Value of the maximum shear stress:

Illustrates that for a given maximum shear stress the minimum diameter needed for a solid circular shaft to transmit P kW at N rpm may be expressed as

d = Constant × ( P/ N )³

What value of the maximum shear stress has been used if the constant equals 84.71, being in millimetre - N units?

Solution

 We know,

P × 10³  = 2πNT /60 watts      

T =60 ×10³/2πN

= (60 ×103 P /2πN )×1000 N mm

= 3 ×10⁷ (P /πN )  N mm

 T =  (π/16) d³ τ_m

d ³  = 16T /π τ_m

d ³  = (16T /π τ_m )× 3 × 107 ×   P = (48 × 10⁷ / π² τ_m  ). (P/ N)

∴ d = K ×(  P /N)^(1/3)

where,

While   K = 84.71, we get,

84.71 = 368 / (τm )^{(1/ 3)}

∴          (τm)^{(1/ 3)}  = 368 /84.71 = 4.344

∴ τ_m = 82 N/mm²