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Chapter 6: Problem 35

Demonstrate that Equation 6.16 , the expression defining true strain, may also be represented by $$\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)$$ when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why?

Short Answer

Expert verified
Answer: The expression for true strain can indeed be represented by $$\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)$$ assuming the volume remains constant during deformation. The expression $$\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)$$ is more valid during necking, as it considers the change in cross-sectional area during deformation, while the general expression for true strain based on length does not account for the rapid reduction in cross-sectional area during necking.

Step by step solution

01

Understand the Expression for True Strain

True strain is the natural logarithm of the instantaneous length over the original length. The general expression for true strain is given by: $$\epsilon_{True}=\ln\left(\frac{L}{L_{0}}\right)$$ where \(L\) is the length after deformation and \(L_{0}\) is the original length.
02

Calculate the Volume Relation

Since the volume remains constant during deformation, we can relate the original area \(A_0\) and original length \(L_0\) to the final area \(A_i\) and final length \(L\). The volume \(V\) before deformation and after deformation can be written as: $$V_{0} = A_{0} L_{0}$$ $$V_i = A_{i} L_i$$ Given that the volume remains constant, we have: $$A_{0} L_{0} = A_{i}L_{i}$$
03

Relate the True Strain Expression to Cross-sectional Area

By substituting the expression relating area and length from Step 2, we can rewrite the true strain expression as follows: $$A_{0} L_{0} = A_{i}L_{i} \Rightarrow \frac{A_{0}}{A_{i}}=\frac{L_{i}}{L_{0}}$$ Then we can substitute this into the true strain formula: $$\epsilon_{T}=\ln\left(\frac{L_{i}}{L_{0}}\right)=\ln\left(\frac{A_{0}}{A_{i}}\right)$$ So we have demonstrated that the expression for true strain can indeed be represented by: $$\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)$$ assuming the volume remains constant during deformation.
04

Determine Validity during Necking

During necking, the material becomes inhomogeneous and an abrupt reduction in the area occurs. This localized deformation results in an increase in engineering strain and stress. The expression: $$\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)$$ is more valid during necking as it considers the change in cross-sectional area during deformation. Meanwhile, the general expression for true strain based on length does not account for the rapid reduction in cross-sectional area during necking. In conclusion, the expression \(\epsilon_{T}=\ln \left(\frac{A_{0}}{A_{i}}\right)\) is more valid during necking because it accounts for the changes in cross-sectional area.

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Most popular questions from this chapter

Taking the logarithm of both sides of Equation 6.19 yields $$\log \sigma_{T}=\log K+n \log \epsilon_{T}\quad\quad\quad\quad(6.27)$$ Thus, a plot of \(\log \sigma_{T}\) versus \(\log \epsilon_{T}\) in the plastic region to the point of necking should yield a straight line having a slope of \(n\) and an intercept (at \(\log \sigma_{T}=0\) ) of \(\log K\). Using the appropriate data tabulated in Problem 6.28 , make a plot of \(\log \sigma_{T}\) versus log \(\epsilon_{T}\) and determine the values of \(n\) and \(K\). It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 6.18 and 6.18 b.

A cylindrical rod of steel \((E=207 \mathrm{GPa}, 30 \times\) \(\left.10^{6} \text { psi }\right)\) having a yield strength of \(310 \mathrm{MPa}\) \((45,000 \mathrm{psi})\) is to be subjected to a load of \(11,100 \mathrm{N}\left(2500 \mathrm{lb}_{\mathrm{f}}\right) .\) If the length of the rod is \(500 \mathrm{mm}(20.0 \text { in. }),\) what must be the diameter to allow an elongation of \(0.38 \mathrm{mm}(0.015 \text { in. }) ?\)

A brass alloy is known to have a yield strength of \(240 \mathrm{MPa}(35,000 \mathrm{psi}),\) a tensile strength of \(310 \mathrm{MPa}(45,000 \mathrm{psi}),\) and an elastic modulus of 110 GPa \(\left(16.0 \times 10^{6} \mathrm{psi}\right) .\) A cylindrical specimen of this alloy \(15.2 \mathrm{mm}(0.60 \mathrm{in.})\) in diameter and \(380 \mathrm{mm}(15.0 \mathrm{in.})\) long is stressed in tension and found to elongate \(1.9 \mathrm{mm}\) \((0.075 \text { in. }) .\) On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.

A cylindrical metal specimen having an original diameter of \(12.8 \mathrm{mm}(0.505 \text { in. })\) and gauge length of \(50.80 \mathrm{mm}(2.000 \text { in. })\) is pulled in tension until fracture occurs. The diameter at the point of fracture is \(8.13 \mathrm{mm}(0.320 \mathrm{in.},\) and the fractured gauge length is \(74.17 \mathrm{mm}\) \((2.920 \text { in. }) .\) Calculate the ductility in terms of percent reduction in area and percent elongation.

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0 \mathrm{mm}(0.39 \text { in. }) .\) A tensile force of \(1500 \mathrm{N}\) \(\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{mm}\left(2.64 \times 10^{-5} \mathrm{in.}\right)\). Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35.
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