Section CB.2.1
Moment Distribution
Last Revised:
11/04/2014
Moment distribution is a great method for quickly computing end
moments on continuous beams. Over the years, several variations of the
method have been presented. These methods take advantage of various
observations made about the process. While this method can be applied to a
variety of indeterminate structures, the discussion here is limited to
continuous beams that may or may not be continuous with supporting columns.
The moment distribution method begins by determining the relative flexural
stiffness, in the plane of loading, of all the elements rigidly connected to each joint.
Rotational stiffness of a member is proportional to material and geometric
stiffness:
Relative Rotational Stiffness = EI/L
Where:
 E is the modulus of elasticity of the material,
 I is the moment of inertia about the axis of bending in the plane of the
frame, and
 L is the length of the member.
The basic premise of the moment distribution method is that any unbalanced
moment on a joint is redistributed to the members rigidly attached to that joint
in proportion to contributions each element makes to the total rotational
stiffness of the joint. Consequently, distribution factors are computed at
each joint, one for each element attached to the joint that provides rotational
stiffness in the considered plane. The equation for computing the
distribution factor of the i^{th} member at a joint with n members is:
Note that if any of the variables is constant for all members of the joint,
then the constant variable will cancel out of the equation.
The distribution factors are computed joint by joint for the ends of each
member connected to the joint.
The moment distribution method assumes that all joints are "fixed" at the
start of the problem. This means that when loads are applied to each span,
fixed end moments are developed at each end of the loaded span. The fixed
end moments are a function of the nature and location of the applied loads on
the span. If there is more than one load source on a span, super position
can be used to determine the total fixed end moment at the end of each span.
Concerning notation, in this method the end moments are considered
to be reactions. As a result, the right hand rule is used to determine the sign of the
moments. Do not use beam internal force notation.
After computing the fixed end moments (FEM) for each span, you will observe
that there is an unbalanced moment at each joint. In other words, the sum
of moments at the joint does not equal zero, a necessary requirement for
equilibrium. To rectify this problem, a joint is "released" and the
unbalanced moment (i.e. the difference of the FEMs) is divided among the members
attached to joint in proportion to their contributions to the joint rotational
stiffness using the distribution factors. Once the distributed moments
are added to each preexisting FEM at the joint, the resulting FEMs are
"balanced". In other words, the sum of the moments at the joint equal zero
and equilibrium is satisfied at that joint (for now!).
The addition of moment to each element attached to the joint induces a moment
on the opposite end of each member. This is called "carry over" (CO).
The carry over moment is added to the preexisting FEM at that joint. The
CO moment tends to unbalance the joints adjacent to newly balanced joint.
These joints are, in turn released, balanced and send back CO moments to
unbalance the adjacent joint.
Joints are successively released and balanced until the CO moments get small
enough to ignore. This means that this is an iterative process that looks
for convergence.
An example may help to illustrate the process.
Consider the beam shown in Figure CB.2.1.1. This beam will have a
constant E and I for all three spans, so the relative stiffness of each can be
computed as 1/L.
Figure CB.2.1.1
Beam Problem Definition
Compute the Distribution Factors
For Joint "A":
Two items contribute to the rotational stiffness at A. One is the
beam AB the other is the infinitely stiff support. The distribution
factor at the left end of beam AB then is:
DF_{AB,L} = (1/7.5) / (infinity + 1/7.5) = 0
The distribution factor for the support at A is:
DF_{A} = infinity / (infinity +
1/7.5) = 1
For Joint "B":
The two items contributing to rotation stiffness at B are the beams that
come in from either side. The support is a roller so it does not
contribute to the rotational support of the joint. The two
distribution factors are:
DF_{AB,R} = (1/7.5) / ( (1/7.5) + (1/5) ) = 0.400
DF_{BC,L} = (1/5) / ( (1/7.5) + (1/5) ) = 0.600
For Joint "C":
The two items contributing to rotation stiffness at C are the beams that
come in from either side. The support is a roller so it does not
contribute to the rotational support of the joint. The two
distribution factors are:
DF_{BC,R} = (1/5) / ( (1/6.25) + (1/5) ) = 0.556
DF_{CD,L} = (1/6.25) / ( (1/6.25) + (1/5) ) =
0.444
For Joint "D":
There is only one item contributing to rotation stiffness at D: Beam CD.
The support is a roller so it does not contribute to the rotational support
of the joint. The two distribution factors are:
DF_{CD,R} = (1/6.25) / ( (1/6.25) + (0) ) = 1.000
DF_{D} = (0) / ( (1/6.25) + (0) ) = 0.000
Table CB.2.1.1 summarizes the progress to this
point.
Table CB.2.1.1
Distribution Factors

