1. Thin Walled Pressurized Tanks (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)

2. Parallel Reading Chapter 9 Section 9.2

3. Consider a Tank for Pressurized Gasses The up and down force componants Cancel each other out – but there is A net force to the side if we slice the Tank. It is the force Produced by The resistance Of the metal of The tank that Resists this force

4. If We Treat the Thickness of the Metal as Small If the metal thickness is small there Will be no significant differences In stress from top to bottom That tensile stress In the tank will be The same no matter What angle we take The slice at.

5. Because that Stress is Uniform Around the Circle We Call it a Hoop Stress

6. The Magnitude of the Hoop Stress P rThe Force must be 2*r*P The resisting Area is the Thickness of The metal 2t Force = 2*r*P Therefore the Hoop Stress Is

7. Hoop Stress Shows Up in Several Designs

8. Of Course the Pressure in the Tank is Uniform in All Directions So there is also a longitudinal stress

9. Longitudinal Stress Magnitude P Force must be P*π*r 2 Resisting area must be 2*π*r*t Thickness t Longitudinal stress must be

10. An Example A 500 gallon propane tank has a length Of 12 feet, a diameter of 61 inches and A wall thickness of 7/16ths of an inch Steel rated for 60 ksi tension. How much pressure can be put in it? Where will it break? Hoop Stress Longitudinal Stress Hoop Stress is twice Longitudinal. The tank will blow first With hoop stress.

11. Working it Out. With a little algebra It will split down the length when It fails.

12. Doggone Hoop Stress Is there some way to get rid of it? How about this?

13. Another Failure Problem (No I’m not referring to your last quiz) Sometimes the easiest place to blow a tank is on one of our connections – rather than tear through the material itself.

14. Shapes Like This are not Naturally Occurring I can weld a spiral of metal to make A tank. Of course I can also Just weld rectangular Plates together.

15. I Wonder Which Design is Better? This design puts the hoop stress directly On a welded joint. This design puts the weld on a diagonal to the hoop stress. Lets consider the case of a compressed air tank 30 inches in diameter made of 3/8 th inch steel plate and pressurized to 180 psi. What kinds of stresses will we Be putting on those welds?

16. Case 1 the weld faces right into the Hoop Stress This hoop stress will be directly applied to the weld Of course the longitudinal stress is ½ the hoop so it is 3510 psi. We all know which weld is likely to go first.

17. Case 2 Let us suppose the angle on the spiral weld is 25 degrees. What is the best way to find the stresses at an angle to the principle stress? 7020 psi3510 psi 5265 Mohr’s Circle to the Rescue!

18. Now How Do We Check Out the State of Stress at 25 degrees? 3510 5265 7020 Mohr’s Circle doubles angles So if I want to look 3510 psi 7020 psi Looks like 25 degrees Down from horizontal 50 ̊ 1755=r

19. Decision Case 1 7020 psi tension 4140 psi tension 1344 psi shear Case 2 I suspect that if you pick Case 1 when strength is Really needed, that you Will need a lot of what is In case 1.

20. Assignment 12 Problem 9.2-2 part a and b Problem 9.2-6 part a, b, and c

21. Into the Thick of Things What happens when the walls of the pressure Vessel are thick enough that we can no longer Call them thin walled? Things get “thick” when the Wall thickness exceeds about 1/20 th of the diameter of the vessel

22. Thin Walls Allow Us to Drop Consideration of Stress and Deformation Changes through Thickness In a thin wall we are concerned about two stresses – stress down the length Longitudinal stress and circumferential or hoop stress

23. Thickwall Means We Must Also Consider Radial Stress

24. Derivation is Tedious (And therefore skipped) Lame’s Equations (Because many thick walled cylinders – think pipe, gun barrel, Mine shaft are open on both ends most developments of Lame’s Equation leaves longitudinal stress out and then adds it by Superposition later if needed) (ok so it’s a bit Lame)

25. Lets Try One

26. We Get Some Simplifications in Lame’s Equations For any radius r But life gets better. We know the maximum stress will Be on the inside of the tank. That’s Dandy

27. Apply to Our Problem Lets do the hard one – if the pressure Inside is 100 MPa -100 Mpa What does the negative number mean? The material at the inside Edge is getting squeezed

