The Slope Deflection Method: Frames Without Sidesway - 2 Civil Engineering (CE) Notes | EduRev

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Civil Engineering (CE) : The Slope Deflection Method: Frames Without Sidesway - 2 Civil Engineering (CE) Notes | EduRev

 Page 1


?
= 0
B
M    ?     0 10 = - +
BC BD
M M    (3) 
 
Substituting the value of  and  and from equation (2) in the above 
equation 
BD
M
BC
M
 
0 10 5 = - + +
B B
EI EI ? ? 
 
EI
B
5 . 2
= ?      (4) 
 
Substituting the values of 
B
? in equation (2), the beam end moments are 
calculated 
 
m kN 5 . 7 · + =
BD
M 
 
m kN 75 . 3 · - =
DB
M 
 
m kN 5 . 2 · + =
BC
M 
 
m kN 25 . 1 · + =
CB
M     (5) 
 
The reactions are evaluated from static equations of equilibrium. The free body 
diagram of each member of the frame with external load and end moments are 
shown in Fig 16.3 (e) 
 
Page 2


?
= 0
B
M    ?     0 10 = - +
BC BD
M M    (3) 
 
Substituting the value of  and  and from equation (2) in the above 
equation 
BD
M
BC
M
 
0 10 5 = - + +
B B
EI EI ? ? 
 
EI
B
5 . 2
= ?      (4) 
 
Substituting the values of 
B
? in equation (2), the beam end moments are 
calculated 
 
m kN 5 . 7 · + =
BD
M 
 
m kN 75 . 3 · - =
DB
M 
 
m kN 5 . 2 · + =
BC
M 
 
m kN 25 . 1 · + =
CB
M     (5) 
 
The reactions are evaluated from static equations of equilibrium. The free body 
diagram of each member of the frame with external load and end moments are 
shown in Fig 16.3 (e) 
 
 
 
( ) ? = kN 9375 . 10
Cy
R   
 
( ) ? - = kN 9375 . 0
Cx
R 
 
( ) ? = kN 0625 . 4
Dy
R 
 
( ) ? = kN 9375 . 0
Dx
R     (6) 
 
 
Bending moment diagram is shown in Fig 16.3(f) 
 
 
Page 3


?
= 0
B
M    ?     0 10 = - +
BC BD
M M    (3) 
 
Substituting the value of  and  and from equation (2) in the above 
equation 
BD
M
BC
M
 
0 10 5 = - + +
B B
EI EI ? ? 
 
EI
B
5 . 2
= ?      (4) 
 
Substituting the values of 
B
? in equation (2), the beam end moments are 
calculated 
 
m kN 5 . 7 · + =
BD
M 
 
m kN 75 . 3 · - =
DB
M 
 
m kN 5 . 2 · + =
BC
M 
 
m kN 25 . 1 · + =
CB
M     (5) 
 
The reactions are evaluated from static equations of equilibrium. The free body 
diagram of each member of the frame with external load and end moments are 
shown in Fig 16.3 (e) 
 
 
 
( ) ? = kN 9375 . 10
Cy
R   
 
( ) ? - = kN 9375 . 0
Cx
R 
 
( ) ? = kN 0625 . 4
Dy
R 
 
( ) ? = kN 9375 . 0
Dx
R     (6) 
 
 
Bending moment diagram is shown in Fig 16.3(f) 
 
 
 
 
The vertical hatching is use to represent the bending moment diagram for the 
horizontal member (beams) and horizontal hatching is used for bending moment 
diagram for the vertical members. 
The qualitative elastic curve is shown in Fig 16.3 (g). 
 
 
 
Page 4


?
= 0
B
M    ?     0 10 = - +
BC BD
M M    (3) 
 
Substituting the value of  and  and from equation (2) in the above 
equation 
BD
M
BC
M
 
0 10 5 = - + +
B B
EI EI ? ? 
 
