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Centrifugal pressure, hoop stress, thick-walled cylinder
3

Centrifugal pressure, hoop stress, thick-walled cylinder

Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
Greetings,

I've been tasked with finding the stresses on a cylindrical part, to be rotated at high speed.  I don't need a precise answer, but a worst case scenario is a good place to start.

I've found a source (patent 5015940) that gives an equation for centrifugal pressure for a thin-walled cylinder, as well as a few sources that provide hoop stress for a thin-walled rotating cylinder.

However, wall thickness is 35% of radius, and I haven't had luck deducing a formula that combines pressure/force from rotation with a thick-walled body.

Intuition tells me that stresses will be greatest on the outer surface, and the stresses of a thick-walled cylinder should be less than or equal to those of a thin-walled equivalent.  Are these assumptions correct?  If not, how should I proceed?

Thanks!  

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Roark's should have formulas for both a spinning cylinder with a hole and a thick walled cylinder. I don't have my copy in front of me, though, so in its absence here is a link that covers spinning cylinders:
http://www.codecogs.com/reference/engineering/materials/rotating_discs_and_cylinders.php

And here is a link covering thick-walled cylinders:
http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html

You'll want to sum up the directional stresses and then you'll probably want to compute the von Mises stress.

Google: a miracle of the modern age. smile

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

RyreInc,

The main stresses of a spinning cylinder are in the radial and hoop directions.  The combined stress is actually highest at the inner diameter and lowest at the outer diameter.  The thickness definitley affects the stresses and a thin wall assumption will give you inacurrate results.  The radial stress is highest at a radius equal to R=(Ro*Ri)^0.5 while the stresses are zero at the inner and outer surfaces.  The hoop stress is extremely high at the inner surface and relatively low on the outside.   

Nick Flores
Mechanical Engineer

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Radial stress = A – B/r^2 – [(3 – ν)/8]*m*ω^2*r^2

Hoop stress = A + B/r^2 – [(1 + 3ν)/8]*m*ω^2*r^2

Where:

A and B are constants that be computed with the boundary conditions on internal and external radius.

ν = Poisson modulus

ω = angular velocità

m = cylinder density

r = baricentric radius
 

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
nanobot, thanks for the description of stresses.  I had just solved for hoop and radial stresses but was confused that the greatest stress was on the inner radius, so its reassuring that my results make sense.

Thanks everyone else too for the links and equations.  the codecogs site has what I need, although it doesn't seem to define what g is...

I did stumble upon a source 9http://www.sussex.ac.uk/engineering/internal/documents/sm-lecture-22) that has similar info to the codecogs site, so between the two I was able to solve the problem.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

"it doesn't seem to define what g is" ... that's a joke comment, no?

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

what codecogs is hiding is rho is weight density (not mass density) ... hence divide by "g" for mass

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
No joke, g is never defined.  As far as I can tell it is the inverse of the cylinder length, since that would create a volume.  The sussex.ac.uk link has a similar equation that uses t instead of 1/g.

The only thing I ever recall seeing g as is gravitational acceleration, and G as conductance, neither of which make sense.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

"...neither of which make sense" ????????????

Well conductance has nothing to do with this problem, but gravity acceleration has (check again rb1957 last post 8 Apr 11 9:43)

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
I'm not so sure about that, dimensional analysis shows that g should have units of 1/m to get kg*m/s^2 (=Newtons).

Plus, having g as an acceleration implies an assumption about the orientation of the object, among other things.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

you've seen the expression for radial force (m*r*w^2) ...
codecogs has this as m = rho*... see their equation (4)
m = mass
rho is weight density (* volume = mass)
g is accel due to gravity (i guess we, who use these expressions everyday, don't need to recall this ... no slight intended)

note also codecogs is "per unit thickness"

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
I understand the equation, the author has rearranged the centripetal force equation F=m*r*w^2

units:

rho: kg*m^-3
r: m
dtheta: (none)
dr: m
r: m
w^2: s^-2
g: ?

rho*r*dtheta*dr*r*w^2: kg*m^-3*m*m*m*s^2 = kg/s^2
we need N=kg*m*s^2, therefore g = m^-1

or if N/m is needed then g is unitless.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Do you know the difference between kg (force) and kg (mass) ?

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
I've heard of lbm and lbf, but kg is always mass (there is kgf, kilogram-force).  Please find an example of kg used as a force, otherwise your claim is groundless.

