Module 16 · Lesson 16.4
Moment distribution and the matrix stiffness method
Balancing joints by hand, and the matrix method that replaced it.
Why this matters
Moment distribution was the workhorse of structural design for decades, and it is still the fastest way to get a feel for how a frame shares moment between members. The matrix stiffness method that superseded it is the same physics organised for a computer — and understanding the hand method is the shortest route to understanding what the software is doing.
By the end of this lesson you should be able to
- Calculate member stiffnesses and distribution factors at a joint
- Carry out a moment distribution cycle
- Explain the role of the carry-over factor
- Describe how element stiffness matrices are assembled and solved
What you should already know
- Fixed-end moments (this module, lesson 2)
- The slope-deflection equation (this module, lesson 3)
Moment distribution starts by telling a comfortable lie and then correcting it.
Step 1 — lock every joint. Imagine clamping every joint so no rotation can occur. Each member is then a fixed-ended beam, and its end moments are just the fixed-end moments.
Step 2 — look at the out-of-balance. At a real joint the moments from the members meeting there will not sum to zero. That imbalance is being held by the imaginary clamp.
Step 3 — release the joint and let it rotate. The out-of-balance moment is distributed among the members in proportion to their stiffness. A stiffer member resists rotation more, so it takes a larger share.
Step 4 — carry over. When one end of a member rotates, a moment appears at its far end too. For a member whose far end is fixed, half of the distributed moment carries over.
Step 5 — repeat. The carried-over moments unbalance the neighbouring joints, so go round again. The corrections shrink rapidly, and two or three cycles are usually enough.
What it calculates: the moment needed at one end to produce unit rotation there
- K
- rotational stiffness of the member at the joint (kNm per radian)
- E
- Young's modulus (N/mm²)
- I
- second moment of area (mm⁴)
- L
- member length (mm)
This assumes
- Prismatic member
- Elastic behaviour
- No sway
In plain terms: A member with a pinned far end is three-quarters as stiff, because the far end rotates freely instead of pushing back. It also has no carry-over — there is nothing at the far end to carry to.
What it calculates: the share of an out-of-balance moment taken by each member at a joint
- Ki
- stiffness of member i at that joint (kNm/rad)
- ΣK
- total stiffness of all members meeting there (kNm/rad)
This assumes
- The joint is rigid, so all members rotate together
In plain terms: The distribution factors at any joint must add to 1.0 — the out-of-balance moment has to go somewhere, and it all goes into the members meeting there. A factor summing to anything else is an arithmetic error.
Predict first
Two members of identical EI meet at a rigid joint. One is 6 m long, the other 4 m, and both have fixed far ends. Which takes the larger share of an out-of-balance moment?
Worked example
Distributing an out-of-balance moment at a joint
Given
- Two members meet at a rigid joint B, both with the same EI
- Member BA is 6.00 m long with a fixed far end at A
- Member BC is 4.00 m long with a fixed far end at C
- The out-of-balance moment at B, after locking, is 50.0 kNm
Find
The distribution factors, the moment each member takes, and the carry-over moments.
The matrix stiffness method is this same idea, written so a computer can do it without judgement.
Each element gets a stiffness matrix relating the forces at its ends to the displacements there. For a plane beam element with four degrees of freedom — a transverse displacement and a rotation at each end, taken in the order v₁, θ₁, v₂, θ₂ — it is:
What it calculates: the end forces of a beam element from its end displacements
- k
- element stiffness matrix, 4 × 4 (N/mm and N·mm/rad)
- v₁, v₂
- transverse displacements of the two ends (mm)
- θ₁, θ₂
- rotations of the two ends (radians)
- EI
- flexural rigidity (N·mm²)
- L
- element length (mm)
This assumes
- Prismatic, linearly elastic element
- Bending deformation only — axial and shear deformation are excluded from this 4 × 4 form
- Small displacements
In plain terms: Every entry is a stiffness: kᵢⱼ is the force at degree of freedom i caused by a unit displacement at j with all others held at zero. The matrix is symmetric, and the 4L² entry multiplied through by EI/L³ gives 4EI/L — exactly the rotational stiffness moment distribution used. The two methods are the same physics.
The procedure is then mechanical:
- 1.Build the element matrix for every member.
- 2.Assemble them into one global matrix K, adding each element's contribution into the rows and columns of the degrees of freedom it connects.
- 3.Apply the loads as a vector f.
- 4.Apply the boundary conditions by removing the restrained degrees of freedom.
- 5.Solve Kd = f for the displacements d.
- 6.Recover member forces from the element matrices.
No choice of redundant, no iteration, no judgement about which release is convenient. That is exactly why it, and not moment distribution, is what runs inside every analysis package.
Practice
Two members of equal EI meet at a rigid joint. One is 6.00 m long, the other 4.00 m, and both have fixed far ends. What is the distribution factor for the 6 m member?
Practice
An out-of-balance moment of 50.0 kNm at that joint is distributed. The 6 m member takes its share, and its far end is fixed. What moment is carried over to that far end, in kNm?
Practice
A member's far end is changed from fixed to pinned, with E, I and L unchanged. By what factor does its rotational stiffness at the near joint change?
Worked example
A propped cantilever solved by the matrix stiffness method
Given
- Propped cantilever, span L, built in at node 1 and propped on a roller at node 2
- One beam element, uniform EI, carrying a UDL of intensity w
- Degrees of freedom in the order v₁, θ₁, v₂, θ₂
Find
The prop reaction, by assembling and solving the element equations rather than by compatibility.
Practice
In the 4 × 4 plane beam element stiffness matrix, the leading entry is 12EI/L³. What is the entry coupling a nodal rotation to itself?
Summary
- Moment distribution: lock the joints, then release and balance them one at a time
- Stiffness K = 4EI/L (fixed far end) or 3EI/L (pinned far end)
- Distribution factors are the share of joint stiffness and must sum to 1.0
- Carry-over factor is 0.5 towards a fixed far end, 0 towards a pinned one
- Stiffness attracts moment — an unintentionally stiff element attracts unintended force
- The matrix stiffness method assembles element matrices and solves Kd = f
- The stiffness matrix is symmetric, which is Maxwell's reciprocal theorem in matrix form
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint