A cylindrical rod of diameter \(6.7 \mathrm{~mm}\) fabricated from a \(70 \mathrm{Cu}-30 \mathrm{Zn}\) brass alloy is subjected to rotating-bending load cycling; test results (as \(S-N\) behavior) are shown in Figure 8.20. If the maximum and minimum loads are \(+120 \mathrm{~N}\) and \(-120 \mathrm{~N}\), respectively, determine its fatigue life. Assume that the separation between loadbearing points is \(67.5 \mathrm{~mm}\).

Short Answer

Expert verified
Answer: The estimated fatigue life of the rod is approximately between 2 million and 4 million cycles.

Step by step solution

01

Calculate the cross-sectional area of the rod

Diameter of the rod is given as \(6.7 \mathrm{~mm}\). We can calculate the cross-sectional area (A) using the formula for the area of a circle: \(A=\pi\left(\frac{d}{2}\right)^{2}=\pi\left(\frac{6.7 \mathrm{~mm}}{2}\right)^{2}\). Calculating the cross-sectional area, we get: \(A \approx 35.29 \mathrm{~mm}^{2}\).
02

Calculate maximum and minimum stresses

Now we will calculate the maximum and minimum stresses (σ_max and σ_min) using the formula for stress: \(\sigma = \frac{F}{A}\), where F is the force and A is the cross-sectional area. For maximum stress (σ_max), F is the maximum load, \( +120 \mathrm{~N}\): \(\sigma_{max} = \frac{120 \mathrm{~N}}{35.29 \mathrm{~mm}^2} \approx 3.4 \mathrm{MPa}\). For minimum stress (σ_min), F is the minimum load, \( -120 \mathrm{~N}\): \(\sigma_{min} = \frac{-120 \mathrm{~N}}{35.29 \mathrm{~mm}^2} \approx -3.4 \mathrm{MPa}\).
03

Calculate stress amplitude and mean stress

We can now calculate the stress amplitude (σ_a) and mean stress (σ_m) using the following formulas: \(\sigma_{a} = \frac{\sigma_{max} - \sigma_{min}}{2} = \frac{3.4 \mathrm{MPa} - (-3.4 \mathrm{MPa})}{2} = 3.4 \mathrm{MPa}\). \(\sigma_{m} = \frac{\sigma_{max} + \sigma_{min}}{2} = \frac{3.4 \mathrm{MPa} + (-3.4 \mathrm{MPa})}{2} = 0 \mathrm{MPa}\).
04

Estimate the fatigue life from the S-N behavior graph

Now that we have the stress amplitude and mean stress values, we can use the S-N behavior graph to find the fatigue life of the rod. We can compare the stress amplitude value with the S-N graph, which shows cycles (N) on the x-axis and stress amplitude (S) on the y-axis. From the graph, we can find the approximate number of cycles that correspond to a stress amplitude of \(3.4 \mathrm{MPa}\). This will give us an estimate of the fatigue life (N) of the rod. Using the \(S-N\) behavior graph given in Figure 8.20, we find that the fatigue life for a stress amplitude of \(3.4 \mathrm{MPa}\) is approximately between 2 million and 4 million cycles. Therefore, the estimated fatigue life of the rod falls within this range.

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Most popular questions from this chapter

The following creep data were taken on an aluminum alloy at \(480^{\circ} \mathrm{C}\left(900^{\circ} \mathrm{F}\right)\) and a constant stress of \(2.75 \mathrm{MPa}\) (400 psi). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included. $$ \begin{array}{cccc} \hline \text { Time } \text { (min) } & \text { Strain } & \text { Time } \text { (min) } & \text { Strain } \\ \hline 0 & 0.00 & 18 & 0.82 \\ \hline 2 & 0.22 & 20 & 0.88 \\ \hline 4 & 0.34 & 22 & 0.95 \\ \hline 6 & 0.41 & 24 & 1.03 \\ \hline 8 & 0.48 & 26 & 1.12 \\ \hline 10 & 0.55 & 28 & 1.22 \\ \hline 12 & 0.62 & 30 & 1.36 \\ \hline 14 & 0.68 & 32 & 1.53 \\ \hline 16 & 0.75 & 34 & 1.77 \\ \hline \end{array} $$

A cylindrical 2014-T6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis; results of these tests are shown in Figure \(8.20 .\) If the bar diameter is \(12.0 \mathrm{~mm}\), calculate the maximum allowable load amplitude (in N) to ensure that fatigue failure will not occur at \(10^{7}\) cycles. Assume a factor of safety of \(3.0\), data in Figure \(8.20\) were taken for reversed axial tensioncompression tests, and that \(S\) is stress amplitude.

List four measures that may be taken to increase the resistance to fatigue of a metal alloy.

Steady-state creep data taken for an iron at a stress level of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) are given here: $$ \begin{array}{cc} \hline \dot{\boldsymbol{\epsilon}}_{s}\left(\boldsymbol{h}^{-\mathbf{1}}\right) & \boldsymbol{T}(\boldsymbol{K}) \\ \hline 6.6 \times 10^{-4} & 1090 \\ \hline 8.8 \times 10^{-2} & 1200 \\ \hline \end{array} $$ If it is known that the value of the stress exponent \(n\) for this alloy is \(8.5\), compute the steady-state creep rate at \(1300 \mathrm{~K}\) and a stress level of \(83 \mathrm{MPa}\) \((12,000 \mathrm{psi}) .\)

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \(82.4 \mathrm{MPa} \sqrt{\mathrm{m}}(75.0 \mathrm{ksi} \sqrt{\mathrm{in} .}) .\) If the plate is exposed to a tensile stress of \(345 \mathrm{MPa}(50,000\) psi) during service use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of \(1.0\) for \(Y\).

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