How Do You Calculate the Stress Limit of a Spinning Flywheel?

• kyrosboi
In summary, the maximum stress on a spinning flywheel can be calculated using the equation ##\sigma = \rho\omega^2r^2##, where ##\rho## is the density of the material, ##\omega## is the angular velocity, and ##r## is the radius of the flywheel. This equation can be modified to include Poisson's ratio for more precise calculations. There is also the added consideration of shear stress and the use of Mohr's Circle to determine the maximum combined stress on an element in the wheel.
kyrosboi
TL;DR Summary
total mega pascals that can be applied to a spinning flywheel before it exceeds its tensile strength
I'm wonder how one could calculate the total mega pascals total mega pascals that can be applied to a spinning flywheel before it exceeds its tensile strength

example: a cast iron flywheel is spinning at 10,000rpm(Tangential velocity = 104.72) it has a radius of 100mm and a weight of 500g

Typical tensile yield strength of cast iron is 428 MPa
Typical ultimate tensile strength of cast iron is 497 MPa

The equation for centrifugal force is F = (m x v²) / r
m = 500g
v = 104.72m/s
r = 100mm
so according to these numbers Force = 54.83kN, Centrifugal Acceleration = 109.66kM/s²
F = (500 x 104.72²) / 100
F = (500 x 10966.28) / 100
F = (5483139.2) / 100
F = 54831.392

Welcome to PF.

kyrosboi said:
example: a cast iron flywheel is spinning at 10,000rpm(Tangential velocity = 104.72) it has a radius of 100mm and a weight of 500g
Thank goodness that it's not a big flywheel, but still, hopefully you will have it contained in a strong enclosure, right? Especially if you are testing this system to failure, you don't want to create any collateral damage in your lab.

kyrosboi said:
Summary: total mega pascals that can be applied to a spinning flywheel before it exceeds its tensile strength

I'm wonder how one could calculate the total mega pascals total mega pascals that can be applied to a spinning flywheel before it exceeds its tensile strength

example: a cast iron flywheel is spinning at 10,000rpm(Tangential velocity = 104.72) it has a radius of 100mm and a weight of 500g

Typical tensile yield strength of cast iron is 428 MPa
Typical ultimate tensile strength of cast iron is 497 MPa

The equation for centrifugal force is F = (m x v²) / r
m = 500g
v = 104.72m/s
r = 100mm
so according to these numbers Force = 54.83kN, Centrifugal Acceleration = 109.66kM/s²
F = (500 x 104.72²) / 100
F = (500 x 10966.28) / 100
F = (5483139.2) / 100
F = 54831.392
The first thing you are going to have to start doing is writing the units. Your units are a mess!
##F = \dfrac{(0.500 \text{ kg} )(104.72 \text{ m/s} )^2}{0.100 \text{ m} } = 54.83 \text{ kN}##

You got the right number, but you were lucky!

Now, this is the force on one end of the flywheel. To find the pressure exerted on the arm you will need to calculate P = F/A, ie. you need the force over the cross-sectional area of the arm.

-Dan

You are talking about what rpm can you spin a flywheel (cylinder - isotropic material) such that is self-destructs?

My gut is telling me it's not as simple as calculating ##F = \frac{m v^2}{R}##.

My first concern is that ##m##. I believe that is the force a test point mass ##m## would feel if it were on the flywheel at some arbitrary location. I don't think the mass of the flywheel is the ##m## in question here. Its mass is not a point mass at ##R## but its mass is instead distributed throughout the wheel as ## \propto R^2##.

I think if we look at an element of the wheel you are going to have both shear stresses ##\tau## and normal stresses ##\sigma## developing as a function of ##R##. I would then believe you would use Mohr's Circle to determine the maximum combined stress on an element in the wheel. I think it's going to be quite the task to get the proper stresses acting on an element.

Last edited:
Lnewqban, kyrosboi and topsquark
erobz said:
You are talking about what rpm can you spin a flywheel (cylinder - isotropic material) such that is self-destructs?

My gut is telling me it's not as simple as calculating ##F = \frac{m v^2}{R}##.

My first concern is that ##m##. I believe that is the force a test point mass ##m## would feel if it were on the flywheel at some arbitrary location. I don't think the mass of the flywheel is the ##m## in question here. Its mass is not a point mass at ##R## but its mass is instead distributed throughout the wheel as ## \propto R^2##.

