Moment Distribution Method (FE Reference Handbook 10.5)

Moment distribution solves statically indeterminate beams and frames by locking every joint, then releasing them one at a time. The FE Reference Handbook 10.5 gives the three pieces you need in its Structural Analysis section: beam stiffness, the carryover factor and member fixed-end moments.

Beam Stiffness and Moment Carryover

A moment $M_{A}$ applied at end A of a prismatic member with end B fixed rotates end A by $θ$ :

M_{A} = \frac{4 E I}{L} θ = k_{A B} θ M_{B} = \frac{M _{A}}{2}

$k_{A B} = 4 E I / L$ is the flexural stiffness of the member. If the far end is pinned, use $3 E I / L$ instead.
The carryover factor from A to B is 1/2 (zero when the far end is pinned).

Distribution Factors

At a joint, an unbalanced moment is shared between the members in proportion to their stiffness:

D F_{i} = \frac{k _{i}}{\sum k}

The factors at a joint add up to 1. A fixed support takes no distributed moment (DF = 0) and a pinned end support keeps none (DF = 1).

Member Fixed-End Moments

The handbook lists the fixed-end moment magnitudes for the two standard load cases:

Load on member AB	FEM at A	FEM at B
Uniform load w over the full span	$\frac{w L ^{2}}{12}$	$\frac{w L ^{2}}{12}$
Point load P at distance a from A (b from B)	$\frac{P a b ^{2}}{L ^{2}}$	$\frac{P a ^{2} b}{L ^{2}}$

With a point load at midspan (a = b = L/2) both reduce to $P L /8$ . In the method, give them signs: here clockwise moments acting on the member end are positive, so a downward load gives a negative FEM at the left end and a positive FEM at the right end.

Procedure

Compute member stiffnesses and the distribution factors at each joint.
Lock all joints and compute the fixed-end moments.
At each free joint, find the unbalanced moment (the sum of member-end moments).
Distribute the negative of that moment to the members using the distribution factors.
Carry half of each distributed moment to the far end of the member (if that end is fixed).
Repeat until the unbalanced moments are negligible, then add each column to get the final end moments.

Worked Example

A two-span beam A–B–C has fixed supports at A and C and a support at B. Both spans are 6 m long with the same EI. Span AB carries a uniform load of 20 kN/m; span BC is unloaded. Find the end moments.

Distribution factors at B: both members have stiffness 4EI/6, so $D F_{B A} = D F_{B C} = 0.5$ .

Fixed-end moments: $w L^{2} /12 = 20 \times 6^{2} /12 = 60$ kN·m, so $F E M_{A B} = - 60$ and $F E M_{B A} = + 60$ kN·m. Span BC has no FEM.

Step	AB	BA	BC	CB
DF	0	0.5	0.5	0
FEM	−60	+60	0	0
Distribute (balance +60 at B)		−30	−30
Carry over (½)	−15			−15
Final (kN·m)	−75	+30	−30	−15

Joint B balances (+30 − 30 = 0), and A and C are fixed, so one cycle is exact. The beam has a 75 kN·m hogging moment at A, 30 kN·m over B and 15 kN·m at C. A slope-deflection check gives the same numbers.

To see the full shear and moment diagrams for a beam like this, model it in the beam calculator, which solves indeterminate beams directly with the stiffness method.

FAQ

Where is moment distribution in the FE Reference Handbook 10.5?

In the Civil Engineering chapter under Structural Analysis: "Beam Stiffness and Moment Carryover" (handbook page 268) and "Member Fixed-End Moments (Magnitudes)" (handbook page 271).

What is the carryover factor?

For a prismatic member with the far end fixed it is 1/2: a moment M applied at the near end induces M/2 at the far end. If the far end is pinned, nothing carries over.

What is the stiffness factor 4EI/L?

It is the moment needed to rotate the near end of a member through one radian when the far end is fixed. With the far end pinned it drops to 3EI/L.

When does the moment distribution method stop?

When the unbalanced moments at every joint are small enough to ignore. For joints next to fixed supports it often converges in one or two cycles.