The Slope-Deflection Method: Beams - 3 | Structural Analysis - Civil Engineering (CE) PDF Download

 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. 
 
      radians 
4
10 9629 . 5
-
× - =
A
?
 
      radians 
5
10 9013 . 7
-
× - =
B
?
 
      radians 
5
10 7653 . 8
-
× - =
C
?
 
      radians   (6) 
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. 
 
      radians 
4
10 9629 . 5
-
× - =
A
?
 
      radians 
5
10 9013 . 7
-
× - =
B
?
 
      radians 
5
10 7653 . 8
-
× - =
C
?
 
      radians   (6) 
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. 
 
      radians 
4
10 9629 . 5
-
× - =
A
?
 
      radians 
5
10 9013 . 7
-
× - =
B
?
 
      radians 
5
10 7653 . 8
-
× - =
C
?
 
      radians   (6) 
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.