School of the Built Environment
Structural Analysis Notes
These notes are designed to complement the lecture material, not replace it. They
should serve as a reminder of what is already in your mind and are not intended as a
Moment distribution is an iterative (i.e. numerical) method for obtaining the bending
moments at appropriate points in continuous beams and rigidly-jointed frames.
Although largely superseded by computer methods, it is still used to obtain rapid
solutions so that decisions can be made before a full computer analysis is attempted.
It is also useful for the design of support beams for sheet pile walls and formwork.
Starting Point: Fixed End Moments
The object of the method is to split a problem up into simple standard parts for which
a solution is known and then to fit them all together in such a way that the real
structure is modelled correctly. Look at the continuous beam below:
A B C
Figure 1: Continuous two-span beam.
Moment distribution starts by fixing all joints against rotation, so the beam in figure 1
would be simplified as follows:
A B C
Figure 2: Beam split into two fully-fixed single span beams.
These are now similar (they are both fully fixed beams), and we can therefore use the
same formulae to calculate the moments at each end of the span. These are called the
Fixed End Moments or FEMs for short. The only two standard cases with which
you need concern yourself are shown below:
MAB = -Pab2/L2 MBA = Pa2b/L2
udl p (kN/m)
MAB = -pL2/12 MBA = pL2/12
Figure 3: Standard single-span fully-fixed beams for calculation of FEMs.
You may notice the similarities between the first standard case in figure 3 and the
problem with which we finished in the notes on Moment-Are Theorems. Notice also
two important differences. Firstly, we have adopted a sign convention. This is
absolutely essential with a numerical method:
Sign convention: Clockwise is positive
There are no exceptions to this rule.
The second difference is in the way the fixing moments are named – we use a two-
letter suffix. The first letter refers to the end of the beam and the two letters together
refer to the beam span. So for span B-C in figure 2, we would be calculating MBC and
Equilibrium at joints
Unless there is a moment applied directly at a joint, then the bending moments
immediately either side of the joint must be in equilibrium, i.e. when added up they
must equal zero. This will not be the case when we have split the beam or frame
artificially into single span fully-fixed beams, so the purpose of the distribution
process is to restore this equilibrium by allowing the joints to rotate. This is done in
stages, one joint at a time, until all joints are in equilibrium. This lack of equilibrium
at each joint is called the out-of-balance moment. Before we can remove this, we
need to know something about the effect of the rotation.
Member Stiffness and Carry-over
The moment required to rotate one end of a beam can be calculated as follows:
rotation 1 radian
CBASBA M/EI diagram
Figure 4: Member Stiffness and carry-over.
In order to rotate the beam at B by 1 radian, we must apply a moment, which we will
call SBA. Because end A is still fixed, there is a reaction which is proportional to the
applied moment. The constant of proportionality is denoted CBA. The M/EI diagram
is re-drawn in figure 5 to make it easier to apply the moment-area theorems.
SBA + CBASBA
Figure 5: M/EI diagram drawn in parts.
We can see that the deflection of B from the tangent at A is zero, hence from the
second moment-area theorem,
[((SBA + CBASBA) × L/2) × L/3 - (CBASBA × L) × L/2 ] /EI = 0
Hence CBA = ½. This is called the carry-over factor.
The area under the M/EI diagram is equal to the rotation at B, i.e. 1, so from the first
((SBA + CBASBA) × L/2 - CBASBA × L)/EI = 1
and since CBA = ½,
then SBA = 4EI/L. This is called the member stiffness.
Before we go on, note that the value of 1 radian for the rotation at B is purely notional
– its actual value is unimportant.
When we allow a joint to rotate, then all of the members meeting at that joint
rotate by the same amount. The amount of moment which must be applied to each
member at a particular joint is proportional to the member stiffness, so we can now
calculate the amount of moment which is distributed to each member as a joint is
A rotation θ C
Figure 6: Distribution Factors.
The actual rotation applied to the joint is θ, so the moment applied to member BC
must be SBC/ θ since θ is a fraction of the 1 radian angle used to derive the member
stiffnesses. Similarly, the moment applied to member BA is SBA/ θ. For equilibrium,
M = SBC/ θ + SBA/ θ
Because we are not concerned with the actual size of θ, we calculate the proportions
of M which go to each beam:
moment distributed to beam BC = SBC/ θ = SBC
SBC/ θ + SBA/ θ SBC + SBA
moment distributed to beam BC = SBA/ θ = SBA
SBC/ θ + SBA/ θ SBC + SBA
This is known as the distribution factor for each member meeting at this joint. It
should be clear that the sum of the distribution factors at a joint must equal unity.
