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 × 10^{3} = 2πNT /60 watts
T =60 ×10^{3}/2πN
= (60 ×103 P /2πN )×1000 N mm
= 3 ×10^{7} (P /πN ) N mm
T = (π/16) d^{3} τ_{m}
d^{ 3} = 16T /π τ_{m}
d ^{3 } = (16T /π τ_{m} )× 3 × 107 × P = (48 × 10^{7} / π^{2} 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/mm^{2}