Question

Select a suitable diameter of a solid shaft of circular section to transmit 112.5kW of power at 200rpm, if the allowable shear stress is 75N/mm² and allowable twist is 1 degree in a length of 3m. G = 0.82 x 105 N/mm²​

Answer 1

the answer is in the pic

please mark as brainliest

Answer 2

Answer:

Explanation:

P = 75 KW = 75 X 103 W

N = 200 RPM

θ = 1º = π/180

radian  L = 2m  τ = 50 MN/m2 = 50 × 106 N/m2

G = 100 GN/m2 = 100 × 109 N/m2  Using the relation, P = 2π.N.Tmax/60 watts  75 × 103 = 2π.200.Tmax/60  Tmax = 3581 N-m ...

(i)  CASE – 1: Shaft diameter when allowable shear stress is considering Tmax = (π/16) τmax.D3  3581 = (π/16).50 × 106.D3  D = 0.0714 m or 71.4 mm CASE – 2 : Shaft diameter when twist angle is considering  T = (πD4/32).G.θ/L  3581 = (πD4/32).[100 × 109(π/180)]/2  D = 80.4 mm ...

(iii)  For suitable value always take larger value of diameter  D = 80.4 mmRead more on Sarthaks.com - https://www.sarthaks.com/511222/solid-circular-shaft-transmits-75kw-power-rpm-calculate-the-shaft-diameter-the-twist-the