How do I calculate the diameter and wall thickness of a round steel tube that needs to support 515lbs across 32ft with a max of 0.03" deflection per inch?

How do I calculate the diameter and wall thickness of a round steel tube that needs to support 515lbs across 32ft with a max of 0.03" deflection per inch?

First we need to compute the absolute deflection at the center of the beam.

Let's assume the beam sags in the shape of a circular chord (i.e. an arc). The maximum slope will be at the outer edges (i.e. at the 0' and 32' marks), where we want it to be no greater than 0.03, and the greatest deflection will occur in the center (at the 16' mark). For very small slopes like this, we can use a small-angle approximation which says that the angle (in radians) subtended by the half the circular chord (from the center point to one edge) is approximately equal to that slope of 0.03. Meanwhile, we also know by definition that the subtended angle (in radians) is equal to the half the length of the chord divided by the radius of curvature. We can use these two ways of computing the angle to write an equation so we can solve for the radius of curvature:

\frac{half length}{radius}=\frac{192}{r} \approx 0.03

r \approx 6400 inches.

Now that we know the radius of curvature and the subtended angle, we can compute the deflection in the center as the difference between the radius and the radius times the cosine of the angle.

D = 6400 - 6400\cos(0.03) \approx 2.88 inches. [Again, remember that the angle is in radians, not degrees. If you want to do the calculation in degrees, convert 0.03 radians \approx 1.72 degrees].

Now that you know the deflection in the center of the beam is about 2.88 inches, you can find a suitable beam using an online calculator such as this one:

Beam Deflection Calculators - Solid Rectangular Beams, Hollow Rectangular Beams, Solid Round Beams (http://www.calculatoredge.com/civil%20engg%20calculator/beam.htm#tube)

For example, I see that an 8" diameter steel tube with 0.75" wall thickness just barely does the trick.

I hope that helps! I didn't spend much time explaining the math up above, so if you want more detail just ask.

Actually, now that I went through all that, there's a much, much simpler way. For a circular arc, if you were to draw a line from the center at 16' (point of maximum deflection) to the edge at 0' (point of maximum slope), the slope of the straight line will be exactly half the slope of the arc at the steepest point. Therefore, the slope of the straight line must be 0.015 or less.

slope=\frac{rise}{run}=\frac{maximum \; deflection}{distance \; to \; center}=\frac{D}{192}=0.015

D=2.88

That's a much easier way to compute it!