Notched bar in axial fatigue — notch factor, stresses and infinite-life margin (20 marks): 22-Mec-B1 Advanced Machine Design, December 2019, Q4

Question 4: Notched bar in axial fatigue — notch factor, stresses and infinite-life margin (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A rectangular bar with a transverse central hole, cycled between a compressive and a tensile axial force.

Given data — Problem 4
Quantity	Symbol	Value
Bar cross-section	$t\times w$	22 mm × 30 mm
Transverse hole diameter	$d$	10 mm
Minimum axial force	$F_{\min}$	−4 kN
Maximum axial force	$F_{\max}$	+12 kN
Ultimate tensile strength	$S_{ut}$	500 MPa
Surface finish / temperature	—	machined, room temperature
Reliability	—	99.999 %

Find. (1) the fatigue stress-concentration factor $K_f$; (2) the worst-case mean and alternating stresses; (3) the factor of safety for infinite life.

Figure 4.1 — Axially loaded bar with a transverse central hole; the critical plane passes through the hole, so the net section is (30 − 10) × 22 = 440 mm².

Approach. Take $K_t$ from the standard net-section chart for a plate with a central transverse hole, reduce it to $K_f$ through Neuber notch sensitivity, apply $K_f$ to both the mean and alternating stresses computed on the net area, build the corrected endurance limit with the Marin factors appropriate to axial loading, and close with the modified Goodman criterion.

Geometric stress-concentration factor. The hole diameter is one third of the width, $$\frac{d}{w}=\frac{10}{30}=0.333$$ For a flat bar in axial tension with a central circular hole, the net-section chart gives $$K_t \approx 2.35$$
Notch sensitivity by the Neuber relation. With $S_{ut}=500$ MPa $=72.5$ ksi, the Neuber constant is $$\sqrt{a}=0.246-3.08\times10^{-3}S_{ut}+1.51\times10^{-5}S_{ut}^{2}-2.67\times10^{-8}S_{ut}^{3}=0.0919\ \sqrt{\text{in}}$$ (with $S_{ut}$ in ksi). The notch radius is the hole radius, $r=5$ mm $=0.1969$ in, so $$q=\frac{1}{1+\dfrac{\sqrt a}{\sqrt r}}=\frac{1}{1+\dfrac{0.0919}{0.4437}}=0.828$$
Fatigue stress-concentration factor. $$K_f = 1+q\left(K_t-1\right)=1+0.828\times 1.35$$ $$\boxed{K_f = 2.12}$$ The notch is only 83 % "effective" in fatigue because at this strength level the material can blunt the notch tip by local plastic flow.
Net section and load components. The critical plane passes through the hole, so the resisting area is the net area: $$A_{net}=\left(w-d\right)t=\left(30-10\right)\times 22 = 440\ \text{mm}^{2}$$ $$F_m=\frac{F_{\max}+F_{\min}}{2}=\frac{12-4}{2}=4\ \text{kN},\qquad F_a=\frac{F_{\max}-F_{\min}}{2}=\frac{12+4}{2}=8\ \text{kN}$$
Worst-case mean and alternating stresses. For a worst-case assessment the notch factor is applied to both components (no relief is claimed from local yielding on the mean stress): $$\sigma_m = K_f\frac{F_m}{A_{net}}=2.12\times\frac{4000}{440}=2.12\times 9.09$$ $$\sigma_a = K_f\frac{F_a}{A_{net}}=2.12\times\frac{8000}{440}=2.12\times 18.18$$ $$\boxed{\sigma_m = 19.3\ \text{MPa},\qquad \sigma_a = 38.5\ \text{MPa}}$$
Endurance limit of the specimen. For wrought steel with $S_{ut}\lt 1400$ MPa, $$S'_e = 0.5\,S_{ut}=0.5\times 500 = 250\ \text{MPa}$$
Marin modifying factors. Machined surface, axial loading, room temperature and the specified reliability give $$k_a = 4.51\,S_{ut}^{-0.265}=4.51\times 500^{-0.265}=0.869$$ $$k_b = 1.0\ \text{(axial loading has no size effect)},\qquad k_c = 0.85\ \text{(axial)},\qquad k_d = 1.0$$ $$k_e = 0.659\ \text{for 99.999\,\% reliability}$$ so the corrected endurance limit is $$S_e = k_ak_bk_ck_dk_e\,S'_e = 0.869\times 1.0\times 0.85\times 1.0\times 0.659\times 250$$ $$\boxed{S_e = 121.7\ \text{MPa}}$$
Fatigue safety factor, modified Goodman. $$\frac{1}{n_f}=\frac{\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}=\frac{38.5}{121.7}+\frac{19.3}{500}=0.3166+0.0385=0.3551$$ $$\boxed{n_f = 2.82\ \text{(infinite life)}}$$ The bar has an ample margin: even the highly conservative combination of a worst-case notch factor on both stress components and a 99.999 % reliability knock-down leaves a factor near three.

