Structural Engineering & Statics
where the load goes
Equilibrium

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

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 Weight is the only thing resisting it.

18 min read 8 figures The load must go somewhereDrawing as calculationThe load that will not hold still

Nothing in this essay explains why the wind does what it does. The pressure coefficients used here are measured data — wind tunnel values that have been stable for sixty years, and which belong to a different subject. What this collection can do with them is the thing it always does: name the free body and add up the forces.

The adding up has three results that a drawing of arrows on the front of a building does not contain.

The wind pushes on one face and pulls on threeA 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to.windward c_p = 0.8leeward c_p = -0.5sidec_p = -0.7sidec_p = -0.7842 kN38% of it fromthe back face30 × 20 m
Fig. 1 A thirty by twenty metre building in plan, with the measured coefficient on each face and the arrows drawn in the direction the pressure acts. One face is pushed and three are pulled.

Which free body produced the number

The whole building, cut at ground level. Every external surface carries a pressure, and the pressures are usually written as a coefficient on the velocity pressure qq:

face cpc_p qcpq c_p
windward +0.80 +0.72 kN/m²
leeward −0.50 −0.45
sides (each) −0.70 −0.63
roof −0.70 to −0.30 −0.63 to −0.27

Positive means pressing inward. One face out of four is pressed and the other three are sucked, which is the first thing the arrows in a hand sketch usually get wrong.

For a 30 m by 24 m windward face at q=0.9q = 0.9 kPa, that is 518 kN pushing. The leeward face is 30 by 24 as well, at 0.45 kPa of suction: 324 kN pulling downwind. A suction on the back pulls the building in exactly the direction the pressure on the front pushes it, so the two add:

V=q(cpw−cpl) bH=0.9×1.3×30×24=842 kNV = q(c_{pw} - c_{pl})\, b H = 0.9 \times 1.3 \times 30 \times 24 = 842\ \text{kN}

and 38.5% of it is supplied by a face nothing is blowing on.

The faces that carry the most and contribute nothing

The side faces carry 0.63 kN/m² each — the second largest external pressure on the building, on the largest areas — and they contribute exactly zero to the horizontal resultant, because they are equal and opposite.

That combination is worth dwelling on, because it produces two different design problems from one figure.

The stability system cares about the resultant. It sees 842 kN and nothing at all from the sides.

The cladding cares about the local pressure. On the side faces it is carrying 302 kN of suction per face, on panels whose fixings are working in tension away from the building.

Those are answered by different people from the same drawing, and the second one is where the failures happen. A cladding panel that departs on a side elevation is not a stability failure and it does not appear anywhere in the base shear.

The two faces carrying most, and carrying none of itThe net pressure on each surface of a 30 × 20 × 24 m building, in kN/m², at a velocity pressure of 0.9 kPa and with the inside at c_pi = -0.3. Inward is positive. The horizontal resultant is the windward face plus the leeward face, because a suction on the back pulls the building downwind exactly as a pressure on the front pushes it — and that comes to 842 kN, of which 38% is supplied by the face the wind never reaches. The two side faces carry 0.36 kN/m², the largest pressure anywhere on the building, and contribute nothing whatever to it: they are equal and opposite. The coefficients are wind tunnel measurements, not results — what is computed here is only the adding up.windward0.99leeward-0.18side (near)-0.36side (far)-0.36roof (windward half)-0.36roof (leeward half)0.00-0.4-0.20.20.40.60.81net pressure (kN/m²) — inward positivebase shear 842 kN · 38% of it from the back face
Fig. 2 The net pressure on every surface, with the two that make up the base shear marked apart from the four that do not. The largest number on the chart contributes nothing to the resultant.

The inside of the building, which nobody draws

A building is not sealed. Whatever pressure exists inside acts outward on every surface at once, and that single fact has two opposite consequences.

On the horizontal resultant it cancels exactly. An internal pressure pushes out on the windward face and out on the leeward face, and those two effects are equal and opposite on the resultant. Sealed, the base shear is 842 kN. With a door blown in on the windward face, it is 842 kN. Not approximately — the same number, because the internal pressure acts on two equal opposed areas.

On the roof it does not cancel at all. The roof has no opposite face; whatever is pushing up on it from inside is simply added to the suction already on it from outside.

cpic_{pi} net roof coefficient uplift
sealed −0.3 −0.4 to 0.0 108 kN
leeward opening −0.5 −0.2 to +0.2 0 kN
windward opening +0.7 −1.4 to −1.0 648 kN

The roof of this building weighs 360 kN. Sealed, it is held down with a comfortable margin. With a dominant opening on the windward face it is 288 kN short, and it leaves.

