The Moment Distribution Method: Frames without Sidesway Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Moment Distribution Method: Frames without Sidesway Civil Engineering (CE) Notes | EduRev

 Page 1


Instructional Objectives 
After reading this chapter the student will be able to 
1. Solve plane frame restrained against sidesway by the moment-distribution 
method. 
2. Compute reactions at the supports. 
3. Draw bending moment and shear force diagrams. 
4. Draw the deflected shape of the plane frame. 
 
 
20.1 Introduction 
In this lesson, the statically indeterminate rigid frames properly restrained against 
sidesway are analysed using moment-distribution method. Analysis of rigid 
frames by moment-distribution method is very similar to that of continuous beams 
described in lesson 18. As pointed out earlier, in the case of continuous beams, 
at a joint only two members meet, where as in case of rigid frames two or more 
than two members meet at a joint. At such joints (for example joint C in Fig. 
20.1) where more than two members meet, the unbalanced moment at the 
beginning of each cycle is the algebraic sum of fixed end beam moments (in the 
first cycle) or the carry over moments (in the subsequent cycles) of the beam 
meeting at C. The unbalanced moment is distributed to members and 
 according to their distribution factors. Few examples are solved to explain 
procedure. The moment-distribution method is carried out on a working diagram.  
CD CB,
CE
 
 
 
 
Page 2


Instructional Objectives 
After reading this chapter the student will be able to 
1. Solve plane frame restrained against sidesway by the moment-distribution 
method. 
2. Compute reactions at the supports. 
3. Draw bending moment and shear force diagrams. 
4. Draw the deflected shape of the plane frame. 
 
 
20.1 Introduction 
In this lesson, the statically indeterminate rigid frames properly restrained against 
sidesway are analysed using moment-distribution method. Analysis of rigid 
frames by moment-distribution method is very similar to that of continuous beams 
described in lesson 18. As pointed out earlier, in the case of continuous beams, 
at a joint only two members meet, where as in case of rigid frames two or more 
than two members meet at a joint. At such joints (for example joint C in Fig. 
20.1) where more than two members meet, the unbalanced moment at the 
beginning of each cycle is the algebraic sum of fixed end beam moments (in the 
first cycle) or the carry over moments (in the subsequent cycles) of the beam 
meeting at C. The unbalanced moment is distributed to members and 
 according to their distribution factors. Few examples are solved to explain 
procedure. The moment-distribution method is carried out on a working diagram.  
CD CB,
CE
 
 
 
 
Example 20.1 
Calculate reactions and beam end moments for the rigid frame shown in Fig. 
20.2a. Draw bending moment diagram for the frame. Assume EI to be constant 
for all the members. 
 
 
 
Solution 
 
In the first step, calculate fixed end moments. 
 
m kN M
m kN M
m kN M
m kN M
F
CB
F
BC
F
DB
F
BD
. 0 . 0
. 0 . 0
. 0 . 5
. 0 . 5
=
=
- =
=
     (1) 
 
Also, the fixed end moment acting at B on BA is clockwise.  
 
m kN M
F
BA
. 0 . 10 - = 
 
 
Page 3


Instructional Objectives 
After reading this chapter the student will be able to 
1. Solve plane frame restrained against sidesway by the moment-distribution 
method. 
2. Compute reactions at the supports. 
3. Draw bending moment and shear force diagrams. 
4. Draw the deflected shape of the plane frame. 
 
 
20.1 Introduction 
In this lesson, the statically indeterminate rigid frames properly restrained against 
sidesway are analysed using moment-distribution method. Analysis of rigid 
frames by moment-distribution method is very similar to that of continuous beams 
described in lesson 18. As pointed out earlier, in the case of continuous beams, 
at a joint only two members meet, where as in case of rigid frames two or more 
than two members meet at a joint. At such joints (for example joint C in Fig. 
20.1) where more than two members meet, the unbalanced moment at the 
beginning of each cycle is the algebraic sum of fixed end beam moments (in the 
first cycle) or the carry over moments (in the subsequent cycles) of the beam 
meeting at C. The unbalanced moment is distributed to members and 
 according to their distribution factors. Few examples are solved to explain 
procedure. The moment-distribution method is carried out on a working diagram.  
CD CB,
CE
 
 
 
 
Example 20.1 
Calculate reactions and beam end moments for the rigid frame shown in Fig. 
20.2a. Draw bending moment diagram for the frame. Assume EI to be constant 
for all the members. 
 
