Calculation of torsion
Calculation of torsion
(OP)
Is there a formula which can be applied to determine the allowable torsion (torque) on a 1" diameter rod of 316 stainless steel, expressed in foot-lbs. ? What is standard safety factor for recommended torque below this value (75%, 50%,...).
Thanks in advance.
Thanks in advance.





RE: Calculation of torsion
RE: Calculation of torsion
Ssr = T/Zp
Zp = pi * D^3 / 16
Ssa = Ssy / Fs
Ssy = Sty / 2
where
Ssr = resulting shear stress (psi)
T = applied torque (in-lb)
Zp = polar section modulus (in^3)
D = diameter = 1 in.
Ssa = allowable shear stress (psi)
Ssy = shear yield strength (psi)
Fs = safety factor
Sty = tensile yield strength = 30000 psi for annealed 316 SS
Combining equations, setting Ssr equal to Ssa, and solving for T gives:
T = (Sty * pi * D^3) / (32 * Fs)
or in this case
T = (30000 * pi * 1^3) / (32 * Fs)
= 2945 / Fs (in-lb)
= 245 / Fs (ft-lb)
The safety factor (Fs) that you use depends on many variabless including the uncertanties in your loads, restraints, and material properties, whether the loads are cyclic, and what harm will occur to people or property if the part fails. I use a safety factor of at least 2 for almost all applications, and more typically 3 or higher.
For your benefit, please check my math.
ATL