slope deflection method used in structure analysis in civil engineering

1. Guided By:- Zalak P. Shah
Subject:- Structural Analysis 2
Topic:- Slope Deflection
Jurik Jariwala 130490106037
Prepared by…

2. Using Slope – Deflection method determine the moments
at points A, B, C, Support B settle 0.08 m in y direction
EI=1000kN.m2
2

3. BA
AB
BABA
AB
AB
BAAB
MF
LL
EI
M
MF
LL
EI
M





 




 

3
2
2
3
2
2


)03.02(0
4
8.03
20
4
2
)03.0(
2
0
4
8.03
0
4
2



 





 

BBBA
BBAB
EI
EI
M
EIEI
M


For member AB
027.0015.0
1000
12
24
03.0224


B
BBA
EI
M


kN.mM
mkN
EIM
BA
BAB
24
.3)03.0027.0(1000
)03.0(


 
3

4. Using Slope – Deflection method determine the moments
at points A, B, C, Support B settle 0.1 ft in y direction
EI=302(10)3 K.ft2
4

5.    
    72)(
68
245.1
020
24
2
72)(
128
245.1
00
24
2
2
2


BBBA
BBAB
EIEI
M
EIEI
M


)015.02(
1020
)1.0(3
2
20
2
)015.02(
1020
)1.0(3
2
20
2












CBCBCB
CBCBBC
EIEI
M
EIEI
M


5

6. )02.0(
5.715
)1.0(3
0
15
2
)02.02(
5.715
)1.0(3
02
15
2


























CCDC
CCCD
EIEI
M
EIEI
M


6

7. 0 CDCB MM ( 2 0.015) (2 0.02) 0
10 7.5
2 2
0.0015 0.00267 0
10 10 7.5
35
0.001167 0
10 75
B C C
CB
C CB
EI EI
 
 


   
         
   
   
       

 
 


0 BCBA MM
3
( ) 72 (2 0.015) 0
6 10
72
0.0015 0
6 5 10
11 72
0.0015 0
11
3
0.00126 0
30 10
0 10 302(10)
B B C
CB B
CB
CB
EI EI
EI

  
 

   
         
   
  
       
  
 






7

8. 00344.0
00438.0


C
B


ftkM
ftkM
ftkM
ftkM
ftkM
ftkM
DC
CD
CB
BC
BA
AB
.667
.529
.529
.292
.292
.2.38






8

9. 9

10. Using Slope – Deflection method determine the moments
at points A, B, C,D
10

11.  
 
2
0 0 0 ( )
12 6
2
0 2 0 0 ( )
12 3
AB B B
BA B B
EI EI
M
EI EI
M
 
 
     
     
 
 
2
2
2 5
2 0 (2 ) 80
8 96 4
2 5
2 0 ( 2 ) 80
8 96 4
BC B C B C
CB B C B C
EI wL EI
M
EI wL EI
M
   
   
       
       
 
 
2
2 0 0 0 ( )
12 3
2
0 0 0 ( )
12 6
CD C C
DC C C
EI EI
M
EI EI
M
 
 
     
     
For member AB
For member BC
For member CD
11

12. 0 BCBA MM
( ) (2 ) 80 0
5
0
6 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 0
4 6
0B C C
B C
EI I
I EI
E
E
 
    
   
    
  
  


 
1
2 137.1
B C
EI
   
12

13. 137.1
B C
EI
   
22.9 .
45.7 .
45.7 .
45.7 .
45.7 .
22.9 .
AB
BA
BC
CB
CD
DC
M kN m
M kN m
M kN m
M kN m
M kN m
M kN m


 

 
 
13

14. Using Slope – Deflection method determine the moments
at points A, B, C,D
14

15. 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 EI
M
EI EI
M
 
 
  
        
  
        
 
 
2 2
2 0 0 (2 )
15 15
2 2
2 0 0 ( 2 )
15 15
BC B C B C
CB B C B C
EI EI
M
EI EI
M
   
   
      
      
2
2 0 3 0 (2 )
18 18 9 6
2
0 3 0 ( )
18 18 9 6
CD C C
DC C C
EI EI
M
EI EI
M
 
 
  
        
  
        
For member AB
For member BC
For member CD
15

16. 0 BCBA MM
0.34 0.0416
0
2
(2
0.607 0.134 0.0416 0
0.267 0.134 0
) (2 )
6 4 15
B
B
B
C
C
B
B
C
EI EI
 

 


 
  
   

    
 
 
 
 
 

0 CDCB MM
0.134 0.49
0.134 0.267 0.22 0.0
2
( 2 ) (2 )
1
0.0185 0
5
5
1
0
6
8 0
9
B
B
B C C
C C
C
EI EI
 

  
 

   
   
   
  

   
   

1
2
16

17. 40
12 18
40
CD DCAB BA
A DV
M
V
MM M 
 
 
( ) (2 ) ( ) (2 )
6 4 6 4 9 6 9 6 40
12 18
3 3
0.009 4
960
0.444 0.21
7 162
6
0
2
B B C C
B
C
C
B
EI EI EI EI
EI E
EI
I
EI
 
 
 
 
   
         
 
    
    3
40 0A DV V  Q
17

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 ft
M k ft
M k ft
M k ft
M k ft
M k ft
 
 


 
 
18

19. 19