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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(20.000\) and \(20.025 \mathrm{~mm}\), respectively, and its final length is \(74.96 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(105 \mathrm{GPa}\) and \(39.7 \mathrm{GPa}\), respectively.

Short Answer

Expert verified
Answer: The original length of the cylindrical metal alloy specimen is approximately 75.185 mm.

Step by step solution

01

Understand the problem and given data

The problem involves finding the original length of a cylindrical metal alloy specimen under compression. We are given the original diameter (\(20.000\,\mathrm{mm}\)), the final diameter (\(20.025\,\mathrm{mm}\)), the final length (\(74.96\,\mathrm{mm}\)), and the mechanical properties, such as the elastic modulus (\(105\,\mathrm{GPa}\)) and shear modulus (\(39.7\,\mathrm{GPa}\)), of the alloy. Since the deformation is elastic, there is no permanent deformation involved.
02

Calculate the initial and final volumes of the specimen

To establish the relationship between the original and final lengths, we first need to calculate the initial and final volumes of the specimen. The volume of a cylinder is given by \(V = \pi r^2 h\), where \(V\) is the volume, \(r\) is the radius, and \(h\) is the height (or length). Since the mass and density of the specimen remain constant during elastic deformation, the volume remains constant as well; that is, \(V_\text{initial} = V_\text{final}\). Calculate the initial and final volumes as: \(V_\text{initial} = \pi (r_\text{initial})^2 (h_\text{initial}) = \pi (10.000\,\mathrm{mm})^2 (h_\text{initial})\) \(V_\text{final} = \pi (r_\text{final})^2 (h_\text{final}) = \pi (10.0125\,\mathrm{mm})^2 (74.96\,\mathrm{mm})\) Using \(V_\text{initial} = V_\text{final}\), we can write: \((10.000\,\mathrm{mm})^2 (h_\text{initial}) = (10.0125\,\mathrm{mm})^2 (74.96\,\mathrm{mm})\)
03

Calculate the original length of the specimen

To find the original length, rearrange the equation from Step 2 as follows: \(h_\text{initial} = \frac{(10.0125\,\mathrm{mm})^2 (74.96\,\mathrm{mm})}{(10.000\,\mathrm{mm})^2}\) Now, plug in the numerical values and compute the original length: \(h_\text{initial} = \frac{(10.0125\,\mathrm{mm})^2 (74.96\,\mathrm{mm})}{(10.000\,\mathrm{mm})^2} \approx 75.185\,\mathrm{mm}\) The original length of the given cylindrical specimen is approximately \(75.185\,\mathrm{mm}\).

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Key Concepts

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

Compressive Stress
Compressive stress is a fundamental concept in materials science and engineering, particularly relevant when studying elastic deformation. It refers to the force applied over the cross-sectional area of a material leading to a decrease in its volume. In simpler terms, it's the 'squeeze' exerted on materials which can cause objects to shorten.

To visualize this, imagine pressing down on a spring with your hand. The force you apply causes the spring to compress—that force, divided by the area over which it's applied, is the compressive stress. Mathematically, it can be expressed as \[\begin{equation}\sigma_c = \frac{F}{A}\end{equation}\]where \(\sigma_c\) is the compressive stress, \(F\) is the force applied, and \(A\) is the area over which the force is distributed. In the cylindrical specimen from our exercise, as the diameter increases due to compression, the area over which the stress is applied changes, influencing the overall stress state of the material.
Elastic Modulus
The elastic modulus, also known as Young's modulus, is a measure of a material's ability to resist changes in length when under tension or compression. It's an important property that indicates how stiff a material is. Elastic modulus is defined as the ratio of stress (force per unit area) to strain (deformation in the material).

For any elastic material, the relationship between the applied stress and the resulting strain is linear within certain limits. This proportionality is represented by the equation\[\begin{equation}E = \frac{\sigma}{\epsilon}\end{equation}\]where \(E\) is the elastic modulus, \(\sigma\) is the stress, and \(\epsilon\) is the strain. A high elastic modulus indicates that a material is rigid and does not deform easily under stress. The hypothetical metal alloy in our exercise, with its elastic modulus of 105 GPa, would be considered quite stiff, which is why it returns to its original length after the compressive force is removed.
Volume Conservation
Volume conservation, otherwise known as volume constancy, is a key principle in understanding elastic deformation in materials. When an object like our cylindrical specimen undergoes elastic deformation, its shape might change—such as getting shorter and wider—but its volume remains constant as long as the deformation is purely elastic and temperature remains consistent.

Elastic deformation is temporary. Remove the force, and the object will revert to its original shape. This is due to the material's molecules rearranging themselves to accommodate the stress without breaking bonds, keeping the same overall volume. This principle allows us to perform calculations like those in the given exercise, where we can equate the initial volume to the final volume and solve for unknown dimensions.

In mathematical terms, volume conservation can be written as\[\begin{equation}V_\text{initial} = V_\text{final}\end{equation}\]This formula, along with the understanding that volume is a product of cross-sectional area and length (or height), is crucial for solving problems involving elastic deformation, such as determining the original length of an elastically deformed object. By maintaining the volume constant, despite changes in shape, we have the necessary relationship to calculate the initial dimensions from the final state after deformation.

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Most popular questions from this chapter

Demonstrate that Equation \(6.16\), the expression defining true strain, may also be represented by $$ \epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right) $$ when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why?

(a) A 10-mm-diameter Brinell hardness indenter produced an indentation \(1.62 \mathrm{~mm}\) in diameter in a steel alloy when a load of \(500 \mathrm{~kg}\) was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of \(450 \mathrm{HB}\) when a \(500-\mathrm{kg}\) load is used?

Upon what three criteria are factors of safety based?

Using the data in Problem \(6.29\) and Equations 6.15, 6.16, and 6.18a, generate a true stress-true strain plot for aluminum. Equation \(6.18\) a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last four data points, which should be used in true stress computations. $$ \begin{array}{cccccc} \hline \text {Load} & & \text {Length} & & \text {Diameter} \\ \hline \boldsymbol{N} & \boldsymbol{l b}_{f} & \boldsymbol{m m} & \text { in. } & {\boldsymbol{m m}} & \text { in. } \\ \hline 46,100 & 10,400 & 56.896 & 2.240 & 11.71 & 0.461 \\ 44,800 & 10,100 & 57.658 & 2.270 & 11.26 & 0.443 \\ 42,600 & 9,600 & 58.420 & 2.300 & 10.62 & 0.418 \\ 36,400 & 8,200 & 59.182 & 2.330 & 9.40 & 0.370 \\ \hline \end{array} $$

Using the solution to Problem \(6.13\), rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate \(A, B\), and \(n\) parameters (Equation \(6.25)\), for these three materials are shown in the following table; they yield \(E_{N}\) in units of electron volts and \(r\) in nanometers: $$ \begin{array}{cccc} \hline \text { Material } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{n} \\\ \hline \mathrm{X} & 2.5 & 2.0 \times 10^{-5} & 8 \\ \mathrm{Y} & 2.3 & 8.0 \times 10^{-6} & 10.5 \\ \mathrm{Z} & 3.0 & 1.5 \times 10^{-5} & 9 \\ \hline \end{array} $$

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