A while back I did a post about counterintuitive math problems. However, I deliberately held one back, since I was using it as my Problem of the Week for that week. So here it is:

Suppose a steel beam, one mile long, is fastened securely to the ground at each end. As the
day heats up, the metal expands. Let us assume that at the hottest part of the day, the metal
is actually one mile and one foot long. Let us further assume that the beam is fastened in such
a way that it can only buckle upward, and not side to side. Your problem is to estimate how
high the beam will rise above the ground. Would you be able to slip a playing card under it?
A pencil? Would you be able to crawl under it? Walk under it? Could an elephant (roughly
13 ft tall) walk under it? How about a person standing on the back of an elephant? That sort
of thing. Be sure to justify your answer!

Keep in mind I’m only asking for an approximation here, so feel free to help yourself to whatever (reasonable) simplifying assumptions you want. Or you can just follow the link above. This one was Problem Nine.

Comments

  1. #1 Mark Hanna
    New Zealand
    May 25, 2013

    I’m new to commenting here, do you mind if we post our answers in the comments or is that discouraged?

    P. S. Is it just me, or does the link to your answer (in the PDF for the question behind that link) no longer work?

  2. #2 Steven Carr
    May 25, 2013

    I make it that it would rise about 50 feet into the air.

    I took it as bucking the highest in the middle, and approximating it as 2 right angled triangles.

  3. #3 Mark Hanna
    New Zealand
    May 25, 2013

    Haha, I guess that basically answers my question. Here’s what I was going to post before my uncertainty made me ask if it was alright first:

    My approximation was to treat the curve as though it were 2 right angle triangles back to back, looking kind of like this: /|\
    While obviously this approximation falls apart for greater angles, for an increase in length as small as 1 part in 5280 I figured it would suffice. With this approximation, I already know the length of the bottom side of a triangle (half a mile), and I know the length of the hypotenuse (half a mile plus half an inch), so I can use pythagoras’ theorem to calculate the length of the vertical side and thus approximate the height at the middle of the beam.

    The answer I came up with was 15.7 m, or about 51.5 feet if I’m to fit with the theme. If an elephant is 13 ft tall, I guess that means you could just about have a tower 4 elephants tall fit under the beam at the middle. Certainly higher than I’d have expected!

  4. #4 Cthulhu
    R'Lyeh
    May 25, 2013

    If one increases the circumfence of a circle by 1ft, the radius increase is about 1.9in. Since the beam attached to the earth is only part of earth circumfence, I take this as a lower bound. For the upper bound, I take triangle constructions. Assuming the highest point to be in the centre of the beam, the construction of two right triangles and Pythagoras’ gives a height of about 103ft. Somewhere between these to bounds, the actual value will be, most likely closer to the 103ft, as the true form of the beam will be close to an cosh, which for these values won’t deviate much from the triangle construction.

  5. #5 Cthulhu
    R'Lyeh
    May 25, 2013

    Sorry, the upper bound is about 51.5ft, I forgot to divide the mile and the foot by 2.

  6. #6 G.
    May 25, 2013

    (Dyslexic with decent conceptual grasp here, about to do something terribly inelegant but none the less practical;-)

    The first thing that came to mind was: This reduces to a civil engineering problem. Span a width W with an arch of length L at its underside: find height H of arch. Think of a freeway underpass.

    Long-span arches are hinged at points in the length of the arch, to allow for expansion & contraction of the materials. An arch with a single hinge at its midpoint can be roughly abstracted to a triangle. Admittedly this is not going to get me a CalTrans contract, but some simple eidetic imagery plus the Pythagorean theorem and a hand calculator gets me the following result:

    Visualize the arch abstracted to the triangular form of a conventional peaked roof, which is an isosceles triangle. The base is 5,280 feet in length. Each side is 1/2 of 5,281 feet in length, thus 2,640.5 feet. Find the height of the triangle. One isosceles triangle is equal to two right triangles stuck together along their respective heights. So we have two right triangles, each of which has a base of 2,640 feet and a hypotenuse of 2,640.5 feet.

    The base (a) 2,640 squared = 6,969,600 feet.
    The hypotenuse (c) 2,640.5 squared = 6,972,240.25 feet.
    a^2 + b^2 = c^2, therefore c^2 – a^2 = b^2, which is 2,640.25 feet.
    Square root of 2,640.25 is 51.38 feet, which is the height (b) of the right triangle that forms half of our peaked roof.

    In reality we’re still dealing with an arc or arch, which is not going to be as high at its peak as a triangle or roof. In my layperson’s guts I suspect this is going to turn out to be somewhere between 40 and 50 feet.