Check: the assumptions that carry this answer. $K_t=2.35$ is read from the net-section chart for a central hole at $d/w=1/3$ (Shigley Table A-15-1); a slightly different chart reading of 2.3 or 2.4 moves $n_f$ by only about 2 %. The compressive half-cycle is assumed not to buckle the bar and not to close the notch, so the full alternating range is retained. No yield strength is given, so the first-cycle (Langer) check cannot be performed; with $\sigma_{\max}=\sigma_m+\sigma_a = 57.8$ MPa against a 500 MPa ultimate, static yielding is clearly not the limiting mode.

Question 4 — results
Quantity	Value
Geometric factor, $K_t$ (at $d/w=1/3$)	2.35
Notch sensitivity, $q$	0.83
Fatigue stress-concentration factor, $K_f$	2.12
Net area, $A_{net}$	440 mm²
Mean / alternating force	4 kN / 8 kN
Mean stress, $\sigma_m$	19.3 MPa
Alternating stress, $\sigma_a$	38.5 MPa
Corrected endurance limit, $S_e$	121.7 MPa
Fatigue safety factor (Goodman), $n_f$	2.82

Topic. Infinite-life fatigue design of a notched member under a fluctuating axial load.

Key relations. $K_f = 1+q(K_t-1)$ with Neuber $q = 1/(1+\sqrt a/\sqrt r)$; net area $(w-d)t$; $F_m,F_a$ from $F_{\max},F_{\min}$; Marin $S_e = k_ak_bk_ck_dk_eS'_e$ with $S'_e = 0.5S_{ut}$; modified Goodman $1/n_f = \sigma_a/S_e+\sigma_m/S_{ut}$.

Why this works. A notch does two separate things. Geometrically it multiplies the nominal stress by $K_t$; metallurgically, the elevated stress exists only in a very small volume, and a material with any ductility redistributes it, so the fatigue penalty is smaller than the elastic one. Neuber's constant packages that size effect: a small notch radius means a steep gradient and a low $q$, while a large radius approaches $q\to1$ and $K_f\to K_t$. The Marin factors then walk the polished-specimen endurance limit down to the real part — surface roughness ($k_a$), load type ($k_c$, lower for axial because there is no stress gradient to help), and the statistical scatter band demanded by the reliability target ($k_e$). Goodman finally accounts for the fact that a tensile mean stress opens cracks and so reduces the alternating stress the part can survive.

Common pitfall. Computing the stresses on the gross area $30\times22 = 660$ mm² instead of the net 440 mm², which understates both stresses by a third; and setting $k_b\lt 1$ for axial loading, when the size factor applies only to bending and torsion. A third trap is dropping the negative $F_{\min}$: the alternating force is half the full range, 8 kN, not half of 12 kN.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §6-9 to §6-12 (Marin factors, notch sensitivity, Goodman) and Table A-15-1.

Question 4 of 6: Notched bar in axial fatigue — notch factor, stresses and infinite-life margin (20 marks)

Question 4: Notched bar in axial fatigue — notch factor, stresses and infinite-life margin (20 marks)