07 Matrix Frame

1

! Simple Frames! Frame-Member Stiffness Matrix! Displacement and Force Transformation Matrices! Frame-Member Global Stiffness Matrix

! Special Frames! Frame-Member Global Stiffness Matrix

FRAME ANALYSIS USING THE STIFFNESSMETHOD

2

Simple Frames

3

Frame-Member Stiffness Matrix

0 0 00 - AE/LAE/L

4EI/L - 6EI/L2 2EI/L6EI/L2 00

6EI/L2 - 12EI/L3 6EI/L212EI/L3 00

0

2EI/L

-6EI/L2

0

- 6EI/L2

12EI/L3

0

-6EI/L2

4EI/L

0

6EI/L2

-12EI/L3

AE/L

0

0

-AE/L

0

0

m

x´y´

i

j

3´

5´

6´

2´

4´

1´

3´ 6´2´ 4´1´ 5´

[k´]

4´5´6´

1´2´3´

6EI/L24EI/L

6EI/L24EI/L

AE/L

AE/L

6EI/L2

6EI/L2

12EI/L3

12EI/L3

2EI/L

d 1́ = 1

AE/L

AE/L

d 2́ = 1

12EI/L3

12EI/L3

d 3́ = 1

2EI/L

6EI/L2

6EI/L2

6EI/L26EI/L2AE/L

2EI/L

6EI/L2

6EI/L2d 6́

= 1

6EI/L2

6EI/L22EI/L

6EI/L2

12EI/L3

12EI/L3

4EI/L4EI/L12EI/L3

12EI/L3

6EI/L2

6EI/L2AE/L

6EI/L2d 4́ = 1

d 5́ = 1AE/L

AE/L

4

m

i

j

m

i

j

x´y´

x

y

Displacement and Force Transformation Matrices

12

3

45

6

θyθx

4´

5´

6´

1´

2´

3´

5

λx

q4 = q4´ cos θx - q5´ cos θy

q5 = q4´ cos θy + q5´ cos θx

q6 = q6´

λy

i

jθy

θx

y´

x´

4´5´

6´

1´

2´

3´

θy

θx

m

i

j

x

y

12

3

45

6

Force Transformation

Lxx ij

x

−=λ

Lyy ij

y

−=λ

−=

6'

5'

4'

6

5

4

10000

qqq

λλλλ

qqq

xy

yx

−

−

=

6'

5'

4'

3'

2'

1'

6

5

4

3

2

1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

xy

yx

xy

yx

[ ] [ ] [ ]'qTq T=

6

[q] = [T]T[q´]

= [T]T ( [k´][d´] + [q´F] )

= [T]T [k´][d´] + [T]T [q´F]

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

Therefore, [k] = [T]T [k´][T]

[qF] = [T]T [q´F]

[q] = [T]T[q´]

[d´] = [T][d]

[k] = [T]T [k´][T]

7

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

Frame Member Global Stiffness Matrix

[k] [qF][ k ] = [ T ]T[ k´ ][T] =

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL

- ( λixλjy- λiyλjx )12EIL3

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

8

5 kN

6 m

6 m

AB

C

Example 1

For the frame shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2

9

5 kN

6 m

6 m

AB

C kN/m666.667(6m)

)m10)(60mkN1012(20012

3

462

6

3 =××

=

−

LEI

kN/m200006m

)mkN10)(200m10(600 2

626

=××

=

−

LAE

kN2000(6m)

)m10)(60mkN106(2006

2

462

6

2 =××

=

−

LEI

mkN80006m

)m10)(60mkN104(2004

462

6

•=××

=

−

LEI

mkN40006m

)m10)(60mkN102(2002

462

6

•=××

=

−

LEI

Global :

AB

C

1

2

78 9

4

6 5

12 3

10

Global :

AB

C

1

2

78 9

4

6 5

12 3

A B14´

5´ 6´

1´

2´ 3´

Local :

5´

2´

1´ 3´

4´

6´

2

Using Transformation Matrix:

� Member Stiffness Matrix

[ ]

−−−−

−−−

−

=

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

Mi

Vj

Mj

Vi

Nj

Ni

θi ∆j θj∆i δjδi

11

A B14´

5´ 6´

1´

2´ 3´

Local :

[q] = [q´]

-> [k]1 = [k´]1

Stiffness Matrix: Member 1

Global:

AB

C

1

2

78 9

4

6 5

12 3

4 6 5 1 2 34

6

5

1

2

3

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[k]1 =

12

Local:

5´

2´

1´ 3´

4´

6´

2

[q]2 = [ T ]T[ q´]2

q1´

q3´

q2´

q4´q5´

q6´

q1

q3

q2

q7

q8q9

[T]T

Stiffness Matrix: Member 2

=

123789

4´0000

0-1

5´000100

6´000001

1´0

0-1

000

2´100000

3´001000

90o

λjx = cos (-90o) = 0λjy = sin (-90o) = -1

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

Global:

AB

C

1

2

78 9

4

6 5

12 3

13

[k]2 = [ T ]T[ k´ ]2[ T ]

1´ 2´ 3´ 4´ 5´ 6´1´

2´

3´

4´

5´

6´

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[k´]2 =

1 2 3 7 8 9

666.667 20001

2

3

7

8

9

2000

0

0

-666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000

-2000

0

0

4000

4000

666.667 0

0

-2000

-2000

20000 0

0 8000

[k]2 =

14

[k]14 6 5 1 2 3

4

6

5

1

2

3

20000

0

0

-20000

0

0

0666.667

20000

-666.667

2000

02000

80000

-2000

4000

-200000

020000

0

0

0-666.667

-20000

666.667

-2000

02000

40000

-2000

8000

1 2 3 7 8 9666.667 20001

2

3

7

8

9

2000

00

666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000-2000

0

0

4000

4000

666.667 0

0

-2000

2000

20000 0

0 8000

[k]2

Global Stiffness Matrix:

20000

0

-20000

0

0

0

8000

0

-2000

4000

0

-2000

0

4000

-20000

0

0

20666.667

-2000

2000

-2000

16000

20666.667

0

2000

[K]

4

5

1

2

3

4 5 1 2 3

Global:

AB

C

1

2

78 9

4

6 5

12 3

15

AB

C

Q4 = 0Q5 = 0

Q1 = 5Q2 = 0

Q3 = 0

D4

D5

D1

D2

D3

+

0 0

0 0

0

D4

D5

D1

D2

D3

=

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

5 kN

6 m

6 m

AB

C

1

2

Global:

1

2

78 9

4

6 5

12 3

5 kN

=

4

5

1

2

3

4 5 1 2 3-20000 0 0

20666.667 00

2000

2000

20666.667 -2000

-2000 16000

0-2000

4000

08000 0 -2000 4000

-200000

20000

0

0

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

16

D4 = 0.01316

D5 = 9.199(10-4)

D6 = 0

D1 = 0.01316

D2 = -9.355(10-5)

D3 = -1.887(10-3)

15 kN

6 m

6 m

AB

C

2

q4

q5

q6

q1

q2

q3

0

0

-1.87

0

1.87

-11.22

Member 1

A B11

2 3

4

6 5

A B1

1.87 kN 11.22 kN�m1.87 kN

4 6 5 1 2 3

4

6

5

1

2

3

20000

0

0

-20000

0

0

0

666.667

2000

0

-666.667

2000

0

2000

8000

0

-2000

4000

-20000

0

0

20000

0

0

0

-666.667

-2000

0

666.667

-2000

0

2000

4000

0

-2000

8000

[q]1 = [k]1[d]1 + [qF]1

17

5 kN

1.87 kN 11.22 kN�m

5 kN1.87 kN

18.77 kN�m

Member 2

q1

q3

q2

q7

q8

q9

D1 = 0.01316

D3 = -1.887(10-3)

D2 = -9.355(10-5)

D7 = 0

D8 = 0

D9 = 0

5

11.22

-1.87

-5

1.87

18.77

1 2 3 7 8 9

666.667 20001

2

3

7

8

9

2000

0

0

-666.667

-666.667

0

0

2000

2000

20000 0

0

0

0

-20000

-20000

0

0

8000 -2000

-2000

0

0

4000

4000

666.667 0

0

-2000

-2000

20000 0

0 8000

[q]2 = [k]2[d]2 + [qF]2

15 kN

6 m

6 m

AB

C

2 1

2 3

78 9

2 2

18

AB

C

Bending moment diagram

A B1

1.87 kN 11.22 kN�m1.87 kN5 kN

6 m

6 m

AB

C

1.87 kN

18.77 kN�m5 kN

1.87 kN

AB

C

Shear diagram

5

5

+

-1.87-

+

18.77

11.22

-11.22

-

5 kN

1.87 kN 11.22 kN�m

5 kN

1.87 kN

18.77 kN�m

2

19

Deflected shape

AB

C

AB

C

Bending moment diagram +

18.77

11.22

-11.22

-

D3=-0.00189 rad

D3=-0.00189 rad

D1=13.16 mmD4=13.16 mm

D4

D5

D1

D2

D3

=

0.01316 m

0.01316 m9.199(10-4) rad

-9.355(10-5) m

-1.887(10-3) rad

Global :

AB

C

1

2

78 9

4

6 5

12 3

D5=0.00092 rad

20

4.5 m

6 m 6 m

3 kN/m

Example 2

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B(b) Determine all the reactions at supports(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.

