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. $$ \begin{array}{cccc} \hline \text { Time } \text { (min) } & \text { Strain } & \text { Time } \text { (min) } & \text { Strain } \\ \hline 0 & 0.00 & 18 & 0.82 \\ \hline 2 & 0.22 & 20 & 0.88 \\ \hline 4 & 0.34 & 22 & 0.95 \\ \hline 6 & 0.41 & 24 & 1.03 \\ \hline 8 & 0.48 & 26 & 1.12 \\ \hline 10 & 0.55 & 28 & 1.22 \\ \hline 12 & 0.62 & 30 & 1.36 \\ \hline 14 & 0.68 & 32 & 1.53 \\ \hline 16 & 0.75 & 34 & 1.77 \\ \hline \end{array} $$

Short Answer

Expert verified
Answer: To determine the steady-state creep rate of an aluminum alloy with given creep data, follow these two steps: 1. Plot the given data points as strain versus time to visualize how the strain changes over time under a constant stress. 2. Examine the plot, identify the linear region, and calculate the slope (strain rate) of that region to find the steady-state (minimum) creep rate.

Step by step solution

01

Plot the data points

First, let's plot the given data points as strain (y-axis) vs time (x-axis). This will give us a visual representation of how the strain changes over time under a constant stress. The data points are: Time (min): [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34] Strain: [0.00, 0.22, 0.34, 0.41, 0.48, 0.55, 0.62, 0.68, 0.75, 0.82, 0.88, 0.95, 1.03, 1.12, 1.22, 1.36, 1.53, 1.77] Plot these data points on a graph using any graphing tool or software you prefer (e.g., Excel, Google Sheets, Desmos, etc.).
02

Determine the steady-state creep rate

Now that we have the plot, we need to determine the steady-state (minimum) creep rate. This can be found by identifying the linear region of the plot and calculating the slope (strain rate) of that region. 1. Examine the plot and identify the linear region where the slope remains constant. This is typically observed in the middle of the curve. 2. Choose two points within the linear region from the given data points and calculate the slope, or strain rate, using the formula: Slope (strain rate) = (strain2 - strain1) / (time2 - time1) 3. The Slope calculated in step 2 represents the steady-state (minimum) creep rate. Note: The exact steady-state (minimum) creep rate may vary depending on the chosen points within the linear region. To minimize error, you can use more than two points and calculate the average slope among those points. Also, you may consider using linear regression to calculate the slope in the linear region.

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Most popular questions from this chapter

Briefly explain the difference between fatigue striations and beachmarks in terms of (a) size and (b) origin.

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.

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.

An aircraft component is fabricated from an aluminum alloy that has a plane- strain fracture toughness of \(40 \mathrm{MPa} \sqrt{\mathrm{m}}\) (36.4 ksi \sqrt{in.). It has been deter- } mined that fracture results at a stress of \(300 \mathrm{MPa}\) (43,500 psi) when the maximum (or critical) internal crack length is \(4.0 \mathrm{~mm}\) (0.16 in.). For this same component and alloy, will fracture occur at a stress level of \(260 \mathrm{MPa}\) ( 38,000 psi) when the maximum internal crack length is \(6.0 \mathrm{~mm}(0.24\) in.) \(?\) Why or why not?

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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