/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The following true stresses prod... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 39

The following true stresses produce the corresponding true plastic strains for a brass alloy: $$\begin{aligned} &\text { True }\\\ &\text {Stress}\\\ &\begin{array}{cc} (p s i) & \text {True Strain} \\ \hline 60,000 & 0.15 \\ 70,000 & 0.25 \end{array} \end{aligned}$$ What true stress is necessary to produce a true plastic strain of \(0.21 ?\)

Short Answer

Expert verified
Answer: The true stress necessary to produce a true plastic strain of 0.21 is 66,000 psi.

Step by step solution

01

Calculate the slope and intercept of the linear relationship between true stress and true plastic strain

We can write the linear relationship between true stress (Y) and true plastic strain (X) as: $$Y = mX + b$$ Where m is the slope and b is the intercept. Using point 1: \((X_{1}, Y_{1}) = (0.15, 60000)\) and point 2: \((X_{2}, Y_{2}) = (0.25, 70000)\), we can calculate the slope and intercept as follows: Slope: $$m = \frac{Y_{2} - Y_{1}}{X_{2} - X_{1}} = \frac{70000 - 60000}{0.25 - 0.15} = \frac{10000}{0.1} = 100000$$ Next, we can substitute the slope and point 1's coordinates to find the intercept: $$60000 = 100000\cdot0.15 + b$$ $$b = 60000 - 100000\cdot0.15 = 45000$$ So the linear relationship looks like this: $$Y = 100000X + 45000$$
02

Use the linear relationship to find the true stress for a true plastic strain of 0.21

Now that we have our linear relationship between true stress and true plastic strain, we can plug in the desired true plastic strain of 0.21 to find the corresponding true stress. $$Y = 100000 \cdot 0.21 + 45000 = 21000 + 45000 = 66000$$ The true stress necessary to produce a true plastic strain of 0.21 is 66,000 psi.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

True Stress
True stress is a crucial concept in understanding how materials deform under load. It represents the actual stress on a material at any given time, considering both the applied force and the current actual area over which it is applied. Unlike engineering stress, which is based upon the original area, true stress takes into account the change in area as the material is stretched or compressed.

To understand how to calculate true stress, remember that it involves the instant cross-sectional area of the material rather than the initial area. This is why true stress is often higher than engineering stress, especially in ductile materials that can undergo significant deformation.

In summary, true stress provides a more accurate measure of stress as it considers the real-time changes in the material's dimensions under load. This helps in predicting the material's behavior under different loading conditions effectively.
True Plastic Strain
True plastic strain represents the permanent deformation of a material after removing the stress. In simpler terms, it shows how much a material has deformed irreversibly.

Plastic deformation happens when a material is stretched beyond its elastic limit. When the stress is removed, the material does not return to its original shape, and this permanent change is referred to as true plastic strain. It is an essential factor in understanding a material's ductility and toughness since it provides insight into the material's ability to withstand potential loads after initial yielding.

Calculating true plastic strain involves measuring the strain after the elastic part of the deformation is subtracted. This gives an accurate depiction of the material's deformation behavior, beyond just temporary changes.
Linear Relationship
In materials science, the relationship between variables often simplifies problem-solving and enhances our understanding. A linear relationship describes a direct proportionality between two variables, often represented as a straight line on a graph.

In this particular exercise, the linear equation used is of the form:
  • \( Y = mX + b \)
Where:
  • \( Y \) is the true stress,
  • \( X \) is the true plastic strain,
  • \( m \) is the slope of the line, depicting how much \( Y \) changes with a change in \( X \), and
  • \( b \) is the intercept, indicating where the line crosses the Y-axis.
The slope \( m \) indicates the rate of change of true stress per unit change in true plastic strain. This straightforward relationship allows easy prediction of one variable if the other is known, making calculations simpler and more intuitive for problems involving these mechanical properties.

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 cylindrical metal specimen having an original diameter of \(12.8 \mathrm{mm}(0.505 \text { in. })\) and gauge length of \(50.80 \mathrm{mm}(2.000 \text { in. })\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(8.13 \mathrm{mm}(0.320 \mathrm{in.},\) and the fractured gauge length is \(74.17 \mathrm{mm}\) \((2.920 \text { in. }) .\) Calculate the ductility in terms of percent reduction in area and percent elongation.

For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\) and the modulus of elasticity is 103 GPa \(\left(15.0 \times 10^{6} \mathrm{psi}\right)\). (a) What is the maximum load that may be applied to a specimen with a cross- sectional area of \(130 \mathrm{mm}^{2}\left(0.2 \text { in. }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{mm}\) \((3.0 \text { in. }),\) what is the maximum length to which it may be stretched without causing plastic deformation?

A steel alloy to be used for a spring application must have a modulus of resilience of at least \(2.07 \mathrm{MPa}(300 \mathrm{psi}) .\) What must be its minimum yield strength?

A specimen of copper having a rectangular cross section \(15.2 \mathrm{mm} \times 19.1 \mathrm{mm}(0.60 \mathrm{in.} \times\) \(0.75 \text { in. })\) is pulled in tension with \(44,500 \mathrm{N}\) \(\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

A cylindrical rod of steel \((E=207 \mathrm{GPa}, 30 \times\) \(\left.10^{6} \text { psi }\right)\) having a yield strength of \(310 \mathrm{MPa}\) \((45,000 \mathrm{psi})\) is to be subjected to a load of \(11,100 \mathrm{N}\left(2500 \mathrm{lb}_{\mathrm{f}}\right) .\) If the length of the rod is \(500 \mathrm{mm}(20.0 \text { in. }),\) what must be the diameter to allow an elongation of \(0.38 \mathrm{mm}(0.015 \text { in. }) ?\)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Force

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism

Read Explanation

Linear Momentum

Read Explanation

Classical Mechanics

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.