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A single strain gage forming an angle \(\beta=18^{\circ}\) with a horizontal plane is used to determine the gage pressure in the cylindrical steel tank shown. The cylindrical wall of the tank is \(6 \mathrm{mm}\) thick, has a \(600-\mathrm{mm}\) inside diameter, and is made of a steel with \(E=200 \mathrm{GPa}\) and \(\nu=0.30 .\) Determine the pressure in the tank indicated by a strain gage reading of \(280 \mu\).

Short Answer

Expert verified
Gauge pressure in the tank is determined using tensile strain and Hooke's law, resulting in a calculated pressure value.

Step by step solution

01

Understand the Problem

The task is to find the gauge pressure inside a cylindrical tank using the reading from a strain gauge placed at an angle of \( \beta = 18^{\circ} \) to the horizontal. The cylindrical tank has specific material and geometric properties given, which will be used in the calculations.
02

Use the Strain Formula in a Plane

The strain measured by the gauge, \( \epsilon_{\beta} \), is related to the normal strains in the longitudinal and circumferential directions, \( \epsilon_x \) and \( \epsilon_y \), by the equation: \[ \epsilon_{\beta} = \epsilon_x \cos^2 \beta + \epsilon_y \sin^2 \beta \]where \( \beta = 18^{\circ} \). Here, \( \epsilon_x \) and \( \epsilon_y \) are the strain components along the longitudinal and circumferential directions respectively.
03

Solve for Circumferential Strain

Given that the strain measured by the gauge is \( \epsilon_{\beta} = 280 \mu = 280 \times 10^{-6} \), substitute this value and \( \beta \) into the equation to solve for the strains. Assume \( \epsilon_x = 0 \) because the tank is assumed not to deform significantly along its length. Hence, \[ 280 \times 10^{-6} = 0 \times \cos^2 18° + \epsilon_y \sin^2 18° \]Solve for \( \epsilon_y \).\[ \epsilon_y = \frac{280 \times 10^{-6}}{\sin^2 18°} \]
04

Relate Strain to Pressure

Using Hooke's Law for cylindrical pressure vessels:\[ \epsilon_y = \frac{1}{E} \left( \sigma_y - u \sigma_x \right) \]For a thin-walled pressure vessel, the hoop stress is \( \sigma_y = \frac{pR}{t} \) and the longitudinal stress \( \sigma_x = \frac{pR}{2t} \). Substitute these expressions into Hooke's Law along with \( E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa} \) and \( u = 0.30 \).
05

Calculate the Pressure

Rearrange Hooke's Law to solve for \( p \):\[ \epsilon_y = \frac{1}{E} \left( (\frac{pR}{t}) - u (\frac{pR}{2t}) \right) \]\[ \epsilon_y = \frac{1}{E} \left( \frac{pR}{t} \cdot (1 - \frac{u}{2}) \right) \]Plug in the values: \[ \epsilon_y = 280 \times 10^{-6} \]\[ p = \frac{280 \times 10^{-6} \times E \times t}{R (1 - \frac{u}{2})} \]with \( t = 0.006 \text{ m} \), \( R = \frac{0.6}{2} \text{ m} \). Solve for \( p \).
06

Final Solution Calculation

Substitute the numerical values\[ p = \frac{280 \times 10^{-6} \times 200 \times 10^9 \times 0.006}{0.3 \times (1 - 0.15)} \]Calculate \( p \) to find the pressure inside the tank.

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Key Concepts

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

Mechanical Properties of Materials
Mechanical properties are fundamental in understanding how materials behave under different types of forces and loads. These properties include elasticity, plasticity, toughness, and hardness, among others. In this exercise, two essential properties are Young's modulus ( E ) and Poisson's ratio ( u).

Young's modulus ( E ) is a measure of the stiffness of a material. It quantifies the relationship between stress (force per unit area) and strain (deformation over original length) in the elastic deformation region. For steel in this problem, E is given as 200 GPa, indicating a very stiff material that does not deform easily under applied stress.

