# (Slope Deflection Method – Displacement Method) ??아대학교 토목공학과 구조역학 2 강의: 박현우 Lecture 4 - 1 4. 처짐각법 – 변위법 (Slope Deflection Method – Displacement Method) Objective of this

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• 2 :

Lecture 4 - 1

4. (Slope Deflection Method Displacement Method)

Objective of this chapter:

.

What will be presented:

Theoretical background

George A. Maney 1915 .

.

(Fixed end moment)

.

.

• 2 :

Lecture 4 - 2

4.1

(1) (Fixed-end moment; FEM): AB

(P3, P4, w) A B

FEMABM , FEMBAM

(2) A B A B

BAAB MM ,

(3) A B

BAAB MM ,

LA B

P1 P2 P5 P3 P4

w

A B FEMABM

FEMBAM

A B

B A ABM

BAM

A B

ABMBAM =/L

• 2 :

Lecture 4 - 3

(4) A B

.

A: )( ABABFEMABAB MMMM

B: )( BABAFEMBABA MMMM

• 2 :

Lecture 4 - 4

A,B ABFEMABM FEM ,

BAFEMBAM FEM :

A B FEMABM

FEMBAM

• 2 :

Lecture 4 - 5

A, B A B

BAAB MM ,

- A, B A B

A B .

-

BAABBAABA MMEILL

EIMLL

EIML

L

2

63232

21

-

ABBAABBAB MMEILL

EIMLL

EIML

L

2

63232

21

)2(2),2(2 BABABAAB LEIM

LEIM

EIABM

EIBAM

L

A B

+

A B

B A ABM

BAM

• 2 :

Lecture 4 - 6

A B

BAAB MM ,

- A, B -

A B .

BAAB

BAABBAAB

A

MMEIL

MMEILL

EIMLL

EIML

L

26

26323

22

1

ABBA

ABBAABBA

B

MMEIL

MMEILL

EIMLL

EIML

L

26

26323

22

1

LEIMM BAAB

6

EIABM

EIBAM

ML

A=- B=-

+

A B

ABM

BAM =/L

• 2 :

Lecture 4 - 7

A, B

.

- A:

)32(2FEM

6)2(2FEM

)(

BAAB

BAAB

ABABFEMABAB

LEI

LEI

LEI

MMMM

- B:

)32(2FEM

6)2(2FEM

)(

ABBA

ABBA

BABAFEMBABA

LEI

LEI

LEI

MMMM

():

( ) .

(matrix analysis) .

• 2 :

Lecture 4 - 8

4.2

:

- 12

FEM2wL

AB , 12FEM

2wLBA

- 8

FEM PLBC , 8FEM PLCB

: 00 BCABCA

(Slope deflection equation):

- 12

2212

2 22 wLLEIMwL

LEIM BBABAB

- 8

28

22 PLLEIMPL

LEIM BCBBBC

B :

12880

2wLPLLEIMM BBCBA EI

PLEI

wLB 6496

23

EI = A B C

w

L L

P

• 2 :

Lecture 4 - 9

:

- 3248

5 2 PLwLM AB 1624

2 PLwLM BA

- 1624

2 PLwLM BC 325

48

2 PLwLMCB

:

- 02 1

2

LBMwLM yBAAB 323

167

1PwLBy

- 323

169

1PwLBwLA yy

- 02

LCMPLM yCBBC 3219

16PwLCy

- 32

13162

PwLCPB yy

- 2221PwLBBB yyy

- ..

A BwP

B

MAB MBA MBC MCB Ay By1 By2 Cy

By

By1 By2

• 2 :

Lecture 4 - 10

4.3

(1) Sidesway

P D E . sidesway . .

( C) sidesway .

• 2 :

Lecture 4 - 11

(axis of symmetry) sidesway .

(2) Sidesway

sidesway

L

L EI = constnat

w

A

B C

D

• 2 :

Lecture 4 - 12

12,

12

022 wLFEMwLFEM

FEMFEMFEMFEM

CBBC

DCCDBAAB

0,0 DACDBCAB

CB , 2.

DCCDDCDC

CDCDDCCD

CBBCCBCB

BCBCCBBC

BAABBABA

ABABBAAB

FEMLEIM

FEMLEIM

FEMLEIM

FEMLEIM

FEMLEIM

FEMLEIM

322

322

322

322

322

322

,

CDCCCDCBCB

CBBCBBABAB

LEIM

LEIMwL

LEIM

wLLEIM

LEIM

LEIM

2,4,12

2212

22,4,2

2

2

• 2 :

Lecture 4 - 13

B C

00

CDCB

BCBA

MMMM

0

1282

012

28

2

2

wLLEI

LEI

wLLEI

LEI

CB

CB

0 CB

EI

wLEI

wLCB 72

,72

33

:

36,

18

18,

18

18,

36

22

22

22

wLMwLM

wLMwLM

wLMwLM

DCCD

CBBC

BAAB

• 2 :

Lecture 4 - 14

:

w

36

2wL

18

2wL 18

2wL

36

2wL

18

2wL 18

2wL

2wL

2wL

12wL

12wL

2wL

12wL

12wL

2wL

2wL

2wL

12wL

12wL

2wL

12wL

[A.F.D.]

+

+

12wL

2wL

12wL

2wL

[S.F.D.]

36

2wL

18

2wL 72

5 2wL

[B.M.D.]

• 2 :

Lecture 4 - 15

(3) Sidesway

0 DCCDCBBCBAAB FEMFEMFEMFEMFEMFEM

CDABBC

DA

,00

,, CB 3.

.

32,322

22,22

322,32

CDCCCD

CBCBCBBC

BBABAB

LEIM

LEIM

LEIM

LEIM

LEIM

LEIM

L

L EI = constnat

A

B C

D

P

• 2 :

Lecture 4 - 16

B C

00

CDCB

BCBA

MMMM

3( ,, CB ) 2??

Sidesway

, sidesway

.

:

0 DA HHP

- B : 0 BAAAB MLHM LMMH BAABA

- , L

MMH DCCDB

PLMMMM DCCDBAAB

MAB

MBA

HA HD MDC

MCD

• 2 :

Lecture 4 - 17

,

PLLEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

CB

CB

CB

2466

0682

0628

53

CB

EI

PLEI

PLEI

PLCB 84

52828

222

:

72,

143

143,

143

143,

72

PLMPLM

PLMPLM

PLMPLM

DCCD

CBBC

BAAB

• 2 :

Lecture 4 - 18

:

72PL

143PL 14

3PL

72PL

143PL

143PL

73P

73P

2P

2P

73P

2P

2P

73P

2P

2P

P

73P

73P

+

73P

2P

[A.F.D.]

+

+

2P

73P

2P

73P

[S.F.D.]

72PL

143PL

[B.M.D.] +

+ 14

3PL

72PL