Transcript
Page 1: Slope Deflection Method

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: Slope Deflection Method

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Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 3: Slope Deflection Method

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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: Slope Deflection Method

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L

A B

1

Stiffness

kBAkAA

LEIkAA

4=

LEIkBA

2=

Page 5: Slope Deflection Method

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L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

Page 6: Slope Deflection Method

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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: Slope Deflection Method

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General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

Page 8: Slope Deflection Method

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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: Slope Deflection Method

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Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

Page 10: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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Example of Beams

Page 27: Slope Deflection Method

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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: Slope Deflection Method

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PPL8 wwL2

30FEM

PL8

wL2

20

MBA MBC

B

Substitute θB in the moment equations:

MBC = 8.8 kN�m

MCB = -10 kN�m

MAB = 10.6 kN�m,

MBA = - 8.8 kN�m,

[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: Slope Deflection Method

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10 kN 6 kN/m

A C

B 6 m4 m4 m

MBC = 8.8 kN�m

MCB= -10 kN�m

MAB = 10.6 kN�m,

MBA = - 8.8 kN�m,

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

10 kN�m8.8 kN�m

10.6 kN�m8.8 kN�m

10 kN

10.6 kN�m

8.8 kN�m

A B

ByLAy

2 m

6 kN/m8.8 kN�m

10 kN�mB

CyByR

18 kN

Page 30: Slope Deflection Method

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10 kN 6 kN/m

A C

B 6 m4 m4 m

10 kN�m10.6 kN�m

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 (kN�m) x (m)

-10.6

10.3

-8.8 -10- -

+-

EIB4.2

Deflected shape x (m)

Page 31: Slope Deflection Method

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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: Slope Deflection Method

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: Slope Deflection Method

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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 kN�m

MCB = - 8.30 kN�m

MBA = - 12.19 kN�m)4(

20)6)(6(

62 2

−−−−= BCBEIM θ

Page 34: Slope Deflection Method

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10 kN 6 kN/m

A C

B 6 m4 m4 m

MBA = - 12.19 kN�m, MBC = 12.19 kN�m, MCB = - 8.30 kN�m

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

12.19 kN�m

12.19 kN�m8.30 kN�m

10 kN

12.19 kN�m

A B

ByLAy

2 m

6 kN/m12.19 kN�m

8.30 kN�mC

CyByR

18 kN

B

Page 35: Slope Deflection Method

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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 (kN�m) x (m)

14

-12.2-8.3

11.35 kN3.48 kN

6.52 + 6.65 = 13.17 kN

EIA74.23−

Page 36: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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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: Slope Deflection Method

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

MBA = - 14.18 kN�m, MBC = 14.18 kN�m, MCB = -10.91 kN�m

14.18

140.18 kN�m

10 kN

A B

ByLAy = 6.73 kN= 3.23 kN

14.18 kN�m

10.91 kN�m

4 kN/m

C

CyByR

24 kN

14.18 10.91

= 11.46 kN= 12.55 kN

Page 40: Slope Deflection Method

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

11.46 kN3.23 kN

10.91 kN�m

V (kN)x (m)

3.23

-6.73

12.55

-11.46

+-

+-

2.86

M (kN�m) 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: Slope Deflection Method

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 kN�m

Page 42: Slope Deflection Method

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kN�m

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

MBA

MBCB

12 kN�mMBA

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: Slope Deflection Method

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10 kN

A B

4 kN/m

C

24 kN

17.45 kN�m5.45 kN�m

15.27 kN�m

2.82 kN 13.64 kN10.36 kN7.18 kN

12 kN�m10 kN 4 kN/m

A C

B13.64 kN2.82 kN

15.27 kN�m

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: Slope Deflection Method

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10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kN�m

13.64 kN2.82 kN

15.27 kN�m

17.54 kN

V (kN)x (m)3.41 m+

-+

-

2.82

-7.18

10.36

-13.64

M (kN�m) 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: Slope Deflection Method

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 kN�m 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

Page 46: Slope Deflection Method

46

12 kN�m 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: Slope Deflection Method

47

12 kN�m 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN� 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: Slope Deflection Method

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 kN�m

MCB = -146.69 kN�m

MBA = 64.58 kN�m,

MBA MBC

B

12 kN�m 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: Slope Deflection Method

49

12 kN�m 10 kN 6 kN/m

A CB

6 m4 m4 m

64.58 kN�m 64.58 kN�m

= -1.57 kN= 11.57 kN

146.69 kN�m

= 47.21 kN= -29.21 kN

10 kN

A B

ByLAy

12 kN�m64.58 kN�m

2 m

6 kN/m

C

CyByR

18 kN

B

64.58 kN�m

146.69 kN�m

MBC = -64.58 kN�m

MCB = -146.69 kN�m

MBA = 64.58 kN�m,

Page 50: Slope Deflection Method

50

12 kN�m 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 (kN�m) x (m)

1258.29 64.58

-146.69

+

-

47.21 kN

146.69 kN�m

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: Slope Deflection Method

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 kN�m

10 mm

Page 52: Slope Deflection Method

52

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

)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: Slope Deflection Method

53

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

)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

−=−

� Equilibrium equation:

Page 54: Slope Deflection Method

54

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

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 kN�m20.08 kN�m

6 kN/m

AC

12 kN�m

40.16 kN�m

18 kN

8.39 kN26.39 kN

Page 55: Slope Deflection Method

55

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

6 kN/m

AC

12 kN�m

40.16 kN�m8.39 kN26.39 kN

B C40.16 kN�m20.08 kN�m

20.08 kN20.08 kN

V (kN)

x (m)

