Solved by passing it around

A continuous beam over four supports cannot be solved by statics. There are more reactions than equations, the answer depends on the beam’s stiffness, and the honest route is to write down a set of simultaneous equations and solve them. For a three-span beam that is three equations; for a twelve-storey frame it is several hundred, and in 1930 there was no way to solve several hundred simultaneous equations at all.

Hardy Cross’s answer was to not solve them.

Every number in that sequence was obtained by arithmetic a person can do: multiply by a fraction, halve, add. No equation was ever assembled, nothing was inverted, and the answer is exact in the limit.

The three rules #

Clamp every joint. With every joint held against rotation, each span is a fixed-ended beam and its end moments are known in closed form — $wL^2/12$ for a uniform load. The structure is now in equilibrium with the load and is not compatible: the joints are being held by moments nobody applied.

Release a joint and share out the imbalance. At a joint where two members meet, the two fixed-end moments do not cancel, and the difference is what the clamp is holding. Take the clamp off and the joint rotates until the members’ resisting moments balance. Since each member’s resisting moment is its rotational stiffness times the same rotation, the imbalance divides between them in proportion to stiffness:

$$\text{DF}_i = \frac{k_i}{\sum k} \quad\text{with}\quad k = \frac{4EI}{L}$$

Carry over half. A rotation applied at one end of a member with its far end fixed produces a moment at that far end of half the near-end value, with the same sign. So each share sent into a member deposits half of itself at the other end, which unbalances the joint there, which is why the process repeats.

That is the entire method. It is short enough to memorise and it is what an engineer of the 1930s to the 1960s used for every frame they ever designed.

The two numbers that make it work #

The factor of a half deserves a second look, because it is the whole reason the process converges rather than wandering.

For a prismatic member with its far end fixed, the near-end stiffness is $4EI/L$ and applying a unit rotation there produces $2EI/L$ at the far end — a ratio of exactly 0.5. Every cycle therefore multiplies the residual imbalance by (a distribution factor below one) times (a half), so the residue is reduced by a factor of at least two per member per cycle, and in practice by three or four.

That is geometric convergence with a comfortable ratio. The table above shows it: 55.0, 28.8, 7.8, 2.4, 0.62, 0.16, 0.04 kNm of imbalance on successive cycles. Anyone can see when to stop, which is a property no matrix solution has — a matrix answer is either obtained or not, and a distribution answer arrives with its own error estimate visible in the last row.

The second number is the $4EI/L$ itself, and it is worth stating what it hides. It is the moment needed at one end of a member to produce a unit rotation there while the far end is held, and it is the same quantity a stiffness matrix carries in its diagonal. Moment distribution is not an alternative to the stiffness method; it is Gauss–Seidel iteration on the stiffness equations, performed by somebody who never wrote the matrix down. Cross did not describe it that way, and the identification came later.

It agrees with a method that shares nothing with it #

The claim that the method converges to the exact answer is checkable, and on this site it is checked by computing the same beam a second way.

The comparison is the point rather than the agreement. A method that is checked against its own last iteration is checking convergence and not correctness, and an iterative scheme can converge beautifully to the wrong answer if its distribution factors are wrong. Agreement between two routes that share no arithmetic is the only evidence that either is right, which is why this site’s gate computes both.

The distribution factors are the design #

The arithmetic gets more interesting when the spans are not equal, because the distribution factors stop being halves and start expressing a design decision.

Those fractions are the whole difference between this beam and the equal-span one, and every line of arithmetic that follows is their consequence. The long span arrives clamped at 83.3 kNm against the short spans’ 30.0, so the first release at an interior joint has 53.3 kNm of imbalance to share out — and it sends the larger part of it back into the shorter member, which is the opposite of where a reader expecting load to follow span would put it.

That the short span attracts more moment is the practical content of the whole method, and it is a fact about stiffness rather than about load. A designer who deepens one member has attracted moment into it; one who lengthens a span has pushed moment away from it. In a hand method those consequences are visible in the second line of the table, before the calculation is finished — which is what a designer wants, because the question being asked is usually what should this be rather than what is this.

There is a lesson here that survived the method. Moment follows stiffness, and stiffness is chosen, so an indeterminate structure’s internal forces are partly a matter of design intent rather than of external circumstance. That statement is easy to say and hard to feel from a computer output; it is unavoidable when the sharing is done by hand.