A

B C

21

Global 2

1

1

2 3

4

56

7

8 9

2 7

8 9

1

2 3

[FEM] 9 kN�m

9 kN

9 kN�m

9 kN

3 kN/m

2

Members1

2 3

4

56 1

4.5 m

6 m 6 m

3 kN/m

22

4.5 m

6 m 6 m

3 kN/m

7.5 m

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

mkNm

mkNmL

AE

/160005.7

)/10200)(10600( 2626

=

××=

−

mkNm

mmkNLEI

/33.341)5.7(

)1060)(/10200(12123

4626

3

=

××=

−

kNm

mmkNLEI

1280)5.7(

)1060)(/10200(662

4626

2

=

××=

−

mkNm

mmkNLEI

•=

××=

−

64005.7

)1060)(/10200(44 4626

mkNm

mmkNLEI

•=

××=

−

32005.7

)1060)(/10200(22 4626

23

[ km ] = [ T ]T[ k´ ][T ] =

λx = cos θx

Member m:

θx

θy Uj

VjMj

Ui

Vi

Mim

λy = cos θy

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL


AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

24

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

4.5 m

6 m 6 m

3 kN/m

7.5 m

=

4

5

6

1

2

3

4 1

-10362.879

-7516.162

768

10362.879

768

7516.162

2

-7516.162

-5978.451

-1024

7516.162

5978.451

-1024

3

-768

1024

3200

768

-1024

6400

10362.879

-768

7516.162

-10362.879

-7516.162

-768

5

7516.162

5978.451

1024

-7516.162

-5978.451

1024

6

-768

1024

6400

768

-1024

3200

[k1]

25

λx = cos 0o = 1.0, λy = cos 90o = 0

Member 2:

2 7

8 9

1

2 3

4.5 m

6 m 6 m

3 kN/m

7.5 mmkN

mmkNm

LAE

/200006

)/10200)(10600( 2626

=

××=

−

mkNm

mmkNLEI

/667.666)6(

)1060)(/10200(12123

4626

3

=

××=

−

kNm

mmkNLEI

2000)6(

)1060)(/10200(662

4626

2

=

××=

−

mkNm

mmkNLEI

•=

××=

−

80006

)1060)(/10200(44 4626

mkNm

mmkNLEI

•=

××=

−

40006

)1060)(/10200(22 4626

26

[k2] =0

2EI/L

-6EI/L2

- 6EI/L2

6EI/L2

0

6EI/L2

-12EI/L3

0

0

0

0

-AE/L

0

0

0

6EI/L2

0

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

2EI/L

0

-6EI/L2

0 - AE/L

0

0

4EI/L

12EI/L3

4EI/L

12EI/L3

AE/L

AE/L

3

8

9

2

7

1

3 92 71 8

[k2] =0

4000

-2000

- 2000

2000

0

2000

-666.667

0

0

0

0

-20000

0

0

0

2000

0

- 2000

- 666.667

0

0

2000

4000

0

-2000

0 - 20000

0

0

8000

666.667

8000

666.667

20000

20000

3

8

9

2

7

1

3 92 71 8

2 7

8 9

1

2 3

27

=

4

5

6

1

2

3

4 1

-10362.879

-7516.162

768

10362.879

768

7516.162

2

-7516.162

-5978.451

-1024

7516.162

5978.451

-1024

3

-768

1024

3200

768

-1024

6400

10362.879

-768

7516.162

-10362.879

-7516.162

-768

5

7516.162

5978.451

1024

-7516.162

-5978.451

1024

6

-768

1024

6400

768

-1024

3200

[k1]

[k2] =0

4000

-2000

- 2000

2000

0

2000

-666.667

0

0

0

0

-20000

0

0

0

2000

0

- 2000

- 666.667

0

0

2000

4000

0

-2000

0 - 20000

0

0

8000

666.667

8000

666.667

20000

20000

3

8

9

2

7

1

3 92 71 8

28

D1

D3

D2 =4.575(10-4) m

-5.278(10-4) rad

-1.794(10-3) m

Global:

1

21

2 3

4

5 6

7

8 9

4.5 m

6 m 6 m

3 kN/m

7.5 m9 kN

9 kN�m

9 kN

0

0

0

Q1

Q3

Q2

D1

D3

D2

0

9

9+1

2

3

1

30362.9

768

7516.16

2

7516.16

6645.12

976

3

768

976

14400=

9 kN�m

9 kN

29

λx = cos θx = 6/7.5 = 0.8

Member 1:

θx

θy 1

2 3

4

56 1

λy = cos θy = 4.5/7.5 = 0.6

0

0

0

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

0

0

0

0

0

0

+ =

q4

q6

q5

q1

q2q3

4

5

6

1

2

3

5 1 2 34 6

k1

1

11.37 kN

11.37 kN

0.09 kN

1.19 kN�m

0.09 kN

0.50 kN�m 1

9.15 kN6.75 kN 1.19 kN�m

9.15 kN

6.75 kN

0.50 kN�m

9.15

0.50

6.75

-9.15

-6.75

-1.19

=

30

Member 2:

2 7

8 9

1

2 3

0

9

9

0

9

-9

+=

q1

q3

q2

q7

q8q9

D1 = 4.575(10-4)

D2 = -1.794(10-3)

D3 = -5.278(10-4)