The door blows in first, and then the roof leavesUplift on the roof against the internal pressure coefficient, with the roof's own weight drawn across it. Internal pressure acts outward on every surface at once, so it cancels exactly on the horizontal resultant and does not cancel at all on the roof, where the external coefficient is already a suction. A sealed building sits at about c_pi = −0.3 and the roof stays down. Open a dominant door on the windward face and the inside goes to +0.7: the net roof coefficient doubles, the uplift reaches 648 kN against 360 kN of roof, and it lifts. The crossing is at c_pi = 0.17.-0.4-0.20.20.40.60.8200400600internal pressure coefficient c_piroof uplift (kN)the roof's own weightsealeda door on the windward faceuplift 648 kN against a roof of 360 kN
Fig. 3 Roof uplift against the internal pressure coefficient, with the roof’s own weight drawn across it. A sealed building sits on the left of the crossing and a building with a windward door blown in sits on the right.

That is the whole mechanism behind the standard failure of a light building in a storm, and the order of events matters: the door goes first, and then the roof. It is a statics result, not a meteorological one, and it is why the wind loading standards treat internal pressure as a range to be bracketed rather than as a value to be computed. It is an envelope in the strict sense: no single state produces every number in it.

It also reverses which surfaces are worst. With cpi=+0.7c_{pi} = +0.7 the windward face carries a net 0.09 kN/m² — almost nothing — while the sides carry 1.26. The cladding design for a building with a large opening is a completely different problem from the one for a sealed building, on the same elevation with the same wind.

Where the resultant sits, which is not mid-height

The coefficients above do not vary with height. The velocity pressure does, because the wind is slower near the ground.

So the load has a constant shape and a varying magnitude, and the resultant of it sits at 59% of the height rather than at 50%. The overturning moment is therefore 18% larger than the same total force applied uniformly would give.

That is a small-looking number with a large consequence, because overturning is checked against a weight and a width and has no material property in it. Eighteen per cent on the demand side of that check is eighteen per cent of the whole answer.

The resultant sits above mid-height, and nothing about the shape says soWindward pressure and leeward suction up the height of a 24 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 59% of the height rather than at a half, and the overturning moment is 1.18 times what a uniform pressure of the same total would give. Base shear 576 kN, overturning 8156 kNm.-0.4-0.20.20.40.60.805101520pressure (kN/m²)height (m)resultant at 59%mid-heightwindwardleeward
Fig. 4 The pressure up the height. The coefficients are constant and the velocity pressure is not, so a load whose shape never changes has a resultant above mid-height and an overturning moment larger than its total would suggest.

The two combinations that have to be checked, and why

A designer handed the table above has a decision to make that the table does not make for them: which internal pressure to use.

The honest answer is both, because they govern different checks and they are not both conservative at once.

cpi=+0.7c_{pi} = +0.7 maximises the roof uplift and the outward pressure on the side and leeward cladding. It is the case that takes roofs off and blows walls out.

cpi=−0.3c_{pi} = -0.3 maximises the inward pressure on the windward face, which is where a cladding panel is pushed against its fixings rather than pulled off them, and where a large glazed panel reaches its own bending limit.

Neither is a worst case for everything, and — importantly — neither affects the base shear, so the stability designer can ignore the whole question while the cladding designer cannot. That division is why internal pressure is so often left out of a project’s load schedule: the person who writes the schedule is usually doing the first job.

The other reason the pair has to be carried is that the internal pressure is not a property of the building as designed. It is a property of the building as operated — which door is open, whether the roller shutter was left up, whether a window failed early in the storm. A building with a large opening on one face has a different internal pressure depending on the wind direction, and the design has to cover a building whose openings are not known.

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. 5 The general form of the same difficulty. A load that can be present or absent produces an envelope rather than a case, and no single arrangement is worst for every member — so the check is against a set of states rather than against a state.

What the plan shape buys

The coefficients are properties of the shape, so changing the shape changes the load — which is the one structural intervention available on the demand side rather than the capacity side.

A deeper plan narrows the wake relative to the building and reduces the leeward suction from −0.5 to about −0.3. The base shear falls from 842 kN to 713, and the back’s share falls from 38.5% to 27.3%. A building twice as deep along the wind is not twice as loaded across it; it is slightly less loaded.

