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Chapter 6: Problem 30

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{tabular}{lcc} \hline & Circular \((\) diameter, \(\boldsymbol{m m})\) & Rectangular (mm) \\ \hline Original dimensions & \(18.0\) & \(20 \times 50\) \\ \hline Deformed dimensions & \multicolumn{1}{c}{\(15.9\)} & \(13.7 \times 55.1\) \\\ \hline \end{tabular} Which of these specimens will be the hardest after plastic deformation, and why?

Short Answer

Expert verified
Answer: The circular specimen with a percentage reduction in cross-sectional area of 34.38% will be harder after plastic deformation compared to the rectangular specimen with a percentage reduction of 24.56%. This is because a greater deformation leads to increased dislocation density and work hardening, making the material harder.

Step by step solution

01

Calculate the original cross-sectional areas of the specimens

For the circular specimen, we will use the formula for the area of a circle (A = πr²), and for the rectangular specimen, we will use the formula for the area of a rectangle (A = length × width). For the circular specimen: Original diameter = 18.0 mm Radius (r) = Diameter / 2 = 18.0 / 2 = 9.0 mm Original area (A1_circular) = πr² = π(9.0)² = 81π mm² For the rectangular specimen: Original dimensions: 20 × 50 mm Original area (A1_rectangular) = length × width = 20 × 50 = 1000 mm²
02

Calculate the deformed cross-sectional areas of the specimens

We will use the same formulas as in step 1 but with the deformed dimensions. For the circular specimen: Deformed diameter = 15.9 mm Radius (r) = Deformed Diameter / 2 = 15.9 / 2 = 7.95 mm Deformed area (A2_circular) = πr² = π(7.95)² ≈ 199.64 mm² For the rectangular specimen: Deformed dimensions: 13.7 × 55.1 mm Deformed area (A2_rectangular) = length × width = 13.7 × 55.1 ≈ 754.37 mm²
03

Calculate the percentage reduction in cross-sectional area for each specimen

To find the percentage reduction in cross-sectional area, we will use the following formula: Percentage reduction = ((Original Area - Deformed Area) / Original Area) × 100 For the circular specimen: Percentage reduction (circular) = ((81π - 199.64) / 81π) × 100 ≈ 34.38% For the rectangular specimen: Percentage reduction (rectangular) = ((1000 - 754.37) / 1000) × 100 ≈ 24.56%
04

Determine which specimen is harder and why

The circular specimen has a higher percentage reduction in cross-sectional area compared to the rectangular specimen (34.38% vs. 24.56%). Generally, when a metal undergoes plastic deformation, the greater the deformation, the harder the material becomes due to the increased dislocation density and the process of work hardening. Therefore, it can be concluded that the circular specimen will be harder after plastic deformation because it experienced a greater percentage reduction in cross-sectional area.

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Most popular questions from this chapter

List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

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 \(d F / d r\).

An undeformed specimen of some alloy has an average grain diameter of \(0.050 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.020 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

(a) Equations \(6.4 a\) and \(6.4 \mathrm{~b}\) are expressions for normal \(\left(\sigma^{\prime}\right)\) and shear \(\left(\tau^{\prime}\right)\) stresses, respectively, as a function of the applied tensile stress \((\sigma)\) and the inclination angle of the plane on which these stresses are taken ( \(\theta\) of Figure 6.4). Make a plot showing the orientation parameters of these expressions (i.e., \(\cos ^{2} \theta\) and \(\sin \theta \cos \theta\) ) versus \(\theta\). (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) At what inclination angle is the shear stress a maximum?

A specimen of copper having a rectangular cross section \(15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}(0.60 \mathrm{in} . \times 0.75 \mathrm{in}\).) 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.
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