DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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1 ! General Case ! Stiffness Coefficients ! Stiffness Coefficients Derivation ! Fixed-End Moments ! Pin-Supported End Span ! Typical Problems ! Analysis of Beams ! Analysis of Frames: No Sidesway ! Analysis of Frames: Sidesway DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

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Page 1: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

1

! General Case! Stiffness Coefficients! Stiffness Coefficients Derivation! Fixed-End Moments! Pin-Supported End Span! Typical Problems! Analysis of Beams! Analysis of Frames: No Sidesway! Analysis of Frames: Sidesway

DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

Page 2: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Slope ñ Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 3: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Degrees of Freedom

L

θΑ

A B

M

1 DOF: θΑ

PθΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

Page 4: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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L

A B

1

Stiffness

kBAkAA

LEIkAA

4=

LEIkBA

2=

Page 5: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

Page 6: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Fixed-End ForcesFixed-End Moments: Loads

P

L/2 L/2

L

w

L

8PL

8PL

2P

2P

12

2wL12

2wL

2wL

2wL

Page 7: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 8: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

Page 10: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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+

1

1

i jMij Mji

θi

θj

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

iθ×

jθ×

Stiffness Coefficients

L

Page 11: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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L

Real beam

Conjugate beam

Stiffness Coefficients Derivation: Fixed-End Support

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: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Conjugate beam

L

Real beam

Stiffness Coefficients Derivation: Pinned-End Support

Mi θi

θj

LM i

LMi

EIM i

EILMi

2

32L

θi θj

0)3

2)(2

(:0' =−=Σ+ LLEI

LMM ii

j θ

i j

0)2

()3

(:0 =+−=Σ↑+ jii

y EILM

EILMF θ

LEIM

EILM

ii

i3)

3(1 =→==θ

)6

(EI

LM ij

−=θ)

3(

EILM i

i =θ

Page 15: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Fixed end moment : Point Load

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

Page 16: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 17: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 18: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 19: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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wP

A B

A B+θA θB

(FEM)AB(FEM)BA

Pin-Supported End Span: Simple Case

BA LEI

LEI θθ 24

+ BA LEI

LEI θθ 42

+

)1()()/2()/4(0 −−−++== ABBAAB FEMLEILEIM θθ

)2()()/4()/2(0 −−−++== BABABA FEMLEILEIM θθ

BABABBA FEMFEMLEIM )()(2)/6(2:)1()2(2 −+=− θ

2)()()/3( BA

BABBAFEMFEMLEIM −+= θ

wP

AB

L

Page 20: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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AB

wP

A B

A BθA θB

(MF AB)load

(MF BA)load

Pin-Supported End Span: With End Couple and Settlement

BA LEI

LEI θθ 24

+ BA LEI

LEI θθ 42

+

L

(MF AB)∆

(MF BA) ∆

)1()()(24−−−+++== ∆

FABload

FABBAAAB MM

LEI

LEIMM θθ

)2()()(42−−−+++= ∆

FBAload

FBABABA MM

LEI

LEIM θθ

2)(

21])(

21)[(3:

2)1()2(2lim AF

BAloadFABload

FBABBAA

MMMMLEIMbyinateE ++−+=

−∆θθ

MA

wP

AB

Page 21: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Fixed-End MomentsFixed-End Moments: Loads

P

L/2 L/2

P

L/2 L/2

8PL

8PL

163)]

8()[

21(

8PLPLPL

=−−+

12

2wL12

2wL

8)]

12()[

21(

12

222 wLwLwL=−−+

Page 22: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Typical Problem

0

0

0

0

A C

B

P1P2

L1 L2

wCB

P

8PL

8PL w12

2wL

12

2wL

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

L L

Page 23: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 24: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 25: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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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 26: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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Example of Beams

Page 27: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 6 kN/m

A C

B 6 m4 m4 m

Example 1

Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

Page 28: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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PPL8 wwL2

30FEM

PL8

wL2

20

MBA MBC

B

Substitute θB in the moment equations:

MBC = 8.8 kNïm

MCB = -10 kNïm

MAB = 10.6 kNïm,

MBA = - 8.8 kNïm,

[M] = [K][Q] + [FEM]

