# DISPLACEMENT METHOD OF ANALYSIS: SLOPE e! ents! on! nts Span! s Beams! ay! ay DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS

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! 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

j

i

i j

• 3

Degrees of Freedom

L

A B

M

1 DOF:

P

A B C 2 DOF: ,

• 4

L

A B

1

Stiffness

kBAkAA

LEIkAA

4=

LEIkBA

2=

• 5

L

A B1

kBBkAB

LEIkBB

4=

LEIkAB

2=

• 6

Fixed-End ForcesFixed-End Moments: Loads

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

j

i

i j

• 8

wP

settlement = j

(MFij) (MFji)

+

Mij Mji

i

j

+

i jwPMijMji

settlement = jj

i

=+ ji LEI

LEI 24

ji LEI

LEI 42 +=

ijF

jiij MMLEI

Fji

Fjiji MML

EILEIM )()()4()2( +++=

L

• 9

Mji Mjk

Pi j kw Cj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+=+ jjkjij CMMM

• 10

+

1

1

i jMij Mji

i

j

LEIkii

4=

LEIk ji

2=

LEIkij

2=

LEIk jj

4=

i

j

Stiffness Coefficients

L

• 11

[ ]

=

jjji

ijii

kkkk

k

Stiffness Matrix

)()2()4( ijFjiij MLEI

LEIM ++=

)()4()2( jiFjiji MLEI

LEIM ++=

+

=

F

ji

Fij

j

iI

ji

ij

MM

LEILEILEILEI

MM

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

Matrix Formulation

• 12

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

Displacementmatrix

Stiffness matrix

+

+

i jwPMijMji

j

i

j

Mij Mji

i

j

(MFij) (MFji)

Fixed-end momentmatrix

][]][[][ FEMKM +=

]][[])[]([ KFEMM =

][][][][ 1 FEMMK =

L

• 13

L

Real beam

Conjugate beam

Stiffness Coefficients Derivation: Fixed-End Support

MjMi

L/3

i

LMM ji +

EIM j

EIMi

EILM j

2

EILMi

2

)1(2

0)3

2)(2

()3

)(2

(:0'

=

=+=+

ji

jii

MM

LEI

LMLEI

LMM

)2(0)2

()2

(:0 =+=+EI

LMEI

LMF jiiy ij

ii

LEIM

LEIM

andFrom

)2(

)4(

);2()1(

=

=

LMM ji +

i j

• 14

Conjugate beam

L

Real beam

Stiffness Coefficients Derivation: Pinned-End Support

Mi i

j

LM i

LMi

EIM i

EILMi

2

32L

i j

0)3

2)(2

(:0' ==+ LLEI

LMM iij

i j

0)2

()3

(:0 =+=+ jiiy EILM

EILMF

LEIM

EILM

ii

i3)

3(1 ===

)6

(EI

LM ij

=)3

(EI

LM ii =

• 15

Fixed end moment : Point Load

Real beam

8,0

162

22:0

2 PLMEI

PLEI

MLEI

MLFy ==+=+

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 ==+=+

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 =+=+ LEI

MLLEI

MLM B

EIM

EIML2

EIMEI

ML2

• 19

wP

A B

A B+A B

(FEM)AB (FEM)BA

Pin-Supported End Span: Simple Case

BA LEI

LEI 24 + BA L

EILEI 42 +

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

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

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

2)()()/3( BABABBA

FEMFEMLEIM +=

wP

AB

L

• 20

AB

wP

A B

A BA B

Pin-Supported End Span: With End Couple and Settlement

BA LEI

LEI 24 + BA L

EILEI 42 +

L

(MF AB) (MF

BA)

FABBAAAB MML

EILEIMM

FBABABA MML

EILEIM

2)(

21])(

21)[(3:

