Matrix beam

148
1 ! Development: The Slope-Deflection Equations ! Stiffness Matrix ! General Procedures ! Internal Hinges ! Temperature Effects ! Force & Displacement Transformation ! Skew Roller Support BEAM ANALYSIS USING THE STIFFNESS METHOD

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Transcript of Matrix beam

Page 1: Matrix beam

1

! Development: The Slope-DeflectionEquations

! Stiffness Matrix! General Procedures! Internal Hinges! Temperature Effects! Force & Displacement Transformation! Skew Roller Support

BEAM ANALYSIS USING THE STIFFNESSMETHOD

Page 2: Matrix beam

2

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 3: Matrix beam

3

� Degrees of Freedom

L

θΑ

A B

M

1 DOF: θΑ

PθΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

Page 4: Matrix beam

4

L

A B

1

� Stiffness Definition

kBAkAA

LEIkAA

4=

LEIkBA

2=

Page 5: Matrix beam

5

L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

Page 6: Matrix beam

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� Fixed-End ForcesFixed-End Forces: Loads

P

L/2 L/2

L

w

L

8PL

8PL

2P

2P

12

2wL12

2wL

2wL

2wL

Page 7: Matrix beam

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� General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 8: Matrix beam

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wP

settlement = ∆j

(MFij)∆ (MF

ji)∆

(MFij)Load (MF

ji)Load

+

Mij Mji

θi

θj

+

i jwPMij

Mji

settlement = ∆jθj

θi

ψ

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

+=

,)()()2()4( LoadijF

ijF

jiij MMLEI

LEIM +++= ∆θθ Loadji

Fji

Fjiji MM

LEI

LEIM )()()4()2( +++= ∆θθ

L

Page 9: Matrix beam

9

Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

� Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

Page 10: Matrix beam

10

+

1

1

i jMij Mji

θi

θj

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

iθ×

jθ×

� Stiffness Coefficients

L

Page 11: Matrix beam

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[ ]

=

jjji

ijii

kkkk

k

Stiffness Matrix

)()2()4( ijF

jiij MLEI

LEIM ++= θθ

)()4()2( jiF

jiji MLEI

LEIM ++= θθ

+

=

F

ji

Fij

j

iI

ji

ij

MM

LEILEILEILEI

MM

θθ

)/4()/2()/2()/4(

� Matrix Formulation

Page 12: Matrix beam

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[D] = [K]-1([Q] - [FEM])

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

ij)Load (MFji)Load

+

+

i jwPMij

Mji

θj

θi

ψ ∆j

Mij Mji

θi

θj

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

L

Page 13: Matrix beam

13

L

Real beam

Conjugate beam

� Stiffness Coefficients Derivation

MjMi

L/3

θi

LMM ji +

EIM j

EIMi

θι

EILM j

2

EILMi

2

)1(2

0)3

2)(2

()3

)(2

(:0'

−−−=

=+−=Σ+

ji

jii

MM

LEI

LMLEI

LMM

)2(0)2

()2

(:0 −−−=+−=Σ↑+EI

LMEI

LMF jiiy θ ij

ii

LEIM

LEIM

andFrom

θ

θ

)2(

)4(

);2()1(

=

=

LMM ji +

i j

Page 14: Matrix beam

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� Derivation of Fixed-End Moment

Real beam

8,0

162

22:0

2 PLMEI

PLEI

MLEI

MLFy ==+−−=Σ↑+

P

M

M EIM

Conjugate beamA

EIM

B

L

P

A B

EIM

EIML2

EIM

EIML2

EIPL

16

2

EIPL4 EI

PL16

2

Point load

Page 15: Matrix beam

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L

P

841616PLPLPLPL

=+−

+−

8PL

8PL

2P

2PP/2

P/2

-PL/8 -PL/8

PL/8

-PL/16-PL/8

-

PL/4+

-PL/16-PL/8-

Page 16: Matrix beam

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Uniform load

L

w

A B

w

M

M

Real beam Conjugate beam

A

EIM

EIM

B

12,0

242

22:0

23 wLMEI

wLEI

MLEI

MLFy ==+−−=Σ↑+

EIwL8

2

EIwL

24

3

EIwL

24

3

EIM

EIML2

EIM

EIML2

Page 17: Matrix beam

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Settlements

M

M

L

MjMi = Mj

LMM ji +L

MM ji +

Real beam

2

6LEIM ∆

=

Conjugate beam

EIM

A B

EIM

,0)3

2)(2

()3

)(2

(:0 =+−∆−=Σ+L

EIMLL

EIMLM B

EIM

EIML2

EIMEI

ML2

Page 18: Matrix beam

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� Typical Problem

0

0

0

0

A C

B

P1P2

L1 L2

wCB

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI

LEIM BABA −++= θθ

128024 2

222

22

wLLPLEI

LEIM CBBC ++++= θθ

128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

P

8PL

8PL

L

w12

2wL

12

2wL

L

Page 19: Matrix beam

19

MBA MBC

A C

B

P1P2

L1 L2

wCB

B

CB

8042 11

11

LPLEI

LEIM BABA −++= θθ

128024 2

222

22

wLLPLEI

LEIM CBBC ++++= θθ

BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0

Page 20: Matrix beam

20

A C

B

P1P2

L1 L2

wCB

Substitute θB in MAB, MBA, MBC, MCB

MABMBA

MBC

MCB

0

0

0

0

8024 11

11

LPLEI

LEIM BAAB +++= θθ

8042 11

11

LPLEI

LEIM BABA −++= θθ

128024 2

222

22

wLLPLEI

LEIM CBBC ++++= θθ

128042 2

222

22

wLLPLEI

LEIM CBCB −

−+++= θθ

Page 21: Matrix beam

21

A BP1

MAB

MBA

L1

A CB

P1P2

L1 L2

wCB

ByR CyAy ByL

Ay Cy

MABMBA

MBC

MCB

By = ByL + ByR

B CP2

MBCMCB

L2

Page 22: Matrix beam

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2EI EI

Stiffness Matrix

� Node and Member Identification

� Global and Member Coordinates

� Degrees of Freedom

�Known degrees of freedom D4, D5, D6, D7, D8 and D9� Unknown degrees of freedom D1, D2 and D3

