Procedures for calculating shoulder-fillet stresses should take multiaxial stresses into account, even if loading is uniaxial.

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“Shoulder-fillet stresses finesse,” Jan. 8, 2009 |

A recent article in MACHINE DESIGN by Steve Tipton, James Sorem, and Curtis Schmidt detailing more accurate calculations for stresses in shoulder fillets (“Shoulder Fillets Stress Finesse,” Jan. 8, 2009, p. 55-60) generated questions from readers looking to put those equations to use.

Reader Lee Ruiz, for example, felt the article was unclear about the uniaxial, biaxial, or multiaxial nature of the stresses in question.

“The article mentions the relationship *K _{q} = σ_{eq} /S_{nom}. σ_{eq} *normally indicates a multiaxial stress. However the article seems to imply a uniaxial stress for

*S*by using it in both

_{nom}*K*and

_{t}*K*ratios,” Ruiz wrote. “Readers might think only one uniaxial load is required for an

_{q}*S*value used to calculate

_{nom}*σ*. Since

_{eq}*K*for a given

_{q}< K_{t}*D/d*and

*r/d*ratio, using a uniaxial

*S*value for calculating both

_{nom}*σ*and

_{eq }*σ*

_{1}will result in

*σ*

_{eq}always less than

*σ*

_{1}— which is not always the case.”

Tipton responded to clarify a few points about the notch root stress state and the definitions of stress-concentration factors (SCFs) *K _{t}* and

*K*.

_{q}"First, nominal stresses, *S _{nom}*, are computed as though a smooth, round bar of diameter

*d*is loaded in bending, tension, or torsion. Under conditions like bending only or tension only, the nominal stress state is uniaxial,” says Tipton. However, when a notch is present, a bar loaded only in bending or tension generates not only an axial stress, which happens to be the maximum principal stress,

*σ*

_{1}, but also a transverse, or circumferential stress,

*σ*

_{2}, he explains.

"For a given geometry and loading in pure bending or pure tension, using the equation constants the article lists for *K _{t}* lets us calculate

*σ*

_{1}, and using the constants for

*K*, the SCF in terms of the von Mises equivalent stress, gives

_{q}*σ*. The two stresses can be combined, using the last equation in the article, to find

_{eq}*σ*,” says Tipton.

_{2}For torsion, to compute the von Mises equivalent stress users should change the nomenclature so *σ _{1}* =

*σ*, and

_{x}*σ*=

_{2}*σ*. To compute torsional engineering shear stress, max, first find

_{y}*K*,

_{t}*using the constants listed in the article. Then*

_{torsion}*τ*=

_{max}*K*,

_{t}*×*

_{torsion}*τ*and

_{nom}*σ*=

_{eq}*√3*×

*τ*.

_{max}In these terms, *τ _{max}* =

*τ*. From here, compute the maximum principal stresses,

_{xy}*σ*and

_{1}*σ*from the three stress components,

_{2}*σ*,

_{x}*σ*, and

_{y}*τ*using

_{xy}*σ*= [(

_{1,2}*σ*+

_{x}*σ*)/2] ± √[((

_{2}*σ*–

_{x}*σ*)/2)

_{y}^{2}+

*τ*

_{xy}^{2}].

"I should note that Kq for either tension or bending is only valid for that particular load case,” says Tipton. “The transverse stress from bending only or tension only is always in the same sense as the maximum axial stress. That is, the transverse stress is tension if the axial stress is tensile, and it is tensile on the tension side of the bending case. That's why *K _{t}* is always greater than

*K*.

_{q}He notes that if tension and bending are applied simultaneously without torsion, engineers can compute the components separately and add them, keeping in mind they may not occur at the same location in the fillet. FEA models show the spots are close but not concurrent. Assuming they act at the same location is conservative.

If torsion is also applied, the resulting equivalent stress is not an arithmetic sum. Instead, *τ _{eq}* = √(

*σ*–

_{x}^{2}*σ*+

_{x}σ_{y}*σ*

_{y}^{2}+ 3

*τ*

_{xy}^{2}).

Tipton also clarified the geometry of the shoulder filet, specifically, the angle at which the maximum stress occurs. Φ = 0 where the shoulder radius r meets the shaft diameter d, and the angle increases up the filet radius. Although it was not noted in the original article, the equations the authors derived return Φ in degrees, not radians.