Tag Archives: steel

Maximum height of an NYC skyscraper, including wind.

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Some months ago, I demonstrated that the maximum height of a concrete skyscraper was 45.8 miles, but there were many unrealistic assumptions. The size of the base was 100 mi2, about that of Sacramento, California; the shape was similar to that of the Eiffel tower, and there was no wind. This height is semi-reasonable; it’s about that of the mountains on Mars where there is a yellow sky and no wind, but it is 100 times taller than the tallest skyscraper on earth. the Burj Khalifa in Dubai, 2,426 ft., shown below. Now I’d like to include wind, and limit the skyscraper to a straight tower of a more normal size, a city-block square of manhattan, New York real-estate. That’s 198 feet on a side; this is three times the length of Gunther’s surveying chain, the standard for surveying in 1800.

Burj Khalifa, the world’s tallest building, Concrete + glass structure. Dubai tourism image.

As in our previous calculation, we can find the maximum height in the absence of windby balancing the skyscrapers likely strength agains its likely density. We’ll assume the structure is made from T1 steel, a low carbon, vanadium steel used in bridges, further assume that the structure occupies 1/10 of the floor area. Because the structure is only 1/10 of the area, the average yield strengthener the floor area is 1/10 that of T1 steel. This is 1/10 x 100,000 psi (pounds per square inch) = 10,000 psi. The density of T1 steel is 0.2833 pounds per cubic inch, but we’ll assume that the density of the skyscraper is about 1/4 this; (a skyscraper is mostly empty space). We find the average is 0.07 pounds per cubic inch. The height, is the strength divided by the density, thus

H’max-tower = 10,000psi / 0.07 p/in3 = 142, 857 inches = 11, 905 feet = 3629 m,

This is 4 1/4 times higher than the Burj Khalifa. The weight of this structure found from the volume of the structure times its average density, or 0.07 pounds per cubic inch x 123 x 1982x 11,905 = 56.45 billion pounds, or, in SI units, a weight of 251 GNt.

Lets compare this to the force of a steady wind. A steady wind can either either tip over the building by removing stress on the upwind side, or add so much extra stress to the down-wind side that the wall fails. The force of the wind is proportionals to the wind’s energy dissipation rate. I’ll assume a maximum wind speed of 120 mph, or 53.5 m/s. The force of the wind equals the area of the building, times a form factor, ƒ, times the rate of kinetic energy dissipation, 1/2ρv2. Thus,

F= (Area)*ƒ* 1/2ρv2, where ρ is the density of air, 1.29kg/m3.

The form factor, ƒ, is found to be 1.15 for a flat plane. I’ll presume that’s also the form factor for a skyscraper. I’ll take the wind area as

Area = W x H,

where W is the width of the tower, 60.35 m in SI, and the height, H, is what we wish to determine. It will be somewhat less than H’max-tower, =3629 m, the non-wind height. As an estimate for how much less, assume H = H’max-tower, =3629 m.
For this height tower, the force of the wind is found to be:

F = 3629 * 60.35* 2123 = 465 MNt.

This is 1/500 the weight of the building, but we still have to include the lever effect. The building is about 60.1 times taller than it is wide, and as a result the 465 MNt sideways force produces an additional 28.0 GNt force on the down-wind side, plus and a reduction of the same amount upwind. This is significant, but still only 1/9 the weight of the building. The effect of the wind therefore is to reduce the maximum height of this New York building by about 9 %, to a maximum height of 2.05 miles or 3300 m.

The tallest building of Europe is the Shard; it’s a cone. The Eiffel tower, built in the 1800s, is taller.

A cone is a better shape for a very tall tower, and it is the shape chosen for “the shard”, the second tallest building in Europe, but it’s not the ideal shape. The ideal, as before, is something like the Eiffel tower. You can show, though I will not, that even with wind, the maximum height of a conical building is three times as high as that of a straight building of the same base-area and construction. That is to say that the maximal height of a conical building is about 6 miles.

In the old days, one could say that a 2 or 6 mile building was inconceivable because of wind vibration, but we’ve found ways to deal with vibration, e.g. by using active damping. A somewhat bigger problem is elevators. A very tall building needs to have elevators in stages, perhaps 1/2 mile stages with exchanges (and shopping) in-between. Yet another problem is fire. To some extent you eliminate these problems by use of pre-mixed concrete, as was used in the Trump tower in New York, and later in the Burj Khalifa in Dubai.

The compressive strength of high-silica, low aggregate, UHPC-3 concrete is 135 MPa (about 19,500 psi), and the density is 2400 kg/m3 or about 0.0866 lb/in3. I will assume that 60% of the volume is empty and that 20% of the weight is support structure (For the steel building, above, I’d assumed 3/4 and 10%). In the absence of wind,

H’max-cylinder-concrete = .2 x 19,500 psi/(0.4 x.0866  lb/in3) = 112,587″ = 9,382 ft = 1.77 miles. This building is 79% the height of the previous, steel building, but less than half the weight, about 22,000,000,000 pounds. The effect of the wind will be to reduce the above height by about 14%, to 1.52 miles. I’m not sure that’s a fire-safe height, but it is an ego-boost height.

Robert Buxbaum. December 29, 2019.

How long could you make a suspension bridge?

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The above is one of the engineering questions that puzzled me as a student engineer at Brooklyn Technical High School and at Cooper Union in New York. The Brooklyn Bridge stood as a wonder of late 1800s engineering, and it had recently been eclipsed by the Verrazano bridge, a pure suspension bridge. At the time it was the longest and heaviest in the world. How long could a bridge be made, and why did Brooklyn bridge have those catenary cables, when the Verrazano didn’t? (Sometimes I’d imagine a Chinese engineer being asked the top question, and answering “Certainly, but How Long is my cousin.”)