 
 
Solution 
 
In the first step, calculate fixed end moments. 
 
m kN M
m kN M
m kN M
m kN M
F
CB
F
BC
F
DB
F
BD
. 0 . 0
. 0 . 0
. 0 . 5
. 0 . 5
=
=
- =
=
     (1) 
 
Also, the fixed end moment acting at B on BA is clockwise.  
 
m kN M
F
BA
. 0 . 10 - = 
 
 
In the next step calculate stiffness and distribution factors. 
  
EI
EI
K
BD
25 . 0
4
= =   and  EI
EI
K
BC
25 . 0
4
= =   
 
At jointB : 
 
EI K 50 . 0 =
?
 
 
5 . 0 ; 5 . 0
5 . 0
25 . 0
= = =
BC BD
DF
EI
EI
DF  (2) 
 
All the calculations are shown in Fig. 20.2b. Please note that cantilever member 
does not have any restraining effect on the joint B from rotation. In addition its 
stiffness factor is zero. Hence unbalanced moment is distributed between 
members and BC BD only. 
 
 
 
In this problem the moment-distribution method is completed in only one cycle, 
as equilibrium of only one joint needs to be considered. In other words, there is 
only one equation that needs to be solved for the unknown 
B
? in this problem. 
This problem has already been solved by slop- deflection method wherein 
reactions are computed from equations of statics. The free body diagram of each 
member of the frame with external load and beam end moments are again 
reproduced here in Fig. 20.2c for easy reference. The bending moment diagram 
is shown in Fig. 20.2d. 
 
 
 
Page 4


Instructional Objectives 
After reading this chapter the student will be able to 
1. Solve plane frame restrained against sidesway by the moment-distribution 
method. 
2. Compute reactions at the supports. 
3. Draw bending moment and shear force diagrams. 
4. Draw the deflected shape of the plane frame. 
 
 
20.1 Introduction 
In this lesson, the statically indeterminate rigid frames properly restrained against 
sidesway are analysed using moment-distribution method. Analysis of rigid 
frames by moment-distribution method is very similar to that of continuous beams 
described in lesson 18. As pointed out earlier, in the case of continuous beams, 
at a joint only two members meet, where as in case of rigid frames two or more 
than two members meet at a joint. At such joints (for example joint C in Fig. 
20.1) where more than two members meet, the unbalanced moment at the 
beginning of each cycle is the algebraic sum of fixed end beam moments (in the 
first cycle) or the carry over moments (in the subsequent cycles) of the beam 
meeting at C. The unbalanced moment is distributed to members and 
 according to their distribution factors. Few examples are solved to explain 
procedure. The moment-distribution method is carried out on a working diagram.  
CD CB,
CE
 
 
 
 
Example 20.1 
Calculate reactions and beam end moments for the rigid frame shown in Fig. 
20.2a. Draw bending moment diagram for the frame. Assume EI to be constant 
for all the members. 
 
 
 
Solution 
 
In the first step, calculate fixed end moments. 
 
m kN M
m kN M
m kN M
m kN M
F
CB
F
BC
F
DB
F
BD
. 0 . 0
. 0 . 0
. 0 . 5
. 0 . 5
=
=
- =
=
     (1) 
 
Also, the fixed end moment acting at B on BA is clockwise.  
 
m kN M
F
BA
. 0 . 10 - = 
 
 
In the next step calculate stiffness and distribution factors. 
  
EI
EI
K
BD
25 . 0
4
= =   and  EI
EI
K
BC
25 . 0
4
= =   
 
At jointB : 
 
EI K 50 . 0 =
?
 
 
5 . 0 ; 5 . 0
5 . 0
25 . 0
= = =
BC BD
DF
EI
EI
DF  (2) 
 
All the calculations are shown in Fig. 20.2b. Please note that cantilever member 
does not have any restraining effect on the joint B from rotation. In addition its 
stiffness factor is zero. Hence unbalanced moment is distributed between 
members and BC BD only. 
 