Regardless, the kg in the dimensional analysis comes from density, which is mass/volume.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

the density in the equation is weight density.

if you have mass density, you don't need the "g" term

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
Thank you rb1957, this would resolve the discrepancy.  

I believe it justifies my position as well, since the site does not mention "weight density", only "density", and also because weight density is not part of the SI.  Additionally, the site claims to solve for (total) centripetal force, then uses the exact same equation directly below for force per unit thickness.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Try this Matlab/Octave code:

CODE

function pipe
clear;clc;close all;
R0 = 1.00;    % Internal Radius (inch)
R  = 1.25;    % External Radius (inch)
r  = linspace(R0,R);

q  = 2500;    % Pressure (psi)
delta = 0.5;    % Weight (lb/in^3)
omega = 5000;    % Angular Velocity (rad/sec)

g  = 32.2;    % Gravitational Acceleration (ft/sec^2)
g  = g.*12;    % Gravitational Acceleration (in/sec^2)

neu = 0.3;    % Poisson's ratio

[SR1 ST1] = pressurized_pipe(r,R0,R,q);
[SR2 ST2] = spinning_pipe(r,R0,R,g,delta,omega,neu);

SR = SR1 + SR2;
ST = ST1 + ST2;

h = figure;
subplot(3,1,1);
 hold on;
 title('Hoop Stress');
 plot(r,ST1,'ro','MarkerSize',2);
 plot(r,ST2,'mo','MarkerSize',2);
 plot(r,ST,'k','LineWidth',2);
 legend('Pressurized Pipe','Rotating Pipe','Combined Stress','Location','NorthEastOutside');
 grid on;
 set(gca,'XTickLabel',' ');
subplot(3,1,2);
  hold on;
  title('Radial Stress');
  plot(r,SR1,'ro','MarkerSize',2);
  plot(r,SR2,'mo','MarkerSize',2);
  plot(r,SR,'k','LineWidth',2);
  ylabel('Mechanical Stress (psi)');
  legend('Pressurized Pipe','Rotating Pipe','Combined Stress','Location','NorthEastOutside');
  grid on;
  set(gca,'XTickLabel',' ');

subplot(3,1,3);
 plot(r,sqrt(.5*(ST.^2+SR.^2+(ST-SR).^2)),'k','LineWidth',2);
 title('Total Stress');
 xlabel('Radius (in)');
 legend('Von Mises Stress', 'Location', 'NorthEastOutside');
 grid on;

filename = ['pipestress.png'];
print(h,filename,'-dpng');

function [sig_rad, sig_tan] = pressurized_pipe(r,b,a,q)
% From Roark's Formulas for Stress and Strain (Seventh Edition)
% (Sec 13.8, pg 683)
% Pipe with pressure differential unrestrained axially (ends not capped)
sig_rad = -q.*b.^2.*(a.^2 - r.^2)./r.^2./(a.^2 - b.^2);
sig_tan =  q.*b.^2.*(a.^2 + r.^2)./r.^2./(a.^2 - b.^2);

function [sig_rad, sig_tan] = spinning_pipe(r,R0,R,g,delta,omega,neu)
% From Roark's Formulas for Stress and Strain (Seventh Edition)
% (Sec 16.2, pg 746)
% Spinning disk or cylinder with center hole

sig_rad = ((3+neu)./8).*(delta.*omega.^2./g).*(R.^2 + R0.^2 - R.^2*R0.^2./r.^2 - r.^2);
sig_tan = (1./8).*(delta.*omega.^2./g).*((3+neu).*(R.^2 + R0.^2 + R.^2*R0.^2./r.^2) - (1+3.*neu).*r.^2);

No warranties, exchanges or refunds.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

RyreInc, the imperial system uses "slugs" as a unit of mass.  The term "lbm" and "lbf" denotes pounds mass and pounds force.  This is why you need the acceleration due to gravity or "g" in the mathematical expression.

Metric gets around this, kg is always mass, N for newtons is always force.  But it is common to give a force as a unit of mass noting the "g" is needed to convert between the two.

A good example of this is with Springs and stated measure of compression by the various manufacturers.  I suppose they rely on the educated audience to understand "kg X g = N" type thing.

But yeah, imperial and metric have that specific gravity thing.  You're running into this rho=0.283 lb/in^3 thing for density of steel, density is mass over volume so you need to divide by "g" in order to correct for "slugs/in^3".