0

0

0

1

2

3

7

8

9

2 7 8 91 3

k2

9.15

1.19

6.75

-9.15

11.25

-14.70

=

9 kN�m

9 kN

9 kN�m

9 kN

3 kN/m

2[FEM]

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

31

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

1

9.15 kN6.75 kN 1.19 kN�m

9.15 kN

6.75 kN

0.50 kN�m

3 kN/m

All Reactions

9.15 kN11.25 kN

14.70 kN�m

9.15 kN

6.75 kN

0.50 kN�m

32

-1.19

Shear diagram (kN)

Deflected shape

-0.09

-0.09

6.75

3 kN/m

29.15 kN

6.75 kN

1.19 kN�m

9.15 kN11.25 kN

14.70 kN�m

1

11.37 kN

11.37 kN

0.09 kN

1.19 kN�m

0.09 kN0.50 kN�m

D3 =-5.278(10-4) rad

D2 = -1.79 mm

D1 = 0.46 mm

D1

D3

D2 =4.575(10-4) m

-5.278(10-4) rad

-1.794(10-3) m -11.25

-

+

Bending-moment diagram (kN�m)

-14.70

0.5

33

Example 3

For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.

4.5 m

6 m 3 m

10 kN

A

BC15 kN

20 kN�m

3 kN/m

3 m

34

[FEM] 13 kN/m 11.25 kN

wL/2 = 11.25 kN

wL2/12 = 14.06 kN�m

14.06 kN�m

Global

21

2 7

8 9

1

2 3

4.5 m

6 m 3 m

10 kN

A

BC15 kN

20 kN�m

3 kN/m

3 m

7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN

λx = cos θx = 6/7.5 = 0.8

1

Members1

2 3

4

56

1

2 3

4

56

7

8 9

θy = 53.13

θx =36.87

λy = cos θy = 4.5/7.5 = 0.6

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

11.25(0.8) = 9 kN

35

Global:

1

21

2 3

4

5 6

7

8 9

Q1 = 15

Q3 = 20

Q2 = 0

D1

D3

D2

-6.75

-14.06 + 7.5

9 + 5+1

2

3

1

30362.9

768

7516.16

2

7516.16

6645.2

976

3

768

976

14400=

[FEM]13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

14.06 kN�m7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

10 kN

A

BC15 kN

20 kN�m

3 kN/m

14.06 kN�m7.5 kN�m

6.75 kN

5 kN9 kN

15 kN20 kN�m

36

D1

D3

D2 =1.751(10-3) m

2.049(10-3) rad

-4.388(10-3) m

1

21

2 3

4

5 6

7

8 9

37

Member 1:

0

0

0

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

-6.75

14.06

9

-6.75

9

-14.06

+ =

q4

q6

q5

q1

q2q3

4

5

6

1

2

3

5 1 2 34 6

k1

6.51

26.46

24.17

-20.01

-6.17

4.89

=

[FEM]

13 kN/m 11.25 kN

11.25 kN

14.06 kN�m

14.06 kN�m

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

θx

θy 1

2 3

4

56 1

λx = cos 36.87o = 0.8, λy = cos 53.13o = 0.6

14.06 kN�m

14.06 kN�m

6.75 kN

9 kN

11.25(0.6) = 6.75 kN

9 kN

38

3 kN/m

7.5 m1

19.71 kN

19.71 kN

7.07 kN

4.89 kN�m

15.43 kN

26.46 kN�m

=

q4

q6

q5

q1

q2q3

6.51

26.46

24.17

-20.01

-6.17

4.89

θx = 36.87o

θy = 53.13o 1

2 3

4

56 1

3 kN/m

7.5 m

20.01 kN

4.89 kN�m

6.51 kN

24.17 kN

26.46 kN�m

6.17 kN

39

Member 2:

2 7

8 9

1

2 3

0

7.5

5

0

5

-7.5

+=

q1

q3

q2

q7

q8q9

D1 = 1.751(10-3)

D2 = -4.388(10-3)

D3 = 2.049(10-3)

0

0

0

1

2

3

7

8

9

2 7 8 91 3

k2

35.02

15.12

6.17

-35.02

3.83

-8.08

=

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

[FEM]

7.5 kN�m

5 kN

7.5 kN�m

5 kN

2

10 kN 2

10 kN

7.5 kN�m

5 kN

7.5 kN�m

5 kN

40

3 kN/m

7.5 m

20.01 kN

4.89 kN�m

6.51 kN

24.17 kN

26.46 kN�m

6.17 kN

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

2

10 kN

10 kN

A

BC15 kN

20 kN�m

3 kN/m

6 m 3 m3 m

6.51 kN

24.17 kN

26.46 kN�m

35.02 kN3.83 kN

8.08 kN�m

41

Shear diagram (kN)


