Thin walled pressure vessel with added forces

In summary: In short, in calculating the direct axial stress due to the 30 kN force, the stress is developed in the area of the 8 mm tank shell, not the entire cross section of the tank. So the area of the material region is pi(.14^2-.008^2) = .06137 m^2. Then the longitudinal stress would be:pr/2t - F/A
  • #1
hideki
6
0

Homework Statement


Hi, this is my first time posting so I hope you can see the attachment.


Homework Equations



Hoop stress = pr/t
Longitudinal stress = pr/2t


The Attempt at a Solution


I am stuck on part a), afterwards I think I should be fine.
I am guessing you can "add" axial stresses induced by the 30kN force...but I'm not sure? In this case I have:

Circumferential stress = pr/t = 26.25MPa [Tension]
Longitudinal stress = pr/2t - 2 * (F/A) = -285.29Mpa. [compression]

The longitudinal stress is large which leads me to believe I have done something wrong.

Any help or clarification would be greatly appreciated.
 

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  • #2
Hi hideki, welcome to PF!

Your reasoning about superposing the stress from the 30 kN force is fine. But your calculations need checking. Where did the factor of 2 come from? What is the cross-sectional area of the material under stress in the cylindrical part of the vessel?
 
  • #3
Hi Mapes, thankyou for your reply.

I used a factor of two because there are two forces on each side of the cylinder. I guessed the cross sectional area was pi * (.14^2)...but this is probably incorrect because the end of the cylinder is curved and not a flat surface.
 
  • #4
hideki: Both concepts in post 3 are incorrect. Try again. As Mapes mentioned, check your second calculation in post 1.

Your circumferential stress in post 1 is correct. Always leave a space between a numeric value and its following unit symbol. E.g., 26.25 MPa, not 26.25MPa. See the international standard for writing units (ISO 31-0). Also, it is spelled MPa, not Mpa.

Do not worry about the stress in the elliptical end caps, for now. Compute the stress in the cylindrical portion of the pressure vessel.
 
  • #5
Hi nvn, thanks for your reply.

So the longitudinal stress due to the pressure is:
longitudinal stress = pr/2t = 13.125 MPa [Tension]

and the longitudinal stress due to the forces is:
longutdinal stress = F/A = (30*10^3)/(pi*.14^2) = 0.487 MPa [Compression]

Is that true?
 
  • #6
In calculating the direct axial stress due to the 30 kN force, the stress is developed in the area of the 8 mm tank shell, not the entire cross section of the tank.
 
  • #7
Sorry I don't understand. What would the area be then?
 
  • #8
hideki said:
Sorry I don't understand. What would the area be then?

If you were to look at a cross section of the cylindrical part, what is the area of the material region that you see? ([itex]\pi r^2[/itex] is the cross-sectional area taken up by air).
 
  • #9
Would the area be pi(.14^2-.008^2) = .06137 m^2?
 
  • #10
Then the longitudinal stress would be:

pr/2t - F/A

where A is the area from the previous post
 
  • #11
hideki said:
Would the area be pi(.14^2-.008^2) = .06137 m^2?

What is the reasoning here? This is equivalent to taking the entire cross-section area and subtracting the area of a circle with radius 8 mm. But there is no circle with radius 8 mm in the problem.

Image unwrapping the wall of the cylindrical cross-section. You'd get approximately a rectangle with width 8 mm and length equal to the perimeter of the circle, right?
 

FAQ: Thin walled pressure vessel with added forces

1. What is a thin walled pressure vessel?

A thin walled pressure vessel is a container that is designed to hold fluids or gases under pressure. It is typically cylindrical in shape with a relatively thin wall, and is used in a variety of industrial and scientific applications.

2. How is a thin walled pressure vessel different from a thick walled pressure vessel?

The main difference between a thin walled pressure vessel and a thick walled pressure vessel is the ratio of the vessel's diameter to its wall thickness. A thin walled pressure vessel has a larger diameter and a thinner wall compared to a thick walled vessel, which has a smaller diameter and a thicker wall.

3. What types of forces can be applied to a thin walled pressure vessel?

There are two main types of forces that can be applied to a thin walled pressure vessel: internal pressure and external forces. Internal pressure is the force exerted by the fluid or gas inside the vessel, while external forces can include things like wind, seismic activity, or mechanical loads.

4. How are added forces accounted for in the design of a thin walled pressure vessel?

Added forces, such as external loads, are taken into consideration during the design of a thin walled pressure vessel by using safety factors and stress analysis. Safety factors ensure that the vessel can withstand the additional forces without failure, while stress analysis helps determine the potential stress and strain on the vessel's walls.

5. What are some common uses for thin walled pressure vessels with added forces?

Thin walled pressure vessels with added forces are commonly used in industries such as oil and gas, chemical processing, and aerospace. They are also used in scientific research for experiments involving high pressure and extreme temperatures. Some specific examples include pressure vessels for storing and transporting compressed gases, or for conducting high pressure reactions in a controlled environment.

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