Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 · 07-Mec-B1 Advanced Machine Design.
Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; candidates answer any three of the four Part II problems (3–6). For study value, complete worked solutions to all six problems are provided below.
Check — assumptions stated per the exam rubric.
(1) “ton” is read as the US short ton (2000 lb), consistent with the inch/ft·min/hp unit set of Problem 2.
(2) In Problem 3 the concentrated couple Mz is taken counter-clockwise (out of the page, per the dot symbol in the figure); a clockwise reading would give RA = 0, RB = 9.5 kN.
(3) In Problem 5 the fluctuating torque is assumed transmitted over the 18 in from the drive end to the load; no stress concentration is used as the problem directs.
Question 6: Notched Bar in Axial Fatigue (20 marks)
Given. Bar $30\times22$ mm; transverse hole $d=10$ mm through the 30-mm width; axial force $F_{min}=8$ kN, $F_{max}=24$ kN; $S_{ut}=500$ MPa; reliability $99.999\%$.
Problem 6: axially loaded bar with a transverse central hole; net section is $(30-10)\times22$ mm.
Find. The fatigue stress-concentration factor $K_f$; the worst-case mean and alternating stresses; and the finite-life fatigue strength at $5\times10^5$ cycles.
Approach. Read $K_t$ for a transverse hole in an axially loaded bar ($d/w=1/3$), reduce it to $K_f$ with the Neuber notch sensitivity, resolve the force into mean/alternating parts on the net area, then build the Marin endurance limit and the $S=aN^b$ line to read $S_f$ at $5\times10^5$ cycles.
Fatigue stress-concentration factor. For a transverse hole in a bar under axial load with $d/w=10/30=0.333$, the net-area geometric factor is $K_t\approx2.35$. The Neuber notch sensitivity for $S_{ut}=500$ MPa and hole radius $r=5$ mm gives $q\approx0.83$, so
$$K_f=1+q(K_t-1)=1+0.83(1.35)=\boxed{2.12}.$$
Load resolution and net section. The net area at the hole is $A=(30-10)(22)=440\ \text{mm}^2$. The mean and alternating forces are $F_m=\tfrac12(24+8)=16$ kN and $F_a=\tfrac12(24-8)=8$ kN, giving nominal net stresses $\sigma_{m,net}=36.4$ MPa and $\sigma_{a,net}=18.2$ MPa.
Worst-case stresses. Applying $K_f$ to both components (no local yielding to relieve the mean),
$$\sigma_a=K_f\sigma_{a,net}=2.12(18.2)=\boxed{38.5\ \text{MPa}},\qquad \sigma_m=K_f\sigma_{m,net}=2.12(36.4)=\boxed{77.0\ \text{MPa}}.$$
Finite-life fatigue strength. With $f=0.9$, the S–N constants are $a=(fS_{ut})^2/S_e=(450)^2/122=1665$ MPa and $b=-\tfrac13\log_{10}(fS_{ut}/S_e)=-0.189$. At $N=5\times10^5$ cycles,
$$S_f=a\,N^{\,b}=1665\,(5\times10^5)^{-0.189}=\boxed{139\ \text{MPa}}.$$
Since the applied alternating stress (38.5 MPa) is well below $S_f$, the notched bar has ample fatigue margin at this life.
