Question

The diameter of a shaft is 20cm. Find the safe maximum torque which can be transmitted by the shaft if the permissible shear stress in the shaft material be 4000 N/cm2. and permissible angle of twist is 0.2 degree per meter length. Take G = 8 × 106 N/cm2. If the shaft rotates at 320 r.p.m. what maximum power can be transmitted by the shaft.

Answer 1

Given that :

d = 20 cm

Tmax = ?

ιmax = 4000 N/cm2

θ = 0.2/meter length = 0.2 × (π/180) = 0.0035 rad

G = 8 × 106 N/cm2

N = 320 r.p.m.

P = ?

CASE – 1: When ιmax = 4000 N/cm2

T =(π/16) (ιmax .d3)

T = (π/16) × 4000 × (20)3

T = 6283185.3 N – cm ...(i)

CASE – 2: When θ = 0.2/meter length

T = G.J. θ/L = (π/32)d4. G.θ/L

Let L = 100cm

T = (π/32)(20)4. 8 × 106.(0.0035)/100

T = 4386490.85 N – cm ...(ii)

The permissible torque is the minimum of (i) and (ii)

i.e.; T = 4386490.85 N – cm = 43.86 KN – m

Power = 2πNT/60 Watt, T in N

= {2π(320)(43.86 × 103)}/60

= 1469762.7 Watt

P = 1469.76 KW