The Moment Distribution Method, developed by Hardy Cross in 1930, is an iterative technique for analysing statically indeterminate structures — continuous beams and frames. It remains the most important hand-calculation method in structural analysis and is a high-weightage topic in GATE and UPSC ESE Civil Engineering.
Basic Concepts
Stiffness (K)
- Far end fixed: K = 4EI/L
- Far end pinned/free: K = 3EI/L
- Far end guided (no rotation, sway only): K = EI/L
Distribution Factor (DF)
At a joint where members meet:
DF_member = K_member / ΣK_all members at joint
ΣDF at any joint = 1.0 (except at fixed support where DF = 0)
Carry-Over Factor (COF)
- Far end fixed: COF = 0.5 (half the distributed moment carries over)
- Far end pinned: COF = 0 (no carry-over; modified stiffness 3EI/L used)
Fixed End Moments (FEM)
| Loading Condition | FEM_AB | FEM_BA |
|---|---|---|
| UDL over full span (w per unit length) | −wL²/12 | +wL²/12 |
| Point load P at midspan | −PL/8 | +PL/8 |
| Point load P at distance a from A (b from B) | −Pab²/L² | +Pa²b/L² |
| UDL over left half only (w per unit) | −5wL²/96 | +11wL²/192 |
| Settlement of support B by Δ | −6EIΔ/L² | −6EIΔ/L² |
Sign convention: Clockwise moment = positive; anti-clockwise = negative (or use your consistent sign convention throughout).
Procedure — Moment Distribution for Continuous Beams
- Calculate stiffness K for each span (note far-end condition)
- Calculate distribution factors at each internal joint
- Calculate fixed end moments for each span under applied loading
- Unbalance at each joint = sum of FEMs at the joint; joints are "unlocked" one at a time
- Distribute the unbalance proportionally to DFs; carry over half to far fixed ends
- Repeat until carry-over moments are negligible (within 1% of original FEM)
- Sum all moments at each end of each member for final end moments
- Draw BMD and SFD using final end moments
Worked Example — Two-Span Continuous Beam
Problem
Beam ABC: AB = 6 m (UDL 20 kN/m), BC = 4 m (Point load 40 kN at midspan). A = fixed, C = pinned. EI = constant.
Step 1: Stiffness
K_AB = 4EI/6 = 0.667EI (far end A = fixed)
K_BC = 3EI/4 = 0.75EI (far end C = pinned → use 3EI/L)
Step 2: Distribution Factors at B
DF_BA = 0.667EI / (0.667EI + 0.75EI) = 0.667/1.417 = 0.471
DF_BC = 0.75EI / 1.417 = 0.529
Step 3: FEM
FEM_AB = −wL²/12 = −20 × 36/12 = −60 kN·m
FEM_BA = +60 kN·m
FEM_BC = −PL/8 = −40 × 4/8 = −20 kN·m
FEM_CB = +20 kN·m → but C is pinned; initially treat as fixed
Step 4: Release Pinned End C
Modified approach: Since C is pinned, modify K_BC = 3EI/L and FEM_BC adjusted:
Modified FEM_BC = −20 − (20/2) × 1 = −20 + 10 = −10 kN·m (release half FEM_CB to B)
FEM_CB = 0 (pinned end released)
Step 5: Moment Distribution Table
| Joint | A (fixed) | B left (AB) | B right (BC) | C (pinned) |
|---|---|---|---|---|
| DF | 0 | 0.471 | 0.529 | — |
| FEM | −60 | +60 | −10 | 0 |
| Balance B | — | −23.6 | −26.4 | — |
| CO to A | −11.8 | — | — | — |
| Balance B (round 2) | — | 0 | 0 | — |
| Final M | −71.8 | +36.4 | −36.4 | 0 |
Check: M_BA + M_BC = 36.4 − 36.4 = 0 ✓ (equilibrium at B)
Sway Analysis of Frames
Frames with asymmetric loading or asymmetric geometry sway (translate horizontally). Sway introduces additional moments proportional to the sway magnitude.
No-Sway Analysis + Sway Correction
- Restrain the frame against sway (add artificial horizontal restraint)
- Perform moment distribution for applied loads (no-sway moments, M₀)
- Find the artificial restraint force R in the restraint
- Remove the restraint; apply equal and opposite force R to the frame
- Assume a unit sway — calculate FEMs from sway (FEM = 6EI Δ/L²)
- Distribute these sway FEMs; find the horizontal force corresponding to unit sway
- Scale the sway moments so horizontal force = R
- Superpose no-sway + sway moments for final moments
Slope Deflection Method (Alternative)
For comparison — slope deflection equations:
M_AB = 2EI/L × (2θ_A + θ_B − 3ψ) + FEM_AB
M_BA = 2EI/L × (2θ_B + θ_A − 3ψ) + FEM_BA
where ψ = Δ/L (chord rotation); solve simultaneous equations for θ and Δ, then back-calculate moments.
Kani's Method
Rotation contribution method — a modified moment distribution approach that directly gives rotation contributions; faster for multi-storey frames. Used in ESE advanced problems but not as common as Hardy Cross.
Frequently Asked Questions
When should the modified stiffness (3EI/L) be used instead of 4EI/L?
Modified stiffness K = 3EI/L is used when the far end of a member is a pin or roller support (free to rotate). In this case, no moment is carried over to the far end (COF = 0). Using 3EI/L eliminates the need to balance and carry over to the pin end, reducing the number of distribution cycles. It cannot be used for interior joints of a continuous beam — only at the boundary pinned ends.
How do you know when the moment distribution has converged?
The iteration converges when the largest carry-over moment in any cycle is less than ~1% of the largest FEM (or less than 0.5–1 kN·m for practical problems). Typically 3–5 cycles are sufficient for most beams; frames with sway may need more. The key check is that the sum of moments at each joint equals zero after each round of distribution.