14140.9899

0.60.800

[k*]5 = AE

4* 83*

4*78

3*

0.0960.128

0.02263-0.1584

0.0720.096

0.01697-0.1188

-0.1188-0.1584-0.0280.196

0.016970.022630.004-0.028

7

θ j = 36.

87o ;

λ jx = cos

(36.87

o ) = 0.8,

λ jy = sin

(36.87

o ) = 0.6

[ ]TL

AETk T

−=

1111

][*][

8.13o

• 550 -0.2357-0.2357

0

[k*]1 = AE63*4*

53*

-0.1250.125

0-0.1768

4*

0.125-0.125

00.1768

5

0.1768-0.1768

00.25

6

0000

[k*]3 = AE

6 25

612

5

0.0960.128-0.096-0.128

0.0720.096-0.072-0.096

-0.096-0.1280.0960.128

-0.072-0.0960.0720.096

1

3 m

4 m 4 m

2

1

3

4

5

Global Stiffness:1

2

5

6 3*4*

78

5

6 4*

78

[k*]2 = AE

2 4*1

23*4*

1

0.16670.1667-0.2357

0

0.16670.1667-0.2357

0

0000

-0.2357-0.23570.3333

03*

[k]4 = AE278

17

00.25

0-0.25

8

0000

1

0-0.25

00.25

2

0000

[k*]5 = AE

4* 83*

4*78

3*

0.0960.128

0.02263-0.1584

0.0720.096

0.01697-0.1188

-0.1188-0.1584-0.0280.196

0.016970.022630.004-0.028

7

[K] = AE 23*

11 2

0.0960.378

0.40530.096

3*

0.4877

• 56

Global :30 kN

4 m 4 m

3 m2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

[Q] = [K][D] + [QF]

Q1 = 30

Q3*= 0

Q2 = 0D1

D3*D2

D1

D3*D2 =

86.612

-13.791-28.535AE

1

= AE 23*

11

00.0960.378

2

-0.23570.40530.096

0.4877-0.2357

03*

• 57

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

D1

D3*D2 =

86.612

-13.791-28.535AE

1

[q´F]1 = AE00

0D3*

-1 0 0.707 -0.7074

= -2.44 kN, (C)

D2

D1

0D3*

[q´F]2 = AE 0 1 -0.707 -0.7073

= -6.26 kN, (C)

00

D2

D1[q´F]3 = AE -0.8 -0.6 0.8 0.6

5 = 10.43 kN, (T)

Member Forces :

Member

#1

λiyλix λjx λjy

#2

#3

1 0 0.707 -0.707

0 -1 -0.707 -0.707

0.8 0.6 0.8 0.6

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 58

Member

#4

λiyλix λjx λjy

#5

3 m

4 m 4 m

2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

D1

D3*D2 =

86.612

-13.791-28.535AE

1

D2

D1

00

[q´F]4 = AE -1 0 1 04

= -21.65 kN, (C)

0D3*

00

[q´F]5 = AE -0.9899 0.141 0.8 0.65

= 2.73 kN, (T)1 0 1 0

0.9899 -0.141 0.8 0.6

Member Forces :

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 59

36.87o 45o45o

81.87o

36.87o

Member

Member Force (kN)

[q´]1 [q´]2 [q´]3 [q´]4 [q´]5

-2.44 -6.26 10.43 -21.65 2.73

Reactions :30 kN

4 m 4 m

3 m2

1

3

4

5

12

5

6 3*4*

78

5

6 4*

78

6.26 kN10.43 kN

21.65 kN

2.73 kN

2.44 kN 2.44 kN5.90 kN

6.26 kN

19.47 kN

1.64 kN

6.54 kN

• 60

AB

C

3 m

D

8 kN

4 kN

4 m36.87o

4 m

λjyλjx

λjxλjx

−λiyλjx

−λixλjx

λjyλjy

λjxλjy

−λiyλjy

−λixλjyλiyλix

λixλix

−λjyλix

−λjxλix

λiyλiy

λixλiy

−λjyλiy

−λjxλiy

Vi Uj VjUi

[ k *]m = AEL

ViUjVj

Ui

Example 7

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint.Take AE = 8(103) kN.