FBABBAA

MMMMLEIMbyinateE +++=

MAwP

AB

• 21

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=+

• 22

Typical Problem

0

0

0

0

A C

B

P1 P2

L1 L2

wCB

P

8PL

8PL w12

2wL

12

2wL

8024 11

11

LPLEI

LEIM BAAB +++=

8042 11

11

LPLEI

LEIM BABA ++=

128024

2222

22

wLLPLEI

LEIM CBBC ++++=

128042

2222

22

wLLPLEI

LEIM CBCB

+++=

L L

• 23

MBA MBC

A C

B

P1 P2

L1 L2

wCB

B

CB

8042 11

11

LPLEI

LEIM BABA ++=

128024

2222

22

wLLPLEI

LEIM CBBC ++++=

BBCBABB forSolveMMCM ==+ 0:0

• 24

A C

B

P1 P2

L1 L2

wCB

Substitute B in MAB, MBA, MBC, MCB

MABMBA

MBCMCB

0

0

0

0

8024 11

11

LPLEI

LEIM BAAB +++=

8042 11

11

LPLEI

LEIM BABA ++=

128024

2222

22

wLLPLEI

LEIM CBBC ++++=

128042

2222

22

wLLPLEI

LEIM CBCB

+++=

• 25

A BP1

MABMBA

L1

A CB

P1 P2

L1 L2

wCB

ByR CyAy ByL

Ay Cy

MABMBA

MBCMCB

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 wwL

2

30FEM

PL8

wL220

MBA MBC

B

Substitute B in the moment equations:

MBC = 8.8 kNm

MCB = -10 kNm

MAB = 10.6 kNm,

MBA = - 8.8 kNm,

[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

8)8)(10(

84

82

+= BABAEIEIM

30)6)(6(

62

64 2

++= CBBCEIEIM

20)6)(6(

64

62 2

+= CBCBEIEIM

0:0 ==+ BCBAB MMM

EI

EIEI

B

B

4.2

030

)6)(6(10)6

48

4(2

=

=++

• 29

10 kN 6 kN/m

A C

B 6 m4 m4 m

MBC = 8.8 kNm

MCB= -10 kNm

MAB = 10.6 kNm,

MBA = - 8.8 kNm,

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

10 kNm8.8 kNm

10.6 kNm8.8 kNm

10 kN

10.6 kNm

8.8 kNm

A B

ByLAy

2 m

6 kN/m8.8 kNm

10 kNmB

CyByR

18 kN

• 30

10 kN 6 kN/m

A C

B 6 m4 m4 m

10 kNm10.6 kNm

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 (kNm) 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 wwL

2

30FEM

PL8

wL220

[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

)2(8

)8)(10(8

48

2+= BABA

EIEIM

)3(30

)6)(6(6

26

4 2++= CBBC

EIEIM

)4(20

)6)(6(6

46

2 2+= CBCB

EIEIM

10

10

0

0

0

308

62:)1()2(2 = BBAEIM

)5(158

3= BBA

EIM

• 33

MBA MBC

B

)5(158

3= BBA

EIM

)3(30

)6)(6(6

4 2+= BBC

EIM

0:0 ==+ BCBAB MMM

EI

EIEI

B

B

488.7

)6(030

)6)(6(15)6

48

3(2

=

=++

EI

EIEIinSubstitute

A

BAB

74.23

108

28

40:)1(

=

+=

Substitute A and B in (5), (3) and (4):

MBC = 12.19 kNm

MCB = - 8.30 kNm

MBA = - 12.19 kNm)4(

20)6)(6(

62 2

= BCBEIM

• 34

10 kN 6 kN/m

A C

B 6 m4 m4 m

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

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

12.19 kNm

12.19 kNm8.30 kNm

10 kN

12.19 kNm

A B

ByLAy

2 m

6 kN/m12.19 kNm

8.30 kNmC

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 (kNm) 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

)2(8

)8)(10(8

)2(48

)2(2+= BABA

EIEIM

0 10

10

)2(8

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

)2(3:2

)1()2(2 aEIM BBA =

)3(12

)6)(4(6

)3(4 2+= BBC

EIM

• 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

==+==

15

12

)2(8

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

)2(3 aEIM BBA =

)3(12

)6)(4(6

)3(4 2+= BBC

EIM

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

• 39

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

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

14.18

140.18 kNm

10 kN

A B

ByLAy = 6.73 kN= 3.23 kN

14.18 kNm

10.91 kNm

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 kNm

V (kN)x (m)

3.23

-6.73

12.55

-11.46

+-

+-

2.86

M (kNm) x (m)

12.91

-14.18

5.53

-10.91

+

-+

-

6.77 + 12.55 = 19.32 kN

B = 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 kNm

• 42

10 kN 4 kN/m

A C

B6 m4 m4 m

2EI 3EI

12 kNm

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

MBA

MBCB

12 kNmMBA

MBC

EIB273.3

=

EIA21.7

=

)1(158

)2(3= BBA

EIM

)2(126

)3(4+= BBC

EIM

)3(126

)3(2= BCB

EIM

8)8)(10(

8)3(2

8)2(4

++= BAABEIEIM

0 -3.273/EI

012:int = BCBA MMBJo

012)122()1575.0( =+ BEIEI

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 kNm5.45 kNm

15.27 kNm

2.82 kN 13.64 kN10.36 kN7.18 kN

12 kNm10 kN 4 kN/m

A C

B13.64 kN2.82 kN

15.27 kNm

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 kNm

13.64 kN2.82 kN

15.27 kNm

17.54 kN

V (kN)x (m)3.41 m+

-+

-

2.82

-7.18

10.36

-13.64

M (kNm) x (m)+

- -+

11.28

-17.45

-5.45-15.27

7.98

Deflected shape x (m)

