The Slope Deflection Method: Beams - 1 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive slope-deflection equations for the case beam with yielding supports. 
2. Estimate the reactions induced in the beam due to support settlements. 
3. Analyse the beam undergoing support settlements and subjected to external 
loads. 
4. Write joint equilibrium equations in terms of moments. 
5. Relate moments to joint rotations and support settlements. 
 
 
15.1 Introduction  
In the last lesson, slope-deflection equations were derived without considering 
the rotation of the beam axis. In this lesson, slope-deflection equations are 
derived considering the rotation of beam axis. In statically indeterminate 
structures, the beam axis rotates due to support yielding and this would in turn 
induce reactions and stresses in the structure. Hence, in this case the beam end 
moments are related to rotations, applied loads and beam axes rotation. After 
deriving the slope-deflection equation in section 15.2, few problems are solved to 
illustrate the procedure. 
 
Consider a beam AB as shown in Fig.15.1.The support B is at a higher 
elevation compared to A by an amount ? . Hence, the member axis has rotated 
by an amount ? from the original direction as shown in the figure. Let L be the 
span of the beam and flexural rigidity of the beamEI , is assumed to be constant 
for the beam. The chord has rotated in the counterclockwise direction with 
respect to its original direction. The counterclockwise moment and rotations are 
assumed to be positive. As stated earlier, the slopes and rotations are derived by 
superposing the end moments developed due to  
 
(1) Externally applied moments on beams. 
(2) Displacements 
A
? ,
B
? and ? (settlement) 
 
Page 2


Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive slope-deflection equations for the case beam with yielding supports. 
2. Estimate the reactions induced in the beam due to support settlements. 
3. Analyse the beam undergoing support settlements and subjected to external 
loads. 
4. Write joint equilibrium equations in terms of moments. 
5. Relate moments to joint rotations and support settlements. 
 
 
15.1 Introduction  
In the last lesson, slope-deflection equations were derived without considering 
the rotation of the beam axis. In this lesson, slope-deflection equations are 
derived considering the rotation of beam axis. In statically indeterminate 
structures, the beam axis rotates due to support yielding and this would in turn 
induce reactions and stresses in the structure. Hence, in this case the beam end 
moments are related to rotations, applied loads and beam axes rotation. After 
deriving the slope-deflection equation in section 15.2, few problems are solved to 
illustrate the procedure. 
 
Consider a beam AB as shown in Fig.15.1.The support B is at a higher 
elevation compared to A by an amount ? . Hence, the member axis has rotated 
by an amount ? from the original direction as shown in the figure. Let L be the 
span of the beam and flexural rigidity of the beamEI , is assumed to be constant 
for the beam. The chord has rotated in the counterclockwise direction with 
respect to its original direction. The counterclockwise moment and rotations are 
assumed to be positive. As stated earlier, the slopes and rotations are derived by 
superposing the end moments developed due to  
 
(1) Externally applied moments on beams. 
(2) Displacements 
A
? ,
B
? and ? (settlement) 
 
 
 
The given beam with initial support settlement may be thought of as 
superposition of two simple cases as shown in Fig.15.1 (b) and in Fig. 15.1(c). In 
Fig.15.1b, the kinematically determinate beam is shown with the applied load. 
For this case, the fixed end moments are calculated by force method. Let 
A
f and 
B
f be the end rotations of the elastic curve with respect to rotated beam axis AB’ 
(see Fig.15.1c) that are caused by end moments  and . Assuming that 
rotations and displacements shown in Fig.15.1c are so small that  
'
AB
M
'
BA
M
 
l
?
= = ? ? tan    (15.1) 
 
Also, using the moment area theorem, 
A
f and 
B
f are written as  
 
    
EI
L M
EI
L M
AB AB
A A
6
'
3
'
- = - = ? ? f  (15.2a) 
 
Page 3


Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive slope-deflection equations for the case beam with yielding supports. 
2. Estimate the reactions induced in the beam due to support settlements. 
3. Analyse the beam undergoing support settlements and subjected to external 
loads. 
4. Write joint equilibrium equations in terms of moments. 
5. Relate moments to joint rotations and support settlements. 
 