• 61

Member 1:

λix = cos 73.74o = 0.28,λiy = sin 73.74o = 0.96

[q*] = [T*]T[q´] + [T*]T[q´F]

λjx = cos 36.87o = 0.8,λjy = sin 36.87o = 0.6

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

x *

y *

73.74o

x

y

36.87o

1

i

j1

2

3*

4*1

i

j

q´i

q´jq3*

q1

q4*

q2

q´iq´j=

00

0.960.28

0.60.8

00

[T*]T

3

2

4

5

1

1

2

56

78

3*

4*

• 62

[k]1 = 8x103

4* 1 23*

4*12

3*

0.0960.128

-0.1536-0.0448

0.0720.096

-0.1152-0.0336

-0.0336-0.04480.053760.01568

-0.1152-0.15360.184320.05376

000.960.280.60.800

[k]1 =00

0.960.28

0.60.8

00

8x1035

1-1

-11

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

1

2

56

78

3*

4*

[ ]TL

AETk T

−=

1111

][][

• 63

x *

y*

36.87x

yMember 2:

λix = cos 36.87o = 0.8,λiy = sin 36.87o = 0.6

[q*] = [T*]T[q´] + [T*]T[q´F]

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λjx = cos 0o = 1,λjy = sin 0o = 0

q3*

q5q4*

q6

q´iq´j

[T*]T

[k] = [TT] AEL

[T]1-1

-11

q´i q´ji j

2

=00

0.60.8

0100

i j2 5

6

3*

4*

[k]2 = 8x103

4* 5 63*

4*56

3*

00.25

-0.15-0.2

0000

0-0.20.120.16

0-0.150.090.12

3

2

4

5

1

1

2

56

78

3*

4*

• 64

x

y

270o 1

2

784

Member 3:

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λx = cos 323.13o = 0.8,λy = sin 323.13o = -0.6

Member 4:

[k]4 = 8x103

2 7 81

278

1

-0.0960.1280.096

-0.128

0.072-0.096-0.0720.096

0.096-0.128-0.0960.128

-0.0720.0960.072

-0.096

λx = cos 270o = 0, λy = sin 270o = -1

[k]3 = 8x103

0000

0.3330

-0.3330

00

00

2 5 61

256

10.333

0

-0.3330

x

y

323.13o3

1

2

56

3

2

4

5

1

1

2

56

78

3*

4*

• 65

x

yMember 5:

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

λx = cos 0o = 1,λy = sin 0o = 0

00.25

0-0.25

0000

00.25

0-0.25

00

00

[k]5 = 8x103

6 7 85

678

5

5 78

56

3

2

4

5

1

1

2

56

78

3*

4*

• 660.17568

-0.2-0.20.5

00

0 0

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

[k]1 = 8x103

4* 1 23*

4*12

3*

0.0960.128

-0.1536-0.0448

0.0720.096

-0.1152-0.0336

-0.0336-0.04480.053760.01568

-0.1152-0.15360.184320.05376

[k]4 = 8x103

2 7 81

278

1

-0.0960.1280.096

-0.128

0.072-0.096-0.0720.096

0.096-0.128-0.0960.128

-0.0720.0960.072

-0.096

00.25

0-0.25

0000

00.25

0-0.25

00

00

[k]5 = 8x103

6 7 85

678

5

[k]2 = 8x103

4* 5 63*

4*56

3*

00.25

-0.15-0.2

0000

0-0.20.120.16

0-0.150.090.12

[k]3 = 8x103

0000

0.3330

-0.3330

00

00

2 5 61

256

10.333

0

-0.3330

[K] = 8x103

2 3* 51

23*5

1

3

2

4

5

1

12

56

78

3*

4* 6

784

*

• 67

3

2

4

5

1

2

3*

4*5

6

78A

BC

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

1

Global: [Q] = [K][D] + [QF]

