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Chapter 6: Problem 8

A cylindrical rod of copper \((E=110 \mathrm{GPa}\), \(16 \times 10^{6}\) psi) having a yield strength of 240 MPa ( \(35,000 \mathrm{psi})\) is to be subjected to a load of \(6660 \mathrm{~N}\left(1500 \mathrm{lb}_{\mathrm{f}}\right)\). If the length of the rod is \(380 \mathrm{~mm}\) (15.0 in.), what must be the diameter to allow an elongation of \(0.50 \mathrm{~mm}\) \((0.020 \mathrm{in} .)\) ?

Short Answer

Expert verified
Question: Determine the required diameter of a copper cylindrical rod to allow an elongation of 0.50 mm under an axial load, given the following properties: elastic modulus (E) = 110 GPa, applied load (P) = 6660 N, and length (L) = 380 mm. Answer: The diameter of the copper cylindrical rod must be approximately 7.67 mm to allow an elongation of 0.50 mm under the given axial load and properties.

Step by step solution

01

Write down the elongation formula

The formula for the elongation of a rod under an axial load is given by: $$ \delta = \frac{PL}{AE} $$ Where, - \(\delta\) is the elongation of the rod - \(P\) is the applied load - \(L\) is the length of the rod - \(A\) is the cross-sectional area of the rod - \(E\) is the elastic modulus of the material
02

Determine the area of the rod

To find the diameter of the cylindrical rod, we need to determine the cross-sectional area of the rod first. We can rewrite the elongation formula to solve for the area: $$ A = \frac{PL}{E \delta} $$
03

Substitute the given values

Plug in the given values into the formula: $$ A = \frac{6660 \, N \times 380 \, mm}{110 \times 10^9 \, Pa \times 0.50 \, mm} $$
04

Calculate the area

Perform the calculation to find the cross-sectional area: $$ A \approx 4.619 \times 10^{-5} \, m^2 $$
05

Find the diameter

The cross-sectional area of a cylindrical rod is given by the formula: $$ A = \frac{\pi d^2}{4} $$ Where \(d\) is the diameter of the rod. Solving for the diameter: $$ d = \sqrt{\frac{4A}{\pi}} $$ Substitute the calculated area value: $$ d = \sqrt{\frac{4 \times 4.619 \times 10^{-5} \, m^2}{\pi}} $$
06

Calculate the diameter

Perform the calculation to find the diameter: $$ d \approx 7.67 \times 10^{-3} \, m \approx 7.67 \, mm $$ The diameter of the copper cylindrical rod must be approximately \(7.67 \, mm\) to allow an elongation of \(0.50 \, mm\).

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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: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ 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. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-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 \(d F / d r\).

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As noted in Section 3.15, for single crystals of some substances, the physical properties are anisotropic; that is, they are dependent 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, \(E_{u v w}\), is described by the relationship $$ \begin{gathered} \frac{1}{E_{u v w}}=\frac{1}{E_{(100}}-3\left(\frac{1}{E_{(100}}-\frac{1}{E_{(111)}}\right) \\\ \left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}\right) \end{gathered} $$ where \(E_{100}\) and \(E_{(111)}\) are the moduli of elasticity in \([100]\) 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 \(E_{\langle 110\rangle}\) values for aluminum, copper, and iron in Table \(3.3\) are correct.

A specimen of aluminum having a rectangular cross section \(10 \mathrm{~mm} \times 12.7 \mathrm{~mm}(0.4\) in. \(\times 0.5\) in.) is pulled in tension with \(35,500 \mathrm{~N}\) \(\left(8000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

A cylindrical specimen of aluminum having a diameter of \(19 \mathrm{~mm}\) (0.75 in.) and length of 200 \(\mathrm{mm}(8.0 \mathrm{in}\).) is deformed elastically in tension with a force of \(48,800 \mathrm{~N}\left(11,000 \mathrm{lb}_{\mathrm{f}}\right)\). Using the data in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen, Will the diameter increase or decrease?
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