determine the maximum torque - shaftdetermine the


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

τmax  = 40 N/mm

2  = 40 × 106 N/m2

θ= 2o  = 2 × (π /180)

= 0.035 radians , l = 10 m

G = 80 N/mm2 = 80 × 109 N-m

(a)    τmax  =16T / πd3

Use      τm = 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

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