    A quick online search doesn’t turn up equations for calculating the height of a circular arc (or if it did, I didn’t understand what I was looking at), but it does turn up “the complete circular arc calculator.” Plugging in the numbers for width and length returns a result that the height of the arc is 44.49846 feet. (Right in the middle of my gut-estimated range, though I don’t deserve credit for that, because the range is so darn large. By analogy, if I was doing science, I couldn’t claim that a hypothesis is supported when the result of an experiment is p < .2. Had I guessed "approx. 45 feet," that would be a creditable answer.)

    If I was building an arched structure using the above methods, I would have used the low end of my wild guess, which was 40 feet. If I still needed to clear 50 feet of height, I would have re-calculated the triangle for a height closer to 60 feet, on the basis that when converted to an arc (arch) that would be "safe" to clear 50 feet of height. That would have gotten me the following:

    The base (a) 2,640 squared = 6,969,600 feet.
    The height (b) 60 squared = 3,600 feet.
    a^2 plus b^ 2 = 6,973,200.
    Square root of 6,973,200 = 2,640.682.
    Double that to get length of our hypothetical peaked roof, 5,281.36 feet.

    So to get the extra approx. 10' of clearance under a triangular roof, only entails increasing the total length of the roof members by 0.36 foot (about 4-1/3 inches) over the entire span. That should be sufficient, when translated back to an arc, to get us an arch of equivalent length with 50' of clearance under it.

    Apologies for the length of this comment… OK, tell me where I screwed up;-)

  7. #7 Reginald Selkirk
    May 25, 2013

    Don’t forget to consider the curvature of the Earth. And is the beam oriented east-west or north-south?

  8. #8 miller
    SF bay
    May 25, 2013

    I treated the beam as a circular arc, and used the first two terms in the Taylor expansions of the trig functions. I solved for r and theta, and then for r(1-cos(theta/2)) ~ r theta^2/8.

    I get sqrt(6/n)/4 miles, where n is 5280. No calculator on hand, but it’s around 45 feet. Interesting that it goes as the square root of the length change for small changes.

  9. #9 Corkscrew
    May 25, 2013

    Assume the steel bar forms a section of a circle. The circle has radius r and the bar is an arc of 2.theta radians. Then r, theta satisfy:
    2.r.sin(theta) = 1 – i.e. fastenings are 1 mile apart
    2.r.theta = 1+1/5280 – i.e. length of bar is 1 mile + 1 foot

    By a bit of trial and error in Excel, we have:
    theta = 0.0337 (to 3 significant figures)
    r = 29.7

    Then the displacement is:
    r – r.cos(theta) = 0.0168 miles = 89.0 feet

  10. #10 Divalent
    May 25, 2013

    Similar to others above, I modeled the situation as a right triangle, with the mile+1ft lenth the hypotenuse and the mile length as one side. Then used the pythagorean theorem to get the other side. Result was ~103 ft.

    So divide by 2 to get height in the middle of ~51 ft.

  11. #11 Tony Sidaway
    London
    May 25, 2013

    The beam increases by 12 inches per mile of circumference so intuitively I compute that the radius increases by about one-sixth (1/2pi) of that, or two inches per mile. The circumference of the earth is about 25,000 miles, so the beam hoists itself up by about 50,000 inches. Around 0.8 miles

  12. #12 Tony Sidaway
    LondonThe beam increases by 12 inches per mile of circumference so intuitively I compute that the radius increases by about one-sixth (1/2pi) of that, or two inches per mile. The circumference of the earth is about 25,000 miles, so the beam hoists itself up by about 50,000 inches. Around 0.8 miles
    May 25, 2013

    So where did I go wrong? A 25,000 mile band around the earth increases by 25,000 feet, so isn’t it going to increase the diameter by about a sixth of that? Taking the beam as an arc of that seems to be the right approach.

  13. #13 RM
    May 25, 2013

    Tony, where you went wrong is to assume that you have a beam that surrounds the earth, all expanding at 1 foot per mile. You don’t. At best you would have one 1 mile section expanding 1 foot, and 24,999 miles of band not expanding (Because the rest of the earth isn’t made of the steel beam) – 1 extra foot in the circumference, not 25,000 extra feet. (And that’s ignoring the fact that the two ends of the 1 mile beam are tacked to the current circumference of the earth.)

  14. #14 Blaine
    May 25, 2013

    Assuming the non-curvature of the earth, the solution is the circle of which the original beam is the chord. Allowing the curvature of the earth, the solution is the lune of Hippocrates. For either one, take the distance at the midpoint.

    Assuming that the earth is flat, and the radius of the earth is 3959 miles, the distance is 2 inches.

    Secant method:
    Distance = ( 3959 – sqrt( 3959^2 – 0.25))*5280 * 12

  15. #15 Blaine
    May 25, 2013

    Whoops – messed up – looks like it ~ 50 ft.