I found the above problem unsolvable with the basic calculus at my disposal. because it was clear that both the angle of the main cable and its tension varied significantly along the length of the cable. Eventually I solved this problem using a big dose of geometry and vectors, as I’ll show.

Vector diagram of forces on the cable at the center-left of the bridge.

Vector diagram of forces on the cable at the center-left of the bridge.

Consider the above vector diagram (above) of forces on a section of the main cable near the center of the bridge. At the right, the center of the bridge, the cable is horizontal, and has a significant tension. Let’s call that T°. Away from the center of the bridge, there is a vertical cable supporting a fraction of  roadway. Lets call the force on this point w. It equals the weight of this section of cable and this section of roadway. Because of this weight, the main cable bends upward to the left and carries more tension than T°. The tangent (slope) of the upward curve will equal w/T°, and the new tension will be the vector sum along the new slope. From geometry, T= √(w2 +T°2).

Vector diagram of forces on the cable further from the center of the bridge.

Vector diagram of forces on the cable further from the center of the bridge.

As we continue from the center, there are more and more verticals, each supporting approximately the same weight, w. From geometry, if w weight is added at each vertical, the change in slope is always w/T° as shown. When you reach the towers, the weight of the bridge must equal 2T Sin Θ, where Θ is the angle of the bridge cable at the tower and T is the tension in the cable at the tower.

The limit to the weight of a bridge, and thus its length, is the maximum tension in the main cable, T, and the maximum angle, that at the towers. Θ. I assumed that the maximum bridge would be made of T1 bridge steel, the strongest material I could think of, with a tensile strength of 100,000 psi, and I imagined a maximum angle at the towers of 30°. Since there are two towers and sin 30° = 1/2, it becomes clear that, with this 30° angle cable, the tension at the tower must equal the total weight of the bridge. Interesting.

Now, to find the length of the bridge, note that the weight of the bridge is proportional to its length times the density and cross section of the metal. I imagined a bridge where the half of the weight was in the main cable, and the rest was in the roadway, cars and verticals. If the main cable is made of T1 “bridge steel”, the density of the cable is 0.2833 lb/in3, and the density of the bridge is twice this. If the bridge cable is at its yield strength, 100,000 psi, at the towers, it must be that each square inch of cable supports 50,000 pounds of cable and 50,000 lbs of cars, roadway and verticals. The maximum length (with no allowance for wind or a safety factor) is thus

L(max) = 100,000 psi / 2 x 0.2833 pounds/in3 = 176,500 inches = 14,700 feet = 2.79 miles.

This was more than three times the length of the Verrazano bridge, whose main span is ‎4,260 ft. I attributed the difference to safety factors, wind, price, etc. I then set out to calculate the height of the towers, and the only rational approach I could think of involved calculus. Fortunately, I could integrate for the curve now that I knew the slope changed linearly with distance from the center. That is for every length between verticals, the slope changes by the same amount, w/T°. This was to say that d2y/dx2 = w/T° and the curve this described was a parabola.

Rather than solving with heavy calculus, I noticed that the slope, dy/dx increases in proportion to x, and since the slope at the end, at L/2, was that of a 30° triangle, 1/√3, it was clear to me that

dy/dx = (x/(L/2))/√3

where x is the distance from the center of the bridge, and L is the length of the bridge, 14,700 ft. dy/dx = 2x/L√3.

We find that:
H = ∫dy = ∫ 2x/L√3 dx = L/4√3 = 2122 ft,

where H is the height of the towers. Calculated this way, the towers were quite tall, higher than that of any building then standing, but not impossibly high (the Dubai tower is higher). It was fairly clear that you didn’t want a tower much higher than this, though, suggesting that you didn’t want to go any higher than a 30° angle for the main cable.

I decided that suspension bridges had some advantages over other designs in that they avoid the problem of beam “buckling.’ Further, they readjust their shape somewhat to accommodate heavy point loads. Arch and truss bridges don’t do this, quite. Since the towers were quite a lot taller than any building then in existence, I came to I decide that this length, 2.79 miles, was about as long as you could make the main span of a bridge.

I later came to discover materials with a higher strength per weight (titanium, fiber glass, aramid, carbon fiber…) and came to think you could go longer, but the calculation is the same, and any practical bridge would be shorter, if only because of the need for a safety factor. I also came to recalculate the height of the towers without calculus, and got an answer that was shorter, for some versions, a hundred feet shorter, as shown here. In terms of wind, I note that you could make the bridge so heavy that you don’t have to worry about wind except for resonance effects. Those are the effects are significant, but were not my concern at the moment.

The Brooklyn Bridge showing its main cable suspension structure and its catenaries.

Now to discuss catenaries, the diagonal wires that support many modern bridges and that, on the Brooklyn bridge, provide  support at the ends of the spans only. Since the catenaries support some weight of the Brooklyn bridge, they decrease the need for very thick cables and very high towers. The benefit goes down as the catenary angle goes to the horizontal, though as the lower the angle the longer the catenary, and the lower the fraction of the force goes into lift. I suspect this is why Roebling used catenaries only near the Brooklyn bridge towers, for angles no more than about 45°. I was very proud of all this when I thought it through and explained it to a friend. It still gives me joy to explain it here.

Robert Buxbaum, May 16, 2019.  I’ve wondered about adding vibration dampers to very long bridges to decrease resonance problems. It seems like a good idea. Though I have never gone so far as to do calculations along these lines, I note that several of the world’s tallest buildings were made of concrete, not steel, because concrete provides natural vibration damping.