(a) DESIGN OF RIM
Stresses in a rim
- Tensile stress due to centrifugal force
σt = ρv2
- Tensile stresses due to bending 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
NUMBER OF ARMS
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 %.
(b) DESIGN OF ARMS
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.
Tensile stress due to centrifugal force σt = 0.75 ρv2
M is due to the torque ‘T’ transmitted, M = = T(R-r)/nR
Where n is the number of arms
σbt1 =Due to torque and hence due to M,
σbt 2=Due to belt tensions =(F1—F2)(R–r)/(n/2)Z
Only half the arms are considered for transmitting power.
σtotal = σt +σbt1 + σbt2
CHECK σtotal < 38 MPa
The cross section of the arms 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
Dimension near the rim are taken 2/3 rd of dimension near the hub.