# Calculating Torsional Capacity of Square HSS

I have having dificulty in determining the torsional capacity of a square HSS section. I am designing a shaft for 9.5 cubic yard mixer powered by a 80hp drive at 6 rpm output.

I have calculated the applied torque to be 70,000 ft-lbs (95 kNm). I would like to determine the size of HSS required if I use both 319 Stainless (Fv = 103 MPa) and 350W Mild Steel (Fv = 210 MPA) [Fv = allowable shear stress]. I will be comparing both materials as both offer advantages.

Is there a way to use the Torsional Constant [J] to determine allowable torque on a section? I have been able to calculate the required capacity for solid round, hollow round, and solid square sections. The method to determine the allowable torque on a hollow square shaft eludes me! It is my hypothesis that a 10x10x1/2" S.S. HSS will suffuce however I am not certain.

Any advise or direction would be greatly appreciated!

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nvn
Homework Helper
anicolajsen: No, torsional constant (It, sometimes called J or K), more aptly named torsional stiffness constant (It), would be for computing torsional deflection. You instead need torsional shear constant (Ct, sometimes called C), more aptly named torsional shear stress constant (Ct).

Your allowable shear stress currently looks incorrect. You listed the shear yield strength (actually Ssy = 0.577*350 = 202 MPa), not allowable shear stress (Ssa). Allowable shear stress involves, e.g., a yield factor of safety (FSy). For ground-based applications subjected to static loading, it is not uncommon to see FSy = 1.50 or 1.67. Because your loading is not static, you might want to use, e.g., FSy = 2.0. Therefore, your mild steel CSA G40.21, grade 350W, allowable shear stress would then be Ssa = Ssy/FSy = 202/2.0 = 101 MPa, not 210 MPa.

(1) For a 250 x 10.0 mm square tube, Ct = 1 065 000 mm^3. Therefore, your torsional shear stress would be tau = T/Ct = (94.91e6 N*mm)/(1 065 000 mm^3) = 89.12 MPa, which does not exceed Ssa = 101 MPa, and is therefore not overstressed, for mild steel grade 350W.

(2) For a 260 x 8.0 mm square tube, Ct = 956 000 mm^3. Therefore, tau = T/Ct = (94.91e6 N*mm)/(956 000 mm^3) = 99.28 MPa ≤ 101 MPa, and is therefore not overstressed, for mild steel grade 350W.

(3) For a 254 x 9.525 mm square tube, Ct = 1 070 000 mm^3. Therefore, tau = T/Ct = (94.91e6 N*mm)/(1 070 000 mm^3) = 88.70 MPa ≤ 101 MPa, and is therefore not overstressed, for mild steel grade 350W.

(4) For a 300 x 6.35 mm square tube, Ct = 1 043 000 mm^3. Therefore, tau = T/Ct = (94.91e6 N*mm)/(1 043 000 mm^3) = 91.00 MPa ≤ 101 MPa, and is therefore not overstressed, for mild steel grade 350W.​

I have not yet heard of stainless steel 319. Did you mean to say stainless steel SAE 316L (UNS S31603)? If not, what is the tensile yield strength (Sty), and tensile ultimate strength (Stu), of your stainless steel 319, and what is the material heat treatment condition?

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Thanks nvn!

Yes - I should not have used the term allowable shear stress for the values I stated as I will be incorporating a factor of safety.

We will be using S.S. 316L. Sources I found state that it has a tensile yeild strength (Sty) of 205 MPa (0.2% offset method). In your experience would it be safe to assume that Ssy=0.6(Sty) for this grade of stainless steel?

Thanks for the equation relating applied shear stress to torque and the torsional shear constant. It is exactly what I was looking for!

nvn
Homework Helper
anicolajsen: I usually use, Ssy = 0.577*Sty. I usually use, Ssu = 0.60*Stu for steel, and Ssu = 0.55*Stu for stainless steel.

Stainless steel SAE 316 (UNS S31600) is Sty = 205, and Stu = 515 MPa. For stainless steel SAE 316L (UNS S31603), Sty = 170, and Stu = 485 MPa.

One more thing nvn, where do you obtain your Ct values from?

I am in a similar situation as anicolajsen. I have a cantilever beam on which a load is creating bending and torsion. I am looking at using 5" x 5" x 3/8" square HHS ASTM A500 Gr B material. This material has a yield of 45,700 psi and ultimate tensile of 58,000 psi. FEA is resulting a 33,690 psi torsional stress. According to Blodgett in Design of Welded Structures, in the absence of test data the ultimate shear stress is assumed to be 75% of the materials ultimate tensile stress (resulting a 43,500 psi ultimate shear stress). This is almost a 1.3 safety factor. Any concerns with this approach?

SteamKing
Staff Emeritus