Moment Distribution Method

1. The Core Concept — Lock, Load, Unlock, Distribute

The Moment Distribution Method is built on a simple physical idea. Imagine all joints in an indeterminate structure are initially locked against rotation — every joint is temporarily fixed. When loads are applied to the locked structure, Fixed End Moments (FEMs) develop at both ends of every loaded member, just as in a fully fixed beam. At most joints, the FEMs from different members do not balance — there is an out-of-balance moment at the joint.

Now, joints are unlocked one at a time. When a joint is released, it rotates until the moments from all connected members balance (ΣM = 0). This rotation distributes the out-of-balance moment among the connected members in proportion to their stiffnesses — stiffer members attract more moment. Simultaneously, rotating one joint end of a member induces a moment at the far end of that member — this is the carry-over effect.

Carrying over moments to far ends changes the balance at those joints — so they need to be released again. The process of releasing joints, distributing moments, and carrying over is repeated iteratively. Each cycle brings the structure closer to equilibrium. For most practical structures, three to five cycles are sufficient to achieve the accuracy required for exam problems.

The elegance of MDM is that it replaces the simultaneous equations of the Slope-Deflection Method with a simple tabular calculation that converges rapidly. Once the distribution factors and carry-over factors are established, the entire solution is a table of additions and multiplications — no algebra required.

2. Key Terms — Stiffness, DF, COF, FEM

3. Distribution Factor — Calculation & Rules

Calculating distribution factors correctly before starting the iterative table is the single most important preparation step. An error in DF invalidates the entire solution.

Procedure to calculate DFs at a joint:

1. List all members meeting at the joint.
2. For each member, identify the far end condition (fixed/rigid or pin/roller/free).
3. Calculate K for each member: 4EI/L (far end fixed/rigid) or 3EI/L (far end pin/roller).
4. Sum all K values at the joint: ΣK.
5. DF for each member = K_member / ΣK.
6. Verify: sum of all DFs at the joint = 1.0.

Effect of relative EI on DFs: If EI is the same for all members, it cancels from the DF calculation and only the L values (and the 3 vs 4 factor) matter. If members have different EI values, the actual EI must be included. For GATE CE problems, EI is usually uniform — but always check the problem statement.

4. Step-by-Step Procedure

For no-sway beams and frames (the standard case):

1. Step 1 — Calculate Fixed End Moments (FEM): For every loaded member, compute FEM at both ends using the standard FEM formulae. Use the sign convention: clockwise moments on the member end are positive (or use the hogging = positive convention — be consistent). FEM at unloaded members = 0.
2. Step 2 — Calculate member stiffnesses (K): For every member at every joint, compute K = 4EI/L or K = 3EI/L depending on the far end condition.
3. Step 3 — Calculate Distribution Factors (DF): At every joint, DF = K / ΣK for each member. Verify ΣDF = 1.0 at each joint.
4. Step 4 — Set up the distribution table: Create a table with one column per member end. Enter the FEMs in the first row.
5. Step 5 — Balance the first joint: Find the out-of-balance moment at a joint = −ΣFEMs at that joint. Distribute this moment to each member in proportion to the DF. The distributed moment for each member = DF × (out-of-balance moment). Write these in the next row.
6. Step 6 — Carry over: For each distributed moment, carry over COF × distributed moment to the far end of the same member. Write these carry-over values in the same row, at the far end column.
7. Step 7 — Repeat for all other joints: Move to the next unbalanced joint. Compute the new out-of-balance moment (including any carry-over moments just received). Distribute and carry over. Continue cycling through all joints.
8. Step 8 — Stop when convergence is achieved: Continue cycles until the carry-over moments are negligibly small (typically less than 0.5 kN·m for exam accuracy). In most GATE CE problems, 3–4 cycles are sufficient.
9. Step 9 — Sum each column: The final end moment at each member end = sum of all values in that column (FEM + all distribution rows + all carry-over rows).
10. Step 10 — Check: At every internal joint, verify ΣM = 0. At fixed ends, the moment is whatever it summed to. At pin/roller ends, the moment must be zero (or very close to zero after convergence).

5. Fixed End Moments — Reference Table

Sign convention reminder: In the standard MDM table convention, a positive FEM means the moment acts clockwise on the member end (which corresponds to hogging at that end for a horizontal beam under downward loads). Some textbooks use the sagging-positive convention — always state which convention you are using and apply it consistently throughout.

6. Worked Example 1 — Two-Span Beam with Simply Supported Ends

Problem: A two-span continuous beam ABC has: A = pin support, B = intermediate roller, C = pin support. Span AB = 4 m carries UDL of 20 kN/m. Span BC = 6 m carries central point load of 60 kN. EI is constant. Find all end moments using MDM.

7. Worked Example 2 — Two-Span Beam with Fixed End

Problem: Continuous beam ABC: fixed at A, roller at B (intermediate), roller at C (end). Span AB = 5 m carries UDL of 10 kN/m. Span BC = 4 m carries point load 40 kN at 1 m from B. EI is constant throughout. Find all end moments.

8. Worked Example 3 — Non-Sway Portal Frame

Problem: A symmetric portal frame ABCD: columns AB and CD each 4 m tall with fixed bases at A and D. Beam BC = 6 m with rigid joints at B and C. UDL of 15 kN/m on beam BC only. No sway (symmetric). EI constant. Find all end moments.

9. Sway Frames — Two-Cycle Method

When a frame can sway (horizontal joint displacement), the standard no-sway MDM gives incorrect results because it ignores the chord rotation in the columns. The sway correction uses the principle of superposition — the actual solution is the sum of a no-sway solution and a sway correction solution scaled by a factor k.

