Equilibrium

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

Assumes The free body is a choice, and choosing it well is the whole skill, The load that is spread out, and the force that replaces it and The envelope is not a structure.

Every other load in this collection is applied to a structure. Snow is placed on one.

The distinction sounds like pedantry and it decides the whole calculation. A wind pressure exists because air is moving past a building now; stop the wind and the load is gone. Snow fell, some time ago, over a region tens of kilometres across, and what is on a particular roof this morning is the history of every wind that has crossed it since. Nothing about that history is in the meteorological number, which is a ground snow load — a water equivalent measured on flat open ground, where nothing gets in the way and the wind has nowhere to put anything.

The snow that left the roof is standing against the wallA roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.wind0.85 m3.4 m of driftbalanced 0.48 kN/m²peak 2.60 kN/m² — 5.4× the balanced load20 m of roof upwindvertical scale exaggerated 4×
Fig. 1 A roof in section with a parapet at its downwind end, drawn with the vertical scale exaggerated because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The wedge against the wall is the snow the roof upwind of it lost.

Which free body produced the number

A column of snow, one square metre in plan, sitting on the roof. Two things touch it: the roof below, pushing up, and gravity. That is the whole free body, and it says the pressure on the roof is the weight of the column and nothing else — depth times unit weight, with no dynamics and no direction in it anywhere.

So the load is a depth, and the entire question is how deep the snow is here rather than somewhere else. Fresh snow is about 1 kN/m³ and drifted snow between 2 and 3, having been broken up, carried and packed by the wind that moved it; the figures here use 2.5. At that density the balanced layer on this roof — 0.48 kN/m² — is 190 mm deep, which is a small number, and the wedge against the parapet is 0.85 m, which is not.

Nothing in that pair required a wind speed. It required only that the snow which is no longer on the upwind roof is now somewhere, and the somewhere is where the wind stopped being able to carry it.

The drift is a conservation of mass

Take the roof upwind of the parapet as the accounting region. Some fraction of the snow that fell on it has been scoured off and carried downwind — call it 30%, which is a judgement and the only one in this calculation. It ran along the roof and it stopped at the parapet, where the flow separated and dropped what it was carrying.

The drift is therefore a wedge whose area in section is the area the fetch lost. If the wedge runs out four times as far as it rises, which is roughly what drifts settle at, then

12λh2=βLudgh=2βLudgλ\tfrac{1}{2}\,\lambda h^2 = \beta L_u d_g \quad\Longrightarrow\quad h = \sqrt{\frac{2\beta L_u d_g}{\lambda}}

and the depth is a square root of the fetch. That single exponent is the most useful thing anyone knows about drifting. It means a roof twice as long upwind produces a drift 1.41 times as deep, not twice; and it means that the cheap intuition — a bigger roof is a proportionally bigger problem — is wrong in the direction that matters, because the problem grows more slowly than the building does.

The drift is a square root of the roof upwind of itDrift height against the length of roof the wind crossed before reaching the obstruction. The wedge holds the snow that fetch lost, so its area grows in proportion to the fetch and its height as the square root — doubling the roof upwind multiplies the drift by 1.41 and not by two. Past a fetch of 28 m the 1.0 m obstruction is full and more roof adds nothing at all, which is the one place on this curve where the answer stops depending on the weather. At the 20 m fetch drawn, the drift is 0.85 m deep and the local load is 5.4 times the balanced one.051015202530354000.20.40.60.811.2roof upwind of the obstruction (m)drift depth (m)the obstruction, 1.0 m0.85 m hereunlimited supplyfull at a fetch of 28 m
Fig. 2 Drift depth against the length of roof upwind of the obstruction. The curve is a square root, and it stops entirely once the obstruction is full.

The curve also stops. A drift cannot pile higher than the thing that is stopping it, so past a certain fetch the parapet is full, the wind carries the rest of the snow over the top, and more roof adds nothing at all. That is the one place in the whole of snow loading where the answer becomes independent of the weather, and it is worth knowing where it is: for a one-metre parapet in this arithmetic it is at 28 m of roof.

The intensity is not the quantity, and this is the trap

Read the two numbers on the section together and they look contradictory. The peak load in the drift is 5.4 times the balanced load. The drift is 27% of the snow on the roof.

Both are true and neither is a rounding of the other. The wedge is short — 3.4 m of a 20 m roof — so a load several times the design intensity, over a sixth of the plan, is a modest fraction of the total weight. A structure that carries its roof snow as a global quantity, like a set of columns or a foundation, barely notices the drift, in the way a whole-body free body never notices anything internal. A member whose tributary area happens to sit inside the wedge carries five times what its neighbour four metres away does.

Where a beam's load comes fromA 7.5 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 13.5 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 45.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.13.5 m²9.00 kN/m13.5 m²9.00 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m7.5 m6 m45.0 m² divided, 45.0 m² of panel — the division closes
Fig. 3 Where a beam’s load comes from. Every drift argument is an argument about which member’s tributary area the wedge lands on, and that is decided by a framing layout rather than by the weather.

So the same event produces two entirely different design problems from one drawing. That pairing is exactly what happens with wind on a building, where the largest pressures sit on the faces that contribute nothing to the base shear — and the resemblance is not a coincidence. Both loads are decided by the shape of the building rather than by the intensity of the weather, and in both cases the quantity that governs stability and the quantity that governs a local member are different quantities.

The snow that left the roof is standing against the wallA roof in section with a 2.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 5 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 40% of the 45 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 1.47 m deep and 5.9 m long, with a peak load of 4.15 kN/m² — 8.7 times the balanced value, and still only 33% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.wind1.47 m5.9 m of driftbalanced 0.48 kN/m²peak 4.15 kN/m² — 8.7× the balanced load45 m of roof upwindvertical scale exaggerated 5×
Fig. 4 The same construction on a longer roof against a taller obstruction. The drift is deeper and still a third of the snow present, which is the ratio this figure exists to make visible.

The load that weighs less and does more

The second half of the argument has nothing to do with drifts and everything to do with continuity.

Wind moves snow from one part of a roof to another. It does not create any. So an unbalanced case — one slope of a pitched roof cleared, the other loaded; one bay of a multi-bay roof scoured, its neighbour buried — weighs the same as the balanced case at most, and usually less. It is very easy to conclude from that fact that the balanced case governs, and on a simply supported member it does.

On a continuous member it does not.

Half the load and a third more momentA two-span continuous beam carrying 1.15 kN/m of balanced snow on both spans, and the same beam with one span cleared by the wind and the other unchanged. The second case weighs half as much — 9 kN against 17 — and produces a span moment of 6.2 kNm against 4.6, which is 1.36 times larger. The support moment moves the other way, halving, so neither case governs everything and both have to be carried. A continuous structure responds to the pattern of a load and not to its total, which is why an unbalanced case that nobody would call severe is the one that sizes the member.both spans17 kN4.6 kNm-8.1one span9 kN6.2 kNm-4.0
Fig. 5 A two-span beam under the balanced load and under half of it, with the moment diagrams drawn to the same scale. The lighter case produces the larger span moment.

Both spans loaded gives a span moment of 9wL2/1289wL^2/128. One span loaded gives 49wL2/51249wL^2/512 — a third larger, for half the weight. The mechanism is the neighbouring span: when it is loaded, it pushes the shared support into hogging and that hogging reaches back into the span being designed, holding its sagging moment down. Clear the neighbour and the help goes away. The support moment halves at the same time, so nothing is uniformly worse and both arrangements have to be carried.

This is the same structure of argument as the imposed load envelope, and it deserves the same warning. The set of cases is not ordered. There is no worst one.