10 kN 6 kN/m

A C

B 6 m4 m4 m

0

0

0

0

8)8)(10(

82

84

++= BAABEIEIM θθ

8)8)(10(

84

82

−+= BABAEIEIM θθ

30)6)(6(

62

64 2

++= CBBCEIEIM θθ

20)6)(6(

64

62 2

−+= CBCBEIEIM θθ

0:0 =−−=Σ+ BCBAB MMM

EI

EIEI

B

B

4.2

030

)6)(6(10)6

48

4(2

=

=+−+

θ

θ

Page 29: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 6 kN/m

A C

B 6 m4 m4 m

MBC = 8.8 kNïm

MCB= -10 kNïm

MAB = 10.6 kNïm,

MBA = - 8.8 kNïm,

= 5.23 kN = 4.78 kN = 5.8 kN = 12.2 kN

10 kNïm8.8 kNïm

10.6 kNïm8.8 kNïm

10 kN

10.6 kNïm

8.8 kNïm

A B

ByLAy

2 m

6 kN/m8.8 kNïm

10 kNïmB

CyByR

18 kN

Page 30: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 6 kN/m

A C

B 6 m4 m4 m

10 kNïm10.6 kNïm

5.23 kN 12.2 kN

4.78 + 5.8 = 10.58 kN

V (kN)x (m)

5.23

- 4.78

5.8

-12.2

+-

+

-

M (kNïm) x (m)

-10.6

10.3

-8.8 -10- -

+-

EIB4.2

Deflected shape x (m)

Page 31: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 6 kN/m

A C

B 6 m4 m4 m

Example 2

Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

Page 32: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

32

PPL8 wwL2

30FEM

PL8

wL2

20

[M] = [K][Q] + [FEM]

10 kN 6 kN/m

A C

B 6 m4 m4 m

)1(8

)8)(10(8

28

4−−−++= BAAB

EIEIM θθ

)2(8

)8)(10(8

48

2−−−−+= BABA

EIEIM θθ

)3(30

)6)(6(6

26

4 2

−−−++= CBBCEIEIM θθ

)4(20

)6)(6(6

46

2 2

−−−−+= CBCBEIEIM θθ

10

10

0

0

0

308

62:)1()2(2 −=− BBAEIM θ

)5(158

3−−−−= BBA

EIM θ

Page 33: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

33

MBA MBC

B

)5(158

3−−−−= BBA

EIM θ

)3(30

)6)(6(6

4 2

−−−+= BBCEIM θ

0:0 =−−=Σ+ BCBAB MMM

EI

EIEI

B

B

488.7

)6(030

)6)(6(15)6

48

3(2

=

−−−=+−+

θ

θ

EI

EIEIinSubstitute

A

BAB

74.23

108

28

40:)1(

−=

−+=

θ

θθθ

Substitute θA and θB in (5), (3) and (4):

MBC = 12.19 kNïm

MCB = - 8.30 kNïm

MBA = - 12.19 kNïm)4(

20)6)(6(

62 2

−−−−= BCBEIM θ

Page 34: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

34

10 kN 6 kN/m

A C

B 6 m4 m4 m

MBA = - 12.19 kNïm, MBC = 12.19 kNïm, MCB = - 8.30 kNïm

= 3.48 kN = 6.52 kN = 6.65 kN = 11.35 kN

12.19 kNïm

12.19 kNïm8.30 kNïm

10 kN

12.19 kNïm

A B

ByLAy

2 m

6 kN/m12.19 kNïm

8.30 kNïmC

CyByR

18 kN

B

Page 35: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

35

Deflected shape x (m)EIB49.7

10 kN 6 kN/m

A C

B 6 m4 m4 m

V (kN)x (m)

3.48

- 6.52

6.65

-11.35

M (kNïm) x (m)

14

-12.2-8.3

11.35 kN3.48 kN

6.52 + 6.65 = 13.17 kN

EIA74.23−

Page 36: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

36

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

Example 3

Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

Page 37: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI(4)(62)/12 (4)(62)/12 (10)(8)/8(10)(8)/8

15

12

)1(8

)8)(10(8

)2(28

)2(4−−−++= BAAB

EIEIM θθ

)2(8

)8)(10(8

)2(48

)2(2−−−−+= BABA

EIEIM θθ

0 10

10

)2(8

)8)(10)(2/3(8

)2(3:2

)1()2(2 aEIM BBA −−−−=− θ

)3(12

)6)(4(6

)3(4 2

−−−+= BBCEIM θ

Page 38: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI(4)(62)/12 (4)(62)/12(3/2)(10)(8)/8

