The weight that has to be known before it can be found

Every load in a schedule can be looked up. Somebody has measured the snow, tabulated the wind, and decided what a floor of an office is worth per square metre. One line in that schedule cannot be looked up, and it is usually the largest: the weight of the structure, which depends on the sizes, which depend on the loads, which include the weight of the structure.

That circularity is the first calculation on any project and it is not a nuisance. It has an exact solution, it converges at a rate that is itself a design quantity, and it stops converging at a span that is a property of the material.

Which free body produced the number #

A simply supported beam of span $L$ and fixed depth $d$, carrying an imposed load $q$ per metre and its own weight. Cut it at midspan. The moment there is $(q + g)L^2/8$ where $g = \rho A$ is the self weight, $\rho$ the unit weight of the material and $A$ the area of the section.

The capacity of the cut is $\sigma Z$, and for a section of a fixed proportion the modulus is $Z = \alpha A d$ — for a plate girder with all its area in the flanges, $\alpha$ is a half; for a rolled beam, about 0.4. Setting the two equal:

$$\sigma \alpha A d = \frac{(q + \rho A) L^2}{8}$$

Every quantity in that equation is known except $A$, which appears on both sides, and that is the whole difficulty stated in one line. Rearranged,

$$A = \frac{qL^2/8}{\sigma\alpha d - \rho L^2/8}$$

The share, which is a ratio of two spans #

The rearrangement contains something better than an answer. Divide the self weight by the total load and almost everything cancels:

$$\frac{g}{q+g} = \frac{\rho L^2}{8\sigma\alpha d} = \left(\frac{L}{L^*}\right)^2, \qquad L^* = \sqrt{8\alpha\, d\, \frac{\sigma}{\rho}}$$

The fraction of a member’s capacity spent carrying itself is exactly the square of its span as a fraction of a limiting span. There is no imposed load in that statement, no section area, and no shape beyond the constant $\alpha$. A beam at a fifth of its limiting span spends 4% of itself on itself; one at half spends a quarter; one at three quarters spends 56%; and at $L^*$ the denominator above is zero and no area solves the equation at all.

The limiting span is worth reading closely, because it is the geometric mean of two lengths. One is the depth, which is a decision. The other is $\sigma/\rho$, the material’s strength divided by its unit weight, which has the dimension of length and is the height of a column of the material that would crush under its own weight. For structural steel that length is about 3,500 m, for reinforced concrete in bending about 600, for structural timber about 2,800.

A 600 mm deep steel beam therefore has a limiting span of $\sqrt{8 \times 0.4 \times 0.6 \times 3500} = 82$ m, and a concrete one of the same depth about 34 m. Neither number is a span anybody would build, and that is the point of computing them: they are the denominators that decide how badly the circularity bites at spans people do build.

What the iteration is actually doing #

Nobody solves that equation as an equation. The universal practice is to guess a self weight, size the member, recompute the self weight from the size, and go round again — which is a fixed-point iteration, and its convergence ratio is the number above.

Each pass removes a fraction $1 - (L/L^*)^2$ of the remaining error. At a fifth of the limiting span the ratio is 0.04 and one pass is exact to four figures. At half it is 0.25 and two passes suffice. At three quarters it is 0.56, each pass fixes less than half of what is left, and the sequence takes eight passes to reach a per cent — on a member that is already spending more than half its capacity on itself and would never be built.

So the practice is safe, and it is safe for a reason worth knowing rather than by luck: the iteration converges quickly exactly where the self weight is a small fraction, and it is a small fraction exactly where the span is short compared with the limiting one. The two facts are the same fact.

Why depth is the answer and area is not #

Look again at where $A$ and $d$ enter the equation. Capacity is $\sigma \alpha A d$ — linear in both. Self weight is $\rho A$ — linear in the area and independent of the depth.

So adding area adds capacity and weight in the same proportion, and adding depth adds capacity and almost none. That is the reason depth is the cheapest strength restated as a statement about self weight, and it is why every long-span structure ever built is deep and hollow rather than large and solid.

The limit case is a structure whose whole load is its own weight, and there the argument becomes visible in a single figure.

A masonry dome under its own weight has a meridional stress of $\gamma R/(1 + \cos\varphi_0)$: a unit weight, a radius and an angle, and nothing about how thick it is. Making it thicker adds exactly as much load as capacity and changes nothing. That is the purest available statement of the rule this essay is about — a structure cannot be thickened out of a self-weight problem, and the only variables that help are the ones that change the geometry.

The material question, which is not about strength #

If a limiting span contains $\sigma/\rho$ rather than $\sigma$, then the ranking of materials for long spans is a ranking on specific strength, and it is not the ranking anybody expects.

Timber is a twentieth of mild steel’s strength and a sixteenth of its weight, so on specific strength the two are within a fifth of each other — and on the stiffness-limited beam index, where the exponent on the modulus is a half rather than one, timber beats steel by 8.56 to one. Neither material moved. That is the whole content of an index: the exponent on the shape decides the ranking and the properties merely fill it in.

