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#30702 - 10/16/09 02:56 AM Principal shear stress theory
Zahid_mech Offline
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Registered: 10/16/09
Posts: 25
Loc: India
For calculating the allowable stress the code B31.3 is based on Max principal stress theory (Max Shear stress theory). According to which, the allowable Stress must be 0.5 time the Yield stress. But in code the allowable stress is 2/3 (0.667) of Yield stress. I am not able to understand why code takes an allowable stress value which is greater than value allowed by Max shear stress theory. Please explain and correct me if I am wrong. Thanks.
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#30713 - 10/16/09 10:57 AM Re: Principal shear stress theory [Re: Zahid_mech]
Dave Diehl Offline
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Registered: 12/14/99
Posts: 2382
Loc: Houston, TX, USA
Yes, maximum shear stress theory is applied but things are simplified in the Codes - Codes built for slide rules. If maximum shear stress is the radius of Mohr's circle, then "stress intensity" is the diameter. Why go through the trouble of dividing the diameter by two and the yield stress by two when you can just compare that stress intensity to yield? So, even though Appendix A may be based on yield stress (or other, lower limitatioins) Tresca Theory of failure (max shear) is applied.
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#31057 - 11/03/09 12:34 AM Re: Principal shear stress theory [Re: Dave Diehl]
Zahid_mech Offline
Member

Registered: 10/16/09
Posts: 25
Loc: India
Thanks Dave,

From your explaination what I have understood is ...

If
Sb = calculated bending stress
St = Calculated torsional stress

As Max shear stress is Tmax = [ (Sb/2)^2 + St^2 ]^0.5
But we calculate 2*Tmax = [ Sb^2 + 4*St^2 ]^0.5 and take this as displacement expansion stress "SE". And this is directly compared with yield stress Sy, instead of comparing Tmax with 0.5*Sy.

Thanks for this clarification.

Regards
Zahid.
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Zahid

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