# บทที่ 2 Slope-Deflection 10 2.2 Slope-Deflection Equations ความสัมพันธ ระหว าง moment ภายในที่เกิดขึ้นที่ ปลายของชิ้นส วน AB (MAB และ MBA

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30-Apr-2018

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• 1

STRUCTURAL ANALYSIS

By

Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

2

2Slope-Deflection Method

1. concept slope-deflection2. frame statically

indeterminate slope-deflection shear diagram, moment diagram elastic curve

3

equilibrium equations force-displacement relationships

displacementsDisplacement method

compatibility equations force-displacement relationships

forcesForce methodUnknown

2.1 Displacement Method: General Procedures

:1. linear elastic2.

4

1. DOF (joint) (node) B C

displacement method

DOF(force-displacement relations)

2 2 3 (FEM)BA B A BAIM EL L

= + +

• 5

2. (DOF)

4. force-displacement relations /

0532 =++ MMM

074 =+MM

3.

6

Degrees of FreedomNodes - Degree of freedom [] node

a.) Degree of indeterminacy = 3 degree of freedom = 1

b.) Degree of indeterminacy = 1 degree of freedom = 3

/

7

2 node DOF 3 2 1 etc. DOF

DOF = 1

DOF = 4

8

DOF = 3 DOF = 9

• 9 10

2.2 Slope-Deflection Equations

moment AB(MAB MBA) DOF (A, B )

AB (continuous beam)

bending moment (joint) rigid joint

11

Sign conventionMAB MBA + A B + tangent

deflection curve + ( ) 12

principle of superposition MAB MBA

1. A(A) B

2. B(B) A

3. B A ()

4. tangent cord

• 13

=

+

+

+

principle of superposition

14

1: A

0;AM + =12 3

ABM LLEI

0;BM + =12 3

BAM LLEI

4AB A

EIML

=

2BA A

EIML

=

1 2 02 3

BAM LLEI

=

1 2 02 3

ABA

M LL LEI

+ =

15

2: B4

BA BEIML

=

2AB B

EIML

=

16

3:

0;BM + =

1 22 3M LLEI

moment M

2

6AB BA

EIM M ML

= = =

12 3M LLEI

0 =

• 17

4: Fixed-End Moment (FEM)

0;yF+ =1 12 02 4 2PL ML LEI EI

=

8PLM =

FEM node A - ()FEM node B + ()

18

Slope-Deflection Equation

2 2 3 (FEM)AB A B ABIM EL L

= + +

2 2 3 (FEM)BA B A BAIM EL L

= + +

4AB A

EIML

= 2BA AEIML

=

4BA B

EIML

= 2AB BEIML

=

2

6AB BA

EIM ML

= =

19

Pin-Supported End Span

2 2 3 (FEM)AB A B ABIM EL L

= + +

0 2 2 3B AIEL L

= +

3 (FEM)AB A ABIM EL L

= +

32 2A

B L = +

2 2 3 (FEM)AB A B ABIM EL L

= + +

2 2 3 (FEM)BA B A BAIM EL L

= + +

20

2 2 3 (FEM)AB A B ABIM EL L

= + +

2 2 3 (FEM)BA B A BAIM EL L

= + +

3 (FEM)AB A ABIM EL L

= +

Slope-Deflection Equation:

• 21

Fixed-End Moment (FEM)

22

Fixed-End Moment ()

23

2.3 1. FEM 2. / (E, I, L, , FEM) slope-

deflection 3. joint / support slope-deflection ( 2) (simultaneous equation) slope deflection joint / support

4. slope deflection slope-deflection ( 2) end-moment M

5. shear diagram, moment diagram elastic curve24

: slope-deflection2.2, 2.7, 2.9 +

• 25

2-1 slope-deflection shear moment diagram elastic curve E = 200 GPa, IAB = 0.006 m4 IBC = 0.008 m4

DOF = 3 A, B, C A C pin roller DOF B

degree of Indeterminacy = 4-3 = 1

26

1. fixed-end moment A C pin roller

2

(FEM)8BAwL

= +

2

(FEM)8BCwL

=

28(5) 25 kN-m8

= =

212(6) 54 kN-m8

= =

FEMBA FEMBC

3 (FEM)AB A ABIM EL L

= +

27

2. slope-deflection A C pin roller

3 (FEM)AB A ABIM EL L

= +

3 ( ) 25ABBA BAB AB

IM EL L

= +

3 ( ) 255 AB BEI = +MBA MBC

3 ( ) 546BC BC B BC

M EIL

=

1 ( ) 542 BC BEI =

28

3. 0BA BCM M+ =

MBA MBC

3 (0.006) 255 BE + +

1 (0.008) 54 02 BE =

3815.79B E

4. 3 (0.006) 255BA B

M E = + 38.74 kN-m=

1 (0.008) 542BC B

M E = 38.74 kN-m=

3 ( ) 255BA AB B

M EI = + 1 ( ) 542BC BC B

M EI =

• 29

5. FBD shear diagram moment diagram

5 m 6 m

30

31

6. elastic curve

32

2-2 slope-deflection shear moment diagram D elastic curve B = 10 mm EIAB = 100 kN-m2 EIBC = 200 kN-m2

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

• 33

2

2(FEM)ABPabL

=

2

2(FEM)ABPbaL

= +

2

(FEM)12BCwL

=

2

(FEM) 4.167 kN-m12CBwL

= + = +

1. 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

= =

slope-deflection 1 why???

