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This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • An independent structural reviewer must check the illustrated models, prediction feedback, geometry and full interactive input ranges.
  • The stated 4EI/L, 3EI/L and carry-over factors apply to prismatic members with the specified far-end restraints and no chord translation. Sway needs additional treatment.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

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Module 13 · Course notes

Moment distribution: balancing joints by hand

Moment distribution is the stiffness method carried out by hand, one joint at a time. It was the workhorse of design offices for decades, and it is still the quickest way to feel how a frame shares moment. You lock the joints, look at the out-of-balance, release a joint and share the correction by stiffness, carry a little over to the far ends, and repeat until it settles.

Begin with restrained joint rotations, then release them in stages to restore joint equilibrium.

1. Lock every joint so no rotation can occur. Each member is fixed-ended, carrying its fixed-end moments. 2. At a real joint the member moments will not sum to zero — that out-of-balance is held by the imaginary clamp. 3. Release the joint. The out-of-balance is shared among the members in proportion to their stiffness — the stiffer member takes more. 4. Carry over: when one end of a member rotates, a moment appears at its far end — half of the distributed moment, for a fixed far end. 5. Repeat. The carried-over moments unbalance the neighbours, so go round again. Continue until the remaining joint imbalance is small enough for the chosen accuracy; the number of cycles depends on the structure.

bending moment (tension side)

the locked state — fixed-end moments
Lock every joint and each span is fixed-ended, carrying wL²/12 hogging at each end. These are the starting moments; releasing the joints and balancing them corrects the diagram to the real one.

Two quantities run the arithmetic. The distribution factor at a joint is a member's share of the total stiffness there, DF = K / ΣK, and the factors at any joint must sum to 1 — the out-of-balance has nowhere else to go. The carry-over factor is ½ towards a fixed far end and 0 towards a pinned one (there is nothing at a pin to carry to).

A member's stiffness is 4EI/L with a fixed far end, or 3EI/L with a pinned one — so a member with a pinned far end is three-quarters as stiff, and attracts a smaller share.

bending moment (tension side)

the balanced result
Distributing and carrying over until the joints balance yields the real continuous-beam moment: hogging over the central support, sagging in the spans — the same answer the stiffness method gives, reached by hand.

Worked example

Worked example — distribute at one joint

Two members meet at a rigid joint, both with fixed far ends: one 6 m long, one 4 m, the same EI. Share an out-of-balance moment there.

  1. Step 1 — distribution factors

    Stiffness is 4EI/L, so K₆ = 0.667 EI and K₄ = 1.000 EI, total 1.667 EI. The distribution factors are 0.40 and 0.60 — they sum to 1.0, and the shorter, stiffer member takes the larger share.

    bending moment (tension side)

  2. Step 2 — distribute and carry over

    A 50 kNm out-of-balance splits as 20 kNm to the 6 m member and 30 kNm to the 4 m member. Half of each carries over to the fixed far end — 10 kNm and 15 kNm. The shorter member receives the larger balancing moment because its rotational stiffness is greater.

    bending moment (tension side)

In the exercises, compute distribution factors, apply carry-over, and connect the hand method to the stiffness method it mimics.

How to read these problems

The three-step method

  1. 1Points of certainty. Match the displacement and rotation that each support prescribes, including any settlement. Infer the displacement direction from the whole structure and load case; it need not be downward at every loaded point.
  2. 2Deflected shape and reaction directions. Sketch the compatible deflected shape. To find a reaction's direction, imagine removing that support: the direction that pushes the structure back to its place is the reaction's sense (it may be a hold-down).
  3. 3Bending moment and contraflexure. Draw the bending-moment diagram on the tension side and check it against the shape: hogging where the curve is convex-up, sagging where convex-down, zero at pins and at every contraflexure.

Rules that must always hold

  • 1.Moment is zero at a genuine moment release and at an unloaded simple end. A continuous member can carry moment over a translational support.
  • 2.A bending-moment diagram crosses the baseline exactly at a point of contraflexure.
  • 3.Under a distributed load the bending-moment diagram is curved; under point loads alone it is straight lines.
  • 4.At a fully fixed support the deflected shape leaves the support with zero rotation (tangent along the member).
  • 5.For an initially straight elastic beam without imposed curvature, zero curvature implies zero bending moment.
  • 6.The moment is drawn on the tension side: sagging below the member, hogging above it.

Now predict for yourself

Moment distribution exercises

5 exercises on distribution factors, carry-over, the starting fixed-end moments and the link to the stiffness method. Predict, then reveal an explanation.

Start the exercises →

This lesson is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions.