The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 3 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


Example 19.3 
 
Analyse the continuous beam shown in Fig. 19.4a by moment-distribution 
method. The support 
ABC
B settles by  below mm 5 and C. Assume A EI to be 
constant for all members ; and . GPa E 200 =
4 6
10 8 mm I × =
 
 
 
Solution: 
 
Calculate fixed end beam moments due to externally applied loads assuming that 
support  and C are locked. B
 
 
m kN M m kN M
m kN M m kN M
F
CB
F
BC
F
BA
F
AB
. 67 . 2 ; . 67 . 2
. 2 ; . 2
- = + =
- = + =
     (1) 
 
 
 
In the next step calculate fixed end moments due to support settlements. In the 
span AB , the chord ' AB rotates in the clockwise direction and in span , the 
chord  rotates in the counterclockwise direction (Fig. 19.4b). 
BC
C B'
 
 
 
 
 
Page 2


Example 19.3 
 
Analyse the continuous beam shown in Fig. 19.4a by moment-distribution 
method. The support 
ABC
B settles by  below mm 5 and C. Assume A EI to be 
constant for all members ; and . GPa E 200 =
4 6
10 8 mm I × =
 
 
 
Solution: 
 
Calculate fixed end beam moments due to externally applied loads assuming that 
support  and C are locked. B
 
 
m kN M m kN M
m kN M m kN M
F
CB
F
BC
F
BA
F
AB
. 67 . 2 ; . 67 . 2
. 2 ; . 2
- = + =
- = + =
     (1) 
 
 
 
In the next step calculate fixed end moments due to support settlements. In the 
span AB , the chord ' AB rotates in the clockwise direction and in span , the 
chord  rotates in the counterclockwise direction (Fig. 19.4b). 
BC
C B'
 
 
 
 
 
 
radians
AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ? 
 
radians
BC
3
3
10 25 . 1
4
10 5
-
-
× =
×
= ?       (2) 
 
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
4
10 5
4
10 8 10 200 6 6
3 6 9
AB
AB
AB S
BA
S
AB
L
EI
M M ? 
 
.         (3)  3 3000 kNm Nm = =
 
m kN M M
S
CB
S
BC
. 0 . 3 - = = 
 
In the next step, calculate stiffness and distribution factors.  
 
EI EI K
EI K K
BC
BA AB
1875 . 0 25 . 0
4
3
25 . 0
= =
= =
          (4) 
 
At joint  :    429 . 0 ; 571 . 0 ; 4375 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    0 . 1 ; 1875 . 0 = =
? CB
DF EI K
 
 
At fixed joint, the joint does not rotate and hence no distribution moments are 
developed and consequently distribution factor is equal to zero. The complete 
moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The 
diagram is self explanatory. In this particular case results are obtained in two 
cycles. In the first cycle joint is balanced and carry over moment is taken to 
joint 
C
. In the next cycle , joint B B is balanced and carry over moment is taken to 
joint . The bending moment diagram is shown in fig. 19.4d.       A
 
 
 
 
 
 
 
 
Page 3


Example 19.3 
 
Analyse the continuous beam shown in Fig. 19.4a by moment-distribution 
method. The support 
ABC
B settles by  below mm 5 and C. Assume A EI to be 
constant for all members ; and . GPa E 200 =
4 6
10 8 mm I × =
 
 
 
Solution: 
 
Calculate fixed end beam moments due to externally applied loads assuming that 
support  and C are locked. B
 
 
m kN M m kN M
m kN M m kN M
F
CB
F
BC
F
BA
F
AB
. 67 . 2 ; . 67 . 2
. 2 ; . 2
- = + =
- = + =
     (1) 
 
 
 
In the next step calculate fixed end moments due to support settlements. In the 
span AB , the chord ' AB rotates in the clockwise direction and in span , the 
chord  rotates in the counterclockwise direction (Fig. 19.4b). 
BC
C B'
 
 
 
 
 
 
radians
AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ? 
 
radians
BC
3
3
10 25 . 1
4
10 5
-
-
× =
×
= ?       (2) 
 
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
4
10 5
4
10 8 10 200 6 6
3 6 9
AB
AB
AB S
BA
S
AB
L
EI
M M ? 
 
.         (3)  3 3000 kNm Nm = =
 
m kN M M
S
CB
S
BC
. 0 . 3 - = = 
 
In the next step, calculate stiffness and distribution factors.  
 
