The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 Civil Engineering (CE) Notes | EduRev

 Page 1


 
Support C ,    .0100m vertically downwards.   
 
 
Assume ; . GPa E 200 =
4 3
10 35 . 1 m I
-
× =
 
 
 
 
 
Solution: 
 
Assume that supports and  are locked and calculate fixed end moments 
due to externally applied load and support settlements. The fixed end beam 
moments due to externally applied loads are,  
D C B A , ,
5 100
41.67 kN.m;
12
F
AB
M
×
== 41.67 kN.m
F
BA
M =- 
41.67 kN.m;
F
BC
M =+  41.67 kN.m
F
BC
M =-
41.67 kN.m;
F
CD
M =+         (1) 41.67 kN.m
F
DC
M =-
 
 
, the chord joining joints and In the spanAB A B rotates in the clockwise direction 
as  moves vertical downwards with respect to (see Fig. 19.3b). B A
 
 
 
Page 2


 
Support C ,    .0100m vertically downwards.   
 
 
Assume ; . GPa E 200 =
4 3
10 35 . 1 m I
-
× =
 
 
 
 
 
Solution: 
 
Assume that supports and  are locked and calculate fixed end moments 
due to externally applied load and support settlements. The fixed end beam 
moments due to externally applied loads are,  
D C B A , ,
5 100
41.67 kN.m;
12
F
AB
M
×
== 41.67 kN.m
F
BA
M =- 
41.67 kN.m;
F
BC
M =+  41.67 kN.m
F
BC
M =-
41.67 kN.m;
F
CD
M =+         (1) 41.67 kN.m
F
DC
M =-
 
 
, the chord joining joints and In the spanAB A B rotates in the clockwise direction 
as  moves vertical downwards with respect to (see Fig. 19.3b). B A
 
 
 
 
 
0.0005 radians
AB
? =- (negative as chord ' AB rotates in the clockwise direction 
from its original position) 
 
0.0005 radians
BC
? =- 
    
0.001 radians
CD
? = (positive as chord  rotates in the counterclockwise 
direction). 
D C'
 
 
Now the fixed end beam moments due to support settlements are, 
 
93
6 6 200 10 1.35 10
( 0.0005)
10
81000 N.m 81.00 kN.m
S AB
AB AB
AB
EI
M
L
?
-
×× × ×
=- =- -
==
   
81.00 kN.m
S
BA
M = 
81.00 kN.m
SS
BC CB
MM == 
162.00 kN.m
SS
CD DC
MM ==-       (3) 
 
 
In the next step, calculate stiffness and distribution factors. For span AB and  
modified stiffness factors are used as supports 
CD
A and  are hinged. Stiffness 
factors are, 
D
 
 
Page 3


 
Support C ,    .0100m vertically downwards.   
 
 
Assume ; . GPa E 200 =
4 3
10 35 . 1 m I
-
× =
 
 
 
 
 
Solution: 
 
Assume that supports and  are locked and calculate fixed end moments 
due to externally applied load and support settlements. The fixed end beam 
moments due to externally applied loads are,  
D C B A , ,
5 100
41.67 kN.m;
12
F
AB
M
×
== 41.67 kN.m
F
BA
M =- 
41.67 kN.m;
F
BC
M =+  41.67 kN.m
F
BC
M =-
41.67 kN.m;
F
CD
M =+         (1) 41.67 kN.m
F
DC
M =-
 
 
, the chord joining joints and In the spanAB A B rotates in the clockwise direction 
as  moves vertical downwards with respect to (see Fig. 19.3b). B A
 
 
 
 
 
0.0005 radians
AB
? =- (negative as chord ' AB rotates in the clockwise direction 
from its original position) 
 
0.0005 radians
BC
? =- 
    
0.001 radians
CD
? = (positive as chord  rotates in the counterclockwise 
direction). 
D C'
 
