Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 6

Some aircraft component 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 \(300 \mathrm{MPa}(43,500 \mathrm{psi})\) when the maximum (or critical) internal crack length is \(4.0 \mathrm{mm}(0.16 \text { in. }) .\) For this same component and alloy, will fracture occur at a stress level of \(260 \mathrm{MPa}(38,000 \mathrm{psi})\) when the maximum internal crack length is \(6.0 \mathrm{mm}(0.24 \mathrm{in.}) ?\) Why or why not?

Short Answer

Expert verified
Answer: Yes, the aluminum alloy will fracture under a stress level of 260 MPa and a crack length of 6.0 mm because the stress intensity factor (60.82 MPa√m) is greater than the alloy's plane strain fracture toughness (40 MPa√m).

Step by step solution

01

Calculation of Stress Intensity Factor

We will use the formula for the stress intensity factor, which is given by: \(K = \sigma \sqrt{\pi a}\) where \(K\) is the stress intensity factor, \(\sigma\) is the applied stress, and \(a\) is the crack length. We are given the values for the stress (\(\sigma = 260 \,\text{MPa}\)) and the crack length (\(a = 6.0 \,\text{mm} \, (\text{0.24 in.})\)). Replace the values in the formula: \(K = 260 \,\text{MPa} \sqrt{\pi (6.0 \,\text{mm})}\)
02

Simplify and Solve for Stress Intensity Factor

Now, we'll calculate the value of the stress intensity factor: \(K = 260 \,\text{MPa} \sqrt{\pi (6.0 \,\text{mm})} = 260 \,\text{MPa} \sqrt{18.85 \,\text{mm}} \approx 60.82 \,\text{MPa}\sqrt{\text{m}}\)
03

Compare Stress Intensity Factor with Plane Strain Fracture Toughness

We are given the plane strain fracture toughness of the aluminum alloy: \(40 \,\text{MPa}\sqrt{\text{m}}\). Now, compare the calculated stress intensity factor with the given plane strain fracture toughness value: \(K = 60.82 \,\text{MPa}\sqrt{\text{m}} > 40 \,\text{MPa}\sqrt{\text{m}}\) Since the stress intensity factor is greater than the plane strain fracture toughness, the fracture will occur at the given stress level (260 MPa) and crack length (6.0 mm).
04

Conclusion

Fracture will occur in the aluminum alloy at a stress level of 260 MPa and a maximum internal crack length of 6.0 mm, because the stress intensity factor (60.82 MPa√m) is greater than the plane strain fracture toughness of the alloy (40 MPa√m).

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

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of \(1.9 \times 10^{-4} \mathrm{mm}\) \(\left(7.5 \times 10^{-6} \text {in. }\right)\) and a crack length of \(3.8 \times\) \(10^{-2} \mathrm{mm}\left(1.5 \times 10^{-3} \mathrm{in.}\right)\) when a tensile stress of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) is applied?

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of \(26 \mathrm{MPa} \sqrt{\mathrm{m}}\) \((23.7 \mathrm{ksi} \sqrt{\mathrm{in.}}) .\) It has been determined that fracture results at a stress of \(112 \mathrm{MPa}\) \((16,240 \text { psi })\) when the maximum internal crack length is \(8.6 \mathrm{mm}(0.34 \text { in. }) .\) For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of \(6.0 \mathrm{mm}(0.24\) in.)

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

Steady-state creep data taken for an iron at a stress level of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) are given here $$\begin{array}{cc}\hline \dot{\epsilon}_{s}\left(h^{-1}\right) & T(K) \\\\\hline 6.6 \times 10^{-4} & 1090 \\\8.8 \times 10^{-2} & 1200 \\\\\hline\end{array}.$$ If it is known that the value of the stress exponent \(n\) for this alloy is \(8.5,\) compute the steady-state creep rate at \(1300 \mathrm{K}\) and a stress level of 83 MPa \((12,000 \text { psi })\).

The fatigue data for a brass alloy are given as follows: (a) Make an \(S-N\) plot (stress amplitude versus logarithm cycles to failure) using these data. (b) Determine the fatigue strength at \(4 \times 10^{6}\) cycles. (c) Determine the fatigue life for 120 MPa.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Nuclear Physics

Read Explanation

Quantum Physics

Read Explanation

Kinematics 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