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The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 1 Civil Engineering (CE) Notes | EduRev

Civil Engineering (CE) : The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 1 Civil Engineering (CE) Notes | EduRev

``` Page 1

Instructional Objectives
After reading this chapter the student will be able to
1. Solve continuous beam with support settlements by the moment-
distribution method.
2. Compute reactions at the supports.
3. Draw bending moment and shear force diagrams.
4. Draw the deflected shape of the continuous beam.

19.1 Introduction

In the previous lesson, moment-distribution method was discussed in the context
of statically indeterminate beams with unyielding supports. It is very well known
that support may settle by unequal amount during the lifetime of the structure.
Such support settlements induce fixed end moments in the beams so as to hold
the end slopes of the members as zero (see Fig. 19.1).

In lesson 15, an expression (equation 15.5) for beam end moments were derived
by superposing the end moments developed due to

1. Externally applied loads on beams
2. Due to displacements
B A
? ? , and ? (settlements).

The required equations are,

?
?
?
?
?
? ?
- + + =
AB
B A
AB
AB F
AB AB
L L
EI
M M
3
2
2
? ?      (19.1a)

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. Solve continuous beam with support settlements by the moment-
distribution method.
2. Compute reactions at the supports.
3. Draw bending moment and shear force diagrams.
4. Draw the deflected shape of the continuous beam.

19.1 Introduction

In the previous lesson, moment-distribution method was discussed in the context
of statically indeterminate beams with unyielding supports. It is very well known
that support may settle by unequal amount during the lifetime of the structure.
Such support settlements induce fixed end moments in the beams so as to hold
the end slopes of the members as zero (see Fig. 19.1).

In lesson 15, an expression (equation 15.5) for beam end moments were derived
by superposing the end moments developed due to

1. Externally applied loads on beams
2. Due to displacements
B A
? ? , and ? (settlements).

The required equations are,

?
?
?
?
?
? ?
- + + =
AB
B A
AB
AB F
AB AB
L L
EI
M M
3
2
2
? ?      (19.1a)

?
?
?
?
?
? ?
- + + =
AB
A B
AB
AB F
BA BA
L L
EI
M M
3
2
2
? ?     (19.1b)

This may be written as,

(19.2a) []
S
AB B A AB
F
AB AB
M K M M + + + = ? ? 2 2

[ ] 22
F
BA BA AB B A BA
S
M MK M ?? =+ + +       (19.2b)

where
AB
AB
AB
L
EI
K =  is the stiffness factor for the beam AB. The coefficient 4 has
been dropped since only relative values are required in calculating distribution
factors.

Note that
2
6
AB
AB S
BA
S
AB
L
EI
M M
?
- = =       (19.3)

S
AB
M is the beam end moments due to support settlement and is negative
(clockwise) for positive support settlements (upwards). In the moment-distribution
method, the support moments  and  due to uneven support settlements
are distributed in a similar manner as the fixed end moments, which were
described in details in lesson 18.
S
AB
M
S
BA
M

It is important to follow consistent sign convention. Here counterclockwise beam
end moments are taken as positive and counterclockwise chord rotation
?
?
?
?
?
? ?
L
is
taken as positive. The moment-distribution method as applied to statically
indeterminate beams undergoing uneven support settlements is illustrated with a
few examples.

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. Solve continuous beam with support settlements by the moment-
distribution method.
2. Compute reactions at the supports.
3. Draw bending moment and shear force diagrams.
4. Draw the deflected shape of the continuous beam.

19.1 Introduction

In the previous lesson, moment-distribution method was discussed in the context
of statically indeterminate beams with unyielding supports. It is very well known
that support may settle by unequal amount during the lifetime of the structure.
Such support settlements induce fixed end moments in the beams so as to hold
the end slopes of the members as zero (see Fig. 19.1).

In lesson 15, an expression (equation 15.5) for beam end moments were derived
by superposing the end moments developed due to

1. Externally applied loads on beams
2. Due to displacements
B A
? ? , and ? (settlements).

