Page 400 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
APPLICATION TO MACHINES
382
For springs made of steel, this value of the free length (L o ) is given by Eq. (9.42), which
is dependent only on the mean diameter (D) and the end-constant (α).
D
L o ≤ 2.63 (9.42)
α
U.S. Customary SI/Metric
Example 8. Determine the critical deflection Example 8. Determine the critical deflection
(y cr ) for a steel compression helical spring (y cr ) for a steel compression helical spring
positioned between two flat parallel surfaces, positioned between two flat parallel surfaces,
where where
L o = 3in L o = 7.5 cm
D = 1in D = 2.5 cm
6
2
2
9
E = 30 × 10 lb/in (steel) E = 210 GPa = 210 × 10 N/m (steel)
9
6
2
2
G = 11.5 × 10 lb/in (steel) G = 80.5 GPa = 80.5 × 10 N/m (steel)
solution solution
Step 1. Using the guidelines in Table 9.2, Step 1. Using the guidelines in Table 9.2,
choose the end-condition (α) as choose the end-condition (α) as
α = 0.5 α = 0.5
Step 2. Using the end-condition (α) from Step 2. Using the end-condition (α) from
step 1 and the given information, calculate step 1 and the given information, calculate
the effective slenderness ratio (λ eff ) using the effective slenderness ratio (λ eff ) using
Eq. (9.38) as Eq. (9.38) as
αL o (0.5)(3in) αL o (0.5)(7.5cm)
λ eff = = = 1.5 λ eff = = = 1.5
D (1in) D (2.5cm)
Step 3. Using the given moduli of elasticities Step 3. Using the given moduli of elasticities
(E) and (G), calculate the elastic constants (C 1 ) (E) and (G), calculate the elastic constants (C 1 )
and (C 2 ) as and (C 2 ) as
E E
C 1 = C 1 =
2 (E − G) 2 (E − G)
6
9
30 × 10 lb/in 2 210 × 10 N/m 2
= =
6
9
2(30 − 11.5) × 10 lb/in 2 2(210 − 80.5) × 10 N/m 2
6
9
30 × 10 lb/in 2 210 × 10 N/m 2
= = 0.81 = = 0.81
6
9
37 × 10 lb/in 2 259 × 10 N/m 2
2
2
2π (E − G) 2π (E − G)
C 2 = C 2 =
2 G + E 2 G + E
9
6
2
2
2π (30 − 11.5) × 10 lb/in 2 2π (210 − 80.5) × 10 N/m 2
= =
9
6
[2(11.5) + 30] × 10 lb/in 2 [2(80.5) + 210] × 10 N/m 2
9
6
365 × 10 lb/in 2 2,556 × 10 N/m 2
= 2 = 6.9 = 2 = 6.9
9
6
53 × 10 lb/in 371 × 10 N/m
Step 4. Using the effective slenderness ratio Step 4. Using the effective slenderness ratio
(λ eff ) found in step 2, the elastic constants (C 1 ) (λ eff ) found in step 2, the elastic constants (C 1 )
and (C 2 ) found in step 3, and the free length and (C 2 ) found in step 3, and the free length
(L o ) in Eq. (9.37) to determine the critical (L o ) in Eq. (9.37) to determine the critical