Centrifugal pressure, hoop stress, thick-walled cylinder 3

[RyreInc]

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!

 

[flash3780]

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.

 

[nanobot29]

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

 

[ione]

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

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

Where:

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

? = Poisson modulus

? = angular velocità

m = cylinder density

r = baricentric radius

 

[RyreInc]

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 9

http://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.

 

[rb1957]

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

 

[rb1957]

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

 

[RyreInc]

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.

 

[ione]

"...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)

 

[RyreInc]

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.

 

[rb1957]

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"

 

[RyreInc]

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.

 

[ione]

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

 

[RyreInc]

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.

 

[rb1957]

the density in the equation is weight density.

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

 

[RyreInc]

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.

 

[flash3780]

Try this Matlab/Octave 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.

 

[Cockroach]

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

 

[RyreInc]

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!

 

[rb1957]

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