# Shear Pressure?

1. Feb 3, 2016

### perky416

Hi guys,

The imagine the picture attached is a cross section of a cylinder. How would I work out the maximum amount of pressure this can be exposed to before the metal shears off at the location labelled "2mm thick"?

Lets say the following material is used:

T6 Aluminium
Ultimate Tensile Strength - 310 MPa
Tensile Yield Strength - 276 MPa
Shear Strength - 207 MPa

Iv googled and googled but can only seem to find information relating to shear force in beams or punching through a material.

All help greatly appreciated.
Thanks

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2. Feb 3, 2016

### Nidum

Imagine the small diameter section being separate and fitting into the larger diameter part like a piston in a cylinder . The contact area between the two parts represents the area carrying the shear force in the one piece object

3. Feb 3, 2016

### perky416

Hi,

I understand the location, its pretty much where iv marked with a black line on the diagram, I just cant for the life of my figure out how to calculate the max force.

Regards

4. Feb 3, 2016

### Nidum

You may be confused by the blind bore . Pressure acts where you have shown it but also on the area at top of the blind bore . Effectively you can ignore the details of the bore and just treat the upper component as if it was solid for purposes of calculating the effective 'piston area' that pressure acts on .

5. Feb 3, 2016

### perky416

Hi,

Sorry reading my post back I understand I wasn't clear, I dont mean where the arrows are I mean where i've done the black line and labeled it 2mm thick i.e. "the contact area between the 2 parts".

6. Feb 3, 2016

### Nidum

It's getting late now . Hopefully someone else will pick up this thread but if not I'll get back to you tomorrow .

7. Feb 3, 2016

### perky416

ok thank you

8. Feb 3, 2016

You know how to calculate the upward force due to pressure on that upward piece correct? Don't forget to include both the very top and the two horizontal surfaces toward which you have the arrows pointing in your picture.

9. Feb 3, 2016

### perky416

Sorry not too sure what you mean. Please could you elaborate a little?

10. Feb 3, 2016

Well if you want to know the shear stress where those black lines are, you need two things: a force and an area. So, do you know how to calculate the force you need? What about the area?

11. Feb 3, 2016

### perky416

Oh i see what you mean.

The total area that the pressure is acting on including the upward area is 3.14cm^2.

For the force cam i simply convert the pressure? For example 100 bar of pressure = 101.972 kgmf/cm^2.

12. Feb 3, 2016

So the inner diameter of your larger part of the cylinder is 2 cm?

You are probably better off doing everything in the MKS units (meter, kilogram, second), so that you do pressure in pascals, force in newtons, and area in m2. What you suggest would work but keeping units straight will be more of a pain.

13. Feb 3, 2016

### perky416

Yes 2cm diameter is correct.

So using MSK....
Pressure = 100,000 Pa
Force = 100,000 N/m^2
Area = 0.0314 m^2

Where do I go from here?

14. Feb 3, 2016

Well first, force is not measured in N/m2. Both of your first two numbers are pressures since Pa = N/m2.

Second, you have to add a couple extra zeros to your area because 1 cm x 1 cm is not 0.01 m; it is 0.0001 m.

Then, how do you feel you should proceed to find the force? What does pressure actually mean and how does it relate to force?

15. Feb 3, 2016

### perky416

Ahh we are getting there...

Pressure is force applied to the surface so to speak.

Pressure = force / area.

Which gives us...

Force = pressure x area.
Force = 100,000Pa x 0.000314m^2
Force = 31.4N

16. Feb 3, 2016

Alright, so now you have a force. Now you have to convert that back into a stress in the region your are interested. Stress is also a force per unit area, but unlike pressure, it can occur in solids and can occur in shear rather than just normal to a surface. Essentially, pressure is the normal component of stress in a fluid.

Anyway, so you need to find the shear stress. You have a force. You need an area.

17. Feb 3, 2016

### perky416

When you say area do you mean the area where the stress will occur?

If so would it be the circumference x thickness?
62.83mm x 2mm =125.66mm^2 or 0.00012566m^2

Or am i completely off the mark?

18. Feb 3, 2016

No you're still heading in the right direction (though I didn't check your math).

19. Feb 4, 2016

### perky416

Ok so if we used Shear stress = force/ area

Shear stress = 31.4N / 0.00012566m^2
Shear stress = =249,880.63 Pa

Should i factor in the strength of the material? For example wouldnt steel be able to handle more pressure than aluminium?

(Ps i really really appreciate all your help, thank you)

20. Feb 4, 2016