Determine the maximum torque - shaft:
Determine the maximum torque that can be safely transmitted by a shaft of 400 mm diameter, if
(a) the maximum allowable shear stress is 40 N/mm^{2}, and
(b) the maximum allowable angle of twist is 2o in a length of 10 m. Take G = 80 kN/mm^{2}.
Solution
d = 400 mm = 0.40 = m
τ_{max} = 40 N/mm
2 = 40 × 106 N/m^{2}
θ= 2^{o} = 2 × (π /180)
= 0.035 radians , l = 10 m
G = 80 N/mm2 = 80 × 10^{9 }N-m
(a) τ_{max} =16T / πd^{3}
Use τ_{m} = 40 × 106 N/m2,
⇒ 40 × 10^{6} = 16T / (π (0.4)^{3})
T = 503 × 103 N-m = 503 kN-m
(b) G θ / l = T / J= 32T / πd ^{4}
⇒ 80 × 109 × 0.035 /10 =32T / π (0.4)^{4}
∴ T = 704 × 103 N-m = 704 kN-m
∴ Safe torque is smaller then above two
∴ T = 503 kN-m