Steel Shaft Torsion Capacity

[Hokie93]

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

 

[JAE]

Why are you using 0.6Fy (which is typically an "allowable" stress) to find an ultimate capacity?

 

[271828]

I'll try to remember to break out Boresi tomorrow. My first guess is that they're doing something less sophisticated than what you're doing. Could simply be wrong too, LOL.

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.

 

[GalileoG]

JAE just meant that the author of this thread said in his post that he used 0.6Fy for "shear yield stress" when he meant to say the allowable yield stress.

 

[271828]

Clansman, I disagree. "Shear yield stress" is correct. He needs the shear yield stress to plug into the 8/3*... equation.

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.

 

[Hokie93]

Thanks to all of you for the prompt responses. I was rounding to 0.6Fy for the shear yield stress, rather than using 0.577Fy - sorry if that was misleading.

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

 

[PSlem]

Seems you are using the wrong number for ultimate shear. Blodgett's Design of Welded Structues and Machinery's Handbook put it at 75% of tensile. Chance's 1-3/4" square shaft is 70 ksi yield and 100 ksi tensile. Plugging 75 ksi in the formula gives you 11,165'#. Chance says you can exceed the yield and twist the shaft without harm.

 

[Ussuri]

Shear yield is (1/[√]3) x tensile yield and is determined from the von Mise yield criterion under a pure shear condition.

 

[271828]

PSlem, that's an interesting idea. It would be very interesting to see if the Chance number can be duplicated with either 0.75*Fu or 0.577*Fu as the shear stress used in the plastic torsion equation. Another idea is to assume that failure = this ultimate shear stress at the extreme fiber and forget the plastic distribution.

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.

 

[rb1957]

maybe the higher numbers (published by the manufacturer) are based on test, which would probably exceed allowables based on calculation. maybe they properly derated their test result for the material strength of their specimen (probably higher than spec minimums).

 

[Tomfh]

271828,

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.