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Chapter 7: Problem 25

A cylindrical rod \(500 \mathrm{~mm}(20.0\) in.) long and having a diameter of \(12.7 \mathrm{~mm}(0.50 \mathrm{in}\).) is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than \(1.3 \mathrm{~mm}(0.05 \mathrm{in} .)\) when the applied load is \(29,000 \mathrm{~N}\left(6500 \mathrm{lb}_{i}\right)\), which of the four metals or alloys listed in the following table are possible candidates? Justify your choice(s). \begin{tabular}{lccc} \hline & Modulus of Material & \begin{tabular}{c} \mathrm{ Yield } \(\\\ {\text { Elasticity }} \\ {\text { (GPa) }}\) & Tensile (MPa) & Strength (MPa) \\ \hline Aluminum alloy & 70 & 255 & 420 \\ \hline Brass alloy & 100 & 345 & 420 \\ \hline Copper & 110 & 210 & 275 \\ \hline Steel alloy & 207 & 450 & 550 \\ \hline \end{tabular} \end{tabular}

Short Answer

Expert verified
Answer: Brass alloy and Steel alloy are the suitable materials as they satisfy both the stress and elongation conditions.

Step by step solution

01

Calculate the stress

First, we need to calculate the stress acting upon the rod when the force of 29000 N is applied. The formula for stress is given by: Stress \(\mathrm{= \dfrac{Force}{Area}}\) The cross-sectional area of the rod can be calculated using the formula for the area of a circle: Area \(\mathrm{= \pi (\dfrac{d}{2})^2}\), where \(d\) is the diameter of the rod. Using the given diameter of 12.7 mm (0.5 in), we can calculate the area: Area \(\mathrm{= \pi (\dfrac{12.7}{2})^2 = 126.7 \mathrm{mm^2}}\) Now, we can calculate the stress: Stress \(\mathrm{= \dfrac{29000 \mathrm{N}}{126.7 \mathrm{mm^2}} = 228.7 \mathrm{MPa}}\)
02

Compare stress to yield tensile strength

Now that we have calculated the stress, we can compare it to the yield tensile strength of each material. The material is suitable if the stress does not exceed the yield tensile strength. 1. Aluminum alloy: Yield tensile strength = 255 MPa (stress < 255 MPa, suitable) 2. Brass alloy: Yield tensile strength = 345 MPa (stress < 345 MPa, suitable) 3. Copper: Yield tensile strength = 210 MPa (stress > 210 MPa, not suitable) 4. Steel alloy: Yield tensile strength = 450 MPa (stress < 450 MPa, suitable)
03

Compare elongation to the given limit

Finally, we need to check whether the elongation of the rod for each suitable material does not exceed the limit of 1.3 mm (0.05 in). The formula for elongation is given by: \(\mathrm{\Delta L = \dfrac{Stress \times L}{E}}\), where \(\Delta L\) is the elongation, \(L\) is the length of the rod, and \(E\) is the modulus of elasticity. 1. Aluminum alloy: \(\mathrm{\Delta L_{Al} = \dfrac{228.7 \times 500}{70} = 1.63~mm}\) (exceeds 1.3 mm, not suitable). 2. Brass alloy: \(\mathrm{\Delta L_{Br} = \dfrac{228.7 \times 500}{100} = 1.14~mm}\) (within 1.3 mm limit, suitable). 3. Steel alloy: \(\mathrm{\Delta L_{St} = \dfrac{228.7 \times 500}{207} = 0.55~mm}\) (within 1.3 mm limit, suitable). From the above calculations, we find that Brass alloy and Steel alloy are the suitable materials for this problem, as they satisfy both the stress and elongation conditions.

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Most popular questions from this chapter

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.

(a) Define a slip system. (b) Do all metals have the same slip system? Why or why not?

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.

The lower yield point for an iron that has an († average grain diameter of \(1 \times 10^{-2} \mathrm{~mm}\) is 230 MPa (33,000 psi). At a grain diameter of \(6 \times 10^{-3}\) \(\mathrm{mm}\), the yield point increases to \(275 \mathrm{MPa}(40,000\) psi). At what grain diameter will the lower yield point be \(310 \mathrm{MPa}(45,000 \mathrm{psi}) ?\)

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?
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