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Chapter 6: Problem 15

A cylindrical specimen of steel having a diameter of \(15.2 \mathrm{~mm}(0.60 \mathrm{in} .)\) and length of 250 \(\mathrm{mm}(10.0 \mathrm{in} .)\) is deformed elastically in tension with a force of \(48,900 \mathrm{~N}\left(11,000 \mathrm{lb}_{\mathrm{f}}\right)\). Using the data contained 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?

Short Answer

Expert verified
Answer: The elongation of the steel specimen in the axial direction is 0.325 mm, and its diameter will decrease by 0.0059 mm.

Step by step solution

01

Calculate the area of the cylinder's cross-section

To calculate the area of the cylinder's cross-section, we use the formula for the area of a circle: \(A = \pi r^2\), where \(r\) is the radius. Given that the diameter is \(15.2 \mathrm{mm}\), the radius will be half of that: \(r = \frac{15.2}{2} \mathrm{mm}\). Now we can calculate the area: \(A = \pi \left(\frac{15.2}{2}\right)^2 = \pi (7.6^2) = 181.54 \mathrm{mm^2}\)
02

Calculate the stress applied to the specimen

To calculate the stress applied to the specimen, we use the formula: \(\sigma = \frac{F}{A}\), where \(\sigma\) is the stress, \(F\) is the applied force, and \(A\) is the area of the cylinder's cross-section. Using the values given in the exercise, we find: \(\sigma = \frac{48,900 \mathrm{N}}{181.54 \mathrm{mm^2}} = 269 \mathrm{MPa}\)
03

Calculate the strain in the axial direction (elongation)

The strain in the axial direction is given by the relation: \(\epsilon_\text{axial} = \frac{\sigma}{E}\), where \(E\) is the modulus of elasticity. Table 6.1 provides the modulus of elasticity for steel as \(E = 207 \mathrm{GPa}\). We have already calculated the stress acting on the specimen, so we can now calculate the strain: \(\epsilon_\text{axial} = \frac{269 \mathrm{MPa}}{207 \mathrm{GPa}} = 1.3 \times 10^{-3}\) Now we can find the elongation by multiplying this strain by the original length of the specimen: \(\Delta L = \epsilon_\text{axial} \times L = 1.3 \times 10^{-3} \times 250 \mathrm{mm} = 0.325 \mathrm{mm}\) So the elongation of the specimen in the axial direction is 0.325 mm.
04

Calculate the strain in the radial direction (change in diameter)

To calculate the strain in the radial direction, we utilize Poisson's ratio. For steel, Table 6.1 provides Poisson's ratio as \(\nu = 0.30\). We use the relation: \(\epsilon_\text{radial} = -\nu \cdot \epsilon_\text{axial}\). With the axial strain already calculated, we can now determine the radial strain: \(\epsilon_\text{radial} = -0.30 \cdot 1.3 \times 10^{-3} = -3.9 \times 10^{-4}\) To determine the change in diameter, multiply this strain by the original diameter of the specimen: \(\Delta d = \epsilon_\text{radial} \times D = -3.9 \times 10^{-4} \times 15.2 \mathrm{mm} = -0.0059 \mathrm{mm}\) The change in diameter is -0.0059 mm.
05

Determine if the diameter will increase or decrease

Since the change in diameter is negative (-0.0059 mm), the diameter will decrease. So the specimen will elongate by 0.325 mm in the axial direction and its diameter will decrease by 0.0059 mm.

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Most popular questions from this chapter

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

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\mathrm{crs}}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\text {crss }}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\mathrm{crss}}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).

Briefly cite the differences between the recovery and recrystallization processes.

Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized.

A non-cold-worked brass specimen of average grain size \(0.01 \mathrm{~mm}\) has a yield strength of \(150 \mathrm{MPa}\) (21,750 psi). Estimate the yield strength of this alloy after it has been heated to \(500^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\), if it is known that the value of \(\sigma_{0}\) is 25 MPa (3625 psi).
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