Moment Distribution Method: A Step-by-Step Guide with Solved Examples
Structural analysis is the bedrock of safe and efficient civil engineering design. For statically indeterminate structures, where basic equilibrium equations are not enough, we need more advanced techniques. The moment distribution method (MDM) is a powerful, classical technique for analyzing these complex structures. Developed by Professor Hardy Cross in 1930, it provides an intuitive, iterative approach to solving indeterminate beams and frames.
This comprehensive guide will walk you through everything you need to know. We will start with the core concepts, such as stiffness and distribution factors. Then, we will detail the step-by-step process in a clear, easy-to-follow format. You will learn to apply the moment distribution method using detailed solved examples for both continuous beams and frames. By the end, you will not only understand the procedure but also the engineering logic behind it.
What is the Moment Distribution Method?
The moment distribution method is a numerical iterative procedure for analyzing statically indeterminate beams and frames. The core idea is simple. We begin by assuming all joints are temporarily locked or fixed. This allows us to calculate the “fixed end moments” caused by external loads.
Of course, in reality, most joints are not fixed and are free to rotate. So, the method systematically “unlocks” and “relocks” each joint, one by one. When a joint is unlocked, the unbalanced moment is distributed among the connecting members. This distribution depends on the relative stiffness of each member. A portion of this distributed moment is then “carried over” to the far end of the member. This process is repeated until the unbalanced moments become negligibly small.
While modern software now handles complex analysis, MDM remains vital in engineering education. It provides an unparalleled conceptual understanding of how indeterminate structures carry loads and how members interact.
Core Concepts You Must Understand Before You Begin
Before diving into the procedure, mastering a few fundamental concepts is essential. These are the building blocks of the entire method.
Fixed End Moments (FEMs)
A Fixed End Moment (FEM) is the moment induced at the ends of a member when its ends are assumed to be completely fixed against rotation. We imagine each loaded span of our structure as an isolated, fixed-fixed beam. Calculating FEMs is the very first step of the analysis.
You can find FEMs using standard formulas. Here is a table of the most common cases:
| Loading Type | Diagram | FEM Formula at A (FEM_AB) | FEM Formula at B (FEM_BA) |
| Uniformly Distributed Load (UDL) | (Visual of beam with UDL) | -wL² / 12 | +wL² / 12 |
| Central Point Load | (Visual of beam with central point load P) | -PL / 8 | +PL / 8 |
| Eccentric Point Load | (Visual of beam with point load P at ‘a’ from A, ‘b’ from B) | -Pab² / L² | +Pa²b / L² |
| Uniformly Varying Load (Triangular) | (Visual of beam with triangular load, max at B) | -wL² / 30 | +wL² / 20 |
Note: Clockwise moments are typically considered positive (+), and counter-clockwise moments are negative (-).
Stiffness Factor (K)
The stiffness factor represents a member’s resistance to rotation. It tells us how much moment is required to rotate one end of a member by a unit angle (one radian). The stiffness depends on the support condition at the member’s far end.
- Case 1: Far End is Fixed
If the far end of the member is fixed, it offers more resistance to rotation.
Stiffness (K) = 4EI / L - Case 2: Far End is Pinned or Roller (Simply Supported)
If the far end is free to rotate (pinned or roller), it offers less resistance.
Stiffness (K) = 3EI / L
This is often called the modified stiffness and can be used as a shortcut.
Where:
- E = Modulus of Elasticity of the material
- I = Moment of Inertia of the member’s cross-section
- L = Length of the member
Distribution Factor (DF)
When we unlock a joint, the unbalanced moment must be distributed among all the members connected to that joint. The distribution factor determines the proportion of the moment that each member receives. Stiffer members attract more moment.
The distribution factor for a member at a specific joint is its stiffness factor divided by the sum of stiffness factors of all members meeting at that joint.
Distribution Factor (DF) = K / ΣK
Key properties of Distribution Factors:
- The sum of DFs for all members at a rigid joint must always be 1.0.
- For a fixed support, the connected member has a DF of 0 (it absorbs all moment).
- For a pinned/roller end support, the member has a DF of 1.0 (it cannot resist any moment).
Carry-Over Factor (COF)
The carry-over principle is a crucial part of the iteration. When a moment is applied to balance a joint, it causes the member to bend. This bending induces a moment at the far end of the member. This induced moment is the “carry-over moment.”