 
Now the fixed end beam moments due to support settlements are, 
 
93
6 6 200 10 1.35 10
( 0.0005)
10
81000 N.m 81.00 kN.m
S AB
AB AB
AB
EI
M
L
?
-
×× × ×
=- =- -
==
   
81.00 kN.m
S
BA
M = 
81.00 kN.m
SS
BC CB
MM == 
162.00 kN.m
SS
CD DC
MM ==-       (3) 
 
 
In the next step, calculate stiffness and distribution factors. For span AB and  
modified stiffness factors are used as supports 
CD
A and  are hinged. Stiffness 
factors are, 
D
 
 
EI
EI
K EI
EI
K
EI
EI
K EI
EI
K
CD CB
BC BA
075 . 0
10 4
3
; 10 . 0
10
10 . 0
10
; 075 . 0
10 4
3
= = = =
= = = =
     (4) 
 
 
At joint :  0 . 1 ; 075 . 0 = =
? AB
DF EI K A
 
At joint  :    571 . 0 ; 429 . 0 ; 175 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    429 . 0 ; 571 . 0 ; 175 . 0 = = =
? CD CB
DF DF EI K
 
At joint :  0 . 1 ; 075 . 0 = =
? DC
DF EI K D
 
The complete procedure of successively unlocking the joints, balancing them and 
locking them is shown in a working diagram in Fig.19.3c. In the first row, the 
distribution factors are entered. Then fixed end moments due to applied loads 
and support settlements are entered. In the first step, release joints A and . The 
unbalanced moments at 
D
A and are 122.67 kN.m, -203.67 kN.m respectively. 
Hence balancing moments at 
D
A and  are -122.67 kN.m, 203.67 kN.m 
respectively. (Note that we are dealing with beam end moments and not joint 
moments). The joint moments are negative of the beam end moments. Further 
leave 
D
and  unlocked as they are hinged joints. Now carry over moments 
 and  to joint 
A D
-61.34 kN.m kN.m 101.84 B and  respectively. In the next cycle, 
balance joints 
C
 and C. The unbalanced moment at joint B B is . 
Hence balancing moment for beam 
100.66 kN.m
BA is 43.19 ( 100.66 0.429) - -× and for  is 
. The balancing moment on  gives a carry over 
moment of  to joint C . The whole procedure is shown in Fig. 19.3c 
and in Table 19.2. It must be noted that there is no carryover to joints 
BC
BC 57.48 kN.m (-100.66 x 0.571) -
26.74 kN.m -
A and  
as they were left unlocked.   
D
 
 
 
 
Page 4


 
Support C ,    .0100m vertically downwards.   
 
 
Assume ; . GPa E 200 =
4 3
10 35 . 1 m I
-
× =
 
 
 
 
 
Solution: 
 
Assume that supports and  are locked and calculate fixed end moments 
due to externally applied load and support settlements. The fixed end beam 
moments due to externally applied loads are,  
D C B A , ,
5 100
41.67 kN.m;
12
F
AB
M
×
== 41.67 kN.m
F
BA
M =- 
41.67 kN.m;
F
BC
M =+  41.67 kN.m
F
BC
M =-
41.67 kN.m;
F
CD
M =+         (1) 41.67 kN.m
F
DC
M =-
 
 
, the chord joining joints and In the spanAB A B rotates in the clockwise direction 
as  moves vertical downwards with respect to (see Fig. 19.3b). B A
 
 
 
 
 
0.0005 radians
AB
? =- (negative as chord ' AB rotates in the clockwise direction 
from its original position) 
 
0.0005 radians
BC
? =- 
    
0.001 radians
CD
? = (positive as chord  rotates in the counterclockwise 
direction). 
D C'
 
 
Now the fixed end beam moments due to support settlements are, 
 
93
6 6 200 10 1.35 10
( 0.0005)
10
81000 N.m 81.00 kN.m
S AB
AB AB
AB
EI
M
L
?
-
×× × ×
=- =- -
==
   
81.00 kN.m
S
BA
M = 
81.00 kN.m
SS
BC CB
MM == 
162.00 kN.m
SS
CD DC
MM ==-       (3) 
 