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

Compute the Fixed End Moments
The fixed end moments are computed span by span using the given loading for
each span and considering the ends of the beam to be fixed.
Figure CB.2.1.2
Fixed End Moments 

Any number of sources can provide equations for fixed end moments.
Maybe, we'll eventually add a table of them here. For now, you can use the
beam tables found in the Steel Construction Manual, the
Fundamentals of Engineering (FE) exam reference manual, or derive your own.
Figure CB.2.1.2 shows each of the spans, their loading, and the definition of
the FEMs. Note that each of the right end FEMs are right hand rule
negative moments. It is important to keep the sign.
For Span AB, the FEMS are:
FEM_{AB,L} = + PL / 8 = + (10 kN) (7.5 m) / 8 = +
9.375 kN/m
FEM_{AB,R} =  PL / 8 =  (10 kN) (7.5 m) / 8 = 
9.375 kN/m
The two FEMs have the same magnitude since the load is centered on the span.
They would be different otherwise.
For Span BC, the FEMS are:
FEM_{BC,L} = + wL^{2} / 12 = + (2 kN/m) (5 m)^{2}
/ 12 = + 4.167 kN/m
FEM_{BC,L} = + wL^{2} / 12 =  (2 kN/m) (5 m)^{2}
/ 12 =  4.167 kN/m
For Span CD, the FEMS are:
FEM_{CD,L} = + wL^{2} / 12 = + (1.5 kN/m)
(6.25 m)^{2} / 12 = + 4.883 kN/m
FEM_{CD,L} = + wL^{2} / 12 =  (1.5 kN/m)
(6.25 m)^{2} / 12 =  4.883 kN/m
Table CB.2.1.2 show our spreadsheet table expanded to include the initial
FEMs.
Table CB.2.1.2
Distribution Factors

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Load 
10 
kN 
2 
kN/m 
1.5 
kN/m 

Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

FEM 
9.375 
9.375 
4.167 
4.167 
4.883 
4.883 
kNm 
Balancing the Joints and Doing the Carry Over
At this time, all joints are said to be fixed. In other words, it is as
if an invisible lock has taken the imbalance at each joint. This explains
why equilibrium is not satisfied at a joint. Lets take each joint and do a
balance.
Joint A:
The table that we are developing does not show the support side of the
joint. This is common. Joint A is always balanced because the
fixed support moment matches (with opposite sign) the beam end moment.
This joint does not require balancing.
Joint B:
The unbalanced moment a this joint equals (9.375 kNm + 4.167 kNm) = 
5.208 kNm. To balance this joint we need to add + 5.208 kNm to the
joint. This is done by adding to each beam end in proportion to
its
contribution to the rotational stiffness of the joint. This is what
the distribution factors are for.
The new FEMs are:
FEM_{AB,R} =  9.375 kN/m + DF_{AB,R}
(+5.208 kNm)
FEM_{AB,R} =  9.375 kN/m + 0.400 (+5.208 kNm)
= 9.375 kNm + 2.083 kNm
FEM_{AB,R} =  7.292 kN/m
FEM_{BC,L} = + 4.167 kN/m + DF_{BC,L}
(+5.208 kNm)
FEM_{BC,L} = + 4.167 kN/m + (0.600) (+5.208
kNm) = + 4.167 kN/m + 3.125 kNm
FEM_{BC,L} = + 7.292 kN/m
Note that the sum of the two FEMs is zero. The joint is balanced.
The result of the balancing creates moments at the far ends of Beams AB
and BC equal to half the balancing moments. Table CB.2.1.3 shows the
balancing moments at joint B and the resulting carry over moments at joints
A and C. The summation line shows the sum of the FEM, balance, and CO
lines to show the current end moments on each span. Note that only
joint B is "balanced" at this point.
Table CB.2.1.3
Joint B Balance and Carry Over

