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Chapter 7: Problem 17

A cylindrical specimen of a metal alloy \(10 \mathrm{~mm}\) \((0.4\) in.) in diameter is stressed elastically in tension. A force of \(15,000 \mathrm{~N}\left(3370 \mathrm{lb}_{\mathrm{f}}\right.\) ) produces a reduction in specimen diameter of \(7 \times 10^{-3} \mathrm{~mm}\) \(\left(2.8 \times 10^{-4}\right.\) in.). Compute Poisson's ratio for this material if its elastic modulus is \(100 \mathrm{GPa}(14.5 \times\) \(\left.10^{6} \mathrm{psi}\right)\)

Short Answer

Expert verified
Answer: The Poisson's ratio for the given metal alloy is approximately 0.366.

Step by step solution

01

Find the axial stress

First, we need to find the axial stress (the stress in the direction of the applied force) using the formula: Axial Stress (蟽) = Force / Area where, Force = 15,000 N Diameter = 10 mm Area = 蟺 * (Diameter/2)^2 The Area of the cylindrical specimen can be calculated as follows: Area = 蟺*(5^2) Area 鈮� 78.54 mm虏 Now we can find the axial stress: Axial stress (蟽) = 15000 N / 78.54 mm虏 Axial stress (蟽) 鈮� 191.04 MPa Note that we can leave the stress in MPa since the elastic modulus is given in GPa.
02

Find the axial strain

Next, we need to find the axial strain (the change in length due to the applied force) using the formula: Axial Strain (蔚) = Axial Stress / Elastic Modulus where, Axial Stress = 191.04 MPa Elastic Modulus = 100 GPa The axial strain can be calculated as follows: Axial Strain (蔚) = (191.04 MPa) / (100 GPa) Axial Strain (蔚) 鈮� 1.9104 x 10^-3
03

Find the lateral strain

Now, we need to find the lateral strain, which is the change in diameter due to the applied force divided by the original diameter: Lateral Strain = (Change in Diameter) / (Original Diameter) where, Change in Diameter = 7 x 10^(-3) mm Original Diameter = 10 mm The lateral strain can be calculated as follows: Lateral Strain = (7 x 10^(-3) mm) / (10 mm) Lateral Strain 鈮� 7 x 10^(-4)
04

Calculate Poisson's Ratio

Finally, we can calculate Poisson's Ratio using the formula: Poisson's Ratio (谓) = Lateral Strain / Axial Strain where, Lateral Strain 鈮� 7 x 10^(-4) Axial Strain 鈮� 1.9104 x 10^-3 The Poisson's Ratio can be calculated as follows: Poisson's Ratio (谓) = (7 x 10^(-4)) / (1.9104 x 10^-3) Poisson's Ratio (谓) 鈮� 0.366 The Poisson's Ratio for the metal alloy is approximately 0.366.

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Key Concepts

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

Axial Stress
Axial stress is a crucial concept in the field of mechanics of materials. It represents the stress component parallel to the axis of the specimen, which in our case is the metal alloy cylinder. When force is applied to an object, it creates stress. To find the axial stress, you use the formula:
  • Axial Stress (\( \sigma \)) = Force / Area.
The force applied here is 15,000 N, and the area is calculated based on the diameter of the cylindrical specimen. The diameter is 10 mm, and the cross-sectional area is \( \pi \times (\text{Diameter}/2)^2 \). After doing the calculations, the axial stress comes out to be approximately 191.04 MPa.
In essence, axial stress helps us understand how force influences material elongation or compression along a specific direction, essential for design and analysis of structural components.
Lateral Strain
Lateral strain occurs as a result of deformation in a direction perpendicular to the applied force. When a material is stretched or compressed, not only does it experience changes in length, but its diameter (or width) changes too. This change is known as lateral strain and can be calculated with:
  • Lateral Strain = Change in Diameter / Original Diameter.
In this exercise, the change in the specimen's diameter was observed to be \( 7 \times 10^{-3} \text{ mm} \), and the original diameter was 10 mm, making the lateral strain approximately \( 7 \times 10^{-4} \).
Lateral strain complements axial strain as they both are used to measure the deformation behavior of materials under stress, which aids in calculating Poisson's Ratio.
Elastic Modulus
The elastic modulus, often referred to as Young's modulus, is a material property that measures its ability to withstand changes in length when subjected to stress. Essentially, it quantifies material stiffness, defined by:
  • Elastic Modulus = Axial Stress / Axial Strain.
For the alloy used in this exercise, the elastic modulus is provided as 100 GPa. It serves as a fundamental constant for predicting elastic behavior under load.
An understanding of the elastic modulus provides insights into the range of deformation a material can undergo before permanent changes occur. In engineering, knowing the elastic modulus helps ensure materials are not pushed beyond their elastic limits, preventing unforeseen failures in applications.

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Most popular questions from this chapter

(a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the \([100]\) direction. If the magnitude of this stress is \(4.0 \mathrm{MPa}\), compute the resolved shear stress in the \([1 \overline{11}]\) direction on each of the \((110),(011)\), and \((10 \overline{1})\) planes. (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented?

Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.

In Section \(2.6\), it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.31\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is,Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system), using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus- \(r\) curve (Figure 2.10b), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\) - 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 and \(12 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(11 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.

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.
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