Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 5

An aluminum bar \(125 \mathrm{mm}(5.0\) in.) long and having a square cross section \(16.5 \mathrm{mm}(0.65 \text { in. })\) on an edge is pulled in tension with a load of \(66,700 \mathrm{N}\left(15,000 \mathrm{lb}_{\mathrm{f}}\right),\) and experiences an elongation of \(0.43 \mathrm{mm}\left(1.7 \times 10^{-2} \text {in. }\right) .\) Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

Short Answer

Expert verified
The modulus of elasticity of the aluminum under the given conditions is approximately 71,155.35 N/mm².

Step by step solution

01

Calculate the cross-sectional area of the aluminum bar

To calculate the stress experienced by the bar, we need to determine its cross-sectional area. The bar is square in cross-section, with an edge length of 16.5 mm. The area \(A\) of a square is the square of the edge length: \(A = (16.5 \, \text{mm})^2 = 272.25 \, \text{mm}^2\)
02

Calculate the stress experienced by the aluminum bar

Stress is defined as the force per unit area. In this case, the applied force is 66,700 N, and we have just calculated the cross-sectional area of the bar. The stress \(\sigma\) can be calculated as: \(\sigma = \frac{\text{Force}}{\text{Area}} = \frac{66,700 \, \text{N}}{272.25 \, \text{mm}^2} = 244.93 \, \text{N/mm}^2\)
03

Calculate the strain experienced by the aluminum bar

Strain is the ratio of the change in length to the original length of the bar. In this case, the change in length is 0.43 mm, and the original length is 125 mm. The strain \(\epsilon\) can be calculated as: \(\epsilon = \frac{\text{Change in Length}}{\text{Original Length}} = \frac{0.43 \, \text{mm}}{125 \, \text{mm}} = 0.00344\)
04

Calculate the modulus of elasticity of the aluminum

Now that we have the stress and strain experienced by the aluminum bar, we can use Hooke's Law to calculate the modulus of elasticity (Young's modulus) \(E\): \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{244.93 \, \text{N/mm}^2}{0.00344} = 71,155.35 \, \text{N/mm}^2\) So, the modulus of elasticity of the aluminum under the given conditions is approximately 71,155.35 N/mm².

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

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 \(0.21 ?\)

Below are tabulated a number of Rockwell G hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. $$\begin{array}{lll} 47.3 & 48.7 & 47.1 \\ 52.1 & 50.0 & 50.4 \\ 45.6 & 46.2 & 45.9 \\ 49.9 & 48.3 & 46.4 \\ 47.6 & 51.1 & 48.5 \\ 50.4 & 46.7 & 49.7 \end{array}$$

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?

Taking the logarithm of both sides of Equation 6.19 yields $$\log \sigma_{T}=\log K+n \log \epsilon_{T}\quad\quad\quad\quad(6.27)$$ Thus, a plot of \(\log \sigma_{T}\) versus \(\log \epsilon_{T}\) in the plastic region to the point of necking should yield a straight line having a slope of \(n\) and an intercept (at \(\log \sigma_{T}=0\) ) of \(\log K\). Using the appropriate data tabulated in Problem 6.28 , make a plot of \(\log \sigma_{T}\) versus log \(\epsilon_{T}\) and determine the values of \(n\) and \(K\). It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 6.18 and 6.18 b.

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 \(103 \mathrm{GPa}\left(15 \times 10^{6} \mathrm{psi}\right),\) and that elastic deformation terminates at a strain of \(0.007 .\) 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 \(1520 \mathrm{MPa}(221,000 \mathrm{psi})\) and \(0.15,\) respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and \(0.60,\) at which point fracture occurs.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Engineering Physics

Read Explanation

Solid State Physics

Read Explanation

Thermodynamics

Read Explanation

Torque and Rotational Motion

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