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Chapter 6: Problem 18

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.

Short Answer

Expert verified
Question: Calculate the elastic modulus of a cylindrical specimen of an unknown metal alloy with the following characteristics: Diameter (D) = 10.0 mm, Tensile force (F) = 1500 N, Elastic reduction in diameter (ΔD) = 6.7 x 10^-4 mm, Poisson's ratio (v) = 0.35. Answer: The elastic modulus of the unknown metal alloy is approximately 99.8 GPa.

Step by step solution

01

Understand the given information and equations

We've got a cylindrical specimen with given dimensions and a tensile force applied. And we know the change in diameter, and Poisson's ratio. Let's list this information: 1. Diameter (D) = \(10.0\:mm\) 2. Tensile force (F) = \(1500\:N\) 3. Elastic reduction in diameter (\(\Delta D\)) = \(6.7 \times 10^{-4}\:mm\) 4. Poisson's ratio (\(\nu\)) = 0.35 We are asked to find the elastic modulus (E). To calculate it, we can use Hooke's Law and Poisson's ratio equations. The relationships are as follows: 1. Hooke's Law equation: \(\sigma = E \cdot \epsilon\) 2. Poisson's ratio equation: \(\nu = -\frac{\epsilon_{transversal}}{\epsilon_{axial}}\) Now, knowing that \(\epsilon_{axial} = \frac{\Delta L}{L}\) and \(\epsilon_{transversal} = \frac{\Delta D}{D}\), we can rewrite the Poisson's ratio equation as: \(\nu = -\frac{\frac{\Delta D}{D}}{\frac{\Delta L}{L}}\) Our first goal will be to find \(\epsilon_{axial}\) and then use Hooke's Law to solve for E.
02

Find \(\epsilon_{axial}\) using Poisson's ratio equation

We can rewrite the Poisson's ratio equation for \(\epsilon_{axial}\): \(\epsilon_{axial} = -\frac{\Delta D}{\nu \cdot D}\) Now, we can plug in the given values: \(\epsilon_{axial} = -\frac{6.7 \times 10^{-4} \: mm}{0.35 \cdot 10.0 \: mm} = -1.914 \times 10^{-4}\)
03

Find Stress using Young's modulus and Hooke's Law

To find the stress, \(\sigma\), we need to calculate the force per unit area (cross-sectional area in this case). The stress formula is: \(\sigma = \frac{F}{A}\) The cross-sectional area A of the cylindrical specimen is given by \(A = \pi ( \frac{D}{2} )^2\): \(A = \pi \left(\frac{10.0\:mm}{2}\right)^2 = 78.54\: mm^2\) Now, we can find the stress: \(\sigma = \frac{1500\:N}{78.54\: mm^2} \approx19.11\:MPa\)
04

Calculate Elastic Modulus using Hooke's Law and known values

Now use Hooke's law equation to find the elastic modulus (E): \(E = \frac{\sigma}{\epsilon_{axial}}\) Plug in the calculated values: \(E = \frac{19.11\:MPa}{-1.914 \times 10^{-4}} \approx -99.8\: GPa\) The elastic modulus of this hypothetical metal alloy is approximately \(99.8\: GPa\).

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Key Concepts

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

Poisson's Ratio
Understanding Poisson's ratio is crucial when we deal with materials under stress. It's a measure of the deformation that occurs in a direction perpendicular to the direction of force. Specifically, it is defined as the negative ratio of transverse to axial strain. When we apply a tensile force to a material like a metal rod, it will stretch and become thinner.

Poisson's ratio, denoted by the Greek letter u, helps us describe this change. For example, if we have a rod with a ratio of 0.35, this would imply that for every unit of length it extends, the sideways contraction is 0.35 units. Thus, if a cylinder exhibits a reduction in diameter, we can use this ratio alongside the change in length to find the axial strain, which is critical in the calculation of the elastic modulus.
Hooke's Law
At the core of elasticity lies Hooke's Law, a principle that states the force needed to extend or compress a spring by some distance scales linearly with respect to that distance. In terms of materials, such as the cylindrical specimen in our example, Hooke's Law is represented as \[\begin{equation}\[\[\begin{align*}\frac{F}{A} = E \frac{\begin{Delta} L}{L}\end{align*}\]\]\end{equation}\]where
  • \begin{Delta} L\end{em}
    • ,,
Tensile Force
Tensile force is the pulling force applied along the length of a body, which tends to stretch the material. This force is a key variable in material science and engineering, leading to a form of deformation characterized by stretching. In our exercise, a cylindrical specimen aligns with this force, causing both a change in length and diameter. Understanding the effect of a given tensile force is vital for calculating stress and, subsequently, determining the elastic properties of the material.
Strain
Strain is an expression of deformation represented by the change in length compared to the initial length (axial strain) or the change in diameter compared to the original diameter (transverse strain). It's dimensionless, being a ratio of lengths.
  • Axial strain: \[\begin{equation}\[\[\begin{align*}\frac{\begin{Delta} L}{L}\end{align*}\]\]\end{equation}\]
  • Transverse strain: \[\begin{equation}\[\[\begin{align*}\frac{\begin{Delta} D}{D}\end{align*}\]\]\end{equation}\]
When we deal with problems like the one at hand, we deal primarily with axial strain resulting from tensile forces. Strain is used along with stress to define the material's response to stress through Hooke's Law.
Stress
Stress in materials is like the physical burden, or pressure, felt by an object. More formally, stress is the force applied over a unit area, represented mathematically as \[\begin{equation}\[\[\begin{align*}\frac{F}{A}\end{align*}\]\]\end{equation}\]where F is the applied force and A is the cross-sectional area through which the force is applied. In engineering and physics, stress is quantified to ensure materials can handle the loads they're subjected to. In the exercise provided, stress is a pivotal factor in calculating the elastic modulus, as we determine the pressure experienced by the specimen given a specific force.

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Most popular questions from this chapter

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) A 10 -mm-diameter Brinell hardness indenter produced an indentation \(2.50 \mathrm{mm}\) in diameter in a steel alloy when a load of \(1000 \mathrm{kg}\) was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of \(300 \mathrm{HB}\) when a \(500-\mathrm{kg}\) load is used?

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 \(0.21 ?\)

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

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.
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