math question

How long do you calculate it will take us to brand you a TROLL?

What did I miss?

Aside from not being worded perfectly, it sounded like an innocent question to me. I assumed he wanted to know how many foot-pounds of torque are needed to bend a 1.25in, 0.125 wall mild steel tubing around a

Actually, I doubted whether it would even be possible without heat, and even then, whether it would leave the tubing still looking like tubing.

Jim

OK, here goes:

Stress = M*C/I

C = 1.25/2 = 0.625 in

I = (Ro^4-Ri^4)/2 Ro = 1.25/2 = .625 in Ri = Ro-.125 = 0.50 in

I = (.635^4 - .5^4)/2 = 0.045 in^4

Yeild stress for AIST 1020 cold rolled steel is 66, 000 psi

so

M = 4752 in-lb = 396 Ft-lb.

A lot of assmptions here, but if you apply about 400 ft-lbs, the round pipe will just start to bend. Figure another 50 or so ft-lbs to account for friction, variability in the steel, etc. Of coarse, if the pipe is other than AISI 1020 you would need to adjust the assumed yield stress. If you are trying to estimate safety of a structure, apply a Factor of Safety of at least 5.0 (10 would be better if people are going to be near it.)

H>

I think you are right there. Bending 1" .010 wall tube around a 6" radius without crimping it is a real challange. At a minimum it would require an internal mandrel

He's more talking 1" water pipe here Glenn... ;)

Tim

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