The Slope Deflection Method: Frames Without Sidesway - 3 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Slope Deflection Method: Frames Without Sidesway - 3 Civil Engineering (CE) Notes | EduRev

 Page 1


 
 
?
= 0
B
M       ?        0 = +
BA BC
M M   (4)  
 
Substituting the value of  and  in the above equation, 
BC
M
BA
M
 
    5 . 12 667 . 0 333 . 2 = +
A B
EI EI ? ?    (5) 
 
Or,     
EI
A B
741 . 18
498 . 3 = + ? ? 
 
Solving equation (3) and (4) 
 
) (
245 . 16
) (
002 . 10
clockwise
EI
ckwise counterclo
EI
B
B
-
=
=
?
?
    (6)  
 
Substituting the value of 
A
? and  
B
?  in equation (2) beam end moments are 
evaluated. 
 
0
002 . 10
667 . 0
245 . 16
333 . 1 15 = ?
?
?
?
?
?
+ ?
?
?
?
?
? -
+ =
EI
EI
EI
EI M
AB
 
 
1
002 . 10
33 . 1 .
245 . 16
667 . 0 15 - =
?
?
?
?
?
?
+
?
?
?
?
?
? -
+ - =
EI
EI
EI
EI M
BA
 
 
kN.m 5 . 12
002 . 10
5 . 2 =
?
?
?
?
?
?
+ =
EI
EI M
BC
 
 
kN.m 5 . 2
002 . 10
5 . 0 5 . 2 =
?
?
?
?
?
?
+ - =
EI
EI M
CB
    (7) 
 
Page 2


 
 
?
= 0
B
M       ?        0 = +
BA BC
M M   (4)  
 
Substituting the value of  and  in the above equation, 
BC
M
BA
M
 
    5 . 12 667 . 0 333 . 2 = +
A B
EI EI ? ?    (5) 
 
Or,     
EI
A B
741 . 18
498 . 3 = + ? ? 
 
Solving equation (3) and (4) 
 
) (
245 . 16
) (
002 . 10
clockwise
EI
ckwise counterclo
EI
B
B
-
=
=
?
?
    (6)  
 
Substituting the value of 
A
? and  
B
?  in equation (2) beam end moments are 
evaluated. 
 
0
002 . 10
667 . 0
245 . 16
333 . 1 15 = ?
?
?
?
?
?
+ ?
?
?
?
?
? -
+ =
EI
EI
EI
EI M
AB
 
 
1
002 . 10
33 . 1 .
245 . 16
667 . 0 15 - =
?
?
?
?
?
?
+
?
?
?
?
?
? -
+ - =
EI
EI
EI
EI M
BA
 
 
kN.m 5 . 12
002 . 10
5 . 2 =
?
?
?
?
?
?
+ =
EI
EI M
BC
 
 
kN.m 5 . 2
002 . 10
5 . 0 5 . 2 =
?
?
?
?
?
?
+ - =
EI
EI M
CB
    (7) 
 
Using these results, reactions are evaluated from equilibrium equations as shown 
in Fig 16.4 (e) 
 
 
 
The shear force and bending moment diagrams are shown in Fig 16.4(g) and 
16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). 
 
Page 3


 
 
?
= 0
B
M       ?        0 = +
BA BC
M M   (4)  
 
Substituting the value of  and  in the above equation, 
BC
M
BA
M
 
    5 . 12 667 . 0 333 . 2 = +
A B
EI EI ? ?    (5) 
 
Or,     
EI
A B
741 . 18
498 . 3 = + ? ? 
 
Solving equation (3) and (4) 
 
) (
245 . 16
) (
002 . 10
clockwise
EI
ckwise counterclo
EI
B
B
-
=
=
?
?
    (6)  
 
Substituting the value of 
A
? and  
B
?  in equation (2) beam end moments are 
evaluated. 
 
0
002 . 10
667 . 0
245 . 16
333 . 1 15 = ?
?
?
?
?
?
+ ?
?
?
?
?
? -
+ =
EI
EI
EI
EI M
AB
 
 
1
002 . 10
33 . 1 .
245 . 16
667 . 0 15 - =
?
?
?
?
?
?
+
?
?
?
?
?
? -
+ - =
EI
EI
EI
EI M
BA
 
 
kN.m 5 . 12
002 . 10
5 . 2 =
?
?
?
?
?
?
+ =
EI
EI M
BC
 
 
kN.m 5 . 2
002 . 10
5 . 0 5 . 2 =
?
?
?
?
?
?
+ - =
EI
EI M
CB
    (7) 
 
Using these results, reactions are evaluated from equilibrium equations as shown 
in Fig 16.4 (e) 
 
 
 
The shear force and bending moment diagrams are shown in Fig 16.4(g) and 
16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). 
 