 
15.1 Introduction  
In the last lesson, slope-deflection equations were derived without considering 
the rotation of the beam axis. In this lesson, slope-deflection equations are 
derived considering the rotation of beam axis. In statically indeterminate 
structures, the beam axis rotates due to support yielding and this would in turn 
induce reactions and stresses in the structure. Hence, in this case the beam end 
moments are related to rotations, applied loads and beam axes rotation. After 
deriving the slope-deflection equation in section 15.2, few problems are solved to 
illustrate the procedure. 
 
Consider a beam AB as shown in Fig.15.1.The support B is at a higher 
elevation compared to A by an amount ? . Hence, the member axis has rotated 
by an amount ? from the original direction as shown in the figure. Let L be the 
span of the beam and flexural rigidity of the beamEI , is assumed to be constant 
for the beam. The chord has rotated in the counterclockwise direction with 
respect to its original direction. The counterclockwise moment and rotations are 
assumed to be positive. As stated earlier, the slopes and rotations are derived by 
superposing the end moments developed due to  
 
(1) Externally applied moments on beams. 
(2) Displacements 
A
? ,
B
? and ? (settlement) 
 
 
 
The given beam with initial support settlement may be thought of as 
superposition of two simple cases as shown in Fig.15.1 (b) and in Fig. 15.1(c). In 
Fig.15.1b, the kinematically determinate beam is shown with the applied load. 
For this case, the fixed end moments are calculated by force method. Let 
A
f and 
B
f be the end rotations of the elastic curve with respect to rotated beam axis AB’ 
(see Fig.15.1c) that are caused by end moments  and . Assuming that 
rotations and displacements shown in Fig.15.1c are so small that  
'
AB
M
'
BA
M
 
l
?
= = ? ? tan    (15.1) 
 
Also, using the moment area theorem, 
A
f and 
B
f are written as  
 
    
EI
L M
EI
L M
AB AB
A A
6
'
3
'
- = - = ? ? f  (15.2a) 
 
EI
L M
EI
L M
AB BA
B B
6
'
3
'
- = - = ? ? f   (15.2b) 
 
Now solving for  and  in terms of
'
A
M
'
B
M
A
? ,
B
? and ? , 
 
) 3 2 (
2
' ? ? ? - + =
B A AB
L
EI
M    (15.3a) 
 
) 3 2 (
2
' ? ? ? - + =
A B BA
L
EI
M    (15.3b) 
 
Now superposing the fixed end moments due to external load and end moments 
due to displacements, the end moments in the actual structure is obtained .Thus 
(see Fig.15.1) 
 
'
AB
F
AB AB
M M M + =     (15.4a) 
 
'     (15.4b) 
BA
F
BA BA
M M M + =
 
Substituting for   and   in equation (15.4a) and (15.4b), the slope-
deflection equations for the general case are obtained. Thus, 
'
AB
M
'
BA
M
 
) 3 2 (
2
? ? ? - + + =
B A
F
AB AB
L
EI
M M    (15.5a) 
) 3 2 (
2
? ? ? - + + =
A B
F
BA BA
L
EI
M M    (15.5b) 
 
In the above equations, it is important to adopt consistent sign convention. In the 
above derivation ? is taken to be negative for downward displacements. 
 
Example 15.1 
Calculate the support moments in the continuous beam (see Fig.15.2a) 
having constant flexural rigidity 
ABC
EI throughout ,due to vertical settlement of the 
support B by 5mm. Assume E=200 GPa and I = .Also plot 
quantitative elastic curve. 
4 4
10 4 m
-
×
 
 
Page 4


Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive slope-deflection equations for the case beam with yielding supports. 
2. Estimate the reactions induced in the beam due to support settlements. 
3. Analyse the beam undergoing support settlements and subjected to external 
loads. 
4. Write joint equilibrium equations in terms of moments. 
5. Relate moments to joint rotations and support settlements. 
 
 
15.1 Introduction  
In the last lesson, slope-deflection equations were derived without considering 
the rotation of the beam axis. In this lesson, slope-deflection equations are 
derived considering the rotation of beam axis. In statically indeterminate 
structures, the beam axis rotates due to support yielding and this would in turn 
induce reactions and stresses in the structure. Hence, in this case the beam end 
moments are related to rotations, applied loads and beam axes rotation. After 
deriving the slope-deflection equation in section 15.2, few problems are solved to 
illustrate the procedure. 
 
