# Thin walled pressure vessel with added forces

1. Jun 11, 2011

### hideki

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.

#### Attached Files:

• ###### MECHproblem.jpg
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2. Jun 12, 2011

### Mapes

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. Jun 12, 2011

### hideki

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. Jun 12, 2011

### nvn

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. Jun 13, 2011

### hideki

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. Jun 13, 2011

### SteamKing

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. Jun 13, 2011

### hideki

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

8. Jun 13, 2011

### Mapes

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. Jun 13, 2011

### hideki

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

10. Jun 13, 2011

### hideki

Then the longitudinal stress would be:

pr/2t - F/A

where A is the area from the previous post

11. Jun 13, 2011

### Mapes

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?