1 21 2 31

2 3

7

8

9

4

5

6

1

2 3

7

8

9

4

5

6

Page 23: Matrix beam

23

i j

E, I, A, L

Beam-Member Stiffness Matrix

d4 = 1

AE/LAE/L AE/L AE/L

64

5

1

2

3

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

d1 = 1

k11 = = k44

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

k41 k14

Page 24: Matrix beam

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d2 = 1

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

6EI/L2

6EI/L2

d5 = 1

6EI/L2

6EI/L2

12EI/L312EI/L3 12EI/L3

12EI/L3

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

k22 = = k55

= k52

k62 =

= k32 = k65

= k25

= k35

i j

E, I, A, L6

4

5

1

2

3

Page 25: Matrix beam

25

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

- AE/LAE/L

-AE/L AE/L

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

4EI/L

6EI/L2 6EI/L2

2EI/L

0

0

2EI/L

-6EI/L2

0

0

-6EI/L2

4EI/L

d6 = 1

d3 = 1 2EI/L 4EI/L

6EI/L26EI/L26EI/L2

4EI/L 2EI/L

6EI/L2

k33 =

k23 = = k53

= k63 = k36

= k56 k26 =

= k66

i j

E, I, A, L6

4

5

1

2

3

Page 26: Matrix beam

26

FFxi

FFyi

MFi

FFxj

FFyj

MFj

+

MFi MF

j

FFxjFF

xi

FFyi FF

yj

1

6EI/L2

6EI/L2

12EI/L312EI/L3

+

x ∆i

12EI/L3

6EI/L2

6EI/L2

12EI/L3

1

+

4EI/L 2EI/L

6EI/L26EI/L2

x θi

6EI/L2

4EI/L 2EI/L

6EI/L2

x δjAE/L

1

AE/L

+

AE/L AE/L

6EI/L2

6EI/L2

1

12EI/L312EI/L3

x ∆j

6EI/L2

6EI/L2

12EI/L3

12EI/L3

1

+

2EI/L

6EI/L26EI/L2

4EI/L

x θj

6EI/L2

2EI/L

6EI/L2

4EI/L

Fxi

Fyi

MiFxj

Fyj

Mj

� Member Equilibrium Equations

Mj

i jFxj

Fxi

Fyi Fyj

=

MiE, I, A, L

x δiAE/L AE/L

1

AE/LAE/L

Page 27: Matrix beam

27

+

−−−−

−−−

=

Fj

Fyi

Fxj

Fi

Fyi

Fxi

j

j

j

i

i

i

j

yj

xj

i

yj

xi

MFFMFF

θ∆δθ∆δ

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

MFFMFF

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

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

00/00/

22

2323

22

2323

Fjjjjiiij

Fyjjjjiiiyj

Fxijjjiiixj

Fijjjiiixi

Fyijjjiiiyi

Fxijjjiiixi

MθLEI∆LEIδθLEI∆LEIδMFθLEI∆δθLEI∆LEIδFFθ∆δLAEθ∆δLAEFMθLEI∆LEIδθLEI∆LEIδMFθLEI∆LEIδθLEI∆LEIδFFθ∆δLAEθ∆δLAEF

)/4()/6()0()/2()/6()0()/6()0()0()/6()/12()0(

)0()0()/()0()0()/()/2()/6()0()/4()/6()0()/6()/12()0()/6()/12()0(

)0()0()/()0()0()/(

22

223

22

2323

−=−−−=

−=−=−+=

+++−++=

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

Fixed-end force matrix

End-force matrix

Stiffness matrix

Displacement matrix

Page 28: Matrix beam

28

6x6 Stiffness Matrix

[ ]

−−−−

−−−

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

k

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

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

00/00/

22

2323

22

2323

66

θi ∆j θj∆i δjδi

Mi

Vj

Mj

Vi

Nj

Ni

4x4 Stiffness Matrix

[ ]

−−−−

−−

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

θi ∆j θj∆i

Mi

Vj

Mj

Vi

Page 29: Matrix beam

29

Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included.

2x2 Stiffness Matrix

θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

Page 30: Matrix beam

30

21 312

1

2

3

4

5

6

General Procedures: Application of the Stiffness Method for Beam Analysis

wP P2y

M2M1

Local

1

2

Global

Page 31: Matrix beam

31

Member

1

2

3

4

1

3

4

5

6

2

1

2

3

4

5

6

3121 2

Global

wP P2y

M2M1

Local

1

2

Page 32: Matrix beam

32

(q´F4 )2

(q´F3 )2

(q´F1)2

(q´F2)2

w

2

P

(q´F4)1

(q´F3)1

(q´F2)1

(q´F1 )1

1[FEF]

1

2

3

4

5

6

3121 2

Global

Local

1

2

wP P2y

M2M1

Page 33: Matrix beam

33

[q] = [T]T[q´]

1

2

3

4

5

6

3121 2

Global

Local

1

[k] = [T]T[q´] [T]

=

'4

'3

'2

'1

4

3

2

1

1000010000100001

qqqq

qqqq

1´ 4´2´ 3´12341

Page 34: Matrix beam

34

[q] = [T]T[q´]

1

2

3

4

5

6

3121 2

Global

Local

2

[k] = [T]T[q´] [T]

=

'4

'3

'2

'1

6

5

4

3

1000010000100001

qqqq

qqqq

1´ 4´2´ 3´34562

Page 35: Matrix beam

35

Stiffness Matrix:

21 312

1

2

3

4

5

6

[k]1 =

1

2

3

4

1 2 3 4

[k]2 =

3

4

5

6

3 4 5 6

Member 1

[K] =

1

23

4

5

6

1 2 3 4 5 6

Member 2

Node 2

1 2

Page 36: Matrix beam

36

+

=

F

F

F

F

F

F

QQQQQQ

DDDDDD

QQQQQQ

6

5

1

4

3

2

6

5

1

4

3

2

6

5

1

4

3

2 234156

5 612 3 4

000

M1z

-P2y

M3z

KBA

KAB

KBB

KAA

21 312

1

2

3

4

5

6

ReactionQu

Joint LoadQk

Du=Dunknown

Dk = DknownQF

B

QFA

[ ] [ ][ ] [ ]FAuAAk QDKQ +=

[ ] [ ] [ ] [ ])(1 FAkAAu QQKD −+= −

[ ] [ ][ ] [ ]FqdkqForceMember +=:

Page 37: Matrix beam

37

+

6

5

4

3

2

1

DDDDDD

Global:

21 312

1

2

3

4

5

6

1 2

2

3

4

51

6

(q´F6 )2

(q´F5 )2

(qF4)2

(qF3)2

w

2

P

(qF4)1

(qF3)1

(qF2)1

(qF1 )1

1[FEF]

=−=

=

24

23

12

MQPQ

MQ

y

4

3

2

DDD

+F

F

F

QQQ

4

3

2

2 3 4

Node 2

=

234

Member 1

123456

1 2 3 4 5 6

Member 2

Node 2=

6

5

4

3

2

1

QQQQQQ

+=+=

F

F

F4

F4

F

F3

F3

F

F

F

QQ

qqQqqQ

QQ

6

5

214

213

2

1

)()()()(

M1

-P2y

M2

0

00

Page 38: Matrix beam

38

1 2 312

1

2

3

4

5

6

4

3

2

DDD

2 3 4

Node 2

=

234

=−=

=

F

F

F

QQQ

MQPQ

MQ

4

3

2

24

2y3

12

Page 39: Matrix beam

39

Member 1:

21 312

1

2

3

4

5

6

P

qF4

qF3

qF2

qF1

1

[FEF]

4

3

2

1

qqqq

====

44

33

22

1 0

DdDdDd

d

+

F

F

F

F

qqqq

4

3

2

11234

1 2 3 4

= 1k

Page 40: Matrix beam

40

21 312

1

2

3

4

5

6

Member 2:

qF6

qF5

qF4

qF3

w

2

[FEM]

6

5

4

3

qqqq

==

==

00

6

5

44

33

dd

DdDd

+

F

F

F

F

qqqq

6

5

4

31234

1 2 3 4

= 1k

Page 41: Matrix beam

41

AC

B

9 m 3 m

10 kN1 kN/m

1.5 m

Example 1

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.

