SNPIT & RC,UMRAKH

Guided By:- Zalak P. Shah

Subject:- Structural Analysis 2

Topic:- Slope Deflection

Jurik Jariwala 130490106037Prepared by…

Example 1Using Slope – Deflection method determine the moments at points A, B, C, Support B settle 0.08 m in y directionEI=1000kN.m2

2

BAAB

BABA

ABAB

BAAB

MFLL

EIM

MFLL

EIM

322

322

)03.02(04

8.03204

2

)03.0(2

04

8.0304

2

BBBA

BBAB

EIEIM

EIEIM

For member AB

027.0015.01000

12

2403.02 24

B

BBA EIM

kN.mMmkN

EIM

BA

BAB

24.3)03.0027.0(1000

)03.0(

3

Example 2Using Slope – Deflection method determine the moments at points A, B, C, Support B settle 0.1 ft in y directionEI=302(10)3 K.ft2

4

5

72)(68

245.102024

2

72)(128

245.10024

2

2

2

BBBA

BBAB

EIEIM

EIEIM

)015.02(1020

)1.0(3220

2

)015.02(1020

)1.0(3220

2

CBCBCB

CBCBBC

EIEIM

EIEIM

6

)02.0(5.715

)1.0(3015

2

)02.02(5.715

)1.0(30215

2

CCDC

CCCD

EIEIM

EIEIM

7

0 CDCB MM ( 2 0.015) (2 0.02) 010 7.5

2 20.0015 0.00267 010 10 7.5

35 0.001167 010 75

B C C

CB

C CB

EI EI

0 BCBA MM

3

( ) 72 (2 0.015) 06 10

72 0.0015 06 5 10

11 720.0015 0

113

0.00126 030 10

0 10 302(10)

B B C

CB B

CB

CB

EI EI

EI

8

00344.000438.0

C

B

ftkMftkMftkMftkMftkMftkM

DC

CD

CB

BC

BA

AB

. 667. 529

. 529

. 292. 292. 2.38

9

(Frame with Sway )

Example 3 (No Sides-Sway Frame)Using Slope – Deflection method determine the moments at points A, B, C,D

10

2 0 0 0 ( )12 6

2 0 2 0 0 ( )12 3

AB B B

BA B B

EI EIM

EI EIM

2

2

2 52 0 (2 ) 808 96 4

2 52 0 ( 2 ) 808 96 4

BC B C B C

CB B C B C

EI wL EIM

EI wL EIM

2 2 0 0 0 ( )12 32 0 0 0 ( )12 6

CD C C

DC C C

EI EIM

EI EIM

For member AB

For member BC

For member CD

11

12

0 BCBA MM

( ) (2 ) 80 0

5 06 4

4

8

3

0

B B C

B C

EI

E I

I

E

E

I

0 CDCB MM

5

( 2 ) 80 ( )4 3

80 04 6

0B C C

B CEI I

I EI

E

E

1

2 137.1B C EI

13

137.1B C EI

22.9 .45.7 .

45.7 .45.7 .

45.7 .22.9 .

AB

BA

BC

CB

CD

DC

M kN mM kN mM kN mM kN mM kN mM kN m

Example 5 (Frame with Sway )Using Slope – Deflection method determine the moments at points A, B, C,D

14

2 0 3 0 ( )12 12 6 4

2 0 2 3 0 (2 )12 12 6 4

AB B B

BA B B

EI EIM

EI EIM

2 22 0 0 (2 )15 15

2 22 0 0 ( 2 )15 15

BC B C B C

CB B C B C

EI EIM

EI EIM

2 2 0 3 0 (2 )18 18 9 62 0 3 0 ( )18 18 9 6

CD C C

DC C C

EI EIM

EI EIM

For member AB

For member BC

For member CD

15

16

0 BCBA MM

0.34 0.0416

02(2

0.607 0.134 0.0416 00.267 0.134 0

) (2 )6 4 15B

B

B

C

C

B

B

C

EI EI

0 CDCB MM

0.134 0.490.134 0.267 0.22 0.0

2 ( 2 ) (2 )1

0.0185 05

51

068 0

9

B

B

B C C

C C

C

EI EI

1

2

17

4012 18

40

CD DCAB BA

A DVM

VMM M

( ) (2 ) ( ) (2 )6 4 6 4 9 6 9 6 40

12 183 3 0.009 4

9600.444 0.21

7 162

6

02

B B C C

B

C

C

B

EI EI EI EI

EI E

EI

I EI

3

40 0A DV V

18

438.From

81 1, 2,

136.18

6 54

3

7 .7

B

C

EI

EI

EI

208 .135 .

135 .94.8 .

94.8 .110 .

AB

BA

BC

CB

CD

DC

M k ftM k ftM k ftM k ftM k ftM k ft

19

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