Moment Distribution Method

Why an iterative method at all?

A simply-supported beam is statically determinate: three equilibrium equations give the two reactions and you are done. Add an extra support, build the ends into walls, or join several beams over continuous supports, and the equilibrium equations alone can no longer find the reactions — the structure is statically indeterminate. The classical exact route is the slope-deflection method, which expresses each member-end moment in terms of the unknown joint rotations and then solves a system of simultaneous equations. For a two-span beam that means solving for one rotation; for a four-storey rigid frame it can mean inverting a 30×30 matrix by hand. Before computers, that was a day's work and an easy place to make a sign error.

Hardy Cross, an engineering professor at the University of Illinois, published the moment distribution method in a ten-page 1930 ASCE paper that is one of the most cited in structural engineering history. His insight: you do not have to solve the equations simultaneously. Instead, satisfy equilibrium at one joint at a time, accept that fixing one joint disturbs its neighbours, and iterate. The disturbances decay geometrically, so a handful of passes gives an answer good to a fraction of a percent. It turned a matrix-inversion problem into an arithmetic tableau a junior engineer could fill in with a slide rule.

The three ingredients

Every moment distribution analysis is built from three quantities. Get these right and the bookkeeping does the rest.

1. Fixed-end moments (FEM)
   The moments that develop at each member end if BOTH ends
   were perfectly clamped. Read from standard load tables, e.g.:
     UDL w over span L:        FEM = ± wL²/12
     Central point load P:     FEM = ± PL/8
     Off-centre load P at a,b: FEM = -Pab²/L²  and  +Pa²b/L²

2. Member stiffness K
     Far end fixed:   K = 4EI/L
     Far end pinned:  K = 3EI/L   (modified stiffness)

3. Distribution factor at a joint
     DF_i = K_i / ΣK        (the DFs at any joint sum to 1.0)

4. Carry-over factor (prismatic members)  = 1/2
   Apply M at the rotating near end → M/2 appears at the far
   fixed end, SAME sign.

Sign convention matters and is the single most common source of error. The usual structural convention takes clockwise moments acting on the member end as positive (the "BM" or rigid-frame convention). Pick one convention and never switch mid-table.

The procedure, step by step

1. Lock every joint. Imagine each joint clamped so no rotation can occur. Each loaded span now behaves as a fixed-fixed beam; write down its fixed-end moments.
2. Compute distribution factors at every joint that is free to rotate. At a fixed support DF = 0; at a free pin end DF = 1.0 for the single member.
3. Release one joint. Sum the fixed-end moments meeting there — the result is the unbalanced moment. To restore equilibrium, apply the negative of that sum, distributed among the members by their DFs.
4. Carry over. Send half of each distributed moment to the member's far end, keeping the sign.
5. Re-lock and move on. Clamp that joint again and repeat for the next joint. The carry-overs you just deposited become new unbalanced moments.
6. Iterate. Keep cycling through the joints. Each pass the unbalanced moments shrink. Stop when the distributions are smaller than your tolerance (often when they round to zero at the precision you care about).
7. Sum each column. The final end moment for each member end is the algebraic sum of its FEM plus all distributions and carry-overs accumulated in its column.

Worked example: two-span continuous beam

Consider a beam continuous over three supports A–B–C. Span AB = 6 m carries a uniformly distributed load of 20 kN/m; span BC = 4 m carries a 60 kN point load at its midpoint. Both spans have the same EI, and A and C are simple (pin/roller) supports while B is an interior support free to rotate. We will treat A and C as pins, so the AB and BC members use modified stiffness 3EI/L on the outer ends.

Fixed-end moments (clamp both ends first):
  Span AB, UDL 20 kN/m, L = 6 m:
     FEM_AB = -wL²/12 = -20·36/12 = -60 kN·m
     FEM_BA = +wL²/12 = +60 kN·m
  Span BC, point load 60 kN at centre, L = 4 m:
     FEM_BC = -PL/8 = -60·4/8 = -30 kN·m
     FEM_CB = +PL/8 = +30 kN·m

Because A and C are real pins, release them first (carry half
their released moment to B), or equivalently use the 3EI/L
modified stiffness so no carry-over ever returns from them.
Using modified stiffness at joint B:
  K_BA = 3EI/6 = 0.5 EI
  K_BC = 3EI/4 = 0.75 EI
  ΣK   = 1.25 EI
  DF_BA = 0.5/1.25  = 0.40
  DF_BC = 0.75/1.25 = 0.60

Adjust the FEMs at B for the released pins (carry over ½ of the
moment released at each pin):
  pin A releases +60 → +30 to BA:  FEM'_BA = +60 + 30 = +90
  pin C releases -30 → -15 to BC:  FEM'_BC = -30 - 15 = -45
  Unbalanced moment at B = +90 - 45 = +45 kN·m

Restore equilibrium: distribute -45 kN·m by DF:
     to BA: 0.40·(-45) = -18 kN·m
     to BC: 0.60·(-45) = -27 kN·m

With pins at A and C there is no carry-over back into B, so a
single balancing of joint B completes the analysis:
  M_BA = +90 - 18 = +72 kN·m
  M_BC = -45 - 27 = -72 kN·m   (equal & opposite, joint B
                                in balance; M_AB = M_CB = 0)

Because both outer supports are released pins, this single balance of joint B is exact — a slope-deflection solve of the same beam returns M_BA = +72 kN·m and M_BC = -72 kN·m identically. Notice the elegance: the stiffer span (BC, shorter and therefore higher 3EI/L) grabs the larger share — 60 percent — of the correction. Stiffness wins. A flexible span sheds its moment onto its stiffer neighbour, exactly as physical intuition demands.

