Page 403 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown˙C09
Brown.cls
U.S. Customary MACHINE ENERGY SI/Metric 385
Step 2. Substitute the term (ρG) found in Step 2. Substitute the term (ρG) found in
step 1 and the other given information in step 1 and the other given information in
Eq. (9.46) to give the critical frequency ( f cr ) as Eq. (9.46) to give the critical frequency ( f cr ) as

Dk 8 (1.5in)(50 lb/in) Dk 8 (0.04 m)(9,000 N/m)
f cr = = f cr = =
πd 3 ρG π(0.125 in) 3 πd 3 ρG π(0.003 m) 3

8 8
2 ×

2
×
8.43 × 10 3 lb · s 2 6.32 × 10 14 N · s 2
in 6 m 6

75 lb in 6 360 N −14 m 6
= 3 9.49 × 10 −4 2 = −8 3 1.3 × 10 2
0.006 in lb · s 2 8.5 × 10 m N · s 2

lb in 9 −7
3 N m 3
= 12,500 3.08 × 10 −2 = 4.2 × 10 3 1.1 × 10
in 3 lb · s m N · s
cycle
cycle = 462 = 462 Hz
= 385 = 385 Hz s
s
Step 3. Divide the critical frequency ( f cr ) Step 3. Divide the critical frequency ( f cr )
found in step 2 by 20 to obtain the limiting found in step 2 by 20 to obtain the limiting
frequency of the cyclic loading. frequency of the cyclic loading.
f cr 385 Hz f cr 462 Hz
f limiting = = ∼ 19 Hz f limiting = = ∼ 23 Hz
=
=
20 20 20 20
Cyclic loading frequencies greater than this Cyclic loading frequencies greater than this
value are unsafe. value are unsafe.
9.2.8 Fatigue Loading
Rarely are helical springs not subjected to fatigue loading. The number of cycles may only
be in hundreds or thousands, but usually they must be designed for millions and millions
of cycles such that an infinite life is desired.
Helical springs may be subjected to completely reversed loading, where the mean shear
stress (τ m ) is zero; however, as this type of spring is installed with a preload, the spring
is usually subjected to fluctuating loading. The fluctuating loading may be compressive or
tensile, but never both.
If the maximum force on the spring is denoted as (F max ) and the minimum force is
denoted as (F min ), whether they are compressive or tensile, then the mean force (F m ) and
alternating force (F a ) are given by the relationships in Eqs. (9.47) and (9.48) as
F max + F min
F m = (9.47)
2
F max − F min
F a = (9.48)
2
In Chap. 7 it was shown that any stress-concentration factors are applied only to the
alternating stresses. Therefore, using Eq. (9.13) the mean shear stress (τ m ) is given by
Eq. (9.49) as
8F m D
τ m = K s 3 (9.49)
πd
where the shear-stress correction factor (K s ) is given by Eq. (9.14).