 
 
 
 
 
 
Page 4


 
 
?
= 0
B
M       ?        0 = +
BA BC
M M   (4)  
 
Substituting the value of  and  in the above equation, 
BC
M
BA
M
 
    5 . 12 667 . 0 333 . 2 = +
A B
EI EI ? ?    (5) 
 
Or,     
EI
A B
741 . 18
498 . 3 = + ? ? 
 
Solving equation (3) and (4) 
 
) (
245 . 16
) (
002 . 10
clockwise
EI
ckwise counterclo
EI
B
B
-
=
=
?
?
    (6)  
 
Substituting the value of 
A
? and  
B
?  in equation (2) beam end moments are 
evaluated. 
 
0
002 . 10
667 . 0
245 . 16
333 . 1 15 = ?
?
?
?
?
?
+ ?
?
?
?
?
? -
+ =
EI
EI
EI
EI M
AB
 
 
1
002 . 10
33 . 1 .
245 . 16
667 . 0 15 - =
?
?
?
?
?
?
+
?
?
?
?
?
? -
+ - =
EI
EI
EI
EI M
BA
 
 
kN.m 5 . 12
002 . 10
5 . 2 =
?
?
?
?
?
?
+ =
EI
EI M
BC
 
 
kN.m 5 . 2
002 . 10
5 . 0 5 . 2 =
?
?
?
?
?
?
+ - =
EI
EI M
CB
    (7) 
 
Using these results, reactions are evaluated from equilibrium equations as shown 
in Fig 16.4 (e) 
 
 
 
The shear force and bending moment diagrams are shown in Fig 16.4(g) and 
16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). 
 
 
 
 
 
 
 
 
 
Example 16.3 
Compute reactions and beam end moments for the rigid frame shown in Fig 
16.5(a). Draw bending moment diagram and sketch the elastic curve for the 
frame. 
 
Solution 
 
 
 
Page 5


 
 
?
= 0
B
M       ?        0 = +
BA BC
M M   (4)  
 
Substituting the value of  and  in the above equation, 
BC
M
BA
M
 
    5 . 12 667 . 0 333 . 2 = +
A B
EI EI ? ?    (5) 
 
Or,     
EI
A B
741 . 18
498 . 3 = + ? ? 
 
Solving equation (3) and (4) 
 
) (
245 . 16
) (
002 . 10
clockwise
EI
ckwise counterclo
EI
B
B
-
=
=
?
?
    (6)  
 
Substituting the value of 
A
? and  
B
?  in equation (2) beam end moments are 
evaluated. 
 
0
002 . 10
667 . 0
245 . 16
333 . 1 15 = ?
?
?
?
?
?
+ ?
?
?
?
?
? -
+ =
EI
EI
EI
EI M
AB
 
 
1
002 . 10
33 . 1 .
245 . 16
667 . 0 15 - =
?
?
?
?
?
?
+
?
?
?
?
?
? -
+ - =
EI
EI
EI
EI M
BA
 
 
kN.m 5 . 12
002 . 10
5 . 2 =
?
?
?
?
?
?
+ =
EI
EI M
BC
 
 
kN.m 5 . 2
002 . 10
5 . 0 5 . 2 =
?
?
?
?
?
?
+ - =
EI
EI M
CB
    (7) 
 
Using these results, reactions are evaluated from equilibrium equations as shown 
in Fig 16.4 (e) 
 
 
 
The shear force and bending moment diagrams are shown in Fig 16.4(g) and 
16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). 
 
 
 
 
 
 
 
 
 
Example 16.3 
Compute reactions and beam end moments for the rigid frame shown in Fig 
16.5(a). Draw bending moment diagram and sketch the elastic curve for the 
frame. 
 
Solution 
 
 
 
The given frame is kinematically indeterminate to third degree so three rotations 
are to be calculated,
C B
? ? ,  and
D
? .  First calculate the fixed end moments (see 
Fig 16.5 b). 
 
 
 
2
54
4 kN.m
20
F
AB
M
×
== 
 
2
54
2.667 kN.m
30
F
BA
M
-×
==- 
 
2
2
10 3 3
7.5 kN.m
6
F
BC
M
××
== 
 
2
2
10 3 3
7.5 kN.m
6
F
CB
M
-× ×
==- 
 
    (1) 0 = = = =
F
EC
F
CE
F
DB
F
BD
M M M M
 
 
The frame is restrained against sidesway. Four spans must be considered for 
rotating slope – deflection equation: AB, BD, BC and CE. The beam end 
 
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