The fatigue data for a brass alloy are given as follows: $$ \begin{array}{cc} \hline \text { Stress Amplitude (MPa) } & \text { Cycles to Failure } \\ \hline 170 & 3.7 \times 10^{4} \\ \hline 148 & 1.0 \times 10^{5} \\ \hline 130 & 3.0 \times 10^{5} \\ \hline 114 & 1.0 \times 10^{6} \\ \hline 92 & 1.0 \times 10^{7} \\ \hline 80 & 1.0 \times 10^{8} \\ \hline 74 & 1.0 \times 10^{9} \\ \hline \end{array} $$ (a) Make an \(S-N\) plot (stress amplitude versus logarithm of cycles to failure) using these data. (b) Determine the fatigue strength at \(4 \times 10^{6}\) cycles. (c) Determine the fatigue life for \(120 \mathrm{MPa}\).

Short Answer

Expert verified
Based on the given fatigue data for a brass alloy, the fatigue strength of the material at 4 × 10^6 cycles is approximately 113.2 MPa. The fatigue life for a stress amplitude of 120 MPa is approximately 1.9 × 10^6 cycles.

Step by step solution

01

Make an S-N plot (stress amplitude vs logarithm of cycles to failure) using the given data

To create the S-N plot, we will plot the stress amplitude given in the table on the vertical axis and the logarithm (to base 10) of the cycles to failure on the horizontal axis. You can use any plotting software or draw it manually on graph paper.
02

Determine the fatigue strength at \(4 \times 10^{6}\) cycles

To find the fatigue strength at \(4 \times 10^{6}\) cycles, we will use our S-N plot. First, locate the point on the horizontal axis corresponding to the logarithm of \(4 \times 10^{6}\) cycles. This is equal to \(\log_{10}(4 \times 10^{6}) \approx 6.6\). Now, interpolate the curve to determine the corresponding stress amplitude on the vertical axis. Assuming a linear relationship, we can interpolate between the points for \(3.0 \times 10^{5}\) cycles and \(1.0 \times 10^{6}\) cycles: The slope of the line: $$m = \frac{S2-S1}{log(N2)-log(N1)} = \frac{130-114}{\log(3.0\times10^{5})-\log(1.0\times10^{6})}$$ Calculating the slope, we get: $$m \approx -53.64$$ Using the slope and any point on the line, such as \((\log(3.0\times10^{5}),130)\), we can derive the equation of the line: $$y = mx + c$$ $$130 = -53.64(\log(3.0 \times 10^{5})) + c$$ Calculating c, we get: $$c \approx 456.72$$ Now that we have the equation of the line (\(y \approx -53.64x + 456.72\)), we can find the stress amplitude at \(\log(4 \times 10^{6})\) cycles: $$S \approx -53.64(\log(4 \times 10^{6})) + 456.72$$ Calculating S, we get: $$S \approx 113.2 \mathrm{MPa} $$ So, the fatigue strength at \(4 \times 10^{6}\) cycles is approximately \(113.2 \mathrm{MPa}\).
03

Determine the fatigue life for \(120 \mathrm{MPa}\)

To find the fatigue life for a stress amplitude of \(120\ \mathrm{MPa}\), we will once again use the S-N plot. First, locate the point corresponding to the stress amplitude of \(120\ \mathrm{MPa}\) on the vertical axis. Now, interpolate the curve to determine the corresponding logarithm of cycles to failure on the horizontal axis. Using the equation of the line found in step 2 (\(y \approx -53.64x + 456.72\)), we can find the number of cycles for a stress amplitude of \(120\ \mathrm{MPa}\) by solving for \(x\): $$120 = -53.64x + 456.72$$ Calculating x, we get: $$x \approx \log(N) \approx 6.28$$ Now take the antilog to find the number of cycles N: $$N = 10^{6.28} \approx 1.9 \times 10^{6}$$ So, the fatigue life for a stress amplitude of \(120 \mathrm{MPa}\) is approximately \(1.9 \times 10^{6}\) cycles.

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Most popular questions from this chapter

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 \(1.0 .\)

A cylindrical component constructed from an S-590 alloy (Figure 8.31) is to be exposed to a tensile load of \(20,000 \mathrm{~N}\). What minimum diameter is required for it to have a rupture lifetime of at least \(100 \mathrm{~h}\) at \(925^{\circ} \mathrm{C} ?\)

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

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.

(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.

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