Consider a beam AB as shown in Fig.15.1.The support B is at a higher 
elevation compared to A by an amount ? . Hence, the member axis has rotated 
by an amount ? from the original direction as shown in the figure. Let L be the 
span of the beam and flexural rigidity of the beamEI , is assumed to be constant 
for the beam. The chord has rotated in the counterclockwise direction with 
respect to its original direction. The counterclockwise moment and rotations are 
assumed to be positive. As stated earlier, the slopes and rotations are derived by 
superposing the end moments developed due to  
 
(1) Externally applied moments on beams. 
(2) Displacements 
A
? ,
B
? and ? (settlement) 
 
 
 
The given beam with initial support settlement may be thought of as 
superposition of two simple cases as shown in Fig.15.1 (b) and in Fig. 15.1(c). In 
Fig.15.1b, the kinematically determinate beam is shown with the applied load. 
For this case, the fixed end moments are calculated by force method. Let 
A
f and 
B
f be the end rotations of the elastic curve with respect to rotated beam axis AB’ 
(see Fig.15.1c) that are caused by end moments  and . Assuming that 
rotations and displacements shown in Fig.15.1c are so small that  
'
AB
M
'
BA
M
 
l
?
= = ? ? tan    (15.1) 
 
Also, using the moment area theorem, 
A
f and 
B
f are written as  
 
    
EI
L M
EI
L M
AB AB
A A
6
'
3
'
- = - = ? ? f  (15.2a) 
 
EI
L M
EI
L M
AB BA
B B
6
'
3
'
- = - = ? ? f   (15.2b) 
 
Now solving for  and  in terms of
'
A
M
'
B
M
A
? ,
B
? and ? , 
 
) 3 2 (
2
' ? ? ? - + =
B A AB
L
EI
M    (15.3a) 
 
) 3 2 (
2
' ? ? ? - + =
A B BA
L
EI
M    (15.3b) 
 
Now superposing the fixed end moments due to external load and end moments 
due to displacements, the end moments in the actual structure is obtained .Thus 
(see Fig.15.1) 
 
'
AB
F
AB AB
M M M + =     (15.4a) 
 
'     (15.4b) 
BA
F
BA BA
M M M + =
 
Substituting for   and   in equation (15.4a) and (15.4b), the slope-
deflection equations for the general case are obtained. Thus, 
'
AB
M
'
BA
M
 
) 3 2 (
2
? ? ? - + + =
B A
F
AB AB
L
EI
M M    (15.5a) 
) 3 2 (
2
? ? ? - + + =
A B
F
BA BA
L
EI
M M    (15.5b) 
 
In the above equations, it is important to adopt consistent sign convention. In the 
above derivation ? is taken to be negative for downward displacements. 
 
Example 15.1 
Calculate the support moments in the continuous beam (see Fig.15.2a) 
having constant flexural rigidity 
ABC
EI throughout ,due to vertical settlement of the 
support B by 5mm. Assume E=200 GPa and I = .Also plot 
quantitative elastic curve. 
4 4
10 4 m
-
×
 
 
 
 
In the continuous beam , two rotations ABC
B
? and 
C
? need to be evaluated. 
Hence, beam is kinematically indeterminate to second degree. As there is no 
external load on the beam, the fixed end moments in the restrained beam are 
zero (see Fig.15.2b). 
 
 
 
For each span, two slope-deflection equations need to be written. In spanAB , 
B is belowA. Hence, the chord AB rotates in clockwise direction. Thus,
AB
? is 
taken as negative. 
 
Page 5


Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive slope-deflection equations for the case beam with yielding supports. 
2. Estimate the reactions induced in the beam due to support settlements. 
3. Analyse the beam undergoing support settlements and subjected to external 
loads. 
4. Write joint equilibrium equations in terms of moments. 
5. Relate moments to joint rotations and support settlements. 
 
 
15.1 Introduction  
In the last lesson, slope-deflection equations were derived without considering 
the rotation of the beam axis. In this lesson, slope-deflection equations are 
derived considering the rotation of beam axis. In statically indeterminate 
structures, the beam axis rotates due to support yielding and this would in turn 
induce reactions and stresses in the structure. Hence, in this case the beam end 
moments are related to rotations, applied loads and beam axes rotation. After 
deriving the slope-deflection equation in section 15.2, few problems are solved to 
illustrate the procedure. 
 
