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Page 1
curve " AB rotates by an angle
AB
? , rotates by " "C B
BC
? and rotates by DC
CD
? as shown in figure. Due to symmetry,
AB CD
? ? = . From the geometry of the
figure,
AB AB
AB
L L
BB
1
" ?
- = = ?
But
a cos
1
?
= ?
Thus,
5 cos
?
- =
?
- =
a
?
AB
AB
L
5
?
- =
CD
?
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
(1)
We have three independent unknowns for this problem
C B
? ? , and . The ends ?
A and are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are,
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M
Now, writing the slope-deflection equations for the six beam end moments,
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- =
? + = EI EI M
B AB
471 . 0 784 . 0 ?
? + = EI EI M
B BA
471 . 0 568 . 1 ?
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? + = EI EI M
C CD
471 . 0 568 . 1 ?
? + = EI EI M
C DC
471 . 0 784 . 0 ? (2)
Now, considering the joint equilibrium of B andC , yields
Page 2
curve " AB rotates by an angle
AB
? , rotates by " "C B
BC
? and rotates by DC
CD
? as shown in figure. Due to symmetry,
AB CD
? ? = . From the geometry of the
figure,
AB AB
AB
L L
BB
1
" ?
- = = ?
But
a cos
1
?
= ?
Thus,
5 cos
?
- =
?
- =
a
?
AB
AB
L
5
?
- =
CD
?
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
(1)
We have three independent unknowns for this problem
C B
? ? , and . The ends ?
A and are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are,
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M
Now, writing the slope-deflection equations for the six beam end moments,
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- =
? + = EI EI M
B AB
471 . 0 784 . 0 ?
? + = EI EI M
B BA
471 . 0 568 . 1 ?
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? + = EI EI M
C CD
471 . 0 568 . 1 ?
? + = EI EI M
C DC
471 . 0 784 . 0 ? (2)
Now, considering the joint equilibrium of B andC , yields
0 0 = + ? =
?
BC
BA B
M M M
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ? (3)
0 0 = + ? =
?
CD
CB C
M M M
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ? (4)
Shear equation for
Column AB
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
(5)
Column CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
(6)
Beam BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M (7)
Page 3
curve " AB rotates by an angle
AB
? , rotates by " "C B
BC
? and rotates by DC
CD
? as shown in figure. Due to symmetry,
AB CD
? ? = . From the geometry of the
figure,
AB AB
AB
L L
BB
1
" ?
- = = ?
But
a cos
1
?
= ?
Thus,
5 cos
?
- =
?
- =
a
?
AB
AB
L
5
?
- =
CD
?
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
(1)
We have three independent unknowns for this problem
C B
? ? , and . The ends ?
A and are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are,
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M
Now, writing the slope-deflection equations for the six beam end moments,
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- =
? + = EI EI M
B AB
471 . 0 784 . 0 ?
? + = EI EI M
B BA
471 . 0 568 . 1 ?
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? + = EI EI M
C CD
471 . 0 568 . 1 ?
? + = EI EI M
C DC
471 . 0 784 . 0 ? (2)
Now, considering the joint equilibrium of B andC , yields
0 0 = + ? =
?
BC
BA B
M M M
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ? (3)
0 0 = + ? =
?
CD
CB C
M M M
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ? (4)
Shear equation for
Column AB
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
(5)
Column CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
(6)
Beam BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M (7)
?
= + = 5 0
2 1
H H F
X
(8)
?
= - - = 0 10 0
2 1
V V F
Y
(9)
From equation (7),
2
10
1
+ +
=
CB BC
M M
V
From equation (8),
2 1
5 H H - =
From equation (9), 10
2
10
10
1 2
-
+ +
= - =
CB BC
M M
V V
Substituting the values of and in equations (5) and (6),
1 1
,H V
2
V
0 2 2 10 60
2
= + + - - -
CB BC BA AB
M M M M H (10)
0 2 2 10 10
2
= + + - - + -
CB BC DC CD
M M M M H (11)
Eliminating in equation (10) and (11),
2
H
25 = - - + + +
CB BC DC CD BA AB
M M M M M M (12)
Substituting the values of in (12) we get the required third
equation. Thus,
DC CD BA AB
M M M M , , ,
+ ? + EI EI
B
471 . 0 784 . 0 ? + ? + EI EI
B
471 . 0 568 . 1 ? + ? + EI EI
C
471 . 0 568 . 1 ?
? + EI EI
C
471 . 0 784 . 0 ? -( ? - + + EI EI EI
C B
6 . 0 2 5 . 2 ? ? )-
( ? - + + - EI EI EI
C B
6 . 0 2 5 . 2 ? ? ) 25 =
Simplifying,
25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI
B C
? ? (13)
Solving simultaneously equations (3) (4) and (13), yields
205 . 1 ; 741 . 0 = - =
C B
EI EI ? ? and 204 . 8 = ? EI .
Substituting the values of
C B
EI EI ? ? , and ? EI in the slope-deflection equation
(3), one could calculate beam end moments. Thus,
3.28 kN.m
AB
M =
Page 4
curve " AB rotates by an angle
AB
? , rotates by " "C B
BC
? and rotates by DC
CD
? as shown in figure. Due to symmetry,
AB CD
? ? = . From the geometry of the
figure,
AB AB
AB
L L
BB
1
" ?
- = = ?
But
a cos
1
?
= ?
Thus,
5 cos
?
- =
?
- =
a
?
AB
AB
L
5
?
- =
CD
?
5
tan
2
tan 2
2
2
?
= ? =
?
=
?
= a
a
?
BC
(1)
We have three independent unknowns for this problem
C B
? ? , and . The ends ?
