In first part of the flywheel design calculation tutorial example, we saw about calculating required mass moment of inertia for a particular application.

Flywheel design doesn’t stop with that. Its size, shape and material density / mass and maximum stresses are also to be taken into account while designing a flywheel. Which we will discuss in this part.

R1 = Flywheel inner radius

R2 = Flywheel outer radius

**Step-1: Material selection**

** **The material of flywheel on most applications will be either cast iron or cast steel.

The Density of the materials (* d*) is as follows.

**Cast iron – 7250 kg/m**^{3 }

**Cast steel – 7800 kg/m**^{3}

**Step-2: Calculation of the flywheel mass**

We already know how to get the mass moment of inertia of a flywheel (I) required for a specific application.

Another equation for calculating the mass moment of inertial (I) is:

I = m* K^{2}

Or, I = m* (R1^{2} + R2^{2})/2………………..eq.2.1

R2 = 400 mm (assumption)

R2= 1.5 * R1 (assumption)………………eq.2.2

Where,

I = Mass Moment of Inertia of flywheel

m = Mass of the flywheel

K = Radius of gyration

R1 = Inner radius of flywheel

R2 = Outer radius of flywheel

From **eq.2.1 **mass of the flywheel can be calculated, how? Will discuss with the example later.

**Step-3: Size and shape calculation**

m = π * (R2^{2} – R1^{2} ) * t * d ……………………..eq.2.3

Where,

m = mass (will be obtained from **step-2)**

d = Density (will be obtained from **Step-1)**

t = Thickness of the flywheel

From the **eq.2.2, 2.3 **the cross section, size and thickness of the flywheel can be calculated.

**Step-4: Stresses in the flywheel**

Maximum stress (tangential) of a flywheel is given by:

Ϭ_{t-max} = (d/4)*(ω^{2})*[(3+v)*R2^{2} + (1-v)* R1^{2}]…………..eq.2.4

FOS = ω_{yield }/ ω ………….eq.2.5

Where,

Ϭ_{t-max} = Max internal stress in the flywheel [ N/m^{2} ]

d = Density of flywheel material [kg /m^{3}]

ω = Angular velocity of the flywheel [rad/sec]

v = Poisson ratio of the flywheel material

FOS = Design factor of safety

ω_{yield} = Possible angular velocity when maximum internal stress equals to yield stress of the material

**Example**

Let’s take the same example discussed in Part-1 and find out the mass, size, shape and design parameters as per the equations discussed above.

**Solution:**

**Step-1: density**

We will select cast steel for our application, so

Density, **d = 7800 kg/m**^{3}

**Step-2: Mass **

From Part-1,

I = 3.43 kg.m^{2}

Using **eq.2.1, eq.2.2 and using R1=0.4 m**

Mass of flywheel, **m= 29.68 kg**

**Step-3: Size & shape**

Outer radius of the flywheel, **R2 = 400 mm** (given)

From **eq.2.2,**

Inner radius of flywheel**, R1= 266.67 mm**

From** eq.2.3,**

Thickness of flywheel, **t=13.62 mm**

**Step-4: Stresses & FOS**

From **eq.2.4 **

Along **with** flywheel angular velocity from Part-1, as **ω=1000 rpm = 2*π*1000/60= 104.67 rad/sec**

And assuming Poisson’s ratio**, v=0.4**

Max. stress in the flywheel, **Ϭ**_{t-max}** = 12533445.38 N/m**^{2}** = 12.53 MPa**

Assuming, **Ϭ**_{yield}** = 150 MPa**

Max speed at yield stress**, ω**_{yield}** = 362.10 rad/sec**

From **eq.2.**5,

**FOS = 362.10 / 104.67 = 3.45 **

The **ω _{yield}** gives the critical speed of the flywheel.

This flywheel design and sizing calculation shows how to find out mass, size, cross section, maximum internal tangential stress and factor of safety for a flywheel.

how you found angular velocity at yield stress??