What it is not good at #

Sway. A frame free to move sideways has an extra unknown per storey that no amount of joint rotation addresses, and the treatment is a separate outer iteration: guess the sway, distribute, find the out-of-balance shear, correct the sway, repeat. It works and it is laborious, and it is where the method’s economy ends. A leaning portal frame is exactly this case.

Imposed displacements. A settlement or a temperature effect enters as a set of fixed-end moments computed from the movement, which is fine, but the physical story the method tells is much less natural — the engineer is now distributing moments that arrived from a support going down rather than from a load going on.

There is a second difficulty in that case which the table hides completely. The distribution factors above contain a ratio of stiffnesses, so EI cancels out of every one of them and never appears in the sharing at all; the movement’s fixed-end moments, by contrast, are proportional to EI outright. A settlement therefore produces moments with no load on the structure whatever, and produces larger ones the stiffer the beam is — the finding that makes settlement dangerous, and one a distribution table cannot express because the quantity responsible has already cancelled everywhere the table looks.

Anything with axial deformation in it. The method assumes members do not change length, which is excellent for beams and frames and useless for a truss. The competing hand method — virtual work with a unit load — has the opposite profile: it handles trusses beautifully and continuous frames laboriously.

The refinements, which are all the same trick #

The method as stated releases every joint in turn, including the pinned ends, and the end joints then bounce the imbalance back and forth with their neighbours for several cycles. Two standard refinements remove most of that work, and both are the same manoeuvre: use a boundary condition that is already known instead of iterating towards it.

A member whose far end is pinned does not need a carry-over at all. Its near-end rotational stiffness is $3EI/L$ rather than $4EI/L$, and its carry-over factor is zero, because the far end is free to rotate and cannot accumulate a moment. Release the end joint once at the very start, use the modified stiffness thereafter, and that joint never appears in the iteration again.

On the three-span beam that changes the arithmetic visibly. The distribution factors at the first interior joint go from a half and a half to $3/(3+4) = 0.429$ toward the end span and $0.571$ toward the interior one, and the number of joints being cycled drops from four to two. The answer is identical; roughly half the lines are gone.

A structure symmetric about its centre, under a symmetric load, has a known rotation at the centre — zero — so the middle span can be cut there and only half the beam analysed. A member cut at a plane of symmetry has a modified stiffness of $2EI/L$ with no carry-over; under an antisymmetric load, where the centre has zero moment instead, it is $6EI/L$. Any load decomposes into a symmetric part and an antisymmetric one, so a symmetric frame is always two half-sized problems rather than one whole one.

What is worth carrying is not the three constants but what they have in common. $4EI/L$, $3EI/L$, $2EI/L$ and $6EI/L$ are the same member’s stiffness under four different statements about what its far end is doing, and each modification replaces an unknown that would have been iterated with a fact that was known in advance.

There is a practical corollary that matters more than the saving in lines. Each refinement requires the analyst to notice a fact about the structure before starting — that this end is pinned, that this frame is symmetric, that this load can be split into two parts one of which the structure barely feels. A method that rewards noticing is a method that teaches, and the engineer who has spotted the symmetry has understood something about the frame that the answer alone would not have told them.

Which is precisely what a modern solver does under the name static condensation: eliminate a degree of freedom whose behaviour is determined, and carry its effect into the stiffness of what remains. The hand method’s refinements are not tricks laid on top of it — they are the same operation, done by choosing a coefficient from a short table instead of by eliminating a row. The essay’s earlier identification of the method with Gauss–Seidel has a companion: its shortcuts are condensation, and between the two there is very little in a structural solver that moment distribution was not already doing by hand.

The tables in every handbook are this method, frozen #

Open any structural handbook at the page of continuous-beam coefficients — 0.080, 0.100, 0.025, the little grid of numbers for two, three, four and five equal spans — and what is printed there is the output of this method, run once by somebody in the 1930s and copied ever since.

The three-span case above is the entry $0.100\,wL^2$ at the interior supports, which is 64.0 kNm for $w = 10$ and $L = 8$, exactly as computed. The two-span case is $0.125\,wL^2$, which is $wL^2/8$ — the same number a simply supported beam carries at midspan, arriving at a support instead. Those coefficients are exact for the case they describe and they are the reason a generation could design continuous beams without doing any of the arithmetic above.

Nothing in those coefficients comes from statics. A single simple span is determinate and its 80 kNm at midspan can be written down from equilibrium alone; every entry beyond it needs the stiffness, and the peak moment falls each time a restraint is added because part of it has been moved to a support. Moment distribution is how the whole grid was obtained before a matrix could be inverted, and the arithmetic above is the two-span row of it reproduced in seven lines.

