Moment Distribution Method Notes

Last reviewed 16 Sept 2026 · 6 min read

The idea

The moment distribution method (Hardy Cross, 1930) solves indeterminate beams and frames by successive corrections instead of simultaneous equations.

Imagine every joint locked against rotation; loaded members develop fixed-end moments.
Release one joint at a time. The unbalanced moment at the joint is shared among the members meeting there in proportion to their stiffnesses (distribution).
Each distributed moment induces half of itself (for a fixed far end) at the far end of the member (carry-over).
Repeat until the carried-over moments are negligible.

Each step is an exact equilibrium statement, and the process converges quickly. It is a displacement method, equivalent to solving slope-deflection equations iteratively.

Definitions

DefinitionStiffness, distribution and carry-over

Stiffness $K$ — moment needed at the near end to rotate it by one radian.
- Far end fixed: $K = \frac{4 E I}{L}$
- Far end pinned/roller (simple end): $K = \frac{3 E I}{L}$
- Symmetric member in a symmetric frame (far end rotates equal and opposite): $K = \frac{2 E I}{L}$
- Antisymmetric case (far end rotates equally in the same sense): $K = \frac{6 E I}{L}$
Relative stiffness — use $I / L$ (fixed far end) and $\frac{3}{4} I / L$ (pinned far end) when $E$ is common.
Distribution factor at a joint: $D F_{i} = \frac{K _{i}}{\sum K}$ ; the factors at a joint add to 1. At a fixed support $D F = 0$ ; at a free pinned end $D F = 1$ .
Carry-over factor — ratio of the moment induced at the far end to the moment applied at the near end: $+ \frac{1}{2}$ when the far end is fixed; $0$ when the far end is pinned.

Sign convention and fixed-end moments

Use clockwise end moments positive (same as slope deflection). Fixed-end moments for a span loaded with a UDL $w$ : $- w L^{2} /12$ at the left end, $+ w L^{2} /12$ at the right end; central point load: $\mp W L /8$ ; eccentric load: $- W a b^{2} / L^{2}$ , $+ W a^{2} b / L^{2}$ .

Procedure

Compute stiffness and distribution factors at every joint.
Compute fixed-end moments.
Balance each joint: unbalanced moment $U$ = algebraic sum of end moments at the joint (plus any external moment); distribute $- U \times D F$ to each member end.
Carry over half of each distributed moment to the far end (if that end is fixed or continuous).
Repeat balancing and carrying over; stop when changes are small.
Sum each column to get final end moments; check that the moments at each internal joint balance.

Exam TipPinned outer ends

Two ways: (a) use stiffness $3 E I / L$ for the member and do not carry over to the pinned end after releasing it once; or (b) use $4 E I / L$ and balance the pinned end to zero in every cycle. Method (a) converges faster.

Overhangs

The moment at the support due to the overhang is statically known ( $=$ load × arm). Treat that support like a pinned end carrying a known external moment; the overhang itself has zero stiffness.

Frames with sway

If a frame can sway (unsymmetrical geometry or loading, lateral loads):

Non-sway analysis: prevent sway with an imaginary horizontal restraint; distribute as usual; find the restraint force $R$ from horizontal equilibrium of column shears.
Sway analysis: allow an arbitrary sway $Δ$ with joints locked. Fixed-end moments due to sway in each column $= - \frac{6 E I Δ}{L ^{2}}$ (with far end fixed) or $- \frac{3 E I Δ}{L ^{2}}$ (far end pinned). Choose convenient round values in the correct ratio; distribute; find the corresponding lateral force $S$ .
Combine: final moments $= M_{n o n - s w a y} + \frac{R}{S} M_{s w a y}$ (with signs so the restraint force is cancelled).

For columns of equal height and fixed feet, sway FEMs are in the ratio of $I / L^{2}$ .

Support settlement

A settlement $δ$ of one support produces fixed-end moments $- 6 E I δ / L^{2}$ at both ends of each affected span (sign by chord rotation). Distribute them like any other fixed-end moments.

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Moment Distribution Method