Consider a beam AB as shown in Fig.15.1.The support B is at a higher 
elevation compared to A by an amount ? . Hence, the member axis has rotated 
by an amount ? from the original direction as shown in the figure. Let L be the 
span of the beam and flexural rigidity of the beamEI , is assumed to be constant 
for the beam. The chord has rotated in the counterclockwise direction with 
respect to its original direction. The counterclockwise moment and rotations are 
assumed to be positive. As stated earlier, the slopes and rotations are derived by 
superposing the end moments developed due to  
 
(1) Externally applied moments on beams. 
(2) Displacements 
A
? ,
B
? and ? (settlement) 
 
 
 
The given beam with initial support settlement may be thought of as 
superposition of two simple cases as shown in Fig.15.1 (b) and in Fig. 15.1(c). In 
Fig.15.1b, the kinematically determinate beam is shown with the applied load. 
For this case, the fixed end moments are calculated by force method. Let 
A
f and 
B
f be the end rotations of the elastic curve with respect to rotated beam axis AB’ 
(see Fig.15.1c) that are caused by end moments  and . Assuming that 
rotations and displacements shown in Fig.15.1c are so small that  
'
AB
M
'
BA
M
 
l
?
= = ? ? tan    (15.1) 
 
Also, using the moment area theorem, 
A
f and 
B
f are written as  
 
    
EI
L M
EI
L M
AB AB
A A
6
'
3
'
- = - = ? ? f  (15.2a) 
 
EI
L M
EI
L M
AB BA
B B
6
'
3
'
- = - = ? ? f   (15.2b) 
 
Now solving for  and  in terms of
'
A
M
'
B
M
A
? ,
B
? and ? , 
 
) 3 2 (
2
' ? ? ? - + =
B A AB
L
EI
M    (15.3a) 
 
) 3 2 (
2
' ? ? ? - + =
A B BA
L
EI
M    (15.3b) 
 
Now superposing the fixed end moments due to external load and end moments 
due to displacements, the end moments in the actual structure is obtained .Thus 
(see Fig.15.1) 
 
'
AB
F
AB AB
M M M + =     (15.4a) 
 
'     (15.4b) 
BA
F
BA BA
M M M + =
 
Substituting for   and   in equation (15.4a) and (15.4b), the slope-
deflection equations for the general case are obtained. Thus, 
'
AB
M
'
BA
M
 
) 3 2 (
2
? ? ? - + + =
B A
F
AB AB
L
EI
M M    (15.5a) 
) 3 2 (
2
? ? ? - + + =
A B
F
BA BA
L
EI
M M    (15.5b) 
 
In the above equations, it is important to adopt consistent sign convention. In the 
above derivation ? is taken to be negative for downward displacements. 
 
Example 15.1 
Calculate the support moments in the continuous beam (see Fig.15.2a) 
having constant flexural rigidity 
ABC
EI throughout ,due to vertical settlement of the 
support B by 5mm. Assume E=200 GPa and I = .Also plot 
quantitative elastic curve. 
4 4
10 4 m
-
×
 
 
 
 
In the continuous beam , two rotations ABC
B
? and 
C
? need to be evaluated. 
Hence, beam is kinematically indeterminate to second degree. As there is no 
external load on the beam, the fixed end moments in the restrained beam are 
zero (see Fig.15.2b). 
 
 
 
For each span, two slope-deflection equations need to be written. In spanAB , 
B is belowA. Hence, the chord AB rotates in clockwise direction. Thus,
AB
? is 
taken as negative. 
 
3
3
10 1
5
10 5
-
-
× - =
× -
=
AB
?   (1) 
 
Writing slope-deflection equation for spanAB , 
 
(
AB B A
AB )
L
EI
M ? ? ? 3 2
2
- + =    
  
For span AB , , 0 =
A
? Hence, 
  ( )
3
10 3
5
2
-
× + =
B AB
EI
M ? 
 
EI EI O M
B AB
0012 . 4 . + = ?   (2) 
 
Similarly, for beam-end moment at endB , in span AB 
 
( )
3
10 3 2 4 . 0
-
× + =
B BA
EI M ? 
EI EI M
B BA
0012 . 0 8 . 0 + = ?   (3) 
 
In span , the support  is above support BC C B, Hence the chord joining C B ' 
rotates in anticlockwise direction. 
 
3
10 1
-
× = =
CB BC
? ?    (4) 
 
Writing slope-deflection equations for span , BC
 
EI EI EI M
C B BC
3
10 2 . 1 4 . 0 8 . 0
-
× - + = ? ?    
   
EI EI EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0
-
× - + = ? ?  (5) 
 
Now, consider the joint equilibrium of support B (see Fig.15.2c)