  16. #16 bad Jim
    May 26, 2013

    I did the quick triangular solution like everyone else, but It wasn’t satisfying, somehow. I don’t want to admit how much time I spent trying, and failing, to find an analytic solution for the circular arch. The approximate solution is a clearance of 44.88 ft, based on a radius of 77,662 ft and an angle of 3.9° or 0.034 radians. Perhaps a catenary curve would have been more appropriate.

    A quick experiment with a sheet of paper demonstrates that tiny displacements open noticeable gaps. I’m grateful that I wasn’t a mechanical engineer.

  17. #17 JimR
    May 26, 2013

    I setup the same equations as bad Jim in a spreadsheet and get 44.50′ clearance at a radius of 78,327′ and angle of 3.863 degrees. I do, however, have the radian value of the angle as 0.0674? I realize the spreadsheet has a solver function, but I am not conversant with it.

  18. #18 Count Iblis
    May 26, 2013

    If you assume that the beam was straight to begin with (which is not so realistic), then using the Euler-Bernoulli beam equation, you find that an increase in beam length from L to L + delta L will lead it to buckle upward in the middle by:

    1/4 sqrt(6 L delta L)

    assuming that deltaL/L is small. This yields 44.5 feet.

    This is not the correct solution, because you need to take into account the fact that the beam wasn’t straight to begin with due to its weight.

  19. #19 Count Iblis
    May 26, 2013

    I missed that the beam is initially resting on the ground. At first if the beam has expanded a bit, it will only rise off the ground at the two ends. If all of the beam is off the ground, then the height y(x) will be of the form:

    y(x) = A [x^2 - (L/2)^2] + B [x^4 - (L/2)^4]

    where we’ve taken the ends at x = +/- L/2. For a steel beam with circular cross section, the constant B is approximately given by:

    B = 6.42*10^(-8)/R^2

    where R is the radius of the beam and all lengths are in meters (the dimension of B is 1/Length^3).

    Then A can de determined by computing the arc lenght and equating that to L + delta L. Ths computation will then only be valid if y(x) > 0 for all x between -L/2 and L/2. Note that when all of the beam is off the ground, you’ll at first have a minimum at the middle at two maxima inbetween the middle and the ends.

  20. #20 bad Jim
    May 26, 2013

    JimR, the value I gave in radians was for the half-angle, not the full angle. Sorry about that. I was solving for x/sin x=1.0002

  21. #21 Greg Esres
    May 26, 2013

    Seems like the weight of the beam would prevent one uniform curve; might it look like a McDonald’s double arch? or even a sin wave?

    Since the problem is supposed to be counter-intuitive, 50 ft can’t be right.

  22. #22 MNb
    May 26, 2013

    Take a Cartesian coordinate system. Draw a circle with the geometrical centre in origin O and radius r.
    Draw a chord PQ parallel to the x-axis. The coordinates of Q are (x,y). Let T be the top of the arc PQ. The length of arc PQ I call l’ and the length of chord PQ I call l. Note that l = 2x.
    Beta is the angle between OQ and OT. Usually we calculate with the angle Alpha between the x-axis and OQ, so we have Alpha + Beta = 90 degrees.

    We can calculate height h, also called sagitta according to Wikipedia by using:

    (1) sin (Beta) / Beta = l/(2l’)
    (2) r = l’/(Beta)
    (3) square (r – h) = square (r) – square (x)
    Beta is expressed in radians.

    The problem is of course calculating Beta, but I guess somebody else can write a program for it. In JR’s problem l = 1 mile and l’ = 1 mile + 1 foot (when will you bloody Americans learn to work with the metric system?!).

    Proof: As Q = (x,y) we know that x = (1/2) l and h = r – y. Because of Pythagoras we also know that square (x) + square (y) = square (r). Hence equation (3).
    Let O be the perimeter of the circle. Then we get l’/O = (Beta)/(2pi). This is btw the reason I use angle Beta and not Alpha. As O = 2pi times r we get equation (2).
    Alas it’s very inconvenient to solve r, so I prefer to use sin (Beta) = x/r. Substitute r and we get (1).
    I suppose that this is the same as @19, but I am too lazy to check.

  23. #23 Omega Centauri
    May 26, 2013

    There is a problem with your description. In mechanical engineering those endpoint constraints would be called single point constraints. The issue is do they apply simply to the position of the ends of the beam? Or do they also apply to the rotational degrees of freedom of the beam at the endpoints? The former case is simpler, as the beam would then be in the shape of part of a circle, with a chord length of 5280feet, and a length along the circumference of 5281feet. If the constraints are also to the rotational degrees of freedom, the “mode” shape is more complicated, with zero slope at the ends……

  24. #24 Count Iblis
    May 26, 2013

    I have found the solution, albeit it in a very rough approximation. If you take the beam radius to be 30 centimeters, it will come off the ground only about 84 meters from both ends. The beam height from one end at x = 0 till x = p = 84 meters is:

    y(x) = c x^2 (x – p)^2

    where c = 7.13*10^(-7) meters^(-3).

