Calculation of torsion 1

[joeswoes]

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.

 

[bobde3]

Thought you might be interested...Find a copy of Raymond Roark, Formulas for Stress and Strain and it will supply you with the info on calculating that stress. A good materials handbook should provide the necessary material properties for the 316 and the safety factor for the recommended torque will be something which is determined by the design of the part and the required LCF life of the part. I know of no standard.

 

[atlarson]

The relevant equations for this problem are (from Mott, Machine Elements in Mechanical Design):

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