Allowable torsional shear stress for steel 1

[Guest090822]

Does anyone know how to determine the allowable torsional shear stress for steel? I don't believe AISC is a good resource as my situation is related to steel shaft design. I can easily determine the applied shear stress, but I have nothing to compare it to. My old college textbook just states that the maximum torsional shear stress at failure of a ductile material is half the yield strength. Since this is an academic book and not a code I'm not sure if there is something more strict.

 

[struct_eeyore]

A purely theoretical approach - find your principal stresses and apply Von Mises criterion.

 

[WARose]

As per AISC's Design Guide 9 (which is based on older codes), using LRFD, F_uv=ø(0.6)Fy. (With ø=0.9 IIRC.)

DG 9 combines the flexural shear & torsional shear by simple addition.

 

[Guest090822]

My situation is pure torsion. I don’t think the AISC building design code deals with members just subject to twisting only.

I will use a theoretical approach like Structee suggests. I thought something would be in a machine handbook but no luck.

 

[WARose]

I will use a theoretical approach like Structee suggests. I thought something would be in a machine handbook but no luck.

I don't see why you would use any "theoretical" approach. DG 9 gives allowable & ultimate stresses. (Per stress type and how to combine them. Whether we are talking "pure" torsion or not.)

I don't see the mystery here. (Last time I looked DG 9 is available on line for free. The codes may be a bit dated.....but the allowables are pretty much the same.)

 

[JoshPlumSE]

What type of members? If it's an HSS section then AISC does have torsional capacities defined in section H3.1, "Round and Rectangular HSS subject to Torsion".

There is a section H3.3 titled "Non-HSS Members subject to Torsion and Combined Stress". But, in my opinion, this section is pretty useless. Why? Because, the capacity of a WF or Channel under combined bending + torsion has a huge drop in capacity when the torsion goes from 0 to 0.000001 lb-in. Though I think it would be useful for member subject ONLY to torsion.

What I prefer is something alluded to in the AISC Design Guide #9.... 1) Take the torsional shear stresses and covert them into an equivalent shear force. Add that to the Flexural Shear force for the particular direction. Then you can compare that "equivalent total shear force" to the Shear Strength of the member. 2) Take the torsional warping stresses in the flange and convert them to an equivalent weak axis bending moment. Add that to the actual weak axis bending moment. Then you compare that "equivalent total weak axis moment" to the Moment Strength for the weak direction.

I think this is more rational than what AISC suggests and it doesn't cause any weird discontinuities in the combined stress equations when torsion, shear, and bending are combined.

 

[azcats]

Blodgett says: "...where, lacking test data, the ultimate shear strength of steel is assumed to be 75% of the material's ultimate tensile strength."

From Section 3.6 of Design of Weldments.

 

[Guest090822]

Thanks for the replies.

I’ll try to hunt down the D9 guide. I’ve been using AISC for over 20 years and I know the manuals inside and out. Chapter H was the first place I went but didn’t seem applicable to my situation.

If it helps, I’m designing a tube that is attached to a large valve that opens and closes very slowly and never rotates more than 90 degrees. The torque applied is 1,000 kip-inches. The tube will be vertical. The self weight is negligible, so we are looking at it as just pure torsion. Using an allowable torsional shear stress of 0.6 Fy with another factor of safety of 1.67 from AISC seemed too low.

I’m just trying to size either a pipe or round HSS section and I do have size constraints so obviously trying to minimize the diameter.

 

[Guest090822]

azcats - that’s interesting. So using AISC’s guidance with an Fy of 35 ksi my allowable would be 0.6x35x1/1.67)=12.57 ksi.

Using Bodgett I’d have .75 Fu, using Fu of 58 ksi then my allowable would be 43.5 ksi.

This is the type of discrepancy I’m concerned about. These numbers aren’t even close to each other.

 

[WARose]

So using AISC’s guidance with an Fy of 35 ksi my allowable would be 0.6x35x1/1.67)=12.57 ksi.

