Respuesta :
Answer:
P = 99000 lb.in / s
T = 630.25 lb-in
dr = 0.9272 in
Explanation:
Given:
- The length of the shaft L= 7.5 ft
- The power transmitted P = 15 hp
- The rotational speed f = 1500 rpm
- Allowable shearing stress of 4.5 ksi
- Allowable angle of twist θ = 4°
Find:
a. The power transmitted by the steel shaft in units of in-lb/sec.
b. The corresponding torque for the steel shaft in units of in-lb.
c. The required diameter of the shaft
Solution:
- The power transmitted can be calculated as:
P = 15 hp
P = 15*6600 lb.in / s
P = 99000 lb.in / s
- The Torque T for the steel shaft can be related to P and w as follows:
T = P / 2*pi*f
T = 99000 / 2*pi*(1500 / 60 )
T = 630.25 lb-in
- Use allowable shearing stress (τ) for design of shaft:
τ = 16*T / pi*d^3
Where, d is the diameter of the shaft:
d^3 = 16*T / pi*τ
d = cbrt ( 16*T / pi*τ )
d = cbrt ( 16*630.25 / pi*4.5*10^3 )
d = 0.89345 in
- Use allowable angle of twist (θ) for design of shaft:
θ = 32*T*L / pi*d^4*G
d^4 = 32*T*L / pi*θ*G
d = (32*T*L / pi*θ*G)^0.25
d = (32*630.25*7.5*12 / pi^2*(4/180)*11.2*10^6)^0.25
d = 0.9272 in
- The required diameter should be kept in lieu to both allowable angle of twist and allowable shear stress so dr
dr = max ( 0.89345 , 0.9272 )
dr = 0.9272 in
