Page 423 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
U.S. Customary MACHINE ENERGY SI/Metric 405
Step 4. Calculate the weight of the four spokes Step 4. Calculate the weight of the four spokes
of the composite flywheel. of the composite flywheel.
W spokes = 4 m spoke g W spokes = 4 m spoke g
where the mass of the one spoke is found from where the mass of the one spoke is found from
Eq. (9.80) as Eq. (9.80) is
1 2 1 2
m spoke = ρπ L spoke d spoke m spoke = ρπ L spoke d spoke
4 4
1 slug 1 kg
= 15.2 π(1ft) = 7,850 π(0.3m)
4 ft 3 4 m 3
×(0.125 ft) 2 ×(0.04 m) 2

slug 2 kg 2
= 11.94 × (0.0156 ft ) = 1,850 × (0.0016 m )
ft 2 m 2
= 0.19 slug = 2.96 kg
Therefore, the weight of the four spokes is Therefore, the weight of the four spokes is
W spokes = 4 m spoke g W spokes = 4 m spoke g
ft m

= 4 (0.19 slug) 32.2 = 4 (2.96 kg) 9.8
s 2 s 2
= 4 (6lb) = 24 lb = 4 (29 N) = 116 N
Step 5. Calculate the mass moment of inertia Step 5. Calculate the mass moment of inertia
of one spoke using Eq. (9.79). of one spoke using Eq. (9.79).

1 2 2 1 2 2
I spoke = m spoke L spoke + L cg I spoke = m spoke L spoke + L cg
12 12
= (0.19 slug) = (2.96 kg)

1 2 2 1 2 2
× (1ft) + (0.75 ft) × (0.3m) + (0.23 m)
12 12
2
2
= (0.19 slug) × (0.646 ft ) = (2.96 kg) × (0.0604 m )
= 0.123 slug · ft 2 = 0.179 kg · m 2
Step 6. Using Eq. (9.81) and the mass Step 6. Using Eq. (9.81) and the mass
moment of inertia from Example 1 and the mass moment of inertia from Example 1 and the mass
moments of inertia for the hub and one spoke moments of inertia for the hub and one spoke
found in steps 3 and 5, respectively, determine found in steps 3 and 5, respectively, determine
the mass moment of inertia needed from the rim. the mass moment of inertia needed from the rim.
I flywheel = I hub + I rim + N spokes I spoke I flywheel = I hub + I rim + N spokes I spoke
2 2 2 2
29.80 slug · ft = (0.013 slug · ft ) + I rim 39.85 kg · m = (0.023 kg · m ) + I rim
2
2
+(4)(0.123 slug · ft ) +(4)(0.179 kg · m )
Solving for (I rim ) gives Solving for (I rim ) gives
I rim = (29.80 − 0.013 − 0.492) slug · ft 2 I rim = (39.85 − 0.023 − 0.716) kg · m 2
= 29.3 slug · ft 2 = 39.1kg · m 2