Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the maximum-minimum stress cycles listed in the following table; the frequency is the same for all three tests. $$ \begin{array}{ccc} \hline \text { Specimen } & \sigma_{\max }(\boldsymbol{M P a}) & \sigma_{\min }(\boldsymbol{M P a}) \\ \hline \mathrm{A} & +450 & -350 \\ \mathrm{~B} & +400 & -300 \\ \text { C } & +340 & -340 \\ \hline \end{array} $$ (a) Rank the fatigue lifetimes of these three specimens from the longest to the shortest. (b) Now justify this ranking using a schematic \(S-N\) plot.

Short Answer

Expert verified
Based on the stress amplitude and mean stress calculations for each specimen, rank the fatigue lifetimes from longest to shortest. Explain your ranking using a schematic S-N plot.

Step by step solution

01

Calculate stress amplitude and mean stress for each specimen

The stress amplitude \((\sigma_a)\) and mean stress \((\sigma_m)\) for each specimen can be calculated using the following formulas: $$ \sigma_a = \frac{(\sigma_{max} - \sigma_{min})}{2} $$ and $$ \sigma_m = \frac{(\sigma_{max} + \sigma_{min})}{2} $$ Calculate the stress amplitude and mean stress for each specimen: Specimen A: $$ \sigma_{aA} = \frac{(450 - (-350))}{2} = \frac{800}{2} = 400~MPa $$ $$ \sigma_{mA} = \frac{(450 + (-350))}{2} = \frac{100}{2} = 50~MPa $$ Specimen B: $$ \sigma_{aB} = \frac{(400 - (-300))}{2} = \frac{700}{2} =350~MPa $$ $$ \sigma_{mB} = \frac{(400 + (-300))}{2} = \frac{100}{2} = 50~MPa $$ Specimen C: $$ \sigma_{aC} = \frac{(340 - (-340))}{2} = \frac{680}{2} = 340~MPa $$ $$ \sigma_{mC} = \frac{(340 + (-340))}{2} = \frac{0}{2} = 0~MPa $$
02

Rank the fatigue lifetimes of the specimens

To rank the fatigue lifetimes of the specimens, we will compare their stress amplitudes and mean stresses. A lower stress amplitude generally results in a longer fatigue life. Also, having a lower mean stress generally results in a longer fatigue life. From our calculations in Step 1, we can rank the specimens' fatigue lifetimes as follows: 1. Specimen C (Longest fatigue life): This specimen has the lowest stress amplitude (340 MPa) and the lowest mean stress (0 MPa). 2. Specimen B: This specimen has a stress amplitude of 350 MPa and a mean stress of 50 MPa. 3. Specimen A (Shortest fatigue life): This specimen has the highest stress amplitude (400 MPa) and a mean stress of 50 MPa.
03

Justify the ranking using a schematic S-N plot

An S-N plot is a graphical representation of the relationship between stress amplitude and the number of cycles to failure (fatigue life) for a given material. In an S-N plot, the stress amplitude is plotted on the vertical axis and the number of cycles to failure (in logarithmic scale) is plotted on the horizontal axis. The plot will usually show a downward curve, meaning that higher stress amplitudes result in a shorter fatigue life. To justify our ranking, we can plot the S-N curves for each specimen: 1. Specimen C's S-N curve will be above Specimens B and A, indicating the longest fatigue life. 2. Specimen B's S-N curve will be between Specimens C and A, indicating a medium fatigue life. 3. Specimen A's S-N curve will be below Specimens B and C, indicating the shortest fatigue life. The ranking determined in Step 2 matches the S-N plot description, which confirms our ranking of the fatigue lifetimes of the three specimens from longest to shortest as Specimen C, Specimen B, and Specimen A.

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Most popular questions from this chapter

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length \(0.25 \mathrm{~mm}(0.01\) in.) and having a tip radius of curvature of \(1.2\) \(\times 10^{-3} \mathrm{~mm}\left(4.7 \times 10^{-5} \mathrm{in} .\right)\) when a stress of \(1200 \mathrm{MPa}(174,000 \mathrm{psi})\) is applied.

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \(55 \mathrm{MPa} \sqrt{\mathrm{m}}(50 \mathrm{ksi} \sqrt{\mathrm{in} .})\). If, during service use, the plate is exposed to a tensile stress of \(200 \mathrm{MPa}(29,000 \mathrm{psi})\), determine the minimum length of a surface crack that will lead to fracture. Assume a value of \(1.0\) for \(Y\).

A fatigue test was conducted in which the mean stress was \(50 \mathrm{MPa}\) (7250 psi) and the stress amplitude was \(225 \mathrm{MPa}(32,625 \mathrm{psi})\). (a) Compute the maximum and minimum stress levels. (b) Compute the stress ratio. (c) Compute the magnitude of the stress range.

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of \(40 \mathrm{MPa} \sqrt{\mathrm{m}}(36.4 \mathrm{ksi} \sqrt{\mathrm{in}}\).). It has been determined that fracture results at a stress of 365 MPa (53,000 psi) when the maximum internal crack length is \(2.5 \mathrm{~mm}(0.10 \mathrm{in} .)\). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of \(4.0 \mathrm{~mm}(0.16 \mathrm{in}\).).

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of \(35 \mathrm{MPa} \sqrt{\mathrm{m}}(31.9 \mathrm{ksi} \sqrt{\text { in. }})\). It has been determined that fracture results at a stress of \(250 \mathrm{MPa}\) (36,250 psi) when the maximum (or critical) internal crack length is \(2.0 \mathrm{~mm}\) (0.08 in.). For this same component and alloy, will fracture occur at a stress level of \(325 \mathrm{MPa}\) (47,125 psi) when the maximum internal crack length is \(1.0 \mathrm{~mm}(0.04\) in.)? Why or why not?

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