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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
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FAQs on The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Moment Distribution Method? |  |
Ans. The Moment Distribution Method is a structural analysis technique used to analyze statically indeterminate beams with support settlements. It involves distributing the moments at the supports based on the stiffness of the members and iterating until the moments converge to a consistent solution.
| 2. How does the Moment Distribution Method handle support settlements? | |
Ans. The Moment Distribution Method takes into account the effects of support settlements by adjusting the stiffness of the beams at the settled supports. The settlements are incorporated into the analysis by modifying the distribution factors used in the iterative process, resulting in more accurate calculations of the moments and reactions.
| 3. What are the advantages of using the Moment Distribution Method for analyzing beams with support settlements? | |
Ans. The Moment Distribution Method is particularly useful for analyzing beams with support settlements because it can accurately account for the effects of settlement on the structural response. This method provides a more realistic representation of the beam's behavior, allowing engineers to make informed design decisions and ensure the safety and stability of the structure.
| 4. How does the Moment Distribution Method handle statically indeterminate beams? | |
Ans. The Moment Distribution Method is specifically designed to analyze statically indeterminate beams. It uses an iterative process to distribute the moments at the supports based on the stiffness of the members. This redistribution of moments allows the method to determine the internal forces and reactions in a statically indeterminate beam, which cannot be calculated using traditional statics methods alone.
| 5. Are there any limitations or considerations when using the Moment Distribution Method for beams with support settlements? | |
Ans. While the Moment Distribution Method is a powerful analysis technique, it does have some limitations. One important consideration is that it assumes linear-elastic behavior of the structure and may not accurately capture the effects of nonlinear material or geometric behavior. Additionally, the method is most effective for beams with small settlements. In cases of large settlements or highly nonlinear behavior, more advanced analysis methods may be required.
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