Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 19

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(8.0 \mathrm{~mm}(0.31 \mathrm{in} .)\). A tensile force of \(1000 \mathrm{~N}\) \(\left(225 \mathrm{lb}_{\mathrm{f}}\right.\) ) produces an elastic reduction in diameter of \(2.8 \times 10^{-4} \mathrm{~mm}\left(1.10 \times 10^{-5} \mathrm{in}\right.\).). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is \(0.30\).

Short Answer

Expert verified
Answer: The modulus of elasticity for the given metal alloy is \(1.7 \times 10^5\ MPa\).

Step by step solution

01

Calculate the Stress on the specimen

To compute the stress, we need to divide the applied force by the initial cross-sectional area of the specimen. First, we will calculate the area and then the stress. The diameter of the cylinder is given as \(8.0mm\). Thus, the initial cross-sectional area \(A_{0}\) can be found using the formula for the area of a circle: \(A_{0} = \pi \cdot \left(\frac{d}{2}\right)^2\) where \(d\) is the diameter. \(A_{0} = \pi \cdot \left(\frac{8.0}{2}\right)^2 = 50.27 mm^2\) Now, we calculate the stress, denoted as \(\sigma\): \(\sigma = \frac{Force}{A_{0}}\) \(\sigma = \frac{1000 N}{50.27 mm^2} = 19.89 \frac{N}{mm^2}\)
02

Calculate the Strain on the specimen

The next step is to calculate the strain. The lateral strain (\(\epsilon_{lateral}\)) can be calculated by dividing the change in diameter by the initial diameter: \(\epsilon_{lateral} = \frac{\Delta d}{d}\) \(\epsilon_{lateral} = \frac{2.8 \times 10^{-4}\ mm}{8.0\ mm} = 3.5 \times 10^{-5}\)
03

Relate Lateral Strain and Axial Strain using Poisson's Ratio

Poisson's ratio, denoted as \(\nu\), is the ratio of lateral strain to axial strain (\(\epsilon_{axial}\)). Therefore, we can write: \(\nu = \frac{\epsilon_{lateral}}{\epsilon_{axial}}\) Given \(\nu = 0.30\), we can find the axial strain by rearranging this equation: \(\epsilon_{axial} = \frac{\epsilon_{lateral}}{\nu}\) \(\epsilon_{axial} = \frac{3.5 \times 10^{-5}}{0.30} = 1.17 \times 10^{-4}\)
04

Calculate Modulus of Elasticity

Finally, we can find the modulus of elasticity, denoted by \(E\), using the following relationship between stress and axial strain: \(E = \frac{\sigma}{\epsilon_{axial}}\) \(E = \frac{19.89\ N/mm^2}{1.17 \times 10^{-4}} = 1.7 \times 10^{5}\ N/mm^2\) Thus, the modulus of elasticity for this alloy is \(1.7 \times 10^5\ N/mm^2\) or \(1.7 \times 10^5\ MPa\).

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

As noted in Section 3.15, for single crystals of some substances, the physical properties are anisotropic; that is, they are dependent on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general \([u v w]\) direction, \(E_{u v w}\), is described by the relationship $$ \begin{gathered} \frac{1}{E_{u v w}}=\frac{1}{E_{(100}}-3\left(\frac{1}{E_{(100}}-\frac{1}{E_{(111)}}\right) \\\ \left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}\right) \end{gathered} $$ where \(E_{100}\) and \(E_{(111)}\) are the moduli of elasticity in \([100]\) and [111] directions, respectively; \(\alpha, \beta\), and \(\gamma\) are the cosines of the angles between \([u v w]\) and the respective \([100],[010]\), and [001] directions. Verify that the \(E_{\langle 110\rangle}\) values for aluminum, copper, and iron in Table \(3.3\) are correct.

For some metal alloy, a true stress of \(415 \mathrm{MPa}\) (60,175 psi) produces a plastic true strain of \(0.475\). How much will a specimen of this material elongate when a true stress of \(325 \mathrm{MPa}\) \((46,125 \mathrm{psi})\) is applied if the original length is \(300 \mathrm{~mm}\) (11.8 in.)? Assume a value of \(0.25\) for the strain-hardening exponent \(n\).

A brass alloy is known to have a yield strength of \(275 \mathrm{MPa}(40,000 \mathrm{psi})\), a tensile strength of \(380 \mathrm{MPa}(55,000 \mathrm{psi})\), and an elastic modulus of \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\). A cylindrical specimen of this alloy \(12.7 \mathrm{~mm}(0.50\) in.) in diameter and \(250 \mathrm{~mm}\) (10.0 in.) long is stressed in tension and found to elongate \(7.6 \mathrm{~mm}(0.30 \mathrm{in}\).). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.

A cylindrical rod \(100 \mathrm{~mm}\) long and having a diameter of \(10.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(27,500 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(7.5 \times 10^{-3} \mathrm{~mm}\). Of the materials listed as follows, which are possible candidates? Justify your choice(s). $$ \begin{array}{lccc} \hline \begin{array}{c} \text { Material } \end{array} & \begin{array}{c} \text { Modulus } \\ \text { of Elasticity } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Poisson's } \\ \text { Ratio } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 200 & 0.33 \\ \text { Brass alloy } & 101 & 300 & 0.34 \\ \text { Steel alloy } & 207 & 400 & 0.30 \\ \text { Titanium alloy } & 107 & 650 & 0.34 \\ \hline \end{array} $$

Using the solution to Problem \(6.13\), rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate \(A, B\), and \(n\) parameters (Equation \(6.25)\), for these three materials are shown in the following table; they yield \(E_{N}\) in units of electron volts and \(r\) in nanometers: $$ \begin{array}{cccc} \hline \text { Material } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{n} \\\ \hline \mathrm{X} & 2.5 & 2.0 \times 10^{-5} & 8 \\ \mathrm{Y} & 2.3 & 8.0 \times 10^{-6} & 10.5 \\ \mathrm{Z} & 3.0 & 1.5 \times 10^{-5} & 9 \\ \hline \end{array} $$
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Atoms and Radioactivity

Read Explanation

Force

Read Explanation

Electricity

Read Explanation

Engineering Physics

Read Explanation

Space 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