Moment Distribution Method Notes for GATE, ESE & Civil Exams

By Deepanshu Rastogi|Updated : December 2nd, 2021

Through Champion Study Plan for GATE Civil Engineering (CE) 2022, we are providing Moment Distribution Method study notes and other important materials on every topic of each subject.

These topic-wise study notes are useful for the preparation of various upcoming exams like GATE CivilIESBARCISROSSC-JEState Engineering Services examinations and other important upcoming competitive exams.

This article contains fundamental notes on the "Moment Distribution Method" topic of the "Structural Analysis" subject.

Moment Distribution Method


  • The moment distribution method is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross.
  • The method only accounts for flexural effects and ignores axial and shear effects.
  • The moment distribution method falls into the category of displacement method of structural analysis.
  • In the slope deflection method, the end moments are computed using the slopes and deflection at the ends. Contrarily in the moment distribution method, as a first step the slopes at the ends are made zero. This is done by fixing the joints.

In the moment distribution method, every joint of the structure to be analysed is fixed so as to develop the fixed-end moments. Then each fixed joint is sequentially released and the fixed-end moments (which by the time of release are not in equilibrium) are distributed to adjacent members until equilibrium is achieved. The moment distribution method in mathematical terms can be demonstrated as the process of solving a set of simultaneous equations by means of iteration.

Important Points 

  1. When the member is fixed at one end and a moment is applied at the other end which is simply supported or hinged, the moment induced at the fixed end is one half of the applied moment. The induced moment at the fixed end is in the same direction as the applied moment.
  2. If a moment is applied in a stiff joint of a structure, the moment is resisted by various members in proportion to their respective stiffnesses (i.e., moment of inertia divided by the length). If the stiffness of the member is more; then it resists more bending moment and it absorbs a greater proportion of the applied moment.
  3. While distributing the moments in a rigid joint, if one end of the member is not restrained then its stiffness should be multiplied by (3/4).
  4. In a fixed beam, if the support settles/subsides/sinks by an amount Δ, the moment required to make the ends horizontal is 6EIΔ/l2

Basic Definition

  1. Stiffness: Rotational stiffness can be defined as the moment required to rotate through a unit angle (radian) without the translation of either end.


    Where, K = Stiffness

    F = Force required to produce deflection Δ

    M = Moment required to produce rotation θ.

  2. Stiffness Factor: 

    (i) It is the moment that must be applied at one end of a constant section member (which is unyielding supports at both ends) to produce a unit rotation of that end when the other end is fixed

, i.e. k = 4EI/l.image026

Where K = Stiffness of BA at joint B. When the farther end is fixed.

El = Flexural rigidity

L = Length of the beam

M = Moment at B.

(ii) It is the moment required to rotate the near end of a prismatic member through a unit angle without translation, the far end being hinged is k = 3EI/l.


  1. Carry Over Factor:
    It is the ratio of the induced moment to the applied moment. The carry-over factor is always (1/2) for members of the constant moment of inertia (prismatic section). If the end is hinged/pin-connected, the carry-over factor is zero. It should be mentioned here that carry over factors values differ for non-prismatic members. For non-prismatic beams (beams with variable moment of inertia); the carryover factor is not half and is different for both ends.

    Carryover factor = image029

    COF may greater than, equal to or less than 1.

    Standard Cases:



    (ii) COF = 0




  2. Distribution Factors: 


Consider a frame with members OA, OB, OC and OD rigidly connected at O as shown below. Let M be the applied moment at joint O in the clockwise direction. Let the joint rotate through an angle θ. The members OA, OB, OC and OD also rotate by the same angle θ


Let kOA , kOB, kOC and kOD be the stiffness values of the members OA, OB, OC and OD respectively; then



5. Relative Stiffness

(i) When farther end is fixed

Relative stiffness for member image037

(ii) When the farther end is hinged

Relative stiffness for member image038



Stiffness of OA image040

Stiffness of OB image041

Stiffness of OC image041

Stiffness of OD = 0


Clockwise moments are considered positive and anticlockwise moments negative

+ve → Sagging

–ve → Hogging

and All clockwise moment → +ve

and All Anti clockwise moment → –ve

Span length is l





Draw the bending moment diagram for the continuous beam ABCD loaded as shown below. The relative moment of inertia of each span of the beam is also shown in the figure. 


Note that joint C is hinged and hence stiffness factor BC gets modified. Assuming that the supports are locked, calculate fixed end moments. They are


In the next step calculate stiffness and distribution factors


Now all the calculations are shown below




Bending Moment Diagram

Note: This problem has also been solved by the slope-deflection method 

Example 2

Analyse the continuous beam by the moment distribution method. Draw the shear force diagram and bending moment diagram.








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All the Best


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Sachin Singh
Actually in fig 2 there should be fixed support on the right and roller/hinge support on the left side.
Thanks for mentioning.
We will update soon!!
Rishabh Kashyap
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hemanth naalla
Sir please give the notes for orifice and venturi meter
Nitesh Mishra
A complete copy of NPTEL, khud se kuch banate yaar.
Chirag Shah

Chirag ShahSep 19, 2019

If any student need study materials for their any exam then do message on
Ankit Rawat

Ankit RawatSep 20, 2019

Kisi Bhi state wala apply kar sakata hai kya plz reply
Kapil Verma

Kapil VermaSep 21, 2019

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Ankit Tamrakar
In BMD of que 2, the BMD shape should be triangular due to point loading.
Abinash Karmakar
Analysis of Frame type of structure by using Moment distribution method sfd bmd
Kunal Verma

Kunal VermaSep 10, 2020

GATE 2021 made easy and iesmaster
Test series material full solved online test series in the PDF format available for 2021
Interested aspirant contact me at whatsapp 8225950652 #GATE2021

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