MOMENT DISTRIBUTION METHOD
 It is known as HARDY CROSS method as well.
 Provides a convenient means of analysing statically indeterminate beams
and frames by simple hand calculations.
 This is an basically an iterative process. The process is repeated a number
of times or cycles till the values obtained within the desired accuracy.
This method of consists of solving slope deflection equations by successive
approximation 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. This method of analysis is both
repetitive and easy to apply.
SIGN CONVENTION
 The sign convention followed in the development of the method is the
same as the one followed for the slope deflection method.
Before developing this method, it is necessary to define certain terms employed
in this method.
1. Absolute Stiffness of a Member
Can be defined as the ‘moment required to produce a unit rotation’.
CE382 (Supplementary notes) Prepared by: Asst.Prof.Dr.Rifat Reşatoğlu
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Figure 1. Absolute stiffness of a member when far end is fixed.
For this beam, the moment Mij thus represents the absolute stiffness of
member i-j. From table , the moment Mij=4EI/L considering i as unity.
The ratio I/L is referred to as the relative stiffness and is denoted by the letter
K.
2. Carry Over Factor (C.O.F.)
If a moment Mij is applied at end I of the member in Fig.1, a specified
amount of moment Mji , is generated at the farther restrained end.
It is defined as the factor by which the moment at simply supported end ‘i’,
Mij , is multiplied to get the moment carried over to the other end, that is Mji,
Mji=C.O.F. x Mij
According to Fig.1 , the C.O.F. is always (+1/2). If the farther end is a hinged
end instead of a fixed one as shown in table, the corresponding stiffness is
known as the modified stiffness of a member and is equal to
Mij=3EI/L
CE382 (Supplementary notes) Prepared by: Asst.Prof.Dr.Rifat Reşatoğlu
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And the relative stiffness
K*=3/4 K
Obviously , the moment carried over to the farther hinged end, Mji=0.
3. Distribution Factor (D.F.)
Consider a joint in a structure where two or more members meet. If an
external moment M is applied to such a joint, the joint undergoes a rotation 
as shown. Since all the members meeting at this point undergo the same
rotation , the applied moment M is distributed to each of the ends of the
members according to their relative stiffness values. The factor by which the
applied moment is multiplied to obtain the end moment of any member is
known as the Distribution Factor (D.F.).
Consider the FBD of joint 1 in Fig2b. For equilibrium of joint 1, we have
M12+M13+M14+M15-M=0
CE382 (Supplementary notes) Prepared by: Asst.Prof.Dr.Rifat Reşatoğlu
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In writing the equilibrium, clockwise moments on a joint are considered
positive. The end-moments of members can be written in terms of the angle
of rotation  as
M12=4EK12
M13=4EK13
M14=4EK14
M15=4EK15
M=4E(K12+K13+K14+K15) or M=4EK or =M / 4EK
where K denotes the sum of the relative stiffness of all the members meeting
at joint 1. Substituting the value for  , we get
M12=4EK12
M
K
 12 M
4E K  K
Ratio K12/K indicates the fraction by which the applied moment M is to
be multiplied to get the moment resisted by member 1-2. This ratio by
definition is the distribution factor. The distribution factor for any member ij is defined in general as
rij = Kij / K
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BASIC STEPS
 Find the distribution factors
 Compute the FEM due to total load and dead load for each member
 Carry over the moments (fixed-end) once and add
 Carry over moments and the FEM are balanced. This gives the
maximum member end moments.
 To find the maximum positive span moment, the carry over moment to
the mid-span is added to the positive fixed-end moment.
MOMENT DISTRIBUTION FOR FRAMES WITHOUT SIDESWAY
It is the same procedure as for continuous beams. (When the joint translations
are prevented).
It is quite usual that in frames more than 2 members meet at a joint. Care must
be taken in such cases to include the stiffness of all members meeting at any
joint evaluating the distribution factors.
For a single bay, single storey frame, it is convenient to spread out legs of the
frame so that the frame lies along a straight line.
