Rail Bend in Hot Weather

Date: 10/13/2002 at 00:40:38
From: Jaysen
Subject: Rail Bend

Hi Dr. Math, my question to you is this:

A 20-ft piece of rail expands 1 in. in length during a hot spell. If 
there are no expansion gaps (spaces found between rails that allow a 
rail to expand without buckling), how high off the ground will the 
rail rise?

I don't have the slightest idea of how or what you can do to attempt 
to solve this problem. The only thing I do know is that it would cause 
the rail to make an arc. Can you help me out?

Date: 10/14/2002 at 07:18:44
From: Doctor Jeremiah
Subject: Re: Rail Bend

Hi Jaysen,

The easiest way to figure this out is to draw a picture:

                         + 20'1" +
                +                         +
       +                                           +
  +-------------------------20'0"-----------------------+

The distance along the rail is 20 feet 1 inch, and the distance along 
the rail is 20 feet 0 inches.

The curved rail is part of a circle, so the ends of the rail must be 
on the circle.  This means that they can be at the ends of a radius:

                         + 20'1" +
                +                         +
       +                                           +
  +------------------------20'0"------------------------+
     +                                               +
        +                                         +
           +                                   +
              +                             +
                 R                       R
                    +                 +
                       +           +
                          +  a  +
                             +

Now we really have two shapes: a triangle and a section of a circle.  
The circle has a circumerence of 2(Pi)R but the section is 
20"1'/2(Pi)R percent of a whole circle. And since a whole circle is 
360 degrees, the section is 360 x 20"1'/2(Pi)R degrees:

          a = 360 x 20"1'/2(Pi)R degrees

The triangle has three sides, but it is symmetric so we could just use 
half a tringle to get one with a right angle:

  +----------20'0"/2---------+
     +                       |
        +                    |
           +                 |
              +              |
                 R           |
                    +        |
                       + a/2 |
                          +  |
                             +

Now the sine of a/2 is (20'0"/2)/R, which means that:

   sin(a/2) = (20'0"/2)/R
        a/2 = arcsin( (20'0"/2)/R )
          a = 2 x arcsin( (20'0"/2)/R )

Now we have two equations for a. Since they are both equal to a, they 
must be equal to each other, so equate them and solve for R. Once you 
know R, find h (from the triangle). The height the rail is off the 
ground is R-h.

                     === + 20'1" +
                +     |                   +
       +              |                            +
  +-------------------|----20'0"---===------------------+
     +                |             |                +
        +             |             |             +
           +          |             |          +
              +       R             h       +
                 R    |             |    R
                    + |             | +
                      |+           +|
                      |   +     +   |
                     ===     +     ===

See if you can get it from here. If you get stuck, please write back.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/