This is a rehash of an article I wrote in my weekly newsletter, “Back of the Envelope” — where I teach you SE-related things in 5 minutes (or less), once a week.
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Have you ever set a pencil down vertically on the table to pretend it's a skyscraper, only to then shake that table violently (or blow on it) to knock it down? (you monster)
A standard #2 pencil has a diameter of 0.28 inches, and a length of 7.48 inches — making your skyscraper's aspect ratio equal to about 1:27.
Turns out that they've actually built one of these pencil towers…
(Today's estimated read time = 4 minutes and 30 seconds)
When I first saw that picture (top of this page) of 111 West 57th Street (aka Steinway Tower), I thought it was perhaps Photoshopped.
Except… it's real life, not fantasy.
With a width of 59 ft and a height of 1,428 ft, its ratio is about 1:24 – making it the slenderest tower ever built (so far).
You probably have a lot of questions, such as:
1/ What is the building for?
Super-high-end residential; we are talking about $8M-$66M per unit.
2/ What does the view look like from up there?
3/ Who designed it?
Struct: WSP – formerly Parsons Brinckerhoff if you are wondering.
4/ How did they make the structure work?
Thick, high strength concrete shear walls on east & west + core; lots of 70-foot long rock anchors; plus a tuned mass damper at the top.
5/ What's the overturning and uplift?
Just kidding… well, we are structural engineers, so we should run some quick numbers for funsies. In today's issue, we will do a back of the envelope calc on 111-West-57-St for the following:
- Dead load
- Seismic load
- Wind load
- Overturning & uplift
Let's do this.
Side note: By the way, obviously they've done tons of serious analyses with this thing (wind tunnel tests, 3d dynamic models etc.); but hey, we are only running calcs with our trusty TI-83 and HP33s, so be easy on me here.
Quick-n-dirty dead load
Area: The width of the tower is constant at 59 feet, but the length sort of tapers back slightly as it gets taller. I scaled off some plans and got 76′. (So rough floor plate = 4484 sqft).
Slab: I have no idea how thick the slab is, but some videos mentioned 15′ floor-to-floor and 13′ ceiling. So maybe 8″ slab sounds about right (Slab weight = 100 psf).
Wall: 36″ thick at the lower levels and 16″ at the top, so average 26″. Guestimate some wall length (288′) x 15′ tall, divide by area, we get (Wall weight = 313 psf).
Terracotta: Estimate 15 psf but smeared into the floor plate (Cladding = 10 psf).
Miscellaneous: (Misc = 10 psf).
Tune mass damper: 800 ton (TMD = 1600 kip)
(Source: New York Times)
Steel structure: This is the unoccupied structure at the pinnacle. Guessing roughly 10 floors at 30 psf, with an average floor area of 2,000 sqft. (Pinnacle = 900 kip).
All in all, DL = 4484 sqft x 84 floors x (100+313+10+10)/1000 + 1600 + 900 = 165,000 kip.
Put that into perspective in terms of compression on the concrete walls: 165,000 kip / (36 in x 70 ft of wall x 2 x 12) = 2,728 psi
Nice. Seems right.
Quick-n-dirty seismic load
Grabbing some essentials from ATC:
- Sds = 0.304
- Sd1 = 0.096
Make some may-or-may-not-be-true assumptions:
- Risk category = III (not just II because the tower is for the super-rich. Jk? I am guessing it should be higher because it could cause serious damage to its surroundings if it falls down.)
- I = 1.25
- R = 5 (special reinforced concrete shear wall)
- h = 1428’
Not going to bore you with all of the math here but if we follow along with ASCE 7 Chapter 12:
- Cs (Eq 12.8-2) = 0.076
- Cs (Eq 12.8-3) = 0.005
- Cs (Eq 12.8-5) = 0.017
And if I put ASCE 7 words into a formula that you and I intuitively understand (from years of Excel-ing):
Cs = max(min(0.076,0.005),0.017) = 0.017
(I'm definitely not used to seeing this since I am in California… did I make a mistake?)
Seismic base shear is then 165000 x 0.017 = 2,759 kip
(For fun: 2759 / (70’ x 2) = 19.7 k/ft per shear wall.)
Not bad at all. So seismic probably not an issue compared to wind. Let’s see…
Quick-n-dirty wind load
Back to good o’ ATC:
V = 125mph
Make some not-so-accurate-but-probably-ok-enough assumptions:
- Kd = 0.85
- Exposure = C (I have no clue – feels like things are different high up there)
- Kzt = 1.0
- Ke = 1.0
- G = 0.85
Chugging along with ASCE 7 Chapter 26:
- Kz (@100’) = 1.27
- Kz (@500’) = 1.77
- Kz (@1000’) = 2.06
- Kz (@1400’) = 2.21
- qz (@100’) = 43 psf
- qz (@500’) = 60 psf
- qz (@1000’) = 70 psf
- qz (@1400’) = 75 psf
(Phew… almost there)
I am probably butchering this last part following ASCE 7 Chapter 27 but remember, this is just for fun – don't put me in timeout.
- GCp (windward) = 0.8
- GCp (leeward) = -0.5
- GCp (net) = 1.3
- p_net (@100’) = 56 psf
- p_net (@500’) = 79 psf
- p_net (@1000’) = 91 psf
- p_net (@1400’) = 98 psf
And the moment of truth:
Wind base shear = (56×100+79×400+91×500+98×400)x76’ = 9,264 kip
Hey look at that, at least three times higher than seismic.
Alright, I am running out of time (and steam) so going to bump this up a notch to “super-quick-n-dirty.
Super-quick-n-dirty overturning and uplift
OT = 9264 * 1400 / 2 = 6,484,800 kip-ft
T = 6484800 / 58’ = 111,807 kip (uplift from wind)
Net uplift = 111807 – 0.9 x (165000×0.5) = 37,557 kip
So many digits.
I saw a video that says there are about 200 x 3″ diameter rock anchors. (Here is the video – it's long but pretty good. Watch it at 2x speed for extra fun).
So let's say half of them take the tension.
Tension per anchor = 37557/100 = 376 kip
Stress per anchor = 376 / 7 in^2 = 54 ksi
Not too bad – seems reasonable.
(original image source: Youtube)
And… I am out. (why did I do that to myself?)
I suppose the point is that anyone, including you, could run a back-of-the-envelope calc to see if things make sense-ish.
There are, of course, wind tunnel tests to make the wind load more accurate. Also, in most cases for these towers, drift and occupancy comfort probably govern more than anything else.
But user “MIStructE_IRE” at Eng-Tips said it best:
“…they didn't have ETABS or spreadsheets when they built the empire state! I believe if you can't explain a fancy computer model using a couple of quick calcs and some free body diagrams – then either there's something wrong or the engineer hasn't a clue what they're doing!”
That's all for now. Hope you liked this, and I'll see you next week.