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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.

Short Answer

Expert verified
(a) 6400 psi, (b) 4700 psi.

Step by step solution

01

Calculate Inner Diameter for Standard Pipe

The wall thickness is 0.375 inches, and the outer diameter of the pipe is 12.75 inches. The formula to calculate the inner diameter \(d_i\) of the pipe is given by subtracting twice the wall thickness from the outer diameter:\[d_i = 12.75 - 2 \times 0.375 = 12.00\] inches.
02

Calculate Maximum Tensile Stress for Standard Pipe

The formula to determine the maximum tensile stress \(\sigma_t\) in the pipe wall is given by:\[\sigma_t = \frac{P \cdot d_i}{2 \cdot t}\]where:- \(P\) is the internal pressure (400 psi),- \(d_i\) is the inner diameter (12.00 inches),- \(t\) is the wall thickness (0.375 inches).Substituting the values, we get:\[\sigma_t = \frac{400 \cdot 12.00}{2 \cdot 0.375} = 6400\] psi.
03

Calculate Inner Diameter for Extra-strong Pipe

The wall thickness for the extra-strong pipe is 0.5 inches, with the outside diameter still being 12.75 inches. The inner diameter is calculated as:\[d_i = 12.75 - 2 \times 0.5 = 11.75\] inches.
04

Calculate Maximum Tensile Stress for Extra-strong Pipe

Using the same formula for tensile stress \(\sigma_t\) with the new inner diameter and wall thickness:\[\sigma_t = \frac{P \cdot d_i}{2 \cdot t}\]where:- \(P\) is 400 psi,- \(d_i\) is 11.75 inches,- \(t\) is 0.5 inches.Substitute the values into the formula:\[\sigma_t = \frac{400 \cdot 11.75}{2 \cdot 0.5} = 4700\] psi.

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Key Concepts

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

Pipe Wall Thickness
The thickness of a pipe's wall is a crucial factor when analyzing its ability to withstand internal pressures. The wall thickness determines how much pressure the pipe can handle before failing. It serves two main purposes:
  • Strength: Thicker walls can support more pressure, enhancing the pipe's structural integrity.
  • Durability: Thicker walls offer resistance to wear and tear, extending the pipe's lifespan.
Pipe wall thickness is usually derived by subtracting the inner diameter from the outer diameter and dividing by two. This relationship between wall thickness and the pipe's structural capability is evident in cases where different wall thicknesses result in varied tensile stresses, as shown in the exercise. For the standard pipe, a wall thickness of 0.375 inches results in a higher tensile stress compared to the extra-strong pipe with a thickness of 0.5 inches.
Internal Pressure
Internal pressure refers to the pressure exerted by the fluid inside the pipe against its walls. In engineering, it's denoted by the symbol \( P \) and is usually measured in psi (pounds per square inch). Internal pressure plays a critical role in calculating the tensile stress experienced by the pipe walls.

Large stresses on pipe walls can lead to material failure, making it essential to design pipes that can handle specific internal pressures safely. The formula used in the exercise to determine tensile stress \( \sigma_t \) is:
\[\sigma_t = \frac{P \cdot d_i}{2 \cdot t}\]This equation highlights how the internal pressure impacts the stress experienced by the pipe. The greater the internal pressure, the higher the tensile stress, necessitating either thicker walls or stronger materials.
Inner Diameter
The inner diameter of a pipe is the measurement across the inside of the pipe, indicating the space available for the fluid to flow. To determine the inner diameter \( d_i \), you subtract twice the wall thickness from the outside diameter:
\[d_i = d_o - 2 \cdot t\]where \( d_o \) is the outer diameter and \( t \) is the wall thickness. For both standard and extra-strong pipes discussed in the exercise, the inner diameter adjustments reflect how wall thickness impacts the pipe's overall capacity to channel fluid.

Inner diameter influences not only the flow capacity but also the tensile stress, as seen in the formula provided for stress calculation. Smaller inner diameters typically lead to reduced tensile stress, assuming a constant internal pressure and wall thickness, as observed in the transition from the standard to the extra-strong pipe in the exercise.

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Most popular questions from this chapter

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.

The 1.5 -in.-diameter shaft \(A B\) is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when \(P=60\) kips.

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 \(280 \mu\).

For the given state of plane strain, use Mohr's circle to determine \((a)\) the orientation and magnitude of the principal strains, \((b)\) the maximum in-plane strain, \((c)\) the maximum shearing strain. \begin{equation}\begin{array}{ccc} \hline \epsilon_{x} & \epsilon_{y} & \gamma_{x y} \\ \hline+60 \mu & +240 \mu & -50 \mu \\ +400 \mu & +200 \mu & +375 \mu \\ +300 \mu & +60 \mu & +100 \mu \\ -180 \mu & -260 \mu & +315 \mu \\ \hline \end{array}\end{equation}

The following state of strain has been measured on the surface of a thin plate. Knowing that the surface of the plate is unstressed, determine \((a)\) the direction and magnitude of the principal strains, ( \(b\) ) the maximum in- plane shearing strain, \(\left.(c) \text { the maximum shearing strain. (Use } \nu=\frac{1}{3}\right)\) \begin{equation}\begin{array}{ccc} \epsilon_{x} & \epsilon_{y} & \gamma_{x y} \\ \hline-260 \mu & -60 \mu & +480 \mu \\ -600 \mu & -400 \mu & +350 \mu \\ +160 \mu & -480 \mu & -600 \mu \\ +30 \mu & +570 \mu & +720 \mu \\ \hline \end{array}\end{equation}

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