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Chapter 6: Problem 20

A brass alloy is known to have a yield strength of \(275 \mathrm{MPa}(40,000 \mathrm{psi})\), a tensile strength of \(380 \mathrm{MPa}(55,000 \mathrm{psi})\), and an elastic modulus of \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\). A cylindrical specimen of this alloy \(12.7 \mathrm{~mm}(0.50\) in.) in diameter and \(250 \mathrm{~mm}\) (10.0 in.) long is stressed in tension and found to elongate \(7.6 \mathrm{~mm}(0.30 \mathrm{in}\).). 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.

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Step by step solution

01

Define the parameters

We are given the following information: - Yield strength: \(275 \mathrm{MPa}\) - Tensile strength: \(380 \mathrm{MPa}\) - Elastic modulus: \(103 \mathrm{GPa}\) - Diameter: \(12.7 \mathrm{mm}\) - Initial Length: \(250 \mathrm{mm}\) - Elongation: \(7.6 \mathrm{mm}\) Let's first convert all units into consistent form by changing all units into meters and pascals. - Diameter: \(0.0127 \mathrm{m}\) - Initial Length: \(0.25 \mathrm{m}\) - Elongation: \(0.0076 \mathrm{m}\) - Elastic Modulus: \(103 \times 10^{9} \mathrm{Pa}\) - Yield Strength: \(275 \times 10^{6} \mathrm{Pa}\) - Tensile Strength: \(380 \times 10^{6} \mathrm{Pa}\)
02

Calculate the stress and strain

Before calculating the required load, let's calculate the specimen's stress and strain. To calculate the strain, we will use the elongation and the initial length: $$ \text{strain} = \frac{\text{elongation}}{\text{initial length}} = \frac{0.0076}{0.25} = 0.0304 $$ Now, let's use Hooke's Law to calculate the stress. Hooke's Law states that the stress is proportional to the strain for a linearly elastic material: $$ \text{stress} = \text{Elastic Modulus} \times \text{strain} = (103 \times 10^{9}) \times 0.0304 = 3.13 \times 10^{9} \mathrm{Pa} $$
03

Compare the calculated stress to the yield and tensile strength

To ensure that we can compute the required load without causing plastic deformation or failure of the material, we must verify that the calculated stress is less than the yield strength and tensile strength: $$ \text{calculated stress} \leq \text{yield strength} \Rightarrow 3.13 \times 10^{9} \mathrm{Pa} \leq 275 \times 10^{6} \mathrm{Pa} $$ This is not true. Therefore, we cannot compute the required load without causing plastic deformation or failure of the material, as the calculated stress surpasses the yield strength of the alloy.

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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 table gives a number of Rockwell B hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. $$ \begin{array}{lll} 83.3 & 80.7 & 86.4 \\ 88.3 & 84.7 & 85.2 \\ 82.8 & 87.8 & 86.9 \\ 86.2 & 83.5 & 84.4 \\ 87.2 & 85.5 & 86.3 \end{array} $$

A cylindrical specimen of a titanium alloy having an elastic modulus of \(107 \mathrm{GPa}\) (15.5 \(\times\) \(10^{6} \mathrm{psi}\) ) and an original diameter of \(3.8 \mathrm{~mm}\) (0.15 in.) will experience only elastic deformation when a tensile load of \(2000 \mathrm{~N}\) \(\left(450 \mathrm{lb}_{\mathrm{f}}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.42\) \(\mathrm{mm}(0.0165\) in.).

A steel bar \(100 \mathrm{~mm}\) (4.0 in.) long and having a square cross section \(20 \mathrm{~mm}(0.8 \mathrm{in} .)\) on an edge is pulled in tension with a load of 89,000 \(\mathrm{N}\left(20,000 \mathrm{lb}_{\mathrm{f}}\right)\) and experiences an elongation of \(0.10 \mathrm{~mm}\left(4.0 \times 10^{-3}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

A cylindrical specimen of aluminum having a diameter of \(19 \mathrm{~mm}\) (0.75 in.) and length of 200 \(\mathrm{mm}(8.0 \mathrm{in}\).) is deformed elastically in tension with a force of \(48,800 \mathrm{~N}\left(11,000 \mathrm{lb}_{\mathrm{f}}\right)\). Using the data in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen, Will the diameter increase or decrease?
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