Notched Bar in Axial Fatigue: 25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1), May 2016, Q6

Question 6: Notched Bar in Axial Fatigue (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. 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), $$\sigma_a=K_f\sigma_{a,net}=2.12(18.2)=\boxed{38.5\ \text{MPa}},\qquad \sigma_m=K_f\sigma_{m,net}=2.12(36.4)=\boxed{77.0\ \text{MPa}}.$$
Endurance limit (Marin). $S_e'=0.5S_{ut}=250$ MPa; machined $k_a=4.51\,S_{ut}^{-0.265}=0.869$; axial size factor $k_b=1$; axial load factor $k_c=0.85$; reliability $99.999\%\Rightarrow k_e=0.659$. Hence $$S_e=k_a k_c k_e S_e'=0.869(0.85)(0.659)(250)=122\ \text{MPa}.$$
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.

Final results — Problem 6
Quantity	Value
Fatigue stress-concentration factor, $K_f$	2.12
Worst-case alternating stress, $\sigma_a$	38.5 MPa
Worst-case mean stress, $\sigma_m$	77.0 MPa
Endurance limit, $S_e$	122 MPa
Fatigue strength at $5\times10^5$ cycles	139 MPa

Topic: Notch fatigue of an axially loaded bar — $K_f$, mean/alternating stresses, and finite-life strength.

Key relations: $K_f=1+q(K_t-1)$; net stress on $(w-d)t$; $\sigma_{a,m}=\tfrac12(F_{max}\mp F_{min})/A$; Marin $S_e$; $S_f=aN^b$ with $a=(fS_{ut})^2/S_e$, $b=-\tfrac13\log(fS_{ut}/S_e)$.

Why this works: The geometric factor $K_t$ over-states the fatigue effect for small notch radii, so notch sensitivity $q$ tempers it to $K_f$. Fatigue is driven by the local stress on the reduced net section. The Basquin (high-cycle) line between $10^3$ and $10^6$ cycles interpolates strength at any intermediate life.

Common pitfall: Using the gross area, or basing $K_t$ on gross stress when the chart is net-referenced; and forgetting the axial load factor $k_c=0.85$ and the strong reliability knock-down ($k_e=0.659$ at 99.999%).

Source: Shigley’s Mechanical Engineering Design, §6-10 (notch sensitivity), Fig. A-15-1 ($K_t$), §6-8/6-9 (Marin factors, S–N line).

Question 6 of 6: Notched Bar in Axial Fatigue

Question 6: Notched Bar in Axial Fatigue (20 marks)