Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 41

Using the data in Problem \(6.29\) and Equations 6.15, 6.16, and 6.18a, generate a true stress-true strain plot for aluminum. Equation \(6.18\) a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last four data points, which should be used in true stress computations. $$ \begin{array}{cccccc} \hline \text {Load} & & \text {Length} & & \text {Diameter} \\ \hline \boldsymbol{N} & \boldsymbol{l b}_{f} & \boldsymbol{m m} & \text { in. } & {\boldsymbol{m m}} & \text { in. } \\ \hline 46,100 & 10,400 & 56.896 & 2.240 & 11.71 & 0.461 \\ 44,800 & 10,100 & 57.658 & 2.270 & 11.26 & 0.443 \\ 42,600 & 9,600 & 58.420 & 2.300 & 10.62 & 0.418 \\ 36,400 & 8,200 & 59.182 & 2.330 & 9.40 & 0.370 \\ \hline \end{array} $$

Short Answer

Expert verified
Answer: To generate a true stress-true strain plot for aluminum with the given data, we would first use Equations 6.15, 6.16, and 6.18a to calculate the true stress and true strain for each data point. Then, we would plot the true stress values on the vertical axis and the true strain values on the horizontal axis to visualize the relationship between true stress and true strain for aluminum.

Step by step solution

01

Recall the equations

We need to use Equations 6.15, 6.16, and 6.18a to compute true stress and true strain. These equations are given by: - True stress: \(σ_T = \frac{P}{A_0} (1 + \frac{l - l_0}{l_0})\) (Equation 6.15) - True strain: \(ε_T = ln (\frac{l}{l_0})\) (Equation 6.16) - Engineering stress: \(σ_T = \frac{P}{A}\) (Equation 6.18a)
02

Perform true stress calculations for the last four data points

For each of the last four data points, apply Equation 6.18a: 1. For load 46,100 N, Final Length 56.896 mm, and diameter 11.71 mm Cross-sectional Area \(A = \frac{π \times (d/2)^2}\) \( σ_T = \frac{46,100}{A}\) 2. For load 44,800 N, Final Length 57.658 mm, and diameter 11.26 mm Cross-sectional Area \(A = \frac{π \times (d/2)^2}\) \( σ_T = \frac{44,800}{A}\) 3. For load 42,600 N, Final Length 58.420 mm, and diameter 10.62 mm Cross-sectional Area \(A = \frac{π \times (d/2)^2}\) \( σ_T = \frac{42,600}{A}\) 4. For load 36,400 N, Final Length 59.182 mm, and diameter 9.40 mm Cross-sectional Area \(A = \frac{π \times (d/2)^2}\) \( σ_T = \frac{36,400}{A}\) Plug the diameters into the equation and calculate the true stress for each data point.
03

Perform true strain calculations for the last four data points

For each of the last four data points, apply Equation 6.16: 1. For Final Length 56.896 mm: \(ε_T = ln (\frac{56.896}{l_0})\) 2. For Final Length 57.658 mm: \(ε_T = ln (\frac{57.658}{l_0})\) 3. For Final Length 58.420 mm: \(ε_T = ln (\frac{58.420}{l_0})\) 4. For Final Length 59.182 mm: \(ε_T = ln (\frac{59.182}{l_0})\) Calculate the true strain for each data point by plugging in the initial length, \(l_0\).
04

Plot the true stress-true strain graph

Now that we have the true stress and true strain for all data points, plot the true stress on the vertical axis against the true strain on the horizontal axis. The graph will allow us to visualize the relationship between true stress and true strain for aluminum. Using a graphing tool, input the calculated true stress and true strain values for the four data points. The resulting true stress-true strain curve will provide important information about the mechanical behavior of aluminum.

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

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?

Figure \(6.22\) shows, for a gray cast iron, the tensile engineering stress- strain curve in the elastic region. Determine (a) the tangent modulus at \(10.3\) MPa (1500 psi) and (b) the secant modulus taken to \(6.9 \mathrm{MPa}\) (1000 psi).

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.

A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure \(6.21\) is subjected to a tensile load; the specimen is 300 mm (12 in.) long and has a square cross section \(4.5 \mathrm{~mm}(0.175 \mathrm{in}\).) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of \(0.45 \mathrm{~mm}\) \((0.018\) in.). (b) What will be the deformation after the load has been released?

A cylindrical rod \(380 \mathrm{~mm}\) (15.0 in.) long, having a diameter of \(10.0 \mathrm{~mm}(0.40\) in.), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than \(0.9 \mathrm{~mm}(0.035\) in.) when the applied load is \(24,500 \mathrm{~N}\left(5500 \mathrm{lb}_{\mathrm{f}}\right)\), which of the four metals or alloys listed in the following table are possible candidates? Justify your choice(s). $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Modulus } \\ \text { Material } \end{array} & \begin{array}{c} \text { \mathrm{ Yield } } \\ \text { of Elasticity } \\ \text { Strength } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Tensile } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { (MPa) } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 255 & 420 \\ \text { Brass alloy } & 100 & 345 & 420 \\ \text { Copper } & 110 & 250 & 290 \\ \text { Steel alloy } & 207 & 450 & 550 \\ \hline \end{array} $$
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Physics of Motion

Read Explanation

Electricity

Read Explanation

Radiation

Read Explanation

Force

Read Explanation

Medical 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