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Chapter 6: Problem 43

For some metal alloy, a true stress of \(415 \mathrm{MPa}\) (60,175 psi) produces a plastic true strain of \(0.475\). How much will a specimen of this material elongate when a true stress of \(325 \mathrm{MPa}\) \((46,125 \mathrm{psi})\) is applied if the original length is \(300 \mathrm{~mm}\) (11.8 in.)? Assume a value of \(0.25\) for the strain-hardening exponent \(n\).

Short Answer

Expert verified
Question: Calculate the elongation of a metal alloy specimen subjected to a lower true stress level of 325 MPa, given the initial length of the specimen as 300 mm, a plastic strain of 0.475 at a higher true stress level of 415 MPa, and a strain-hardening exponent of 0.25. Answer: __________ (Include your calculation based on the steps mentioned earlier.)

Step by step solution

01

Determine the true stress-true strain relationship

In this problem, we have a metal alloy specimen subjected to true stresses, and a strain-hardening exponent is given. We assume that the material obeys the power-law relationship between true stress and true strain: $$\sigma = K\varepsilon^n$$ Where \(\sigma\) is the true stress, \(\varepsilon\) is the plastic true strain, \(K\) is the strength coefficient, and \(n\) is the strain-hardening exponent. We are given the true stress, the plastic strain, and the exponent. Therefore, we can determine the strength coefficient, \(K\), using the power-law relationship.
02

Calculate the strength coefficient, K

Given that \(\sigma = 415 \mathrm{MPa}\), \(\varepsilon = 0.475\), and \(n = 0.25\), we will substitute these values into the power-law relationship to calculate the strength coefficient, \(K\): $$K = \frac{\sigma}{\varepsilon^n}$$ $$K = \frac{415 \mathrm{MPa}}{(0.475)^{0.25}}$$ Calculate the value of \(K\).
03

Obtain the plastic true strain at the lower true stress level

Now that we have the strength coefficient \(K\), we can use the power-law relationship to obtain the plastic true strain, \(\varepsilon_2\), at the lower true stress level, \(\sigma_2 = 325 \mathrm{MPa}\): $$\varepsilon_2 = \left(\frac{\sigma_2}{K}\right)^\frac{1}{n}$$ Plug in the values for \(\sigma_2\), \(K\), and \(n\) in the equation to obtain \(\varepsilon_2\).
04

Calculate the elongation of the specimen

Once we have the plastic true strain, \(\varepsilon_2\), we can calculate the elongation of the metal alloy specimen. We have the initial length of the specimen, \(L_0 = 300 \mathrm{mm}\). Elongation, \(\Delta L\), is the product of the initial length and the plastic true strain: $$\Delta L = L_0 \times \varepsilon_2$$ Substitute the values of \(L_0\) and \(\varepsilon_2\) into the expression to calculate the elongation of the specimen when subjected to the lower true stress level.

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Most popular questions from this chapter

Figure \(6.22\) shows, for a gray cast iron, the tensile engineering stress- strain curve in the elastic region. Determine (a) the tangent modulus at \(10.3\) MPa (1500 psi) and (b) the secant modulus taken to \(6.9 \mathrm{MPa}\) (1000 psi).

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 \(0.25\) ?

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

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.

(a) A 10-mm-diameter Brinell hardness indenter produced an indentation \(1.62 \mathrm{~mm}\) in diameter in a steel alloy when a load of \(500 \mathrm{~kg}\) was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of \(450 \mathrm{HB}\) when a \(500-\mathrm{kg}\) load is used?
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