Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 9

Consider a cylindrical specimen of a steel alloy (Figure \(6.21) 8.5 \mathrm{mm}(0.33 \text { in. })\) in diameter and \(80 \mathrm{mm}(3.15 \text { in. })\) long that is pulled in tension. Determine its elongation when a load of \(65,250 \mathrm{N}\left(14,500 \mathrm{lb}_{\mathrm{f}}\right)\) is applied.

Short Answer

Expert verified
Answer: The elongation of the cylindrical steel alloy specimen when a load of 65,250 N is applied is approximately 0.4599 mm.

Step by step solution

01

Calculate the stress in the specimen

To calculate the stress on the cylindrical specimen, we need to determine the force applied and the cross-sectional area of the specimen. The force applied is given as \( F = 65,250 \, \mathrm{N}\), and the diameter of the specimen is given as \(d = 8.5 \, \mathrm{mm}\). The cross-sectional area of the cylindrical specimen is given by the formula \(A = \pi \frac{d^2}{4}\), where d is the diameter. Calculate the cross-sectional area: $$ A = \pi \frac{8.5^2}{4} = 56.749 \, \mathrm{mm^2} $$ Now we can calculate the stress: $$ \sigma = \frac{F}{A} = \frac{65,250 \, \mathrm{N}}{56.749 \, \mathrm{mm^2}} = 1,149.75 \, \mathrm{MPa} $$
02

Convert the applied stress to strain

To calculate the strain, we need to know the Young's modulus (E) of the steel alloy. The problem statement does not provide this value; however, we can use the approximate value for steel: \(E \approx 200 \, \mathrm{GPa} = 200,000 \, \mathrm{MPa}\). Now, we can calculate the strain using the applied stress and Young's modulus of the steel alloy: $$ \epsilon = \frac{\sigma}{E} = \frac{1,149.75 \, \mathrm{MPa}}{200,000 \, \mathrm{MPa}} = 5.7487 \times 10^{-3} $$
03

Calculate the elongation of the specimen

Now that we have the strain, we can calculate the elongation (ΔL) of the specimen using the initial length (L) and the strain: $$ \Delta L = \epsilon \cdot L = (5.7487 \times 10^{-3}) \cdot (80 \, \mathrm{mm}) = 0.4599 \, \mathrm{mm} $$ The elongation of the cylindrical steel alloy specimen when a load of 65,250 N is applied is approximately 0.4599 mm.

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

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0 \mathrm{mm}(0.39 \text { in. }) .\) A tensile force of \(1500 \mathrm{N}\) \(\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{mm}\left(2.64 \times 10^{-5} \mathrm{in.}\right)\). Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35.

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. $$\begin{array}{lll} 47.3 & 48.7 & 47.1 \\ 52.1 & 50.0 & 50.4 \\ 45.6 & 46.2 & 45.9 \\ 49.9 & 48.3 & 46.4 \\ 47.6 & 51.1 & 48.5 \\ 50.4 & 46.7 & 49.7 \end{array}$$

Cite five factors that lead to scatter in measured material properties.

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.

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 and \(30.04 \mathrm{mm}\) respectively, and its final length is \(105.20 \mathrm{mm}\) compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 65.5 and $25.4 \mathrm{GPa}, respectively.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Circular Motion and Gravitation

Read Explanation

Kinematics Physics

Read Explanation

Work Energy and Power

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