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ANALYSIS OF FRAMES WITH LATERAL TRANSLATION OF
JOINTS (moment distribution for frames with sidesway)
So far we considered frames in which the joints are not allowed to translate
laterally. However in frames, the translation of some joints is common due to
forces
i-) acting in the lateral direction.
or due to
ii-) unsymmetrical forces
or due to
iii-) unsymmetry in the make of the frame even though the load is symmetrical
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In frames undergoing lateral translation, the analysis is carried out in 2
stages.
1-) moment distribution for the frame without sway (Mno sidesway)
2-) moment distribution for the sway effect only (M sidesway)
The final result can be obtained by the superposition of the separate cases
MFINAL=(Mno sidesway)+ (M sidesway)
Procedure example
*In the first stage, the frame is prevented from undergoing any lateral translation
by applying an artificial restraint.
*The value of artificial restraining force X can be obtained by first evaluating
the shear at the bases of the columns.
*Then from the equilibrium condition, FH = 0, the value of X can be evaluated.
* At this stage the end moments obtained are true only when restraining force X
were acting.To achieve the true condition of the structure, the frame has to be
analysed again by applying a force equal and opposite to artificial restraining
force X.
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*The member end-moments resulting from this condition of loading will be
combined with the moments obtained from the earlier restrained condition to
obtain the true values of moments in the frame.
*The moments in the members of the frame due to application of the force (-X)
are obtained in an indirect manner. The frame is assumed to be subjected to an
arbitrary loading, say X/.
*The true values of moments under a horizontal force (-X) can be obtained by
multiplying the moments caused by the X/ by the ratio, X /X’
*The moments thus obtained are added to the moments obtained in the first
stage of moment distribution.
Example : 1
Determine the end-moments of the members of the frame shown in figure. E is
constant and relative I values are indicated on the frame. (Let E=20)
* As a first step we artificially restrain joint 3 from undergoing lateral
translation. The moment distribution for this restrained condition is already
worked out before.
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The artificial restraining force X is worked out considering the FBD of the
columns as in figure shown below. Summing the moments about joint 2 we have
;
M2=0
4V12-(50x2)+(54.77)=0
V12=11.31 kN
Similarly summing the moments about joint 3 we get;
Applying equilibrium condition FH=0 for the entire structure
50-11.31-13.39-X=0
X=25.3 kN (This is the force, the artificial restraint, that prevent
lateral translation).
To obtain the true condition, the artificial constraint has to be removed by
applying a force equal but opposite to force X. This needs a second distribution
of moments. However, this has to be worked out in an indirect manner.
Apply an unknown force, X/. The FEM’s in the columns can be worked out
considering that joints 2 and 3 only translate and do not rotate. If tha axial
deformation in the beam is neglected, the tanslations  are the same at both
joints 3 and 4. Therefore the FEM’s can be written as ;
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The FEM’s are entered, as usual, in the table shown below. The distribution
factors are same as found before. The moment distr. İs carried out as usual and
the final moments are shown in the last row of the table.
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The lateral force X/ which produced these moments can be evaluated by
considering the free body diagram of the columns as in figure shown below. The
summation of the moments about joint 2 gives,
V12=11.52/4=2.88 kN
Similarly, the summation of the moments about joint 3 gives
V43=16.15+23.11/4=9.82 kN
From considerations of equilibrium of horizontal forces FH=0, we have
X/=2.88+9.82=12.70 kN
The true value of the horizontal force to be applied to the frame is 25.30 kN.
Therefore, the moments due to lateral force of 25.30 kN are obtained by
proportion,
that
is,
by
multiplying
the
moments
in
by
a
factor
25.30/12.70=1.992.
These moments are added to the values of the moments found. Thus, the true
moments in the frame are ;
M12=0
M21=-54.77+11.52x1.992=-31.82
M32=-35.73-16.15x1.992=-67.90=-M34
M43=17.80+23.11x1.992=63.84
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As a check it is seen that the sum of the shears in the columns is equal to the
external lateral force as shown below.
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