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# The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 3 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 3 Civil Engineering (CE) Notes | EduRev

``` 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'

AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ?

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'

AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ?

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
(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'

AB
3
3
10 25 . 1
4
10 5
-
-
× - =
×
- = ?

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
(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.

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## Structural Analysis

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