The Slope Deflection Method: Frames with Sidesway - 2 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


? + = EI EI M
C DC
3
2
3
2
?    (3) 
 
Now, consider the joint equilibrium of B andC (vide Fig. 17.3c). 
 
0 0 = + ? =
?
BC
BA B
M M M  (4) 
 
      (5) 0 0 = + ? =
?
CD
CB C
M M M
 
 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame   (vide Fig. 17.3d). . .e i
?
= 0
X
F
 
 
    0 10
2 1
= - + - H H 
       
    10
2 1
= + ? H H .   (6) 
 
 
 
 
Page 2


? + = EI EI M
C DC
3
2
3
2
?    (3) 
 
Now, consider the joint equilibrium of B andC (vide Fig. 17.3c). 
 
0 0 = + ? =
?
BC
BA B
M M M  (4) 
 
      (5) 0 0 = + ? =
?
CD
CB C
M M M
 
 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame   (vide Fig. 17.3d). . .e i
?
= 0
X
F
 
 
    0 10
2 1
= - + - H H 
       
    10
2 1
= + ? H H .   (6) 
 
 
 
 
 
 
Considering the equilibrium of the column AB andCD , yields 
 
3
1
AB BA
M M
H
+
=      
   
and  
 
     
3
2
DC CD
M M
H
+
=      (7) 
 
The equation (6) may be written as, 
 
30 = + + +
DC CD AB BA
M M M M     (8) 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(6) 
 
10 667 . 0 5 . 0 333 . 2 - = ? + + EI EI EI
C B
? ?  (9) 
 
10 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?  (10) 
  
 
Page 3


? + = EI EI M
C DC
3
2
3
2
?    (3) 
 
Now, consider the joint equilibrium of B andC (vide Fig. 17.3c). 
 
0 0 = + ? =
?
BC
BA B
M M M  (4) 
 
      (5) 0 0 = + ? =
?
CD
CB C
M M M
 
 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame   (vide Fig. 17.3d). . .e i
?
= 0
X
F
 
 
    0 10
2 1
= - + - H H 
       
    10
2 1
= + ? H H .   (6) 
 
 
 
 
 
 
Considering the equilibrium of the column AB andCD , yields 
 
3
1
AB BA
M M
H
+
=      
   
and  
 
     
3
2
DC CD
M M
H
+
=      (7) 
 
The equation (6) may be written as, 
 
30 = + + +
DC CD AB BA
M M M M     (8) 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(6) 
 
10 667 . 0 5 . 0 333 . 2 - = ? + + EI EI EI
C B
? ?  (9) 
 
10 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?  (10) 
  
 
30
3
8
2 2 = ? + + EI EI EI
C B
? ?    (11) 
 
Equations (9), (10) and (11) indicate symmetry and this fact may be noted. This 
may be used as the check in deriving these equations. 
 
Solving equations (9), (10) and (11), 
 
355 . 1 ; 572 . 9 = - =
C B
EI EI ? ?     and    417 . 17 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
5.23 kN.m (counterclockwise)
AB
M = 
 
1.14 kN.m(clockwise)
BA
M =- 
 
1.130 kN.m
BC
M = 
 
13.415 kN.m
CB
M =- 
 
13.406 kN.m
CD
M = 
 
12.500 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.3 e. And the 
elastic curve is shown in Fig 17.3 f. the bending moment diagram is drawn on the 
compression side. Also note that the vertical hatching is used to represent 
bending moment diagram for the horizontal members (beams). 
 
 
Page 4


? + = EI EI M
C DC
3
2
3
2
?    (3) 
 
Now, consider the joint equilibrium of B andC (vide Fig. 17.3c). 
 
