In this solution some basic concepts and formulas of Fatigue Mechanical Design are used. For more information, refer to any standard textbook or drop a comment below. Please give a Thumbs Up, if solution is helpful.
Solution :
a.)
The Matlab Code to create the plot would be as follows,
N = [1 1000 1000000 10000000000];
Su = 1200; % Ultimate Strength
Se = 216.15; % Endurance Limit
S(1) = log10(Su);
S(2) = log10(0.9*Su)
S(3) = log10(Se);
S(4) = S(3);
semilogx(N,S)
xlabel('Numbe of Cycles (N) ')
ylabel('Fatigue Stress')
xlim([0 100000000])
ylim([2.2 3.1])
grid on
b.)
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5. (20 pts) A notched circular shaft shown is subjected to fully reversed bending. D-12.7mm, d-10.16 mm, and r-0.5mm. The steel material has the following mechanical and fatigue properties: yield str...
A notched circular shaft shown is subjected to fully reversed bending. D=12.7mm, d=10.16 mm, and r=0.5mm. The steel material has the following mechanical and fatigue properties: yield strength (σ0)=1000 MPa, ultimate strength (σu)=1200 MPa, the fatigue strength coefficient (σf’)=1600 MPa, the fatigue strength exponent (b)=-0.1, and the fatigue limit (σe)=400 MPa. 5. (20 pts) A notched circular shaft shown is subjected to fully reversed bending. D-12.7mm, d-10.16 mm, and r-0.5mm. The steel material has the following mechanical and fatigue properties:...