Page 422 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
Brown˙C09
Brown.cls
404
U.S. Customary 15:28 APPLICATION TO MACHINES SI/Metric
slug 2 kg 2

m solid disk = 2.98 × (8.94 ft ) m solid disk = 493.2 × (0.8036 m )
ft 2 m 2
= 26.7 slug = 396.4kg
Therefore, the weight of the flywheels is Therefore, the weight of the flywheels is
W solid disk = m solid disk g W solid disk = m solid disk g
ft
m
= (26.7 slug) 32.2 = (396.4kg) 9.8 s 2
s 2
= 3,884 N
= 860 lb
Step 2. Calculate the weight of the hub of the Step 2. Calculate the weight of the hub of the
composite flywheel. composite flywheel.
W hub = m hub g W hub = m hub g
where the mass of the hub is found from where the mass of the hub is found from
Eq. (9.76) as Eq. (9.76) as
1 2 2 1 2 2
m hub = ρπw hub d − d i m hub = ρπw hub d − d i
o
o
4 4
1 slug 1 kg
= 15.2 π(0.25 ft) = 7,850 π(0.08 m)
4 ft 3 4 m 3
2
2
2
2
×((0.5ft) − (0.25 ft) ) ×((0.16 m) − (0.08 m) )
slug 2 kg 2

= 2.98 × (0.1875 ft ) = 493.2 × (0.0192 m )
ft 2 m 2
= 0.56 slug = 9.47 kg
Therefore, the weight of the hub is Therefore, the weight of the hub is
W hub = m hub g W hub = m hub g
m
ft
= (0.56 slug) 32.2 = (9.47 kg) 9.8 2
s 2 s
= 93 N
= 18 lb
Step 3. Calculate the mass moment of inertial Step 3. Calculate the mass moment of inertia
of the hub using Eq. (9.75). of the hub using Eq. (9.75).
1 2 2 1 2 2
I hub = m hub d − d i I hub = m hub d − d i
o
o
8 8
1 1
= (0.56 slug) = (9.47 kg)
8 8
2
2
2
2
×((0.5ft) − (0.25 ft) ) ×((0.16 m) − (0.08 m) )
1 2 1 2
= (0.56 slug) × (0.1875 ft ) = (9.47 kg) × (0.0192 m )
8 8
= 0.013 slug · ft 2 = 0.023 kg · m 2