Page 392 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
Brown˙C09
Brown.cls
374
U.S. Customary 15:28 APPLICATION TO MACHINES SI/Metric
Try a guess of (0.125) Try a guess of (0.003)
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(1,570)d − d − 2 = 0 (9.8 × 10 )d − d − 2 = 0
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(1,570)(0.125) − (0.125) − 2 = 0 (9.8 × 10 )(0.003) − (0.003) − 2 = 0
? ?
(3.066) − (0.125) − 2 = 0 (2.646) − (0.003) − 2 = 0
0.941 > 0 (guess slightly too high) 0.643 > 0 (guess slightly too high)
Try a guess of (0.125 − 0.005 = 0.12) Try a guess of (0.003 − 0.0001 = 0.0029)
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(1,570)d − d − 2 = 0 (9.8 × 10 )d − d − 2 = 0
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(1,570)(0.12) − (0.12) − 2 = 0 (9.8 × 10 )(0.0029) − (0.0029) − 2 = 0
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(2.713) − (0.12) − 2 = 0 (2.3901) − (0.0029) − 2 = 0
0.593 > 0 (guess still too high) 0.3872 > 0 (guess still too high)
Try a guess of (0.12 − 0.01 = 0.11) Try a guess of (0.0029 − 0.0002 = 0.0027)
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(1,570)d − d − 2 = 0 (9.8 × 10 )d − d − 2 = 0
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(1,570)(0.11) − (0.11) − 2 = 0 (9.8 × 10 )(0.0027) − (0.0027) − 2 = 0
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(2.09) − (0.11) − 2 = 0 (1.9289) − (0.0027) − 2 = 0
−0.02 ∼ 0 (close enough) −0.074 ∼ 0 (close enough)
=
=
So the required wire diameter (d) is So the required wire diameter (d) is
d = 0.11 in d = 0.0027 m = 2.7mm
Step 5. Though not required, calculate the Step 5. Though not required, calculate the
spring index (C) using Eq. (9.12). spring index (C) using Eq. (9.12).
D 1in D 0.025 m
C = = = 9.09 C = = = 9.26
d 0.11 in d 0.0027 m
which is in the range 6 ≤ C ≤ 12. which is in the range 6 ≤ C ≤ 12.
Consider an extension of Example 2, where the deflection (y) and the shear modulus of
elasticity (G) are also given and the number of active coils (N a ) is required.
U.S. Customary SI/Metric
Example 3. Suppose that the force (F) given Example 3. Suppose that the force (F) given
in Example 2 causes a deflection (y). Determine in Example 2 causes a deflection (y). Determine
the number of active coils (N a ) required, where the number of active coils (N a ) required, where
y = 1.25 in y = 3cm = 0.03 m
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G = 11.5 × 10 lb/in 2 G = 80 GPa = 80 × 10 N/m 2
F = 50 lb (given in Example 2) F = 225 N (given in Example 2)
D = 1 in (given in Example 2) D = 0.025 m (given in Example 2)
d = 0.11 in (determined in Example 2) d = 0.0027 m (determined in Example 2)