EIEIMM

B

BBCBA

/091.13151275.2:0

==+−==−−

θθ

15

12

)2(8

)8)(10)(2/3(8

)2(3 aEIM BBA −−−−= θ

)3(12

)6)(4(6

)3(4 2

−−−+= BBCEIM θ

mkNEIM

mkNEI

EIM

mkNEI

EIM

BCB

BC

BA

•−=−=

•=−=

•−=−=

91.10126

)3(2

18.1412)091.1(6

)3(4

18.1415)091.1(8

)2(3

θ

Page 39: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

39

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

MBA = - 14.18 kNïm, MBC = 14.18 kNïm, MCB = -10.91 kNïm

14.18

140.18 kNïm

10 kN

A B

ByLAy = 6.73 kN= 3.23 kN

14.18 kNïm

10.91 kNïm

4 kN/m

C

CyByR

24 kN

14.18 10.91

= 11.46 kN= 12.55 kN

Page 40: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

40

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

11.46 kN3.23 kN

10.91 kNïm

V (kN)x (m)

3.23

-6.73

12.55

-11.46

+-

+-

2.86

M (kNïm) x (m)

12.91

-14.18

5.53

-10.91

+

-+

-

6.77 + 12.55 = 19.32 kN

θB = 1.091/EIDeflected shape x (m)

Page 41: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

41

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

Example 4

Draw the quantitative shear , bending moment diagrams and qualitativedeflected curve for the beam shown. EI is constant.

12 kNïm

Page 42: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

42

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kNïm

wL2/12 = 12 wL2/12 = 121.5PL/8 = 15

MBA

MBCB

12 kNïmMBA

MBC

EIB273.3

−=θ

EIA21.7

−=θ

)1(158

)2(3−−−−= BBA

EIM θ

)2(126

)3(4−−−+= BBC

EIM θ

)3(126

)3(2−−−−= BCB

EIM θ

8)8)(10(

8)3(2

8)2(4

++= BAABEIEIM θθ

0 -3.273/EI

012:int =−−− BCBA MMBJo

012)122()1575.0( =−+−−− BEIEI θ

mkNEI

EIM BA •−=−−= 45.1715)273.3(75.0

mkNEI

EIM BC •=+−= 45.512)273.3(2

mkNEI

EIM CB •−=−−= 27.1512)273.3(

Page 43: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

43

10 kN

A B

4 kN/m

C

24 kN

17.45 kNïm5.45 kNïm

15.27 kNïm

2.82 kN 13.64 kN10.36 kN7.18 kN

12 kNïm10 kN 4 kN/m

A C

B13.64 kN2.82 kN

15.27 kNïm

17.54 kN

mkNEI

EIM BA •−=−−= 45.1715)273.3(75.0

mkNEI

EIM BC •=+−= 45.512)273.3(2

mkNEI

EIM CB •−=−−= 27.1512)273.3(

Page 44: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

44

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kNïm

13.64 kN2.82 kN

15.27 kNïm

17.54 kN

V (kN)x (m)3.41 m+

-+

-

2.82

-7.18

10.36

-13.64

M (kNïm) x (m)+

- -+

11.28

-17.45

-5.45-15.27

7.98

Deflected shape x (m)

EIB273.3

=θEIA

21.7−=θ

EIB273.3

−=θ

EIA21.7

−=θ

Page 45: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

45

Example 5

Draw the quantitative shear, bending moment diagrams, and qualitativedeflected curve for the beam shown. Support B settles 10 mm, and EI isconstant. Take E = 200 GPa, I = 200x106 mm4.

12 kNïm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

Page 46: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

46

12 kNïm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

[FEM]∆

A

B

2

6LEI∆

2

6LEI∆

BC

2

6LEI∆

2

6LEI∆

P w

[FEM]load

8PL

8PL

30

2wL30

2wL

)1(8

)8)(10(8

)01.0)(2(68

)2(28

)2(42 −−−+++=

EIEIEIM BAAB θθ

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 −−−−++=

EIEIEIM BABA θθ

)3(30

)6)(6(6

)01.0)(3(66

)3(26

)3(4 2

2 −−−+−+=EIEIEIM CBBC θθ

)4(30

)6)(6(6

)01.0)(3(66

)3(46

)3(2 2

2 −−−−−+=EIEIEIM CBCB θθ

-12

0

0

Page 47: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

47

12 kNïm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kNï m2 :

)1(8

)8)(10(8

)01.0)(2(68

)2(28

)2(42 −−−+++=

EIEIEIM BAAB θθ

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 −−−−++=

EIEIEIM BABA θθ

)1(10758

)2(28

)2(4−−−+++= BAAB

EIEIM θθ

)2(10758

)2(48

)2(2−−−−++= BABA

EIEIM θθ

)2(2/12)2/10(10)2/75(758

)2(3:2

)1()2(2 aEIM BBA −−−−−−−+=− θ

16.5

Page 48: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

48

+ ΣMB = 0: - MBA - MBC = 0 (3/4 + 2)EIθB + 16.5 - 192.8 = 0

θB = 64.109/ EI

Substitute θB in (1): θA = -129.06/EI

Substitute θA and θB in (5), (3), (4):

