Interactive Lecture
How Geometric Discontinuities Amplify Local Stress — and Why This Governs Fatigue Failure
by Dr Arnaldo Delli Carri
Whenever a loaded component changes cross-section — at a hole, a notch, a fillet, a keyway — the load-carrying stress trajectories must crowd through the reduced area. The result is a local stress spike that can be many times the average stress calculated from the cross-section. That average is the nominal stress \(\sigma_\text{nom}\); the peak at the discontinuity is \(\sigma_\text{max}\). The ratio between them is the stress concentration factor:
\[K_t = \frac{\sigma_\text{max}}{\sigma_\text{nom}}\]
A tiny notch root radius causes a disproportionately large stress spike. Halving the notch root radius multiplies \((K_t - 1)\) by \(\sqrt{2}\). Making a notch sharper always hurts far more than blunting it by the same factor helps.
A circular hole of radius \(R\) in an infinite plate under remote tension \(\sigma\) is exactly solvable in plane elasticity. Along the transverse centreline (perpendicular to the applied load), the tangential stress \(\sigma_{\theta\theta}\) — which here acts along the loading direction — is:
\[\sigma_{\theta\theta}(r) = \frac{\sigma}{2}\!\left(2 + \frac{R^2}{r^2} + \frac{3R^4}{r^4}\right), \quad r \geq R\]
At the hole surface (\(r = R\)): \(\sigma_{\theta\theta} = 3\sigma\), giving \(K_t = 3\) exactly. The stress decays back to \(\sigma\) within a few hole radii — at \(r = 3R\) the excess is still about 7%, falling below 5% only by roughly \(r = 3.5R\).
For a finite-width plate the Howland solution, curve-fitted by Peterson, gives:
\[K_t \approx 3 - 3.13\!\left(\frac{d}{W}\right) + 3.66\!\left(\frac{d}{W}\right)^{\!2} - 1.53\!\left(\frac{d}{W}\right)^{\!3}, \quad \frac{d}{W} \leq 0.5\]
where \(d = 2R\) is the hole diameter and \(W\) is the plate width. \(\sigma_\text{nom}\) here is the remote applied (gross-section) stress.
An elliptical hole with semi-axis \(a\) perpendicular to the load and semi-axis \(b\) parallel to it gives:
\[K_t = 1 + \frac{2a}{b}\]
For \(a = b\) (circle): \(K_t = 3\). For \(a \gg b\) (flat crack-like slot): \(K_t \to \infty\). The notch tip radius is \(\rho = b^2/a\), so an equivalent form is \(K_t = 1 + 2\sqrt{a/\rho}\).
For a U-notch of depth \(t\) and root radius \(\rho\) cut into a bar or plate, Neuber's approximation is:
\[K_t \approx 1 + 2\sqrt{\frac{t}{\rho}}\]
As \(\rho \to 0\) (sharp notch), \(K_t \to \infty\). Linear elastic theory predicts an unbounded stress at a perfectly sharp crack tip; fracture mechanics (LEFM, with its \(1/\sqrt{r}\) singularity) must replace the \(K_t\) approach at that limit. Here \(\sigma_\text{nom}\) is the net-section stress.
A stepped change in section width (or shaft diameter) introduces a \(K_t\) that depends on the size ratio \(D/d\) and the fillet radius \(r\). Here \(D\) and \(d\) are the larger and smaller section diameters and \(r\) is the fillet radius — reusing the letters that stood for the hole diameter and the radial distance in the Kirsch sections above. Peterson's tabulated polynomial fit gives:
\[K_t \approx C_1 + C_2\!\left(\frac{2r}{D-d}\right) + C_3\!\left(\frac{2r}{D-d}\right)^{\!2} + C_4\!\left(\frac{2r}{D-d}\right)^{\!3}\]
The coefficients \(C_1\)–\(C_4\) (Peterson, 1974) depend on the loading mode (tension, bending, or torsion) and on \(D/d\). For \(D/d = 1.5\), \(K_t\) ranges from about 1.1 with a generous fillet (\(r/d \approx 0.3\)) to around 2.5 with a sharp transition (\(r/d \approx 0.05\)).