A circular plan halves the drag for the same projected width, which is why very tall towers are round or heavily chamfered. It also raises the side suction to about −1.0, so the cladding gets worse while the frame gets better — the two are not optimised together.

Corners are where the local suctions are severe and where every cladding failure starts, and rounding or chamfering them is the cheapest available intervention on the worst local pressure in the building.

None of that is derivable here. It is measured, and the honest description is that the structural engineer takes a table of numbers from another discipline and does arithmetic with them — which is worth stating plainly rather than dressing up.

Where a beam's load comes fromA 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 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.15.0 m²9.38 kN/m15.0 m²9.38 kN/m9.0 m²7.50 kN/m9.0 m²7.50 kN/m8 m6 m48.0 m² divided, 48.0 m² of panel — the division closes
Fig. 6 The other half of turning a pressure into a member force. A coefficient gives a pressure; a tributary area gives a load; and which member carries which piece of the elevation is a decision about the framing rather than about the wind.
The pressure under a base that cannot pullBearing pressure under a 4 × 3 m base carrying 900 kN, at four eccentricities. Inside the middle third — ±0.67 m here — the pressure is a trapezoid and the whole base is working. Beyond it the base lifts: at e = 1.00 m only 3.00 m of the 4 m is in contact and the peak pressure is 200 kPa against 75 kPa at no eccentricity.e = 0.00 m75 kPawholly in bearinge = 0.33 m113 kPawholly in bearinge = 0.67 m150 kPawholly in bearingthe middle third, exactlye = 1.00 m200 kPa1.00 m lifted
Fig. 7 Where the horizontal resultant ends up. A base shear applied above mid-height is an overturning moment against a weight and a width, and nothing about the strength of the structure enters that comparison.

What is not in any of this

Three things, and the first is the one this essay’s neighbours are about.

The wind is not steady. Everything above is a static pressure applied to a static building. The energy is spread across four decades of frequency, a tall building takes most of its response from a narrow band of it, and the gust factor that turns a mean pressure into a design one is a dynamic calculation wearing a static number’s clothes.

The wind is not one direction. The coefficients above are for wind normal to a face. At forty-five degrees the pressures redistribute, the resultant acquires a component the frame was not designed for, and — the case that actually governs many buildings — the whole thing acquires a torsion about its plan which a symmetric analysis never sees.

The building is not rigid. A structure that sways moves through the air it is being loaded by, which is where galloping and flutter live. For ordinary buildings this is a small correction; for very slender ones it is the design.

Where this model stops

The pressure coefficients are averages over a face, and cladding is not designed on averages. The local peak near a corner, a parapet or a roof edge can be three times the face average, over a small area, and every code carries a separate set of much larger coefficients for local pressures on small tributary areas.

That is not a refinement of the same calculation; it is a different calculation with a different set of data. The face averages here are for overall stability and are wrong for anything smaller than a face.

The tabulated coefficients also assume the building stands alone in open country. A building in a group is shielded, funnelled, or both, and the interference effects between neighbouring towers can double a local pressure — which is why the very tall ones are tested in a tunnel with their neighbours modelled around them rather than taken from a table at all.

Where the wind's energy is, and where the structure can reach itThe gust spectrum at a mean speed of 25 m/s, plotted as n·S(n) against frequency on a logarithmic axis so that equal areas are equal energy. Its total variance is 18.75 m²/s², which the closed form 6·K·U² gives as 18.75. Most of the energy is below a hundredth of a hertz — gusts lasting a minute or more. A structure at 0.25 Hz sits far out on the tail, and still takes 86% of its response from there, because the resonant part is amplified by π·f₀/(4·ζ) and the damping is 1.2%. 10⁻³10⁻²10⁻¹10¹00.20.40.60.81frequency (Hz)n·S(n), scaled to its own peakthe structure, 0.25 Hzbackground — the structure following the gusts: 74% of the turesonant — the structure ringing: 86% of the responsegust factor on the mean response: 4.01
Fig. 8 Why the static number is a stand-in. The wind’s energy is spread across frequencies, and a design pressure is a statement about a peak within an averaging time — which is a convention rather than a measurement.

Reading the coefficients honestly

There is a habit of language worth resisting here, because it hides where the knowledge comes from.

It is common to say that the wind “applies” a pressure of 0.72 kN/m² to the windward face. What is actually true is that a model of this shape, in a boundary-layer tunnel, at a reference wind speed, produced a mean pressure of 0.8 times the dynamic pressure over that face — averaged over some seconds, at some scale, in some approach terrain.

