Principle of moment distribution method

Principle

The moment distribution method (Hardy Cross, 1930) is an iterative method, based on the slope-deflection equations, for analysing continuous beams and frames without solving simultaneous equations. The idea is:

1. Lock every joint against rotation. Each loaded member then behaves as a fixed-ended beam and develops fixed-end moments (FEM). At a joint these are generally unequal, so there is an unbalanced moment = $\sum$FEM of the members meeting at the joint.
2. Release one joint at a time. It rotates until the joint is in equilibrium. The balancing moment (equal to the unbalanced moment with reversed sign) is distributed to the members in proportion to their stiffness.
3. A part of each distributed moment is carried over to the far end of the member (if the far end is fixed against rotation).
4. The carried-over moments disturb the neighbouring joints, so the balancing and carry-over are repeated until the carry-over moments are negligibly small.
5. The final end moment = FEM + all distributed moments + all carried-over moments.

Equilibrium of every joint and slope compatibility of the members are therefore satisfied when the iteration converges.

Distribution factor (DF)

The fraction of the unbalanced moment taken by a member at a joint:

$$DF_{jk}=\frac{K_{jk}}{\sum K},\qquad K=\frac{4EI}{L}\ (\text{far end fixed}),\quad K=\frac{3EI}{L}\ (\text{far end hinged/roller})$$

$\sum DF=1$ at every joint. For a fixed support $DF=0$ (it takes no rotation); for a simple end support with no member beyond it, $DF=1$.

Carry-over factor (COF)

A moment $M$ applied at the near end of a member whose far end is fixed produces $M/2$ at the far end, so $COF=\tfrac12$. It is 0 if the far end is hinged or roller (no moment can develop there).

Example

Beam ABC: AB = 6 m, BC = 4 m, EI constant; A is fixed, B and C are simple supports; AB carries 10 kN/m. Since the far end C is a simple (hinged) support, the stiffness of BC is reduced by the factor 3/4.

Distribution factors at B: $K_{BA}=\dfrac{EI}{6}=0.1667EI$, $K_{BC}=\dfrac{3}{4}\cdot\dfrac{EI}{4}=0.1875EI$ (relative values). $DF_{BA}=0.471$, $DF_{BC}=0.529$.

Fixed-end moments: $FEM_{AB}=-\dfrac{wL^2}{12}=-\dfrac{10\times6^2}{12}=-30$ kNm, $FEM_{BA}=+30$ kNm; BC has no load. (C is hinged: $FEM_{BC}=FEM_{CB}=0$.)

Distribution:

• Unbalanced moment at B = $+30$ kNm; balancing moment = $-30$ kNm.
• $BA$ takes $-30\times0.471=-14.12$; $BC$ takes $-30\times0.529=-15.88$.
• Carry-over to A: $\tfrac12\times(-14.12)=-7.06$. No carry-over to C (hinged).
• Joint B is now balanced, and A (fixed) and C (hinged, already at zero moment) need no balancing, so the process ends in one cycle.

Joint check: $M_{BA}+M_{BC}=0.00$ ✓. The final moments equal the exact slope-deflection result, hence the method gives exact answers for this beam (several cycles are needed only when there are more joints to be balanced).