Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
Moment-Distribution Method
The moment-distribution method was originally presented by Prof.
Hardy cross in 1930 and is considered one of the most important
contributions ever made to the structural analysis of continuous beams and
rigid frames.
The moment-distribution is a method of successive approximations
that may be carried out to any desired degree of accuracy. Essentially, the
method begins by assuming each joint of a structure is fixed. Then, by
unlocking and locking each joint in succession, the internal moments at the
joints are distributed and balanced until the joints have rotated to their
final or nearly final positions.
Sign Convention
1. The moments acting on the ends of the members will be considered positive
when they act clockwise.
2. The rotations of the ends of the members will be considered positive when
they act clockwise.
Fixed-End Moments (F.E.Ms)
The moments at the fixed end of a loaded member are called fixed-end
moments. These moments can be determined from Table below, depending upon the
type of loading on the member.
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
Member Stiffness Factor (kij)
Consider a prismatic beam AB, which is hinged at end A and fixed at end B,
as shown in Fig. . If we apply a moment M at the end A, the beam rotates by an
angle θ at the hinged end A and develops a moment MBA at the fixed end B, as
shown in the figure. In previous we related M to θ using the conjugate beam method.
This resulted in Equation below,
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
= carryover
moment
= applied moment
B
A
L
EI= constant
𝑀=
4 𝐸𝐼
𝐿
𝜃
The term
𝐾𝐴𝐵 =
4 𝐸𝐼
𝐿
is referred to as the stiffness factor at A and can be defined as the moment that must
be applied at the end A of the member to cause a unit rotation (𝜃 = 1 ) at A.
Now, suppose that the far end B of the beam is hinged, as shown in Figure below.
The relationship between the applied moment M and the rotation θ of the end A of
the beam can be determined by using the conjugate beam method, as illustrated in
Figure, that is
= applied moment
A
B
L
EI= constant
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
𝑀=
3 𝐸𝐼
𝐿
𝜃
This expression indicates that the stiffness factor of the member at A for this case is
𝐾𝐴𝐵 =
3 𝐸𝐼
𝐿
Thus, it may summarized that the stiffness factor of a member ij at the end i is given
by
𝐾𝑖𝑗 =
𝐾𝑖𝑗 =
4 𝐸𝐼
𝐿
3 𝐸𝐼
𝐿
if far end j of member is fixed
if far end j of member is hinged
Carry-Over Moment (C.O.M.)
Let us consider again the hinged-fixed beam in Figure. When a moment M is applied
at the hinged end A of the beam, a moment MBA develops at the fixed end B, as
shown in the figure. The moment MBA is termed the carryover moment. It was shown
in slope deflection method that:
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
𝑀=
4 𝐸𝐼
𝐿
𝜃
And
𝑀𝐵𝐴 =
2 𝐸𝐼
𝜃
𝐿
Or
𝑀𝐵𝐴 =
𝑀
2
This equation indicates that, when a moment of magnitude M is applied at one end
of a beam, one-half of the applied moment is carried over to the far end, provided
that the far end is fixed. Note that the direction of the carryover moment, MBA, is the
same as the applied moment, M.
When the far end of the beam is hinged, as shown in Figure below, the carry over
moment MBA is zero. Thus we can express the carryover moment as
𝑀𝐵𝐴
𝑀
={2
0
𝑖𝑓 𝑡ℎ𝑒 𝑓𝑎𝑟 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑖𝑠 𝑓𝑖𝑥𝑒𝑑
𝑖𝑓 𝑡ℎ𝑒 𝑓𝑎𝑟 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑖𝑠 ℎ𝑖𝑛𝑔𝑒𝑑
5
Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
The ratio of the carryover moment to the applied moment (
𝑀𝐵𝐴
𝑀
) is called the
carryover factor of the member. It represents the fraction of the applied moment M
that is carried over to the far end of the member. Thus, we can express the carryover
factor (COF) as
1
𝐂𝐎𝐅 = { 2
0
𝑖𝑓 𝑡ℎ𝑒 𝑓𝑎𝑟 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑖𝑠 𝑓𝑖𝑥𝑒𝑑
𝑖𝑓 𝑡ℎ𝑒 𝑓𝑎𝑟 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟 𝑖𝑠 ℎ𝑖𝑛𝑔𝑒𝑑
Distribution Factor (D.F.)
When an external moment is applied to a joint of a structure where two or more
members meet, an important question that arises is how to distribute this moment
among the various members connected at that joint. Consider the joint B show in
Figure below at which three members meet, and suppose that a moment M is applied
to this joint, causing it to rotate by an angle θ. To determine what fraction of the
applied moment M is resisted by each of the three members connected to the joint,
we draw the free-body diagram of joint B as show in Fig. (10.2-b). By considering
the moment equilibrium of joint B (that is, ∑ 𝑀𝐵 = 0 ), we have
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
𝑀 − 𝑀𝐵𝐴 − 𝑀𝐵𝐶 − 𝑀𝐵𝐷 = 0
𝑀 = 𝑀𝐵𝐴 + 𝑀𝐵𝐶 + 𝑀𝐵𝐷
We have,
𝑀𝐵𝐴 = 𝐾𝐵𝐴 𝜃
𝑀𝐵𝐶 = 𝐾𝐵𝐶 𝜃
(1)
𝑀𝐵𝐷 = 𝐾𝐵𝐷 𝜃
𝑀 = (𝐾𝐵𝐴 + 𝐾𝐵𝐶 + 𝐾𝐵𝐷 ) 𝜃 → 𝑀 = 𝜃 ∑ 𝐾
𝑜𝑟
𝜃=
𝑀
∑𝐾
(2)
The above equations can now be written as
𝑀𝐵𝐴 =
𝐾𝐵𝐴
𝑀
∑𝐾
𝑀𝐵𝐶 =
𝐾𝐵𝐶
𝑀
∑𝐾
𝑀𝐵𝐷 =
𝐾𝐵𝐷
𝑀
∑𝐾
𝑲𝒊𝒋
The ratio ∑ for a member ij is termed the distribution factor (D.F) of the member
𝑲
and it will be given the symbol d, that is
𝑑𝑖𝑗 =
7
𝐾𝑖𝑗
∑𝐾
Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
The distribution factor for a member is thus equal to the stiffness of the member
divided by the sum of the stiffnesses of all members meeting at the joint.
Notes:
1. The distribution factor is zero for a fixed end and one for an end roller or
hinged support.
2. The sum of all the distribution factors at a joint must equal one.
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
Example: Draw the bending moment diagram for the beam shown in Fig.
by using the moment-distribution method.
18 kN/m
1
38 kN
2
2I
3
I
6m
2m
2m
Solution:
Fixed -End Moment
𝐹
𝑀12
−𝑤𝐿2
−18 ∗ (6)2
=
=
= −54 𝑘𝑁. 𝑚
12
12
𝐹
𝑀21
𝑤𝐿2
18 ∗ (6)2
=
=
= +54 𝑘𝑁. 𝑚
12
12
𝐹
𝑀23
=
−𝑃𝐿
−38 ∗ 4
=
= −19 𝑘𝑁. 𝑚
8
8
𝐹
𝑀32
=
+𝑃𝐿
38 ∗ 4
=
= +19 𝑘𝑁. 𝑚
8
8
Stiffness Factor (Kij)
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
Distribution Factor (dij)
For joints 1 and 3 (fixed support)
And for member ends connected at joint 2 we have
Make table
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Lecture 18
Theory of Structures
Third Stage
Dr. Abdulkhaliq A. Jaafer
Misan University
11
Lecture 18