The envelope is not a state of the structureEvery arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 0e+0 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.024681012141618202224-200-100100200distance along the beam (m)bending moment (kNm, sagging up)the sagging envelopethe hogging envelopeeach case exact to 0e+0 · the envelope out by 23%
Fig. 6 Every arrangement of an imposed load on three spans, with the envelope over them. No single arrangement produces the envelope, which is why a load that can be present or absent is a set of states rather than a state.

Why the pattern cases are the ones that get missed

There is a practical reason the unbalanced case is skipped more often than any other in structural design, and it is worth naming because it is not laziness.

A balanced load is a number. It goes into a load schedule, it multiplies a tributary area, and every member gets the same one. An unbalanced case is a decision about geometry — which bay, which slope, which side of the ridge — and there are as many of them as there are ways to divide the roof. Nothing in the load schedule prompts anyone to make that decision, and a computer model handed the balanced case produces a complete, plausible, internally consistent set of results in which nothing is wrong except that the worst case was never run.

The check that finds it is the same check that finds every other pattern problem: draw the influence line for the quantity being designed, and put load only where it does harm.

Influence line for the bending moment at x = 6The bending moment at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 6.00, giving 3.600.the station being watched, x = 6unit load, at its worst position3.600shaded: where a spread load must stand to make this quantity worstthe horizontal axis is where the load is, not where the beam is cut
Fig. 7 The influence line for a bending moment at one station. Where it is positive, load makes the moment worse; where it is negative, load helps. An unbalanced snow case is nothing more than obeying that instruction.

That is a general instrument and it makes the snow question mechanical rather than imaginative. Load where the influence line is positive, leave the rest bare, and the arrangement that governs falls out without anybody having to guess which one it is.

Where the load acts, which is not where its total acts

There is a third consequence of the wedge shape, quieter than the other two.

A triangular load and a uniform load of the same total do not produce the same moment diagram, because their resultants sit in different places. Replacing a distributed load by its resultant is exact for the reactions and wrong for everything internal, which is the point the distributed load essay makes in general and which a drift makes concrete: the wedge’s centroid is a third of its length from the wall, so its effect on the member nearest the wall is larger than its weight suggests, and on the second member smaller.

A triangular load and the force that replaces itA triangular distributed load with its resultant computed by integration: an area of 15.6 acting at 8.00 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 15.6at x = 8.00, the centroid of the areamomentspread: 24.0replaced: 41.6reactions agree exactly (5.20 and 5.20); the peak moment does not
Fig. 8 A triangular load, its resultant, and what the substitution costs. The reactions are identical; the peak moment is not.

The roof next door, which is where this is worst

There is one arrangement that combines every effect above and it is common enough to be worth naming on its own: a low roof beside a tall building.

The tall wall is an obstruction of unlimited height, so the drift never saturates and the square-root curve runs as far as the fetch allows. The low roof is downwind of a large area of high roof, which supplies the fetch. And a steep upper roof sheds its own snow onto the lower one by sliding, which adds a quantity the mass balance above does not contain at all, arriving as a single event rather than accumulating.

The result is that the highest snow load on almost any project is on its lowest roof, at the point where it meets something taller, and that the member carrying it is usually a modest one chosen from a table of spans. Nothing about the geometry looks severe. A canopy, a link corridor, a plant enclosure roof at the base of a tower: each is a small structure whose load case is decided by a large one standing beside it.

The design consequence is uncomfortable and worth stating plainly. The load on the low roof depends on a building that may not be in the same ownership, may be built later, and may be extended upwards. A roof designed correctly for an open site becomes wrong on the day a neighbour builds tall, and nothing in the low building has changed. Of the loads a structure carries, that dependence on somebody else’s geometry belongs almost uniquely to snow.

Where the model stops

Three places, and the first is the one every number above rests on.

The moved fraction is a judgement. Nothing here computes how much snow the wind takes off a roof; 30% is a plausible value and the honest description of the whole calculation is that a measured ground load and an assumed scour fraction are combined by arithmetic that is exact. Every published drifting rule contains the same free parameter, dressed as a coefficient fitted to case records rather than as an assumption, and the fitting is what turns a mass balance into a design rule.

