Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-Mec-B1 Advanced Machine Design, May 2017. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; only three of the four Part II problems (Problems 3–6) are required. All six problems are solved as a study resource. “State all assumptions clearly… assume any missing data and properly state it” is an explicit exam instruction, invoked in Problem 2.
Given. Rectangular bar $b=22$ mm (thickness) × $w=30$ mm (width), transverse central hole $d=10$ mm through the 30-mm face. Axial load cycles $F_{min}=-4$ kN to $F_{max}=12$ kN. Machined, room temperature, $S_{ut}=500$ MPa, reliability 99.999%.
Given data
Quantity
Symbol
Value
Width across hole
$w$
30 mm
Thickness
$b$
22 mm
Hole diameter
$d$
10 mm
Min / max load
$F_{min}$ / $F_{max}$
−4 kN / 12 kN
Ultimate strength
$S_{ut}$
500 MPa
Reliability
$R$
99.999%
Find. (1) fatigue stress-concentration factor $K_f$; (2) worst mean and alternating stresses; (3) infinite-life fatigue factor of safety.
Approach. Read the geometric factor $K_t$ for a transverse hole (net-section basis) at $d/w=1/3$, apply notch sensitivity to get $K_f$; compute mean and alternating loads, divide by the net area and scale by $K_f$; build the corrected endurance limit with the Marin factors for axial loading and 99.999% reliability, then apply the Goodman line.
Stress-concentration factor. For a flat bar with a transverse central hole in axial tension, at $d/w=10/30=0.333$, the net-section chart gives $K_t\approx2.35$. Notch sensitivity from the Neuber constant for steel ($\sqrt{a}=0.062$ in$^{1/2}$ at $S_{ut}=500$ MPa, hole radius $r=5$ mm) is $$q=\frac{1}{1+\sqrt{a}/\sqrt{r}}=0.83,\qquad K_f=1+q(K_t-1)=\boxed{2.12}.$$
Mean and alternating loads. $$F_m=\tfrac12(F_{max}+F_{min})=4\text{ kN},\qquad F_a=\tfrac12(F_{max}-F_{min})=8\text{ kN}.$$
Net area & nominal stresses. The hole removes material on the loaded section: $A_{net}=(w-d)\,b=(30-10)(22)=440\ \text{mm}^2$. Nominal $\sigma_{m0}=F_m/A_{net}=9.09$ MPa, $\sigma_{a0}=F_a/A_{net}=18.18$ MPa.
Worst-case (notch-peak) stresses. Applying $K_f$ to both components, $$\boxed{\sigma_m=K_f\sigma_{m0}=19.3\text{ MPa},\qquad \sigma_a=K_f\sigma_{a0}=38.5\text{ MPa}.}$$
Infinite-life factor of safety (Goodman). $$\frac{1}{n_f}=\frac{\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}=\frac{38.5}{121.6}+\frac{19.3}{500}=0.317+0.039,$$ $$\boxed{n_f=2.82}.$$ The factor exceeds unity by a wide margin, so the notched bar has infinite life with a comfortable reserve.