3 kN/m

7.5 m1

19.71 kN

19.71 kN

7.07 kN

4.89 kN�m

15.43 kN

26.46 kN�m

35.02 kN

6.17 kN

15.12 kN�m

35.02 kN3.83 kN

8.08 kN�m

2

10 kN

15.43D1

D3

D2 =1.751(10-3) m

2.049(10-3) rad

-4.388(10-3) m

Deflected shape

D3 = 2.05(10-3) rad

D2 = -4.39 mm

D1 =1.75 mm

-26.46

4.89

-15.12 -8.08

-7.07

6.17

-3.83

42

Special Frames

43

i j

θjθi 4 *

5 * 6*

1*

2*3 *

λix = cos θi

λiy = sin θi

λjx = cos θj

λjy = sin θj

[ q* ] = [ T ]T[ q´ ]

1

Stiffness matrix

i j

4´

5´ 6´

1´

2´3´

[ T ]T

−

−

=

6'

5'

4'

3'

2'

1'

*6

5*

*4

*3

2*

*1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

jxjy

jyjx

ixiy

iyix1*2*3*4*5*6*

1´ 2´ 3´ 4´ 5´ 6´

44

[ ]

−

−

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

T

1´2´3´4´5´6´

1* 2* 3* 4* 5* 6*

� Member Stiffness Matrix

1´2´3´4´5´6´

1´ 2´ 3´ 4´ 5´ 6´

[ ]

−−−−

−−−

−

=



LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

45

[ k ] = [ T ]T[ k´ ][T] =

Ui

Vi

Mi

Uj

Vj

Mj

Vj Mj

- λiy6EIL2

λix6EIL2

2EIL

λjy6EIL2

- λjx6EIL2

4EIL

Ui Vi Mi

- λiy6EIL2

λix6EIL2

4EIL

λjy6EIL2

λjx6EIL2

-

2EIL

Uj

AEL

- λixλiy)(12EIL3

AEL

λiy2 +

12EIL3

λix2 )(

λix6EIL2

λix6EIL2

AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

λixλjx +

12EIL3

λiyλjy)-(

λjy6EIL2

AEL

λjx2 +

12EIL3

λjy2 )(

λjy6EIL2

λjx6EIL2

-

AEL

- λjxλjy)(12EIL3

- λjx6EIL2

AEL

λixλjy -

12EIL3

λiyλjx)-(

AEL

- λixλiy)(12EIL3

- λiy6EIL2

- λiy6EIL2

AEL

λixλjx +

12EIL3

λiy λjy)-(

)(AEL

λix2 +

12EIL3

λiy2

AEL

λiyλjy +

12EIL3

λix λjx )-(

AEL

λiyλjy +

12EIL3

λixλjx)-(

12EIL3

λjx2 )

AEL


AEL

λiyλjx -

12EIL3

λixλjy)-(

AEL

- λjxλjy)(12EIL3

AEL λjy

2 + (

46

40 kN

4 m 4 m

3 m

7.416 m

22.02 o

20 kN200 kN�m

6 kN/m

Example 4

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagrams and qualitativedeflected shape.Take I = 200(106) mm4 , A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

47

1Local

12

40 kN

4 m 4 m

3 m

7.416 m

22.02 o

20 kN200 kN�m

6 kN/m

4

5 6

7

8 922.02 o

Global

2

1´

2´ 3´

4´

5´ 6´

1´

2´3´

4´

5´ 6´

1 *

2* 3 *

48

mkNm

mkNmL

AE /101508

)/10200)(006.0( 3262

×=×

=

mkNm

mmkNLEI

•×=×

= 3426

10208

)0002.0)(/10200(44

mkNLEI

•×= 310102

kNm

mmkNLEI 3

2

426

2 1075.3)8(

)0002.0)(/10200(66×=

×=

mkNm

mmkNLEI /109375.0

)8()0002.0)(/10200(1212 3

3

426

3 ×=×

=

[ ]

−−−−

−−−

−

=



LAEAE/L

/4/60/2/60/6/120/6/120

00/00//2/60/4/60/6/120/6/120

00/00

k'

22

2323

22

2323

Mi

Vj

Mj

Vi

Nj

Ni


49

2

1´

2´ 3´

4´

5´ 6´

1´

2´3´

1 4´

5´ 6´

Local

8 m

8 m

03750

0

- 3750- 937.5

0

0375010000

0-3750

0 - 150000

00

0

10000

-3750

- 3750

3750

0

3750

-937.5

0

0

00

-150000

0

0

[ k´]1 = [ k´]2 =20000

937.5

20000

937.5

150000

150000


Mi

Vj

Mj

Vi

Nj

Ni

50

θi =22.02o

Member 1:

1

1 *

2 * 3 *

4

5 6

Global1´

2´3´

1 4´

5´ 6´Local

θj = 0o

λjx = cos (0o) = 1,λjy = sin (0o) = 0

λix = cos (22.02o) = 0.927,λiy = sin (22.02o) = 0.375

40 kN40 kN�m40 kN�m

20 kN20 kN

1

[FEM]

q1´

q3´

q2´

q4´

q5´

q6´

q*1

q*3

q*2

q4

q5q6

[ T ]1T

00

0

00

0

00

0

000

000

000

1 0

1

01

00

0 0

0.9270.375

- 0.375 0.927

10 0

00

[ q* ] = [ T ]T[ q´ ]