34

2. slope-deflection

MBA MBCMAB MCB

2 2 3 (FEM)AB A B ABIM EL L

= + + 2 2 3 (FEM)BA B A BA

IM EL L

= + +

ABEI EI= 2BCEI EI=

2 0.0102 3 4.4443 3AB A BEIM = +

2 0.020 4.4443 3BEI =

2 0.0102 3 2.2223 3BA B AEIM = + +

2(2 ) 0.0102 3 4.1675 5BC B CEIM = +

8 0.120 4.1675 25BEI = +

2(2 ) 0.0102 3 4.1675 5CB C BEIM = + +

4 0.120 4.1675 25BEI = + +

4 0.020 2.2223 3BEI = +

35

3.

MBA MBCMAB MCB

0BA BCM M+ =

[ ]2.933 0.001867 1.945BEI =

4 0.020 2.2223 3B

BAM EI = + 8 0.120 4.1675 25B

BCM EI = +

2100 kN-mEI =

36

2 0.020 4.4443 3B

ABM EI =

4 0.020 2.2223 3B

BAM EI = +

4 0.120 4.1675 25B

CBM EI = + +

8 0.120 4.1675 25B

BCM EI = +

4.626 kN-mABM =

2.525 kN-mBAM = +

2.525 kN-mBCM =

5.229 kN-mCBM = +

2.525 2.5254.626 5.229

2100 kN-mEI =

• 37

5. FBD shear diagram moment diagram 4.626 kN-mABM = 2.525 kN-mBAM = + 2.525 kN-mBCM = 5.229 kN-mCBM = +

1 m 2 m 5 m

38

39

6. elastic curve

40

7. D

0;DM =1 2.741 0.372(0.372)2 3D

MEI

= =

1 4.626 2(0.628) 0.372 (0.628)2 3EI

+ 1.085EI

= 0.011 m=

D conjugate beam D'

• 41

2-3 slope-deflection shear moment diagram elastic curve EI = 10,000 kN-m2

DOF = 7 A, B, C, D, E, F F EF A DOF 4 B, C, D E

degree of Indeterminacy = 6-3 = 3

42

1. fixed-end moment A

2

(FEM)8BAwL

= +23(3) 3.375 kN-m

8= =

FEMBA FEMBC

2

(FEM)12BCwL

= 23(3) 2.25 kN-m

12= =

FEMCB

(FEM) 2.25 kN-mCB =

FEMCD

(FEM)8CDPL

= 8(3) 3.0 kN-m8

= =

FEMDC

(FEM) 3.0 kN-mDC =

FEMDE

2

(FEM)12DEwL

= 23(3) 2.25 kN-m

12= =

FEMED

(FEM) 2.25 kN-mED =

43

2. slope-deflection2 2 3 (FEM)AB A B AB

IM EL L

= + + 2 2 3 (FEM)BA B A BA

IM EL L

= + +

MBA MBC MCB

3 ( 0) 3.375ABBA BAB

IM EL

= + 3.375BEI= +

2 (2 0) 2.25BCBC B CBC

IM EL

= + 4 2 2.253 3B CEI EI = +

2 (2 0) 2.25BCCB C BBC

IM EL

= + +

MCD

2 (2 0) 3.0CDCD C DCD

IM EL

= +

7.5 kN-m

2 4 2.253 3B CEI EI = + +

4 2 3.03 3C DEI EI = +

44

MBA MBC MCB MCD

4 2 3.03 3CD C D

M EI EI = +

MDC

2 (2 0) 3.0CDDC D CCD

IM EL

= + +

MDE

4 2 2.253 3DE D E

M EI EI = +

MED

2 4 2.253 3ED D E

M EI EI = + +

7.5 kN-m

2 4 3.03 3C DEI EI = + +

• 45

3.

MBA MBC MCB MCD

B

C

0BA BCM M+ =

7 2 1.1253 3B CEI EI + = 3.375BA BM EI= +

4 2 2.253 3BC B C

M EI EI = +

0CB CDM M+ =

2 8 2 0.753 3 3B C DEI EI EI + + =

2 4 2.253 3CB B C

M EI EI = + +

4 2 33 3CD C D

M EI EI = +

7.5 kN-m

46

MDC MDE MED

D

E

7.5 kN-m

0DC DEM M+ =2 4 33 3DC C D

M EI EI = + +

4 2 2.253 3DE D E

M EI EI = +

2 8 2 0.753 3 3C D EEI EI EI + + =

7.5 0EDM =

2 4 5.253 3D EEI EI + =2 4 2.25

3 3ED D EM EI EI = + +

47

simultaneous equation

0.7366B EI

= 0.8906C EI =

1.7009D EI

= 4.7879E EI =

7 2 0 0 1.1253 3B CEI EI + + + =

2 8 2 0 0.753 3 3B C DEI EI EI + + + =

2 8 20 0.753 3 3C D EEI EI EI + + + =

2 40 0 5.253 3D EEI EI + + + =

210000 kN-mEI =

48

=

3.375BA BM EI= +

4 2 2.253 3BC B C

M EI EI = + 2 4 2.253 3CB B C

M EI EI = + +

4 2 33 3CD C D

M EI EI = +

2.638 kN-mBAM = 2.638 kN-mBCM = 2.946 kN-mCBM =

2.946 kN-mCDM = 1.326 kN-mDCM =

2 4 33 3DC C D

M EI EI = + +

4 2 2.253 3DE D E

M EI EI = + 2 4 2.253 3ED D E

M EI EI = + +

1.326 kN-mDEM = 7.50 kN-mEDM =

MBA MBC MCB MC