martindevans
Is there a standard equation used to determine the stress at the edge of a rotating wheel?
kaneda
Guest_Martindevans

thanks for that, however I want stress at the rim of a moving whell, where obviously the stress is different to stress at the rim of a simply spinning wheel.
boit
QUOTE (martindevans+Oct 27 2006, 05:26 PM)
Is there a standard equation used to determine the stress at the edge of a rotating wheel?

QUOTE
http://www.convertit.com/Board/view.asp?mo...forumID=&catID=

Unfortunately that link provided by kaneda does not work. Anybody has an answer? People seem to forget a spinning wheel that is fixed or suspended in space if fundamentally different from a moving, rotating, wheel. Just take an example of a ball spinning on your finger and one kicked and flying in air (not in vacuum). Most remarkable difference is on a car's wheel on the road and one on a dynomometer.
flipmastacash
I think the force of a spinning wheel would be the same as that of a wheel that is both spinning and translating as long as the velocity is constant. Only if it is accelerating will there be an additional stress.
flipmastacash

The max hoop stress in a rotating cylinder is:

p_(y(max))=(ρ ω^2)/4∙((3m-2)(R_1)^2/(m-1) +(m-2)(R_2)^2/(m-1))

Where:
ρ= mass density of wheel rim
m=inverse of poisson's ratio (1/v)

Morley. Strength of Materials. Art. 127.

or

Timoshenko. Theory of Elasticity. Art. 119
boit
QUOTE (flipmastacash+Aug 17 2010, 07:45 PM)
The max hoop stress in a rotating cylinder is:

p_(y(max))=(ρ ω^2)/4∙((3m-2)(R_1)^2/(m-1) +(m-2)(R_2)^2/(m-1))

Where:
ρ= mass density of wheel rim
m=inverse of poisson's ratio (1/v)

Morley. Strength of Materials. Art. 127.

or

Timoshenko. Theory of Elasticity. Art. 119

How could they be the same? Even when a vehicle has constant speed, a point on a translating wheel is always in a state of positive and negative linear acceleration. A wheel merely spinning in space has only arbitrary direction hence the only momentum it has is angular as you rightly pointed out. The speed on a point on that spinning wheel never comes to zero. There is no rolling resistance either. Only the driver of the vehicle may be tempt to view the wheel on a moving vehicle as one merely spinning but we know there is a difference. Am suprised this has on bearing on stress on the wheel. What's the important effect then? Just rolling resistance? Can we modify that formula to get anything?
flipmastacash
Consider a tire rolling without slip and a tire spinning in place. When a point on the rotating tire is at the top of its arc it is moving with velocity r*w+v where r is the radius, w is the angular acceleration, and v is the velocity of the center of the wheel (or the car). At the bottom of its arc the velocity is v-r*w. So the change in speed is 2*r*w. On the tire spinning in place, the velocity at the top is r*w and the velocity at the bottom is -r*w. Thus the change in velocity is also 2*r*w. Because acceleration is a change in velocity, and the change in velocity is the same in the two wheels, we come to the conclusion that the acceleration (and therefore the force) is the same.

The only thing that would be different in a wheel on a car driving with constant velocity and a wheel on a dyno is the normal force on the tire of the car. This would be a constant force perpendicular to the ground.
boit
I think I have to agree with your explanation. Only the way I thought was on a moving wheel we were to assume twice the speed a similar stationary wheel makes for a point on top of the wheel that is in linear motion. This is because the OP did not ask about stress on the whole wheel per se but on the edge. But since he never said the wheel is in contact with the ground i don't see why it has to be any different from one just spinning fixed in space relative to a stationary observer.
dhcracker
Are we considering a wheel that is rotating in space or rolling along the road? A wheel spinning freely only has to overcome the centrifugal force of its own weight, whereas one with a load on it is subject to different force equations because the force acting against it isn't uniform and homogenous, there will be more stress at certain points in the wheel, less stress on the ground where the weight of the car is holding against centrifugal force. However this also would put slightly more stress on the rest of the wheel with the most force being at the left and right edges.

Also the greater force would be on the inner diameter of the wheel which has less area, these cracks from stress related to rotation always go from the inside out when caused by only the rotation.

