Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 38

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 \(925^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The minimum diameter required for the cylindrical component is approximately 25.1 mm.

Step by step solution

01

1. Calculate stress for 100-hour rupture lifetime

Firstly, the graph for the S-590 alloy shows the stress value to be estimated around \(100 \mathrm{~MPa}\) to maintain 100-hour rupture lifetime at the specified temperature of \(925^{\circ} \mathrm{C}\). Keep in mind that \(100 \mathrm{~MPa} = 100 * 10^{6} \mathrm{~N/m^2}\).
02

2. Use the stress formula to find the minimum diameter

We will use the stress formula to find the minimum diameter: Stress = Force / Area Let d be the diameter of the cylindrical component. The area of a cylinder is as follows: Area = π * (diameter / 2)^2 Replacing Area in the stress formula: 100 * 10^{6} \mathrm{~N/m^2} = \(20,000 \mathrm{~N}\) / (π * (d / 2)^2) Now, we will solve for the diameter (d):
03

3. Solving for diameter d

First, isolate the term containing diameter (d): 100 * 10^{6} \mathrm{~N/m^2} * (π * (d / 2)^2) = \(20,000 \mathrm{~N}\) Divide both sides by the constants: (d / 2)^2 = \(\frac{20000}{100*10^6*π}\) Now, we can calculate the value of the square root of both sides: d / 2 = \(\sqrt{\frac{20000}{100*10^6*π}}\) Finally, multiply both sides by 2 to get the minimum diameter: d = \(2 * \sqrt{\frac{20000}{100*10^6*π}}\) d ≈ 0.0251 meters (25.1 mm) So, the minimum diameter required for the cylindrical component to have a rupture lifetime of at least 100 hours at \(925^{\circ} \mathrm{C}\) is about 25.1 mm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(120 \mathrm{MPa}\).

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

(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.

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 \(0.5 \mathrm{~mm}\) (0.02 in.) and a tip radius of curvature of \(5 \times 10^{-3} \mathrm{~mm}\left(2 \times 10^{-4}\right.\) in.), when a stress of \(1035 \mathrm{MPa}(150,000 \mathrm{psi})\) is applied.

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 \(1.0 .\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Electricity

Read Explanation

Solid State Physics

Read Explanation

Engineering Physics

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free