Why it converges so fast

Each time you balance a joint and carry over, only half of the correction leaves the joint, and of that half only a fraction (set by the neighbour's distribution factor) bounces back on the next cycle. The residual unbalanced moments therefore form a geometric series with a ratio typically well below 0.25. After three cycles the residual is below 0.25³ ≈ 1.6 percent of the first imbalance; after four it is below 0.4 percent. In practice engineers stopped when the distributed increments dropped under about 1 kN·m, which usually meant three or four passes for a beam and a few more for a frame.

The geometric decay is also why fixed supports and pins behave so differently. A fixed support has DF = 0 — it never rotates, so it swallows carry-overs without ever sending a distribution back. A pin with the modified 3EI/L stiffness is even friendlier: it can be released once at the start and then ignored, because a pin cannot store moment to carry over. Both choices kill feedback loops and accelerate convergence.

Moment distribution vs. the alternatives

	Moment distribution (Hardy Cross)	Slope-deflection	Matrix stiffness (FEM software)	Three-moment theorem
Year / origin	1930, Hardy Cross	1914–1918, Maney	1950s–60s, computer era	1857, Clapeyron
Approach	Iterative relaxation, one joint at a time	Simultaneous equations in joint rotations	Assemble & invert global K matrix	Simultaneous equations in support moments
Solve by hand?	Yes — arithmetic tableau	Yes, but heavy for many joints	No (needs a computer)	Yes, for continuous beams only
Handles rigid frames	Yes (sway needs extra step)	Yes	Yes, any geometry	No — beams only
Handles sidesway	Awkward — superpose sway cases	Yes, directly	Yes, automatically	No
Exact or approximate	Converges to exact	Exact	Exact	Exact
Best for	Hand analysis, building intuition	Small frames, derivations	Everything in practice today	Continuous beams, quick checks

Member stiffness in detail

The rotational stiffness K is the moment needed to rotate a member end through unit angle while the far end is held in its prescribed condition. The two cases you meet constantly:

• Far end fixed: K = 4EI/L. Apply a unit rotation at the near end and a moment 4EI/L resists it, with 2EI/L (the carry-over) induced at the fixed far end — hence carry-over factor 2EI/L ÷ 4EI/L = ½.
• Far end pinned: K = 3EI/L. With the far end free to rotate, the member is softer in bending, so it absorbs less of any unbalanced moment. There is no moment at a pin, so the carry-over to it is zero.
• Symmetric / antisymmetric structures: you can analyse half the structure using modified stiffnesses (2EI/L for symmetric, 6EI/L for antisymmetric central members), halving the work for symmetric frames under symmetric load.
• Non-prismatic members: haunched beams and tapered columns have stiffness and carry-over factors that are no longer 4EI/L and ½; they come from published stiffness coefficient tables (e.g. the PCA handbooks for haunched concrete members).

Extending to rigid frames and sway

For a no-sway frame — one braced against horizontal translation, or symmetric under symmetric load — the procedure is identical to a continuous beam: columns simply add more members to each joint and contribute their own 4EI/L stiffness to ΣK. The distribution factors automatically split the joint moment between beams and columns by stiffness.

When a frame can sway, the joints translate as well as rotate, and translation produces additional fixed-end moments proportional to the chord rotation ψ = Δ/L. The standard treatment is superposition: (1) solve the structure restrained against sway, finding the artificial horizontal holding force; (2) release that force by applying an arbitrary sway, generating a second set of FEMs, and distribute again; (3) scale the sway solution so the sum of the two horizontal shears equals the real applied load, then add the two analyses. Every independent sway freedom adds one auxiliary distribution, which is why hand analysis of multi-storey frames becomes impractical and matrix software wins.

Pitfalls and failure modes of the method

• Sign-convention drift. The most common error by far. Mixing the clockwise-positive member convention with the sagging-positive beam convention corrupts every carry-over. Fix the convention before the first FEM is written.
• Forgetting to release real pins. Using 4EI/L for a span whose far end is a genuine pin overstates its stiffness and gives the wrong distribution factors. Use the 3EI/L modified stiffness or carry over once and re-balance the pin to zero.
• Stopping too early. Truncating after one cycle on a frame with several flexible joints can leave several percent of error. Watch the magnitude of the distributed increments, not just the cycle count.
• Ignoring sway. Applying the no-sway tableau to an unbraced frame silently omits the sidesway moments, which can dominate under lateral wind or seismic load. Always check whether the frame can translate.
• Non-prismatic members with default factors. Haunched members do not have a ½ carry-over; using the prismatic value introduces a systematic bias toward the wrong end.
• Axial and shear deformation neglected. Like slope-deflection, moment distribution assumes flexure dominates. For short, deep members or trusses where axial effects matter, the bending-only model under-predicts deflections.

Legacy and where it stands today

From 1930 until roughly 1970, moment distribution was how continuous bridges, building frames, and ringbeams were actually designed — by hand, on lined paper, in a tableau that fit on one sheet. The matrix stiffness method, identical in its physics, replaced it for production work the moment computers became cheap, because a machine can invert the global stiffness matrix exactly and handle sway, three dimensions, and thousands of members without fatigue. Yet moment distribution survives in every structures curriculum, because nothing else makes the redistribution of moment so tangible: you can see the stiffer member grab the larger share, watch the corrections decay, and feel why a fixed support stiffens a span. It is the method that teaches the matrix method's answer before you ever build the matrix.