MBC = -64.58 kNïm

MCB = -146.69 kNïm

MBA = 64.58 kNïm,

MBA MBC

B

12 kNïm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

MBC = (4/6)(3EI)θB - 192.8

MBA = (3/4)(2EI)θB + 16.5

Page 49: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

49

12 kNïm 10 kN 6 kN/m

A CB

6 m4 m4 m

64.58 kNïm 64.58 kNïm

= -1.57 kN= 11.57 kN

146.69 kNïm

= 47.21 kN= -29.21 kN

10 kN

A B

ByLAy

12 kNïm64.58 kNïm

2 m

6 kN/m

C

CyByR

18 kN

B

64.58 kNïm

146.69 kNïm

MBC = -64.58 kNïm

MCB = -146.69 kNïm

MBA = 64.58 kNïm,

Page 50: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

50

12 kNïm 10 kN 6 kN/m

A C

B6 m4 m4 m

2EI 3EI

V (kN)x (m)

11.57 1.57

-29.21-47.21

-+

M (kNïm) x (m)

1258.29 64.58

-146.69

+

-

47.21 kN

146.69 kNïm

11.57 kN

1.57 + 29.21 = 30.78 kN

10 mmθA = -129.06/EI

θB = 64.109/ EIθA = -129.06/EI

Deflected shape x (m)

θB = 64.109/ EI

Page 51: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

51

Example 6

For the beam shown, support A settles 10 mm downward, use the slope-deflectionmethod to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape. (3 points)Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

Page 52: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

52

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

)1(3

)2(4−−−= CCB

EIM θ

)2(1005.43

)5.1(23

)5.1(4−−−+++= ACCA

EIEIM θθ

)2(2

122

1002

)5.4(33

)5.1(3:2

)2()2(2 aEIM CCA −−−+++=− θ

)3(1005.43

)5.1(43

)5.1(2−−−+−+= ACAC

EIEIM θθ12

0.01 m

C

AmkN •=

××

1003

)01.0)(502005.1(62

mkN •100MF

10 mm

6 kN/m

A C

5.412

)3(6 2

= 4.5MF

w

Page 53: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

53

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

)1(3

)2(4−−−= CCB

EIM θ

)2(2

122

1002

)5.4(33

)5.1(3 aEIM CCA −−−+++= θ

10 mm

MCBMCA

C

0=+ CACB MM

02

122

1002

)5.4(33

)5.48(=+++

+C

EI θ

radEIC 0015.006.15

−=−

)3(1005.43

)5.1(4)06.15(3

)5.1(212 −−−+−+−

= AEI

EIEI θSubstitute θC in eq.(3)

radEIA 0034.022.34

−=−

ï Equilibrium equation:

Page 54: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

54

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

radEIC 0015.006.15

−=−

=θ radEIA 0034.022.34

−=−

mkNEI

EIEIM CBC •−=−

== 08.20)06.15(3

)2(23

)2(2 θ

mkNEI

EIEIM CCB •−=−

== 16.40)06.15(3

)2(43

)2(4 θ

kN08.203

08.2016.40=

+kN08.20

B C40.16 kNïm20.08 kNïm

6 kN/m

AC

12 kNïm

40.16 kNïm

18 kN

8.39 kN26.39 kN

Page 55: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

55

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

6 kN/m

AC

12 kNïm

40.16 kNïm8.39 kN26.39 kN

B C40.16 kNïm20.08 kNïm

20.08 kN20.08 kN

V (kN)

x (m)

26.398.39

-20.08

+-

M (kNïm)

x (m)20.08

-40.16

12

radC 0015.0−=θ

radA 0034.0−=θ

Deflected shapex (m)

radC 0015.0=θ

radA 0034.0=θ

Page 56: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

56

Example 7

For the beam shown, support A settles 10 mm downward, use theslope-deflection method to(a)Determine all the slopes at supports(b)Determine all the reactions at supports(c)Draw its quantitative shear, bending moment diagrams, and qualitativedeflected shape.Take E= 200 GPa, I = 50(106) mm4.