0 0 = + ? =
?
BC
BA B
M M M  (4) 
 
      (5) 0 0 = + ? =
?
CD
CB C
M M M
 
 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame   (vide Fig. 17.3d). . .e i
?
= 0
X
F
 
 
    0 10
2 1
= - + - H H 
       
    10
2 1
= + ? H H .   (6) 
 
 
 
 
 
 
Considering the equilibrium of the column AB andCD , yields 
 
3
1
AB BA
M M
H
+
=      
   
and  
 
     
3
2
DC CD
M M
H
+
=      (7) 
 
The equation (6) may be written as, 
 
30 = + + +
DC CD AB BA
M M M M     (8) 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(6) 
 
10 667 . 0 5 . 0 333 . 2 - = ? + + EI EI EI
C B
? ?  (9) 
 
10 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?  (10) 
  
 
30
3
8
2 2 = ? + + EI EI EI
C B
? ?    (11) 
 
Equations (9), (10) and (11) indicate symmetry and this fact may be noted. This 
may be used as the check in deriving these equations. 
 
Solving equations (9), (10) and (11), 
 
355 . 1 ; 572 . 9 = - =
C B
EI EI ? ?     and    417 . 17 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
5.23 kN.m (counterclockwise)
AB
M = 
 
1.14 kN.m(clockwise)
BA
M =- 
 
1.130 kN.m
BC
M = 
 
13.415 kN.m
CB
M =- 
 
13.406 kN.m
CD
M = 
 
12.500 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.3 e. And the 
elastic curve is shown in Fig 17.3 f. the bending moment diagram is drawn on the 
compression side. Also note that the vertical hatching is used to represent 
bending moment diagram for the horizontal members (beams). 
 
 
 
 
Page 5


? + = EI EI M
C DC
3
2
3
2
?    (3) 
 
Now, consider the joint equilibrium of B andC (vide Fig. 17.3c). 
 
0 0 = + ? =
?
BC
BA B
M M M  (4) 
 
      (5) 0 0 = + ? =
?
CD
CB C
M M M
 
 
 
The required third equation is written considering the horizontal equilibrium of the 
entire frame   (vide Fig. 17.3d). . .e i
?
= 0
X
F
 
 
    0 10
2 1
= - + - H H 
       
    10
2 1
= + ? H H .   (6) 
 
 
 
 
 
 
Considering the equilibrium of the column AB andCD , yields 
 
3
1
AB BA
M M
H
+
=      
   
and  
 
     
3
2
DC CD
M M
H
+
=      (7) 
 
The equation (6) may be written as, 
 
30 = + + +
DC CD AB BA
M M M M     (8) 
 
Substituting the beam end moments from equation (3) in equations (4), (5) and 
(6) 
 
10 667 . 0 5 . 0 333 . 2 - = ? + + EI EI EI
C B
? ?  (9) 
 
10 667 . 0 5 . 0 333 . 2 = ? + + EI EI EI
B C
? ?  (10) 
  
 
30
3
8
2 2 = ? + + EI EI EI
C B
? ?    (11) 
 
Equations (9), (10) and (11) indicate symmetry and this fact may be noted. This 
may be used as the check in deriving these equations. 
 
Solving equations (9), (10) and (11), 
 
355 . 1 ; 572 . 9 = - =
C B
EI EI ? ?     and    417 . 17 = ? EI . 
 
Substituting the values of 
C B
EI EI ? ? , and ? EI in the slope-deflection equation 
(3), one could calculate beam end moments. Thus, 
 
5.23 kN.m (counterclockwise)
AB
M = 
 
1.14 kN.m(clockwise)
BA
M =- 
 
1.130 kN.m
BC
M = 
 
13.415 kN.m
CB
M =- 
 
13.406 kN.m
CD
M = 
 
12.500 kN.m
DC
M = . 
 
The bending moment diagram for the frame is shown in Fig. 17.3 e. And the 
elastic curve is shown in Fig 17.3 f. the bending moment diagram is drawn on the 
compression side. Also note that the vertical hatching is used to represent 
bending moment diagram for the horizontal members (beams). 
 
 
 
 
 
Example 17.2 
Analyse the rigid frame as shown in Fig. 17.4a and draw the bending moment 
diagram. The moment of inertia for all the members is shown in the figure. 
Neglect axial deformations.