=

1´ 4´ 5´ 6´2´ 3´

1*

2*

3*

4*

5*

6*

51

θi =22.02o1

1 *

2 * 3 *

4

5 6

Global1´

2´3´

1 4´

5´ 6´Local

θj = 0o

[ k* ]1 = [ T ]1T[ k´ ]1[ T ]1

[ k* ]1 = 103

1*2*3*456

1*

129.04651.811-1.406

-139.0580.351-1.406

2*

51.811

21.892

3.476-56.240

-0.8693.476

3*

-1.4063.47620.000.00-3.7510.00

4

-139.058-56.240

01500

0

5

0.351-0.869-3.75

00.938-3.75

6

-1.4063.47610.00

0-3.75

20

52

θi =22.02o1

1 *

2 * 3 *

4

5 6

Globalθj = 0o

40 kN40 kN�m40 kN�m

20 kN20 kN

1

[FEM]

[ qF* ] = [ T ]T[ qF´]

[ qF*] = [ T ]1T

0

40

20

0

20

-40

40 kN40 kN�m

20 kN

1

40 kN�m

18.547.5

-7.5018.5440020-40

=

1*

2*

3*

4

5

6

53

Member 2

q1´

q3´

q2´

q4´

q5´

q6´

q4

q6

q5

q7

q8q9

[ T ]2T

0.927-0.375

0.3750.927

1

00

0

00

0

00

0

000

000

000

1

0.927-0.375

0.3750.927

00

0 0

0 0

00

[ q ] = [ T ]T[ q´ ]

=

1´ 4´ 5´ 6´2´ 3´

456

78

9

6 kN/m2

24 kN

24 kN

32 kN�m

32 kN�m

[ q´F]

2

1´

2´ 3´

4´

5´ 6´

[ q´]

λix = λjx = cos (-22.02o) = 0.927,λiy = λjy = sin (-22.02o) = -0.375

2

45 6

[ q ]7

8 922.02o

22.02o

54

2

1´

2´ 3´

4´

5´ 6´

[ q´ ]

[ k ]2 = [ T ]2T[ k´]2[ T ]2

2

45 6

[ q ]7

8 922.02o

22.02o

[ k ]2 = 103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

55

6 kN/m2

24 kN

24 kN

32 kN�m

32 kN�m

[ q´F ]

[ qF* ] = [ T ]T[ qF´ ]

[ qF ] = [ T ]2T

0

32

24

0

24

-32

8.99822.249328.99822.249-32

=

4

5

6

7

8

9

6 kN/m2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998kN

22.249 kN

8.998 kN

2

45 6

[ q ]7

8 922.02o

22.02o

56

[ k* ]1 = 103

1*2*3*456

1*129.04651.811-1.406

-139.0580.351-1.406

2*51.811

21.892

3.476-56.240-0.8693.476

3*-1.4063.47620.000.00-3.7510.00

4-139.058-56.240

01500

0

50.351-0.869-3.75

00.938-3.75

6-1.4063.47610.00

0-3.75

20

Global Stiffness:

12

1 *

2* 3 *

4

5 6

7

8 9

2*

7

8 9

[ k ]2 = 103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

57

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

1 *

2 * 3 *

4

5 6

7

8 9

6 kN/m2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998 kN

22.249 kN

8.998 kN

40 kN40 kN�m

20 kN

1

40 kN�m

18.54 kN7.5

[ q*F ]

2 *

7

8 9

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

40 kN�m

7.5

40 kN�m 32 kN�m

20 kN 22.249 kN

8.998 kN

20 kN200 kN�m

D1*

D4

D3*

D5

D6

= 0.0

= 0.0

= 0.0

= -20

= -200

Q1*

Q4

Q3*

Q5

Q6

= 103

1*

3*

4

5

6

1*

129.046

-1.406

-139.058

0.352

-1.406

3*

-1.406

20

0

-3.75

10

4

-139.058

0

279.046

-51.811

1.406

5

0.351

-3.75

-51.811

22.829

-0.274

6-1.406

10

1.406

-0.274

40

+ 8.998

-7.5

40

20 +22.249

-40 + 32

58

Global: 40 kN 20 kN200 kN�m

6 kN/m 12

1 *

2 * 3 *

4

5 6

7

8 9

2 *

7

8 9

D1*

D4

D3*

D5

D6

=

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 rad

59

1

1 *

2 * 3 *

4

5 6

[ q ]

