# Can a building get taller at night?

UberFacts recently tweeted "Office buildings are taller at night—a 1,300-foot-tall skyscraper shrinks about 1.5 millimeters under the weight of 50,000 occupants."

Is what they are saying valid? It seems unlikely, but 1.5 is relatively small compared to the actual height of the building. What is the most the building can shrink and under what circumstances?

• Going with Dave Peters suggestion maybe you want to rather restate the last part of your question like "Is there a way to calculate a rough estimate to support the claim in UberFacts?" – docscience Jul 24 '15 at 23:50
• I guess the other question to UberFacts is - did they measure this difference, or did they estimate it based on physical models. – docscience Jul 24 '15 at 23:52
• While it may seem counter intuitive, flexibility makes for a better structure. Too rigid is bad. The Empire State building's steel frame was built to be somewhat flexible, and (true story, I used to work there), office buildings would often develop cracks and we were told "that's supposed to happen, the building is designed to sway in the wind". It's completely reasonable that a building (as each person goes to work every day) gets a teeny tiny bit shorter from that person's weight. How much shorter? That would depend on the building. – userLTK Jul 25 '15 at 0:39
• Technically the building could shrink to a pile of rubble, e.g. if an earthquake hit, but presumably that's not what you meant. I guess you're talking about elastic deformation: how much can the building's size change without permanently altering its structure? A clarification to the question along those lines would probably make it better, and help clarify the relevant physics. – David Z Jul 25 '15 at 3:15
• This suggests the followup question: How does the height difference caused by different weight loads compare to height differences by different (outside) temperatures? – Hagen von Eitzen Jul 25 '15 at 10:29

I don't think this sounds unreasonable as an estimate at all. Let's check it.

One designs a building as a compromise between two competing factors:

1. One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ultimately failing catastrophically;

2. However, mildly stressed load bearing members mean underutilized members: if we overdesign things too much we push the cost up enormously.

Let's for simplicity consider the compressive load bearing strength of the building against its own weight: it needs to bear shear stresses from wind as well, but considering the weight alone will get us to a ballpark figure.

Typical materials behave under stress in a way described by a curve with the following shape:

This curve is one I took from the Wikipedia "Compressive Stress" page.

So we want all the load bearing members to undergo stress which is in the low-stress, linear part of this curve. The upper point of the linear region is called the yield point and we need to be well lower than this, because this is where the supporting material will undergo plastic, irreversible deformation and ultimately be at risk of failing. But we don't want to be too far below this point, otherwise we are adding material, and cost, needlessly to the building.

I don't know what civil engineers use, but a factor of one quarter of the yield stress seems reasonable. For concrete, the curve is more linear than the above up to the point where the concrete fails: it does so when it reaches a compression of about $40{\rm MPa}$ (See the "Engineering Toolbox here ). In its linear working region, its Young's modulus is $E = 30{\rm GPa}$, so at a stress of one quarter its strength (i.e. at $10{\rm MPa}$) its compressive strain is

$$\epsilon = \frac{\sigma}{E} = \frac{10\times 10^6{\rm Pa}}{30\times 10^9{\rm Pa}}\approx 3.3\times 10^{-4}$$

For steel, high strength structural steel has a yield strength of somewhere in the neighborhood of $300{\rm MPa}$ and its Young's modulus is $200{\rm GPa}$. So when it is "working" at one quarter of its yield strength, we have:

$$\epsilon = \frac{\sigma}{E} = \frac{75\times 10^6{\rm Pa}}{200\times 10^9{\rm Pa}}\approx 3.8\times 10^{-4}$$

i.e. rather near the value for concrete. So whether the load is borne by concrete or steel, at the same "utilization", i.e. same fraction of yield strength, both materials show about the same strain.

Now, the cost optimal configuration is with all the load bearing material in the building working at the same fraction of its yield strength. This means that load bearing strength will decrease with height: the upper load bearing members only have to hold up the building above them. This is not quite right - for Taipei 101, for example, there is a huge pendulum high in the building to counteract wind swaying - but an assumption that the whole structure is in roughly the same compressive state is probably a good working estimate.

This then is our key to our estimate:

The whole structure along its whole height undergoes a constant compressive strain of $\epsilon = 3.5\times10^{-4}$

50 000 people put a load of about $5\times 10^4\times 60{\rm kg}$ or about 3000 tonnes on the building. From here I get an estimate of 700 000 tonnes for the total mass of Taipei 101. The building's mass is much greater than that of the people, so we assume that at 700 000 tonnes, the building's strain is $\epsilon = 3.5\times10^{-4}$. This means that 3000 tonnes spread uniformly throughout the building adds a further strain of:

$$\Delta\epsilon \approx \frac{3\times 10^3}{7\times 10^5}\times 3.5\times 10^{-4}\approx 1.5\times 10^{-6}$$

For a building of the height of Taipei 101 (500m), this corresponds to a height shrinking of $1.5\times 10^{-6}\times 500 = 750{\rm \mu m}=0.75{\rm mm}$.

