Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 45

For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking: $$ \begin{array}{cc} \hline \begin{array}{c} \text { Engineering Stress } \\ \text { (MPa) } \end{array} & \text { Engineering Strain } \\ \hline 235 & 0.194 \\ 250 & 0.296 \\ \hline \end{array} $$ On the basis of this information, compute the engineering stress necessary to produce an \(e n\) gineering strain of \(0.25\).

Short Answer

Expert verified
Answer: The engineering stress necessary to produce an engineering strain of 0.25 is approximately 243.27 MPa.

Step by step solution

01

Assign the given data to coordinates

Let's assign the given data to coordinates as follows: $$ Point \: A (x_1, y_1) = (0.194, 235) \\ Point \: B (x_2, y_2) = (0.296, 250) $$ where x is the engineering strain and y is the engineering stress.
02

Find the slope (m)

The slope (m) of the line passing through points A and B can be found using the formula: $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$ Substitute the values of the coordinates A and B: $$ m = \frac{250 - 235}{0.296 - 0.194} = \frac{15}{0.102} $$ Calculate the slope: $$ m \approx 147.06 $$
03

Find the y-intercept (b)

The y-intercept (b) can be found using the formula: $$ b = y - mx $$ Substitute the values of the slope and any of the points A or B. Let's use point A: $$ b = 235 - 147.06(0.194) $$ Calculate the y-intercept: $$ b \approx 206.61 $$
04

Write the linear equation

Now that we have the slope (m) and y-intercept (b), we can write the linear equation in the form of \(y = mx + b\): $$ y = 147.06x + 206.61 $$
05

Find the engineering stress for the desired strain

Finally, we need to find the engineering stress (y) corresponding to an engineering strain (x) of 0.25. Substitute the desired strain into the linear equation we found: $$ y = 147.06(0.25) + 206.61 $$ Calculate the engineering stress: $$ y \approx 243.27 \: \text{MPa} $$ The engineering stress necessary to produce an engineering strain of 0.25 is approximately 243.27 MPa.

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

A steel bar \(100 \mathrm{~mm}\) (4.0 in.) long and having a square cross section \(20 \mathrm{~mm}(0.8 \mathrm{in} .)\) on an edge is pulled in tension with a load of 89,000 \(\mathrm{N}\left(20,000 \mathrm{lb}_{\mathrm{f}}\right)\) and experiences an elongation of \(0.10 \mathrm{~mm}\left(4.0 \times 10^{-3}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equation \(6.5\) for elastic deformation, that the modulus of elasticity is 172 GPa \(\left(25 \times 10^{6} \mathrm{psi}\right)\), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for \(K\) and \(n\) are \(6900 \mathrm{MPa}\left(1 \times 10^{6} \mathrm{psi}\right)\) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of \(0.01\) and \(0.75\), at which point fracture occurs.

Cite five factors that lead to scatter in measured material properties.

A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{~mm}(0.505\) in.) and gauge length of \(50.80 \mathrm{~mm}(2.000 \mathrm{in} .)\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(6.60 \mathrm{~mm}(0.260 \mathrm{in} .)\), and the fractured gauge length is \(72.14 \mathrm{~mm}\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

For a bronze alloy, the stress at which plastic deformation begins is \(275 \mathrm{MPa}\) (40,000 psi), and the modulus of elasticity is \(115 \mathrm{GPa}\) \(\left(16.7 \times 10^{6} \mathrm{psi}\right) .\) (a) What is the maximum load that may be applied to a specimen with a cross- sectional area of \(325 \mathrm{~mm}^{2}\left(0.5 \mathrm{in} .^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(115 \mathrm{~mm}\) (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Mechanics and Materials

Read Explanation

Linear Momentum

Read Explanation

Particle Model of Matter

Read Explanation

Dynamics

Read Explanation

Translational Dynamics

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