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

Thin Walled Pressurized Tanks

(Credit for many illustrations is given to McGraw Hill publishers

and an array of internet search results)

Parallel Reading

Chapter 9 Section 9.2

Consider a Tank for Pressurized Gasses

The up and down force componantsCancel each other out – but there isA net force to the side if we slice theTank.

It is the forceProduced byThe resistanceOf the metal ofThe tank thatResists this force

If We Treat the Thickness of the Metal as Small

If the metal thickness is small thereWill be no significant differencesIn stress from top to bottom

That tensile stressIn the tank will beThe same no matterWhat angle we takeThe slice at.

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

The Magnitude of the Hoop Stress

P

r The Force must be2*r*P

The resistingArea is theThickness ofThe metal

2t

Force = 2*r*P

t

Pr

t

Pr

A

F *

*2

**2

Therefore the Hoop Stress Is

Hoop Stress Shows Up in Several Designs

Of Course the Pressure in the Tank is Uniform in All Directions

So there is also a longitudinal stress

Longitudinal Stress Magnitude

P

Force must be P*π*r2

Resisting area must be2*π*r*t

Thickness t

Longitudinal stress must be

t

rP

tr

P

A

F r*2

*

***2

**2

An Example

A 500 gallon propane tank has a lengthOf 12 feet, a diameter of 61 inches andA wall thickness of 7/16ths of an inchSteel rated for 60 ksi tension.

How much pressure can be put in it?Where will it break?

Hoop Stress Longitudinal Stress Hoop Stress is twiceLongitudinal.The tank will blow firstWith hoop stress.

Working it Out.

psir

tp 873)4375.0

261(

4375.0*000,60*max

With a little algebra

It will split down the length whenIt fails.

Doggone Hoop Stress

Is there some way to get rid of it?

How about this?

Another Failure Problem(No I’m not referring to your last quiz)

Sometimes the easiest place to blow a tank is on one ofour connections – rather than tear through the materialitself.

Shapes Like This are not Naturally Occurring

I can weld a spiral of metal to makeA tank.

Of course I can alsoJust weld rectangularPlates together.

I Wonder Which Design is Better?

This design puts the hoop stress directlyOn a welded joint.

This design puts the weld on a diagonalto the hoop stress.

Lets consider the case of a compressed air tank 30 inches in diameter made of3/8th inch steel plate and pressurized to 180 psi. What kinds of stresses will weBe putting on those welds?

Case 1 the weld faces right into the Hoop Stress

psi7020375.0

625.14*180

This hoop stress will be directly applied to the weld

Of course the longitudinal stress is½ the hoop so it is 3510 psi.

We all know which weld is likely to go first.

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!

Now How Do We Check Out the State of Stress at 25 degrees?

3510 5265 7020

Mohr’s Circle doubles anglesSo if I want to look

3510 psi

7020 psiLooks like25 degreesDown from horizontal

50 ̊

1755=r

psir

tensionpsir

weld

centerweld

1344)50sin(*1755)50sin(*

)(4140)50cos(*17555265)50cos(*

Decision

Case 1

7020 psitension

4140 psitension

1344 psishear

Case 2

I suspect that if you pickCase 1 when strength isReally needed, that youWill need a lot of what isIn case 1.

Assignment 12

Problem 9.2-2 part a and bProblem 9.2-6 part a, b, and c

Into the Thick of ThingsWhat happens when the walls of the pressureVessel are thick enough that we can no longerCall them thin walled?

Things get “thick” when theWall thickness exceeds about1/20th of the diameter of thevessel

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 lengthLongitudinal stress and circumferential or hoop stress

Thickwall Means We Must Also Consider Radial Stress

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’sEquation leaves longitudinal stress out and then adds it bySuperposition later if needed)

(ok so it’s a bit Lame)

Lets Try One

We Get Some Simplifications in Lame’s Equations

For any radius r

But life gets better. We know the maximum stress willBe on the inside of the tank.

That’s Dandy

Apply to Our Problem

Lets do the hard one – if the pressureInside is 100 MPa

-100 Mpa

What does the negative number mean?

The material at the insideEdge is getting squeezed

What’s Happening at the Outside Edge?