If you would like an example I can supply one, I have lots of equations relating to stress and tensil strength, shear strength and so on. I have used these on my own dynamos.
boit
Thanks alot dhcracker. Just supply a formula for calculating stress at the edge of said wheel. We can start with stress at 9 and 3 o'clock.
dhcracker
I'll just spell it out for you, I need a math thing so I can actually type equations out I haven't done that yet normally I just perform calculations, so you'll have to just take the formula from plain english.

First you use the formula posted above and get the hoop stress, then you just add the force from the weight of the car that is loaded on this wheel and divide it by the area of the wheel.. you add this to the hoop stress (though its not loaded evenly over the rim this is close enough to get a rough estimate unless you are planning on reaching the sound barrier or something).. howver there is also a force placed on a rim by the tire itself the inflated PSI is pressing against the inside of the rim, on the flange its pressing that out parallel to the axis however over the rest of the rim it actually presses against the hoop stress, so over the area of the inner facing rim you could subtract the tire pressure from the stress and add some stress over the area of the flange.

Talk about details, but normally there is no need to do all this unless you plan on running your car over 200 mph, then simply get a carbon fiber rim or aircraft alloy. There is no way you could shear that stuff, it would help if I knew exactly what your calculating this for?. Also what kind of rim? Are you meaning a flywheel rim or cylinder? The two are very very different figures and formulas.
boit
Let's put the wheel aside for a moment. Let me show where my reasoning is heading (a bit off what the OP wanted, it's him that wanted the formula remember?)
This is my scenario: A merry-go-round is on board a Cruise ship. The tangential speed of the merry-go-round equals the ship's velocity. For illustration purpose let's say
b]Merry-go-round is rotating clockwise viewed from above.
This thought experiments tells me that:
*The seats arriving at position 9 O'clock will hang perpendicularly down.
*The seats arriving at position 3 O'clock will be thrown up higher than usual
*The seats arriving at 6 & 12 O'clock will be thrown no higher than usual
NB: Usual is when ship is stationary.

If I understand Flipmastacash correctly, there will not be additional stress at 3 O'clock. The additional amplitude is accounted for by the ship's motion. Why I agreed with him is cause a vehicle moving at say a constant speed of 100 m.p.h is not experiencing any more force than one traveling at 50 mph (assuming no air resistance). In this respect the tug on the chain holding suspending the seat is not double or magnitude of force increased. This is cause it is not centrifugal force per se causing this. It is the linear speed of the sheep which can be subtracted from the equation. The end result is there is no additional tension (centripetal pull) on the chain.
Did I understand you correctly in view of what Ive said?
boit
A simpler explanation why the wouldn't be additional stress as long as the vehicle is not accelerating is: assume you're on a train doing 50 mph. You throw a rock dead ahead at 66 mph. You don't feel more stress on your shoulder as you would if you threw the rock at 116 mph at rest. Now back to my merry-go-sound. Maybe the highest swing will be at 12 o'clock and lowest at 6 o'clock. Also i should have said the rotation is counter-clockwise. Also I wrote sheep instead of ship. Too late to edit the earlier post.
dhcracker
QUOTE (boit+Aug 22 2010, 01:57 PM)
Let's put the wheel aside for a moment. Let me show where my reasoning is heading (a bit off what the OP wanted, it's him that wanted the formula remember?)
This is my scenario: A merry-go-round is on board a Cruise ship. The tangential speed of the merry-go-round equals the ship's velocity. For illustration purpose let's say
b]Merry-go-round is rotating clockwise viewed from above.
This thought experiments tells me that:
*The seats arriving at position 9 O'clock will hang perpendicularly down.
*The seats arriving at position 3 O'clock will be thrown up higher than usual
*The seats arriving at 6 & 12 O'clock will be thrown no higher than usual
NB: Usual is when ship is stationary.