A and are fixed. Hence, D . 0 = =
D A
? ? Fixed end moments are,
. 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = =
F
DC
F
CD
F
CB
F
BC
F
BA
F
AB
M M m kN M m kN M M M
Now, writing the slope-deflection equations for the six beam end moments,
[]
AB A AB
I E
M ? ? 3
1 . 5
) 2 ( 2
- =
? + = EI EI M
B AB
471 . 0 784 . 0 ?
? + = EI EI M
B BA
471 . 0 568 . 1 ?
? - + + = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? - + + - = EI EI EI M
C B BC
6 . 0 2 5 . 2 ? ?
? + = EI EI M
C CD
471 . 0 568 . 1 ?
? + = EI EI M
C DC
471 . 0 784 . 0 ? (2)
Now, considering the joint equilibrium of B andC , yields
0 0 = + ? =
?
BC
BA B
M M M
5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI
C B
? ? (3)
0 0 = + ? =
?
CD
CB C
M M M
5 . 2 129 . 0 568 . 3 = ? - + EI EI EI
B C
? ? (4)
Shear equation for
Column AB
0 ) 1 ( 5
1 1
= + - - V M M H
BA AB
(5)
Column CD
0 ) 1 ( 5
2 2
= + - - V M M H
DC CD
(6)
Beam BC
0 10 2 0
1
= - - - =
?
CB BC C
M M V M (7)
?
= + = 5 0
2 1
H H F
X
(8)
?
= - - = 0 10 0
2 1
V V F
Y
(9)
From equation (7),
2
10
1
+ +
=
CB BC
M M
V
From equation (8),
2 1
5 H H - =
From equation (9), 10
2
10
10
1 2
-
+ +
= - =
CB BC
M M
V V
Substituting the values of and in equations (5) and (6),
1 1
,H V
2
V
0 2 2 10 60
2
= + + - - -
CB BC BA AB
M M M M H (10)
0 2 2 10 10
2
= + + - - + -
CB BC DC CD
M M M M H (11)
Eliminating in equation (10) and (11),
2
H
25 = - - + + +
CB BC DC CD BA AB
M M M M M M (12)
Substituting the values of in (12) we get the required third
equation. Thus,
DC CD BA AB
M M M M , , ,
+ ? + EI EI
B
471 . 0 784 . 0 ? + ? + EI EI
B
471 . 0 568 . 1 ? + ? + EI EI
C
471 . 0 568 . 1 ?
? + EI EI
C
471 . 0 784 . 0 ? -( ? - + + EI EI EI
C B
6 . 0 2 5 . 2 ? ? )-
( ? - + + - EI EI EI
C B
6 . 0 2 5 . 2 ? ? ) 25 =
Simplifying,
25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI
B C
? ? (13)
Solving simultaneously equations (3) (4) and (13), yields
205 . 1 ; 741 . 0 = - =
C B
EI EI ? ? and 204 . 8 = ? EI .
Substituting the values of
C B
EI EI ? ? , and ? EI in the slope-deflection equation
(3), one could calculate beam end moments. Thus,
3.28 kN.m
AB
M =
2.70 kN.m
BA
M =
2.70 kN.m
BC
M = -
5.75 kN.m
CB
M = -
5.75 kN.m
CD
M =
4.81 kN.m
DC
M = . (14)
The bending moment diagram for the frame is shown in Fig. 17.5 d.
Summary
In this lesson, slope-deflection equations are derived for the plane frame
undergoing sidesway. Using these equations, plane frames with sidesway are
analysed. The reactions are calculated from static equilibrium equations. A
couple of problems are solved to make things clear. In each numerical example,
the bending moment diagram is drawn and deflected shape is sketched for the
plane frame.
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FAQs on The Slope Deflection Method: Frames with Sidesway - 4 - Structural Analysis - Civil Engineering (CE)
| 1. What is the slope deflection method in civil engineering? |  |
Ans. The slope deflection method is a structural analysis technique used in civil engineering to determine the bending moments and rotations at various points in a frame structure. It can be used to analyze frames with sidesway, where the lateral deflection of the frame influences the distribution of internal forces.
| 2. How does the slope deflection method handle frames with sidesway? | |
Ans. The slope deflection method handles frames with sidesway by considering the influence of lateral deflection on the distribution of internal forces. It allows for the calculation of bending moments and rotations at various points in the frame, taking into account the sidesway effects. This method is particularly useful for analyzing tall and slender structures that are prone to lateral deflection.
| 3. What are the advantages of using the slope deflection method for frames with sidesway? | |
Ans. The slope deflection method offers several advantages for analyzing frames with sidesway. It allows for the accurate determination of internal forces, including bending moments and rotations, considering the effects of lateral deflection. This method also takes into account the stiffness and flexibility of the structure, providing a more realistic analysis. Additionally, the slope deflection method is suitable for both statically determinate and indeterminate frames.
| 4. Are there any limitations or assumptions associated with the slope deflection method? | |
Ans. Yes, there are some limitations and assumptions associated with the slope deflection method. One of the main assumptions is that the frame is linearly elastic, meaning that it behaves within the linear range of its material properties. Additionally, the method assumes that the frame is subjected to static loading conditions and neglects any dynamic effects. It also assumes that the frame members are connected rigidly at their joints, without any rotational or translational movements.
| 5. Can the slope deflection method be used for all types of frame structures? | |
Ans. The slope deflection method can be used for a wide range of frame structures, including both statically determinate and indeterminate frames. However, its applicability may depend on the complexity and size of the structure. For highly complex or large-scale frames, other analysis methods such as the matrix stiffness method or finite element method may be more suitable. It is important to consider the specific characteristics of the frame and consult structural engineering guidelines to determine the most appropriate analysis method.
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