MACHINE DESIGN–Flywheel design, Design of Rim , Design of Arms

MACHINE DESIGN–Flywheel design, Design of Rim , Design of Arms



  • Tensile stress due to centrifugal force

σt = ρv2

  • Tensile stresses due to bending due to restrain of arms

σb = M/Z

M = wL2/12

w = UDL =Centrifugal force between two arms per unit length

w = mRω2 =ρbtLR ω2/L

w = mRω2 =ρbtR ω2


L = length between two arms

    = Behaves as a fixed beam

    L =2πR/n

n = number of arms (normally 4,6 and8)

M = wL2/12= ρbtLR ω2L2/12

M = (ρbtLR ω2/12)( 2πR/n)2

M = π2ρbt v2R/3n2

Z = bt2/6

σb = M/Z =2 π2ρv2R/n2 t

NOTE: Arms stretch by ¾ of free expansion due to the centrifugal force

Total stress in the rim

σtotal = (3/4) σt + σb/4

σtotal = (3/4) ρv2 +  π2ρv2R/2n2 t

σtotal = ρv2(0.75 + 4.935R/n2 t )

For a safe design for a cast iron flywheel, total stress should be less than the allowable stress of 38 MPa.

CHECK  σtotal <38 MPa.


  • Shrinkage stresses due to uneven rate of of cooling of the casting, it is difficult to calculate but can be accounted for a certain factor say 10 %.


Stresses in arms

  • Tensile stress due to centrifugal force acting on the rim
  • Bending stresses due to torque transmitted from the rim to the shaft
  • Shrinkage stresses—Cannot be determined. To account for this, 10% stress is added.


σt = 0.75 ρv2

σb = M/Z

σbt1 =Due to torque   MT = T(R-r)/nR

σbt 2=Due to belt tensions Mbelt =(F1—F2)(R0r)/(nz/2)

σtotal = σtbt1 +  σbt2

CHECK       σtotal < 38 MPa


Design of Arms

The cross section is normally elliptical.

Let a = Major axis near the hub

B= Minor axis near the hub

Zy = section modulus = πba2/32

Taking b = a/2

a= (64Z/π)1/3

Dimension near the rim are taken 2/3 rd of near the hub.