Choose the geometry, then drag the radius and stress sliders. The left panel shows the specimen and the right panel plots the stress distribution along the critical axis. Try reducing the radius close to its minimum — watch how \(\sigma_\text{max}\) spikes far beyond \(\sigma_\text{nom}\).
| Concept | Equation | Notes |
|---|---|---|
| Definition | \(K_t = \sigma_\text{max}/\sigma_\text{nom}\) | Purely geometric; no material dependence in the elastic range |
| Circular hole, infinite plate (Kirsch) | \(K_t = 3\) | Exact; \(\sigma_\text{nom}\) = remote applied stress |
| Circular hole, finite plate (Howland/Peterson) | \(K_t \approx 3 - 3.13(d/W) + 3.66(d/W)^2 - 1.53(d/W)^3\) | Valid for \(d/W \leq 0.5\); gross-section \(\sigma_\text{nom}\) |
| Elliptical hole (Inglis) | \(K_t = 1 + 2a/b\) | \(a\) perpendicular to load, \(b\) parallel; tip radius \(\rho = b^2/a\) |
| U-notch (Neuber) | \(K_t \approx 1 + 2\sqrt{t/\rho}\) | Depth \(t\), root radius \(\rho\); net-section \(\sigma_\text{nom}\) |
| Fatigue notch factor | \(K_f = 1 + q(K_t - 1)\) | Notch sensitivity \(q \in [0,1]\); notched fatigue limit = \(\sigma_e / K_f\) |
A flat steel plate is 300 mm wide and 10 mm thick. It contains a central circular hole of diameter 30 mm and carries a tensile load \(F = 180\) kN. Since \(d/W = 0.10 \approx 0\), treat the plate as effectively infinite. Calculate \(\sigma_\text{nom}\) and \(\sigma_\text{max}\).
Show Solution
Given: \(W = 300\) mm, thickness \(t_p = 10\) mm, \(d = 30\) mm, \(F = 180\) kN.
Step 1 — gross cross-sectional area: the infinite-plate result \(K_t = 3\) is defined against the remote (gross-section) stress, so use the full width:
\[A = W\,t_p = 300 \times 10 = 3{,}000 \text{ mm}^2\]
Step 2 — nominal (gross-section) stress:
\[\sigma_\text{nom} = \frac{F}{A} = \frac{180{,}000}{3{,}000} = 60 \text{ MPa}\]
Step 3 — stress concentration factor:
\(d/W = 0.10\), so the infinite-plate approximation applies: \(K_t = 3\).
Step 4 — peak stress:
\[\sigma_\text{max} = K_t\,\sigma_\text{nom} = 3 \times 60 = 180 \text{ MPa}\]
Answer: \(\sigma_\text{nom} = 60\) MPa, \(\sigma_\text{max} = 180\) MPa.
A flat bar carries a tensile load giving a net-section nominal stress of 80 MPa. Two symmetric U-notches of depth \(t = 8\) mm are machined in the sides. The material has a yield strength \(\sigma_y = 320\) MPa. Using the Neuber formula, find the minimum notch root radius \(\rho\) that keeps \(\sigma_\text{max} \leq \sigma_y\).
Show Solution
Given: \(t = 8\) mm, \(\sigma_\text{nom} = 80\) MPa, \(\sigma_y = 320\) MPa.