This is remarkably near to your estimate. Given than I am no civil engineer, it is altogether possible that different strength materials are used, and that it may be safe practice to design nearer to the yield strength than I have assumed. A figure of 1.5mm implies a constant "utilization" of materials at one half of their yield strength: this too sounds reasonable (a little bit scary for someone untrained such as I, but still believable). It is probably also cheaper to use higher cost, higher strength materials in the lower sections of the building and lower cost, lower strength in the upper part, thus deviating from my constant utilization / constant strain with height assumption.

I therefore find your quoted shrinking to be highly believable. It's definitely in the right ballpark.

• How much of a factor would expansion due to temperature be? I'd expect it to be much more than 1.5 millimeters. – Davor Jul 25 '15 at 12:46
• @Davor That's a very good point. Both steel and concrete have $\alpha$ about $10^{-5}{\rm m\,K^{-1}}$, so temperature effects are about ten times as big as this strain for each degree kelvin. I was a little surprised at how small the effect came out to be, but then 50 000 people is a very much smaller mass than the mass of the building. So we find that the peoples' weight effect is about a hundredth of the temperature effect (one part in $10^{4}$ for 10K change) which in turn is of the same order of the strain begotten by the building's own weight. – Selene Routley Jul 25 '15 at 13:17
• OK, so the claim is essentially false: skyscrapers are shorter at night than during the day, because they're cooler. On the other hand, they're taller at weekends than during the week. – David Richerby Jul 25 '15 at 16:24
• @DavidRicherby: if they are taller at weekends than during the week, it will be due to the effect of vehicle exhaust fumes on urban temperature ;-) Or in some cases they might be taller at weekends in summer (because the air-con is switched off) and shorter at weekends in winter (because the heating is switched off). – Steve Jessop Jul 25 '15 at 17:49

Young's modulus is the ratio of tensile stress to tensile strain for a material:

E = (F/A)/(∆L/L) = (F * L) / (A * ∆L)

F/A is force per area, and (∆L/L) is change in length per original length

For structural steel, Young's modulus is 200 gigapascals. This quantity can be used to predict how much the steel will compress under a given weight per unit area.

The ratio of a skyscraper's height to its width typically is 5/1 or 7/1. I would expect a 1300 foot skyscraper to be near the low end, so the side of a floor would likely be 1300/5=260, which would make the footprint 260^2 = 67,600 sf. Wildly guessing at the cross sectional area of each steel post and the number of posts bearing the building, I'm going to say that the combined steel cross section is 5% of the footprint, or 3,380 sf = 314 square meters.

The weight of 50,000 people is roughly 7,000,000 pounds, or 31,138,000 newtons of force applied to 314 square meters of steel, or 99,165 newtons per square meter. However, because the upper posts bear progressively less weight than the lower posts, I am going to say that this is borne by half of 1,300 = 650 linear feet, or 198 meters of steel.

You can use Young's modulus to compute the amount of compression during the day and expansion at night when the people are gone (ignoring the effect of temperature variation between night and day):

200 gigapascals = (31,138,000 * 198) / (314 * ∆L)

∆L = 0.103 meters = 103 millimeters, which is a great deal more than 1.5 millimeters tweeted by UberFacts. The discrepancy between me and UberFacts is large. Perhaps due to my assumptions?

• I think most of the load is on concrete, not steel. – LDC3 Jul 25 '15 at 3:59
• @LDC3 Actually, surprisingly it doesn't matter much whether the building be mainly borne up by steel or concrete. At the cost optimal point, all materials are at the same "utilization", i.e. fraction of their yield strength. At a quarter of their yield strength, both steel and concrete are remarkably alike: they undergo a strain at this stress of about $3.5\times 10^{-4}$, as I found in my answer. – Selene Routley Jul 25 '15 at 4:14
• What are these "feet" and "pounds"? (Choice of units aside, I've never understood why Americans quote huge weights in pounds. It's like saying that the distance from New York to Chicago is fifty million inches. Having said that, fifty million inches is such a beautifully round number, and correct to 3sf if we take Google Maps' 789 miles as the actual distance, that I might have to start using it.) – David Richerby Jul 25 '15 at 16:29

Let us make an estimate. Let the skyscraper be 400 m tall, each storey 4 m high, 500 people per storey, 20 m2 per person, 10000 m2 per storey, let us assume that the building is a 100x100 m2 square in the plan and that it only has 10 cm thick structural walls in a 25x25m2 grid. So the cross-section area of the structural walls is 2x5x100x0.1 m2=100 m2. Let us assume that an average person's weight is 1000 N, and the modulus of elasticity of concrete is 50 GPa. Thus, a weight of 100 m2 x 50 GPa= 5x10^12 N would contract the concrete walls by 400 m (sounds stupid, but never mind). Then the weight of the people 1000 N x 50000=5x10^7 N would contract the concrete walls by 400 m/10^5=4mm. So your figure of 1.5 mm does not seem outrageous. Let me note that 4 mm is additional contraction with respect to contraction under the dead weight of the building. Let me also note that the actual structure may have structural columns , rather than structural walls.

EDIT: @Ernie mentioned the effect of thermal expansion/contraction. Looks like this effect is much more important. The thermal expansion coefficient of concrete is about 10^(-5)/deg. C, therefore, the expansion/ contraction is 400 m x 10^(-5)/deg. C = 4 mm/deg. C.