EIB273.3

=EIA

21.7=

EIB273.3

=

EIA21.7

=

• 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 kNm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

• 46

12 kNm 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

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 ++=

EIEIEIM BABA

)3(30

)6)(6(6

)01.0)(3(66

)3(26

)3(4 22 ++=

EIEIEIM CBBC

)4(30

)6)(6(6

)01.0)(3(66

)3(46

)3(2 22 +=

EIEIEIM CBCB

-12

0

0

• 47

12 kNm 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

)2(8

)8)(10(8

)01.0)(2(68

)2(48

)2(22 ++=

EIEIEIM BABA

)1(10758

)2(28

)2(4+++= BAAB

EIEIM

)2(10758

)2(48

)2(2++= BABA

EIEIM

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

)2(3:2

)1()2(2 aEIM BBA +=

16.5

• 48

+ MB = 0: - MBA - MBC = 0 (3/4 + 2)EIB + 16.5 - 192.8 = 0

B = 64.109/ EI

Substitute B in (1): A = -129.06/EI

Substitute A and B in (5), (3), (4):

MBC = -64.58 kNm

MCB = -146.69 kNm

MBA = 64.58 kNm,

MBA MBC

B

12 kNm 10 kN 6 kN/m

A CB

6 m4 m4 m

2EI 3EI10 mm

MBC = (4/6)(3EI)B - 192.8

MBA = (3/4)(2EI)B + 16.5

• 49

12 kNm 10 kN 6 kN/m

A CB

6 m4 m4 m

64.58 kNm 64.58 kNm

= -1.57 kN= 11.57 kN

146.69 kNm

= 47.21 kN= -29.21 kN

10 kN

A B

ByLAy

12 kNm64.58 kNm

2 m

6 kN/m

C

CyByR

18 kN

B

64.58 kNm

146.69 kNm

MBC = -64.58 kNm

MCB = -146.69 kNm

MBA = 64.58 kNm,

• 50

12 kNm 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 (kNm) x (m)

1258.29 64.58

-146.69

+

-

47.21 kN

146.69 kNm

11.57 kN

1.57 + 29.21 = 30.78 kN

10 mmA = -129.06/EI

B = 64.109/ EIA = -129.06/EI

Deflected shape x (m)

B = 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 kNm

10 mm

• 52

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNm

)1(3

)2(4= CCB

EIM

)2(1005.43

)5.1(23

)5.1(4+++= ACCA

EIEIM

)2(2

122

1002

)5.4(33

)5.1(3:2

)2()2(2 aEIM CCA +++=

)3(1005.43

)5.1(43

)5.1(2++= ACAC

EIEIM 12

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= 4.5

MFw

• 53

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNm

)1(3

)2(4= CCB

EIM

)2(2

122

1002

)5.4(33

)5.1(3 aEIM CCA +++=

10 mm

MCB MCA

C

0=+ CACB MM

02

122

1002

)5.4(33

)5.48(=+++

+C

EI

0015.006.15 ==

)3(1005.43

)5.1(4)06.15(3

)5.1(212 ++= AEI

EIEI Substitute C in eq.(3)

0034.022.34 ==

Equilibrium equation:

• 54

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNm

10 mm

0034.022.34 ==

mkNEI

EIEIM CBC =

== 08.20)06.15(3

)2(23

)2(2

mkNEI

EIEIM CCB =

== 16.40)06.15(3

)2(43

)2(4

kN08.203

08.2016.40=

+kN08.20

B C40.16 kNm20.08 kNm

6 kN/m

AC

12 kNm

40.16 kNm

18 kN

8.39 kN26.39 kN

• 55

6 kN/m

B A C

3 m 3 m2EI 1.5EI

12 kNm

10 mm

6 kN/m

AC

12 kNm

40.16 kNm8.39 kN26.39 kN

B C40.16 kNm20.08 kNm

20.08 kN20.08 kN

V (kN)

x (m)

26.398.39

-20.08

+-

M (kNm)

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 kNm

1