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

Page 57: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

57

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

5.412

)3(6 2

=

6 kN/m

AC4.5

mkN •=

××

1003

)01.0)(502005.1(620.01 m

C

A

100

34 CEI∆

∆C

CB

34

3)2(6

2CC EIEI ∆

=∆

)2(3

43

)2(4−−−∆−= CCCB

EIEIM θ

)3(1005.43

)5.1(23

)5.1(4−−++∆++= CACCA EIEIEIM θθ

)4(1005.43

)5.1(43

)5.1(2−−+−∆++= CACAC EIEIEIM θθ

12

)1(3

43

)2(2−−−∆−= CCBC

EIEIM θ

)3(2

122

1002

)5.4(323

)5.1(3:2

)4()3(2 aEIEIM CCCA −−−+++∆+=− θ

CC EIEI

∆=∆

23)5.1(6∆C

C

A

CEI∆

Page 58: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

58

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

ï Equilibrium equation:

)3

()( CBBCCBy

MMC +−=

6 kN/m

AC

12 kNïm

MCA

18 kN

AyB C

MCBMBC

By3

393

)5.1(1812)( +=

++= CACA

CAyMMC

C

MCB MCA

(Cy)CA(Cy)CB

*)1(0:0 −−−=+=Σ CACBC MMM

*)2(0)()(:0 −−−=+=Σ CAyCByy CCC

)5(15.628333.0167.4*)1( −−−−=∆− CC EIEIinSubstitute θ

)6(75.101167.35.2*)2( −−−−=∆+− CC EIEIinSubstitute θ

mmEIradEIandFrom CC 227.5/27.5200255.0/51.25)6()5( −=−=∆−=−=θ

Page 59: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

59

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

10 mm

ï Solve equation

radEIC 00255.051.25

−=−

mmEIC 227.527.52

−=−

=∆

Substitute θC and ∆C in (4)

radEIA 000286.086.2

−=−

Substitute θC and ∆C in (1), (2) and (3a)

mkNM BC •= 68.35

mkNMCB •= 67.1

mkNMCA •−= 67.1

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

35.68 kNïm

kN

Ay

55.56

68.3512)5.4(18

=

−−=

kNBy

45.12

55.518

=

−=

Page 60: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

60

10 mm

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNïm

35.68 kNïm

12.45 kN 5.55 kN

Deflected shape

x (m)

radC 00255.0−=θ

mmC 227.5−=∆

radA 000286.0−=θ

radC 00255.0=θradA 000286.0=θ

mmC 227.5=∆

M (kNïm)

x (m)1214.57

1.67

-35.68

-+

V (kN)

x (m)0.925 m12.45

-5.55

+

Page 61: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

61

Example of Frame: No Sidesway

Page 62: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

62

Example 6

For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame.

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EI

Page 63: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

63

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EIPL/8 = 30

PL/8 = 30

ï Equilibrium equations

10

MBC

MBA

)3(18366

)2(3−−−++= BBC

EIM θ

*)1(010 −−−=−− BCBA MM

05430310 =−+− BEIθ

EIEIB667.4

)3(14 −

=−

=θ)1(30

6)3(2

−−−+= BABEIM θ

mkNMmkNM

mkNM

BC

BA

AB

•=•−=

•=

33.4933.39

33.25

36/2 = 18

(wL2/12 ) =3636

ï Slope-Deflection Equations

)2(306

)3(4−−−−= BBA

EIM θ

Substitute (2) and (3) in (1*)

)3()1(667.4 toinEI

Substitute B−

Page 64: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

64

10 kN

C

12 kN/m

A

B

6 m

40 kN

3 m

3 m

1 m

2EI

3EI39.33

25.33

49.33

A

B

40 kN

39.33

25.33

C

12 kN/m

B49.33

17.67 kN

27.78 kN

Bending moment diagram

-25.33

27.7

-39.3

Deflected curve

-49.33

20.58

10

θB = -4.667/EIθB

θB

MAB = 25.33 kNïm

MBA = -39.33 kNïm

MBC = 49.33 kNïm

Page 65: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

65

A

B

C

D

E

25 kN

5 m

5 kN/m

60(106) mm4

240(106) mm4 180(106)

120(106) mm4

3 m 4 m3 m

Example 7

Draw the quantitative shear, bending moment diagrams and qualitativedeflected curve for the frame shown. E = 200 GPa.

