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Chapter 6: Problem 34

For the (plain) carbon steel alloy, whose stressstrain behavior may be observed in the "Tensile Tests" module of Virtual Materials Science and Engineering (VMSE), determine the following: (a) the approximate yield strength, (b) the tensile strength, and (c) the approximate ductility, in percent elongation.

Short Answer

Expert verified
Answer: The steps to find the yield strength, tensile strength, and approximate ductility in percent elongation are as follows: 1. Study the stress-strain curve of the plain carbon steel alloy and observe its distinct regions and characteristics. 2. Identify the yield strength by locating the stress value on the curve when it transitions from the linear elastic region to the plastic region. 3. Identify the tensile strength by locating the maximum stress value on the curve, which is the peak of the curve. 4. Determine the approximate ductility in percent elongation by observing the strain value on the curve when the material breaks and calculating the percentage of elongation using the formula: Percent Elongation = (Final Length - Original Length) / Original Length * 100%.

Step by step solution

01

Study the stress-strain curve of the plain carbon steel alloy

First, access the "Tensile Tests" module of Virtual Materials Science and Engineering (VMSE) and analyze the stress-strain curve of the plain carbon steel alloy. The stress-strain curve is a graphical representation of the material's behavior under tensile loading. Observe the distinct regions in the curve and their characteristics.
02

Identify the yield strength

The yield strength is the stress at which a material begins to deform plastically (non-recoverable deformation). On the stress-strain curve, this point is typically represented by the intersection of the elastic (linear) region and the beginning of the plastic (non-linear) deformation region. To determine the approximate yield strength, examine the stress-strain curve to locate the stress value (on the y-axis) at which the curve transitions from the linear elastic region to the plastic region.
03

Identify the tensile strength

The tensile strength is the maximum stress a material can withstand before failure (fracture) under tensile loading. On the stress-strain curve, this point is the highest stress value reached, usually represented by the peak of the curve. To determine the tensile strength, examine the stress-strain curve and identify the maximum stress value (on the y-axis).
04

Identify the approximate ductility in percent elongation

Ductility is a material's ability to deform plastically before fracturing. In tensile testing, it is typically represented by the percent elongation, which is the percentage increase in the sample's length (strain) before it fractures. To determine the approximate ductility in percent elongation, examine the stress-strain curve and observe the strain value (on the x-axis) at which the material breaks. Calculate the percentage of elongation as follows: Percent Elongation = (Final Length - Original Length) / Original Length * 100% Keep in mind that the accuracy of this value may somewhat depend on the resolution of the provided stress-strain curve.

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Most popular questions from this chapter

A steel bar \(100 \mathrm{~mm}\) (4.0 in.) long and having a square cross section \(20 \mathrm{~mm}(0.8 \mathrm{in} .)\) on an edge is pulled in tension with a load of 89,000 \(\mathrm{N}\left(20,000 \mathrm{lb}_{\mathrm{f}}\right)\) and experiences an elongation of \(0.10 \mathrm{~mm}\left(4.0 \times 10^{-3}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Using the data in Problem \(6.29\) and Equations 6.15, 6.16, and 6.18a, generate a true stress-true strain plot for aluminum. Equation \(6.18\) a becomes invalid past the point at which necking begins; therefore, measured diameters are given in the following table for the last four data points, which should be used in true stress computations. $$ \begin{array}{cccccc} \hline \text {Load} & & \text {Length} & & \text {Diameter} \\ \hline \boldsymbol{N} & \boldsymbol{l b}_{f} & \boldsymbol{m m} & \text { in. } & {\boldsymbol{m m}} & \text { in. } \\ \hline 46,100 & 10,400 & 56.896 & 2.240 & 11.71 & 0.461 \\ 44,800 & 10,100 & 57.658 & 2.270 & 11.26 & 0.443 \\ 42,600 & 9,600 & 58.420 & 2.300 & 10.62 & 0.418 \\ 36,400 & 8,200 & 59.182 & 2.330 & 9.40 & 0.370 \\ \hline \end{array} $$

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 172 GPa \(\left(25 \times 10^{6} \mathrm{psi}\right)\), and that elastic deformation terminates at a strain of 0.01. 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 \(6900 \mathrm{MPa}\left(1 \times 10^{6} \mathrm{psi}\right)\) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of \(0.01\) and \(0.75\), at which point fracture occurs.

A cylindrical rod \(100 \mathrm{~mm}\) long and having a diameter of \(10.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(27,500 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(7.5 \times 10^{-3} \mathrm{~mm}\). Of the materials listed as follows, which are possible candidates? Justify your choice(s). $$ \begin{array}{lccc} \hline \begin{array}{c} \text { Material } \end{array} & \begin{array}{c} \text { Modulus } \\ \text { of Elasticity } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Poisson's } \\ \text { Ratio } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 200 & 0.33 \\ \text { Brass alloy } & 101 & 300 & 0.34 \\ \text { Steel alloy } & 207 & 400 & 0.30 \\ \text { Titanium alloy } & 107 & 650 & 0.34 \\ \hline \end{array} $$

A cylindrical specimen of an alloy \(8 \mathrm{~mm}\) (0.31 in.) in diameter is stressed elastically in tension. A force of \(15,700 \mathrm{~N}\) (3530 lb_{f } \()\) produces a reduction in specimen diameter of \(5 \times 10^{-3} \mathrm{~mm}\left(2 \times 10^{-4}\right.\) in.). Compute Poisson's ratio for this material if its modulus of elasticity is \(140 \mathrm{GPa}\left(20.3 \times 10^{6} \mathrm{psi}\right)\).
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