A cylindrical 1045 steel bar (Figure 8.34) is subjected to repeated tension- compression stress cycling along its axis. If the load amplitude is \(22,000 \mathrm{~N}\left(4950 \mathrm{lb}_{\mathrm{f}}\right)\), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of \(2.0\).

First, we need to determine the stress amplitude. This can be calculated using the formula: \(\text{Stress amplitude} = \frac{\text{Load amplitude}}{\text{Area}}\) As we are dealing with a cylindrical bar, the area can be computed using the formula for the area of a circle, which is: \(\text{Area} = \pi \left(\frac{d}{2}\right)^2\) Combining the two formulas, we get: \(\text{Stress amplitude} = \frac{\text{Load amplitude}}{\pi \left(\frac{d}{2}\right)^2}\) We have the load amplitude, which is \(66,700\,N\), and we want to find the minimum allowable diameter (\(d\)) for the cylindrical bar.

Since we have a safety factor of 2.0, the minimum allowable stress amplitude will be: \(\text{Allowable stress amplitude} = \frac{\text{Stress amplitude}}{\text{Safety factor}}\)

With the equations for Stress amplitude and Allowable stress amplitude, we can now set them equal to each other and solve for the minimum diameter \(d\): \(\frac{\text{Load amplitude}}{\pi \left(\frac{d}{2}\right)^2} = \frac{\text{Stress amplitude}}{\text{Safety factor}}\) To find the diameter, we rearrange the equation as: \(d = 2\sqrt{\frac{\text{Load amplitude} \times \text{Safety factor}}{\pi \times \text{Stress amplitude}}}\) We can plug in the values for load amplitude (\(66,700\,N\)) and the safety factor (2.0). According to the problem, 1045 steel has an endurance limit. We use this value as the stress amplitude, which we can find in the literature or in engineering handbooks. For this example, let's assume that the endurance limit for 1045 steel is 290 MPa or \(290 \times 10^6\, \mathrm{N/m^2}\). \(d = 2\sqrt{\frac{66,700\, \mathrm{N} \times 2.0}{\pi \times 290 \times 10^6\, \mathrm{N/m^2}}}\) Now, we can solve for the minimum diameter \(d\): \(d \approx 0.0323\, \mathrm{m}\) or \(32.3\, \mathrm{mm}\) Therefore, the minimum allowable diameter for the cylindrical 1045 steel bar to ensure that fatigue failure will not occur is approximately \(32.3\, \mathrm{mm}\).