Two-step procedure for sway frames:

1. Step 1 — No-sway analysis: Artificially prevent sway by adding a temporary prop at the sway level (e.g., a horizontal force at the beam level). Carry out the full MDM analysis as if no sway occurs. After finding all end moments, calculate the horizontal force P in the temporary prop using the shear equation (horizontal equilibrium of the storey).
2. Step 2 — Sway-only analysis: Remove all applied loads and remove the temporary prop. Apply an arbitrary unit sway displacement Δ = 1 (or any convenient value) to the frame by applying chord rotations ψ = Δ/h to all columns. This introduces FEMs in the columns: FEM = ±6EIψ/L = ±6EIΔ/(hL). Run the full MDM analysis for this sway-only case. After finding all end moments, calculate the horizontal restoring force P’ (the prop force needed to hold the sway in that configuration).
3. Step 3 — Combine: The actual frame is the no-sway case (Step 1) plus k times the sway case (Step 2), where k = −P/P’. The actual end moments = (no-sway moments) + k × (sway moments).

10. Worked Example 4 — Sway Portal Frame

Problem: Portal frame ABCD: columns AB and CD, each 4 m, fixed at A and D. Beam BC = 6 m, rigid joints at B and C. Horizontal load H = 24 kN at joint B (to the right). No vertical load. EI constant. Find all end moments by MDM.

11. Effect of Support Settlement

Support settlement introduces additional FEMs in any member connected to the settling support. For a member AB where end B settles by δ (downward) while end A remains in place:

12. Common Mistakes Students Make

• Using K = 4EI/L for a member with a pin at the far end: If the far end of a member is a pin or roller support, the correct stiffness is K = 3EI/L, not 4EI/L. Using 4EI/L overestimates the stiffness, gives wrong distribution factors, and produces incorrect end moments throughout. This error is very common because students apply the 4EI/L formula universally without checking the far end condition. Always check: is the far end a fixed/rigid joint (use 4EI/L) or a pin/roller (use 3EI/L)?
• Distribution factors not summing to 1.0 at a joint: The sum of all distribution factors at any joint must equal exactly 1.0 — no exceptions. If your DFs at a joint do not sum to 1.0, the stiffness calculation is wrong. Recheck the K values and the sum before proceeding — an error here propagates through every cycle.
• Applying distribution at a fixed end: A fixed support has DF = 0 — it never distributes any moment, regardless of what carry-over moments arrive there. Students sometimes distribute at fixed ends thinking the moments must balance there. The fixed end absorbs carry-over moments permanently (it is the wall resisting rotation) — no further distribution or carry-over comes from a fixed end.
• Wrong carry-over direction or sign: The carry-over moment has the same sign as the distributed moment in the standard MDM convention (both clockwise or both anticlockwise). Some students carry over with the opposite sign. A consistent sign error here causes the moments to diverge rather than converge during iteration, which is a clear signal something is wrong.
• Stopping iteration too early: The Moment Distribution Method is iterative — it converges to the exact answer only after infinite cycles (though accuracy is typically sufficient after 3–5 cycles for practical problems). Stopping after just one cycle gives an approximate answer that may differ significantly from the correct one for structures where carry-overs are large (e.g., when a fixed end receives large carry-over moments, which then influence the next distribution cycle significantly).

13. Frequently Asked Questions

Why does the Moment Distribution Method converge?

Each cycle of distribution and carry-over redistributes a fraction of the remaining out-of-balance moment. The carry-over factor of 0.5 means each cycle reduces the residual out-of-balance moment by at least a factor of 0.5 × (maximum DF). Since DF ≤ 1 and COF = 0.5, the residual moment decreases geometrically each cycle — it is a convergent geometric series. Mathematically, MDM converges to the exact solution because it is equivalent to solving the slope-deflection equations by a successive approximation (Gauss-Seidel iteration). The speed of convergence depends on the distribution factors — structures with very stiff members at some joints and very flexible at others converge more slowly.

How do you handle a beam with an internal hinge in MDM?

An internal hinge introduces a condition that the moment at the hinge location is zero. In MDM, treat the portion of the beam beyond the internal hinge as a separate “span” with the hinge end as a pin support (moment = 0 there). Use K = 3EI/L for any member whose far end is the internal hinge (since the hinge cannot transmit moment — exactly like a pin support). The carry-over factor to the hinge end is zero. This effectively splits the problem — the hinge creates an “end condition” that simplifies the distribution.

What is the difference between distribution factor and stiffness factor?

The stiffness factor K (= 4EI/L or 3EI/L) is an absolute measure of a member’s rotational resistance. The distribution factor DF is a relative measure — it normalises the stiffness of each member at a joint by the total stiffness at that joint. DF = K / ΣK. The stiffness factor tells you how stiff a member is; the distribution factor tells you what fraction of the unbalanced moment that member will attract. Two members with very different absolute stiffnesses can have the same distribution factor if they are in the same ratio to the total joint stiffness.

Can MDM be used for structures with varying EI?

Yes — the only change is in the stiffness calculation. For a member with varying EI, compute the effective stiffness using the actual stiffness integral (rather than 4EI/L or 3EI/L which assume constant EI). The carry-over factor also changes for non-prismatic members. Once the correct K and COF are established for each member, the MDM procedure (distribution table, iteration, summation) proceeds identically. For GATE CE, EI is almost always constant unless the problem explicitly states otherwise — in that case, treat the given EI values carefully in the K calculation.