Steel Shaft Torsion Capacity
Steel Shaft Torsion Capacity
(OP)
I'm looking for assistance in determining the ultimate torsion capacity of a square steel shaft (1.75" x 1.75", ASTM A29 material). The minimum yield stress is 90 ksi. According my advanced mechanics of materials textbook (Boresi et al, 5th edition), the plastic torsion capacity for this shape is 8/3*shear yield stress*(1.75"/2)^3. The result is approximately 8000 ft-lbs. I am using 0.6Fy as the shear yield stress. When comparing my result to published values from A.B. Chance Company for their helical anchor products, the maximum listed torque for a 1.75" square shaft is 10,000 ft-lbs or 11,000 ft-lbs (both values appear in their literature).
Can anyone shed light on the apparent discrepancy? Is my formual incorrect or might the A.B. Chance torque limits be based on actual testing rather than theoretical formulas?
Hokie93
Can anyone shed light on the apparent discrepancy? Is my formual incorrect or might the A.B. Chance torque limits be based on actual testing rather than theoretical formulas?
Hokie93






RE: Steel Shaft Torsion Capacity
RE: Steel Shaft Torsion Capacity
JAE, as for 0.6Fy, I think that's from the von Mises failure criterion? Shear yield stress = 0.577*Axial yield stress. Same as in AISC steel design, Chapter J shear yielding for example.
RE: Steel Shaft Torsion Capacity
RE: Steel Shaft Torsion Capacity
Hokie93, I dug out Boresi & Schmidt and I agree with you 8 kip-ft number. I would've used 0.577Fy, but that's small stuff.
RE: Steel Shaft Torsion Capacity
271828: thanks for the plastic capacity confirmation.
I'll give A.B. Chance a call to see if they can provide some insight into their published capacities.
Hokie93
RE: Steel Shaft Torsion Capacity
RE: Steel Shaft Torsion Capacity
RE: Steel Shaft Torsion Capacity
In principle, I don't agree with using the ultimate shear stress with a plastic stress distribution because there's no "yield plateau" at the ultimate stress. I doubt that the areas close to the centroid can have the ultimate shear stress simultaneously with the areas near the perimeter, which is a fundamental assumption when using the plastic distribution.
RE: Steel Shaft Torsion Capacity
RE: Steel Shaft Torsion Capacity
You are right that you won't develop fu across the entire section before the extreme fibres fails.
For plastic bending capacity of plates I have seen this equation for uniform plastic stress:
fp = (2fu + fy)/3
This makes sense because it is equivalent to taking fu at extreme fibre and fy at the centre.