Transcript Thin Walled Pressurized Tanks

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 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
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.
Because that Stress is Uniform Around the
Circle We Call it a Hoop Stress
The Magnitude of the Hoop Stress
r
P
The resisting
Area is the
Thickness of
The metal
2t
The Force must be
2*r*P
Force = 2*r*P
Therefore the Hoop Stress Is
F 2*r * P r * P
 

A
2*t
t
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
Force must be P*π*r2
Resisting area must be
2*π*r*t
Thickness
t
P
Longitudinal stress must be
F P * * r
P*r
 

A 2 * * r * t 2 * t
2
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.
Working it Out.
 *t
60,000* 0.4375
 873psi
pmax  r  61
(  0.4375)
2
With a little algebra
It will split down the length when
It 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 of
our connections – rather than tear through the material
itself.
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.
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
3/8th inch steel plate and pressurized to 180 psi. What kinds of stresses will we
Be putting on those welds?
Case 1 the weld faces right into the
Hoop Stress
180 *14.625

 7020 psi
0.375
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.
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?
3510 psi
5265
7020 psi
Mohr’s Circle to the Rescue!
Now How Do We Check Out the
State of Stress at 25 degrees?
3510
5265
50̊


weld
weld
7020
1755=r
Mohr’s Circle doubles angles
So if I want to look
  center  r * cos(50)  5265 1755* cos(50)  4140psi(tension)
 r * sin(50)  1755* sin(50)  1344psi
Looks like
25 degrees
Down from horizontal
7020 psi
3510 psi
Decision
7020 psi
tension
Case 1
1344 psi
shear
4140 psi
tension
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.
Assignment 12
Problem 9.2-2 part a and b
Problem 9.2-6 part a, b, and c
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
1/20th of the diameter of the
vessel
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
Thickwall Means We Must Also
Consider Radial Stress
Derivation is Tedious
(And therefore skipped)
Lame’s Equations
(ok so it’s a bit Lame)
(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)
Lets Try One
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
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
What’s Happening at the
Outside Edge?
This python has
Sort or run out of
Squeeze.
Human Interest
What happened to the radial stress between the inside and
outside?
It decays by a second order curve
Now for Circumferential or
Tangential Stress

(0.05  0.07 )
2
h max
 100*
2
(0.07  0.05 )
2
2
 308MPa
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 plug
In for the outside edge
Well that Sucks
2
 houtside  2 * p
i
ri
2
(r o  r i )
2
2
 200*
0.05
 208MPa
(0.07  0.05 )
2
2
How does this result compare to
a Thin walled vessel?
Picking the largest value of t that still
Qualifies as thin wall.
100 * 0.05

 1000 MPa
0.005
308 MPa inside
208 MPa outside
Example
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?
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.
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 Mohr
Pie!
Substituting

 max
p
i
2
(r i  r o )
2
*(
2
(r o  r i )
2
2
 1)
A Bit of Plug and Chug
 *2
4000 * 2
pi  12.52  12.52  639 psi
Inside radius = 1.375 in
Outside radius = 1.5 in
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?
What Happened Here?
What if the Pressure is Outside?
The radial stress maximum
Is at the outside edge
The hoop stress maximum
Is still on the inside.
Watch Out
Note the stress is compressive?
The foot looks fine to me
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.
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.
(30  27 )
2
p
o
 10,000*
2
2
30
About 4,300 ft of depth
 1900psi
Now to Check Hoop Stress
Maximum on inside of cylinder
(30  27 )
2
p
o
 10,000*
2
2
2 * 30
 950psi
This looks familiar – the hoop stress
Is twice to longitudinal.
Well we can still make it to
2,150 ft of depth.
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
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.
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
1
τmax
 max  2 ( biggest   littlest)
Radial Stress
Hoop
Stress
Longitudinal
Stress
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.
I think this
Might leak.
Oh Nut’s – We just lost our
First test subject!
So Where Are We With the FE
Book?
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
Bottom of Page 1
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).