D1

D3*D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

Q1 = 4

Q3*= 0

Q2 = -8

Q5 = 0

D1

D3*

D2

D50.17568

-0.2-0.20.5

00

0 0

-0.0336-0.0448

-0.0448 -0.03360.0

0.2560.4740.0

= 8x103

2 3* 51

23*

5

1

8 kN

4 kN

• 68

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

2

3*

4*5

6

78

1

Member forces

D1

D3*D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

0D3*

D2

D1[q´F]1 = 8x103 -0.28 -0.96 0.8 0.6

5 = 0.46 kN, (T)

0D3*

0D5

[q´F]2 = 8x103 -0.8 -0.6 1 04

= -0.16 kN, (C)

Member

#1

#2

λiyλix λjx λjy

0.28 0.96 0.8 0.6

0.8 0.6 1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 69

D1

D*3D2

D5

1.988x10-3 m

1.996x10-4 m-2.0824x10-3 m

7.984x10-5 m

=

3

2

4

5

1

2

3*

4*5

6

78

1

[q´F]3 = 8x103 D2

D1

0D5

0 1 0 -13

= -5.55 kN

D2

D1

00

[q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65

= -4.54 kN

0D5

00

[q´F]5 = 8x103 -1 0 1 04

= -0.16 kN

Member

#3

λiyλix λjx λjy

#4

#5

0 -1 0 -1

0.8 -0.6 0.8 -0.6

1 0 1 0

[ ] [ ]F

yj

xj

yi

xi

xxyxmF q

DDDD

LAEq ']'[ +

−−= λλλλ

• 70

y*

x *

36.87o

36.87o 36.87o

AB

C

3 m

D

8 kN

4 kN

4 m

36.87o

4 m

3

2

4

5

1

2

3*

4*5

6

78

1

0.46 -0.16 -5.55 -4.54 -0.16

0.46 kN

5.55 kN0.36 kN

4.54 kN

3.79 kN

2.72 kN

Member

Member Force (kN)

[q]2[q]1 [q]3 [q]4 [q]5

0.16 kN 0.16 kN

5.55 kN

• 71

Space-Truss Analysis

• 72

+

−=

Fj

Fi

j

i

j

i

qq

dd

LEA

qq

''

''

1111

''

[ ] [ ]F

jz

jy

jx

iz

iy

ix

q

dddddd

TL

EA '1111

+

−=

[ ]

=

zyx

zyxT

where

λλλλλλ000

000,

[q´] = [k´][d´] + [q´F]

= [k´][T][d] + [q´F]

Member Local Stiffness [k´]:

• 73

Member Global Stiffness [km]:

[km]= [T]T[k´] [T]

[ ]

=zyx

zyx

z

y

x

z

y

x

m LEAk

λλλλλλ

λλλ

λλλ

000000

1111

000

000

• 74

Global equilibrium matrix:

[Q] = [K][D] + [QF]

QI

QII

Du

Dk

KI,II

KII,II

KI,I

KII,I

Reaction Support Boundary Condition

QFI

QFII= +

Fixed End Forces

• 75

q´j

q´ii

j

m

i

j

m

=EAL

λxλx

λzλx

λyλx

λxλyλyλyλzλy

λxλzλyλzλzλz

λxλx

λzλx

λyλx

λxλyλyλyλzλy

λxλzλyλzλzλz

−λxλx

−λzλx

−λyλx

−λxλy−λyλy−λzλy

−λxλz−λyλz−λzλz

−λxλx

−λzλx

−λyλx

−λxλy−λyλy−λzλy

−λxλz−λyλz−λzλz

diydix

djxdiz

djydjz

qiyqix

qjx

qiz

qjyqjz

qjy

qjx

qjzqiy

qix

qiz

+

−=

Fj

Fi

j

i

j

i

qq

dd

LEA

qq

''

''

1111

''

[ ] [ ] [ ]F

jz

jy

jx

iz

iy

ix

zyxzyxmjq

dddddd

LEAq '' +

−−−= λλλλλλ

• 76

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

Example 6

For the truss shown, use the stiffness method to:(b) Determine the end forces of each member.(a) Determine the deflections of the loaded joint.Take E = 200 GPa, A = 1000 mm2.

