# Homework Help: What pressure would a strain of 40 µε indicate?

1. Jan 4, 2017

### Al_Pa_Cone

Q6. A cylindrical vessel 2m internal diameter and 4m long has a wall thickness of 6mm. Strain gauges are installed on the vessel to measure hoop strain (see FIG 6).

The Young's Modulus of the Material = 290 GN m^-2
The Yield Stress of the Material = 500 MPa

The Questions are
(i) What is the maximum allowable pressure if a factor of safety of 4 is to be used?

(ii) What pressure would a strain of 40με indicate?

For (i) I get:

Does this look correct?

for part 2, I cannot find the relevant equation to suit the pressure vessel. I have the hoop equation as
σ = pr/t
Where:
p = Pressure
r = radius of the vessel at 1000mm
t = Thickness of the vessel at 6mm

I am aware the strain is represented as a change in length by 0.000040 but I am not sure how to use this in my working out, given the fact I haven't used the Young's modulus of 290 GN m^-2 either.

Can anyone help me with this question???

2. Jan 4, 2017

### mjc123

(i) No. As your hoop equation indicates, the hoop stress is not equal to the pressure. What pressure would give a hoop stress of 500 MPa?
(ii) What is the definition of Young's modulus? What stress gives a strain of 40 $\mu \epsilon$ ? What pressure causes this stress?

3. Jan 4, 2017

### Al_Pa_Cone

Ok for part (i) This is what I have come up with?

Does this look ok?

I will work on part 2 next.

Thanks

File size:
10.8 KB
Views:
225
4. Jan 4, 2017

### mjc123

On the right lines, but the radius isn't 2, is it?
So what is the max allowable pressure?

5. Jan 4, 2017

### Al_Pa_Cone

No it isn't you are right, So with a correct radius of 1meter the answer gives 3 MPa. If I follow my original post method that the maximum allowable pressure is 1/4 of this I would get 0.75 MPa or 750000 Pa

6. Jan 4, 2017

7. Jan 4, 2017

### Staff: Mentor

How does the axial stress affect the hoop strain?

8. Jan 4, 2017

### Al_Pa_Cone

I think the axial stress is twice the hoop stress therefore twice the hoop strain?

9. Jan 4, 2017

### Staff: Mentor

What is the exact equation for the hoop strain as a function of the combination of the hoop stress and the axial stress.

10. Jan 5, 2017

### Al_Pa_Cone

Would this be the relevant equation I am looking for?

11. Jan 5, 2017

### Staff: Mentor

No. You need Hooke's law in 3D for conditions of plane stress. Google key words Hooke's Law plane stress.

Are you trying to learn this stuff on your own? Are you using a textbook?

12. Jan 5, 2017

### Al_Pa_Cone

Yes I am a distance learning student at Tees University. I can call a tutor to ask a question over the phone which is not ideal, but basically I try to learn all this in my spare time without assistance.

I covered hookes law several months ago amongst many other things. I have found this forum to be very helpful in providing the assistance I need to arrive at the correct answers.
I will have another look over my previous notes and see what I can dig out. Thanks for your help

13. Jan 6, 2017

### Al_Pa_Cone

I have followed the guidelines in the lessons and obtained this figure for pressure?

If this answer is still wrong I will persue the Hookes law in 3D method although it is not in the methods provided to me through the University.

Thanks

14. Jan 6, 2017

### Al_Pa_Cone

For part one of this Question, Am I right in dividing the answer I obtained (3MPa) by 4 to get the safety factor maximum pressure? or is 3 MPa the answer? I think I need to divide

Thanks

15. Jan 9, 2017

### Al_Pa_Cone

For This part question I have spent some time going back through my lessons provided and found this:

does this look like the correct method to follow in obtaining the pressure value which would cause a strain of 40μ?

This is the last question in my assignment and I am determined to get the highest possible marks. Anyone who can offer help please do.

Thanks

16. Jan 9, 2017

### Staff: Mentor

Yes. But, of course, you need to use the corresponding equations involving the stresses and strains in cylindrical coordinates. And don't forget that the stress in the radial direction is much much smaller than the stresses in the axial- and hoop directions, and should thus be taken as zero.

17. Jan 9, 2017

### Al_Pa_Cone

Just to Check, Is part (i) ok now or should I be dividing by 4 for the safety factor? I think I should divide my answer by 4 to get 0.75 MPa maximum allowed pressure?

18. Jan 9, 2017

### Staff: Mentor

It depends on how they are defining the yield stress.

19. Jan 9, 2017

### Al_Pa_Cone

I can only attempt my answer from the information provided in the initial question. It says the yield stress of the material is 500MPa?

When it mentions the safety factor of 4 I am sure I need to divide the final pressure by 4 to get my result. I just needed a verification I was doing the correct thing?

20. Jan 9, 2017

### Staff: Mentor

I was asking about the yield criterion that was being used. Some of the choices are Von Mises, Tresca, or maximum tensile stress over all planes of arbitrary orientation. Which criterion do you think they wanted you to use?

21. Jan 9, 2017

### Al_Pa_Cone

I have no idea! I am not that clued up on yield stress sorry.

22. Jan 9, 2017

### Staff: Mentor

23. Jan 10, 2017

### Al_Pa_Cone

Thank for providing that link, it was very helpful. According to the information given in the link with regards to Anisotropic yield, This occurs when a metal is subjected to large plastic deformations. As the question states 'What is the maximum allowable pressure if a factor of safety of 4 is to be used?' It doesnt reach the large plastic deformation to use the Anisotropic yield criteria which rules out the use of the von Mises yield criterion. I have copied the details of this link at the bottom of this thread.

So from my lessons on this subject:

So I have the wall Thickness = 6 mm or 0.006 m
The Radius = 1000 mm or 1 m
And the Yeild Stress = 500 MPa

Inuptting these known figures into the above equation and transposing for (P) I have an answer. P = 3 MPa.

In the lesson notes It mentions ' In the absence of a code, we will use N > 4 (i.e the design stress will be at most a quater of the allowable stress for the material.)'
Well the yeild stress is 500 MPa and the pressure I have worked out to be 3 MPa. So if a 3 MPa pressure would cause a yeild of 500 MPa then a safety factor of 4 would require me to divide this firgure by 4 giving Pressure = 0.75 to create a yeild of 125 MPa?

Does this look like it could be a correct answer or am I missing something? from my previous questions, my history would suggest I am missing something.

Anisotropic yield criteria
When a metal is subjected to large plastic deformations the grain sizes and orientations change in the direction of deformation. As a result, the plastic yield behavior of the material shows directional dependency. Under such circumstances, the isotropic yield criteria such as the von Mises yield criterion are unable to predict the yield behavior accurately. Several anisotropic yield criteria have been developed to deal with such situations. Some of the more popular anisotropic yield criteria are:

24. Jan 10, 2017

### Al_Pa_Cone

Also, This is what I have obtained as a final answer to part b?
with the 3dimentional method I would need the poissons ratio and also I came up against a complication in measuring the x and y stresses when they would be acting on the rounded edges of the cylinder? what surface area would I use in the equation provided stress = force/ cross sectional area?

I have gone with what I think is the correct method. Can anyone verify this?

25. Jan 10, 2017

### Staff: Mentor

You are not interpreting it correctly. The von Mises yield criterion is fine for determining the onset of yield even for small deformations.
OK. So they are indicating that you should use the maximum principal stress criterion to establish the yield point.
Correct
No. The yield stress is a property of the material, and is independent of the loading. At 125 MPa, the material will not yield.