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Chapter 6: Problem 25

A cylindrical specimen of a brass alloy having a length of \(100 \mathrm{mm}\) (4 in.) must elongate only \(5 \mathrm{mm}(0.2\) in.) when a tensile load of \(100,000 \mathrm{N}\left(22,500 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress-strain behavior shown in Figure 6.12.

Short Answer

Expert verified
Answer: The process involves three main steps: 1. Determine the stress using Hooke's law, considering the elongation and Young's modulus of the material. 2. Calculate the cross-sectional area by relating the stress and applied tensile load. 3. Determine the radius of the specimen using the cross-sectional area and the formula for the area of a circle.

Step by step solution

01

Determine the stress

The stress on the specimen can be calculated using Hooke's law, which is given by \(\sigma = E\epsilon\), where \(\sigma\) is the stress, \(E\) is the modulus of elasticity (Young's modulus), and \(\epsilon\) is the strain. Since we know the elongation is 5mm, the strain can be calculated as the ratio of elongation to the original length, \(\epsilon = \frac{5 \mathrm{mm}}{100 \mathrm{mm}} = 0.05\). Assuming a given value of Young's modulus, we can calculate the stress using this formula: \(\sigma = E * 0.05\).
02

Calculate the cross-sectional area

We know the tensile load applied on the specimen is \(100,000 \mathrm{N}\). The equation for stress is given by \(\sigma = \frac{F}{A}\), where \(F\) is the force and \(A\) is the cross-sectional area. From step 1, we already have the stress value, and we can rearrange the equation to solve for the cross-sectional area: \(A = \frac{F}{\sigma}\). Substituting the values, we can find the cross-sectional area of the specimen.
03

Determine the radius of the specimen

The cross-sectional area of a cylinder is given by \(A = \pi r^2\), where \(r\) is the radius. We already have the value of A from step 2. Rearranging the equation to solve for the radius, we get \(r = \sqrt{\frac{A}{\pi}}\). By plugging in the calculated area, we can determine the required radius to meet the specified elongation under the tensile load.

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Most popular questions from this chapter

A steel alloy to be used for a spring application must have a modulus of resilience of at least \(2.07 \mathrm{MPa}(300 \mathrm{psi}) .\) What must be its minimum yield strength?

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 and \(30.04 \mathrm{mm}\) respectively, and its final length is \(105.20 \mathrm{mm}\) compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 65.5 and $25.4 \mathrm{GPa}, respectively.

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.

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 \(103 \mathrm{GPa}\left(15 \times 10^{6} \mathrm{psi}\right),\) and that elastic deformation terminates at a strain of \(0.007 .\) 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 \(1520 \mathrm{MPa}(221,000 \mathrm{psi})\) and \(0.15,\) respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and \(0.60,\) at which point fracture occurs.

A specimen of copper having a rectangular cross section \(15.2 \mathrm{mm} \times 19.1 \mathrm{mm}(0.60 \mathrm{in.} \times\) \(0.75 \text { in. })\) is pulled in tension with \(44,500 \mathrm{N}\) \(\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.
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