Page 42: Matrix beam

42

21Global

123

1Members

23

2

12

AC

B

9 m 3 m

10 kN1 kN/m

1.5 m

wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75

10 kN

1.5 m1.5 m

2

1 kN/m

9 m 1[FEF]

Page 43: Matrix beam

43

4/3

2/3

2/3

21

123

9 m 3 m

[k]1= EI4/9

2/9

3

4/9

2/92

3

2[k]2 = EI

4/3

4/3

2/3

2/3

2 12

1

[K] = EI

2 12

1

(4/9)+(4/3)

21

123

Stiffness Matrix: θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

Page 44: Matrix beam

44

A CB

10 kN1 kN/m

123

Equilibrium equations: MCB = 0MBA + MBC = 0

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

=2.423/EI

0.779/EI

θC

θB

9 m 1.5 m1.5 m6.756.75

+-3.75

-6.75 + 3.75 = -3MBA + MBC = 0

MCB = 01

2

θC

θB

4/3

2/3

2/3 = EI

2 12

1

(4/9)+(4/3)

3.753.756.75

Page 45: Matrix beam

45

= -6.40

6.92

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

MBA

MAB

2

3

= 0.779/EIθB

θA

0

+ -6.75

6.75

Substitute θB and θC in the member matrix,

1

23

wL2/12 = 6.75wL2/12 = 6.75

1 kN/m

9 m 1[FEF]

4.56 kN 4.44 kN

6.92 kN�m 6.40 kN�m

wL2/12 = 6.75wL2/12 = 6.75

= EI4/9

2/9

3

4/9

2/92

1 kN/m

19 m

Page 46: Matrix beam

46

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

Substitute θB and θC in the member matrix,

MCB

MBC

1

2 = EI

4/3

4/3

2/3

2/3

2 1

= 0

6.40

θC

θB

= 2.423/EI

= 0.779/EI

2

12

[FEF]PL/8 = 3.75PL/8 = 3.75

10 kN

1.5 m1.5 m

2

7.13 kN 2.87 kN

6.40 kN�m

10 kN

2

PL/8 = 3.75PL/8 = 3.75

+-3.75

3.75

Page 47: Matrix beam

47

A CB

10 kN1 kN/m

9 m 1.5 m1.5 m

6.92 kN�m

11.57 kN 2.87 kN4.56 kN

4.56 kN 4.44 kN 7.13 kN 2.87 kN

1 kN/m6.92 6.40

6.40

10 kN

M (kN�m) x (m)

-6.92 -6.40

3.48 4.32

V (kN) x (m)

4.56

-4.44 -2.874.56 m

7.13

θB = +0.779/EIθC = +2.423/EI

Page 48: Matrix beam

48

20 kN

A CBEI2EI

4 m 4 m

9kN/m 40 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.

Page 49: Matrix beam

49

1 2 3

0.5625

3

-0.375

-0.375

1

2

3

4

5

6

[K] = EI

3 4

3

4

Use 4x4 stiffness matrix,

[k]1

0.375EI

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

[k]2

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

2EI EI

1 2

1

2

5

6

θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

=

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

Page 50: Matrix beam

50

20 kN9kN/m 40 kN�m

4 m 4 m

12 kN�m18 kN

12 kN�m

18 kN 18 kN

D4

D3

Global:

D4

D3=

9.697/EI

-61.09/EI

Q4 = 40

Q3 = -203

4 + -12

18 3

4

20 kN

40 kN�m

12 kN�m

1 2

1

2

3

4

5

6

1 2 3

2EI EI

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

0.5625

3

-0.375

-0.375= EI

3 4

3

4

Page 51: Matrix beam

51

Member 1:

1

9 kN/m

A B12 kN�m

[qF]1

12 kN�m

18 kN 18 kN

1

9 kN/m

A B

[q]1

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN

1

1

2

3

4

1 2

2EI 12 kN�m 12 kN�m

18 kN 18 kN

q2

q1

q4L

q3L

67.51

48.18

53.21

-12.18

12

18

-12

18

12(2EI)/43

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

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

d3 = -61.09/EI

d4 = 9.697/EI

d2 = 0

d1 = 0

Page 52: Matrix beam

52

2B C

[q]2

Member 2:

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

2

3

4

5

6

2 3

EI

q4R

q3R

q6

q5

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0

0

0

0

d3 =-61.09/EI

d4 = 9.697/EI

d6 = 0

d5 = 0

18.06 kN�m

7.82 kN

13.21 kN�m

7.82 kN

-13.21

-7.818

-18.06

7.818

Page 53: Matrix beam

53

V (kN) x (m)

-7.818

48.18

-7.818-

M (kN�m) x (m)

13.21

-18.08-

-67.51

-

1

9 kN/m

A B

[q]1

53.21 kN�m

48.18 kN67.51 kN�m

12.18 kN2

B C

[q]2

18.06 kN�m

7.82 kN

13.21 kN�m

7.82 kN

12.18+

53.21

+

20 kN

4 m 4 m

9kN/m 40 kN�m

7.818 kN

18.06 kN�m67.51 kN�m

48.18 kN

D3 = ∆B = -61.09/EI

D4 = θB = +9.697/EI

Page 54: Matrix beam

54

20 kN

A CBEI2EI

4 m 2 m

9kN/m 40 kN�m10 kN

2 m

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.

Page 55: Matrix beam

55

Global1 2

1

2

3

4

5

6

1 2 3

20 kN

A CB EI2EI

9kN/m 40 kN�m10 kN

4 m 2 m 2 m

10 kN5 kN�m

5 kN

5 kN�m

5 kN

212 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

Page 56: Matrix beam

56

1 2

1

2

3

4

5

6

1 2 3 4 m 2 m 2 m

1

2

5

1 2

0.375

0.50.375 0.5 1

[K] = EI

3 4 6

3

46

0.5625

3

-0.375

-0.375

12(2EI)/43

[k]1 =-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

[k]2 =0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

Page 57: Matrix beam

57

MBA+MBC = 40

VBL+VBR = -20

MCB = 0θB

∆B

θC

-12 + 5 = -7-5

θB

∆B

θC

= -7.667/EI

-116.593/EI

52.556/EI

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

= EI

0.5625

3

-0.375

-0.375

0.375

0.50.375 0.5 1

3 4 6

3

46

20 kN

A CB

9kN/m 40 kN�m10 kN

21

1

2

3

4

5

6

1 2 3 10 kN5 kN�m

5 kN

5 kN�m

5 kN

212 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

20 kN

40 kN�m

5 kN�m5 kN�m

5 kN

12 kN�m

18 kN

+18 + 5 = 23

Page 58: Matrix beam

58

0.375EI

= -0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

34

=91.78

55.97

60.11

-19.97

91.78 kN�m

55.97 kN

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

MAB

VA

MBA

VBL+

12

18

-12

18=-116.593/EI

=-7.667/EI

0

0

θB

∆B

1

1

2

3

4

1 2

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1

[FEF]

19.97 kN

60.11 kN�m4 m

9kN/m

1

12 kN�m

18 kN

12 kN�m

18 kN18 kN

Page 59: Matrix beam

59

VC

MBC

VBR

MCB

+ 5

5

5 -5

=-20.11

-0.0278

0

10.03=52.556/EI

=-116.593/EI

=-7.667/EIθB

∆B

θC

0

10 kN5 kN�m

5 kN

5 kN�m

5 kN

2

[FEM]

10.03 kN

2 m

10 kN

2 m2

= 0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0.0278 kN

20.11 kN�m

Member 2: [q]1 = [k]1[d]1 + [qF]1

223

5

6

3

45 kN�m

5 kN

5 kN�m

5 kN

Page 60: Matrix beam

60

20 kN

4 m 2 m

9kN/m 40 kN�m10 kN

2 m

10.03 kN

91.78 kN�m

55.97 kN

V (kN) x (m)

55.97

-0.0278

19.97+

-10.03 -10.03-

91.78 kN�m

55.97 kN 19.97 kN60.11 kN�m4 m

9kN/m

110.03 kN0.0278 kN

20.11 kN�m2 m

10 kN2 m

2

91.78 kN�m

60.11 kN�m

20.11 kN�m

M (kN�m) x (m)

-91.78

-

+20.11

60.11

θB = -7.667/EI

θC = 52.556/EI

∆B = -116.593/EI

Page 61: Matrix beam

61

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

AB

C∆B = -10 mm

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 shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm.