The required equations are,

?
?
?
?
?
? ?
- + + =
AB
B A
AB
AB F
AB AB
L L
EI
M M
3
2
2
? ?      (19.1a)

?
?
?
?
?
? ?
- + + =
AB
A B
AB
AB F
BA BA
L L
EI
M M
3
2
2
? ?     (19.1b)

This may be written as,

(19.2a) []
S
AB B A AB
F
AB AB
M K M M + + + = ? ? 2 2

[ ] 22
F
BA BA AB B A BA
S
M MK M ?? =+ + +       (19.2b)

where
AB
AB
AB
L
EI
K =  is the stiffness factor for the beam AB. The coefficient 4 has
been dropped since only relative values are required in calculating distribution
factors.

Note that
2
6
AB
AB S
BA
S
AB
L
EI
M M
?
- = =       (19.3)

S
AB
M is the beam end moments due to support settlement and is negative
(clockwise) for positive support settlements (upwards). In the moment-distribution
method, the support moments  and  due to uneven support settlements
are distributed in a similar manner as the fixed end moments, which were
described in details in lesson 18.
S
AB
M
S
BA
M

It is important to follow consistent sign convention. Here counterclockwise beam
end moments are taken as positive and counterclockwise chord rotation
?
?
?
?
?
? ?
L
is
taken as positive. The moment-distribution method as applied to statically
indeterminate beams undergoing uneven support settlements is illustrated with a
few examples.

Example 19.1

Calculate the support moments of the continuous beam  (Fig. 19.2a) having
constant flexural rigidity
ABC
EI throughout, due to vertical settlement of support B
by 5mm. Assume ; and . 200 GPa E =
44
410 m I
-
=×

Solution

There is no load on the beam and hence fixed end moments are zero. However,
fixed end moments are developed due to support settlement of B by 5mm. In the
spanAB , the chord rotates by
AB
? in clockwise direction. Thus,
5
10 5
3 -
×
- =
AB
?
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
5
10 5
5
10 4 10 200 6 6
3 4 9
AB
AB
AB S
BA
S
AB
L
EI
M M ?

96000 Nm 96 kNm. ==                 (1)

In the span , the chord rotates by BC
BC
? in the counterclockwise direction and
hence taken as positive.
5
10 5
3 -
×
=
BC
?

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. Solve continuous beam with support settlements by the moment-
distribution method.
2. Compute reactions at the supports.
3. Draw bending moment and shear force diagrams.
4. Draw the deflected shape of the continuous beam.

19.1 Introduction

In the previous lesson, moment-distribution method was discussed in the context
of statically indeterminate beams with unyielding supports. It is very well known
that support may settle by unequal amount during the lifetime of the structure.
Such support settlements induce fixed end moments in the beams so as to hold
the end slopes of the members as zero (see Fig. 19.1).

In lesson 15, an expression (equation 15.5) for beam end moments were derived
by superposing the end moments developed due to

1. Externally applied loads on beams
2. Due to displacements
B A
? ? , and ? (settlements).

The required equations are,

?
?
?
?
?
? ?
- + + =
AB
B A
AB
AB F
AB AB
L L
EI
M M
3
2
2
? ?      (19.1a)

?
?
?
?
?
? ?
- + + =
AB
A B
AB
AB F
BA BA
L L
EI
M M
3
2
2
? ?     (19.1b)

This may be written as,

(19.2a) []
S
AB B A AB
F
AB AB
M K M M + + + = ? ? 2 2

[ ] 22
F
BA BA AB B A BA
S
M MK M ?? =+ + +       (19.2b)

where
AB
AB
AB
L
EI
K =  is the stiffness factor for the beam AB. The coefficient 4 has
been dropped since only relative values are required in calculating distribution
factors.

Note that
2
6
AB
AB S
BA
S
AB
L
EI
M M
?
