2020-05-26T17:46:13-04:00
A solid shaft of 10cm diameter transmits 74 kW at 150 rev/min. Calculate (a) the
torque on the shaft, (b) the maximum shear stress developed, (c) the angle of twist in a
length of 1.50m, and (d) the shear stress at a radius of 3cm. Take G = 8 MN/cm2
1
2020-05-29T09:48:35-0400
a)
P = 2 π N T 60 → 74000 = 2 π ( 150 ) T 60 P=\frac{2\pi NT}{60}\to 74000=\frac{2\pi (150)T}{60} P = 60 2 π NT → 74000 = 60 2 π ( 150 ) T
T = 4711 N m T=4711\ Nm T = 4711 N m
b)
T = π 16 τ d 3 → 4711 = π 16 τ ( 0.1 ) 3 T=\frac{\pi}{16}\tau d^3\to 4711=\frac{\pi}{16}\tau (0.1)^3 T = 16 π τ d 3 → 4711 = 16 π τ ( 0.1 ) 3
τ = 24 ⋅ 1 0 6 N m 2 \tau=24\cdot10^6\frac{N}{m^2} τ = 24 ⋅ 1 0 6 m 2 N
c)
T J = G θ l → T π 32 d 4 = G θ l \frac{T}{J}=\frac{G\theta }{l}\to \frac{T}{ \frac{\pi}{32} d^4}=\frac{G\theta }{l} J T = l Gθ → 32 π d 4 T = l Gθ
4711 π 32 ( 0.1 ) 4 = 8 ⋅ 1 0 10 θ 1.5 \frac{4711}{ \frac{\pi}{32} (0.1)^4}=\frac{8\cdot 10^{10}\theta }{1.5} 32 π ( 0.1 ) 4 4711 = 1.5 8 ⋅ 1 0 10 θ
θ = 0.00899735 r a d = 0.5155 ° \theta=0.00899735\ rad=0.5155\degree θ = 0.00899735 r a d = 0.5155°
d)
T = π 16 τ ′ d 4 − ( 2 r ) 4 d → 4711 = π 16 τ ′ 0. 1 4 − ( 0.06 ) 4 0.1 T=\frac{\pi}{16}\tau' \frac{d^4-(2r)^4}{d}\to 4711=\frac{\pi}{16}\tau' \frac{0.1^4-(0.06)^4}{0.1} T = 16 π τ ′ d d 4 − ( 2 r ) 4 → 4711 = 16 π τ ′ 0.1 0. 1 4 − ( 0.06 ) 4
τ ′ = 2.76 ⋅ 1 0 7 N m 2 \tau'=2.76\cdot10^7\frac{N}{m^2} τ ′ = 2.76 ⋅ 1 0 7 m 2 N
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