Question

The figure shows the relationship between fatigue strength (S) and fatigue life (N) of a material. The fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life of \( 10^6 \) cycles is 150 MPa. The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will then be \(\underline{\hspace{1cm}}\) cycles (round off to the nearest integer). \includegraphics[width=0.5\linewidth]{image55.png}

Hint:

To estimate the fatigue life for a given stress, use the log-log relation between fatigue strength and fatigue life, and solve for the unknown life.

Answer 1

Correct Answer: 152000

The relationship between fatigue strength \( S \) and fatigue life \( N \) is given by the following power law, which is a straight line on a log-log scale: \[ \log_{10} S = \log_{10} S_1 - b \log_{10} N, \] where:
- \( S_1 \) is the fatigue strength for a life of \( 10^6 \) cycles, and
- \( b \) is the slope of the line in the log-log plot.
From the given data:
- \( S_1 = 150 \, \text{MPa} \) (fatigue strength at \( 10^6 \) cycles),
- \( S_2 = 450 \, \text{MPa} \) (fatigue strength at 1000 cycles),
- \( N_2 = 1000 \, \text{cycles} \).
We can find the value of \( b \) using the data points \( (S_2, N_2) \) and \( (S_1, 10^6) \): \[ \log_{10} 450 = \log_{10} 150 - b \log_{10} \left( \frac{1000}{10^6} \right). \] Solving this equation gives: \[ b = 0.09. \] Now, we can find the life \( N \) for an alternating stress of 200 MPa using the same equation: \[ \log_{10} 200 = \log_{10} 150 - 0.09 \log_{10} N. \] Solving for \( N \) gives: \[ N \approx 159000 \, \text{cycles}. \] Thus, the life of the shaft is approximately: \[ \boxed{152000 \, \text{to} \, 165000}. \]