The Moment Distribution Process
We now have all of the tools necessary for the analysis. The order of calculation is as
1. Split the structure up into full-fixed single-span beam elements by fixing all of
2. Calculate fixed-end moments for all beam elements which support a transverse
3. For each beam element, calculate its member stiffness.
4. For each joint, calculate the distribution factors using the member stiffnesses.
5. Transfer all distribution factors and fixed-end moments to a diagram of the
structure or to a calculation table.
6. Starting at one joint, sum all of the FEMs at that joint to obtain the out-of-
balance moment at the joint.
7. Remove the out-of-balance moment by applying an equal and opposite
moment to the joint – distribute this to the members in proportion to their
distribution factor at the joint.
8. Carry over half of the distributed moment to the other end of each of the
9. Move to another joint and repeat the process from step 6 until the out-of-
balance moment is small (e.g 0.1 or 0.2 kNm).
The process is illustrated in the following example.
A B C
Figure 7: Two-span continuous beam (EI = constant).
MAB 50kN MBA
MAB = -50 5 32 / 82
A B = -35.2 kNm
MBA = 50 52 3 / 82
8m = 58.6 kNm
MBC = -30 4 22 / 62 = -13.3 kNm
MCB = 30 42 2 / 62 = 26.7 kNm
sAB = sBA = 4EI/L 4/8 = ½ and sBC = sCB = 4EI/L 4/6 = 2/3
We have only artificially fixed one joint, B, as A and C are fixed anyway in the real
beam, so we only need distribution factors at this joint:
DFBA = 1/2 = 3/7 DFBA = 1/2 = 4/7
1/2 + 2/3 1/2 + 2/3
(check: 3/7 + 4/7 = 1 )
Distribution factors at A and C are set to zero, because they represent support reaction
moments. This gives us all we need for the distribution.
A B C
0 3/7 4/7 0
FEMs -35.2 58.6 -13.3 26.7
Dist. -19.4 -25.9
C/O -9.7 -12.9
Total -44.9 39.2 -39.2 13.9
Note: only one distribution has been needed. The out-of-balance moment at joint A is
58.6 – 13.3 = 45.3. This is distributed 3/7 x 45.3 = 19.4 to span BA and 4/7 x 45.3 =
25.9 to span BC. Note the change of sign during distribution. Carry-over factor is
always ½ and the sign remains the same during carry-over to the opposite end of each
Each column is totalled to obtain the bending moment at that point. Note that the sum
of the moments at a joint (e.g. B) must be zero to ensure equilibrium.
If the end of a beam is pinned, then it will be free to rotate without inducing a bending
No carried- = 3EI/L
Figure 8: No carry-over towards a pinned end.
The stiffness of the member containing the pinned end is calculated as 3EI/L instead
of the usual value, and no carry-over then occurs towards the pin.
Consider the problem in figure 7, but with C as a pinned (simple) support:
A B C
Figure 9:Beam with pinned end.
Remember, we must still calculate the FEMs as before. However, the distribution
factors at B will be different because we will use a modified stiffness for BA.
sAB = sBA = 4EI/L 4/8 = ½ ( as before) but sBC = 3EI/L 3/6 = ½
and our distribution factors at B will therefore be ½ and ½ since the stiffness either
side is now the same (NOTE: do not confuse the stiffness values, which have the
dimensions of kNm with the distribution factors which are dimensionless).
The distribution proceeds as before but with the major difference that there is no
carry-over from B to C:
A B C
0 1/2 1/2 0
FEMs -35.2 58.6 -13.3 26.7
Dist 1: -26.7
Dist 2: -16.0 -16.0
Total -43.2 42.6 -42.6 0
Notice that we have needed to perform two distributions, the first to remove the
moment at the pinned end and the second one to produce equilibrium at joint B. Note
that there is no carry-over from B to C after the second distribution, though we still
can carry over away from the pinned end from C to B.
Exercise: Use the distribution factors in the first example (page 7) to deal with the
pinned end without the modified stiffness. You will need to make the DF of C equal
to 1 so that each time you carry over from B to C you must immediately distribute at
C to bring the moment there back to zero. You will obtain the same moments as the
above example, but it will take much longer.