Where is this 1/1.67 coming from? I gave the allowable (as per LRFD) above.

As per ASD (again, following DG 9's recommendations), F_v=0.4(F_y).

Using a Fy of 35 ksi......that works out to 14 ksi.

 

[ESPcomposites]

A few comments, which explains some of background and also why you might see some discrepancies:

- This is a yield criterion (not ultimate), but for Tresca, Fsy = 0.50*Fty

- This is a yield criterion (not ultimate), but for VM, Fsy = 0.577*Fty (0.577 = sqrt(3)/3)

- Sometimes the yield criteria are applied to ultimate conditions (an extrapolation of sorts), so a number between 0.5 and 0.577 would be "about reasonable" (lacking any other information).

- You may find some information in the free MIL-HDBK-5 (or MMPDS if you have it). I looked at some steels and Ftu/Fsu (ultimate, not yield) ranged from 0.53 to 0.63. So the yield criteria solutions are fair approximations, but to be accurate, you will need actual material properties (as one would expect). Using 0.50 does seem to be conservative so if you can show acceptance with that, it might be acceptable if you do not have material property data. Lacking any other information, using a value greater than 0.53 would be unconservative for some steels.

- Side note: It is not required to determine the principle stresses to use VM (just apply it directly). In this case, its a moot point, but I wanted to explain that point.

Brian

http://www.espcomposites.com

 

[Guest090822]

WARose - DG9 is written the old way. AISC now written in both LRFD and ASD so the capacity equations are the same. The only difference is if you are doing and LRFD design you multiply by the resistance factor, if it's ASD you divide by "omega" which is the factor of safety. The 1.67 is the factor of safety for shear in both Chapter G and H in the 14th Edition of the AISC manual. Rewritten in the old ASD format would get 0.60/1.67 = 0.36 Fy for the allowable Fv which is pretty close to the 0.4 Fy (0.33 Fy in AASHTO) that you are probably more familiar with.

I've actually used DG9 many moons ago. I'm actually more curious as to where the 0.6 comes from in every shear equation, but that's a topic for a new thread!

ESPcomposites - the ultimate was what is suggested by Bloodgett based on what azcats posted. I agree that this should be based on yield. My textbook refers to the 1/2 Fy as the "Tresca yield criterion" so we are on the same page (no pun intended).

I'm going to go with 0.5 Fy and call it a day. That's a minim factor of safety of 2.0 which should be fine for this application.

 

[Hokie93]

Since the member in question is a closed tube (round HSS or pipe) and therefore the applied torsion will generate only shear stresses, I would say 0.5Fy does not provide a factor of safety equal to 2.0. The shear yield for most structural steel material is F_y/√3, or 0.577F_y, which most references round up to 0.60F_y. AISC then applies a factor of safety equal to 1.67 (for ASD) and, historically, has set the allowable shear stress equal to 0.40F_y. So I would recommend you set the allowable shear stress equal to 0.40F_y and call it a day or, if you want to use a factor of safety equal to 2.0, use 0.30F_y as the allowable shear stress. An allowable shear stress of 0.5Fy provides a factor of safety equal to 1.2 against shear yield.

 

[Guest090822]

Hokie93 - that’s a great point. I always wondered where the 0.6 came from! That makes perfect sense. I will use the 0.4 Fy since it the only thing I can find from a reputable code.

Thanks!

 

[JStephen]

My old "Design of Machine Elements" book from college uses Maximum Shear Theory of Failure for circular shafts, which gives 0.5Fy/FS for allowable torsional shear stress.
Beam bending and torsion are combined by figuring total torsional stress from Mohr's circle.
For non-static situations, factors ranging from 1.0 to 3.0 are applied to bending and torsion for fatigue and shock effects with "Stationary shafts, gradually applied load" getting a 1.0, up to 3.0 for "rotating shafts, suddenly applied loads, heavy shocks".
Reference in that book is to ASME B17c which was apparently replaced by ASME B106.1.