Page 66: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

66

A

B

C

D

E

25 kN

5 m

5 kN/m

60(106) mm4

240(106) mm4 180(106)

120(106) mm4

3 m 4 m3 m0

BAABEIEIM θθ

5)2(2

5)2(4

+=0

BABAEIEIM θθ

5)2(4

5)2(2

+=

75.186

)4(26

)4(4++= CBBC

EIEIM θθ

75.186

)4(46

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)(3=

104

)3(3+= CCE

EIM θ

0=+ BCBA MM

)1(75.18)68()

616

58( −−−−=++ CB EIEI θθ

0=++ CECDCB MMM

)2(75.8)49

53

616()

68( −−−=+++ CB EIEI θθ

EIEIandFrom CB

86.229.5:)2()1( =−

= θθ

PL/8 = 18.75 18.75

6.667 (wL2/12 ) = 6.667+ 3.333

Page 67: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

67

MAB = −4.23 kNïm

MBA = −8.46 kNïm

MBC = 8.46 kNïm

MCB = −18.18 kNïm

MCD = 1.72 kNïm

MCE = 16.44 kNïm

Substitute θB = -1.11/EI, θc = -20.59/EI below

0

0BAAB

EIEIM θθ5

)2(25

)2(4+=

BABAEIEIM θθ

5)2(4

5)2(2

+=

75.186

)4(26

)4(4++= CBBC

EIEIM θθ

75.186

)4(46

)4(2−+= CBCB

EIEIM θθ

CCDEIM θ5

)(3=

104

)3(3+= CCE

EIM θ

Page 68: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

68

MAB = -4.23 kNïm, MBA = -8.46 kNïm, MBC = 8.46 kNïm, MCB = -18.18 kNïm,MCD = 1.72 kNïm, MCE = 16.44 kNïm

A

B

5 m

C E

20 kN

A

B

5 m

4.23 kNïm

2.54 kN

(8.46 + 4.23)/5 = 2.54 kN

8.46 kNïm

C

25 kN

B 3 m 3 m2.54 kN 2.54 kN

8.46 kNïm(25(3)+8.46-18.18)/6 = 10.88 kN

14.12 kN18.18 kNïm

16.44 kNïm

(20(2)+16.44)/4= 14.11 kN

5.89 kN

0.34 kN

28.23 kN

14.12+14.11=28.23 kN1.72 kNïm

(1.72)/5 = 0.34 kN

10.88 kN

10.88 kN

2.54-0.34=2.2 kN

2.2 kN

Page 69: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

69

Shear diagram

-2.54

-2.54

10.88

-14.12

14.11

-5.89

0.34

+

-

-

+-

+

2.82 m

1.18 m

Deflected curve

1.18 m

0.78 m2.33 m1.29 m

1.67m

1.29 m

0.78 m

Moment diagram

1.18 m

3.46

24.18

1.72

-18.18 -16.44

4.23

-8.46-8.46

2.33 m

+

-

+

- -

+

θB = −5.29/EI

θC = 2.86/EI1.67m

Page 70: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

70

Example of Frames: Sidesway

Page 71: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

71

A

B C

D

3m

10 kN

3 m

1 m

Example 8

Determine the moments at each joint of the frame and draw the quantitativebending moment diagrams and qualitative deflected curve . The joints at A andD are fixed and joint C is assumed pin-connected. EI is constant for each member

Page 72: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

72

A

B C

D

3m

10 kN

3 m

1 m

MABMDC

Ax Dx

Ay Dy

ï Boundary Conditions

θA = θD = 0

ï Equilibrium Conditions

ï Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

ï Overview

B- Joint B

MBCMBA

Page 73: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

73

)3(3

)(3−−−= BBC

EIM θ

)4(1875.0375.021375.0 −−−∆=∆−∆= EIEIEIM DC

)1(4

64

)1)(3(104

)(222

2

−−−∆

++=EIEIM BAB θ

5.625 0.375EI∆

A

B C

D3m

10 kN

3 m

1 m

(5.625)load

(1.875)load

(0.375EI∆)∆

(0.375EI∆)∆

(0.375EI∆)∆

(0.375EI∆)∆

:0=Σ+ CM

MDC

D

C

4 m

Dx

MAB

MBA

A

B

4 m

Ax

10 kN

∆ ∆

(1/2)(0.375EI∆)∆

:0=Σ+ BM

)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ4

)( BAABx

MMA +=

)6(0468.04

−−−∆== EIMD DCx

)2(4

64

)1)(3(104

)(422

2

−−−∆

+−=EIEIM BBA θ

5.625 0.375EI∆

ï Slope-Deflection Equations

Page 74: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

74

ï Solve equation

*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

)3(3

)(3−−−= BBC

EIM θ

)4(1875.0 −−−∆= EIM DC

)1(375.0625.54

)(2−−−∆++= EIEIM BAB θ

)5(563.11875.0375.0 −−−+∆+= EIEIA Bx θ

)6(0468.0 −−−∆= EIDx

)2(375.0625.54

)(4−−−∆+−= EIEIM BBA θ

Equilibrium Conditions:

