Moment distribution method

The moment distribution method (not to be confused with moment redistribution) is a structural analysis method for statically indeterminate beams and frames developed by Hardy Cross. It was published in 1930 in an ASCE journal.[1] The method only accounts for flexural effects and ignores axial and shear effects. From the 1930s until computers began to be widely used in the design and analysis of structures, the moment distribution method was the most widely practiced method.

Contents

Introduction

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.

Implementation

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

When a joint is released and begins to rotate under the unbalanced moment, resisting forces develop at each member framed together at the joint. Although the total resistance is equal to the unbalanced moment, the magnitudes of resisting forces developed at each member differ by the members' flexural stiffness. Distribution factors can be defined as the proportions of the unbalanced moments carried by each of the members. In mathematical terms, distribution factor of member k framed at joint j is given as:

D_{jk} = \frac{\frac{E_k I_k}{L_k}}{\sum_{i=1}^{i=n} \frac{E_i I_i}{L_i}}

where n is the number of members framed at the joint.

Carryover factors

When a joint is released, balancing moment occurs to counterbalance the unbalanced moment which is initially the same as the fixed-end moment. This balancing moment is then carried over to the member's other end. The ratio of the carried-over moment at the other end to the fixed-end moment of the initial end is the carryover factor.

Determination of carryover factors

Let one end (end A) of a fixed beam be released and applied a moment M_A while the other end (end B) remains fixed. This will cause end A to rotate through an angle \theta_A. Once the magnitude of M_B developed at end B is found, the carryover factor of this member is given as the ratio of M_B over M_A:

C_{AB} = \frac{M_B}{M_A}

In case of a beam of length L with constant cross-section whose flexural rigidity is EI,

M_A = 4 \frac{EI}{L} \theta_A %2B 2 \frac{EI}{L} \theta_B = 4 \frac{EI}{L} \theta_A
M_B = 2 \frac{EI}{L} \theta_A %2B 4 \frac{EI}{L} \theta_B = 2 \frac{EI}{L} \theta_A

therefore the carryover factor

C_{AB} = \frac{M_B}{M_A} = \frac{1}{2}

Sign convention

Once a sign convention has been chosen, it has to be maintained for the whole structure. The traditional engineer's sign convention is not used in the calculations of the moment distiribution method although the results can be expressed in the conventional way.

Framed structures

Framed structures with or without sidesway can be analysed using the moment distribution method.

Example

The statically indeterminate beam shown in the figure is to be analysed.

In the following calcuations, counterclockwise moments are positive.

Fixed end moments

M _{AB} ^f = \frac{Pb^2a }{L^2} = \frac{10 \times 7^2 \times 3}{10^2} = %2B 14.700 \ kN\cdot m
M _{BA} ^f = - \frac{Pa^2b}{L^2} = - \frac{10 \times 3^2 \times 7}{10^2} = - 6.300 \ kN\cdot m
M _{BC} ^f = \frac{qL^2}{12} = \frac{1 \times 10^2}{12} = %2B 8.333 \ kN\cdot m
M _{CB} ^f = - \frac{qL^2}{12} = - \frac{1 \times 10^2}{12} = - 8.333 \ kN\cdot m
M _{CD} ^f = \frac{PL}{8} = \frac{10 \times 10}{8} = %2B 12.500 \ kN\cdot m
M _{DC} ^f = - \frac{PL}{8} = - \frac{10 \times 10}{8} = - 12.500 \ kN\cdot m

Flexural stiffness and distribution factors

The flexural stiffness of members AB, BC and CD are \frac{3EI}{L}, \frac{4\times 2EI}{L} and \frac{4EI}{L}, respectively. Therefore:

D_{BA} = \frac{\frac{3EI}{L}}{\frac{3EI}{L}%2B\frac{4\times 2EI}{L}} = \frac{\frac{3}{10}}{\frac{3}{10}%2B\frac{8}{10}} = 0.2727
D_{BC} = \frac{\frac{4\times 2EI}{L}}{\frac{3EI}{L}%2B\frac{4\times 2EI}{L}} = \frac{\frac{8}{10}}{\frac{3}{10}%2B\frac{8}{10}} = 0.7273
D_{CB} = \frac{\frac{4\times 2EI}{L}}{\frac{4\times 2EI}{L}%2B\frac{4EI}{L}} = \frac{\frac{8}{10}}{\frac{8}{10}%2B\frac{4}{10}} = 0.6667
D_{CD} = \frac{\frac{4EI}{L}}{\frac{4\times 2EI}{L}%2B\frac{4EI}{L}} = \frac{\frac{4}{10}}{\frac{8}{10}%2B\frac{4}{10}} = 0.3333

The distribution factors of joints A and D are D_{AB} = 1 and D_{DC} = 0 .

Carryover factors

The carryover factors are  \frac{1}{2} , 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 balanced moments; arrows ( → / ← ) represent the carry-over of moment from one end to the other end of a member.

As joint A is released, balancing moment of magnitude equal to the fixed end moment M_{AB}^{f} = 14.700 \mathrm{\,kN \,m} develops and is carried-over from joint A to joint B.
The unbalanced moment at joint B now is the summation of the fixed end moments M_{BA}^{f}, M_{BC}^{f} and the carry-over moment from joint A. This unbalanced moment is distributed to members BA and BC in accordance with the distribution factors D_{BA} = 0.2727 and D_{BC} = 0.7273. Step 2 ends with carry-over of balanced moment M_{BC}=3.867 \mathrm{\,kN \,m} to joint C. Joint A is a roller support which has no rotational restraint, so moment carryover from joint B to joint A is zero.
The unbalanced moment at joint C now is the summation of the fixed end moments M_{CB}^{f}, M_{CD}^{f} and the carryover moment from joint B. As in the previous step, this unbalanced moment is distributed to each member and then carried over to joint C and back to joint B. Joint D is a fixed support and carried-over moments to this joint will not be distributed nor be carried over to joint C.
Joint B still has unbalanced moment which was carried over from joint C in step 3. Joint B is released once again to induce moment distribution and to achieve equilibrium.
Joints are released and fixed again until every joint has unbalanced moments of size zero or neglectably small in required precision. Arithmetically summing all moments in each respective columns gives the final moment values.

Result

M_A = 0 \ kN \cdot m
M_B = -11.569 \ kN \cdot m
M_C = -10.186 \ kN \cdot m
M_D = -13.657 \ kN \cdot m
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.

M_A = 0 \ kN \cdot m
M_B = -11.569 \ kN \cdot m
M_C = -10.186 \ kN \cdot m
M_D = -13.657 \ kN \cdot m

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.

Notes

  1. ^ Cross, Hardy (1930). "Analysis of Continuous Frames by Distributing Fixed-End Moments". Proceedings of the American Society of Civil Engineers (ASCE): pp. 919–928 

References

See also