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Chapter 6: Problem 3

A specimen of copper having a rectangular cross section \(15.2 \mathrm{mm} \times 19.1 \mathrm{mm}(0.60 \mathrm{in.} \times\) \(0.75 \text { in. })\) is pulled in tension with \(44,500 \mathrm{N}\) \(\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

Short Answer

Expert verified
Answer: The strain produced in the copper specimen is approximately \(1.39404 \times 10^{-3}\).

Step by step solution

01

Calculate the cross-sectional area of the copper specimen

To calculate the area, multiply the width and height of the rectangular cross-section, considering the given dimensions are in millimeters (mm). \(A = width \times height\) Convert dimensions to meters for consistency in units: 1mm = 0.001m \(A = (15.2 \times 10^{-3}m)(19.1 \times10^{-3}m)= 2.9032 \times 10^{-4} m^2\)
02

Calculate the stress acting on the specimen

Stress can be calculated by dividing the applied force by the cross-sectional area. \(\sigma = \frac{Force}{Area}\) \(\sigma = \frac{44500 N}{2.9032\times 10^{-4}m^2} = 1.53344 \times 10^8 N/m^2\)
03

Use Hooke's law to calculate strain

Hooke's law states that stress is proportional to strain: \(\sigma = E \times \epsilon\) where \(\sigma\) is the stress, \(\epsilon\) is the strain, and \(E\) is Young's modulus. The value of Young's modulus for copper is approximately \(110 \times 10^9 N/m^2\). Rearrange the equation to solve for strain and substitute the given values: \(\epsilon = \frac{\sigma}{E}\) \(\epsilon = \frac{1.53344 \times 10^8 N/m^2}{110 \times 10^9 N/m^2}\) \(\epsilon = 1.39404 \times 10^{-3}\) The resulting strain in the copper specimen is \(1.39404 \times 10^{-3}\).

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Most popular questions from this chapter

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: $$E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}\quad\quad\quad (6.25)$$ where \(A, B,\) and \(n\) are constants for the particular ion pair. Equation 6.25 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, $$E \propto\left(\frac{d F}{d r}\right)_{r_{0}}$$ 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. since \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N^{-} \text {versus- } r}\) curve (Figure \(2.8 b\) ), 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 \(d F / d r\).

For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\) and the modulus of elasticity is 103 GPa \(\left(15.0 \times 10^{6} \mathrm{psi}\right)\). (a) What is the maximum load that may be applied to a specimen with a cross- sectional area of \(130 \mathrm{mm}^{2}\left(0.2 \text { in. }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{mm}\) \((3.0 \text { in. }),\) what is the maximum length to which it may be stretched without causing plastic deformation?

Consider a cylindrical specimen of a steel alloy (Figure \(6.21) 8.5 \mathrm{mm}(0.33 \text { in. })\) in diameter and \(80 \mathrm{mm}(3.15 \text { in. })\) long that is pulled in tension. Determine its elongation when a load of \(65,250 \mathrm{N}\left(14,500 \mathrm{lb}_{\mathrm{f}}\right)\) is applied.

Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equation 6.5 for elastic deformation, that the modulus of elasticity is \(103 \mathrm{GPa}\left(15 \times 10^{6} \mathrm{psi}\right),\) and that elastic deformation terminates at a strain of \(0.007 .\) For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19 , in which the values for \(K\) and \(n\) are \(1520 \mathrm{MPa}(221,000 \mathrm{psi})\) and \(0.15,\) respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and \(0.60,\) at which point fracture occurs.

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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