Page 62: Matrix beam

62

[k]1 =4

8

8

4

1 2

1

2

EI8

1 2

1 2 3

2EI EI

1

2

42

2 3

2

3[K] = EI

8

12

2

4

4

2

2 3

2

3[k]2 = EI

8

Use 2x2 stiffness matrix:θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

Page 63: Matrix beam

63

75 kN�m

6(2EI)∆/L2 = 75 kN�m10 mm

1 37.5 kN�m

6(EI)∆/L2 = 37.5 kN�m10 mm

2[FEM]∆∆∆∆

6(EI)∆/L2 = 37.5 kN�m

D2

D3

=-61.27/EI rad

185.64/EI rad

40 kN

2EI EI

6 kN/m

A

B C

∆B = -10 mm 1 2

1 2 3

[FEM]load

32 kN�mwL2/12 = 32 kN�m

6 kN/m

1 40 kN�mPL/8 = 40 kN�m 2

40 kN8 m 4 m 4 m

32 kN�m 40 kN�mPL/8 = 40 kN�m

75 kN�m 37.5 kN�m

+-32 + 40 + 75 -37.5 = 45.5Q2 = 0

Q3 = 0 -40 - 37.5 = -77.5

D2

D3

2

42

2 3

2

3

12

8EI

=

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

Page 64: Matrix beam

64

6(2EI)∆/L2 = 75 kN�m

[FEM]load

32 kN�mwL2/12 = 32 kN�m

6 kN/m

1

75 kN�m

6(2EI)∆/L2 = 75 kN�m10 mm

1[FEM]∆∆∆∆

6 kN/m

18 m

1

2

2EI

1

q1

q2

d1 = 0

d2 = -61.27/EI=

76.37 kN�m

-18.27 kN�m

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

32 kN�mwL2/12 = 32 kN�m

75 kN�m

+32 + 75 = 107

-32 + 75 = 43

31.26 kN

76.37 kN�m

16.74 kN

18.27 kN�m

4

8

8

4

1 2

1

28EI

=

Page 65: Matrix beam

65

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

2

2 3

40 kN�mPL/8 = 40 kN�m 2

40 kN

37.5 kN�m

6(EI)D/L2 = 37.5 kN�m10 mm

2

[FEM]load

[FEM]∆∆∆∆

40 kN�m

37.5 kN�m

6(EI)D/L2 = 37.5 kN�m

PL/8 = 40 kN�m

=18.27 kN�m

0 kN�m

q2

q3

+40 - 37.5 = 2.5

-40 - 37.5 = -77.5

d2 = -61.27/EI

d3 = 185.64/EI

17.72 kN

18.27 kN�m

22.28 kN

40 kN

2

2

4

4

2

2 3

2

38EI

=

Page 66: Matrix beam

66

Alternate method: Use 4x4 stiffness matrix

2EI EI

1 2

2

5

6

3

4

1

8EI

= 1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

[k]1

8EI

=0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

[k]2

2-0.75

-0.1875-0.75

0.1875-0.75

0.752

-0.754

-0.18750.75

1.5 4

1.54

0.375

-0.3751.5

1.58

-1.5

-0.375-1.5

00

00

00

00

-0.7512

0.5625-0.758

EI=[K]

123456

1 4 5 62 3

Page 67: Matrix beam

67

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

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

A

B C

∆B = -10 mm 1 2

2

5

6

3

4

1

40 kN40 kN�m

20 kN

40 kN�m

20 kN

232 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1[FEF]

53

2

1

D4

D6=

-61.27/EI = -1.532x10-3 rad

185.64/EI = 4.641x10-3 rad

+(200x200/8)(-0.75)(-0.01) = 37.5

(200x200/8)(0.75)(-0.01) = -37.5

D4

D6

Q4 = 0

Q6= 0

2

4

12

2+

8

-40

4 6

4

68EI

=

-0.75-0.75 D5 = -0.01)

8200200( ×

+ 46

5

D4

D6

Q4 = 0

Q6= 0

2

4

12

2

4 6

4

68EI

= +8

-40

Page 68: Matrix beam

68

+32

24

-32

24

q1

q2

q4

q3

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

= (200x200)8

q1

q2

q4

q3=

76.37 kN�m

31.26 kN

-18.27 kN�m

16.74 kN

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

1

2

3

4

1∆B = -10 mm 32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1[FEF]load

d2 = 0

d1 = 0

d4 = -1.532x10-3

d3 = -0.01

6 kN/m

18 m

31.26 kN

76.37 kN�m

16.74 kN

18.27 kN�m

Page 69: Matrix beam

69

Member 2:

+40

20

-40

20

q3

q4

q6

q5

d4 = -1.532x10-3

d3 = -0.01

d6 = 4.641x10-3

d5 = 0

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

= (200x200)8

q3

q4

q6

q5=

18.27 kN�m

22.28 kN

0 kN�m

17.72 kN17.72 kN

18.27 kN�m

22.28 kN

40 kN

2

2

5

6

3

4

-10 mm = ∆B

40 kN40 kN�m

20 kN

40 kN�m

20 kN

2[FEF]

Page 70: Matrix beam

70

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

A

B C

∆B = -10 mm31.26 kN

76.37 kN�m

16.74 + 22.28 kN 17.72 kN

V (kN)

x (m)+ +--

31.26

-16.74

22.28

-17.725.21 m

M (kN�m) x (m)

-76.37

5.06

-18.27

70.85

+

--

D6 = θC = 4.641x10-3 rad

d4 = θB = -1.532x10-3 rad

Deflected Curve

∆B = -10 mm

Page 71: Matrix beam

71

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 shear diagram, bending moment diagramand qualitative deflected shape.Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm.

4 kN

8 m 4 m 4 m

2EI EI

0.6 kN/m

AB

C∆C = -10 mm

20 kN�m

Page 72: Matrix beam

72

2EI EI

1 2-10 mm

2

5

6

3

4

1

2

51

Q3

Q6

Q4

0.75 2

0.7524

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

[k]1

8EI

=

8EI

=0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

[k]2

D3

D6

D4

-0.1875

-0.75-0.75 D5 = -0.01

� Global: [Q] = [K][D] + [QF]

�Member stiffness matrix [k]4x4

0.5625-0.75

-0.7512

346

3 4 6

8EI

= )8

200200( ×+

346

5QF

3

QF6

QF4+

Page 73: Matrix beam

73

4 kN4 kN�m

2 kN

4 kN�m

2 kN

23.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]

4 kN

8 m 4 m 4 m2EI EI

0.6 kN/m

A

BC

10 mm

20 kN�m2EI EI

1 2

6

3

42

51

346

0.5625

0.75-0.75

-0.75122

0.752

3 4 6

48EI

=

D3

D6

D4 +9.375

37.537.5

2.4+2 = 4.4

-4.0-3.2+4 = 0.8+

D3

D6

D4

-377.30/EI = -9.433x10-3 m

+74.50/EI = +1.863x10-3 rad-61.53/EI = -1.538x10-3 rad=

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

Q3 = 0

Q6 = 20Q4 = 0

Page 74: Matrix beam

74

+3.2

2.4

-3.2

2.4

q1

q2

q4

q3

q1

q2

q4

q3=

43.19 kN�m

8.55 kN

6.03 kN�m

-3.75 kN

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

d2 = 0

d1 = 0

d4 = -1.538x10-3

d3 = -9.433x10-3

0.6 kN/m

18 m

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

1

2

3

4

13.2 kN�m

2.4 kN

3.2 kN�m

2.4 kN

0.6 kN/m

1[FEF]load 8 m

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 41

2

3

4

8200200×

=

Page 75: Matrix beam

75

Member 2:

+4

2

-4

2

q3

q4

q6

q5

d4 = -1.538x10-3

d3 = -9.433x10-3

d6 = 1.863x10-3

d5 = -0.01

q3

q4

q6

q5=

-6.0 kN�m

3.75 kN

20.0 kN�m

0.25 kN0.25 kN

6.0 kN�m

3.75 kN

4 kN

2

10 mm2

5

6

3

4 4 kN4 kN�m

2 kN

4 kN�m

2 kN

2[FEF]

8 m

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

8200200×

=

20 kN�m

Page 76: Matrix beam

76

4 kN

8 m 4 m 4 m2EI EI

0.6 kN/m

A

BC

10 mm

20 kN�m

Deflected Curve

0.6 kN/m

18 m

8.55 kN

43.19 kN�m

3.75 kN

6.03 kN�m

V (kN)x (m)

+

8.553.75

-0.25

0.25 kN

6.0 kN�m

3.75 kN

4 kN

2

20 kN�m

M (kN�m) x (m)

-43.19

621+

-

20

D3 = -9.433 mm

D6 = +1.863x10-3 radD4 = -1.538x10-3 rad

D5 = -10 mm

Page 77: Matrix beam

77

30 kN

A CB

EI2EI

4 m 2 m

9kN/mHinge

2 m

Example 6

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.