- Rule: For a prismatic member, the moment carried over to the far end is one-half (1/2) of the distributed moment at the near end.
- Direction: The carry-over moment has the same sign as the distributed moment.
Carry-Over Moment = (1/2) x Distributed Moment
The carry-over happens only when the far end is fixed. If the far end is pinned or a roller, the carry-over factor is zero, as it cannot resist any moment.
The Step-by-Step Process of the Moment Distribution Method
Now that we have the core concepts, let’s outline the systematic procedure. We will use a tabular format to keep the calculations organized.
Step 1: Calculate Stiffness Factors (K)
For each member in the structure, determine its stiffness factor (K). Use 4EI/L for members with a fixed far end and 3EI/L for members with a pinned/roller far end (if using the stiffness modification). For simplicity, you can often work with relative stiffness values (e.g., if EI is constant, K is just 4/L or 3/L).
Step 2: Calculate Distribution Factors (DF)
At each internal joint and support that can rotate, calculate the distribution factor for each connecting member using the formula DF = K / ΣK. Remember, the sum of DFs at any joint must equal 1.
Step 3: Calculate Fixed End Moments (FEMs)
For each member that carries an external load, calculate the fixed end moments using the standard formulas. Pay close attention to the sign convention (counter-clockwise is negative, clockwise is positive).
Step 4: Create the Distribution Table
Set up a table with columns for Joint, Member, Distribution Factor (DF), and rows for FEM, Distribution (or Balance), Carry-Over, and their subsequent iterations.
Step 5: Perform the First Distribution (Balancing)
At each joint, calculate the total unbalanced moment by summing the FEMs. To balance the joint, apply a moment of equal magnitude but opposite sign. Distribute this balancing moment to each member based on its DF.
Distributed Moment = – (Unbalanced Moment) x DF
Step 6: Perform the Carry-Over Step
For each distributed moment from Step 5, carry over half of its value to the far end of the respective member. Write these values in the “Carry-Over” row.
Step 7: Repeat the Cycle
Repeat Steps 5 and 6. The new unbalanced moment at each joint will be the sum of the carry-over moments received from the previous step. Continue this “Distribute -> Carry-Over” cycle.
Step 8: Sum the Final Moments
The iterations stop when the distributed moments become very small. Sum all the moment values in each column (initial FEM + all distributed moments + all carry-over moments). This sum gives the final end moment for that member at that joint.
Solved Example 1: Continuous Beam
Let’s apply the moment distribution method to a practical problem.
Problem: A two-span continuous beam ABC is fixed at A and supported by rollers at B and C. Span AB is 8m long and carries a UDL of 10 kN/m. Span BC is 6m long and carries a central point load of 40 kN. Assume EI is constant for both spans. Find the final support moments.
Step 1: Stiffness Factors (K)
- Member BA: The far end A is fixed.
K_BA = 4EI / L = 4EI / 8 = 0.5 EI - Member BC: The far end C is a roller. We can use the modified stiffness.
K_BC = 3EI / L = 3EI / 6 = 0.5 EI
Step 2: Distribution Factors (DF)
We only need to find DFs for the internal joint B.
- ΣK at B = K_BA + K_BC = 0.5 EI + 0.5 EI = 1.0 EI
- DF_BA = K_BA / ΣK = (0.5 EI) / (1.0 EI) = 0.5
- DF_BC = K_BC / ΣK = (0.5 EI) / (1.0 EI) = 0.5
- Note: Support A is fixed (DF=0), and C is a roller (DF=1).
Step 3: Fixed End Moments (FEMs)
- Span AB (UDL):
FEM_AB = -wL² / 12 = -10 * (8)² / 12 = -53.33 kNm
FEM_BA = +wL² / 12 = +10 * (8)² / 12 = +53.33 kNm - Span BC (Central Point Load):
FEM_BC = -PL / 8 = -40 * 6 / 8 = -30.00 kNm
FEM_CB = +PL / 8 = +40 * 6 / 8 = +30.00 kNm
Step 4: The Moment Distribution Table
Now we set up and populate the table.
| Joint | A | B | C | ||
| Member | AB | BA | BC | CB | |
| DF | 0 | 0.5 | 0.5 | 1 | |
| FEM | -53.33 | +53.33 | -30.00 | +30.00 | |
| Balance C | -30.00 | ||||
| Carry-Over | -15.00 | ||||
| Initial Moments | -53.33 | +53.33 | -45.00 | 0 | |
| Unbalanced Moment at B | (+53.33 – 45.00) = +8.33 | ||||
| Balance B | -4.17 | -4.17 | |||
| Carry-Over | -2.09 | ||||
| Unbalanced Moment | (0) | (0) | |||
| Final Moments (Σ) | -55.42 | +49.16 | -49.17 | 0 |
Explanation of the Table:
- FEM Row: We list the calculated Fixed End Moments.