The same reordering is what makes the limiting-span arithmetic worth doing rather than quoting. A designer asking “what material spans furthest” has asked an incomplete question, because the answer depends on whether the span is stopped by strength, by stiffness or by buckling, and the three orderings are different.

The practical consequence for self weight is that the material with the highest limiting span is very rarely the strongest one, and that a structure whose difficulty is its own weight is being asked a question about density.

Where the weight goes afterwards #

A beam’s self weight is a small problem solved once. A building’s is a running total, and the accumulation is what makes a tall structure a different kind of object.

Every floor adds the same tributary load, so the axial force in a column grows linearly with the number of storeys above it, and the column’s own weight grows with the same count. In a steel frame that self weight is a few per cent of the total and the linearity is nearly exact. In a masonry structure it is most of the load, which is why the walls of a tall masonry building thicken toward the base in a way a steel building’s columns do not — and why a structure too tall for nothing but itself has a height limit that has no applied load in it whatever.

The form that carries its own weight for nothing #

The other escape from the fixed point is not to bend at all.

A cable or an arch of the funicular shape carries its load axially. Axial capacity is $\sigma A$ rather than $\sigma \alpha A d/L$, which is larger by roughly $L/(\alpha d)$ — a factor of forty on an ordinary beam and several hundred on a long one. That is the entire reason long spans are cables and arches rather than beams, and it is a statement about self weight rather than about elegance.

The catch is that the funicular shape depends on the load, and a structure whose dominant load is its own weight has a shape decided by itself — which is a fixed point again, one level up. A hanging chain finds it by hanging; an arch has to be given it, and is given it by drawing the chain and inverting it.

The one place the fixed point is not benign #

There is a class of structure where the iteration above genuinely does not settle quickly, and it is not the long-span roof anybody would expect.

It is the very lightly loaded one. The share $(L/L^*)^2$ is a fraction of the total load, so it says nothing about whether the imposed load is large. A member carrying almost no imposed load is a member almost all of whose load is itself, and its size is decided by a quantity that depends on its size with nothing external to anchor it.

Two familiar cases: a long-span roof over an unoccupied volume, where the imposed load is snow and maintenance access and the structure is most of what it carries; and a mast, a tower or a bridge pylon, where the applied load is a wind pressure on a very small area and the weight of the thing is the design case.

In both, small changes in the assumed section move the answer, and the sequence of sizes a designer walks through is genuinely a sequence rather than a correction. It is also where the choice of form stops being an aesthetic decision. A structure whose own weight is its principal load is a structure whose form has been chosen by that weight — which is why long-span roofs converge on shells, cables and arches from every direction, in every material, in every century.

Where the model stops #

The section is not similar to itself at every size. The constant $\alpha$ in the derivation assumes a family of sections with the same proportions, and real sections do not scale that way — plate thicknesses come in steps, webs have to be thick enough not to buckle, and a very large member is a plate girder with a different $\alpha$ from a small rolled beam.

Nothing here is a stability check. Everything above compares a moment with a section capacity, and a member long enough for its self weight to matter is nearly always governed by something else first — lateral buckling, deflection, or a vibration limit. The limiting spans computed above are therefore ceilings that nothing reaches, and their value is as denominators rather than as limits.

Self weight is not only the structure. Screed, finishes, services, ceilings, façade and partitions are dead load too, they are frequently larger than the frame, and none of them participates in the fixed point because none of them depends on the size of the frame. That is a useful asymmetry: the part of the dead load that is circular is usually the smaller part.

What the picture cannot show #

The equation at the top treats self weight as a load applied to a finished structure. It is not applied to anything; it is present from the first moment a member exists, which on a construction site is before the structure exists.

That distinction has consequences no static calculation contains. A beam carries its own weight while it is being lifted, at supports that are not its final ones. A concrete slab carries wet concrete on formwork that carries it to props that carry it to a slab poured a week earlier. A cantilevered bridge carries a self weight that grows segment by segment, and the moment diagram at the end of construction is not the one the finished structure would have. The structure that was never complete is about exactly that, and self weight is the only load that is present in every one of those states.

Nor does the picture show that self weight is the one load whose magnitude the designer chooses. Every other line in the schedule is imposed by the world. This one is an output of the design fed back as an input, and the good designs are the ones that notice.

The generalisation #

The habit worth carrying is about which quantities in a design are free and which are not.

A structural calculation looks like a function from loads to sizes. Self weight makes it a function of itself, and the general treatment of such a thing is to ask two questions: does it converge, and how fast. Both answers here are the same number, $(L/L^*)^2$, and that number is a ratio of a span to a length made out of a depth and a material property.

The pattern recurs whenever a structure’s own response changes the demand on it. Second-order effects are the same shape of problem — a deflection that increases the moment that produced it — and they converge by the same kind of geometric series, with $P/P_{cr}$ in the place of $(L/L^*)^2$. Ponding is the same again with rainwater doing the work. In each case there is a ratio below one for which the sequence closes, a value at which it does not, and a designer whose first estimate is the first term of a series nobody wrote down.

Knowing that the series exists is most of the benefit. It turns “add something for self weight” into an arithmetic with a convergence rate, and it says exactly when the estimate can be made once and when it cannot be made at all.