Page 417 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown˙C09
Brown.cls
U.S. Customary MACHINE ENERGY SI/Metric 399
Step 3. Substitute the rated power (P rated ) Step 3. Substitute the rated power (P rated )
from step 1 and the rated angular velocity from step 1 and the rated angular velocity
(ω rated ) from step 2 in Eq. (9.65) to give the (ω rated ) from step 2 in Eq. (9.65) to give the
rated torque (T rated ) as rated torque (T rated ) as
ft · lb N · m
2,750 4,000
P rated s P rated s
T rated = = T rated = =
ω rated rad ω rated rad
180.6 180.6
s s
= 15.2ft · lb = 22.1N · m
Step 4. For a punch press system, the rated Step 4. For a punch press system, the rated
torque (T rated ) is the torque (T 1 ) and the rated torque (T rated ) is the torque (T 1 ) and the rated
angular velocity (ω rated ) is the angular velocity angular velocity (ω rated ) is the angular velocity
(ω 1 ). Therefore, (ω 1 ). Therefore,
T 1 = 15.2ft · lb T 1 = 22.1N · m
and and
ω 1 = 1,725 rpm ω 1 = 1,725 rpm
Step 5. If the punching interval is 4 percent of Step 5. If the punching interval is 4 percent of
the total cycle time, then the total cycle time, then
t 1 = 0.04 t 2 t 1 = 0.04 t 2
Step 6. Using the relationship in step 5, Step 6. Using the relationship in step 5,
calculate the recovery time to punching time calculate the recovery time to punching time
ratio (τ) using Eq. (9.68). ratio (τ) using Eq. (9.68).
t 2 − t 1 t 2 − 0.4 t 2 1 − 0.04 0.96 t 2 − t 1 t 2 − 0.4t 2 1− 0.04 0.96
τ = = = = τ = = = =
t 1 0.04 t 2 0.04 0.04 t 1 0.04t 2 0.04 0.04
= 24 = 24
Step 7. Substitute the given punching torque Step 7. Substitute the given punching torque
(T punch ), the torque (T 1 ) from step 4, and the (T punch ), the torque (T 1 ) from step 4, and the
exponent (τ) in Eq. (9.67). exponent (τ) in Eq. (9.67).
τ τ

T 2 T punch − T 1 T 2 T punch − T 1
= =
T 1 T punch − T 2 T 1 T punch − T 2
24 24

T 2 150 − 15.2 T 2 225 − 22.1
= =
15.2 150 − T 2 22.1 225 − T 2
24 24
134.8 202.9

T 2 = 15.2 T 2 = 22.1
150 − T 2 225 − T 2
Step 8. Solve for the torque (T 2 ) in the ex- Step 8. Solve for the torque (T 2 ) in the ex-
pression from step 7 by trial and error. Try a pression from step 7 by trial and error. Try a
first guess of 5 ft · lb. first guess of 9 N · m.
24 24

? 134.8 ? 202.9
5 = 15.2 9 = 22.1
150 − 5 225 − 9
? 24 ? 24
5 = 15.2 (0.930) 9 = 22.1 (0.939)
? ?
5 = 15.2 (0.174) = 2.64 9 = 22.1 (0.223) = 4.92
(too high) (too high)