MOMENT DISTRIBUTION METHOD OF ANALYSIS
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.
The moment distribution method falls into the category of displacement method of structural analysis.
In order to apply the moment distribution method to analyse a structure, the following things must be considered.
Fixed end moments
Fixed end moments are the moments produced at member ends by external loads when the joints are fixed.
Flexural stiffness
The flexural stiffness (EI/L) of a member is represented as the product of the modulus of elasticity (E) and the second moment of area (I) divided by the length (L) of the member. What is needed in the moment distribution method is not the exact value but the ratio of flexural stiffness of all members.
Distribution factors
Distribution factors can be defined as the proportions of the unbalanced moments carried by each of the members.
Carryover factors
Unbalanced moments are carried over to the other end of the member when the joint is released. The ratio of the carried-over moment at the other end to the fixed-end moment of the initial end is the carryover factor.
Sign convention
Any moment acting clockwise is considered to be positive. This differs from the usual engineer’s sign convention, which employs a Cartesian coordinate system with positive x-axis to the right and positive y-axis up, resulting in positive moment about the z-axis being counterclockwise.
Framed structures
Framed structures with or without sidesway can be analysed using the moment distribution method.
The statically indeterminate beam shown in the figure is to be analysed.
- Members AB, BC, CD have the same length .
- Flexural rigidities are EI, 2EI, EI respectively.
- Concentrated load of magnitude acts at a distance from the support A.
- Uniform load of intensity acts on BC.
- Member CD is loaded at its midspan with a concentrated load of magnitude .
In the following calcuations, counterclockwise moments are positive.
Fixed-end moments
Distribution factors
The distribution factors of joints A and D are D_{AB} = D_{DC} = 1.
Carryover factors
The carryover factors are , except for the carryover factor from D (fixed support) to C which is zero.
Moment distribution
Joint | A | Joint | B | Joint | C | Joint | D | ||||
Distrib. factors | 0 | 1 | 0.2727 | 0.7273 | 0.6667 | 0.3333 | 0 | 0 | |||
Fixed-end moments | 14.700 | -6.300 | 8.333 | -8.333 | 12.500 | -12.500 | |||||
Step 1 | -14.700 | ? | -7.350 | ||||||||
Step 2 | 1.450 | 3.867 | ? | 1.934 | |||||||
Step 3 | -2.034 | ? | -4.067 | -2.034 | ? | -1.017 | |||||
Step 4 | 0.555 | 1.479 | ? | 0.739 | |||||||
Step 5 | -0.246 | ? | -0.493 | -0.246 | ? | -0.123 | |||||
Step 6 | 0.067 | 0.179 | ? | 0.090 | |||||||
Step 7 | -0.030 | ? | -0.060 | -0.030 | ? | -0.015 | |||||
Step 8 | 0.008 | 0.022 | ? | 0.011 | |||||||
Step 9 | -0.004 | ? | -0.007 | -0.004 | ? | -0.002 | |||||
Step 10 | 0.001 | 0.003 | |||||||||
Sum of moments | 0 | -11.569 | 11.569 | -10.186 | 10.186 | -13.657 |
Numbers in grey are balaced moments; arrows ( ? / ? ) represent the carry-over of moment from one end to the other end of a member.
Result
- Moments at joints determined by the moment distribution method
The conventional engineer’s sign convention is used here, i.e. positive moments cause elongation at the bottom part of a beam member.
For comparison purposes, the following are the results generated using a matrix method. Note that in the analysis above, the iterative process was carried to >0.01 precision. The fact that the matrix analysis results and the moment distribution analysis results match to 0.001 precision is mere coincidence.
- Moments at joints determined by the matrix method
The complete shear and bending moment diagrams are as shown. Note that the moment distribution method only determines the moments at the joints. Developing complete bending moment diagrams require additional calculations using the determined joint moments and internal section equilibrium.
- SFD and BMD
Shear force diagram | Bending moment diagram |
STEPS:
- Restrain all possible displacements.
- Calculate Distribution Factors:
The distribution factor DFi of a member connected to any joint J is
where S is the rotational stiffness , and is given by - Determine carry-over factors
The carry-over factor to a fixed end is always 0.5, otherwise it is 0.0. - Calculate Fixed End Moments. (Table 3.1).
These could be due to in-span loads, temperature variation and/or relative displacement between the ends of a member. - Do distribution cycles for all joints simultaneously
- Each cycle consists of two steps:
- Distribution of out of balance moments Mo, where
Then;
where
- Distribution of out of balance moments Mo, where
Me is the external moment applied to the joint (if any) Mo is the total out of balance moment at the joint FEMi the fixed-end moment Mi moment distributed to any member DFi distribution factor of member i | at the member end |
- Calculation of the carry over moment at the far end of each member.
The procedure is stopped when, at all joints, the out of balance moment is a negligible value. In this case, the joints should be balanced and no carry-over moments are calculated.
Calculate the final moment at either end of each member.
This is the sum of all moments (including FEM) computed during the distribution cycles.
Nagaraju Kamaraj
Dec 25, 2011 @ 12:14 pm
good
Molay Reja
May 25, 2012 @ 20:14 pm
very good