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

## 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

AB
? =- (negative as chord ' AB rotates in the clockwise direction
from its original position)

BC
? =-

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

AB
? =- (negative as chord ' AB rotates in the clockwise direction
from its original position)

BC
? =-

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

AB
? =- (negative as chord ' AB rotates in the clockwise direction
from its original position)

BC
? =-

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

AB
? =- (negative as chord ' AB rotates in the clockwise direction
from its original position)

BC
? =-

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
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

```
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## Structural Analysis

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