- Balance C: Joint C is a roller support. It cannot take any moment. The initial FEM of +30.00 kNm must be balanced. We apply -30.00 kNm to make it zero. Since we used the modified stiffness for BC (3EI/L), there is no carry-over back from C to B.
- Carry-Over (Initial): This step is a bit of a trick with the modified stiffness method. Some engineers prefer to balance the end support first, then do a carry-over. A more classic approach with 4EI/L would have a carry-over, but here we simplify. Let’s re-run the table using the standard 4EI/L for clarity, which is a more robust method for beginners.
Revised Example Using Standard Stiffness (4EI/L)
- K_BA = 4EI/8 = 0.5EI
- K_BC = 4EI/6 = 0.667EI
- ΣK_B = 1.167EI
- DF_BA = 0.5 / 1.167 = 0.428
- DF_BC = 0.667 / 1.167 = 0.572
| Joint | A | B | C | ||
| Member | AB | BA | BC | CB | |
| DF | 0 | 0.428 | 0.572 | 1 | |
| FEM | -53.33 | +53.33 | -30.00 | +30.00 | |
| Balance C | -30.00 | ||||
| CO to B | -15.00 | ||||
| Cycle 1: Balance B | -3.57 | -4.76 | |||
| Unbalanced at B = (+53.33 – 30.00 – 15.00) = +8.33 | |||||
| Balancing Moment = -8.33 | |||||
| Distributed = -8.33 x 0.428 = -3.57 | |||||
| Distributed = -8.33 x 0.572 = -4.76 | |||||
| Cycle 1: Carry-Over | -1.79 | -2.38 | |||
| Cycle 2: Balance B | +1.02 | +1.36 | |||
| Unbalanced at B = -2.38 | |||||
| Balancing Moment = +2.38 | |||||
| Cycle 2: Carry-Over | +0.51 | +0.68 | |||
| Cycle 3: Balance B | -0.29 | -0.39 | |||
| Final Moments (Σ) | -54.61 | +50.50 | -50.50 | 0 |
This second table shows the full iterative process. We balance joint C, carry over half to B. Then we balance the new total at B, and carry over to A and C. We repeat until the numbers are small. The final moments are the sum of each column.
Solved Example 2: Non-Sway Frame
Let’s analyze a simple non-sway frame. A sway frame is one that can displace laterally. A non-sway frame is restrained against this movement, either by its supports or by symmetry.
Problem: Analyze the frame shown below. It is fixed at supports A and D. A UDL of 20 kN/m acts on beam BC. EI is constant for all members.
- Column AB: Height = 4m
- Beam BC: Length = 6m
- Column CD: Height = 4m
Step 1: Stiffness & Distribution Factors
- Joint B:
- K_BA = 4EI / 4 = 1.0 EI
- K_BC = 4EI / 6 = 0.667 EI
- ΣK_B = 1.667 EI
- DF_BA = 1.0 / 1.667 = 0.6
- DF_BC = 0.667 / 1.667 = 0.4
- Joint C:
- K_CB = 4EI / 6 = 0.667 EI
- K_CD = 4EI / 4 = 1.0 EI
- ΣK_C = 1.667 EI
- DF_CB = 0.667 / 1.667 = 0.4
- DF_CD = 1.0 / 1.667 = 0.6
Step 2: Fixed End Moments (FEMs)
Only member BC has load.
