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Chapter 6: Problem 46

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.

Short Answer

Expert verified
Question: Calculate the toughness of a metal experiencing both elastic and plastic deformation, given the provided information in the problem. Answer: To calculate the toughness (total energy absorbed) of the metal, first calculate the energy absorbed during elastic deformation (W_{elastic}) and the energy absorbed during plastic deformation (W_{plastic}). Then, find the total energy absorbed by summing these two values: $W_{total} = W_{elastic} + W_{plastic}$. Finally, plug in the values from Steps 2 and 4 in order to determine the toughness for the given metal.

Step by step solution

01

Calculate stress at elastic limit

Use Hooke's Law (\(\sigma = E \epsilon\)) to calculate the stress at the elastic limit, where \(\sigma\) is stress, \(E\) is the modulus of elasticity, and \(\epsilon\) is the strain. In this case, \(\epsilon = 0.01\). $$ \sigma = E \epsilon = (172 \times 10^3\,MPa) \times 0.01 $$
02

Calculate energy for elastic deformation

The energy absorbed during elastic deformation is half the product of stress and strain. This can be represented as: $$ W_{elastic} = \frac{1}{2} \sigma \epsilon $$ Substitute the values from Step 1 and calculate the energy: $$ W_{elastic} = \frac{1}{2} \times \sigma \times 0.01 $$
03

Calculate stress for a given strain during plastic deformation

In the plastic deformation region, the relationship between stress and strain is given by the strain hardening equation (\(\sigma = K\epsilon^n\)). So, for any strain value \(\epsilon_p\), $$ \sigma_p = 6900\,MPa \times \epsilon_p^{0.30} $$
04

Calculate energy for plastic deformation

The energy absorbed during plastic deformation can be estimated by integrating stress with respect to the strain from \(\epsilon_1=0.01\) to \(\epsilon_2=0.75\). $$ W_{plastic} = \int_{0.01}^{0.75} \sigma_p d\epsilon_p $$ Substitute the expression for \(\sigma_p\) from Step 3 and integrate with respect to \(\epsilon_p\): $$ W_{plastic} = \int_{0.01}^{0.75} (6900\,MPa \times \epsilon_p^{0.30}) d\epsilon_p $$
05

Find the total energy absorbed (toughness)

To find the total energy absorbed (toughness), sum the energies from the elastic and plastic deformations: $$ W_{total} = W_{elastic} + W_{plastic} $$ Plug in the values from Steps 2 and 4, and find the total energy absorbed (toughness) for the given metal.

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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).

A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{~mm}(0.505\) in.) and gauge length of \(50.80 \mathrm{~mm}(2.000 \mathrm{in} .)\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(6.60 \mathrm{~mm}(0.260 \mathrm{in} .)\), and the fractured gauge length is \(72.14 \mathrm{~mm}\) (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

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

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 \(d F / d r\).

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(20.000\) and \(20.025 \mathrm{~mm}\), respectively, and its final length is \(74.96 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(105 \mathrm{GPa}\) and \(39.7 \mathrm{GPa}\), respectively.
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