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Chapter 7: Problem 19

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0\) \(\mathrm{mm}(0.39 \mathrm{in} .) .\) A tensile force of \(1500 \mathrm{~N}\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{~mm}\left(2.64 \times 10^{-5}\right.\) in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is \(0.35\).

Short Answer

Expert verified
Answer: The elastic modulus of the metal alloy is approximately 294261 N/mm².

Step by step solution

01

Calculate the cross-sectional area of the cylindrical specimen

We will first determine the cross-sectional area (A) of the cylindrical specimen using the diameter (d) given, which is 10.0 mm. The formula for the area of a circle is: \(A = \frac{\pi d^2}{4}\) Plug in the diameter value: \(A = \frac{\pi (10.0~mm)^2}{4} = 78.54~mm^2\)
02

Calculate the stress applied to the material

We are given the tensile force applied to the specimen, which is 1500 N. With the cross-sectional area calculated in step 1, we can now calculate the stress (\(\Delta \sigma\)) using the formula: \(\Delta \sigma = \frac{F}{A}\) Plug in the values for force (F) and area (A): \(\Delta \sigma = \frac{1500~N}{78.54~mm^2} = 19.10 \mathrm{~N/mm^2}\)
03

Calculate the strain experienced by the material

The strain (\(\Delta \epsilon\)) can be calculated using the formula: \(\Delta \epsilon = \frac{\Delta d}{d}\) Given the elastic reduction in diameter (\(\Delta d\)) is \(6.7 * 10^{-4}~mm\) and the original diameter (\(d\)) is \(10.0~mm\): \(\Delta \epsilon = \frac{6.7 \times 10^{-4} \mathrm{~mm}}{10.0 \mathrm{~mm}} = 6.7 \times 10^{-5}\)
04

Calculate the elastic modulus using Poisson's ratio

As we have the Poisson's ratio (\(\nu\)) which is 0.35, we can use the formula for finding the Elastic modulus using Poisson's ratio, stress (\(\Delta \sigma\)) and strain (\(\Delta \epsilon\)): \(E = \frac{\Delta \sigma}{(1-\nu)\Delta \epsilon}\) Plug in the values for Poisson's ratio (\(\nu\)), stress (\(\Delta \sigma\)) and strain (\(\Delta \epsilon\)): \(E = \frac{19.10 \mathrm{~N/mm^2}}{(1-0.35) \times (6.7 \times 10^{-5})} = 294261 \mathrm{~N/mm^2}\) The elastic modulus of the metal alloy is approximately 294261 \(\mathrm{N/mm^2}\).

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Most popular questions from this chapter

The following yield strength, grain diameter, and heat treatment time (for grain growth) data were gathered for an iron specimen that was heat treated at \(800^{\circ} \mathrm{C}\). Using these data, compute the yield strength of a specimen that was heated at \(800^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\). Assume a value of 2 for \(n\), the grain diameter exponent. $$ \begin{array}{lcc} \hline \begin{array}{l} \text { Grain } \\ \text { diameter } \\ (\mathbf{m m}) \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Heat } \\ \text { Treating } \\ \text { Time (h) } \end{array} \\ \hline 0.028 & 300 & 10 \\ \hline 0.010 & 385 & 1 \\ \hline \end{array} $$

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.

(a) Compare planar densities (Section \(3.11\) and Problem 3.60) for the \((100),(110)\), and (111) planes for FCC. (b) Compare planar densities (Problem 3.61) for the (100), (110), and (111) planes for BCC.

As noted in Section \(3.15\), for single crystals of some substances, the physical properties are anisotropic that is, they depend on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general \([u v w]\) direction, \(E_{\text {uww }}\) is described by the relationship $$ \begin{aligned} \frac{1}{E_{u v w}}=& \frac{1}{E_{(100)}}-3\left(\frac{1}{E_{(100)}}-\frac{1}{E_{\\{111)}}\right) \\\ &\left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}\right) \end{aligned} $$ where \(E_{\\{100)}\) and \(E_{\langle 111\rangle}\) are the moduli of elasticity in the \([100]\) and \([111]\) directions, respectively; \(\alpha, \beta\), and \(\gamma\) are the cosines of the angles between \([u v w]\) and the respective [100], [010], and [001] directions. Verify that the \(E_{\\{110\rangle}\) values for aluminum, copper, and iron in Table \(3.4\) are correct.

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\text {crss }}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\mathrm{crss}}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\text {crss }}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).
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