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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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FAQs on The Slope Deflection Method: Frames Without Sidesway - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the slope deflection method? |  |
The slope deflection method is a structural analysis technique used to calculate the moments and rotations at various points in a frame structure. It is based on the assumption that the frame members do not undergo any sidesway, meaning that they remain perfectly straight during loading.
| 2. How does the slope deflection method work for frames without sidesway? | |
In frames without sidesway, the slope deflection method involves analyzing each member individually. The method considers the applied loads, member properties, and support conditions to determine the internal moments and rotations at each joint. These values are then used to calculate the member forces and displacements.
| 3. What are the advantages of using the slope deflection method for frames without sidesway? | |
The slope deflection method has several advantages for analyzing frames without sidesway. Firstly, it provides accurate results for determining member forces and displacements. Secondly, it accounts for the stiffness of the members and their connections, resulting in more realistic predictions. Finally, it allows for the consideration of different load patterns and support conditions, making it versatile for different frame configurations.
| 4. Are there any limitations to using the slope deflection method for frames without sidesway? | |
Yes, there are a few limitations to using the slope deflection method for frames without sidesway. Firstly, it assumes that the members remain perfectly straight, which may not be the case in reality. Secondly, it can become complex and time-consuming for frames with a large number of members and joints. Lastly, the method may not be suitable for analyzing frames with significant sidesway or lateral displacement.
| 5. How is the slope deflection method different from other structural analysis techniques? | |
The slope deflection method differs from other structural analysis techniques, such as the moment distribution method or the direct stiffness method, in its approach and assumptions. Unlike the moment distribution method, which redistributes moments at each joint until equilibrium is reached, the slope deflection method calculates member moments and rotations directly. Additionally, the slope deflection method assumes no sidesway, while the direct stiffness method considers lateral deflections in the analysis.
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Dec 19, 2024 Last updated |
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