The tables have two limitations that the method itself does not, and both are worth naming because the tables outlived the understanding. They assume equal spans, and the section on distribution factors above shows how quickly unequal ones change the answer. And they assume a pattern of loading — usually all spans loaded — where the governing case for a support moment is often alternate spans loaded, which is the same question an influence line answers and which the coefficients handle by having a second row.

Continuity is bought and paid for #

It is worth being explicit about what all this arithmetic is for, because the answer is not obvious from the numbers.

A simply supported 8 m beam under 10 kN/m carries 80 kNm at midspan. Make three of them continuous and the interior spans carry 40 kNm at midspan and 64 at the supports — so the largest moment anywhere has fallen from 80 to 64, a reduction of a fifth, for no material and no change of span. That is the return on continuity, and it is why the method existed.

What was paid for it is threefold, and each cost is somewhere else in this collection. The structure is now indeterminate, so a support that settles produces moments with no load at all. The peak moment is at a support, where the beam is usually shallowest in a haunched design and where the connection has to carry it across a gap. And the beam is now sensitive to the pattern of loading rather than only to its magnitude, which doubles the number of cases to check.

Continuity is a trade of one large moment for two smaller ones plus a set of new sensitivities, and moment distribution is the instrument that made the trade computable by hand. That it also gave engineers an intuition for where moment goes was a side effect, and it was the part that was lost when the arithmetic moved to a machine.

What it was really for #

Cross published the method in 1930 in a paper of ten pages, and it was in universal use within five years. The usual account is that it saved arithmetic, and it did — but the deeper reason it took over is the one visible in the tables above.

The method produces intermediate results that mean something. The first line is the structure with every joint clamped; the second is what one joint does when released; the third is the effect of that release travelling to its neighbours. An engineer following it is watching a moment move through a structure, and the number they end with is one they have watched arrive. It is a calculation with a narrative, and the narrative is the mechanism.

That is worth stating precisely because it is the property a matrix solution lacks, and the loss was real. A generation of engineers who had distributed moments by hand had an intuition for which member attracts what, and it came from the arithmetic rather than from a course. The replacement of the method by matrix analysis in the 1970s was correct on every technical ground — sway, three dimensions, thousands of members — and it removed the only part of the calculation a person could see.

There is one more thing the method quietly taught, and it is visible in the very first line of every table. The fixed-end moments are the answer for a structure with every joint clamped, which is the stiffest possible arrangement, and the distribution process only ever takes moment away from the supports and gives it to the spans. So the first line is a bound: the support moment of a real continuous beam is never larger than its fixed-end moment, and a designer in a hurry could stop at line one and be conservative at the supports. Half the rules of thumb in continuous-beam design are that observation, and the count of unknowns is what says why the bound exists — clamping every joint makes the structure maximally redundant, and every release moves it back toward statics.

What the picture cannot show #

Everything above is elastic and linear. The carry-over factor of a half assumes a prismatic member behaving elastically; a member that has cracked, yielded or has a variable section has different stiffness and carry-over factors, which is arithmetic rather than a difficulty and which every textbook tabulates.

The convergence rate depends on the structure. The three-span beam above converges in four cycles because its distribution factors are around a half. A frame with many members at a joint, or with very unequal stiffnesses, converges more slowly, and a structure with a near-mechanism in it converges very slowly indeed — the same ill-conditioning a matrix solver would meet, showing up as patience rather than as a warning.

Nothing here is a check. The method produces a set of moments in equilibrium and compatible, and it has no opinion about whether they are acceptable. Everything about capacity, deflection and stability is a separate question asked afterwards.

Where the ladder goes #

The first rung is sway, which is where the method’s outer iteration lives and which is the case every real frame has.

The second is the identification with Gauss–Seidel, and what it says about the structure: iterative methods converge fast when a system is diagonally dominant, and a structure’s stiffness matrix is diagonally dominant exactly when each joint is better connected to itself than to its neighbours. The method converges quickly on ordinary structures because ordinary structures are locally stiff, which is a statement about buildings rather than about arithmetic.

The third is the one this collection keeps returning to: what happens when the moments obtained above exceed what a section can carry. The elastic distribution is one equilibrium solution among many, and the plastic one redistributes it entirely — which means the careful sharing above describes the structure only until first yield, and the collapse load does not depend on it at all.