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

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

**DESIGN OF RIM**

**STRESSES IN RIM**

**Tensile stress due to centrifugal force**

**σ _{t} = ρv^{2}**

**Tensile stresses due to bending due to restrain of arms**

**σ _{b} = M/Z**

**M = wL ^{2}/12**

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

**w = mRω ^{2 }=ρbtLR ω^{2}/L**

**w = mRω ^{2 }=ρbtR ω^{2}**

** **

**L = length between two arms**

** = Behaves as a fixed beam**

** L =2πR/n**

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

**M = wL ^{2}/12= ρbtLR ω^{2}L^{2}/12**

**M = (ρbtLR ω ^{2}/12)( 2πR/n)^{2}**

**M = π ^{2}ρbt v^{2}R/3n^{2}**

**Z = bt ^{2}/6**

**σ _{b} = M/Z =2 π^{2}ρv^{2}R/n^{2} t**

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

**Total stress in the rim**

**σ _{total} = (3/4) σ_{t} + σ_{b}/4**

**σ _{total} = (3/4) ρv^{2} + π^{2}ρv^{2}R/2n^{2} t **

**σ _{total} = ρv^{2}(0.75 + 4.935R/n^{2} t )**

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

**CHECK σ _{total} <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 %.**

**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 σ _{t} = 0.75 ρv^{2}**

**M is due to the torque ‘T’ transmitted, M = = T(R-r)/nR**

Where n is the number of arms

**σ _{bt1} =Due to torque and hence due to M, **

**σ _{bt1 }=M/Z**

**σ _{bt} _{2}=Due to belt tensions =(F_{1}—F_{2})(R–r)/(n/2)Z**

**Only half the arms are considered for transmitting power.**

**σ _{total} =**

**σ**

_{t}+σ_{bt1 + }

**σ**

_{bt2}**CHECK σ _{total} < 38 MPa**

** ****Design of Arms**

**The cross section is normally elliptical.**

**Let a = Major axis near the hub**

**B= Minor axis near the hub**

**Z _{y} = section modulus = πba^{2}/32**

**Taking b = a/2**

**a= (64Z/π) ^{1/3}**

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