- = =       (19.3)

S
AB
M is the beam end moments due to support settlement and is negative
(clockwise) for positive support settlements (upwards). In the moment-distribution
method, the support moments  and  due to uneven support settlements
are distributed in a similar manner as the fixed end moments, which were
described in details in lesson 18.
S
AB
M
S
BA
M

It is important to follow consistent sign convention. Here counterclockwise beam
end moments are taken as positive and counterclockwise chord rotation
?
?
?
?
?
? ?
L
is
taken as positive. The moment-distribution method as applied to statically
indeterminate beams undergoing uneven support settlements is illustrated with a
few examples.

Example 19.1

Calculate the support moments of the continuous beam  (Fig. 19.2a) having
constant flexural rigidity
ABC
EI throughout, due to vertical settlement of support B
by 5mm. Assume ; and . 200 GPa E =
44
410 m I
-
=×

Solution

There is no load on the beam and hence fixed end moments are zero. However,
fixed end moments are developed due to support settlement of B by 5mm. In the
spanAB , the chord rotates by
AB
? in clockwise direction. Thus,
5
10 5
3 -
×
- =
AB
?
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
5
10 5
5
10 4 10 200 6 6
3 4 9
AB
AB
AB S
BA
S
AB
L
EI
M M ?

96000 Nm 96 kNm. ==                 (1)

In the span , the chord rotates by BC
BC
? in the counterclockwise direction and
hence taken as positive.
5
10 5
3 -
×
=
BC
?

?
?
?
?
?
?
?
? × × × × ×
- = - = =
- -
5
10 5
5
10 4 10 200 6 6
3 4 9
BC
BC
BC S
CB
S
BC
L
EI
M M ?

. 96 96000 kNm Nm - = - =      (2)

Now calculate stiffness and distribution factors.

EI
L
EI
K
AB
AB
BA
2 . 0 = =   and  EI
L
EI
K
BC
BC
BC
15 . 0
4
3
= =    (3)

Note that, while calculating stiffness factor, the coefficient 4 has been dropped
since only relative values are required in calculating the distribution factors. For
span , reduced stiffness factor has been taken as support C is hinged.  BC
AtB :

EI K 35 . 0 =
?

571 . 0
35 . 0
2 . 0
= =
EI
EI
DF
BA

429 . 0
35 . 0
15 . 0
= =
EI
EI
DF
BC
(4)

At support C :

EI K 15 . 0 =
?
; . 0 . 1 =
CB
DF

Now joint moments are balanced as discussed previously by unlocking and
locking each joint in succession and distributing the unbalanced moments till the
joints have rotated to their final positions. The complete procedure is shown in
Fig. 19.2b   and also in Table 19.1.

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. Solve continuous beam with support settlements by the moment-
distribution method.
2. Compute reactions at the supports.
3. Draw bending moment and shear force diagrams.
4. Draw the deflected shape of the continuous beam.

19.1 Introduction

In the previous lesson, moment-distribution method was discussed in the context
of statically indeterminate beams with unyielding supports. It is very well known
that support may settle by unequal amount during the lifetime of the structure.
Such support settlements induce fixed end moments in the beams so as to hold
the end slopes of the members as zero (see Fig. 19.1).

In lesson 15, an expression (equation 15.5) for beam end moments were derived
by superposing the end moments developed due to

1. Externally applied loads on beams
2. Due to displacements
B A
? ? , and ? (settlements).

The required equations are,

?
?
?
?
?
? ?
- + + =
AB
B A
AB
AB F
AB AB
L L
EI
M M
3
2
2
? ?      (19.1a)

?
?
?
?
?
? ?
- + + =
AB
A B
AB
AB F
BA BA
L L
EI
M M
3
2
2
? ?     (19.1b)

This may be written as,

(19.2a) []
S
AB B A AB
F
AB AB
M K M M + + + = ? ? 2 2

[ ] 22
F
BA BA AB B A BA
S
M MK M ?? =+ + +       (19.2b)

where
AB
AB
AB
L
EI
K =  is the stiffness factor for the beam AB. The coefficient 4 has
been dropped since only relative values are required in calculating distribution
factors.