40 kN 40 kN�m

20 kN

1

40 kN�m

18.547.5

[ qF* ]Member 1: [ q*] = [ k*][ d*] + [ qF*]

q1*

q3*

q2*

q4

q5q6

D1*=-0.0205

D3*=-0.0112D2*= 0.0

D4= -0.0191D5= -0.0476D6=-0.0024

-7.50

40

18.54

0

20

-40

+

q1*

q3*

q2*

q4

q5

q6

0

022.63 kN

-8.49 kN19.02 kN7.87 kN�m

=

40 kN

22.63 kN

8.49 kN

19.02 kN

7.87 kN�m

= 103

1*2*3*456

1*

129.04651.811-1.406

-139.0580.351-1.406

2*

51.811

21.892

3.476-56.240

-0.8693.476

3*

-1.4063.47620.000.00-3.7510.00

4

-139.058-56.240

01500

0

5

0.351-0.869-3.75

00.938-3.75

6

-1.4063.47610.00

0-3.75

20

60

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

q4

q6

q5

q7

q8q9

8.998

3222.249

8.99822.249

-32

+

2

4

5 6

[ q ] 7

8 96 kN/m

2

32 kN�m

32 kN�m

[ qF ]

22.249 kN

8.998 kN

22.249 kN

8.998 kN

D4= -0.0191

D6=-0.0024D5= -0.0476

D7 = 0D8 = 0D9 = 0

q4

q6

q5

q7

q8

q9

8.49 kN

-207.87 kN�m-39.02 kN

9.51 kN83.51 kN-248.04 kN�m

= 8.49 kN

39.02 kN

207.87 kN�m

9.51 kN83.51 kN

248.04 kN�m

6 kN/m

=103

456

78

9

4

129.046-51.811

1.406-129.046

51.8111.4056

5

-51.81121.8923.476

51.811-21.892

3.476

6

1.4063.476

20-1.406-3.476

10

7

-129.04651.811-1.406

129.046-51.811-1.406

8

51.811-21.892-3.476

-51.81121.892-3.476

9

1.4063.476

10-1.406-3.476

20

61

40 kN

22.63 kN

8.49 kN

19.02 kN

7.87 kN�m

8.49 kN

39.02 kN

207.87 kN�m

9.51 kN

83.51 kN

248.04 kN�m6 kN/m

22.5 kN81 kN

22.5 kN

33 kN

8.49 kN

20.98 kN


-

+

207.87

-248.04

83.56

+7.87

Deflected shape

D3*=-0.0112 rad

D1*=- 20.5 mm

D6=-0.0024 rad

D1*

D4

D3*

D5

D6

=

-0.0205 m

-0.0112 rad

-0.0191 m

-0.0476 m

-0.0024 radD4=-19.1 mmD5=-47.6 mm

62

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

Example 5

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members.

63

Global

1

2 3

4

5 6

7*9*

8*

1

2 2

1´2´

3´

4´5´

6´

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

Local

1´

2´ 3´

4´

5´ 6´1

76.31o

64

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

1

2 3

4

5 61

Member 1:

kN/m103000m4

)kN/m10)(200m10(60

3

2623

×=

××=

−

LAE

mkN1080m4

)m10)(400kN/m104(2004

3

4626

•×=

××=

−

LEI

mkN1040m4

)m10)(400kN/m102(2002

3

4626

•×=

××=

−

LEI

kN1030m)(4

)m10)(400kN/m106(2006

3

2

4626

2

×=

××=

−

LEI

kN/m1015m)(4

)m10)(400kN/m1012(20012

3

3

4626

3

×=

××=

−

LEI

65

d1

d3

d2

d4

d5

d6

q1

q3

q2

q4

q5q6

0

26.67

40

040

-26.67

+= 103

123

45

6

1 4 5 6

030

40

0-30

80

3000

0

0

-30000

0

2

015

30

0-15

30

3

030

80

0-30

40

-30000

0

3000

0

0

0-15

-30

015

-30

3 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C

26.67 kN�m26.67 kN�m

40 kN40 kN

1

20 kN/m

Member 1: [ q ] = [ k ][ d ] + [ qF ]

1

2 3

4

5 61

56.31o

76.31o

66

Member 2:

2

1´2´

3´

4´5´

6´

45 6

7*9*

8*

2

i

j

kN/m103324m3.61

)kN/m10)(200m10(60

3

2623

×=

××=

−

LAE

mkN1088.64m3.61

)m10)(400kN/m104(2004

3

4626

•×=

××=

−

LEI

kN1036.83m)(3.61

)m10)(400kN/m106(2006

3

2

4626

2

×=

××=

−

LEI

kN/m1020.41m)(3.61

)m10)(400kN/m1012(20012

3

3

4626

3

×=

××=

−

LEI

mkN1044.32m3.61

)m10)(400kN/m102(2002

3

4626

•×=

××=

−

LEI

[ k´ ]2 = 103

1´2´3´4´5´6´

1´ 4´ 5´ 6´3324

00

-332400

2´0

20.4136.83

0-20.4136.83

3´0

36.8388.64

0-36.8344.32

-332400

3324

00

0-20.41-36.83

020.41

-36.83

036.8344.32

0-36.8388.64

67

45 6

7*9*

8*

2

i

j 76.31oλjx = cos (-76.31o) = 0.24,λjy = sin (-76.31o) = -0.97

[ q* ] = [ T ]T[ q´ ]