Poisson's ratio ( u) refers to the ratio of lateral strain to axial strain in a material subjected to uniaxial stress. It indicates how much a material will expand or contract in directions perpendicular to the direction of compression or stretching. Here, for steel, u is 0.30, suggesting moderate lateral expansion when stretched longitudinally. Together, these properties help us understand how the steel tank will behave when internal pressure is applied.
Thin-Walled Pressure Vessels
Thin-walled pressure vessels are structures such as tanks or pipes where the wall thickness is considerably small compared to the vessel's radius. A general rule of thumb is that the thickness ( t) should be less than one-tenth of the radius ( R) of the vessel. In this exercise, the tank's wall thickness is 6 mm, and the inside radius is 300 mm, qualifying it as a thin-walled structure.

Due to their geometry, thin-walled pressure vessels experience stress mainly in two directions: circumferential (or hoop) stress and longitudinal stress. Hoop stress acts along the circumference, trying to "burst" the vessel open, while longitudinal stress acts along the length of the vessel. In cylindrical vessels, hoop stress ( \sigma_y) is typically higher than longitudinal stress ( \sigma_x), and is given by the formula:

\[ \sigma_y = \frac{pR}{t} \]

Where p is the internal pressure. This helps in understanding the different forces at play in thin-walled vessels and is crucial for safe design and analysis.
Stress-Strain Relations
Stress-strain relations describe how materials deform under stress, essentially linking mechanical stress with the resulting strain. This relation is central to understanding the design and analysis of engineering systems.

For materials like steel, we use Hooke's Law in the context of axial loading, which implies the stress-strain relation as follows:

\[ \epsilon_y = \frac{1}{E} \left( \sigma_y - u \sigma_x \right) \]

Where \epsilon_y is strain in the circumferential direction, E is Young’s modulus, and u is Poisson’s ratio. \sigma_y and \sigma_x are the hoop and longitudinal stresses, respectively.

This equation allows engineers to compute the strain in a material, given known stresses and material properties. By rearranging this equation, as done in the original solution, one can solve for unknown variables like internal pressure when other parameters are available. This practical use of stress-strain relationships is key in calculating the pressures that thin-walled vessels can safely withstand. Understanding and applying these relationships helps ensure that the designs of such vessels are both efficient and reliable.

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Most popular questions from this chapter

A steel pipe of 12 -in. outer diameter is fabricated from \(\frac{1}{4}\) -in.-thick plate by welding along a helix that forms an angle of \(22.5^{\circ}\) with a plane perpendicular to the axis of the pipe. Knowing that a 40 -kip axial force \(\mathbf{P}\) and an 80 -kip \(\cdot\) in. torque T, each directed as shown, are applied to the pipe, determine the normal and in- plane shearing stresses in directions, respectively, normal and tangential to the weld.

A brass ring of 5 -in. outer diameter and 0.25 -in. thickness fits exactly inside a steel ring of 5 -in. inner diameter and 0.125 -in. thickness when the temperature of both rings is \(50^{\circ} \mathrm{F}\). Knowing that the temperature of both rings is then raised to \(125^{\circ} \mathrm{F}\), determine \((a)\) the tensile stress in the steel ring, \((b)\) the corresponding pressure exerted by the brass ring on the steel ring.

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A spherical gas container having an inner diameter of \(5 \mathrm{m}\) and a wall thickness of \(24 \mathrm{mm}\) is made of steel for which \(E=200 \mathrm{GPa}\) and \(\nu=0.29 .\) Knowing that the gage pressure in the container is increased from zero to \(1.8 \mathrm{MPa}\), determine ( \(a\) ) the maximum normal stress in the container, (b) the corresponding increase in the diameter of the container.

A standard-weight steel pipe of 12 -in. nominal diameter carries water under a pressure of 400 psi. ( \(a\) ) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum tensile stress in the pipe. ( \(b\) ) Solve part \(a\), assuming an extra-strong pipe is used, of 12.75 -in. outside diameter and 0.5 -in. wall thickness.

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