Using the load conditions at the beginning of this thread, 12" steel pipe deflects about 6 arc-sec. Too much. So, you either go bigger or just don't bump it in the first place.

This has been gnawing at me. So I redid my numbers. Hopefully this won't be too boring for the denizens of this thread. I thought about sending speedster a PM, but then thought some might find this interesting.

For a cantilevered beam (or column) with a force applied to the end:

d = PL^{3} / 3EI (sorry, no Greek font, that "d" should be delta)

With a bit of rearranging, since speedster calculated 6" arc (or 0.007", 240" * TAN(6" arc) ):

P = 3dEI / L^{3}

P = 3(0.007 in)(30x10^{6} lb/in^{2})(279 in^{4}) / (240 in)^{3} (note, I= 279 is for 12" x 0.375" wall pipe)

P = 12.7 lb (the load at the top of the 20'-0" tall pier pulling sideways).

Since speedster's criterion is 0.5" arc (0.0006 inches) deflection, I reorganized the deflection equation to find I:

I = PL^{3} / 3dE

I = (12.7 lb)(240 in)^{3} / 3(0.0006 in)(30x10^{6} lb/in^{2})

I = 3251 in^{4}

The area moment of inertia for a "hollow" circle is:

I = PI(R^{4} - r^{4}) / 4

Since, at least in the US, "standard weight" pipe is available, I'll assume the wall thickness will be 0.375":

I = PI(R^{4} - (R - 0.375 in)^{4}) / 4

Therefore (I wish I hadn't thrown my scratch paper away, I don't remember how I calculated this without expanding the polynomial above):

3251 __<__ PI[ (15 in)^{4} - (14.625 in)^{4} ] / 4

3251 __<__ 3829; therefore, 30" x 0.375" wall thickness pipe will be suitable. For reference, the moment of inertia of 24" x 0.375" pipe is 1942 in^{4}, so it isn't stiff enough to meet the 0.5" arc deflection criterion.

Those were the calculations behind my recommendation of 30" x 0.375" pipe for a 20'-0" tall pier with a maximum deflection of 0.5" arc under a 12.7 lb side load. Or, in metric: 762mm x 9.5mm pipe for a 6.1m tall pier with a maximum deflection of 0.5" arc under a 5.8kg side load.