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Chapter 6: Problem 24

A cylindrical rod \(380 \mathrm{~mm}\) (15.0 in.) long, having a diameter of \(10.0 \mathrm{~mm}(0.40\) in.), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than \(0.9 \mathrm{~mm}(0.035\) in.) when the applied load is \(24,500 \mathrm{~N}\left(5500 \mathrm{lb}_{\mathrm{f}}\right)\), which of the four metals or alloys listed in the following table are possible candidates? Justify your choice(s). $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Modulus } \\ \text { Material } \end{array} & \begin{array}{c} \text { \mathrm{ Yield } } \\ \text { of Elasticity } \\ \text { Strength } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Tensile } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { (MPa) } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 255 & 420 \\ \text { Brass alloy } & 100 & 345 & 420 \\ \text { Copper } & 110 & 250 & 290 \\ \text { Steel alloy } & 207 & 450 & 550 \\ \hline \end{array} $$

Short Answer

Expert verified
Solution: Based on the step-by-step analysis, calculate the cross-sectional area (A), allowable stress, elongation (∆L), and compare the allowable stress with the yield strength of each material. Choose the suitable materials that meet both the elongation and plastic deformation requirements.

Step by step solution

01

Calculate the cross-sectional area of the rod

First, let's find the cross-sectional area (A) of the cylindrical rod using the formula: $$ A = \pi(\frac{D}{2})^2 $$ Where D is the diameter of the rod. In our case, it's 10.0 mm.
02

Calculate the maximum allowable stress

Next, we need to determine the maximum allowable stress, which is the stress that would cause a 0.9 mm elongation in the rod. We can use the formula: $$ \text{Allowable Stress} = \frac{\text{Applied Load}}{\text{Cross-sectional Area}} $$ Using the values 24,500 N as the applied load and the cross-sectional area calculated in Step 1.
03

Calculate elongation

To meet the elongation requirement, the elongation of the rod (∆L) must not exceed 0.9 mm. With the modulus of elasticity (E) and the allowable stress, we can find the elongation from the formula: $$ \Delta L = \frac{\text{Allowable Stress} \times L}{E \times A} $$ Where L is the length of the rod (380 mm). Calculate the elongation for each material using their respective modulus of elasticity.
04

Check for plastic deformation

To ensure there is no plastic deformation, the applied stress must be less than the yield strength of each material. Compare the calculated allowable stress in Step 2 to the yield strengths provided in the table for each material.
05

Determine the suitable materials

Based on the above calculations and comparisons, choose the materials that meet both the elongation and the plastic deformation requirements. These materials will be the suitable candidates for the cylindrical rod under the given conditions. Following the steps outlined above, you will be able to determine which materials are suitable for the given scenario.

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Most popular questions from this chapter

For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking: $$ \begin{array}{cc} \hline \begin{array}{c} \text { Engineering Stress } \\ \text { (MPa) } \end{array} & \text { Engineering Strain } \\ \hline 235 & 0.194 \\ 250 & 0.296 \\ \hline \end{array} $$ On the basis of this information, compute the engineering stress necessary to produce an \(e n\) gineering strain of \(0.25\).

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(8.0 \mathrm{~mm}(0.31 \mathrm{in} .)\). A tensile force of \(1000 \mathrm{~N}\) \(\left(225 \mathrm{lb}_{\mathrm{f}}\right.\) ) produces an elastic reduction in diameter of \(2.8 \times 10^{-4} \mathrm{~mm}\left(1.10 \times 10^{-5} \mathrm{in}\right.\).). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is \(0.30\).

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 a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(20.000\) and \(20.025 \mathrm{~mm}\), respectively, and its final length is \(74.96 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(105 \mathrm{GPa}\) and \(39.7 \mathrm{GPa}\), respectively.

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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