EI EI K
EI K K
BC
BA AB
1875 . 0 25 . 0
4
3
25 . 0
= =
= =
          (4) 
 
At joint  :    429 . 0 ; 571 . 0 ; 4375 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    0 . 1 ; 1875 . 0 = =
? CB
DF EI K
 
 
At fixed joint, the joint does not rotate and hence no distribution moments are 
developed and consequently distribution factor is equal to zero. The complete 
moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The 
diagram is self explanatory. In this particular case results are obtained in two 
cycles. In the first cycle joint is balanced and carry over moment is taken to 
joint 
C
. In the next cycle , joint B B is balanced and carry over moment is taken to 
joint . The bending moment diagram is shown in fig. 19.4d.       A
 
 
 
 
 
 
 
 
Table 19.3 Moment-distribution for continuous beam ABC 
 
Joints A B C 
 
Member AB BA BC CB 
Stiffness factor 0.25 EI 0.25 EI 0.1875 EI 0.1875 EI 
Distribution Factor  0.571 0.429 1.000 
      
Fixed End Moments 
due to applied loads 
(kN.m) 
2.000 -2.000 2.667 -2.667 
Fixed End Moments 
due to support 
settlements (kN.m) 
3.000 3.000 -3.000 -3.000 
Total 5.000 1.000 -0.333 -5.667 
      
Balance joint C and 
C.O. 
  2.835 5.667 
      
Total 5.000 1.000 2.502 0.000 
      
Balance joint B and 
C.O. to A 
-1.00 -2.000 -1.502  
      
Final Moments (kN.m) 4.000 -1.000 1.000 0.000 
 
 
 
 
Page 4


Example 19.3 
 
Analyse the continuous beam shown in Fig. 19.4a by moment-distribution 
method. The support 
ABC
B settles by  below mm 5 and C. Assume A EI to be 
constant for all members ; and . GPa E 200 =
4 6
10 8 mm I × =
 
 
 
Solution: 
 
Calculate fixed end beam moments due to externally applied loads assuming that 
support  and C are locked. B
 
 
m kN M m kN M
m kN M m kN M
F
CB
F
BC
F
BA
F
AB
. 67 . 2 ; . 67 . 2
. 2 ; . 2
- = + =
- = + =
     (1) 
 
 
 
In the next step calculate fixed end moments due to support settlements. In the 
span AB , the chord ' AB rotates in the clockwise direction and in span , the 
chord  rotates in the counterclockwise direction (Fig. 19.4b). 
BC
C B'
 
 
 
 
 
 
radians
AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ? 
 
radians
BC
3
3
10 25 . 1
4
10 5
-
-
× =
×
= ?       (2) 
 
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
4
10 5
4
10 8 10 200 6 6
3 6 9
AB
AB
AB S
BA
S
AB
L
EI
M M ? 
 
.         (3)  3 3000 kNm Nm = =
 
m kN M M
S
CB
S
BC
. 0 . 3 - = = 
 
In the next step, calculate stiffness and distribution factors.  
 
EI EI K
EI K K
BC
BA AB
1875 . 0 25 . 0
4
3
25 . 0
= =
= =
          (4) 
 
At joint  :    429 . 0 ; 571 . 0 ; 4375 . 0 = = =
? BC BA
DF DF EI K B
 
At joint C :    0 . 1 ; 1875 . 0 = =
? CB
DF EI K
 
 
At fixed joint, the joint does not rotate and hence no distribution moments are 
developed and consequently distribution factor is equal to zero. The complete 
moment-distribution procedure is shown in Fig. 19.4c and Table 19.3. The 
diagram is self explanatory. In this particular case results are obtained in two 
cycles. In the first cycle joint is balanced and carry over moment is taken to 
joint 
C
. In the next cycle , joint B B is balanced and carry over moment is taken to 
joint . The bending moment diagram is shown in fig. 19.4d.       A
 
 
 
 
 
 
 
 
Table 19.3 Moment-distribution for continuous beam ABC 
 
Joints A B C 
 
Member AB BA BC CB 
Stiffness factor 0.25 EI 0.25 EI 0.1875 EI 0.1875 EI 
Distribution Factor  0.571 0.429 1.000 
      
Fixed End Moments 
due to applied loads 
(kN.m) 
2.000 -2.000 2.667 -2.667 
Fixed End Moments 
due to support 
settlements (kN.m) 
3.000 3.000 -3.000 -3.000 
Total 5.000 1.000 -0.333 -5.667 
      
Balance joint C and 
C.O. 
  2.835 5.667 
      
Total 5.000 1.000 2.502 0.000 
      
Balance joint B and 
C.O. to A 
-1.00 -2.000 -1.502  
      
Final Moments (kN.m) 4.000 -1.000 1.000 0.000 
 
 
 
 
 
 
 
 
Summary 
 
The moment-distribution method is applied to analyse continuous beam having 
support settlements. Each step in the numerical example is explained in detail. 
All calculations are shown at appropriate locations. The deflected shape of the 
continuous beam is sketched. Also, wherever required, the bending moment 
diagram is drawn. The numerical examples are explained with the help of free-
body diagrams.