Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 5

A specimen of a 4340 steel alloy having a plane strain fracture toughness of \(45 \mathrm{MPa} \sqrt{\mathrm{m}}\) (41 ksi \(\sqrt{\text { in. }})\) is exposed to a stress of 1000 MPa (145,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is \(0.75 \mathrm{~mm}(0.03\) in.) long? Why or why not? Assume that the parameter \(Y\) has a value of \(1.0\).

Short Answer

Expert verified
Answer: Yes, the 4340 steel alloy specimen will experience fracture under the given stress as the critical crack length (203.8 µm) is lower than the half crack length (375 µm).

Step by step solution

01

List the known values and calculate the half crack length

We are given the following values: - Plane strain fracture toughness (K_IC): \(45 \mathrm{MPa} \sqrt{\mathrm{m}}\) - Stress (σ): \(1000 \mathrm{MPa}\) - Largest surface crack length: \(0.75 \mathrm{mm}\) First, we need to calculate the half crack length (a) which is half of the crack length: \(a = \frac{0.75}{2} = 0.375 \mathrm{mm}\)
02

Rearrange the formula to solve for the critical crack length

We are using the formula \(K_\text{IC} = Y \sigma \sqrt{\pi a}\) and solving for the critical crack length (a). Rearranging the formula, we get: \(a = \frac{K_\text{IC}^2}{Y^2 \sigma^2 \pi}\)
03

Substitute the given values and calculate the critical crack length

Now, we will plug in the given values into the formula and calculate the critical crack length (a): \(a = \frac{(45 \mathrm{MPa} \sqrt{\mathrm{m}})^2}{(1.0)^2 (1000 \mathrm{MPa})^2 \pi}\) \(a \approx 2.038 \times 10^{-7} \mathrm{m} = 203.8 \mathrm{µm}\)
04

Compare the critical crack length to the half crack length and determine if the specimen will fracture

Now, we will compare the critical crack length (a) to the calculated half crack length from Step 1 (\(0.375 \mathrm{mm}\) or \(375 \mathrm{µm}\)). Since the critical crack length (\(203.8 \mathrm{µm}\)) is lower than the half crack length (\(375 \mathrm{µm}\)), the specimen will experience fracture under the given stress.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

4340 Steel Alloy
The 4340 steel alloy is a high-strength material commonly used in the aerospace and automotive industries. It is renowned for its excellent toughness, high tensile strength, and good fatigue resistance. This alloy is composed primarily of iron, with significant amounts of nickel, chromium, and molybdenum, enhancing its hardenability and corrosion resistance.

The performance of 4340 steel in demanding environments makes it suitable for critical components such as gears, aircraft landing gear, and structural parts. When dealing with such a robust material, it's crucial to understand its limitations, especially in the presence of flaws like surface cracks, which can lead to sudden catastrophic failures if the applied stress exceeds the material's fracture toughness.
Plane Strain Fracture Toughness
Plane strain fracture toughness (\( K_{IC} \)) is a critical property of materials which describes their resistance to fracture in the presence of a crack under plane strain conditions, typically found in thick sections. It is measured in terms of stress intensity, considering a unit of crack length, usually expressed as Megapascals times the square root of meters (\( \text{MPa}\root{}\text{m} \)).

For a given material, the higher the fracture toughness, the more resistant it is to the growth of a crack when stressed. The value of plane strain fracture toughness helps engineers predict the performance of materials under specific load conditions and can guide the selection of suitable materials for applications where mechanical failure must be avoided.
Stress Intensity Factor
The stress intensity factor (\( K \) ) quantifies the severity of the stress field near the tip of a crack and is used to predict the growth of cracks in materials under stress. The value of the stress intensity factor depends on the applied load, the size of the crack, and the geometry of the specimen. The critical value at which a material will fracture is known as the fracture toughness (\( K_{IC} \) ).

Mathematically, it is expressed with the formula \( K = Y \( \( \sigma \) \) \sqrt{ \( \pi a \) } \) where \( \sigma \) denotes the applied stress, \( a \) the crack length, and \( Y \) a geometric factor related to the shape and size of the specimen. If the calculated stress intensity factor for a material exceeds its plane strain fracture toughness, the material is likely to experience fracture.

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

Suppose that the fatigue data for the cast iron in Problem \(8.20\) were taken for bendingrotating tests, and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of 750 revolutions per minute. Give maximum lifetimes of continuous driving that are allowable for the following stress levels: (a) \(250 \mathrm{MPa}(36,250\) psi), (b) \(215 \mathrm{MPa}(31,000 \mathrm{psi})\), (c) \(200 \mathrm{MPa}\) \((29,000\) psi). and (d) \(150 \mathrm{MPa}(21,750 \mathrm{psi})\)

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

Following is tabulated data that were gathered from a series of Charpy impact tests on a tempered 4140 steel alloy. $$ \begin{array}{rc} \hline \text { Temperature }\left({ }^{\circ} \boldsymbol{C}\right) & \text { Impact Energy }(\boldsymbol{J}) \\ \hline 100 & 89.3 \\ 75 & 88.6 \\ 50 & 87.6 \\ 25 & 85.4 \\ 0 & 82.9 \\ -25 & 78.9 \\ -50 & 73.1 \\ -65 & 66.0 \\ -75 & 59.3 \\ -85 & 47.9 \\ -100 & 34.3 \\ -125 & 29.3 \\ -150 & 27.1 \\ -175 & 25.0 \\ \hline \end{array} $$ (a) Plot the data as impact energy versus temperature. (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is \(70 \mathrm{~J}\).

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.

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?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Nuclear Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

Read Explanation

Physical Quantities And Units

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