This python hasSort or run out ofSqueeze.

Human InterestWhat happened to the radial stress between the inside and

outside?

It decays by a second order curve

Now for Circumferential or Tangential Stress

MPah

308)(

)(*100

05.007.007.005.0

22

22

max

What does the positive sign mean?

The tank is being pulled apart

What About Outer Edge Circumferential Stress?

No preset simplified formula – we have to plugIn for the outside edge

Well that Sucks

MPai

rrr

pio

ihoutside208

)(*200

)(*2

05.007.005.0

22

2

22

2

How does this result compare to a Thin walled vessel?

MPa1000005.0

05.0*100

Picking the largest value of t that stillQualifies as thin wall.

308 MPa inside

208 MPa outside

Example

Yipes!

I know my maximum stress is at the inside wall of the pipeBut none of these equations are for shear stress!

I don’t like theLooks of this!

Am I cooked?

Then We Remember Mohr’s Circle

Just because you don’t seeShear in your first measurements,Does not mean it is not there.

When material is in stress – all sorts ofCombinations of shear and tensileAnd compressive stresses becomePossible at different angles.

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

Quick Consideration of 3D Mohr’s Circle

Our hoop stress (tension)

Our radial stress (compression)

Some MohrPie!

Substituting

)1)(

)((*

2 22

22

max

rrrrpio

oii

A Bit of Plug and Chug

psipi

63952.12

2*4000

52.12

2*

Inside radius = 1.375 inOutside radius = 1.5 in

Gun Barrels are a Thick Wall Cylinder Application

A new kind of high powerAmmunition is called +P

It reaches higher pressures and sendsThe bullet out at higher speed.(But not all guns are made to handle+P ammunition)

What is the mode of failure in these cases?

What Happened Here?

What if the Pressure is Outside?

The radial stress maximumIs at the outside edge

The hoop stress maximumIs still on the inside.

Watch Out

Note the stress is compressive?

The foot looks fine to me

Lets Apply

Inspired by the concrete canoe competitionStudents at SIU decide to have aConcrete submarine competition.

Connie Concrete wants to decide howDeep her submarine can go.

Pressure outside the vessel increasesBy 0.44 psi for every foot of depth.

To actually crush Connie’s concrete itWill take 10,000 psi. The pipe is 5 ft inOutside Diameter and 6 inches thick.

Connie Crunches Limits

To do a radial crush will take 10,000 psiAt 0.44 psi per foot of depth it will takeAbout 22,700 ft of depth.

Because of end caps the submarine will alsoHave longitudinal stress. psip

o1900

)(*000,103027302

22

About 4,300 ft of depth

Now to Check Hoop StressMaximum on inside of cylinder

psipo

950*2

)(*000,10

3027302

22

This looks familiar – the hoop stressIs twice to longitudinal.

Well we can still make it to2,150 ft of depth.

Want a Ride in Connie’s Sub?

Can you think of anything that Connie andHer team might have missed?

Proposed testSubject forsubmarine

Does this Make You Worry?

I wonder if my concrete could fail inShear?

To actually crush the concrete takes 10,000 psi, but the specimen in aUniaxial compression test (like you ran) fails much sooner because theShear limit for Connie’s Concrete is 2,500 psi.

So How do You Get Max Shear?Pick our spot to check – our most critical hoop stress is on the inside of theConcrete cylinder.

Arrange our principle stresses in order from largest to smallest

1- Largest = hoop stress - compression2- longitudinal stress - compression (1/2 of hoop stress)3- radial stress – 0 on the inside edge of the concrete

HoopStress

LongitudinalStress

Radial Stress

τmax )(2

1max littlestbiggest

Implications

The largest hoop stress will can take without triggering shear failure is twiceThe 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 ourFirst test subject!

I think thisMight leak.

So Where Are We With the FE Book?

Bottom of Page 1

Here are the thick wallCylinder equationsWe have been talkingAbout.

They may be usefulOn class quizes –But they are unlikelySubjects for the FEExam itself

And on Page 2

These are the thin wall vesselEquations – they are more likely thanThick wall vessels, but still unlikelyOn the F.E.(But very much fair game for classQuizes).