I think if we look at an element of the wheel you are going to have both shear stresses ##\tau## and normal stresses ##\sigma## developing as a function of ##R##. I would then believe you would use Mohr's Circle to determine the maximum combined stress on an element in the wheel. I think it's going to be quite the task to get the proper stresses acting on an element.
this was really helpful, exactly the kind of info I needed thanks

berkeman and erobz
And it gets even more complicated. There are many designs of flywheels. The old fashioned low speed flywheels with spokes, solid disks, tapered disks, and constant stress designs. The stress in each different design is calculated differently.

Attaching flywheels to shafts becomes challenging at high speeds. A high speed disk flywheel can expand enough to come loose on the shaft. Careful calculations are required.

High speed flywheels have a lot of kinetic energy. If the flywheel fails, that energy is released in the form of metal chunks flying in all directions at high speed. High speed flywheels require properly designed containment systems.

I fixed the title to better describe this thread.

Lnewqban, topsquark and erobz
Even though the mass is only 500g, the flying fragments after a breakup could certainly be lethal. If you plan to actually do this, you need extraordinary safety precautions.

Are you planning to do it physically, or is this just a though exercise? Will it be done in a supervised laboratory?

Welcome!
Other factors to consider are fabrication deviations from ideal, such as non-homogeneous casting of the material, excentricities and lack of perpendicularity of machined surfaces.
All of the above contribute to static and dynamic lack of perfect balance, inducing additional loads and vibrations, which are difficult to calculate and are commonly covered by a safety factor.

kyrosboi, topsquark and jack action
anorlunda said:
Even though the mass is only 500g, the flying fragments after a breakup could certainly be lethal. If you plan to actually do this, you need extraordinary safety precautions.

Are you planning to do it physically, or is this just a though exercise? Will it be done in a supervised laboratory?
this is mainly just a thought experiment, basically I'm creating a roots supercharger, and want to know roughly the upper-limits of the material, not going use the max rpm but the project gave me the question. the example flywheel was completely hypothetical

Lnewqban
What's a roots supercharger?

kyrosboi said:
this is mainly just a thought experiment, basically I'm creating a roots supercharger, and want to know roughly the upper-limits of the material, ...
In that case, it is important to consider the high temperatures in which the material of the lobes and gears are going to be working.
Higher inter-teeth space of poor quality gears introduce much hitting and grinding of the lobes, as those don’t use lubrication.

berkeman said:
What's a roots supercharger?
Wikipedia

Lnewqban and berkeman
berkeman said:
What's a roots supercharger?
Something YOU should have on a dirt bike? :)

Tom.G, kyrosboi and berkeman
Averagesupernova said:
Something YOU should have on a dirt bike? :)
making a 2l roots supercharger for my 2004 2l inline 4 VW Jetta... horse power bout to double

Lnewqban said:
In that case, it is important to consider the high temperatures in which the material of the lobes and gears are going to be working.
Higher inter-teeth space of poor quality gears introduce much hitting and grinding of the lobes, as those don’t use lubrication.
yeah already calculated that, just had to figure out how much more it would expand because of centrifugal force

kyrosboi said:
making a 2l roots supercharger for my 2004 2l inline 4 VW Jetta... horse power bout to double
How will you drive the two shafts of the supercharger?

kyrosboi said:
making a 2l roots supercharger for my 2004 2l inline 4 VW Jetta... horse power bout to double
How are you going to reprogram the ECU as part of this change?

1. What is a spinning flywheel and how does it relate to stress?

A spinning flywheel is a mechanical device that stores rotational energy. It is often used in machines to maintain a constant speed or to smooth out fluctuations in power. The faster a flywheel spins, the more energy it stores, which can result in increased stress on the flywheel's components.

2. What factors contribute to stress in a spinning flywheel?

There are several factors that can contribute to stress in a spinning flywheel, including the speed of rotation, the weight and balance of the flywheel, and the materials used in its construction. Other factors such as friction and external forces can also impact the level of stress on a flywheel.

3. How does stress affect the performance of a spinning flywheel?

High levels of stress in a spinning flywheel can cause it to deform or even fail, resulting in decreased performance or complete breakdown of the machine it is powering. It can also lead to increased energy consumption and reduced efficiency.

4. How can stress in a spinning flywheel be measured and monitored?

Stress in a spinning flywheel can be measured using strain gauges, which detect changes in the shape or deformation of the flywheel. This data can then be monitored and analyzed to determine the level of stress and make adjustments to prevent potential failures.

5. What are some ways to reduce stress in a spinning flywheel?

One way to reduce stress in a spinning flywheel is to decrease the speed of rotation, which can be achieved by adding more flywheels or using a larger flywheel. Another method is to improve the balance and weight distribution of the flywheel, as well as using stronger and more durable materials in its construction.