Internal Hinges

Page 78: Matrix beam

78

[K] = EI 1

2

1 2

0.0

0.0

1.0

2.0

A C

B

2EI

4 m 2 m 2 m

EI1 2

43

1

2

[k]1 =EI

2EI

2EI

EI

3 1

3

1[k]2 =

0.5EI

1EI

1EI

0.5EI

2 4

2

4

3 4

Use 2x2 stiffness matrix,θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

Page 79: Matrix beam

79

0.0

0.0

D1

D2

15 kN�mB C

15 kN�m30 kN

2

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]

= EI 1

2

1 2

0.0

0.0

1.0

2.0

D1

D2=

0.0006 rad

-0.0015 rad

-12

15+

Global matrix:

30 kN9kN/m

Hinge

A C

B1 2

43

12 kN�m 15 kN�m1

2

Page 80: Matrix beam

80

Member 1:

q3

q1

12

-12+=

EI

2EI

2EI

EI

3 1

3

1

18

0.0=

d3

d1

= 0.0

= 0.0006

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]A

B1

31

18 kN�m22.5 kN 13.5 kN

A B

9 kN/m

1

Page 81: Matrix beam

81

B C30 kN

CB2

4

2

15 kN�mB C

15 kN�m 30 kN

2

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15+=

0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5=

Member 2:

22.5 kN�m

20.63 kN9.37 kN

Page 82: Matrix beam

82

30 kN

A CB

EI2EI

4 m 2 m

9kN/mHinge

2 m

-20.63

V (kN) x (m)

22.59.37

-13.5

+

2.5 m

M (kN�m) x (m)

-18

10.13

-22.5

18.75

+-

+--

θBL= 0.0006 rad θBR= -0.0015 rad

13.5 kN22.5 kN

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

A B18 kN�m

9 kN/m

9.37 kN13.5 kN

22.87 kN

Page 83: Matrix beam

83

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge

Example 7

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.E = 200 GPa, I = 50x10-6 m4.

Page 84: Matrix beam

84

A CB1 2

2EI EI 6

7

4

5

4

5

6

7

0.375

0.375

1.0

-0.75

-0.75 2.0

0.0

0[K] = EI 1

2

3

1 2 3

0.5625

[k]1 =0.75EI

EI-0.75EI

2EI/L

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

51

2

0.375EI

0.375EI-0.1875EI

0.5EI -0.375EI

-0.375EI

[k]2 =

0.1875EI

-0.375EIEI0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI EI

1 3 6 71

36

7

1

2

3

Page 85: Matrix beam

85

Q1 = -20

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =-0.02382 m

0.008933 rad

-0.008333 rad

[FEM]

12 kN�mA B

12 kN�m

18 kN18 kN1

9kN/m

20 kN

A CB

9kN/m

A CB1 2

6

7

4

5

2EI EI

Global:

= EI 1

2

3

1

0.5625

0.375

-0.75

2

-0.75

2.0

0.0

3

0.375

0

1.0

12 kN�m

18 kN

20 kN

1

2

3

Page 86: Matrix beam

86

A B1

2EI1

24

5

[FEF]

12 kN�mA B

12 kN�m

18 kN18 kN1

9kN/m

+12

18

-12

18

Member 1:

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI/L

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

= 0.0

= 0.0

= -0.02382

= -0.00833

d5

d4

d2

d1

= 107.32

44.83

0.0

-8.83

A B

107.32 kN�m

8.83 kN44.83 kN

9kN/m

1

Page 87: Matrix beam

87

Member 2:

CB2

EI1

3

6

7

44.66 kN�mB C

11.16 kN11.16 kN

2

+0

0

0

0

= -0.02382

= 0.008933

= 0.0

= 0.0

d3

d1

d7

d6

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 71

3

6

7

= 0.0

-11.16

-44.66

11.16

Page 88: Matrix beam

88

20 kN

A CB4 m 4 m

9kN/m

V (kN) x (m)

44.83

8.83

-11.16 -11.16

+

-

M (kN�m) x (m)

-107.32

-44.66-

-

A B

107.32 kN�m

8.83 kN44.83 kN

9kN/m

144.66 kN�m

B C

11.16 kN11.16 kN

2

θBL= -0.008333 rad θBR= 0.008933 rad

Page 89: Matrix beam

89

30 kN

A C

BEI2EI

4 m 2 m

9kN/mHinge

40 kN�m

2 m

Example 8

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4.

Page 90: Matrix beam

90

[K] = EI 1

2

1 2

0.0

0.0

1.0

2.0

A C

B

2EI

4 m 2 m 2 m

EI1 2

43

[k]1 =EI

2EI

2EI

EI

3 1

3

1[k]2 =

0.5EI

1EI

1EI

0.5EI

2 4

2

4

3 4

1

2

Use 2x2 stiffness matrix: θi θj

Mi

Mj[ ]

=× LEILEI

LEILEIk

/4/2/2/4

22

Page 91: Matrix beam

91

15 kN�mB C

15 kN�m30 kN

2

12 kN�m

A B

12 kN�m 9kN/m

1[FEM]

Global matrix:

A C

B1 2

43

1

2

12 kN�m 15 kN�m

30 kN

A C

BEI2EI

9kN/mHinge

40 kN�m

Q1 = 40

Q2 = 0.0

D1

D2

-12

15+

D1

D2=

0.0026 rad

-0.0015 rad

= EI 1

2

1 2

0.0

0.0

1.0

2.0

40 kN�m

Page 92: Matrix beam

92

12 kN�m

A B

12 kN�m 9kN/m

1[FEF]A

B1

31

40 kN�mA B38 kN�m

9 kN/m

1.5 kN37.5 kN

Member 1:

38

40=

q3

q1

= 0.0

= 0.0026

d3

d1

12

-12+=

EI

2EI

2EI

EI

3 1

3

1

Page 93: Matrix beam

93

CB2

4

2

15 kN�mB C

15 kN�m 30 kN

2

Member 2:

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15+=

0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5=

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

Page 94: Matrix beam

94

9.37 kN

1.5 kN

30 kN

A C

B

4 m 2 m

9kN/mHinge

40 kN�m

2 m

-20.63

V (kN) x (m)

37.5

9.371.5+

M (kN�m) x (m)

-38

40

-22.5

+-

18.75+

--

θBL= 0.0026 radθBR=- 0.0015 rad

7.87 kN40 kN�mA B

38 kN�m

9 kN/m

1.5 kN37.5 kN

22.5 kN�mB C

30 kN

20.63 kN9.37 kN

Page 95: Matrix beam

95

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge40 kN�m

Example 9

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.40 kN�m at the end of member AB. E = 200 GPa, I = 50x10-6 m4

Page 96: Matrix beam

96

A CB1 2

6

7

4

51

2

3

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge40 kN�m

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEM]θi ∆j θj

Mi

∆i

Vj

Mj

Vi

[ ]