 
 
In this problem the moment-distribution method is completed in only one cycle, 
as equilibrium of only one joint needs to be considered. In other words, there is 
only one equation that needs to be solved for the unknown 
B
? in this problem. 
This problem has already been solved by slop- deflection method wherein 
reactions are computed from equations of statics. The free body diagram of each 
member of the frame with external load and beam end moments are again 
reproduced here in Fig. 20.2c for easy reference. The bending moment diagram 
is shown in Fig. 20.2d. 
 
 
 
 
 
 
 
 
Page 5


Instructional Objectives 
After reading this chapter the student will be able to 
1. Solve plane frame restrained against sidesway by the moment-distribution 
method. 
2. Compute reactions at the supports. 
3. Draw bending moment and shear force diagrams. 
4. Draw the deflected shape of the plane frame. 
 
 
20.1 Introduction 
In this lesson, the statically indeterminate rigid frames properly restrained against 
sidesway are analysed using moment-distribution method. Analysis of rigid 
frames by moment-distribution method is very similar to that of continuous beams 
described in lesson 18. As pointed out earlier, in the case of continuous beams, 
at a joint only two members meet, where as in case of rigid frames two or more 
than two members meet at a joint. At such joints (for example joint C in Fig. 
20.1) where more than two members meet, the unbalanced moment at the 
beginning of each cycle is the algebraic sum of fixed end beam moments (in the 
first cycle) or the carry over moments (in the subsequent cycles) of the beam 
meeting at C. The unbalanced moment is distributed to members and 
 according to their distribution factors. Few examples are solved to explain 
procedure. The moment-distribution method is carried out on a working diagram.  
CD CB,
CE
 
 
 
 
Example 20.1 
Calculate reactions and beam end moments for the rigid frame shown in Fig. 
20.2a. Draw bending moment diagram for the frame. Assume EI to be constant 
for all the members. 
 
 
 
Solution 
 
In the first step, calculate fixed end moments. 
 
m kN M
m kN M
m kN M
m kN M
F
CB
F
BC
F
DB
F
BD
. 0 . 0
. 0 . 0
. 0 . 5
. 0 . 5
=
=
- =
=
     (1) 
 
Also, the fixed end moment acting at B on BA is clockwise.  
 
m kN M
F
BA
. 0 . 10 - = 
 
 
In the next step calculate stiffness and distribution factors. 
  
EI
EI
K
BD
25 . 0
4
= =   and  EI
EI
K
BC
25 . 0
4
= =   
 
At jointB : 
 
EI K 50 . 0 =
?
 
 
5 . 0 ; 5 . 0
5 . 0
25 . 0
= = =
BC BD
DF
EI
EI
DF  (2) 
 
All the calculations are shown in Fig. 20.2b. Please note that cantilever member 
does not have any restraining effect on the joint B from rotation. In addition its 
stiffness factor is zero. Hence unbalanced moment is distributed between 
members and BC BD only. 
 
 
 
In this problem the moment-distribution method is completed in only one cycle, 
as equilibrium of only one joint needs to be considered. In other words, there is 
only one equation that needs to be solved for the unknown 
B
? in this problem. 
This problem has already been solved by slop- deflection method wherein 
reactions are computed from equations of statics. The free body diagram of each 
member of the frame with external load and beam end moments are again 
reproduced here in Fig. 20.2c for easy reference. The bending moment diagram 
is shown in Fig. 20.2d. 
 
 
 
 
 
 
 
 
Example 20.2 
Analyse the rigid frame shown in Fig. 20.3a by moment-distribution method. 
Moment of inertia of different members are shown in the diagram. 
 
 
Solution:  
Calculate fixed end moments by locking the joints  and D C B A , , , E  
 
2
54
4.0 kN.m
20
F
AB
M
×
== 
 
kN.m 667 . 2 - =
F
BA
M 
 
kN.m 5 . 7 =
F
BC
M 
 
kN.m 5 . 7 - =
F
CB
M   
     
0 = = = =
F
EC
F
CE
F
DB
F
BD
M M M M    (1) 
 
The frame is restrained against sidesway. In the next step calculate stiffness and 
distribution factors. 
 
EI K
BA
25 . 0 =   and  EI
EI
K
BC
333 . 0
6
2
= = 
 
 
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