Step 1 — maximum allowable \(K_t\):
\[K_{t,\text{max}} = \frac{\sigma_y}{\sigma_\text{nom}} = \frac{320}{80} = 4.0\]
Step 2 — rearrange Neuber formula:
\[1 + 2\sqrt{\frac{t}{\rho}} \leq 4.0 \implies 2\sqrt{\frac{t}{\rho}} \leq 3.0 \implies \frac{t}{\rho} \leq 2.25\]
Step 3 — minimum radius:
\[\rho \geq \frac{t}{2.25} = \frac{8}{2.25} = 3.56 \text{ mm}\]
Verification with \(\rho = 4\) mm (rounded up to next practical size):
\[K_t = 1 + 2\sqrt{\frac{8}{4}} = 1 + 2\sqrt{2} = 3.83\]
\[\sigma_\text{max} = 3.83 \times 80 = 306 \text{ MPa} < 320 \text{ MPa} \checkmark\]
Answer: Use \(\rho \geq \mathbf{4 \text{ mm}}\).
A plate under remote tension \(\sigma = 40\) MPa contains either (a) a circular hole of radius 20 mm, or (b) an elliptical hole with semi-axis \(a = 20\) mm perpendicular to the load and \(b = 4\) mm parallel to it. Calculate \(\sigma_\text{max}\) in each case and compare. What is the notch root radius of the ellipse?
Show Solution
Given: \(\sigma = 40\) MPa; (a) circular hole \(R = 20\) mm; (b) ellipse \(a = 20\) mm (⊥ load), \(b = 4\) mm (∥ load).
Part (a) — circular hole (Kirsch):
\[K_t = 3 \implies \sigma_\text{max} = 3 \times 40 = 120 \text{ MPa}\]
Part (b) — elliptical hole (Inglis):
\[K_t = 1 + \frac{2a}{b} = 1 + \frac{2 \times 20}{4} = 1 + 10 = 11\]
\[\sigma_\text{max} = 11 \times 40 = \mathbf{440 \text{ MPa}}\]
Comparison: The ellipse gives 440/120 = 3.7 times the peak stress of the circle, despite having the same perpendicular half-width.
Notch tip radius of the ellipse:
\[\rho = \frac{b^2}{a} = \frac{4^2}{20} = \frac{16}{20} = 0.8 \text{ mm}\]
Compared to 20 mm for the circle, the ellipse's tip is 25 times sharper — and it is this geometric sharpness that drives almost the entire stress amplification.
Answer: (a) \(\sigma_\text{max} = 120\) MPa; (b) \(\sigma_\text{max} = 440\) MPa; ellipse tip radius \(\rho = 0.8\) mm.
A shaft has a circumferential U-groove with \(K_t = 3.0\). Under fully reversed bending the nominal stress amplitude is \(\sigma_\text{nom} = 140\) MPa. The material's unnotched fatigue limit is \(\sigma_e = 350\) MPa and the notch sensitivity index is \(q = 1.0\). (a) Find \(\sigma_\text{max}\). (b) Find the notched fatigue limit \(\sigma_{e,n}\). (c) Is fatigue failure expected?
Show Solution
Given: \(K_t = 3.0\), \(\sigma_\text{nom} = 140\) MPa, \(\sigma_e = 350\) MPa, \(q = 1.0\).
(a) Peak stress amplitude:
\[\sigma_\text{max} = K_t\,\sigma_\text{nom} = 3.0 \times 140 = \mathbf{420 \text{ MPa}}\]
(b) Fatigue notch factor and notched fatigue limit:
With \(q = 1.0\) (fully notch-sensitive material):
\[K_f = 1 + q(K_t - 1) = 1 + 1.0 \times (3.0 - 1) = 3.0\]
\[\sigma_{e,n} = \frac{\sigma_e}{K_f} = \frac{350}{3.0} = \mathbf{116.7 \text{ MPa}}\]
(c) Fatigue assessment:
Since \(\sigma_\text{nom} = 140\) MPa \(> \sigma_{e,n} = 116.7\) MPa, fatigue failure is expected. Options to fix this include reducing the applied load, increasing the groove root radius (lowering \(K_t\)), or selecting a less notch-sensitive material (reducing \(q\) below 1.0).
Answer: (a) \(\sigma_\text{max} = 420\) MPa; (b) \(\sigma_{e,n} = 116.7\) MPa; (c) yes — fatigue failure is expected.