 
In the next step, calculate stiffness and distribution factors. For span AB and  
modified stiffness factors are used as supports 
CD
A and  are hinged. Stiffness 
factors are, 
D
 
 
EI
EI
K EI
EI
K
EI
EI
K EI
EI
K
CD CB
BC BA
075 . 0
10 4
3
; 10 . 0
10
10 . 0
10
; 075 . 0
10 4
3
= = = =
= = = =
     (4) 
 
 
At joint :  0 . 1 ; 075 . 0 = =
? AB
DF EI K A
 
At joint  :    571 . 0 ; 429 . 0 ; 175 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    429 . 0 ; 571 . 0 ; 175 . 0 = = =
? CD CB
DF DF EI K
 
At joint :  0 . 1 ; 075 . 0 = =
? DC
DF EI K D
 
The complete procedure of successively unlocking the joints, balancing them and 
locking them is shown in a working diagram in Fig.19.3c. In the first row, the 
distribution factors are entered. Then fixed end moments due to applied loads 
and support settlements are entered. In the first step, release joints A and . The 
unbalanced moments at 
D
A and are 122.67 kN.m, -203.67 kN.m respectively. 
Hence balancing moments at 
D
A and  are -122.67 kN.m, 203.67 kN.m 
respectively. (Note that we are dealing with beam end moments and not joint 
moments). The joint moments are negative of the beam end moments. Further 
leave 
D
and  unlocked as they are hinged joints. Now carry over moments 
 and  to joint 
A D
-61.34 kN.m kN.m 101.84 B and  respectively. In the next cycle, 
balance joints 
C
 and C. The unbalanced moment at joint B B is . 
Hence balancing moment for beam 
100.66 kN.m
BA is 43.19 ( 100.66 0.429) - -× and for  is 
. The balancing moment on  gives a carry over 
moment of  to joint C . The whole procedure is shown in Fig. 19.3c 
and in Table 19.2. It must be noted that there is no carryover to joints 
BC
BC 57.48 kN.m (-100.66 x 0.571) -
26.74 kN.m -
A and  
as they were left unlocked.   
D
 
 
 
 
 
 
 
 
 
Page 5


 
Support C ,    .0100m vertically downwards.   
 
 
Assume ; . GPa E 200 =
4 3
10 35 . 1 m I
-
× =
 
 
 
 
 
Solution: 
 
Assume that supports and  are locked and calculate fixed end moments 
due to externally applied load and support settlements. The fixed end beam 
moments due to externally applied loads are,  
D C B A , ,
5 100
41.67 kN.m;
12
F
AB
M
×
== 41.67 kN.m
F
BA
M =- 
41.67 kN.m;
F
BC
M =+  41.67 kN.m
F
BC
M =-
41.67 kN.m;
F
CD
M =+         (1) 41.67 kN.m
F
DC
M =-
 
 
, the chord joining joints and In the spanAB A B rotates in the clockwise direction 
as  moves vertical downwards with respect to (see Fig. 19.3b). B A
 
 
 
 
 
0.0005 radians
AB
? =- (negative as chord ' AB rotates in the clockwise direction 
from its original position) 
 
0.0005 radians
BC
? =- 
    
0.001 radians
CD
? = (positive as chord  rotates in the counterclockwise 
direction). 
D C'
 
 
Now the fixed end beam moments due to support settlements are, 
 
93
6 6 200 10 1.35 10
( 0.0005)
10
81000 N.m 81.00 kN.m
S AB
AB AB
AB
EI
M
L
?
-
×× × ×
=- =- -
==
   
81.00 kN.m
S
BA
M = 
81.00 kN.m
SS
BC CB
MM == 
162.00 kN.m
SS
CD DC
MM ==-       (3) 
 
 
In the next step, calculate stiffness and distribution factors. For span AB and  
modified stiffness factors are used as supports 
CD
A and  are hinged. Stiffness 
factors are, 
D
 
