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Chapter 6: Problem 17

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.

Short Answer

Expert verified
Answer: To find the original length of the cylindrical specimen, follow these steps: 1. Calculate the original and final cross-sectional area (A1 and A2) using the given diameters and the formula for the area of a circle. 2. Calculate the volumetric strain, Poisson's ratio, and axial strain using the provided elastic and shear moduli and the cross-sectional area change. 3. Use the axial strain and the given final length to determine the original length of the specimen using the rearranged axial strain equation. By following these steps, you can calculate the original length of the cylindrical specimen.

Step by step solution

01

Calculate the original and final cross-sectional area

We need to find the cross-sectional area of the specimen before and after deformation. This can be done using the formula for the area of a circle, which is given by the equation: \(A = \pi r^2\) where A is the area and r is the radius of the circle. Since the original diameter is 30.00 mm, the original radius is \(15.00 \ mm\). The final diameter is 30.04 mm, so the final radius is \(15.02 \ mm\). Original cross-sectional area (A1): \(A1 = \pi (15.00)^2\) Final cross-sectional area (A2): \(A2 = \pi (15.02)^2\)
02

Calculate the volumetric strain

The volumetric strain is the fractional change in volume of an object when subjected to deformation, and is given by the following formula: \(\epsilon_v = \frac{A2 * L2}{A1 * L1} - 1\) where \(\epsilon_v\) is the volumetric strain, \(L1\) and \(L2\) are the original and final lengths, and \(A1\) and \(A2\) are the original and final cross-sectional areas. We know that the volumetric strain is equal to the sum of the axial and the lateral strains. The axial (or longitudinal) strain is defined as the change in length due to the applied stress, while the lateral (transverse) strain is the change in diameter. The ratio of axial to lateral strain is called Poisson's ratio, which can be calculated using the elastic and shear moduli: \(Poisson's \ ratio(\nu) = \frac{E - 2G}{2G}\) where E is the elastic modulus and G is the shear modulus. Given that E = 65.5 GPa and G = 25.4 GPa: \(\nu = \frac{65.5 - 2*25.4}{2*25.4}\) By analyzing the cross-sectional area change (\(A1\) and \(A2\)), we can get the lateral strain (\(\epsilon_t\)): \(\epsilon_t = \frac{A2}{A1} - 1\) And by knowing the lateral strain and Poisson's ratio, we can compute the axial strain (\(\epsilon_a\)): \(\epsilon_a = \frac{\epsilon_t}{\nu}\)
03

Calculate the original length

Knowing the axial strain, we can calculate the original length. The axial strain is defined as follows: \(\epsilon_a = \frac{L2 - L1}{L1}\) where \(L1\) and \(L2\) are the original and final lengths, respectively. Now we can rearrange this equation to solve for the original length: \(L1 = \frac{L2}{1 + \epsilon_a}\) Given that the final length is \(105.20 \ mm\) and the axial strain is calculated in step 2, we can compute the original length of the cylindrical specimen.

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Key Concepts

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

Understanding Volumetric Strain
When a material is deformed, its volume changes in response to the applied force. The measure of this change in volume relative to its original volume is known as volumetric strain. It's an essential concept in material science, helping to understand how materials compress or expand under different conditions.

For example, in the given exercise, the cylindrical specimen changes in diameter and length when compressed. To quantify these changes, we calculate the volumetric strain using the initial and final dimensions of the specimen. It's important to remember that volumetric strain represents a ratio and is therefore dimensionless. Knowing the volumetric strain helps engineers and material scientists predict how a material will behave under various loads, ensuring the structural integrity of the components they're designing.
Poisson's Ratio and Its Significance
Elastic deformation often results in materials shrinking or expanding in directions perpendicular to the applied force. The Poisson's ratio is a fundamental material property that describes this behavior. It's defined as the ratio of transverse (lateral) strain to axial (longitudinal) strain in the direction of the stretching force.

Materials with a high Poisson's ratio tend to exhibit larger lateral deformation when stretched, while those with a low ratio have minimal change. In our exercise, Poisson's ratio plays a crucial role in connecting the lateral changes (diameter) to the longitudinal changes (length) that the cylindrical specimen undergoes. As we analyze deformation in design and materials engineering, Poisson's ratio provides insight into how the material will behave under stress, affecting our choices of materials for specific applications.
Elastic Modulus: Defining Material Stiffness
The elastic modulus, or Young's modulus, measures a material's elasticity, specifically its tendency to return to its original shape after deformation. Think of it like this: the higher the elastic modulus, the stiffer the material. In the context of our exercise, understanding the alloy's elastic modulus allows us to describe how much it resists deformation under a given load.

The elastic modulus is a critical parameter in determining how much a material will deform when a force is applied. In technical terms, it's defined as the ratio of stress (force per unit area) to strain (deformation relative to the original dimension). For any application where rigidity is required, such as bridges or building supports, materials with high elastic modulus values are often selected.
Exploring Shear Modulus in Materials
While the elastic modulus deals with stretching or compressing, the shear modulus is concerned with the concept of shear stress, which involves forces acting parallel to the surface of a material. The shear modulus describes a material's ability to resist these parallel forces that tend to cause one part of a body to slide past another.

In our exercise, though not directly applied, the shear modulus complements the elastic modulus and Poisson's ratio to fully characterize the material's behavior under elastic deformation. It helps predict how the material will react when subjected to forces that cause twisting or sliding, which is crucial in many practical engineering scenarios, like the design of shafts and couplings in machinery.

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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.

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 \(0.21 ?\)

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.
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