Slope-Deflection Equations:

Horizontal reaction at supports:

Substitute (2) and (3) in (1*)

2EI θB + 0.375EI ∆ = 5.625 ----(7)

Substitute (5) and (6) in (2*)

)8(437.8235.0375.0 −−−−=∆−− EIEI Bθ

From (7) and (8) can solve;

EIEIB8.446.5

=∆−

)6()1(8.446.5 toinEI

andEI

Substitute B =∆−

MAB = 15.88 kNïmMBA = 5.6 kNïmMBC = -5.6 kNïmMDC = 8.42 kNïmAx = 7.9 kNDx = 2.1 kN

Page 75: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

75

A

BC

D

Deflected curve

A

BC

D

Bending moment diagram

MAB = 15.88 kNïm, MBA = 5.6 kNïm, MBC = -5.6 kNïm, MDC = 8.42 kNïm, Ax = 7.9 kN, Dx = 2.1 kN,

A

B C

D

3m

10 kN

3 m

1 m

15.88

5.6

8.42

7.9 kN 2.1 kN

15.88

5.6

8.42

∆ = 44.8/EI ∆ = 44.8/EI

θB = -5.6/EI5.6

7.8

Page 76: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

76

B C

DA

3m4 m

10 kN4 m

2 m

pin

2 EI

2.5 EI EI

Example 9

From the frame shown use the slope-deflection method to:(a) Determine the end moments of each member and reactions at supports(b) Draw the quantitative bending moment diagram, and also draw thequalitative deflected shape of the entire frame.

Page 77: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

77

B

C

DA

3m4 m

10 kN4 m

2 m

2EI

2.5EI EI

Ax Dx

Ay Dy

MDCMAB

∆ CD C¥∆BC

ï Overview

ï Boundary Conditions

θA = θD = 0

ï Equilibrium Conditions

ï Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

Page 78: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

78

B

C

DA 3m4 m

10 kN4 m

2 m

2EI

2.5EI EI36.87c = ∆ tan 36.87c = 0.75 ∆

= ∆ / cos 36.87c = 1.25 ∆

∆∆BC

∆ CD C¥

∆BC

∆CD

C

36.87c

)1(5375.04

)(2−−−+∆+= EIEIM BAB θ

)2(5375.04

)(4−−−−∆+= EIEIM BBA θ

)3(2813.04

)2(3−−−∆−= EIEIM BBC θ

)4(375.0 −−−∆= EIM DC

C

BB¥

∆BC= 0.75 ∆

0.375EI∆

6EI∆/(4) 2 = 0.375EI∆

B

A

∆¥ B¥

(6)(2EI)(0.75∆)/(4) 2 = 0.5625EI∆0.5625EI∆

D

C

C¥∆CD= 1.25 ∆

(1/2) 0.75EI∆

(1/2) 0.5625EI∆

PL/8 = 5

5

(6)(2.5EI)(1.25∆)/(5)2 = 0.75EI∆

0.75EI∆

ï Slope-Deflection Equation

Page 79: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

79

B

C

DA

3m4 m

10 kN4 m

2 m

pin2 EI

2.5 EI EI

Dx= (MDC-(3/4)MBC)/4 ---(6)

MBC

4

Ax = (MBA+ MAB-20)/4 -----(5)

B CMBC

C

D

MBC/4MDC

MBC

4

A

B

MBA

10 kN

MAB

ï Horizontal reactions

Page 80: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

80

ï Solve equations

*)2(010 −−−=−− xx DA*)1(0 −−−=+ BCBA MM

Equilibrium Conditions:

Slope-Deflection Equation:

Horizontal reactions at supports:

Substitute (2) and (3) in (1*)

Substitute (5) and (6) in (2*)

From (7) and (8) can solve;

EIEIB56.1445.1 −

=∆=θ

)6()1(56.1445.1 toinEI

andEI

Substitute B−

=∆=θ

MAB = 15.88 kNïmMBA = 5.6 kNïmMBC = -5.6 kNïmMDC = 8.42 kNïmAx = 7.9 kNDx = 2.1 kN