If I understand Flipmastacash correctly, there will not be additional stress at 3 O'clock. The additional amplitude is accounted for by the ship's motion. Why I agreed with him is cause a vehicle moving at say a constant speed of 100 m.p.h is not experiencing any more force than one traveling at 50 mph (assuming no air resistance). In this respect the tug on the chain holding suspending the seat is not double or magnitude of force increased. This is cause it is not centrifugal force per se causing this. It is the linear speed of the sheep which can be subtracted from the equation. The end result is there is no additional tension (centripetal pull) on the chain.
Did I understand you correctly in view of what Ive said?

you could use a flywheel formula though you don't need a cylinder formula a merry go round has a floor and a ceiling.
boit
QUOTE (dhcracker+Aug 24 2010, 12:47 AM)

you could use a flywheel formula though you don't need a cylinder formula a merry go round has a floor and a ceiling.

What if I use a helicopter's rotar in my example? The blades get equal stress while hovering as well as when the chopper is moving, right? Dissimilarity of flight has little effect on stress other than on the vertical axis. If said rotar was not curved (air foil) there will be absolutely on difference. Someones equivalance principle dictates that. I still feel somehow my above example holds true. I don't know where i could be wrong. Position 9 o'clock will behave as a stationary axle would, right? I feel dizzy already.
boit
I feel i've to correct myself again. They may be indistinguishable after all. Didn't Einstein say so. Inertial frame and all.
boit
QUOTE (flipmastacash+Aug 17 2010, 05:59 PM)
I think the force of a spinning wheel would be the same as that of a wheel that is both spinning and translating as long as the velocity is constant. Only if it is accelerating will there be an additional stress.

Does this also apply to also apply to Helicopters roters in normal flight. Or sice the roters accelerates air downwards and a little on the leading edge, this also means it is accelerating?
dhcracker
QUOTE (boit+Aug 24 2010, 02:32 AM)
What if I use a helicopter's rotar in my example? The blades get equal stress while hovering as well as when the chopper is moving, right? Dissimilarity of flight has little effect on stress other than on the vertical axis. If said rotar was not curved (air foil) there will be absolutely on difference. Someones equivalance principle dictates that. I still feel somehow my above example holds true. I don't know where i could be wrong. Position 9 o'clock will behave as a stationary axle would, right? I feel dizzy already.

WEll they are under way more stress in flight because remember air is still a fluid really. As the rpm increases and attack angle there are more forces applied. You have alot of tension and shear stress acting at the hub of the rotor, not so much stress at the end of it it will all be up at the axis from the metal trying to pull itself apart. The blades angle of attack is modified during its cycle thats how it pitches forward and moves in the X and Y dimensions pretty much. Moving forward you would make the pressure at the front drop a hair then equalize it to hole the angle and move forward.

I don't like choppers, too complex.. complexity leads to failures.. Planes now I like those. Ultralights rock. Not so much calculus lol. You don't even want to try to work out calculations on helicopter rotors... but I make you a deal I'll give you the equations if you perform the calculations lol

read up here this is cool

boit
[EIz/m Ω 2R4]. I'll not even try. I flopped maths real bad. Just how do you articulate that equation? Seems it requires more courage to board a helicopter than a fixed wing plane. OK, let's try Auto-gyro then. Hope it is less complex with equally less maths
dhcracker
QUOTE (boit+Aug 29 2010, 08:30 AM)
[EIz/m Ω 2R4]. I'll not even try. I flopped maths real bad. Just how do you articulate that equation? Seems it requires more courage to board a helicopter than a fixed wing plane. OK, let's try Auto-gyro then. Hope it is less complex with equally less maths

Eh I have no idea I haven't seen that equation. The equations I have come out of my book on rotor dynamics and its brainstormingly complex and carries over several pages. It would take me all day to write them down and define them.

Helicopter blades are generaly used in an equation of motion where simply a sum of forces show its efficiency and strength. Frankly I wouldn't dare use a generalization equation on such a thing, too dangerous and I haven't ever seen a simple form to calculate all the stresses and resonances and motions. That may be a simple equation for a rectangular rotor or something like that.
boit
QUOTE (flipmastacash+Aug 17 2010, 10:45 PM)
The max hoop stress in a rotating cylinder is:

p_(y(max))=(ρ ω^2)/4∙((3m-2)(R_1)^2/(m-1) +(m-2)(R_2)^2/(m-1))

Where:
ρ= mass density of wheel rim
m=inverse of poisson's ratio (1/v)