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• 1

STRUCTURAL ANALYSIS

By

Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

2

3Moment Distribution

1. concept moment distribution2. frame statically

indeterminate moment distribution shear diagram, moment diagram elastic curve

3.1 moment distribution simultaneous slope-deflection ( step 3)

3

: moment distribution

3.1, 3.7 3.9 +

4

Sign Convention

slope-deflection

Fixed-End Moments (FEM)

slope-deflection

• 5

1. stiffness factors

3K - Far End PinnedEIL

=

4 K - Far End FixedEIL

=

2. distribution factors KDF=K

DF = 1 - For pinned end

DF = 0 - For fixed end

3. fixed-end moment FEM

4. moment distribution5. FBD shear diagram moment diagram

6. elastic curve 6

Member Stiffness Factor Carry-Over Factormember stiffness factor moment 1 radian

2 (2 3 ) FEMAB A B ABIM EL L

= + +

2 (2 3 ) FEMBA B A BAIM EL L

= + +

4A

EIL

=

2A

EIL

=

4 K

Far End Fixed

EIL

=

7

Carry-over factor moment moment

20.54

ABA

AB A

EIM L

EIML

= = +

Joint Stiffness Factor member stiffness factor joint

K KT =

8

Joint Stiffness Factor

K K K KT AD AB AC= + + 10000 kN-m=1000 4000 5000= + +

JSF = moment joint A 1 radian

EI

=

4K

ABAB L

EI

=

4K

EI

=

4K

• 9

Distribution Factor (DF) moment M joint A AD, AB, AC moment

AD AB ACM M M M= + +

EIML

= = KAB AB AM = KAC AC AM =

moment joint A

EIML L

= + +

10

(K K K )AD AB AC AM = + +

(K K K )A AD AB AC

M =+ +

A moment joint A K DF

KAB

AB ABM M M= =K DF

KAC

AC ACM M M= =

distribution factor

KDF=K

KM

KAB AB AM =

KAC AC AM =

11

DF 4000 /10000 0.4AB = =

DF 5000 /10000 0.5AC = =

= 1.0

0.4(2000) 800 N-mABM = =

0.5(2000) 1000 N-mACM = =

= 2,000 N-m

12

3.2 Moment Distribution1. member stiffness factor K

4( )K ABABAB

EIL

=

4( )K BCBCBC

EIL

=

A C fixed support K KA C = =

6 440 (10 ) m /mE =

6 460 (10 ) m /mE =

64 [30(10 )]3

E =

64 [60(10 )]4

E =

4 K

Far End Fixed

EIL

=

• 13

2. distribution factor DF joint

DF ABBAAB BC

KK K

=+

A:

B:

C:

DF BCBCAB BC

KK K

=+

6

6

40 (10 )DF 040 (10 )ABE

E

= =+

6

6

60 (10 )DF 060 (10 )CBE

E

= =+

KDF=K

6

6

40 10 0.4040 60 10

EE E

= =+6

6

60 10 0.6040 60 10

EE E

= =+

14

3. fixed-end moment FEM

AB (FEM) (FEM) 0AB BA= =

BC w = 3 kN/m 2

(FEM)12BCwL

=

2

(FEM)12CBwL

= +

23(4) 4 kN-m12

= =

23(4) 4 kN-m12

= =

FEMCB

FEMBC

15

4. moment4.a joint B AB BC

fixed-end moment BC

0 0

16

moment joint B BC =

moment joint B BA = 0.4( 4) 1.6 kN-m+ = +

0.6( 4) 2.4 kN-m+ = +

4.b joint B MB = -4 kN-m joint B MB = + 4 kN-m AB BC (distribution)

4 kN-m

1.6 kN-m 2.4 kN-m

• 17

4.c joint B moment joint B (carry-over) carry-over factor +0.5

moment BC CB =

moment BA AB =0.5( 2.4) 1.2 kN-m+ = +

0.5( 1.6) 0.8 kN-m+ = +

18

4.d A C fixed support moment joint B

4.e moment joint moment

19

5. shear diagram moment diagram

20

• 21

6. elastic curve

22

3.3 Stiffness-Factor Modifications1.) Member Pin-Supported at Far End

2 (2 )AB A BIM EL

= +

2 ( 2 ) 0BA A BIM EL = + =

2A

B =

3AB A

EIML

=

3K EIL

=

MBA

MAB

A B

23

B C = BC CBM M =

2 (2 )BC B CIM EL

= +

2 (2 )CB C BIM EL

= +

2EIM

L=

2K EIL

=

2 ( 2 )IEL

= + 2EI ML

= =

2 (2 )IEL

= 2EI ML

= =

MBC

MCB

24

6EIM

L=

6K EIL

=

B C = = BC CBM M M= =

2 (2 ) 2 (2 )BC B CI IM E EL L

= + = +

2 (2 ) 2 (2 )CB C BI IM E EL L

= + = +

MBC

MCB

6EI ML

= =

6EI ML

= =

• 25

3-1 moment distribution shear diagram, moment diagram elastic curve rotation B EIAB = EI EIBC = 2EI

DOF = 1 B degree of indeterminacy = 7-3 = 4

26

1. stiffness factors4( )K ABAB

AB

EIL

=

4( )K BCBCBC

EIL

=

43EI

=

85EI

=

2. distribution factors A C fixed end

4 / 3(DF) 04 / 3ABEI

EI= =

+8 / 5(DF) 0

8 / 5CBEI

EI= =

+

4 / 3(DF)4 / 3 8 / 5BA

EIEI EI

=+

(DF) 1 0.4545 0.5455BC = =

0.4545=

27

2

2(FEM)ABPab

L=

2

2(FEM)BAPba

L= +

2

(FEM)12BCwL

=

(FEM) 4.167 kN-mCB = +

3. fixed-end moment

FEMBA FEMBCFEMAB FEMCB

2

210(1)2 4.444 kN-m

3= =

2

210(2)1 2.222 kN-m

3= = +

22(5) 4.167 kN-m12

= =

28

4. moment distribution

+4.697-3.106+3.106-4.003M+0.5304+0.4419COM

+1.0607+0.8838DM+4.1667-4.1667+2.2222-4.4444FEM

00.54550.45450DFCBBCBAABMemberCBAJoint

(DF) (DF) 0AB BC= = (DF) 0.4545CB =(DF) 0.5455BC =

(FEM) 4.4444 kN-mAB = (FEM) 2.2222 kN-mBA = +

(FEM) 4.1667 kN-mBC = (FEM) 4.1667 kN-mCB = +

• 29

5. FBD shear diagram moment diagram

+4.697-3.106+3.106-4.003 MCBBCBAABMember

30

2.341 m

31

6. elastic curve

32

rotation B

slope-deflection

2 (2 3 ) (FEM)ABBA B A BAABAB

EIML L

= + +

23.106 (2 ) 2.22223 B

EI = +

0.663B EI

=

• 33

moment distribution shear diagram, moment diagram elastic curve rotation B EIAB = 0.75EIBC

DOF = 3 A, B C degree of indeterminacy = 4-3 = 1

34

1. stiffness factors3K ABAB

AB

EIL

=

3K BCBCBC

EIL

=

3(0.75 )5

0.45

BC

BC

EI

EI

=

=

0.5 BCEI=2. distribution factors

A C pinned end 0.45(DF) 1.0

0 0.45BC

ABBC

EIEI

= =+

0.5(DF) 1.00 0.5

BCCB

BC

EIEI

= =+

0.45(DF)0.45 0.5

BCBA

BC BC

EIEI EI

=+

(DF) 1 0.4737BC =

0.4737=

0.5263=

0.75AB BCEI EI=

35

2

(FEM)8BC

wL=

3. fixed-end moment

FEMBA FEMBC

212(6) 54.0 kN-m8

= =

2

(FEM)8BA

wL= +

28(5) 25.0 kN-m8

= + = +

36

4. moment distribution

0-38.737+38.7370M+15.263+13.737DM

0-54.0+25.00FEM1.00.52630.47371.0DFCBBCBAABMemberCBAJoint

(DF) (DF) 1.0AB CB= = (DF) 0.4737BA = (DF) 0.5263BC =

(FEM) 40.0 kN-mBA = +

(FEM) 54.0 kN-mBC =

• 37

5. FBD shear diagram moment diagram

0-38.737+38.7370 MCBBCBAABMember

38.737 38.737

12.25327.747 42.456

29.5445 m 6 m

38

39

6. elastic curve

40

rotation B

slope-deflection3 ( ) (FEM)ABBA B BA

ABAB

EIML L

= +

338.737 ( ) 25.05 B

EI + = +

22.895B EI

= +

• 41

3-2 moment distribution shear diagram, moment diagram elastic curve

AB moment AB

1.2EI EI

42

4K BCBCBC

EIL

=

4K CDCDCD

EIL

=

1. stiffness factors4(1.2 ) 0.96

5EI EI= =

44EI EI= =

1.2EI EI

43

2. distribution factors

D fixed end

AB (DF) 0BA = (DF) 1.00

BCBC

BC

KK

= =+

(DF) 0CDDCCD

K EIK EI

= = =+ +

0.96(DF)0.96CB

EIEI EI

=+

(DF) 1 (DF) 0.5102CD CB= =0.4898=

K 0.96BC EI=

KCD EI=

44

3. fixed-end moment

FEMCB

FEMBA

FEMBC

(FEM) 20(2) 40 kN-mBA = = +2

(FEM)12BCwL

= 240(5) 83.333 kN-m

12= =

AB

2

(FEM)12CBwL

= + 83.333 kN-m= +

C FEM FEM

• 45-2.336+2.653-2.154COM-4.672-4.308+5.306DM

-5.528+8.980-5.306COM-11.055-10.612+17.960DM

-18.707+21.667-17.960COM-37.414-35.919+43.333DM

-10+83.333-83.333+40FEM00.51020.48981.00DF

DCCDCBBCBAMemberDCBJoint

4. moment distribution(DF) 0BA = (DF) 1BC = (DF) 0DC = (DF) 0.4898CB = (DF) 0.5102CD =

(FEM) 40 kN-mBA = + (FEM) 83.333 kN-mBC = (FEM) 83.333 kN-mCB = +

46-27.639-10-55.297+65.297-4040M-0.020-0.020+0.033DM

-0.033+0.040-0.033COM-0.067-0.065+0.080DM

-0.083+0.132-0.080COM-0.166-0.159+0.264DM

-0.275+0.325-0.264COM-0.549-0.528+0.650DM

-0.667+1.077-0.650COM-1.354-1.299+2.154DM

-2.336+2.653-2.154COM00.51020.489810DF

DCCDCBBCBAMember

47

5. FBD shear diagram moment diagram

-27.639-10-55.297+65.297-4040MDC

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