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Chapter 8: Problem 19

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.

Short Answer

Expert verified
Answer: The maximum allowable load amplitude for the cylindrical aluminum alloy bar is 2262 N.

Step by step solution

01

Determine the diameter and cross-sectional area of the bar

The diameter of the aluminum bar is \(12.0 \ mm\). To continue, we need to find the cross-sectional area \(A\) of the bar, which can be found using the formula for the area of a circle: \(A = \pi \frac{d^2}{4}\), where \(d\) is the diameter of the bar. Plug in the given diameter value: \(A = \pi\frac{(12)^2}{4} = 113.1 \ mm^2\).
02

Analyze the data from Figure 8.20 for the given cycles

To find the stress amplitude at \(10^{7}\) cycles, interpolate the data in Figure 8.20 and find the corresponding value of \(S\) (stress amplitude) for \(10^7\) cycles, which in our case is approximately \(S_N = 60 \ MPa\).
03

Apply the factor of safety

As we are asked to assume a factor of safety of \(3.0\), we will divide the stress amplitude at \(10^{7}\) cycles (\(S_N = 60 \ MPa\)) by the factor of safety (\(3.0\)): \(S_A = \frac{S_N}{FOS} = \frac{60 \ MPa}{3} = 20 \ MPa\).
04

Calculate the maximum allowable load amplitude

Now that we have the stress amplitude \(S_A\) and the area \(A\), we can find the maximum allowable load amplitude \(P_A\) using the following equation: \(P_A = S_A \cdot A\). Plug in the values of \(S_A = 20 \ MPa\) and \(A = 113.1 \ mm^2\), and convert the result from \(kN\) to \(N\): \(P_A = 20 \times 10^6 \ N/m^2 \cdot 113.1 \times 10^{-6} \ m^2 = 2262 \ N\). Thus, the maximum allowable load amplitude is \(2262 \ N\).

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Most popular questions from this chapter

(a) Using Figure 8.31, compute the rupture lifetime for an \(S-590\) alloy that is exposed to a tensile stress of \(400 \mathrm{MPa}\) at \(815^{\circ} \mathrm{C}\). (b) Compare this value to the one determined from the Larson-Miller plot of Figure \(8.33\), which is for this same S-590 alloy.

A specimen of a 4340 steel alloy with a plane strain fracture toughness of \(54.8 \mathrm{MPa} \sqrt{\mathrm{m}}(50 \mathrm{ksi} \sqrt{\mathrm{in}}\).) is exposed to a stress of \(1030 \mathrm{MPa}\) (150,000 psi). Will this specimen experience fracture if the largest surface crack is \(0.5 \mathrm{~mm}\) (0.02 in.) long? Why or why not? Assume that the parameter \(Y\) has a value of \(1.0 .\)

The fatigue data for a steel alloy are given as follows: $$ \begin{array}{|c|c} \hline \text { Stress Amplitude [MPa (ksi)] } & \text { Cycles to Failure } \\\ \hline 470(68.0) & 10^{4} \\ \hline 440(63.4) & 3 \times 10^{4} \\ \hline 390(56.2) & 10^{5} \\ \hline 350(51.0) & 3 \times 10^{5} \\ \hline 310(45.3) & 10^{6} \\ \hline 290(42.2) & 3 \times 10^{6} \\ \hline 290(42.2) & 10^{7} \\ \hline 290(42.2) & 10^{8} \\ \hline \end{array} $$ (a) Make an \(S-N\) plot (stress amplitude versus logarithm of cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of \(415 \mathrm{MPa}(60,000 \mathrm{psi})\) and \(275 \mathrm{MPa}(40,000 \mathrm{psi})\). (d) Estimate fatigue strengths at \(2 \times 10^{4}\) and \(6 \times\) \(10^{5}\) cycles.

The following tabulated data were gathered from a series of Charpy impact tests on a commercial low-carbon steel alloy. $$ \begin{array}{|cc|} \hline \text { Temperature }\left({ }^{\circ} \boldsymbol{C}\right) & \text { Impact Energy (J) } \\ \hline 50 & 76 \\ \hline 40 & 76 \\ \hline 30 & 71 \\ \hline 20 & 58 \\ \hline 10 & 38 \\ \hline 0 & 23 \\ \hline-10 & 14 \\ \hline-20 & 9 \\ \hline-30 & 5 \\ \hline-40 & 1.5 \\ \hline \end{array} $$ (a) Plot the data as impact energy versus temperature. (b) Determine a ductile-to-brittle transition temperature as the temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as the temperature at which the impact energy is \(20 \mathrm{~J}\).

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 \(67.5 \mathrm{~mm}\).
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