26.398.39

-20.08

+-

M (kN�m)

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: Slope Deflection Method

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 kN�m

10 mm

Page 57: Slope Deflection Method

57

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

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: Slope Deflection Method

58

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

� Equilibrium equation:

)3

()( CBBCCBy

MMC +−=

6 kN/m

AC

12 kN�m

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: Slope Deflection Method

59

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

� 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 kN�m

35.68 kN�m

kN

Ay

55.56

68.3512)5.4(18

=

−−=

kNBy

45.12

55.518

=

−=

Page 60: Slope Deflection Method

60

10 mm

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

35.68 kN�m

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 (kN�m)

x (m)1214.57

1.67

-35.68

-+

V (kN)

x (m)0.925 m12.45

-5.55

+

Page 61: Slope Deflection Method

61

Example of Frame: No Sidesway

Page 62: Slope Deflection Method

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: Slope Deflection Method

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

� 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

� 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: Slope Deflection Method

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 kN�m

MBA = -39.33 kN�m

MBC = 49.33 kN�m

Page 65: Slope Deflection Method

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: Slope Deflection Method

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: Slope Deflection Method

67

MAB = −4.23 kN�m

MBA = −8.46 kN�m

MBC = 8.46 kN�m

MCB = −18.18 kN�m

MCD = 1.72 kN�m

MCE = 16.44 kN�m

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: Slope Deflection Method

68

MAB = -4.23 kN�m, MBA = -8.46 kN�m, MBC = 8.46 kN�m, MCB = -18.18 kN�m,MCD = 1.72 kN�m, MCE = 16.44 kN�m

A

B

5 m

C E

20 kN

A

B

5 m

4.23 kN�m

2.54 kN

(8.46 + 4.23)/5 = 2.54 kN

8.46 kN�m

C

25 kN

B 3 m 3 m2.54 kN 2.54 kN

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

14.12 kN18.18 kN�m

16.44 kN�m

(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 kN�m

(1.72)/5 = 0.34 kN

10.88 kN

10.88 kN

2.54-0.34=2.2 kN

2.2 kN

Page 69: Slope Deflection Method

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: Slope Deflection Method

70

Example of Frames: Sidesway

Page 71: Slope Deflection Method

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: Slope Deflection Method

72

A

B C

D

3m

10 kN

3 m

1 m

MABMDC

Ax Dx

Ay Dy

� Boundary Conditions

θA = θD = 0

� Equilibrium Conditions

� Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

� Overview

B- Joint B

MBCMBA

Page 73: Slope Deflection Method

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∆

� Slope-Deflection Equations

Page 74: Slope Deflection Method

74

� 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 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 7.9 kNDx = 2.1 kN

Page 75: Slope Deflection Method

75

A

BC

D

Deflected curve

A

BC

D

Bending moment diagram

MAB = 15.88 kN�m, MBA = 5.6 kN�m, MBC = -5.6 kN�m, MDC = 8.42 kN�m, 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: Slope Deflection Method

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: Slope Deflection Method

77

B

C

DA

3m4 m

10 kN4 m

2 m

2EI

2.5EI EI

Ax Dx

Ay Dy

MDCMAB

∆ CD C´∆BC

� Overview

� Boundary Conditions

θA = θD = 0

� Equilibrium Conditions

� Unknowns

θB and ∆

- Entire Frame

*)2(010:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

Page 78: Slope Deflection Method

78

B

C

DA 3m4 m

10 kN4 m

2 m

2EI

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

= ∆ / cos 36.87° = 1.25 ∆

∆∆BC

∆ CD C´

∆BC

∆CD

C

36.87°

)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∆

� Slope-Deflection Equation

Page 79: Slope Deflection Method

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

� Horizontal reactions

Page 80: Slope Deflection Method

80

� 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 kN�mMBA = 5.6 kN�mMBC = -5.6 kN�mMDC = 8.42 kN�mAx = 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: Slope Deflection Method

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 kN�m MBA = 1.91 kN�m MBC = -1.91 kN�m MDC = 5.46 kN�m

Ax = 8.28 kN�m Dx = 1.72 kN�m

B

C

DA

3m4 m

10 kN4 m

2 m

pin2 EI

2.5 EI EI11.19 kN�m

8.27 kN

5.46

1.91

1.91

0.478 kN1.73

0.478 kN

Page 82: Slope Deflection Method

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: Slope Deflection Method

83

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

[FEM]load

� Overview

∆ ∆

� Boundary Conditions

θA = θD = 0

� Equilibrium Conditions

� Unknowns

θB and ∆

- Entire Frame

*)2(060:0 −−−=−−=Σ→+

xxx DAF

*)1(0:0 −−−=+=Σ BCBAB MMM

B- Joint B

MBCMBA

Page 84: Slope Deflection Method

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∆)

� Slope-Deflection Equation

Page 85: Slope Deflection Method

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

� Horizontal reactions

Page 86: Slope Deflection Method

86

� 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: Slope Deflection Method

87

A

C

DMoment diagram

D

A

B C

Deflected shape

∆ ∆

A

BC

D

3 m

3m

4m20 k

N/m 16.52 kN�m

53.87 kN�m

53.48 kN

6.52 kN26 kN�m

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


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