Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 23

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). $$ \begin{array}{lccc} \hline \begin{array}{c} \text { Material } \end{array} & \begin{array}{c} \text { Modulus } \\ \text { of Elasticity } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Poisson's } \\ \text { Ratio } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 200 & 0.33 \\ \text { Brass alloy } & 101 & 300 & 0.34 \\ \text { Steel alloy } & 207 & 400 & 0.30 \\ \text { Titanium alloy } & 107 & 650 & 0.34 \\ \hline \end{array} $$

Short Answer

Expert verified
Answer: The suitable materials for the cylindrical rod are brass alloy, steel alloy, and titanium alloy, as they do not experience plastic deformation and can maintain the required diameter reduction limit.

Step by step solution

01

Calculate the cross-sectional area of the rod

Calculate the cross-sectional area (\(A\)) of the cylindrical rod using the formula: \(A=\pi\left(\frac{d}{2}\right)^2\), where \(d\) is the diameter. \[A = \pi \left(\frac{10 \times 10^{-3}}{2}\right)^2 = 7.853 \times 10^{-5} \mathrm{~m}^2\]
02

Calculate the stress in the rod

Determine the tensile stress (\(\sigma\)) in the rod using the formula: \(\sigma = \frac{F}{A}\), where \(F\) is the tensile load and \(A\) is the cross-sectional area. \[\sigma = \frac{27{,}500~\text{N}}{7.853 \times 10^{-5}~\text{m}^2} = 349{,}934~\text{Pa} = 349.93~\text{MPa}\]
03

Check for plastic deformation

Compare the calculated stress with the yield strength of each material to ensure that the rod will not undergo plastic deformation. - Aluminum alloy: \(\sigma\) = 349.93 MPa, Yield Strength = 200 MPa - Brass alloy: \(\sigma\) = 349.93 MPa, Yield Strength = 300 MPa - Steel alloy: \(\sigma\) = 349.93 MPa, Yield Strength = 400 MPa - Titanium alloy: \(\sigma\) = 349.93 MPa, Yield Strength = 650 MPa The brass, steel, and titanium alloys are all capable of handling the induced stress without undergoing plastic deformation.
04

Calculate the maximum strain

The maximum allowed deformation corresponds to a certain maximum strain (\(\varepsilon_\text{max}\)). Calculate the maximum strain using the limit of the diameter reduction. \[\varepsilon_\text{max} = \frac{7.5 \times 10^{-3}~\text{mm}}{10~\text{mm}} = 0.00075\]
05

Calculate the induced strain in each material

Determine the induced strain (\(\varepsilon\)) in each material using the formula: \(\varepsilon = \frac{ \sigma}{E} - \nu \cdot \frac{\sigma}{E}\), where \(E\) is the modulus of elasticity and \(\nu\) is Poisson's ratio. - Aluminum alloy: \(\varepsilon_\mathrm{Al} = \frac{349.93}{70{,}000} - 0.33 \cdot \frac{349.93}{70{,}000} = 0.00029\) - Brass alloy: \(\varepsilon_\mathrm{Br} = \frac{349.93}{101{,}000} - 0.34 \cdot \frac{349.93}{101{,}000} = 0.00034\) - Steel alloy: \(\varepsilon_\mathrm{St} = \frac{349.93}{207{,}000} - 0.30 \cdot \frac{349.93}{207{,}000} = 0.00035\) - Titanium alloy: \(\varepsilon_\mathrm{Ti} = \frac{349.93}{107{,}000} - 0.34 \cdot \frac{349.93}{107{,}000} = 0.00033\)
06

Check diameter reduction requirements

Compare the calculated strains with the maximum allowed strain (\(\varepsilon_\text{max}\)) to check if they meet the diameter reduction requirements. - Aluminum alloy: \(\varepsilon_\mathrm{Al}=0.00029 \leq \varepsilon_\text{max}=0.00075\) - Brass alloy: \(\varepsilon_\mathrm{Br}=0.00034 \leq \varepsilon_\text{max}=0.00075\) - Steel alloy: \(\varepsilon_\mathrm{St}=0.00035 \leq \varepsilon_\text{max}=0.00075\) - Titanium alloy: \(\varepsilon_\mathrm{Ti}=0.00033 \leq \varepsilon_\text{max}=0.00075\) All four materials have strain values lower than the maximum strain, which means that they can maintain the required diameter reduction limit. #Conclusion# Based on the stress and strain calculations, the suitable candidates for the cylindrical rod are brass alloy, steel alloy, and titanium alloy, as they do not experience plastic deformation and can maintain the required diameter reduction limit.

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 the (plain) carbon steel alloy, whose stressstrain behavior may be observed in the "Tensile Tests" module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength, (b) the tensile strength, and (c) the approximate ductility, in percent elongation.

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.

Upon what three criteria are factors of safety based?

In Section \(2.6\) it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.25\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, $$ E \propto\left(\frac{d F}{d r}\right)_{r_{0}} $$ Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system) using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus-r curve (Figure \(2.8 b\) ), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\). 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

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 \(n\).
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Fluids

Read Explanation

Force

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electricity and Magnetism

Read Explanation

Wave Optics

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