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/mm², and

(b) the maximum allowable angle of twist is 2o in a length of 10 m. Take G = 80 kN/mm².

Solution

d = 400 mm = 0.40 = m

τ_max  = 40 N/mm

2  = 40 × 106 N/m²

θ= 2^o  = 2 × (π /180)

= 0.035 radians , l = 10 m

G = 80 N/mm2 = 80 × 10⁹ N-m

(a)    τ_max  =16T / πd³

Use      τ_m = 40 × 106 N/m2,

 ⇒        40 × 10⁶  = 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