Every one of those qualifications is a place the number can be wrong for a particular building, and the standards handle them by being conservative in the aggregate rather than accurate in each. That is a different kind of number from, say, a section’s second moment, which is exact for the section drawn.

The practical consequence is that refining the structural arithmetic past a couple of significant figures is refining the wrong end of the calculation, which is the same warning the soil modulus issues in another problem. The base shear here is 842 kN; it is not 842.4, and the third digit is a statement about the multiplication rather than about the wind.

What the picture cannot show

The plan view draws pressures as arrows on faces, which makes the load look like something applied to a surface. It is applied to a surface that has to be attached to something, and the load path from a suction on a side elevation runs through the cladding fixing, into the rail, into the column, into the floor plate acting as a diaphragm, and only then into the stability system. Four load paths, of which the base shear calculation contains the last one. Each of the first three is a member whose tributary area was decided by a layout drawing rather than by a load case.

Nor does the picture show that these are coefficients. They are dimensionless numbers derived from a model in a tunnel, transferred to a building by an assumption that pressure scales with dynamic pressure and shape and not with size. That assumption is very good and it is an assumption, and the places it fails — very small buildings, very large ones, unusual proportions — are exactly the places wind tunnel testing is required.

The other resultant, which is a torsion

The plan drawing has one more thing in it that the arithmetic above passed over.

The windward pressure and the leeward suction are both uniform across the width of their faces, so their resultants act at the centre of the plan and produce no torsion. That is true of the idealised pressure distribution and it is not true of any real one: the coefficients vary across a face, the two side faces are rarely loaded identically because the wind is rarely exactly normal, and a building with an asymmetric plan has no reason for its pressure resultant to pass through anything in particular.

Standards handle it by requiring the resultant to be applied at an eccentricity — typically a tenth of the plan width — whether or not the shape suggests one. That is not a refinement of the pressure calculation; it is an admission that the pressure calculation is an average of something that fluctuates, and that the fluctuations do not have to be symmetric at any instant.

The consequence lands on the torsional stiffness of the plan rather than on its lateral stiffness, and the two are decided by different features of the layout. A building with all its stiffness in one core is a poor torsional structure with a good lateral one, and the eccentricity requirement is what makes that visible.

The generalisation

The habit worth carrying is about signs on a free body.

The mistake this essay is really about is not “forgetting the leeward face”. It is drawing a free body and then thinking about only the parts of it that look like loads. A suction is a load. A pressure on the far side of a body pushing towards the near side is a load in the near-side direction. Two equal opposite pressures on opposite faces are not a load at all, however large each of them is.

Those are three separate statements about the same figure, and the only reliable way to get all three right is the one this site keeps returning to: draw the body, put everything that touches it on the drawing, and then add. Reasoning about which contributions “matter” before the sum has been done is how a third of a wind load goes missing.

Where this goes next

2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs The free body is a choice, and choosing it well is the whole skill Cutting a structure open is not a step in the method. It is the method — and where the cut is made decides whether the answer takes one line or twenty. 48 kNW = 120 kNmiddle third: ±0.42 mresultant at 1.20 mrestoring 150 kNmoverturning 144 kNmfactor 1.04 Weight is the only thing resisting it A structure that is strong enough everywhere can still be blown over, and nothing in its material properties has any part in whether it is. The whole answer is a weight and a width — and the failure begins long before anything tips, at the moment one edge stops pressing down. 10⁻³10⁻²10⁻¹10¹00.20.40.60.81frequency (Hz)n·S(n), scaled to its own peakthe structure, 0.2 Hza stiffer one, 2 Hzbackground — the structure following the gusts: 70% of the turesonant — the structure ringing: 90% of the responsegust factor on the mean response: 4.39 The wind is a spectrum A wind load is quoted as a pressure, which suggests something steady. It is not — the energy is spread across four decades of frequency, almost all of it in gusts lasting minutes, and a tall building takes ninety per cent of its response from the sliver of that energy sitting at its own frequency. wall 146 mmframe 140 mmtogether 58 mm20 storeys at 3.5 m · 40 kN per floor How a tall building stands still A shear wall bends and a framed tube shears, and the two deflected shapes are the wrong way up for each other. Tie them together at every floor and the pair is stiffer than the sum of their stiffnesses — because near the base the wall holds the frame back and near the top the frame holds the wall.

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.

Base shearCladdingDistributed loadEquilibriumFree bodyInternal pressureLateral systemLoad pathOverturningPressure coefficientResultantSuctionUpliftVelocity pressureWind pressure