28. What’s Happening at the Outside Edge? This python has Sort or run out of Squeeze.

29. Human Interest What happened to the radial stress between the inside and outside? It decays by a second order curve

30. Now for Circumferential or Tangential Stress What does the positive sign mean? The tank is being pulled apart

31. What About Outer Edge Circumferential Stress? No preset simplified formula – we have to plug In for the outside edge Well that Sucks

32. How does this result compare to a Thin walled vessel? Picking the largest value of t that still Qualifies as thin wall. 308 MPa inside 208 MPa outside

33. Example

34. Yipes! I know my maximum stress is at the inside wall of the pipe But none of these equations are for shear stress! I don’t like the Looks of this! Am I cooked?

35. Then We Remember Mohr’s Circle Just because you don’t see Shear in your first measurements, Does not mean it is not there. When material is in stress – all sorts of Combinations of shear and tensile And compressive stresses become Possible at different angles.

36. What Do We Know About Stresses? The 3 stresses calculated for a pressure vessel are all principle stresses! Lets see the longitudinal stress must be 0 – this pipe is not closed at the end That leaves radial stress – a compression And Hoop stress – a tension

37. Quick Consideration of 3D Mohr’s Circle Our hoop stress (tension) Our radial stress (compression) Some Mohr Pie!

38. Substituting

39. A Bit of Plug and Chug Inside radius = 1.375 in Outside radius = 1.5 in

40. Gun Barrels are a Thick Wall Cylinder Application A new kind of high power Ammunition is called +P It reaches higher pressures and sends The bullet out at higher speed. (But not all guns are made to handle +P ammunition) What is the mode of failure in these cases?

41. What Happened Here?

42. What if the Pressure is Outside? The radial stress maximum Is at the outside edge The hoop stress maximum Is still on the inside.

43. Watch Out Note the stress is compressive? The foot looks fine to me

44. Lets Apply Inspired by the concrete canoe competition Students at SIU decide to have a Concrete submarine competition. Connie Concrete wants to decide how Deep her submarine can go. Pressure outside the vessel increases By 0.44 psi for every foot of depth. To actually crush Connie’s concrete it Will take 10,000 psi. The pipe is 5 ft in Outside Diameter and 6 inches thick.

45. Connie Crunches Limits To do a radial crush will take 10,000 psi At 0.44 psi per foot of depth it will take About 22,700 ft of depth. Because of end caps the submarine will also Have longitudinal stress. About 4,300 ft of depth

46. Now to Check Hoop Stress Maximum on inside of cylinder This looks familiar – the hoop stress Is twice to longitudinal. Well we can still make it to 2,150 ft of depth.

47. Want a Ride in Connie’s Sub? Can you think of anything that Connie and Her team might have missed? Proposed test Subject for submarine

48. Does this Make You Worry? I wonder if my concrete could fail in Shear? To actually crush the concrete takes 10,000 psi, but the specimen in a Uniaxial compression test (like you ran) fails much sooner because the Shear limit for Connie’s Concrete is 2,500 psi.

49. So How do You Get Max Shear? Pick our spot to check – our most critical hoop stress is on the inside of the Concrete cylinder. Arrange our principle stresses in order from largest to smallest 1- Largest = hoop stress - compression 2- longitudinal stress - compression (1/2 of hoop stress) 3- radial stress – 0 on the inside edge of the concrete Hoop Stress Longitudinal Stress Radial Stress τmax

50. Implications The largest hoop stress will can take without triggering shear failure is twice The shear limit 2,500 psi (shear limit) *2 = 5,000 psi maximum allowable hoop stress That’s only half what we thought we could do. The sub will fail in shear at 1,075 ft. Oh Nut’s – We just lost our First test subject! I think this Might leak.

51. So Where Are We With the FE Book? Bottom of Page 1 Here are the thick wall Cylinder equations We have been talking About. They may be useful On class quizes – But they are unlikely Subjects for the FE Exam itself

52. And on Page 2 These are the thin wall vessel Equations – they are more likely than Thick wall vessels, but still unlikely On the F.E. (But very much fair game for class Quizes).