Note that
2
6
AB
AB S
BA
S
AB
L
EI
M M
?
- = =       (19.3)

S
AB
M is the beam end moments due to support settlement and is negative
(clockwise) for positive support settlements (upwards). In the moment-distribution
method, the support moments  and  due to uneven support settlements
are distributed in a similar manner as the fixed end moments, which were
described in details in lesson 18.
S
AB
M
S
BA
M

It is important to follow consistent sign convention. Here counterclockwise beam
end moments are taken as positive and counterclockwise chord rotation
?
?
?
?
?
? ?
L
is
taken as positive. The moment-distribution method as applied to statically
indeterminate beams undergoing uneven support settlements is illustrated with a
few examples.

Example 19.1

Calculate the support moments of the continuous beam  (Fig. 19.2a) having
constant flexural rigidity
ABC
EI throughout, due to vertical settlement of support B
by 5mm. Assume ; and . 200 GPa E =
44
410 m I
-
=×

Solution

There is no load on the beam and hence fixed end moments are zero. However,
fixed end moments are developed due to support settlement of B by 5mm. In the
spanAB , the chord rotates by
AB
? in clockwise direction. Thus,
5
10 5
3 -
×
- =
AB
?
?
?
?
?
?
?
?
? ×
-
× × × ×
- = - = =
- -
5
10 5
5
10 4 10 200 6 6
3 4 9
AB
AB
AB S
BA
S
AB
L
EI
M M ?

96000 Nm 96 kNm. ==                 (1)

In the span , the chord rotates by BC
BC
? in the counterclockwise direction and
hence taken as positive.
5
10 5
3 -
×
=
BC
?

?
?
?
?
?
?
?
? × × × × ×
- = - = =
- -
5
10 5
5
10 4 10 200 6 6
3 4 9
BC
BC
BC S
CB
S
BC
L
EI
M M ?

. 96 96000 kNm Nm - = - =      (2)

Now calculate stiffness and distribution factors.

EI
L
EI
K
AB
AB
BA
2 . 0 = =   and  EI
L
EI
K
BC
BC
BC
15 . 0
4
3
= =    (3)

Note that, while calculating stiffness factor, the coefficient 4 has been dropped
since only relative values are required in calculating the distribution factors. For
span , reduced stiffness factor has been taken as support C is hinged.  BC
AtB :

EI K 35 . 0 =
?

571 . 0
35 . 0
2 . 0
= =
EI
EI
DF
BA

429 . 0
35 . 0
15 . 0
= =
EI
EI
DF
BC
(4)

At support C :

EI K 15 . 0 =
?
; . 0 . 1 =
CB
DF

Now joint moments are balanced as discussed previously by unlocking and
locking each joint in succession and distributing the unbalanced moments till the
joints have rotated to their final positions. The complete procedure is shown in
Fig. 19.2b   and also in Table 19.1.

Table 19.1 Moment-distribution for continuous beam ABC

Joint A B C
Member  BA BC CB
Stiffness factor  0.2EI 0.15EI 0.15EI
Distribution Factor  0.571 0.429 1.000
Fixd End Moments
(kN.m) 96.000 96.000 -96.000 -96.000
Balance joint C and
C.O. to B   48.00 96.000
Balance joint B and
C.O. to A  -13,704 -27.408 -20.592

Final Moments
(kN.m) 82.296 68.592 -68.592 0.000

Note that there is no carry over to joint as it was left unlocked. C

Example 19.2

A continuous beam  is carrying uniformly distributed load  as
shown in Fig. 19.3a. Compute reactions and draw shear force and bending
moment diagram due to following support settlements.
ABCD m kN / 5

,    0.005m vertically downwards. Support B

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