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Moment distribution

Intermediate Structural Analysis Reviewed 24 Oct 2026 2 min read

Hardy Cross's moment distribution method solves continuous beams and frames by successive corrections, without ever writing simultaneous equations. Imagine every joint first locked against rotation, so each span carries its fixed-end moments. Then release the joints one at a time. The unbalanced moment at a released joint is shared among the members meeting there according to their relative stiffness, and half of each share is carried over to the far end of the member. Repeat until the corrections are small.

The member stiffness factor is 4EI/L for a far end fixed and 3EI/L for a far end pinned (with no carry-over to the pin). Distribution factors at a joint are each member's stiffness divided by the total, and they add to one. The carry-over factor is one half for a prismatic member with a fixed far end. Sway in frames is handled by a second distribution with an arbitrary sway, then scaling to satisfy storey shear.

PR_AR_B
Diagram: Indiacon original, not to scale.

The method converges quickly for ordinary beams, gives designers a feel for how moments flow towards stiffer members, and is still useful for checking software output. It also explains why continuous slabs and beams have hogging moments over supports and smaller sagging moments in spans than simply supported ones.

Key formulas
  • Stiffness: k = 4EI/L (far end fixed); 3EI/L (far end pinned)
  • Distribution factor: DF = k / Σk at the joint
  • Carry-over factor: 1/2 to a fixed far end, 0 to a pinned end
  • Fixed-end moments: UDL wL²/12; point load at centre PL/8
Common mistake

Carrying over to a pinned end, which should receive nothing.

Solved example

Problem. A beam ABC: AB = 4 m, BC = 6 m, same EI, A and C fixed, B a support. AB carries 30 kN/m, BC carries 10 kN/m. Find the moment at B.

  1. FEMs: AB ±40 kNm (30 × 16/12); BC ±30 kNm (10 × 36/12).
  2. Stiffness at B: AB 4EI/4 = EI; BC 4EI/6 = 0.667EI. DFs: 0.6 and 0.4.
  3. Unbalance at B: +40 (from BA) − 30 (BC) = 10 kNm. Distribute −6 to BA and −4 to BC.
  4. M_BA = 40 − 6 = 34 kNm; M_BC = −30 − 4 = −34 kNm. Carry-over: −3 to A, −2 to C.

Answer. Moment at B ≈ 34 kNm hogging; M_AB ≈ −43 kNm and M_CB ≈ +28 kNm after carry-over

Quick quiz

1What do the distribution factors at a joint add to?

One.

2Carry-over factor to a fixed far end?

One half.

3Stiffness of a member with its far end pinned?

3EI/L.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
  • C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
  • Devdas Menon, Structural Analysis, Narosa, latest edition.
  • A. K. Chopra, Dynamics of Structures, Pearson, latest edition.

Free and official sources

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