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Chapter 8: Problem 14

A fatigue test was conducted in which the mean stress was \(70 \mathrm{MPa}(10,000 \mathrm{psi}),\) and the stress amplitude was \(210 \mathrm{MPa}(30,000 \mathrm{psi})\) (a) Compute the maximum and minimum stress levels (b) Compute the stress ratio. (c) Compute the magnitude of the stress range.

Short Answer

Expert verified
Question: Determine the maximum and minimum stress levels, stress ratio, and stress range magnitude for a given mean stress of 70 MPa and stress amplitude of 210 MPa. Answer: The maximum stress level is 280 MPa, the minimum stress level is -140 MPa, the stress ratio is -0.5, and the stress range magnitude is 420 MPa.

Step by step solution

01

Find max. and min. stress levels

To compute the maximum and minimum stress levels, we need to use the formulas for the mean stress (m) and stress amplitude (a): Mean stress, \(m = \frac{\sigma_{max} + \sigma_{min}}{2}\) and Stress amplitude, \(a = \frac{\sigma_{max} - \sigma_{min}}{2}\). From the problem, \(m = 70\) MPa and \(a = 210\) MPa. We can rearrange the formulas for \(\sigma_{max}\) and \(\sigma_{min}\): \(\sigma_{max} = m + a\) and \(\sigma_{min} = m - a\). From here, we plug in the given values: \(\sigma_{max} = 70 + 210 = 280\mathrm{MPa}\) \(\sigma_{min} = 70 - 210 = -140\mathrm{MPa}\).
02

Compute stress ratio

The stress ratio (R) is defined as the ratio of the minimum stress level to the maximum stress level: \(R = \frac{\sigma_{min}}{\sigma_{max}}\). Using the values of \(\sigma_{max}\) and \(\sigma_{min}\) we found in Step 1: \(R = \frac{-140}{280} = -0.5\).
03

Compute stress range magnitude

The stress range (S) is defined as the difference between the maximum and minimum stress levels: \(S = \sigma_{max} - \sigma_{min}\). Using the values of \(\sigma_{max}\) and \(\sigma_{min}\) we found in Step 1: \(S = 280 - (-140) = 420\mathrm{MPa}\). Thus, the solutions are: (a) Maximum stress level: 280 MPa Minimum stress level: -140 MPa (b) Stress ratio: -0.5 (c) Stress range magnitude: 420 MPa

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Most popular questions from this chapter

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

Cite five factors that may lead to scatter in fatigue life data.

Cite three metallurgical/processing techniques that are employed to enhance the creep resistance of metal alloys.

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 below; the frequency is the same for all three tests.$$\begin{array}{ccc} \hline \text {Specimen} & \sigma_{\max }(M P a) & \sigma_{\min }(M P a) \\\\\hline \mathrm{A} & +450 & -150 \\\\\mathrm{B} & +300 & -300 \\\\\mathrm{C} & +500 & -200\end{array},$$ (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.

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 \(1030 \mathrm{MPa}\) \((150,000 \mathrm{psi}) .\) Will this specimen experience fracture if it is known that the largest surface crack is \(0.5 \mathrm{mm}(0.02 \text { in. })\) long? Why or why not? Assume that the parameter \(Y\) has a value of 1.0.
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