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Chapter 2: Problem 83

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.

Short Answer

Expert verified
(a) The decrease in diameter is 0.0014 in., (b) the decrease in volume is 0.0819 in鲁, (c) the density increases by 0.0723%.

Step by step solution

01

Calculate Deformation in Diameter Due to Pressure

The formula for volumetric strain in a sphere under pressure is \( \epsilon = \frac{-p}{E} \left(1 - 2u\right) \) where \( p \) is the pressure, \( E \) is Young's modulus, and \( u \) is Poisson's ratio. Plug in \( p = 7.1 \text{ ksi} = 7100 \text{ psi} \), \( E = 29 \times 10^6 \text{ psi} \), and \( u = 0.30 \) to get \( \epsilon = \frac{-7100}{29 \times 10^6} (1 - 2 \times 0.3) = -0.0002414 \). Decrease in diameter \( \Delta D = \epsilon \times D = -0.0002414 \times 6 \text{ in.} = -0.0014484 \text{ in.} \).
02

Calculate the Decrease in Volume of the Sphere

The initial volume \( V_0 \) of the sphere is \( V_0 = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 \), where \( D = 6 \text{ in.} \). So, \( V_0 = \frac{4}{3}\pi \left(3\right)^3 = 113.097 \text{ in}^3 \). The volumetric strain due to pressure \( \epsilon_v = 3\epsilon = 3(-0.0002414) = -0.0007242 \). The decrease in volume \( \Delta V = V_0 \times \epsilon_v = 113.097 \times (-0.0007242) = -0.0819 \text{ in}^3 \).
03

Calculate the Percent Increase in Density of the Sphere

Density \( \rho \) of a sphere is defined as \( \rho = \frac{Mass}{Volume} \). As volume decreases, density increases. Initial density \( \rho_0 = \frac{M}{V_0} \) and new density \( \rho_1 = \frac{M}{V_0 - \Delta V} \). Percent increase in density is \( \frac{\rho_1 - \rho_0}{\rho_0} \times 100 = \frac{\left( \frac{M}{V_0 - \Delta V} - \frac{M}{V_0} \right)}{\frac{M}{V_0}} \times 100 = \frac{V_0}{V_0 - \Delta V} - 1 \times 100 \). With \( V_0 = 113.097 \) and \( \Delta V = 0.0819 \), percent increase in density \( = \left( \frac{113.097}{113.097 - 0.0819} - 1 \right) \times 100 = 0.0723\% \).

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Key Concepts

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

Volumetric Strain
When a material is subjected to pressure, it experiences a deformation that changes its volume. This change in volume relative to the original volume is known as volumetric strain. For a sphere under uniform pressure, the formula for volumetric strain \( \epsilon_v \) can be obtained by multiplying the linear strain by three, considering changes in all three dimensions. It is given by
  • \( \epsilon_v = \frac{-p}{E} \left(1 - 2u\right) \)
Where:
  • \( p \) is the applied pressure
  • \( E \) is Young's modulus
  • \( u \) is Poisson鈥檚 ratio

In the given exercise, the pressure is 7100 psi and the values of Young's modulus and Poisson's ratio are provided. Hence, the volumetric strain can be calculated and used to find the decrease in the diameter and volume of the sphere by applying this strain measure.
Young's Modulus
Young's modulus \( E \) is a fundamental property of materials that measures their stiffness or rigidity. It measures how much a material will deform under a given load. High values of \( E \) indicate that a material is very stiff and doesn't deform much, whereas lower values suggest it is more elastic and will deform easily. Young's modulus is defined as
  • \( E = \frac{\text{stress}}{\text{strain}} \)
Where stress is considered as force per unit area and strain is the deformation per unit length.
In the context of the sphere, Young's modulus of 29 million psi shows that steel is very rigid, meaning the sphere will not deform significantly under pressure. Knowing \( E \) allows us to calculate how much the sphere's diameter and volume decrease when it is subjected to the given underwater pressure, which is a key part of solving the exercise.
Poisson's Ratio
Poisson鈥檚 ratio \( u \) is a measure of how much a material becomes thinner or thicker in one direction when it is stretched or compressed in another. It is defined as:
  • \( u = \frac{-\epsilon_{ ext{lateral}}}{\epsilon_{\text{axial}}} \)
Where \( \epsilon_{ ext{lateral}} \) is the strain in the directions perpendicular to the applied stress, and \( \epsilon_{\text{axial}} \) is the strain in the direction of the applied stress.
In the exercise, Poisson鈥檚 ratio is given as 0.30, which indicates that for the steel sphere, a noticeable degree of lateral contraction occurs when axial stress is applied. This ratio helps us understand the relationship between the deformation in diameter and the volumetric strain, which in turn aids us in calculating the decrease in the sphere鈥檚 diameter and volume.

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Most popular questions from this chapter

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.

Two tempered-steel bars, each \(\frac{3}{16}\) in. thick, are bonded to a \(\frac{1}{2}\) -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude \(P\). Both steels are elastoplastic with \(E=29 \times 10^{6} \mathrm{psi}\) and with yield strengths equal to \(100 \mathrm{ksi}\) and \(50 \mathrm{ksi},\) respectively, for the tempered and mild steel. The load \(P\) is gradually increased from zero until the deformation of the bar reaches a maximum value \(\delta_{m}=0.04\) in. and then decreased back to zero. Determine \((a)\) the maximum value of \(P\), (b) the maximum stress in the tempered-steel bars, \((c)\) the permanent set after the load is removed.

A 2 -m length of an aluminum pipe of 240 -mm outer diameter and 10-mm wall thickness is used as a short column to carry a 640 -kN centric axial load. Knowing that \(E=73 \mathrm{GPa}\) and \(\nu=0.33,\) determine \((a)\) the change in length of the pipe, \((b)\) the change in its outer diameter, (c) the change in its wall thickness.

A block of 10 -in. length and \(1.8 \times 1.6\) -in. cross section is to support a centric compressive load \(\mathbf{P}\). The material to be used is a bronze for which \(E=14 \times 10^{6}\) psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 lssi and that the decrease in length of the block should be at most \(0.12 \%\) of its original length.

The cylindrical rod \(A B\) has a length \(L=6 \mathrm{ft}\) and a 1.25 -in. diameter; it is made of a mild steel that is assumed to be elastoplastic with \(E=29 \times 10^{6} \mathrm{psi}\) and \(\sigma_{Y}=36 \mathrm{ksi} .\) A force \(\mathbf{P}\) is applied to the bar until end \(A\) has moved down by an amount \(\delta_{m}\) Determine the maximum value of the force \(\mathbf{P}\) and the permanent set of the bar after the force has been removed, knowing \((a) \delta_{m}=0.125\) in. \((b) \delta_{m}=0.250\) in.
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