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Load 
10 
kN 
2 
kN/m 
1.5 
kN/m 

Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

FEM 
9.375 
9.375 
4.167 
4.167 
4.883 
4.883 
kNm 
Balance 
0.000 
2.083 
3.125 



kNm 
CO 
1.042 
0.000 

1.563 


kNm 
Sum 
10.417 
7.292 
7.292 
2.604 
4.883 
4.883 
kNm 
Joint C:
Ignoring the carry over moment from the prior joint, the unbalanced
moment at this joint equals (4.167 kNm + 4.883 kNm) = + 0.716 kNm.
To balance this joint we need to add  0.716 kNm to the joint. This
is done by adding to each beam end in proportion to its contribution to the
rotational stiffness of the joint. This is what the distribution
factors are for.
The new FEMs are:
FEM_{BC,R} =  4.167 kN/m + DF_{BC,R}
(0.716 kNm)
FEM_{BC,R} =  4.167 kN/m + 0.556 (0.716 kNm)
= 4.167 kNm  0.398 kNm
FEM_{BC,R} =  4.565 kN/m
FEM_{CD,L} = + 4.883 kN/m + DF_{CD,L}
(0.716 kNm)
FEM_{CD,L} = + 4.883 kN/m + (0.444) (0.716
kNm) = + 4.883 kN/m  0.318 kNm
FEM_{BC,L} = + 4.565 kN/m
Note that the sum of the two FEMs is zero. The joint is balanced,
if you continue to ignore the carryover from joint B. Table CB.2.1.4
shows the balancing moments and carry over moments (highlighted in yellow)
resulting from this operation.
Table CB.2.1.4
Joint B Balance and Carry Over

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Load 
10 
kN 
2 
kN/m 
1.5 
kN/m 

Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

FEM 
9.375 
9.375 
4.167 
4.167 
4.883 
4.883 
kNm 
Balance 
0.000 
2.083 
3.125 
0.398 
0.318 

kNm 
CO 
1.042 
0.000 
0.199 
1.563 

0.159 
kNm 
Sum 
10.417 
7.292 
7.093 
3.002 
4.565 
5.042 
kNm 
Joint D:
As we did for Joint C, we ignore the carry over moment from the prior
joint. As we did at joint A, we did not include support moment in the
table. For this case the joint is a pin, so no moment can be there
when we end. The unbalanced moment a this joint equals 4.883 kNm.
To balance this joint we need to add +4.883 kNm to the joint.
This is done by adding to each beam end in proportion to its contribution
to the rotational stiffness of the joint. This is what the
distribution factors are for.
The new FEM is:
FEM_{CD,R} =  4.883 kN/m + DF_{CD,R}
(+4.883 kNm)
FEM_{CD,R} =  4.883 kN/m + 1.000 (+4.883 kNm)
= 4.167 kNm + 4.883 kNm
FEM_{CD,R} = 0 kN/m
Note that the FEM is zero. The joint is balanced, if you continue
to ignore the carry over from joint C. Table CB.2.1.5 shows the
balancing moment and carry over moment (highlighted in yellow) resulting
from this operation.
Table CB.2.1.5
Joint B Balance and Carry Over

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Load 
10 
kN 
2 
kN/m 
1.5 
kN/m 

Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

FEM 
9.375 
9.375 
4.167 
4.167 
4.883 
4.883 
kNm 
Balance 
0.000 
2.083 
3.125 
0.398 
0.318 
4.883 
kNm 
CO 
1.042 
0.000 
0.199 
1.563 
2.441 
0.159 
kNm 
Sum 
10.417 
7.292 
7.093 
3.002 
7.006 
0.159 
kNm 
Adding the original FEMs to the corresponding balancing moments and carry
over moments we arrive at a new FEM state. Notice that the imbalance at
each joint is significantly less than what the original FEMs provided. We
are starting to converge on the solution.
At this point we repeat the balance, carry over and sum steps until the
imbalances become small enough for our purposes. Generally, this occurs
when the unbalanced moments are about 1% of the end moments. With a
spreadsheet, repeating the steps is simply a matter of copying the balance,
carry over and sum steps once the formulas are all set up. Table CB.2.1.6
shows the results of nine balancing cycles.
Table CB.2.1.6
The Full Solution

Span AB 

Span BC 

Span CD 

Units 
Span 
7.5 

5 

6.25 

m 
1/L 
0.133333 

0.2 

0.16 


Load 
10 
kN 
2 
kN/m 
1.5 
kN/m 

Joint 
A 
B 
B 
C 
C 
D 

DF 
0.000 
0.400 
0.600 
0.556 
0.444 
1.000 

FEM 
9.375 
9.375 
4.167 
4.167 
4.883 
4.883 
kNm 
Balance 
0.000 
2.083 
3.125 
0.398 
0.318 
4.883 
kNm 
CO 
1.042 
0.000 
0.199 
1.563 
2.441 
0.159 
kNm 
Sum 
10.417 
7.292 
7.093 
3.002 
7.006 
0.159 
kNm 
Balance 
0.000 
0.080 
0.119 
2.224 
1.780 
0.159 
kNm 
CO 
0.040 
0.000 
1.112 
0.060 
0.080 
0.890 
kNm 
Sum 
10.456 
7.212 
6.100 
5.167 
5.306 
0.890 
kNm 
Balance 
0.000 
0.445 
0.667 
0.077 
0.062 
0.890 
kNm 
CO 
0.222 
0.000 
0.039 
0.334 
0.445 
0.031 
kNm 
Sum 
10.679 
6.767 
6.729 
4.910 
5.689 
0.031 
kNm 
Balance 
0.000 
0.015 
0.023 
0.433 
0.346 
0.031 
kNm 
CO 
0.008 
0.000 
0.216 
0.012 
0.015 
0.173 
kNm 
Sum 
10.687 
6.752 
6.535 
5.331 
5.358 
0.173 
kNm 
Balance 
0.000 
0.087 
0.130 
0.015 
0.012 
0.173 
kNm 
CO 
0.043 
0.000 
0.008 
0.065 
0.087 
0.006 
kNm 
Sum 
10.730 
6.665 
6.658 
5.282 
5.433 
0.006 
kNm 
Balance 
0.000 
0.003 
0.005 
0.084 
0.067 
0.006 
kNm 
CO 
0.002 
0.000 
0.042 
0.002 
0.003 
0.034 
kNm 
Sum 
10.731 
6.662 
6.620 
5.363 
5.369 
0.034 
kNm 
Balance 
0.000 
0.017 
0.025 
0.003 
0.002 
0.034 
kNm 
CO 
0.008 
0.000 
0.001 
0.013 
0.017 
0.001 
kNm 
Sum 
10.740 
6.645 
6.644 
5.354 
5.383 
0.001 
kNm 
Balance 
0.000 
0.001 
0.001 
0.016 
0.013 
0.001 
kNm 
CO 
0.000 
0.000 
0.008 
0.000 
0.001 
0.007 
kNm 
Sum 
10.740 
6.645 
6.637 
5.370 
5.371 
0.007 
kNm 
Balance 
0.000 
0.003 
0.005 
0.001 
0.000 
0.007 
kNm 
CO 
0.002 
0.000 
0.000 
0.002 
0.003 
0.000 
kNm 
Sum 
10.742 
6.642 
6.641 
5.368 
5.373 
0.000 
kNm 
Notice that the changes to the end moments have become "small". It is
time to stop.
With these end moments, the remaining shears and moments can be found using
equilibrium equations.