    The maximum height is then reached at x = p/2 where it is about 2.2 meters.

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  26. #26 Roy
    May 27, 2013

    Using a linear approximation the maximum deflection is the height of a triangle 5,280 ft at the base and (5,280 + 1) ft the sum of the other two sides. Treating it as two right-triangles and applying Pythagorean’s theorem yields 51 ft 5 in.

    The true deflected beam would take the shape of a catenary curve (hyperbolic cosine), the same as an actual arch, sagging cable, etc. supported at their two ends and not the shape of a circular arc- thus making the circular arc the arch-rival :-)

  27. #27 lordaxil
    May 27, 2013

    The equation for the deflected beam is a quartic (as in #19) not a catenary, since the beam can support a bending moment and a freely hanging chain cannot. As #23 points out, if the ends are clamped without free rotation, the solution is different, but only by a constant.

  28. #28 eric
    May 27, 2013

    Huh, everyone’s saying circle but I figured it would settle into a parabolic curve…isn’t that the more stable end point of deformation?

  29. #29 bad Jim
    May 28, 2013

    It turns out that rails for high-speed trains are welded together into a continuous track. They’re laid when the temperature is roughly average, and maintain dimensional stability by being tightly fastened to the road bed.

    One solution to the original problem is a six-inch high step on either end (or a sigmoid for those of us compulsively committed to continuously differentiable functions); another is an evenly distributed series of wrinkles, which could be very, very small.

  30. #30 Steve Jones
    Bedford, UK
    May 28, 2013

    Let d be the height, and theta the angle it makes to the horizontal; it is reasonable to assume that theta is small and is approximately equal to its tangent. Therefore, approximately, theta equals d divided by 2640 (in ft!). Also, since theta is small, cos(theta) = sq.rt(1 – theta squared), or approximately 1 – theta(squared)/2. But cos(theta) also equals 5280 / (5280+1), or approximately 1-5280(reciprocal). Therefore, theta(sq) = 1/2640, making theta approximately 1/51. Thus d is approximately 2640/51, or approximately 51 ft. You could therefore ride an elephant under the bar with plenty to spare!

  31. #31 William Hendrixson
    May 28, 2013

    Let L be the length of curved steel/2
    Let l be the length of straight steel/2
    radius r, angle theta, height h

    L= theta*r
    l=r*sin theta

    (L/theta)sin theta = l

    (sin theta)/theta = l/L

    2640/2640.5=0.9998

    Graphing sin theta/theta gives theta= 0.0375 rad

    r=l/sin theta r=2640/.03749 r=70419

    2640/(r-h)=tan theta 2640/.037518=70419-h

    70419-70366=h= 53 feet

  32. #32 Anthony
    May 28, 2013

    It doesn’t rise above the ground at all. Young’s modulus for a steel bar is typically around 200 GPa, and strain is 1/5280, so the pressure is 37 MPa. If we assume a steel bar that is 1m cross-section, its total weight under Earth gravity is ~125 million newtons, which is more than sufficient to prevent it from rising.

  33. #33 Max
    May 28, 2013

    Seems to me that at best it would rise barely enough for a playing card. 1 foot spread over a mile is practically nothing.

  34. #34 Jim Thomerson
    May 29, 2013

    Perhaps this has some practical application. When they were finishing up the Arch in St. Louis, they wanted to have a ceremony with installation of the last small apical section. The mayor wanted to do it in early afternoon for maximum publicity. This was turned down because the Arch would have heated to the point that the last section could not be inserted. They had the insertion and ceremony at 9:00AM.

  35. #35 Thad
    May 30, 2013

    Funny how most people here did a more precise calculation than the answer given in the source! Anyways if we add some physics to the problem it gets much more complicated! The force exerted on the two ends of the beam due to the heat expanding the beam has to counteract the force of gravity downward along the length of the beam. Because the beam can bend (a normal beam would be analogues to a long string) it will be bent back down to the ground as you move away from the anchor point. Because this would not slow for enough expansion the beam would rise again for a distance allowed by the strength of the beam. Here we get a wave like pattern and because each rise of the beam allows for a portion of the required expansion each can be added to the overall foot it needs to expand. The rise of each section depends on the number of rises which depends on the ratio of the strength of the beam and its weight per unit length. But that’s just my guess about what would actually happen. Maybe it’s a good question for XKCD!

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