• 77

λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k

= -0.3578 i + 0.2683 j - 0.8944 k

λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k= +0.3578 i + 0.2683 j - 0.8944 k

λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k

= +0.3578 i - 0.2683 j - 0.8944 k

= -0.3578 i - 0.2683 j - 0.8944 k

λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k

Member

#1

#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944

+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944

-0.3578 -0.2683 -0.8944

λm = λxi + λyj + λzk

12

34

12

3

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

1

2

53

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

• 78

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

Member Stiffness Matrix [k]6x6

[k]m =[k12]3x3

[k22]3x3

[k11]3x3

[k21]3x3

2 31

23

1

+0.80-0.240

+0.320

+0.320-0.096

+0.128

-0.240+0.072

-0.096

[k11]1 =AEL

2 31

23

1

+0.80-0.240

-0.320

-0.320+0.096

+0.128

-0.240+0.072

+0.096

[k11]2 =AEL

2 31

23

1

+0.80+0.240

-0.320

-0.320-0.096

+0.128

+0.240+0.072

-0.096

[k11]3 =AEL

2 31

23

1

+0.80+0.240

+0.320

+0.320+0.096

+0.128

+0.240+0.072

+0.096

[k11]4 =AEL

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

[KI,I] =AEL

• 79

12

34

12

3

(0, 0, 10)

(-4, 3, 0)

(4, 3, 0)

(4, -3, 0)

(-4, -3, 0)4

1

2

53

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

0.0-80

60

D3

D2

D1

0.00.0

0.0+

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

=AEL

[Q] = [K][D] + [QF]

• 80

Global equilibrium matrix:

[Q] = [K][D] + [QF]

QI

QII

Du

Dk

KI,II

KII,II

KI,I

KII,I

Reaction Support Boundary Condition

QFI

QFII= +

Fixed End Forces

(AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN

0.0-80

60

D3

D2

D1

0.00.0

0.0+

2 31

23

1

3.20.0

0.0

0.00.0

0.512

0.00.288

0.0

=AEL

0.0 mm-15.53 mm

6.551 mm=

D3

D2

D1= LAE

0.0-277.8

+117.2

• 81

Member forces:

q´F+

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

dyi

dxi

dxj

dzi

dyjdzj

[q´j]m = AEL −λx −λy −λz λx λy λz

D3D2D1

[ 0 ]

= +116.5 kN (T)

+0.3578 -0.2683 +0.8944[q´j]1 =AEL

0.0

117.2

-277.8L

AE

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

• 82

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

= +32.61 kN (T)-0.3578 -0.2683 +0.8944[q´j]2 =AEL

0.0

117.2

-277.8L

AE

= -116.5 kN (T)-0.3578 +0.2683 +0.8944[q´j]3 =AEL

0.0

117.2

-277.8L

AE

= -32.61 kN (T)+0.3578 +0.2683 +0.8944[q´j]4 =AEL

0.0

117.2

-277.8L

AE

x

z

y

4 m4 m

3 m3 m

10 m

80 kN60 kN

O

• 83

5

32.6 kN116.5 kN

116.5 kN

32.6 kN

Member

#1#2

#3

λx λy λz

#4

-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944

+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944

[q´j]m116.5 32.6

-32.6-116.5

12

34

80 kN60 kN

R5x = (-32.6)(-0.3578) = 11.66 kN

R5y = (-32.6)(-0.2683) = 8.75 kN

R5z = (-32.6)(-0.8944) = 29.16 kN

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