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Chapter 6: Problem 44

The following true stresses produce the corresponding true plastic strains for a brass alloy: $$ \begin{array}{cc} \hline \begin{array}{c} \text { True Stress } \\ \text { (psi) } \end{array} & \text { True Strain } \\ \hline 50,000 & 0.10 \\ 60,000 & 0.20 \\ \hline \end{array} $$ What true stress is necessary to produce a true plastic strain of \(0.25\) ?

Short Answer

Expert verified
Answer: The true stress required to produce a true plastic strain of 0.25 for the given brass alloy is 65,000 psi.

Step by step solution

01

Identify the linear relationship

Using the given data points, we can assume that there is a linear relationship between true stress (\(\sigma\)) and true plastic strain (\(\epsilon_p\)) as follows: $$ \sigma = m \cdot \epsilon_p + c $$
02

Find the slope (m) and the intercept (c)

We will now use the given data points to find the slope (m) and the intercept (c) of the linear relationship. Let's consider the given data points as (\(\epsilon_{p1}\), \(\sigma_{1}\)) and (\(\epsilon_{p2}\), \(\sigma_{2}\)). $$ m = \frac{\sigma_{2} - \sigma_{1}}{\epsilon_{p2} - \epsilon_{p1}} $$ Plug in the values: $$ m = \frac{60,000 - 50,000}{0.20 - 0.10} = \frac{10,000}{0.10} = 100,000 $$ Now, we will use one of the data points to find the value of the intercept (c). $$ c = \sigma_{1} - m \cdot \epsilon_{p1} $$ Plug in the values: $$ c = 50,000 - 100,000 \cdot 0.10 = 50,000 - 10,000 = 40,000 $$ So, the linear relationship can be written as: $$ \sigma = 100,000 \cdot \epsilon_p + 40,000 $$
03

Calculate the true stress for the desired true plastic strain

We will now use the above linear relationship to find the true stress required to produce a true plastic strain of \(0.25\). $$ \sigma = 100,000 \cdot 0.25 + 40,000 $$ Calculate the true stress: $$ \sigma = 25,000 + 40,000 = 65,000 $$ Thus, the true stress necessary to produce a true plastic strain of \(0.25\) for the brass alloy is \(65,000\) psi.

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Most popular questions from this chapter

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. $$ \begin{array}{cccccc} \hline \text {Load} & & \text {Length} & & \text {Diameter} \\ \hline \boldsymbol{N} & \boldsymbol{l b}_{f} & \boldsymbol{m m} & \text { in. } & {\boldsymbol{m m}} & \text { in. } \\ \hline 46,100 & 10,400 & 56.896 & 2.240 & 11.71 & 0.461 \\ 44,800 & 10,100 & 57.658 & 2.270 & 11.26 & 0.443 \\ 42,600 & 9,600 & 58.420 & 2.300 & 10.62 & 0.418 \\ 36,400 & 8,200 & 59.182 & 2.330 & 9.40 & 0.370 \\ \hline \end{array} $$

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). $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Modulus } \\ \text { Material } \end{array} & \begin{array}{c} \text { \mathrm{ Yield } } \\ \text { of Elasticity } \\ \text { Strength } \\ \text { (GPa) } \end{array} & \begin{array}{c} \text { Tensile } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { (MPa) } \end{array} \\ \hline \text { Aluminum alloy } & 70 & 255 & 420 \\ \text { Brass alloy } & 100 & 345 & 420 \\ \text { Copper } & 110 & 250 & 290 \\ \text { Steel alloy } & 207 & 450 & 550 \\ \hline \end{array} $$

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.

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 \(0.25\).

Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stressstrain behavior is shown in Figure \(6.12\). (b The steel alloy for which the stress-strain behavior is shown in Figure \(6.21\).
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