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

## 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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## Structural Analysis

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