Stress Concentration Factor Calculator: Kt and Kf for Shaft Shoulders, Grooves and Holes

Method

Geometry ratios. For a shoulder or groove the step height is h = (D − d)/2, where D is the larger diameter, d the smaller and r the fillet or groove radius. The fits use h/r and 2h/D. Shigley's charts use D/d and r/d, which describe the same geometry (2h/D = 1 − d/D, h/r = (D/d − 1)/(2 r/d)). For holes the ratio is the hole diameter over the shaft diameter or plate width.

Theoretical factor Kt. Each case uses a published polynomial Kt = C1 + C2 (2h/D) + C3 (2h/D)² + C4 (2h/D)³, with C1 to C4 given as a + b √(h/r) + c (h/r) over one or two h/r ranges. Stepped round shaft: Pilkey and Pilkey, Peterson's Stress Concentration Factors, 3rd ed. (2008), Charts 3.4, 3.10 and 3.12. U-groove, transverse hole and flat bar shoulder: Young, Budynas and Sadegh, Roark's Formulas for Stress and Strain, 8th ed. (2012), Table 17.1 cases 15, 18 and 5. Plate with a central hole: Peterson 3rd ed. Chart 4.1, Kt = 2 + 0.284 (1 − d/w) − 0.600 (1 − d/w)² + 1.32 (1 − d/w)³. Outside a fit's range no value is given.

Nominal and peak stress. The nominal stress uses the net section, with F the axial force, M the bending moment, T the torque, t the thickness and w the plate width. It is 4F/(πd²), 32M/(πd³) and 16T/(πd³) for round shafts, F/(dt) and 6M/(td²) for the flat bar, and F/((w − d)t) for the plate. The transverse hole follows Roark case 18 and uses the gross section, 32M/(πD³) and 16T/(πD³). In torsion its Kt gives the peak normal stress at the hole edge, which tends to 4 times the shear stress for a small hole. Peak stress = Kt × nominal stress.

Fatigue factor Kf. For steel the Neuber constant √a comes from Shigley, 10th ed. For bending and axial load, Eq. (6-35a) gives √a = 0.246 − 3.08×10⁻³ Sut + 1.51×10⁻⁵ Sut² − 2.67×10⁻⁸ Sut³. For torsion, Eq. (6-35b) gives √a = 0.190 − 2.51×10⁻³ Sut + 1.35×10⁻⁵ Sut² − 2.67×10⁻⁸ Sut³. Sut is the ultimate tensile strength in kpsi and √a is in √in, valid from 50 to 250 kpsi. The notch sensitivity is q = 1/(1 + √a/√r) and Kf = 1 + q (Kt − 1), Eq. (6-32). The torsion factor Kfs uses the torsion constant. For other materials enter q directly.

Assumptions

• Linear elastic, isotropic material. Kt is purely geometric and does not apply once the notch root yields.
• The curve fits reproduce their source charts to within about ±5 % over most of their range (Roark 8th ed., Table 17.1 preamble). Charts from different sources can differ by more.
• U-groove in bending: the Roark and Peterson 3rd ed. fits agree with each other but sit up to about 10 % above the 1951 Peterson chart reproduced in Shigley (Fig. A-15-14). The calculator uses the higher value, which is conservative.
• Each load case is treated alone. For combined loads the peak stresses do not occur at exactly the same point, so combining them is conservative.
• Shoulders are long enough for the stress to spread out (Roark case 5 requires L/D above 3/(r/d)^0.25 in tension, 0.8/(r/d)^0.25 in bending). Short shoulders and closely spaced grooves raise Kt.
• The Neuber constant fits are for steel with Sut from 345 to 1,724 MPa (50 to 250 ksi). Notch sensitivity also depends on surface finish and heat treatment.
• Not included, because the fits could not be verified against two independent sources: U-groove in axial load, transverse hole in axial load, and flat bars with U-notches.

Standards and references

Worked example: Stepped shaft, D 38 mm, d 32 mm, r 3 mm, steel Sut 690 MPa

The geometry of Shigley 10th ed. Example 6-6 (D/d = 1.1875, r/d = 0.094), with an axial force of 10 kN, a bending moment of 200 N·m and a torque of 150 N·m added. Shigley reads Kt = 1.65 in bending from Fig. A-15-9 and gets Kf = 1.55. This is the default case in the calculator.

Frequently asked questions

What is the difference between Kt and Kf?

Kt is the theoretical, purely geometric factor: peak elastic stress over nominal stress. Kf is the fatigue factor actually seen in tests, which is lower because a sharp notch in a real material does not feel its full peak. They are linked by the notch sensitivity q: Kf = 1 + q (Kt − 1). With q = 0 the notch has no effect on fatigue. With q = 1, Kf = Kt.

Can I ignore Kt for a ductile part under static load?

Usually yes. A ductile part yields locally at the notch and redistributes the load, so static strength is checked on the nominal stress (Shigley does this). Keep Kt for brittle materials, for fatigue, for low temperature where the material may be brittle, and where local yielding itself is unacceptable, such as a seal face or a precision fit.

Why does the calculator refuse some geometries?

Each curve fit is valid only over the range of data it was fitted to, for example 0.25 ≤ h/r ≤ 4 for a shoulder fillet in torsion. Outside that range a polynomial can give nonsense, so no value is shown and the message says which ratio is out of range. Use a chart, a finite element model or a test for those cases.

Why is my Kt a few percent different from the Shigley chart?

The fits reproduce Peterson's 3rd edition and Roark's tables, which include data newer than the 1951 charts in Shigley. A chart read by eye is also only good to a few percent. For the Shigley Example 6-6 shoulder the fit gives 1.682 against 1.65 read from the chart. For U-grooves in bending the newer data is up to about 10 % higher.

How do I reduce the stress concentration at a shaft shoulder?

Increase the fillet radius first: it has the largest effect. Reducing the step D/d also helps. Where a bearing needs a sharp shoulder, use a relief groove or an undercut fillet. An elliptical fillet is another option. The Kt curve in the results shows how quickly Kt falls as r/d grows for your D/d.

Which notch sensitivity should I use for aluminium or cast iron?

The built-in Neuber constant fits are for steel only. Switch to custom q and enter a value from test data or a reference chart. Shigley recommends q = 0.2 for cast iron as a conservative value. For aluminium alloys use the aluminium curves in Shigley Fig. 6-20 or published Neuber constants.

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