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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