q1´

q3´

q2´

q4´q5´q6´

q4

q6

q5

q7*q8*q9*

=

4567*8*9*

1´ 4´000

0.24

0-0.97

5´000

0.970.24

0

6´000001

0.55

0-0.83

000

2´0.830.55

0000

3´001000

λix = cos (-56.31o) = 0.55,λiy = sin (-56.31o) = -0.83

56.31o

2

i

j

1´2´

3´

4´5´

6´

[ k* ]2 = [ T ]T[ k´]2[ T ] = 103

4567*8*9*

4 7* 8* 9*5 6-452.884643.585-35.786205.452

-35.786-759.668

1787.474-2689.923

-8.717-759.6683139.053

-8.717

30.64620.43144.321

-35.786-8.71788.643

1036.923

30.646-1524.780

-452.8841787.474

30.646

-1524.7802307.582

20.431643.585

-2689.92320.431

30.64620.43188.643

-35.786-8.71744.321

68

[k]1 = 103

123456

1 4 5 60

30400

-3080

3000

00

-300000

20

1530

0-1530

30

30800

-3040

-300000

3000

00

0-15-300

15-30

1

2 3

4

5 6

7*9*

8*

1

2

[k*]2 = 103

4567*8*9*

4 7* 8* 9*5 6-452.884643.585-35.786205.452

-35.786-759.668

1787.474-2689.923

-8.717-759.6683139.053

-8.717

30.64620.43144.321

-35.786-8.71788.643

1036.923

30.646-1524.780

-452.8841787.474

30.646

-1524.7802307.582

20.431643.585

-2689.92320.431

30.64620.43188.643

-35.786-8.71744.321

69

Global:

= -50= 0= -80= 0= 0

Q4

Q6

Q5

Q7*

Q9*

D4

D6

D5

D7*

D9*

0

-26.6740

00

+

D4

D6

D5

D7*

D9*

-2.199x10-5 m

-2.840x10-4 rad-3.095x10-4 m

0.979x10-3 m6.161x10-4 rad

=

1

2 3

4

5 6

7*9*

8*

1

23 m

80 kN�m20 kN/m

20o

50 kN

4 m 2 m

A B

C 26.67 kN�m

26.67 kN�m

40 kN40 kN

1

20 kN/m

80 kN�m50 kN

= 103

45679

4 7* 9*5 6

205.452

-452.884643.585-35.786

-35.786

30.646-9.56944.321

-35.78688.643

-452.884

4036.923

30.646-1524.780

30.646

-1524.780232.582

-9.569643.58520.431

30.646-9.569

168.643-35.78644.321

70

q1

q3

q2

q4

q5q6

0

26.67

40

040

-26.67

+= 103

123

45

6

1 4 5 6

030

40

0-30

80

3000

0

0

-30000

0

2

015

30

0-15

30

3

030

80

0-30

40

-30000

0

3000

0

0

0-15

-30

015

-30

Member 1:

q1

q3

q2

q4

q5q6

65.97 kN

24.59 kN�m

36.12 kN

-65.97 kN

43.88 kN-40.11 kN�m

=

0

0

0

D4

D5

D6

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

1

20 kN/m

65.97 kN65.97 kN

1

2 3

4

5 61

71

q4

q6

q5

q7*

q8*q9*

Member 2:

q4

q6

q5

q7*

q8*q9*

15.97 kN

-39.89 kN�m

-43.88 kN

0 kN

46.69 kN0 kN�m

=

= 103

456

7*8*

9*

4 7* 8* 9*5 6-455.21651.27

-35.73

210.67

-35.73

-769.1

1769.34-2678.93

-8.84

-769.13128.82

-8.84

30.5720.26

44.32

-35.73-8.84

88.64

1036.923

30.57-1508.14

-455.211769.34

30.57

-1508.142296.15

20.26

651.27-2678.93

20.26

30.5720.26

88.64

-35.73-8.84

44.32

D4

D6

D5

D*7

0

D*9

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

2

45 6

7*9*

8*

2

i

j

72


+

40.11 kN�m24.59 kN�m

43.88 kN36.12 kN

1

20 kN/m

65.97 kN65.97 kN

15.97 kN43.88 kN 39.89 kN�m

46.69 kN

2

-24.59-

-40.11

- +

39.89

D4

D6

D5

D7*D9*

-2.199x10-5 m

-2.840x10-4 rad-3.095x10-4 m

0.979x10-3 m6.161x10-4 rad

=

Deflected shape

D4=-2.2x10-5 m

D5=-3.1x10-4 m

D7*=0.979x10-3 mD6 = -2.84x10-4 rad

D9*=6.161x10-4 rad

1

2 3

4

5 6

7*9*

8*

1

2

07 Matrix Frame

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