Complete part of Question:
Assume a factor of safety of 3.0
The Maximum stress versus logarithm of the number of cycles to fatigue failure curve for 2014-T6 Aluminium bar is attached to this solution
Answer:
The maximum allowable load amplitude is 6403.33 N
Step-by-step explanation:
diameter of the bar, d = 12 mm
d = 0.012 m
Cross sectional area of the bar, A = πd²/4
A = π*0.012²/4
A = 0.000113 m²
From the Maximum stress versus logarithm of the number of cycles to fatigue failure for 2014-T6 Aluminium bar attached to this solution;
At 10⁷ cycles, Maximum stress, S = 170 MPa
S = 170 * 10⁶ Pa
Factor of safety, N = 3.0
The tensile stress is given by the formula:
[tex]\sigma = S/N\\\sigma = \frac{170 * 10^{6} }{3} \\\sigma = 56.67 * 10^{6} N/m^{2}[/tex]
The tensile stress is also given by:
[tex]\sigma = F/A\\\sigma = F/0.000113[/tex]
F/0.000113 = 56.67 * 10⁶
F = 56.67 * 10⁶ * 0.000113
F = 6403.33 N
The maximum allowable load amplitude is 6403.33 N