EI
B
5 . 2
= ?      (4) 
 
Substituting the values of 
B
? in equation (2), the beam end moments are 
calculated 
 
m kN 5 . 7 · + =
BD
M 
 
m kN 75 . 3 · - =
DB
M 
 
m kN 5 . 2 · + =
BC
M 
 
m kN 25 . 1 · + =
CB
M     (5) 
 
The reactions are evaluated from static equations of equilibrium. The free body 
diagram of each member of the frame with external load and end moments are 
shown in Fig 16.3 (e) 
 
 
 
( ) ? = kN 9375 . 10
Cy
R   
 
( ) ? - = kN 9375 . 0
Cx
R 
 
( ) ? = kN 0625 . 4
Dy
R 
 
( ) ? = kN 9375 . 0
Dx
R     (6) 
 
 
Bending moment diagram is shown in Fig 16.3(f) 
 
 
 
 
The vertical hatching is use to represent the bending moment diagram for the 
horizontal member (beams) and horizontal hatching is used for bending moment 
diagram for the vertical members. 
The qualitative elastic curve is shown in Fig 16.3 (g). 
 
 
 
Example 16.2 
Compute reactions and beam end moments for the rigid frame shown in Fig 16.4 
(a). Draw bending moment and shear force diagram for the frame and also 
sketch qualitative elastic curve. 
 
Solution 
 
 
 
In this frame rotations 
A
? and 
B
? are evaluated by considering the equilibrium of 
joint A and B.  The given frame is kinematically indeterminate to second 
degree. Evaluate fixed end moments. This is done by considering the 
kinematically determinate structure. (Fig 16.4 b) 
 
 
Page 5


?
= 0
B
M    ?     0 10 = - +
BC BD
M M    (3) 
 
Substituting the value of  and  and from equation (2) in the above 
equation 
BD
M
BC
M
 
0 10 5 = - + +
B B
EI EI ? ? 
 
EI
B
5 . 2
= ?      (4) 
 
Substituting the values of 
B
? in equation (2), the beam end moments are 
calculated 
 
m kN 5 . 7 · + =
BD
M 
 
m kN 75 . 3 · - =
DB
M 
 
m kN 5 . 2 · + =
BC
M 
 
m kN 25 . 1 · + =
CB
M     (5) 
 
The reactions are evaluated from static equations of equilibrium. The free body 
diagram of each member of the frame with external load and end moments are 
shown in Fig 16.3 (e) 
 
 
 
( ) ? = kN 9375 . 10
Cy
R   
 
( ) ? - = kN 9375 . 0
Cx
R 
 
( ) ? = kN 0625 . 4
Dy
R 
 
( ) ? = kN 9375 . 0
Dx
R     (6) 
 
 
Bending moment diagram is shown in Fig 16.3(f) 
 
 
 
 
The vertical hatching is use to represent the bending moment diagram for the 
horizontal member (beams) and horizontal hatching is used for bending moment 
diagram for the vertical members. 
The qualitative elastic curve is shown in Fig 16.3 (g). 
 
 
 
Example 16.2 
Compute reactions and beam end moments for the rigid frame shown in Fig 16.4 
(a). Draw bending moment and shear force diagram for the frame and also 
sketch qualitative elastic curve. 
 
Solution 
 
 
 
In this frame rotations 
A
? and 
B
? are evaluated by considering the equilibrium of 
joint A and B.  The given frame is kinematically indeterminate to second 
degree. Evaluate fixed end moments. This is done by considering the 
kinematically determinate structure. (Fig 16.4 b) 
 
 
 
 
kN.m 15
12
6 5
2
=
×
=
F
DB
M 
 
kN.m 15
12
6 5
2
- =
× -
=
F
BA
M 
 
kN.m 5 . 2
4
2 2 5
2
2
=
× ×
=
F
BC
M 
 
kN.m 5 . 2
4
2 2 5
2
2
- =
× × -
=
F
CD
M   (1) 
 
Note that the frame is restrained against sidesway. The spans must be 
considered for writing slope-deflection equations viz, A , B and . The beam 
end moments are related to unknown rotations 
AC
A
? and 
B
? by following slope-
deflection equations. (Force deflection equations). Support  is fixed and hence C
. 0 =
C
? 
 
( )
()
B A
AB
F
ABL AB
L
I E
M M ? ? + + = 2
2 2
 
 
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