−−−−

−−

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

/4/6/2/6/6/12/6/12/2/6/4/6/6/12/6/12

22

2323

22

2323

44

Page 97: Matrix beam

97

1 22A CB

6

7

4

5

[K] = EI 1

2

3

1 2 3

6

7

4

5

0.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

[k]1 = 0.75EI

EI-0.75EI

2EI

0.375EI0.75EI

0.75EI-0.375EI

2EI0.75EI

EI-0.75EI

-0.75EI- 0.375EI

0.375EI-0.75EI

4 5 1 24

51

2

1

0.375EI

0.375EI-0.1875EI

0.5EI -0.375EI

-0.375EI

[k]2 =

0.1875EI-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI EI

1 3 6 71

36

7

2EI EI1

2

3

Page 98: Matrix beam

98

Q1 = -20

Q3 = 0.0

Q2 = 40

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =-0.01316 m

0.0049333 rad

-0.002333 rad

A CB1 2

6

7

4

5

2EI EI

20 kN

EI

9kN/m

40 kN�m

20 kN

40 kN�m

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

= EI 1

2

3

1 2 3

0.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

12 kN�m

18 kNGlobal:

1

2

3

Page 99: Matrix beam

99

1

2EI

[q]1

1

12

1

2EI4

5

+12

18

-12

18

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

= 87.37

49.85

40

-13.85

12 kN�m

18 kN

12 kN�m

18 kN

9 kN/m

1[FEF]

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

13.85 kN

40 kN�m87.37 kN�m

49.85 kN

= 0.0

= 0.0

= -0.01316

= -0.002333

d5

d4

d2

d1

12 kN�m

18 kN

12 kN�m

18 kN1

Page 100: Matrix beam

100

22 CB

1

3

EI 6

7

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

+0

0

0

0

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 71

3

6

7

= 0.0

-6.18

-24.69

6.18

24.69 kN�m

6.18 kN6.18 kN

= -0.01316

= 0.004933

= 0.0

= 0.0

d3

d1

d7

d6

22

[q]2

Page 101: Matrix beam

101

20 kN

A CB

EI2EI

4 m 4 m

9kN/m 40 kN�m

24.69 kN�mB C

6.18 kN6.18 kN

A B

87.37 kN�m 13.85 kN49.85 kN

40 kN�m

M (kN�m) x (m)

-

-87.37

+40

--24.69

θBL= -0.002333 rad θBR = 0.004933 rad

49.85

-6.18

V (kN) x (m)-6.18

-13.85+

Page 102: Matrix beam

102

� Fixed-End Forces (FEF)- Axial- Bending

� Curvature

Temperature Effects

Page 103: Matrix beam

103

Tm> TR

� Thermal Fixed-End Forces (FEF)

Tl

Tt

2bt

mTTT +

=

Tt

Tb > Tt

βF

AFF

BF

BFBMF

AM

Rmaxial TTT −=∆ )(

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

y β

)(tandT

dTT tb ∆

=−

A

A B

σaxialσaxial

A B

σy

y

TR Tm

Tt

Tb

Tb - Tt

Room temp = TR

ct

cbd

TR

NA

Page 104: Matrix beam

104

- Axial

A B

σaxialσaxial

FAF F

BF

dAFA

axialF

A ∫= σ

dAE axial∫= ε

dATE axial∫ ∆= )(α

axialTEA )(∆= α

AETTF RmAF

axial )()( −= α

dATE axial ∫∆= )(α

Tt

Tb > Tt

Tm> TR

Rmaxial TTT −=∆ )(

TR Tm

Page 105: Matrix beam

105

- Bending

A B

σy

y

dAyMA

yFA ∫= σ

dATyE y∫ ∆= )(α

dAdTyyE∫

∆= )(α

∫∆

= dAydTE 2)(α

EId

TTF ulA

Fbending )()( −

= α

FBMF

AM

dAyE∫= ε

tbbending TTT −=∆ )(

)()(dTyTy

∆=∆

y β Tt

Tb

Page 106: Matrix beam

106

ds´

dx

Afterdeformation

Before deformation

y dx

ds = dx y

ρ)()( θρ ddx =

)( θdy )(1dxdθ

ρ=

� Elastic Curve: Bending

Page 107: Matrix beam

107

dxdTyyd )()( ∆

= αθ

dxdTd )()( ∆

= αθ

Tb

Tt

yy

dT∆

� Bending Temperature

dx

Tt

Tb

Tl > Tu

y

Tl > Tu

Tu

O

2bt

mTTT +

= dθ

MM

Tb

cb

ct

Tt

∆T = Tb - Tt

dT∆

EIM

dT

dxd

=∆

== )(1)( αρ

θ

Page 108: Matrix beam

108

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

Example 10

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 32.5oC, α = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.

Page 109: Matrix beam

109

12 3

1 2

Mean temperature = (40+25)/2 = 32.5Room temp = 32.5oC

FEM+7.5 oC-7.5 oC

a(∆T /d)EI19.78 kN�m = 19.78 kN�m

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

mkNEIdTF F

bending •=××−

×=∆

= − 78.19)502002)(182.0

2540)(1012()2)(( 6α

Page 110: Matrix beam

110

[M1] = 3EIθ1 + (1.978x10-3)EI0

Element 1:

M1

M2

q1

q2

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

+ 1.978

-1.978(10-3) EI= 1

2

2

1

2 1

2

1EI

Element 2:

M3

M1

q3

q1 + 0

0= 0.5

1

1

0.5

1 3

1

3EI

θ1 = -0.659x10-3 rad

12 3

A C

BEI2EI

4 m 4 m

1 2

19.78 kN�m19.78 kN�m182 mm

Page 111: Matrix beam

111

Element 2:

M3

M1

q3

q1= + 0

0

0.5

1

1

0.5

1 3

1

3EI

= -0.659x10-3

= 0= 6.59 kN�m

-26.37 kN�m

= -0.659x10-3

= 0= -3.30 kN�m

-6.59 kN�m

Element 1:

M1

M2 = q1

q2 + 19.78

-19.78

1

2

2

1

2 1

2

1EI

4.95 kN4.95 kN

26.37 kN�m 6.59 kN�m

A B

6.59 kN�m 3.3 kN�m

B C

2.47 kN2.47 kN

12 3

A C

BEI2EI

4 m 4 m

1 2

19.78 kN�m19.78 kN�m182 mm

Page 112: Matrix beam

112

A CB

4 m 4 m

+7.5 oC

-7.5 oC 3.3 kN�m

26.37 kN�m

4.95 kN2.47 kN

2.47 kN

M (kN�m) x (m)

26.37

6.59

-3.30

+-

V (kN) x (m)

-4.95-2.47

- -

θB = -0.659x10-3 radDeflected curve x (m)

Page 113: Matrix beam

113

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

Example 11

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.(b) Draw the quantitative shear and bending moment diagramsand qualitative deflected shape.Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4,A = 20(10-3) m2

Page 114: Matrix beam

114

A

CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC1

2 3

7

8

9

1 24

5

6

Element 1:

mkNm

mkNmLAE /)10(2

)4()/10200)(1020(2)2( 6

2623

=××

=−

mkNm

mmkNLEI

•=×××

=−

)10(20)4(

)1050)(/10200(24)2(4 34626

mkNm

mmkNLEI

•=×××

=−

)10(10)4(

)1050)(/10200(22)2(2 34626

kNm

mmkNLEI )10(5.7

)4()1050)(/10200(26)2(6 3

2

4626

2 =×××

=−

mkNm

mmkNL

EI /)10(75.3)4(

)1050)(/10200(212)2(12 33

4626

3 =×××

=−

Page 115: Matrix beam

115

Fixed-end forces due to temperatures

Mean temperature(Tm) = (40+25)/2 = 32.5 oC ,TR = 28 oC

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

mkNEIdTF F

bending •=××−

×=∆

= − 78.19)502002)(182.0

2540)(1012()2)(( 6α

19.78 kN�m19.78 kN�m

432 kN432 kNA B

+12 oC

-3 oC

kNmkNmAETF Faxial 432)/10200)(310202)(285.32)(1012()( 2626 =×−××−×=∆= −α

Page 116: Matrix beam

116

FEM

+12 oC

-3 oC

19.78 kN�m

A B

19.78 kN�m

432 kN432 kN

Element 1:

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

0

0

0

d1

d2

d3

q4

q6

q5

q1

q2q3

=

4

5

6

1

2

3

4 1

- 2x106

0.00

0.00

2x106

0.00

0.00

2

0.00

-3750

-7500

0.00

3750

-7500

3

0.00

7500

10x103

0.00

-7500

20x103

2x106

0.00

0.00

-2x106

0.00

0.00

5

0.00

3750

7500

0.00

-3750

7500

6

0.00

7500

20x103

0.00

-7500

10x103

432

-19.78

0.00

-432

0.00

19.78

+

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

Page 117: Matrix beam

117

Element 2:

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

d1

d3

d2

0

0

0

0

0

0

0

0

0

+

q1

q3

q2

q7

q8q9

- 1x106

0.00

0.00

1x106

0.00

0.00

0.00

-1875

-3750

0.00

1875

-3750

0.00

3750

5x103

0.00

-3750

10x103

1x106

0.00

0.00

-1x106

0.00

0.00

0.00

1875

3750

0.00

-1875

3750

0.00

3750

10x103

0.00

-3750

5x103

7 8 91 2 3

=

1

2

3

7

8

9

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

Page 118: Matrix beam

118

D1

D3

D2 =0.000144 m

-719.3x10-6 rad

-0.0004795 m

Q1 = 0.0

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

-432

19.78

0.0+= 1

2

3

1

3x106

0.0

0.0

2

0.0

5625

-3750

3

0.0

-3750

30x103

Global:

A CB

EI, AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm1

2 3

7

8

9

1 24

5

6

Page 119: Matrix beam

119

= 144x10-6

= -479.5x10-6

= -719.3x10-6

Element 1:

0

0

0

d1

d2

d3

q4

q6

q5

q1

q2q3

=

4

5

6

1

2

3

4 1

- 2x106

0.00

0.00

2x106

0.00

0.00

2

0.00

-3750

-7500

0.00

3750

-7500

3

0.00

7500

10x103

0.00

-7500

20x103

2x106

0.00

0.00

-2x106

0.00

0.00

5

0.00

3750

7500

0.00

-3750

7500

6

0.00

7500

20x103

0.00

-7500

10x103

432

-19.78

0.00

-432

0.00

19.78

+

=

q4

q6

q5

q1

q2q3

144.0 kN

-23.38 kN�m

-3.60 kN

-144.0 kN

3.60 kN

9.00 kN�m

A B 1

2 3

4

5

6

A B144 kN

3.60 kN

23.38 kN�m

144 kN

3.60 kN

9 kN�m

Page 120: Matrix beam

120

= 144x10-6

= -479.5x10-6

= -719.3x10-6

144 kN

-9 kN�m

-3.6 kN

-144 kN

3.6 kN

-5.39 kN�m

=

q1

q3

q2

q7

q8q9

B C 7

8 9

1

2

3

144 kNB C

3.60 kN

9 kN�m

144 kN

3.60 kN

5.39 kN�m

Element 2:

d1

d3

d2

0

0

0

0

0

0

0

0

0

+

q1

q3

q2

q7

q8q9

=

- 1x106

0.00

0.00

1x106

0.00

0.00

0.00

-1875

-3750

0.00

1875

-3750

0.00

3750

5x103

0.00

-3750

10x103

1x106

0.00

0.00

-1x106

0.00

0.00

0.00

1875

37500

0.00

-1875

3750

0.00

3750

10x103

0.00

-3750

5x103

7 8 91 2 3

1

2

3

7

8

9

Page 121: Matrix beam

121

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

Isolate axial part from the system

RCRA

RA RA = RC+

Compatibility equation: dC/A = 0

RA = 144 kN

RC = -144 kN

0)4)(285.32(1012)4(2

)4( 6 =−×++ −

AER

AER AA

Page 122: Matrix beam

122

A B

144 kN

3.60 kN

23.38 kN�m

144 kN

3.60 kN

9 kN�m

144 kN

B C3.60 kN

9 kN�m

144 kN

3.60 kN

5.39 kN�m

V (kN) x (m)

-3.6

-

M (kN�m) x (m)

23.388.98

-5.39

+-

D3 = -719.3x10-6 radDeflected curve x (m)

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

D2 = -0.0004795 mm

Page 123: Matrix beam

123

Skew Roller Support

- Force Transformation- Displacement Transformation- Stiffness Matrix

Page 124: Matrix beam

124

m

i

j

m

i

j

x´y´

x

y

� Displacement and Force Transformation Matrices

12

3

45

6

θyθx

Page 125: Matrix beam

125

λx λy

i

jθy

θx

4´5´

θy

θx

m

i

j

x

y

12

3

45

6

Force Transformation

Lxx ij

x

−=λ

Lyy ij

y

−=λ

yx qqq θθ coscos 5'4'4 −=

xy qqq θθ coscos 5'4'5 −=

6'6 qq =

6

5

4

qqq

−=

10000

xy

yx

λλλλ

6'

5'

4'

qqq

=

6'

5'

4'

3'

2'

1'

6

5

4

3

2

1

1000000000000000010000000000

qqqqqq

λλλλ

λλλλ

qqqqqq

xy

yx

xy

yx

[ ] [ ] [ ]'qTq T=

Page 126: Matrix beam

126x

y

θyθx

x

y

θyθx

d4 d 4 cos θ x

d 4 cos θ yθy

θx

d5

θx

θy

d 5 cos θ x

d 5 cos θ y

Displacement Transformation

j4´

6´j4

5

6λx λy

yx θdθdd coscos' 544 +=

xy θdθdd' coscos 545 +−=

66' dd =

6'

5'

4'

ddd

−=

10000

xy

yx

λλλλ

6

5

4

ddd

=

6

5

4

3

2

1

6'

5'

4'

3'

2'

1'

1000000000000000010000000000

dddddd

λλλλ

λλλλ

dddddd

xy

yx

xy

yx

[ ] [ ][ ]dTd ='

Page 127: Matrix beam

127

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ] [ ])'( FT qd'k'T +=

[ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+=

[ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= '

[ ] [ ] [ ][ ]TkTkTherefore T ', =

[ ] [ ] [ ]FTF qTq '=

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ]dTd ='

[ ] [ ] [ ][ ]TkTk T '=

Page 128: Matrix beam

128

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

5´ 6´

2´3´

*6

5*

*4

*3

2*

*1

qqqqqq

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

1 4 5 62 31*2*3*4*5*6*

[ T ]T

6'

5'

4'

3'

2'

1'

qqqqqq

Page 129: Matrix beam

129

[ ]

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

λλλλ

λλλλ

T

1 4 5 62 3123456

[ ]

−−−−

−−−

=

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

k

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

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

00/00/

'

22

2323

22

2323

1

34

6

2

5

1 2 3 4 5 6

Page 130: Matrix beam

130

[ 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 + (

Page 131: Matrix beam

131

40 kN

4 m4 m 22.02 o

Example 12

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 = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

Page 132: Matrix beam

132

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ][FEF]

1Global

i j

1

i j

Local

40 kN

4 m4 m 22.02o

20sin22.02=7.520cos22.02=18.54

4

5

6

x*

x´1 *

2* 3 *

λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,

λjy = cos 67.98o = 0.3749

22.02 o

Page 133: Matrix beam

133

x*

x´1

4

5

6

1 *

2* 3 *

� Transformation matrix

Member 1: [ q ] = [ T ]T[q´]

1

i j1´

Global Local

λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,

λjy = cos 67.98o = 0.3749

*3

2*

*1

6

5

4

qqqqqq

6'

