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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 (σ) = Force / Area where, Force = 15,000 N Diameter = 10 mm Area = π * (Diameter/2)^2 The Area of the cylindrical specimen can be calculated as follows: Area = π*(5^2) Area ≈ 78.54 mm² Now we can find the axial stress: Axial stress (σ) = 15000 N / 78.54 mm² Axial stress (σ) ≈ 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 (ε) = Axial Stress / Elastic Modulus where, Axial Stress = 191.04 MPa Elastic Modulus = 100 GPa The axial strain can be calculated as follows: Axial Strain (ε) = (191.04 MPa) / (100 GPa) Axial Strain (ε) ≈ 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 ≈ 7 x 10^(-4)
04

Calculate Poisson's Ratio

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

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Most popular questions from this chapter

A bar of a steel alloy that exhibits the stressstrain behavior shown in Figure \(6.22\) is subjected to a tensile load; the specimen is \(375 \mathrm{~mm}(14.8 \mathrm{in}\).) long and has a square cross section \(5.5 \mathrm{~mm}(0.22 \mathrm{in}\).) on a side.(a) Compute the magnitude of the load necessary to produce an elongation of \(2.25 \mathrm{~mm}(0.088\) in.). (b) What will be the deformation after the load has been released?

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation, the circular cross section is to remain circular, and the rectangular is to remain rectangular. Their original and deformed dimensions are as follows: $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Circular } \\ (\text { diameter, } \boldsymbol{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \begin{array}{l} \text { Original } \\ \text { dimensions } \end{array} & 18.0 & 20 \times 50 \\ \hline \begin{array}{l} \text { Deformed } \\ \text { dimensions } \end{array} & 15.9 & 13.7 \times 55.1 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?

An undeformed specimen of some alloy has an average grain diameter of \(0.050 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.020 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is \(0.5 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [101], [10\overline{1} ] \text { , and } [ 0 \overline { 1 1 } ] \text { } directions.

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a \([\overline{101}]\) direction and is initiated at an applied tensile stress of \(13.9 \mathrm{MPa}\) (2020 psi), compute the critical resolved shear stress.
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