Question

The stress in the material of a pipe subject to internal pressure varies directly with the internal pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 pounds per square inch when the diameter is 5 inches, the thickness is 0.75 inch, and the internal pressure is 25 pounds per square inch. Find the stress when the internal pressure is 40 pounds per square inch if the diameter is 8 inches and the thickness is 0.50 inch.

Short answer

The stress is 384 psi.

Step-by-step solution

- Understand the Relationship

Stress (\text{S}) varies directly with internal pressure (\text{P}) and the internal diameter (\text{d}) and inversely with the thickness (\text{t}) of the pipe. This relationship is given by the formula: \[\text{S} = k \frac{\text{P} \times \text{d}}{\text{t}}\]where \( k\) is a proportionality constant.

- Determine the Proportionality Constant

Use the given values to find the constant \( k \). The given values are: stress \( S = 100 \text{ psi} \), internal pressure \( P = 25 \text{ psi} \), diameter \( d = 5 \text{ inches} \), and thickness \( t = 0.75 \text{ inches} \).Plug these values into the formula: \[100 = k \frac{25 \times 5}{0.75}\]Solve for \( k \).

- Solve for the Constant

Calculate as follows: \[100 = k \frac{125}{0.75} \]\[100 = k \times \frac{125}{0.75} \]\[100 = k \times 166.67 \]\[ k = \frac{100}{166.67} \approx 0.6 \]

- Calculate the Desired Stress

Now, use the determined constant \( k \) to find the new stress with the new conditions: internal pressure \( P = 40 \text{ psi} \), diameter \( d = 8 \text{ inches} \), and thickness \( t = 0.50 \text{ inches} \).Substitute these values into the formula: \[\text{S} = 0.6 \times \frac{40 \times 8}{0.50}\]

- Solve for the New Stress

Calculate as follows: \[\text{S} = 0.6 \times \frac{320}{0.50} \]\[\text{S} = 0.6 \times 640 \]\[\text{S} = 384 \text{ psi} \]

Key concepts

direct variation

Direct variation describes a relationship between two variables where an increase in one results in a proportional increase in the other. This is a linear relationship and is often represented by the equation \( y = kx \), where \( k \) is the proportionality constant.
In the context of the exercise, stress in the pipe material varies directly with internal pressure and pipe diameter. This means if the internal pressure or diameter increases, the stress also increases proportionally. Such relationships are crucial in understanding and designing systems to withstand specific forces.

inverse variation

Inverse variation occurs when one variable increases as another decreases. This type of relationship can be represented by the equation \( y = \frac{k}{x} \), where \( k \) is the proportionality constant.
In our problem, the stress varies inversely with the thickness of the pipe. This means if the thickness of the pipe increases, the stress decreases. Understanding this inverse relationship helps in materials engineering when deciding the appropriate thickness to mitigate stress effectively.

proportionality constant

The proportionality constant \( k \) is a factor that links two variables in direct or inverse variation. It remains unchanged across cases, serving as a scale factor for the relationship.
To find \( k \) in this exercise, we used known values of stress, internal pressure, diameter, and thickness. The determined constant allows us to apply the relationship under new conditions. Specifically, we calculated \( k \) as follows: \[ 100 = k \frac{25 \times 5}{0.75} \ \Rightarrow k = \frac{100}{166.67} \approx 0.6 \]
With \( k \) known, we can predict stress for different pressures, diameters, and thicknesses.

internal pressure

Internal pressure is the force exerted per unit area within the pipe. It directly influences the stress on the pipe material, following direct variation.
For example, in the problem, the internal pressure values of 25 psi and 40 psi directly affect stress levels. The initial formula shows that as the internal pressure increases, the stress increases proportionally: \[ \text{S} = k \frac{P \times d}{t} \]
In engineering, controlling internal pressure is critical for maintaining the structural integrity of pipes and preventing failures.

pipe diameter

Pipe diameter is a measure of the width across the inside of the pipe. Like internal pressure, it affects the stress on the pipe material directly.
For this problem, the diameters of 5 inches and 8 inches were key variables. Increasing the diameter increased the stress, impacting the overall integrity of the pipe: \[ \text{S} = k \frac{P \times d}{t} \]
Engineers must consider diameter when designing pipes to ensure they can handle expected stress without compromising material strength or safety.

pipe thickness

Pipe thickness is the measure of the wall's width. It has an inverse relationship with stress, meaning thicker walls result in lower stress.
In this example, we used thickness values of 0.75 inches and 0.5 inches. This inverse relationship is crucial for designing pipes to withstand high internal pressures without failing: \[ \text{S} = k \frac{P \times d}{t} \]
By increasing thickness, engineers can effectively reduce stress, enhancing durability and preventing breakage, especially in high-pressure environments.