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Chapter 6: Problem 16

A cylindrical specimen of some metal alloy \(10 \mathrm{mm}(0.4 \text { in. })\) 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 \(7 \times 10^{-3} \mathrm{mm}\left(2.8 \times 10^{-4} \text {in. }\right)\) . Compute Poisson's ratio for this material if its elastic modulus is \(100 \mathrm{GPa}\left(14.5 \times 10^{6} \mathrm{psi}\right)\).

Short Answer

Expert verified
Answer: The Poisson's ratio for this material is approximately 0.366.

Step by step solution

01

Calculate stress

First, we need to calculate the stress experienced by the specimen. Stress can be calculated using the formula: Stress (σ) = Force (F) / Area (A) The area of the cylindrical specimen is given by the formula: A = π(d/2)^2, where d is the diameter of the specimen. In this case, d = 10 mm, and F = 15,000 N. A = π(10/2)^2 = π(5)^2 = 25π mm² Now, we can calculate the stress: σ = F / A = 15,000 N / 25π mm² ≈ 190.99 N/mm²
02

Calculate strain in the axial direction

Next, we will calculate the strain experienced by the specimen in the axial direction. Strain (ε) can be calculated using the formula: Strain (ε) = Stress (σ) / Elastic modulus (E) In this case, E is given as 100 GPa or \(100 * 10^{3}\) N/mm². ε = σ / E = 190.99 N/mm² / (100 * 10^3 N/mm²) ≈ 1.9099 × 10^{-3}
03

Calculate strain in the lateral direction

The reduction in the diameter of the specimen is given as \(7 * 10^{-3} \text{ mm}\). We can calculate the strain in the lateral direction (ε_lat) using the formula: ε_lat = (Reduction in diameter) / (Original diameter) ε_lat = (\(7 * 10^{-3} \text{ mm}\)) / (10 mm) = \(7 * 10^{-4}\)
04

Compute Poisson's ratio

Now, we can compute Poisson's ratio (ν) using the formula: ν = -(ε_lat / ε) ν = -(\(7 * 10^{-4}\) / \(1.9099 * 10^{-3}\)) ≈ 0.366 Poisson's ratio for this material is approximately 0.366.

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Most popular questions from this chapter

An aluminum bar \(125 \mathrm{mm}(5.0\) in.) long and having a square cross section \(16.5 \mathrm{mm}(0.65 \text { in. })\) on an edge is pulled in tension with a load of \(66,700 \mathrm{N}\left(15,000 \mathrm{lb}_{\mathrm{f}}\right),\) and experiences an elongation of \(0.43 \mathrm{mm}\left(1.7 \times 10^{-2} \text {in. }\right) .\) Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

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. $$\begin{array}{lll} 47.3 & 48.7 & 47.1 \\ 52.1 & 50.0 & 50.4 \\ 45.6 & 46.2 & 45.9 \\ 49.9 & 48.3 & 46.4 \\ 47.6 & 51.1 & 48.5 \\ 50.4 & 46.7 & 49.7 \end{array}$$

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 \(0.21 ?\)

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 cylindrical specimen of a nickel alloy having an elastic modulus of \(207 \mathrm{GPa}\left(30 \times 10^{6}\right.\) psi) and an original diameter of \(10.2 \mathrm{mm}\) \((0.40 \text { in. })\) will experience only elastic deformation when a tensile load of \(8900 \mathrm{N}\left(2000 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{mm}\) \((0.010 \text { in. })\).
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