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The Slope-Deflection Method: Beams - 3 Civil Engineering (CE) Notes | EduRev

Civil Engineering (CE) : The Slope-Deflection Method: Beams - 3 Civil Engineering (CE) Notes | EduRev

``` Page 1

) 001 . 0 2 ( 2 . 0 667 . 41 - + + =
D C CD
EI M ? ?
) 001 . 0 2 ( 2 . 0 667 . 41 - + + - =
C D DC
EI M ? ?   (3)

For the present problem, four joint equilibrium equations can be written, one each
for each of the supports. They are

0 0 = ? =
? AB A
M M

0 0 = + ? =
? BC BA B
M M M

0 0 = + ? =
? CD CB C
M M M

(4) 0 0 = ? =
? DC D
M M

Substituting the values of beam end moments from equations (3) in equation (4),
four equations are obtained in four unknown rotations
A
? ,
B
? ,
C
? and
D
? .They
are,

( )
2 6 3
kN.m 000 , 270 10 35 . 1 10 200 = × × × =
-
EI

3
10 2716 . 1 2
-
× - = +
B A
? ?

001 . 0 4 - = + +
C B A
? ? ?

0005 . 0 4 = + +
D C B
? ? ?

(5)
3
10 7716 . 1 2
-
× = +
D C
? ?

Solving the above sets of simultaneous equations, values of
A
? ,
B
? ,
C
? and
D
? are
evaluated.

4
10 9629 . 5
-
× - =
A
?

5
10 9013 . 7
-
× - =
B
?

5
10 7653 . 8
-
× - =
C
?

4
10 2963 . 9
-
× =
D
?

Page 2

) 001 . 0 2 ( 2 . 0 667 . 41 - + + =
D C CD
EI M ? ?
) 001 . 0 2 ( 2 . 0 667 . 41 - + + - =
C D DC
EI M ? ?   (3)

For the present problem, four joint equilibrium equations can be written, one each
for each of the supports. They are

0 0 = ? =
? AB A
M M

0 0 = + ? =
? BC BA B
M M M

0 0 = + ? =
? CD CB C
M M M

(4) 0 0 = ? =
? DC D
M M

Substituting the values of beam end moments from equations (3) in equation (4),
four equations are obtained in four unknown rotations
A
? ,
B
? ,
C
? and
D
? .They
are,

( )
2 6 3
kN.m 000 , 270 10 35 . 1 10 200 = × × × =
-
EI

3
10 2716 . 1 2
-
× - = +
B A
? ?

001 . 0 4 - = + +
C B A
? ? ?

0005 . 0 4 = + +
D C B
? ? ?

(5)
3
10 7716 . 1 2
-
× = +
D C
? ?

Solving the above sets of simultaneous equations, values of
A
? ,
B
? ,
C
? and
D
? are
evaluated.

4
10 9629 . 5
-
× - =
A
?

5
10 9013 . 7
-
× - =
B
?

5
10 7653 . 8
-
× - =
C
?

4
10 2963 . 9
-
× =
D
?

Substituting the values in slope-deflection equations the beam end moments are
evaluated.

0 )} 0005 . 0 ) 10 9013 . 7 ( ) 10 9629 . 5 ( 2 { 000 , 270 2 . 0 667 . 41
5 4
= + × - + × - × + =
- -
AB
M

kN.m 40 . 55 } 0005 . 0 10 9629 . 5 ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41
4 5
- = + × - × - × + - =
- -
BA
M

kN.m 40 . 55 } 0005 . 0 ) 10 7653 . 8 ( ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41
5 5
= + × - + × - × + =
- -
BC
M

kN.m 40 . 28 } 0005 . 0 10 9013 . 7 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41
5 5
- = + × - × - × + - =
- -
CB
M

kN.m 40 . 28 } 001 . 0 10 2963 . 9 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41
4 5
= - × + × - × × + =
- -
CD
M

kN.m 0 } 001 . 0 10 7653 . 8 10 2963 . 9 2 { 000 , 270 2 . 0 667 . 41
5 4
= - × - × × × + - =
- -
DC
M  (7)

Reactions are obtained from equilibrium equations. Now consider the free body
diagram of the beam with end moments and external loads as shown in
Fig.15.3d.

) ( kN 46 . 19 ? =
A
R

Page 3

) 001 . 0 2 ( 2 . 0 667 . 41 - + + =
D C CD
EI M ? ?
) 001 . 0 2 ( 2 . 0 667 . 41 - + + - =
C D DC
EI M ? ?   (3)

For the present problem, four joint equilibrium equations can be written, one each
for each of the supports. They are

0 0 = ? =
? AB A
M M

0 0 = + ? =
? BC BA B
M M M

0 0 = + ? =
? CD CB C
M M M

(4) 0 0 = ? =
? DC D
M M

Substituting the values of beam end moments from equations (3) in equation (4),
four equations are obtained in four unknown rotations
A
? ,
B
? ,
C
? and
D
? .They
are,

( )
2 6 3
kN.m 000 , 270 10 35 . 1 10 200 = × × × =
-
EI

3
10 2716 . 1 2
-
× - = +
B A
? ?

001 . 0 4 - = + +
C B A
? ? ?

0005 . 0 4 = + +
D C B
? ? ?

(5)
3
10 7716 . 1 2
-
× = +
D C
? ?

Solving the above sets of simultaneous equations, values of
A
? ,
B
? ,
C
? and
D
? are
evaluated.

4
10 9629 . 5
-
× - =
A
?

5
10 9013 . 7
-
× - =
B
?

5
10 7653 . 8
-
× - =
C
?

4
10 2963 . 9
-
× =
D
?

Substituting the values in slope-deflection equations the beam end moments are
evaluated.

0 )} 0005 . 0 ) 10 9013 . 7 ( ) 10 9629 . 5 ( 2 { 000 , 270 2 . 0 667 . 41
5 4
= + × - + × - × + =
- -
AB
M

kN.m 40 . 55 } 0005 . 0 10 9629 . 5 ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41
4 5
- = + × - × - × + - =
- -
BA
M

kN.m 40 . 55 } 0005 . 0 ) 10 7653 . 8 ( ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41
5 5
= + × - + × - × + =
- -
BC
M

kN.m 40 . 28 } 0005 . 0 10 9013 . 7 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41
5 5
- = + × - × - × + - =
- -
CB
M

kN.m 40 . 28 } 001 . 0 10 2963 . 9 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41
4 5
= - × + × - × × + =
- -
CD
M

kN.m 0 } 001 . 0 10 7653 . 8 10 2963 . 9 2 { 000 , 270 2 . 0 667 . 41
5 4
= - × - × × × + - =
- -
DC
M  (7)

Reactions are obtained from equilibrium equations. Now consider the free body
diagram of the beam with end moments and external loads as shown in
Fig.15.3d.

) ( kN 46 . 19 ? =
A
R

) ( kN 54 . 30 ? =
BL
R

) ( kN 7 . 27 ? =
BR
R

) ( kN 3 . 22 ? =
CL
R

) ( kN 84 . 27 ? =
CR
R

) ( kN 16 . 22 ? =
D
R

The shear force and bending moment diagrams are shown in Fig.15.5e.

Summary
In this lesson, slope-deflection equations are derived for the case of beam with
yielding supports. Moments developed at the ends are related to rotations and
support settlements. The equilibrium equations are written at each support. The
continuous beam is solved using slope-deflection equations. The deflected shape
of the beam is sketched. The bending moment and shear force diagrams are
drawn for the examples solved in this lesson.

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Structural Analysis

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