# slope deflection method

87
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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DESCRIPTION

a method to solve indeterminate statics

TRANSCRIPT

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

2

Slope � Deflection Equations

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

3

Degrees of Freedom

L

θΑ

A B

M

1 DOF: θΑ

PθΑ

θΒΒΒΒ

A BC 2 DOF: θΑ , θΒΒΒΒ

4

L

A B

1

Stiffness

kBAkAA

LEIkAA

4=

LEIkBA

2=

5

L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

6

P

L/2 L/2

L

w

L

8PL

8PL

2P

2P

12

2wL12

2wL

2wL

2wL

7

General Case

settlement = ∆j

Pi j kw Cj

Mij MjiwP

θj

θi

ψ

i j

8

wP

settlement = ∆j

(MFij)∆ (MF

ji)∆

+

Mij Mji

θi

θj

+

i jwPMij

Mji

settlement = ∆jθj

θi

ψ

=+ ji LEI

LEI θθ 24

ji LEI

LEI θθ 42

+=

ijF

jiij MMLEI

Fji

Fjiji MM

LEI

LEIM )()()4()2( +++= ∆θθ

L

9

Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+−−=Σ+ jjkjij CMMM

10

+

1

1

i jMij Mji

θi

θj

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

iθ×

jθ×

Stiffness Coefficients

L

11

[ ]

=

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

12

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

Displacementmatrix

Stiffness matrix

Force matrixwP(MF

+

+

i jwPMij

Mji

θj

θi

ψ ∆j

Mij Mji

θi

θj

(MFij)∆ (MF

ji)∆

Fixed-end momentmatrix

][]][[][ FEMKM += θ

]][[])[]([ θKFEMM =−

][][][][ 1 FEMMK −= −θ

L

13

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

14

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

15

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

16

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-

17

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

18

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

19

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

20

AB

wP

A B

A BθA θB

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−−−+++== ∆

FABBAAAB MM

LEI

LEIMM θθ

)2()()(42−−−+++= ∆

FBABABA MM

LEI

LEIM θθ

2)(

21])(

21)[(3:

2)1()2(2lim AF

FBABBAA

MMMMLEIMbyinateE ++−+=

−∆θθ

MA

wP

AB

21

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=−−+

22

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

23

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

24

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

25

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

26

Example of Beams

27

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.

28

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

29

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

30

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)

31

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.

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

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 kN�m

MCB = - 8.30 kN�m

MBA = - 12.19 kN�m)4(

20)6)(6(

62 2

−−−−= BCBEIM θ

34

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

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 (kN�m) x (m)

14

-12.2-8.3

11.35 kN3.48 kN

6.52 + 6.65 = 13.17 kN

EIA74.23−

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.

37

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

38

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

39

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

40

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)

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

42

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(

43

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(

44

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

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

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

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

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

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

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,

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

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

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

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

−=−

)3(1005.43

)5.1(4)06.15(3

)5.1(212 −−−+−+−

= AEI

EIEI θSubstitute θC in eq.(3)

−=−

� Equilibrium equation:

54

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

−=−

−=−

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

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

Deflected shapex (m)

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

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∆

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

59

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kN�m

10 mm

� Solve equation

−=−

mmEIC 227.527.52

−=−

=∆

Substitute θC and ∆C in (4)

−=−

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

=

−=

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)

mmC 227.5−=∆

mmC 227.5=∆

M (kN�m)

x (m)1214.57

1.67

-35.68

-+

V (kN)

x (m)0.925 m12.45

-5.55

+

61

Example of Frame: No Sidesway

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

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−

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

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.

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

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

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

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

70

Example of Frames: Sidesway

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

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

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

(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

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

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

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.

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

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

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

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

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

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

83

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

� 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

84

A

BC

D

3 m

3m

4m

20 k

N/m

2EI

3EI

4EI

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

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

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

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