Courses

# The Slope Deflection Method: Frames Without Sidesway - 3 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Slope Deflection Method: Frames Without Sidesway - 3 Civil Engineering (CE) Notes | EduRev

``` 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

```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Structural Analysis

30 videos|122 docs|28 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;