- FEM_BC = -wL² / 12 = -20 * (6)² / 12 = -60 kNm
- FEM_CB = +wL² / 12 = +20 * (6)² / 12 = +60 kNm
Step 3: The Moment Distribution Table
| Joint | A | B | C | D |
| :— | :— | :- | :- | :- | :- |
| Member | AB | BA | BC | CB | CD | DC |
| DF | 0 | 0.6 | 0.4 | 0.4 | 0.6 | 0 |
| FEM | 0 | 0 | -60.0 | +60.0 | 0 | 0 |
| Cycle 1 | | | | | | |
| Balance | | +36.0 | +24.0 | -24.0 | -36.0 | |
| Carry-Over | +18.0 | | -12.0 | +12.0 | | -18.0 |
| Cycle 2 | | | | | | |
| Balance | | +7.2 | +4.8 | -4.8 | -7.2 | |
| Carry-Over | +3.6 | | -2.4 | +2.4 | | -3.6 |
| Cycle 3 | | | | | | |
| Balance | | +1.44 | +0.96 | -0.96 | -1.44 | |
| Carry-Over | +0.72 | | -0.48 | +0.48 | | -0.72 |
| Cycle 4 | | | | | | |
| Balance | | +0.29 | +0.19 | -0.19 | -0.29 | |
| Final Moments (Σ) | +22.32 | +44.93| -44.93| +44.93| -44.93 | -22.32|
Explanation of Frame Table:
- Balance (Cycle 1): At joint B, the unbalanced moment is -60. The balancing moment is +60. It is distributed as +600.6=+36 to BA and +600.4=+24 to BC. At joint C, the unbalanced moment is +60. The balancing moment is -60. It is distributed as -600.4=-24 to CB and -600.6=-36 to CD.
- Carry-Over (Cycle 1): Half of each balanced moment is carried over. +36 from BA goes to AB (+18). +24 from BC goes to CB (+12). -24 from CB goes to BC (-12). -36 from CD goes to DC (-18).
- Balance (Cycle 2): The new unbalanced moment at B is -12 (from CB). The balancing moment is +12, which is then distributed. The new unbalanced moment at C is +12 (from BC), which is balanced and distributed.
- Repeat: The process continues until the numbers are negligible. The final moments are the sum of each column.
Connecting to the Slope-Deflection Method
It is insightful to know that the moment distribution method is not a standalone concept. It is actually a physical relaxation and iterative solution of the slope-deflection equations.
The fundamental slope-deflection equation for the moment at the ‘A’ end of a member AB is:
M_AB = FEM_AB + (2EI/L) * (2θ_A + θ_B – 3ψ)
Where:
- θ_A, θ_B are the rotations at ends A and B.
- ψ is the chord rotation (sway).
At any joint, the sum of moments must be zero (ΣM_joint = 0). The slope-deflection method sets up these equilibrium equations for each joint and solves them simultaneously for the unknown rotations (θ).
The moment distribution method does the exact same thing but avoids solving simultaneous equations. The “balancing” step is equivalent to satisfying the ΣM_joint = 0 equation at that specific joint for that one cycle. The “carry-over” step accounts for the influence of θ_A on M_BA. By iterating, we converge to the final answer that satisfies all joint equilibrium equations simultaneously.
Frequently Asked Questions (FAQ)
Why is the moment distribution method used?
It is used to find the end moments in statically indeterminate beams and frames. Its primary value today is educational. It provides a strong conceptual foundation for understanding how forces and moments are distributed in a structure, which is a skill that complements modern software use.
What is the difference between moment distribution and slope deflection?
Slope deflection is an “exact” method that forms and solves a set of simultaneous algebraic equations to find unknown joint rotations. Moment distribution is an iterative numerical method that achieves the same result by repeatedly balancing joints and carrying over moments until equilibrium is reached. MDM is often faster for manual calculation.
What is the carry-over factor?
The carry-over factor is the ratio of the moment induced at the fixed far end of a member to the moment applied at the near end. For a prismatic member (constant EI), this factor is always 1/2.
When is the stiffness factor 3EI/L used?
The modified stiffness factor K = 3EI/L is used for a member when its far end is a pin or a roller support. Using this “shortcut” simplifies the distribution table because you do not need to balance the final pinned/roller support; its final moment is already known to be zero.
Conclusion
The moment distribution method, though a classical technique, remains a cornerstone of structural engineering theory. Its step-by-step, iterative process demystifies the behavior of indeterminate structures, revealing the elegant interplay of stiffness, distribution, and equilibrium. By working through the examples, you can see how an initially complex problem is broken down into a series of simple, manageable calculations.
Mastering this method gives you more than just a calculation tool. It equips you with a profound, intuitive sense of structural behavior that is indispensable for any design engineer. It builds the foundational knowledge needed to confidently use and interpret the results from modern structural analysis software.