5'

4'

3'

2'

1'

qqqqqq

−=

10000009271.03749.000003749.09271.0000000100000010000001

1´ 4´ 5´ 6´2´ 3´4561*

2*

3*

[ ]

=

1000000000000000010000000000

jxjy

jyjx

ixiy

iyix

TT

λλλλ

λλλλ

Page 134: Matrix beam

134

[k´]1 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

1

i j1´

Local

� Local stiffness matrix

[ ]

−−−−

−−−

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

k

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

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

00/00/

22

2323

22

2323

66

θi ∆j θj∆i δjδi

Mi

Vj

Mj

Vi

Nj

Ni

Page 135: Matrix beam

135

Stiffness matrix [k´]:

[k´]1 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T]

[k*]1 = 103

4561*

2*

3*

4 1* 2* 3*5 6-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

x*

x´1

4

5

6

1 *

2* 3 * 1

i j1´

Global Local

4

5

6

2*

Page 136: Matrix beam

136

Global Equilibrium:

Q1 = 0.0

Q3 = 0.0

D1*

D3*=

36.37x10-6 m

0.002 rad

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

= 1031*

3*

1*

129

1.406

3*

1.406

20.0

D1*

D3*

-7.5

-40+

x*

x´14

5

6

1 *

2* 3 *

4 m4 m 22.02 o

40 kN

x*

x´4

5

6

2*

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ]20sin22.02=7.5

20cos22.02=18.54

Page 137: Matrix beam

137

Member Force : [q] = [k*][D] + [qF]

0

40.020

-7.5418.54-40.0

+

q4

q6

q5

q1*

q2*q3*

= 103

4561*

2*

3*

4 1* 2* 3*5 6-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

0

00

36.4x10-6

02x10-3

-5.06

6027.5

013.48

0

=

40 kN

13.48 kN

60 kN�m

27.5 kN

5.06 kN

4 m4 m 22.02 o

40 kN

x*

x´1

4

5

6

1 *

2* 3 *

40 kN40 kN�m

7.518.54

40 kN�m

20 kN

4

5

6

2*

Page 138: Matrix beam

138

4 m4 m 22.02 o

40 kN

5.05 kN27.51 kN

60.05 kN�m

13.47 kN

40 kN

+

Bending moment diagram (kN�m)

Deflected shape

50.04

-

-60.05

0.002 rad

Page 139: Matrix beam

139

40 kN

22.02o

8 m 4 m 4 m2EI, 2AE EI, AE

6 kN/m

Example 13

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 = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.

Page 140: Matrix beam

140

40 kN

22.02 o8 m 4 m 4 m

2EI, 2AE EI, AE

6 kN/m

1 21

2

3

4

5

6

7 *

8 * 9 *

Global

21´

2´3´

5´6´

11´

Local

mkNm

mkNmL

AE

•×=

×=

3

262

10150)8(

)/10200)(006.0(

mkNm

mmkNLEI

•×=

×=

3

426

1020)8(

)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

2

×=

×=

Page 141: Matrix beam

141

[k]1 = [k´]1 = 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

1 21

2

3

4

5

6

7 *

8 * 9 *

Global Local : Member 1

14

5

6

1

2

3

[ q ]

11´

[ q´]

[q] = [q´] Thus, [k] = [k´]

Page 142: Matrix beam

142

Member 2: Use transformation matrix, [q*] = [T]T[q´]

q1´

q3´

q2´

q4´

q5´

q6´

q1

q3

q2

q7*

q8*q9*

0.92710.3749

-0.37490.9271

000

000

000

0 0

001

10

01

=

1237*

8*

9*

1´ 4´ 5´ 6´2´ 3´

0 0

001

000

000

000

1 21

2

3

4

5

6

7 *

8 * 9 *

λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271,

λjy = cos 67.98o = 0.3749

21

2

3

x*

x´7*

8 * 9* [ q* ]

21´

2´3´

5´6´

[ q´ ]

Page 143: Matrix beam

143

Stiffness matrix [k´]:

[k´]2 = 103

1´2´ 3´4´5´6´

1´ 4´ 5´ 6´2´ 3´-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

Stiffness matrix [k*]: [k*] = [T]T[k´][T]

[k*]2 = 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.47751.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

21

23

x*

x´7 *

8 * 9 * [ q* ]

21´

2´3´

5´6´

[ q´ ]

Page 144: Matrix beam

144

1 21

2

3

4

5

6

7 *

8 * 9 *

1

2

3

4

5

6

8 *

[k]1= 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

[k*]2 = 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.47751.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.47710.00

Page 145: Matrix beam

145

Global:

=-509.84x10-6 rad

18.15x10-6 m

0.00225 rad

58.73x10-6 mD3

D1

D9*

D7*

00

00 +

-32 + 40 = 80

-40-7.5

D3

D1

D9*

D7*

32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

1

40 kN6 kN/m

1 21

2

3

4

5

6

7*

8 *

9*

1

2

3

4

5

6

8 *

= 1030

0-139

450

101.406

600

1.406

1.406-139

129

010

1.40620

1 3 7* 9*

137*

9*

40 kN40 kN�m

7.518.54

40 kN�m

20 kN

40 kN�m

7.5

40 kN�m32 kN�m

Page 146: Matrix beam

146

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

0

3224

024-32

+

0

00

d1=18.15x10-6

0d3=-509.84 x10-6

q4

q6

q5

q1

q2q3

= 2x103

456123

4 1 2 35 6-150.00.0000.000150.0

0.0000.000

0.000-0.9375-3.7500.000

0.9375-3.750

0.0003.75010.000.000

-3.75020.00

150.0

0.0000.000

-150.00.0000.000

0.0000.93753.7500.000

-0.93753.750

0.0003.75020.000.000

-3.75010.00

-5.45 kN

21.80 kN�m20.18 kN

5.45 kN27.82 kN-52.39 kN�m

=

q4

q6

q5

q1

q2q3

6 kN/m

5.45 kN

20.18 kN

21.80 kN�m

5.45 kN

27.82 kN

52.39 kN�m

32 kN�m

24 kN

32 kN�m

24 kN

6 kN/m

114

5

6

1

2

3

Page 147: Matrix beam

147

q1

q3

q2

q7*

q8*q9*

21

3

x*

x´7*

9 *

= 103

1237*

8*

9*

1 7* 8* 9*2 3-139.00.3511.406129.0

1.40651.82

-56.25-0.869-3.75051.8221.90

-3.476

0.0003.75010.001.406

-3.47620.00

150.0

0.0000.000

-139.0-56.250.000

0.0000.93753.7500.351

-0.8693.750

0.0003.75020.001.406

-3.75010.00

0

4020

-7.518.54-40

+

18.15x10-6

-509.84x10-6

0

58.73x10-6

00.00225

q1

q3

q2

q7*

q8*q9*

-5.45 kN

52.39 kN�m26.55 kN

0 kN14.51 kN0 kN�m

= 5.45 kN

26.55 kN

52.39 kN�m

14.51 kN

40 kN

20sin22.02=7.520cos22.02=18.54

40 kN40 kN�m

20 kN

40 kN�m

20 kN

[ q´F ]

2

2 8 *

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

Page 148: Matrix beam

148

5.45 kN26.55 kN

52.39 kN�m

14.51 kN

40 kN6 kN/m

5.45 kN

27.82 kN

52.39 kN�m

5.45 kN

20.18 kN

21.80 kN�m

40 kN

22.02o

8 m 4 m 4 m

6 kN/m

5.45 kN

20.18 kN

21.80 kN�m

14.51 kN54.37 kN

3.36 m

M (kN�m) x (m)

53.81

-21.8-

V (kN)x (m)

26.55

+20.18

+- -

-27.82

14.51cos 22.02o

=13.45 kN

-52.39-

12.14 +

-13.45