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In a standard tensile test a steel rod of 22 -mm diameter is subjected to a tension force of 75 kN. Knowing that \(\nu=0.30\) and \(E=200 \mathrm{GPa},\) determine \((a)\) the elongation of the rod in a \(200-\mathrm{mm}\) gage length, ( \(b\) ) the change in diameter of the rod.

Short Answer

Expert verified
Elongation: 0.1971 mm, Change in diameter: -0.0652 mm.

Step by step solution

01

Given Information

We have a steel rod subjected to a tensile force of 75 kN. The diameter of the rod is 22 mm, the gage length is 200 mm, and the material properties given are Poisson's ratio \(u = 0.30\) and Young's modulus \(E = 200 \text{ GPa} \).
02

Calculate Stress

Firstly, calculate the stress \( \sigma \) using the formula:\[ \sigma = \frac{P}{A} \]where \( P = 75 \text{ kN} = 75,000 \text{ N} \) and \( A = \frac{\pi d^2}{4} \). Substituting \( d = 22 \text{ mm} = 0.022 \text{ m} \):\[ A = \frac{\pi \times (0.022)^2}{4} = 3.801327 \times 10^{-4} \text{ m}^2 \]\[ \sigma = \frac{75,000}{3.801327 \times 10^{-4}} = 197,261.31 \text{ Pa} \approx 197.26 \text{ MPa} \]
03

Calculate Elongation

Using Hooke's Law, the elongation \( \Delta L \) can be calculated using:\[ \Delta L = \frac{\sigma L}{E} \]where \( L = 200 \text{ mm} = 0.2 \text{ m} \). Substituting the known values:\[ \Delta L = \frac{197,261.31 \times 0.2}{200 \times 10^9} = 1.971 \times 10^{-4} \text{ m} = 0.1971 \text{ mm} \]
04

Calculate Change in Diameter

Use Poisson's ratio to calculate the change in diameter \( \Delta d \) using:\[ \Delta d = -u \frac{\Delta L}{L} \times d \]Substituting the known values:\[ \Delta d = -0.30 \times \frac{1.971 \times 10^{-4}}{0.2} \times 22 = -0.0652 \text{ mm} \]This negative sign indicates a reduction in diameter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elongation
Elongation refers to the increase in length of a material when it is subjected to a tensile force. Essentially, it's how much a material stretches when pulled. When conducting a tensile test, elongation is a crucial measure as it tells us about the material's ductility, or how easily it can be stretched into a wire. For example, in our case with the steel rod, we calculated the elongation by looking at the change in the rod's length from its original length due to the applied force.
  • To calculate elongation (\( \Delta L \)), we use Hooke's Law, expressed as \( \Delta L = \frac{\sigma L}{E} \), where \( \sigma \) is stress, \( L \) is the original length, and \( E \) is Young's modulus.
  • In our exercise, we found that the rod elongated by \( 0.1971 \text{ mm} \).
This tells us how the steel rod behaves under tension and is crucial for determining its reliability and safety in practical applications.
Poisson's Ratio
Poisson's Ratio is the constant that describes the relationship between transversal and axial strain in materials. When a material is stretched, not only does its length increase, but its cross-sectional area tends to reduce as well. Poisson's ratio (\( u \)) captures this phenomenon and is defined as the negative ratio of transverse to axial strain.
  • It helps us understand the volume change of a material under tension.
  • For most metals, \( u \) ranges from 0.25 to 0.35.
In our scenario, the Poisson's ratio value was given as 0.30, indicating that for every unit of elongation, the rod's diameter decreases by 0.30 units proportionately. We used \( u \) to calculate the change in diameter (\( \Delta d \)) of the rod when it was subjected to the tensile force, which amounted to \( -0.0652 \text{ mm} \). This change is negative, signifying a contraction.
Young's Modulus
Young's Modulus, denoted as \( E \), is a material property that measures the stiffness of a solid material. It is a fundamental parameter that defines the relationship between stress and strain in the elastic region of the material's deformation.
  • It is given as the ratio between stress (\( \sigma \)) and strain.
  • Mathematically, \( E = \frac{\sigma}{\epsilon} \), where \( \epsilon \) is the strain experienced by the material.
High \( E \) values indicate a material is stiff and requires a larger force to deform, whereas low \( E \) values suggest the material is more easily deformable. In the context of the exercise, the steel rod has a Young’s Modulus of 200 GPa, marking it as quite stiff. This modulus was crucial for calculating the elongation of the rod as it defines how much the rod extends under a given stress, helping us understand the material's behavior under mechanical forces.

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Most popular questions from this chapter

The steel rails of a railroad track \(\left(E_{3}=200 \mathrm{GPa}, \alpha_{5}=11.7 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) were laid at a temperature of \(6^{\circ} \mathrm{C}\). Determine the normal stress in the rails when the temperature reaches \(48^{\circ} \mathrm{C}\), assuming that the rails \((a)\) are welded to form a continuous track, \((b)\) are \(10 \mathrm{m}\) long with \(3-\) mm gaps between them.

A homogenous cable of length \(L\) and uniform cross section is suspended from one end. (a) Denoting by \(\rho\) the density (mass per unit volume) of the cable and by \(E\) its modulus of elasticity, determine the elongation of the cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal and if a force equal to half of its weight were applied at each end.

A 6 -in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is \(7.1 \mathrm{ksi}\) (about 3 miles below the surface \() .\) Knowing that \(E=29 \times 10^{6}\) psi and \(\nu=0.30,\) determine (a) the decrease in diameter of the sphere, (b) the decrease in volume of the sphere, \((c)\) the percent increase in the density of the sphere.

A control rod made of yellow brass must not stretch more than \(3 \mathrm{mm}\) when the tension in the wire is \(4 \mathrm{kN}\). Knowing that \(E=105 \mathrm{GPa}\) and that the maximum allowable normal stress is \(180 \mathrm{MPa}\), determine \((a)\) the smallest diameter rod that should be used, ( \(b\) ) the corresponding maximum length of the rod.

A 250 -mm-long aluminum tube \((E=70 \mathrm{GPa})\) of 36 -mm outer diameter and 28 -mm inner diameter can be closed at both ends by means of single- threaded screw-on covers of 1.5 -mm pitch. With one cover screwed on tight, a solid brass rod \((E=105 \mathrm{GPa})\) of 25 -mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.

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