)1(4

654

)(22 −−−∆

++=EIEIM BAB θ

)2(4

654

)(42 −−−∆

+−=EIEIM BBA θ

)3(4

)75.0)(2(34

)2(32 −−−

∆−=

EIEIM BBC θ

)4(5

)25.1)(5.2(32 −−−

∆=

EIM DC

)5(4

)20(−−−

−+= ABBA

xMMA

)6(443

−−−−

=BCDC

x

MMD

)7(050938.05.2 −−−=−∆+ EIEI Bθ

)8(05334.00938.0 −−−=−∆+ EIEI Bθ

Page 81: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

81

BC

DA

Deflected shape

θB=1.45/EI

θB=1.45/EI

Bending-moment diagram

BC

DA 11.19

1.91

5.35

5.46

1.91

∆ ∆

MAB = 11.19 kNïm MBA = 1.91 kNïm MBC = -1.91 kNïm MDC = 5.46 kNïm

Ax = 8.28 kNïm Dx = 1.72 kNïm

B

C

DA

3m4 m

10 kN4 m

2 m

pin2 EI

2.5 EI EI11.19 kNïm

8.27 kN

5.46

1.91

1.91

0.478 kN1.73

0.478 kN

Page 82: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

82

Example 10

From the frame shown use the moment distribution method to:(a) Determine all the reactions at supports, and also(b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve.

A

BC

D

3 m

3m

4m

20 kN/mpin

2EI

3EI

4EI

Page 83: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

83

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

[FEM]load

ï Overview

∆ ∆

ï Boundary Conditions

θA = θD = 0

ï Equilibrium Conditions

ï Unknowns

θB and ∆

- Entire Frame

*)2(060:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

Page 84: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

84

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

[FEM]load

wL2/12 = 15

wL2/12 = 15

A

B C

D

∆ ∆

[FEM]∆∆∆∆

6(2EI∆)/(3) 2= 1.333EI∆

6(2EI∆)/(3) 2= 1.333EI∆ 6(4EI∆)/(4) 2

= 1.5EI∆

1.5EI∆

BBCEIM θ3

)3(3=

∆++= EIEI B 333.115333.1 θ ----------(1)

∆+−= EIEI B 333.115667.2 θ ----------(2)

BEIθ3= ----------(3)

∆= EI75.0 ----------(4)

∆+++= EIEIEIM BAAB 333.1153

)2(23

)2(4 θθ0

∆+−+= EIEIEIM BABA 333.1153

)2(43

)2(2 θθ0

∆+= EIEIM DDC 75.04

)4(3 θ0

(1/2)(1.5EI∆)

ï Slope-Deflection Equation

Page 85: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

85

C

Dx

MDC

4 m

D

MAB

A

B

60 kN

MBA

1.5 m

1.5 m

Ax + ΣMC = 0:

:0=Σ+ BM

)6(188.04

−−−∆== EIMD DCx

)5(30889.0333.13

)5.1(60

−−−+∆+=

++=

EIEIA

MMA

Bx

ABBAx

θ

ï Horizontal reactions

Page 86: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

86

ï Solve equation

*)1(0 −−−=+ BCBA MM

Equilibrium Conditions

Equation of moment

Horizontal reaction at support

Substitute (2) and (3) in (1*)

Substitute (5) and (6) in (2*)

From (7) and (8), solve equations;

EIEIB67.3451.5

=∆−

)6()1(67.3451.5 toinEI

andEI

Substitute B =∆−

)7(15333.1667.5 −−−=∆+ EIEI Bθ

)8(30077.1333.1 −−−−=∆−− EIEI Bθ

*)2(060 −−−=−− xx DA

)1(333.115333.1 −−−∆++= EIEIM BAB θ

)2(333.115667.2 −−−∆+−= EIEIM BBA θ

)3(3 −−−= BBC EIM θ

)4(75.0 −−−∆= EIM DC

)6(188.0 −−−∆= EIDx

)5(30889.0333.1 −−−+∆+= EIEIA Bx θ

mkNM AB •= 87.53mkNM BA •= 52.16

mkNM BC •−= 52.16mkNM DC •= 0.26

Ax = 53.48 kNDx = 6.52 kN

Page 87: DISPLACEMENT METHOD OF ANALYSIS: SLOPE ...Equations:: ∆ = − ∆ = −

87

A

C

DMoment diagram

D

A

B C

Deflected shape

∆ ∆

A

BC

D

3 m

3m

4m20 k

N/m 16.52 kNïm

53.87 kNïm

53.48 kN

6.52 kN26 kNïm

5.55 kN

5.55 kN

26.

16.52

mkNM AB •= 87.53mkNM BA •= 52.16

mkNM BC •−= 52.16mkNM DC •= 0.26

Ax = 53.48 kNDx = 6.52 kN

53.87

B16.52