The Slope Deflection Method: Frames Without Sidesway - 4 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


 
 
                         ?     
?
= 0
B
M 0 = + +
BD BC BA
M M M   (3) 
 
                               
?
= 0
D
M ? 0 =
DB
M     (4)  
 
                         ?      
?
= 0
C
M 0 = +
CE CB
M M    (5) 
 
Substituting the values of and in the equations (3), 
(4), and (5) 
CB DB BD BC BA
M M M M M , , , ,
CE
M
                                         
3.333 0.667 0.5 4.833
BC D
EI EI EI ? ?? + +=- 
 
0.5 0
BD
EI EI ? ? += 
 
2.333 0.667 7.5
CB
EI EI ? ? + =    (6)  
 
Solving the above set of simultaneous equations, 
C B
? ? , and  
D
? are evaluated. 
 
2.4125
B
EI ? = - 
 
Page 2


 
 
                         ?     
?
= 0
B
M 0 = + +
BD BC BA
M M M   (3) 
 
                               
?
= 0
D
M ? 0 =
DB
M     (4)  
 
                         ?      
?
= 0
C
M 0 = +
CE CB
M M    (5) 
 
Substituting the values of and in the equations (3), 
(4), and (5) 
CB DB BD BC BA
M M M M M , , , ,
CE
M
                                         
3.333 0.667 0.5 4.833
BC D
EI EI EI ? ?? + +=- 
 
0.5 0
BD
EI EI ? ? += 
 
2.333 0.667 7.5
CB
EI EI ? ? + =    (6)  
 
Solving the above set of simultaneous equations, 
C B
? ? , and  
D
? are evaluated. 
 
2.4125
B
EI ? = - 
 
3.9057
C
EI ? = 
 
1.2063
D
EI ? =     (7) 
 
Substituting the values of 
C B
? ? , and 
D
? in (2), beam end moments are 
computed.  
 
2.794 kN.m
AB
M = 
 
5.080 kN.m
BA
M = - 
 
1.8094 kN.m
BD
M = - 
 
0
DB
M = 
 
6.859 kN.m
BC
M = 
 
3.9028 kN.m
CB
M = - 
 
3.9057 kN.m
CE
M = 
 
1.953 kN.m
EC
M =     (8) 
 
The reactions are computed in Fig 16.5(d), using equilibrium equations known 
beam-end moments and given loading. 
 
 
Page 3


 
 
                         ?     
?
= 0
B
M 0 = + +
BD BC BA
M M M   (3) 
 
                               
?
= 0
D
M ? 0 =
DB
M     (4)  
 
                         ?      
?
= 0
C
M 0 = +
CE CB
M M    (5) 
 
Substituting the values of and in the equations (3), 
(4), and (5) 
CB DB BD BC BA
M M M M M , , , ,
CE
M
                                         
3.333 0.667 0.5 4.833
BC D
EI EI EI ? ?? + +=- 
 
0.5 0
BD
EI EI ? ? += 
 
2.333 0.667 7.5
CB
EI EI ? ? + =    (6)  
 
Solving the above set of simultaneous equations, 
C B
? ? , and  
D
? are evaluated. 
 
2.4125
B
EI ? = - 
 
3.9057
C
EI ? = 
 
1.2063
D
EI ? =     (7) 
 
Substituting the values of 
C B
? ? , and 
D
? in (2), beam end moments are 
computed.  
 
2.794 kN.m
AB
M = 
 
5.080 kN.m
BA
M = - 
 
1.8094 kN.m
BD
M = - 
 
0
DB
M = 
 
6.859 kN.m
BC
M = 
 
3.9028 kN.m
CB
M = - 
 
3.9057 kN.m
CE
M = 
 
1.953 kN.m
EC
M =     (8) 
 
The reactions are computed in Fig 16.5(d), using equilibrium equations known 
beam-end moments and given loading. 
 
 
 
 
( )
6.095 kN
Ay
R = ? 
 
( )
9.403 kN
Dy
R = ? 
 
( )
4.502 kN
Ey
R = ? 
 
( ) 1.013 kN
Ax
R = ? 
 
( ) 0.542 kN
Dx
R = ? 
 
( ) 1.465 kN
Ex
R =- ?     (9) 
 
The bending moment diagram is shown in Fig 16.5.(e)  and the elastic curve is  
shown in Fig 16.5(f). 
 
 
Page 4


 
 
                         ?     
?
= 0
B
M 0 = + +
BD BC BA
M M M   (3) 
 
                               
?
= 0
D
M ? 0 =
DB
M     (4)  
 
                         ?      
?
= 0
C
M 0 = +
CE CB
M M    (5) 
 
Substituting the values of and in the equations (3), 
(4), and (5) 
CB DB BD BC BA
M M M M M , , , ,
CE
M
                                         
3.333 0.667 0.5 4.833
BC D
EI EI EI ? ?? + +=- 
 
0.5 0
BD
EI EI ? ? += 
 
2.333 0.667 7.5
CB
EI EI ? ? + =    (6)  
 
Solving the above set of simultaneous equations, 
C B
? ? , and  
D
? are evaluated. 
 
2.4125
B
EI ? = - 
 
3.9057
C
EI ? = 
 
1.2063
D
EI ? =     (7) 
 
Substituting the values of 
C B
? ? , and 
D
? in (2), beam end moments are 
computed.  
 
2.794 kN.m
AB
M = 
 
5.080 kN.m
BA
M = - 
 
1.8094 kN.m
BD
M = - 
 
0
DB
M = 
 
6.859 kN.m
BC
M = 
 
3.9028 kN.m
CB
M = - 
 
3.9057 kN.m
CE
M = 
 
1.953 kN.m
EC
M =     (8) 
 
The reactions are computed in Fig 16.5(d), using equilibrium equations known 
beam-end moments and given loading. 
 
 
 
 
( )
6.095 kN
Ay
R = ? 
 
( )
9.403 kN
Dy
R = ? 
 
( )
4.502 kN
Ey
R = ? 
 
( ) 1.013 kN
Ax
R = ? 
 
( ) 0.542 kN
Dx
R = ? 
 
( ) 1.465 kN
Ex
R =- ?     (9) 
 
The bending moment diagram is shown in Fig 16.5.(e)  and the elastic curve is  
shown in Fig 16.5(f). 
 
 
 
 
 
 
 
Summary 
In this lesson plane frames restrained against sidesway are analysed using 
slope-deflection equations. Equilibrium equations are written at each rigid joint of 
the frame and also at the support. Few problems are solved to illustrate the 
procedure. The shear force and bending moment diagrams are drawn for the 
plane frames.