# Thin walled pressure vessel with added forces

1. ### hideki

6
1. The problem statement, all variables and given/known data
Hi, this is my first time posting so I hope you can see the attachment.

2. Relevant equations

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

3. 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. ### Mapes

2,532
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. ### hideki

6

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. ### nvn

2,124
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. ### hideki

6

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. ### SteamKing

10,399
Staff Emeritus
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. ### hideki

6
Sorry I don't understand. What would the area be then?

8. ### Mapes

2,532
If you were to look at a cross section of the cylindrical part, what is the area of the material region that you see? ($\pi r^2$ is the cross-sectional area taken up by air).

9. ### hideki

6
Would the area be pi(.14^2-.008^2) = .06137 m^2?

10. ### hideki

6
Then the longitudinal stress would be:

pr/2t - F/A

where A is the area from the previous post

11. ### Mapes

2,532
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?