 
EI
EI
K EI
EI
K
EI
EI
K EI
EI
K
CD CB
BC BA
075 . 0
10 4
3
; 10 . 0
10
10 . 0
10
; 075 . 0
10 4
3
= = = =
= = = =
     (4) 
 
 
At joint :  0 . 1 ; 075 . 0 = =
? AB
DF EI K A
 
At joint  :    571 . 0 ; 429 . 0 ; 175 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    429 . 0 ; 571 . 0 ; 175 . 0 = = =
? CD CB
DF DF EI K
 
At joint :  0 . 1 ; 075 . 0 = =
? DC
DF EI K D
 
The complete procedure of successively unlocking the joints, balancing them and 
locking them is shown in a working diagram in Fig.19.3c. In the first row, the 
distribution factors are entered. Then fixed end moments due to applied loads 
and support settlements are entered. In the first step, release joints A and . The 
unbalanced moments at 
D
A and are 122.67 kN.m, -203.67 kN.m respectively. 
Hence balancing moments at 
D
A and  are -122.67 kN.m, 203.67 kN.m 
respectively. (Note that we are dealing with beam end moments and not joint 
moments). The joint moments are negative of the beam end moments. Further 
leave 
D
and  unlocked as they are hinged joints. Now carry over moments 
 and  to joint 
A D
-61.34 kN.m kN.m 101.84 B and  respectively. In the next cycle, 
balance joints 
C
 and C. The unbalanced moment at joint B B is . 
Hence balancing moment for beam 
100.66 kN.m
BA is 43.19 ( 100.66 0.429) - -× and for  is 
. The balancing moment on  gives a carry over 
moment of  to joint C . The whole procedure is shown in Fig. 19.3c 
and in Table 19.2. It must be noted that there is no carryover to joints 
BC
BC 57.48 kN.m (-100.66 x 0.571) -
26.74 kN.m -
A and  
as they were left unlocked.   
D
 
 
 
 
 
 
 
 
 
Table 19.2 Moment-distribution for continuous beam ABCD 
 
Joint A B C D 
 
Members AB  BA BC CB CD DC 
Stiffness factors 0.075 EI 0.075 EI 0.1 EI 0.1 EI 0.075 EI 0.075 EI
Distribution 
Factors 
1.000 0.429 0.571 0.571 0.429 1.000 
        
FEM due to 
externally 
applied loads 
41.670 -41.670 41.670 -41.670 41.670 -41.670 
        
FEM due to 
support 
settlements 
81.000 81.000 81.000 81.000 -
162.000 
-
162.000 
        
Total 122.670 39.330 122.670 39.330 -
120.330 
-
203.670 
        
Balance A and D 
released  
-
122.670 
    203.670 
Carry over  -61.335   101.835  
        
Balance B and C  -43.185 -57.480 -11.897 -8.94  
Carry over   -5.95 -26.740   
       
Balance B and C  2.552 3.40 16.410 12.33  
Carry over to B 
and C 
  8.21 1.70   
        
Balance B and C  -3.52 -4.69 -0.97 -0.73  
C.O. to B and C   -0.49 -2.33   
        
Balance B and C  0.21 0.28 1.34 1.01  
Carry over   0.67 0.14   
        
Balance B and C  -0.29 -0.38 -0.08 -0.06  
        
Final Moments 0.000 -66.67 66.67 14.88 -14.88 0.000 
 
 
 
 
 
 
 
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Sample Paper

,

shortcuts and tricks

,

The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 Civil Engineering (CE) Notes | EduRev

,

Exam

,

The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 Civil Engineering (CE) Notes | EduRev

,

Previous Year Questions with Solutions

,

The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 Civil Engineering (CE) Notes | EduRev

,

study material

,

Objective type Questions

,

Viva Questions

,

mock tests for examination

,

Important questions

,

Semester Notes

,

past year papers

,

Summary

,

Extra Questions

,

practice quizzes

,

video lectures

,

Free

,

pdf

,

ppt

,

MCQs

;