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 10⁶ 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 ____________ cycles (round off to the nearest integer).

The correct answer is 163840.

Step-by-Step Explanation:

From the provided S-N diagram in the figure, we observe a linear relationship between the log-fatigue strength (${\log }_{10}S$ plotted on the vertical y-axis) and log-fatigue life (${\log }_{10}N$ plotted on the horizontal x-axis) in the high-cycle fatigue region (between N = 10³ and N = 10⁶ cycles).

Let us define the coordinate system as:
$x={\log }_{10}N$
$y={\log }_{10}S$

Using the data points provided in the problem description and labeled in the figure:
For Point 1:
$N_1=1000={10}^3\Rightarrow x_1={\log }_{10}\left({10}^3\right)=3$
$S_1=450\text{MPa}\Rightarrow y_1={\log }_{10}\left(450\right)$

For Point 2:
$N_2={10}^6\Rightarrow x_2={\log }_{10}\left({10}^6\right)=6$
$S_2=150\text{MPa}\Rightarrow y_2={\log }_{10}\left(150\right)$

For the target operating condition:
Alternating stress, $S=200\text{MPa}\Rightarrow y={\log }_{10}\left(200\right)$
We need to find the fatigue life $N$, where $x={\log }_{10}N$.

Applying the linear interpolation formula for the straight line segment between Point 1 and Point 2:
$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$$

Substituting the values into the equation:

$$\frac{x-3}{6-3}=\frac{{\log }_{10}\left(200\right)-{\log }_{10}\left(450\right)}{{\log }_{10}\left(150\right)-{\log }_{10}\left(450\right)}$$

Simplifying using logarithm division rules:

$$\frac{x-3}{3}=\frac{{\log }_{10}\left(\frac{200}{450}\right)}{{\log }_{10}\left(\frac{150}{450}\right)}$$

Simplifying the fractions:

$$\frac{x-3}{3}=\frac{{\log }_{10}\left(0.444444\right)}{{\log }_{10}\left(0.333333\right)}$$

Calculating the values of the logarithms:
${\log }_{10}\left(0.444444\right)\approx -0.352183$
${\log }_{10}\left(0.333333\right)\approx -0.477121$

Therefore, we have:

$$\frac{x-3}{3}=\frac{-0.352183}{-0.477121}\approx 0.73814$$

Solving for $x$:
$x-3=3\times 0.73814=2.21442$
$x=3+2.21442=5.21442$

Since $x={\log }_{10}N$, we obtain $N$ by taking the base-10 exponent:
$N={10}^{5.21442}\approx 163840.58\text{cycles}$

Rounding off to the nearest integer, the fatigue life of the shaft is:
$N\approx 163840\text{cycles}$