Work the problem through in metric, convert back to imperial, problem solved.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
Yep, the source of all confusion was weight density vs. mass density, and the codecogs reference implicitly working in the imperial system.

I always convert to SI and do all calculations in that system, so I hadn't considered the possibility of non-standard units.

Summary:  Imperial system blows!

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

the imperial system is fine, if you know what you're doing.

the equation doesn't implicitly assume imperial units.  it does assume weight density (which is abit odd for a text book), but you could deduce that from the equation mass = volume*density/g

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Doesn't matter what units you work in, you have to keep everything straight. I prefer to report stress in stones/furlong^2. smile

Aerospace design is still done (with the exception of NASA) almost entirely in imperial units. The reason is the expense of changing unit systems. You force tons of subcomponents into obsolescence (nuts & bolts, etc). All of your empirical correlations developed from years of engine building have to be converted (with the chance of error, mind you). Not to mention, all of your engineers have a feel for a particular unit system: Experience might tell me the approximate h-value along a turbine blade is in imperial units, for example. For SI units, you have to convert. It makes reviewing designs and analysis difficult for chief engineers who've spent a career learning those sorts of things. You can't exactly stop a design review meeting to run a conversion.

That's not to say that change can't happen. There just has to be a good reason for it. If, for example, foreign sources charged a premium when prints showed up in inches vs mm, engine manufacturers would think long and hard about making the switch.

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
Indeed, the real problem was that I had never heard of weight density before!  

Back on topic: I'm expanding this problem to a cylinder made from two different materials (steel shaft, surrounded by impregnated nylon).  My boundary conditions are As, Bs, An, and Bn (A & B for steel and nylon respectively).  I need to define constraints on the system in order to solve for the four unknowns, but so far I can only think of three:

Bs=0, otherwise stresses are infinity at r=0.
radial stress at outside radius = 0.
radial stresses of steel and nylon are equal at boundary.

What am I missing?

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

"Aerospace design is still done (with the exception of NASA) almost entirely in imperial units." ... on this side of 'the pond'.  on the 'ther side they've managed to convert to metric

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

Here is the formulas used in a program that covers the whole
gambit, thermal, spin, interference, and axial  for shrink fits.
This code was written for a program call Symphony but with a few Minor corrections it can run in Excel.
In your case set the inner diameter to 0.0 or a very, very small number.

http://my.engineering.com/pg/file/unclesyd/read/444170/spin

http://my.engineering.com/pg/file/unclesyd/read/444171/spin2
 

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
unclesyd,

Thank you for these attachments; the form of your equations are more generalized and therefore more useful.

However, after using these formulas I get the same answers as when I treat each cylinder separately.  The radial stresses at the middle radius (boundary between materials) are zero, and it seems like this wouldn't be the case...

Also, for F2 of outer cylinder, you continue to use Poisson's ratio for the inner cylinder.  A typo I assume?

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

It does treat the cylinder separately as there is nothing to impose any stress, as in this case the spin.  As long as the concentric cylinder are at zero clearance or a cylinder standing alone the stress are calculated at the inner and outer diameter or can be evaluated at in point between the two boundaries.

I'm trying to get all the essential equations to where you can download them.    
 

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

RyreInc,

Further to the boundary conditions necessary to solve your problem you have to consider that you do realize a shrink fit and a pressure generates at the interface between the two cylinders. The inner cylinder is subjected to an external pressure and the outer cylinder is subjected to an internal pressure. Now the radial displacement comes into play, that is what you lack in your boundary consitions.
You have to consider that the interference Δ, necessary to realize the shrink fit, is equal to reduction of the external radius of the inner cylinder  plus the increase of the internal radius of the outer cylinder.
 

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

(OP)
ione,

Thanks, that is indeed the information I wasn't considering.  I did think it had something to do with this, but hadn't been able to resolve the particulars.  You've defined it quite nicely.

This is where the problem gets tricky, since the nylon was injection molded over the steel, and I don't know the details of this aspect.  I might have to call it good enough at this point!

RE: Centrifugal pressure, hoop stress, thick-walled cylinder

I would try to use the Nylon properties in the Spin program and look at the results. i would take the boundary conditions as if the Nylon was pushed on withe a very small interference.  You might also plug your data into the [a} and [c} equations on page 21 of the essential equations.

 http://www.tamshell.com/mds-nylon6-30gf.htm

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