Snow slides, melts and refreezes. A slippery pitched roof sheds its snow at somewhere around 15° to 30°, so the load falls off a steep roof rather than accumulating — and lands on whatever is below it, which is how a low roof beside a tall one acquires the highest snow load anywhere in a project. A snow layer that partly melts and refreezes has a much higher density than the one that fell, and ice at the eaves can dam meltwater into a pond, which is a different load with a different failure mode behind it.

The obstruction is not always a parapet. A step in roof level, a plant enclosure, a row of photovoltaic panels, a wall of an adjacent building, a solid balustrade added by a later tenant: all of them are obstructions, and the last of those is the one that matters most, because it arrives after the structure was designed. A drift load is one of very few load cases that a change of use can create out of nothing without anybody applying for anything.

The resultant sits above mid-height, and nothing about the shape says soWindward pressure and leeward suction up the height of a 8 m building. The pressure coefficients do not vary with height at all; the velocity pressure does, because the wind is slower near the ground. So the resultant of a load whose SHAPE is constant sits at 57% of the height rather than at a half, and the overturning moment is 1.13 times what a uniform pressure of the same total would give. Base shear 157 kN, overturning 710 kNm.-0.4-0.20.20.40.602468pressure (kN/m²)height (m)resultant at 57%mid-heightwindwardleeward
Fig. 9 The other load the same wind brings. Snow and wind are usually treated as separate cases, and the wind that built the drift is not the wind the structure is checked for.

What the picture cannot show

The section shows a wedge with a clean triangular face, standing still. Neither is true.

Drifts are built and destroyed repeatedly through a winter. The one that governs is the deepest that ever existed, which nobody observed, and the profile is a succession of layers of different densities laid down in different storms, each with its own crust. A single triangle at a single unit weight is a stand-in for a stratified deposit whose density varies by a factor of two through its own depth.

Nor does the drawing show the plan. Snow does not pile in a two-dimensional section; it piles into a corner, and the deepest snow on any roof is nearly always at a re-entrant corner where two obstructions meet and the flow separates in both directions. The section is the easy case, drawn because it is the one with an arithmetic.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear14.9moment27.9 at x = 3.74the moment peaks exactly where the shear passes through zero
Fig. 10 The chain the load feeds. Whatever the depth of snow turns out to be, everything after it is the ordinary integration — load to shear to moment — and none of the difficulty in this essay lives in that part.

The generalisation

The habit worth carrying out of this is about where a load’s shape comes from.

Most loads in structural design inherit their shape from the structure: a floor load is uniform because floors are used uniformly, a wind pressure follows a face because faces are flat. Snow inherits its shape from the flow field around the building, which is a property of the shape of the roof and of nothing the designer chose deliberately. The consequence is that the interesting quantity is never the intensity in the schedule; it is the geometry of everything standing above roof level.

That makes snow the clearest available example of a rule this collection keeps arriving at from different directions. A load case is not a number. It is a statement about a structure, and two structures given the same number are given different loads. Earth pressure makes the point one way, by depending on how much the wall moves; silo pressure makes it another, by depending on the friction of the wall it acts on. Snow makes it geometrically, and more visibly than either: the load is literally in a different place on two roofs that received identical weather.

And the corollary is the one that closes the argument about patterns. A structure with more supports than equations has bought robustness at the price of caring where its load is, because the internal forces in a redundant structure depend on the arrangement of the load and not merely on its total. Continuity, which buys a moment over the support, is what makes the unbalanced case govern. It is the same purchase, read from the load’s side rather than the structure’s.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Bending momentConservation of massContinuityDistributed loadDriftEnvelopeFetchFree bodyLoad arrangementLoad pathServiceabilityShape coefficientSnow loadTributary areaUnbalanced load