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 \(25.4\) GPa, respectively.

Short Answer

Expert verified
The original length of the cylindrical metal specimen is approximately 105.34 mm.

Step by step solution

01

Calculate the change in diameter and longitudinal strain

First, we need to calculate the change in diameter as follows: \(\Delta d = d_{final} - d_{initial} = 30.04 \mathrm{~mm} - 30.00 \mathrm{~mm} = 0.04 \mathrm{~mm}\) Now, compute the lateral (transverse) strain: \(\varepsilon_{lateral} = \frac{\Delta d}{d_{initial}} = \frac{0.04}{30} = \frac{1}{750}\)
02

Calculate Poisson's ratio

Next, we need to find the Poisson's ratio for the alloy. The elastic and shear moduli for this alloy are given as \(65.5\) and \(25.4\) GPa, respectively. The formula to relate these values to Poisson's ratio is: \(\nu = \frac{E}{2G} - 1\) where \(\nu\) is Poisson's ratio, \(E\) is the elastic modulus, and \(G\) is the shear modulus. Let's compute Poisson's ratio: \(\nu = \frac{65.5}{2 \times 25.4} - 1 = 0.29\)
03

Calculate the longitudinal strain

Using the Poisson's ratio, we can compute the longitudinal strain as follows: \(\varepsilon_{longitudinal} = -\nu \times \varepsilon_{lateral} = -0.29 \times \frac{1}{750} = -\frac{29}{7500}\)
04

Calculate the original length

Now that we have the longitudinal strain, we can compute the original length of the cylinder. We are given the final length as \(105.20\,\mathrm{mm}\). Let's denote the original length by \(L_{initial}\). We have: \(\varepsilon_{longitudinal} = \frac{L_{initial} - L_{final}}{L_{initial}} = -\frac{29}{7500}\) Solve for the original length: \(L_{initial} = \frac{L_{final}}{1 - \varepsilon_{longitudinal}} = \frac{105.20\,\mathrm{mm}}{1 + \frac{29}{7500}} = \frac{105.20\,\mathrm{mm} \times 7500}{7529} \approx 105.34 \mathrm{~mm}\) Hence, the original length of the cylindrical specimen was approximately \(105.34\,\mathrm{mm}\).

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.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson's Ratio
Poisson's ratio, symbolized by ug, is a measure that tells us how much a material will expand or contract in directions perpendicular to the direction of compression or stretching. Consider a rubber band being stretched; not only does it become longer, but also thinner. The same occurs in reverse when you compress a material - it can expand outwards. Poisson's ratio is calculated using the formula: \[u = \frac{E}{2G} - 1\], where \(E\) stands for the elastic modulus and \(G\) represents the shear modulus. A common value of Poisson's ratio for many materials is around 0.3, indicating that materials typically shrink by about 0.3 times their stretching amount transversely when stretched longitudinally.
Longitudinal Strain
The term 'longitudinal strain' refers to the deformation experienced by a material along its length when subjected to stress. It is a dimensionless quantity defined as the change in length per unit length and can be positive (elongation) or negative (contraction). Mathematically, it's represented as:\[\varepsilon_{longitudinal} = \frac{L_{final} - L_{initial}}{L_{initial}}\]. In the context of the exercise, longitudinal strain has a crucial part in finding the original length of the cylinder before its compression. Since materials respond to stress differently, accurately calculating longitudinal strain is essential for engineers when designing structures and predicting their behavior under various load conditions.
Shear Modulus
The shear modulus, or modulus of rigidity, symbolized as \(G\), describes how a material reacts to shearing forces. It quantifies the material's response by measuring the ratio of shear stress to the shear strain. The formula is given by:\[G = \frac{{\text{shear stress}}}{{\text{shear strain}}}\]. It helps us understand the relationship between the force applied parallel to a surface and the deformation it causes from the original shape. A higher shear modulus value indicates a material that resists shear deformation well, which is desired in applications where rigidity is necessary, such as in the beams of a building to maintain structural integrity.
Elastic Modulus
Elastic modulus, also known as Young's modulus and represented by \(E\), is a fundamental property of materials reflecting their stiffness and is defined as the ratio of tensile stress to longitudinal (tensile) strain. The expression for calculating it is:\[E = \frac{{\text{tensile stress}}}{{\text{tensile strain}}}\]. It gives us an insight into how much a material will deform under a certain load and is crucial for guaranteeing that structures won't fail under stress. The elastic modulus plays a pivotal role in determining the degree of compression or extension possible before a material reaches its elastic limit and suffers permanent deformation.

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

6.24 A cylindrical rod \(120 \mathrm{~mm}\) long and having a (- diameter of \(15.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(35,000 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(1.2 \times 10^{-2} \mathrm{~mm}\). Of the following materials listed, which are possible candidates? Justify your choice(s). \begin{tabular}{lccc} \hline & Modulus of Material & \begin{tabular}{c} \mathrm{ Yield } \(\\\ {\text { Elasticity }} \\ {\text { (GPa) }}\) & Strength (MPa) & Poisson's Ratio \\ \hline Aluminum alloy & 70 & 250 & \(0.33\) \\ \hline Titanium alloy & 105 & 850 & \(0.36\) \\ \hline Steel alloy & 205 & 550 & \(0.27\) \\ \hline Magnesium alloy & 45 & 170 & \(0.35\) \\ \hline \end{tabular} \end{tabular}

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30\times10^{6 psi) and an } original diameter of \(10.2 \mathrm{~mm}(0.40\) in.) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{i}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010\) in.).

Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.

For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.

A cylindrical specimen of stainless steel having a diameter of \(12.8 \mathrm{~mm}(0.505 \mathrm{in}\).) and a gauge length of \(50.800 \mathrm{~mm}(2.000 \mathrm{in}\) ) is pulled in tension. Use he load-elongation characteristics shown in the ollowing table to complete parts (a) through (f). (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of \(0.002\). (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience.

See all solutions

Recommended explanations on Physics Textbooks

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