Page 317 - Marks Calculation for Machine Design

P1: Shashi
January 4, 2005
15:4
Brown.cls
Brown˙C07
FATIGUE AND DYNAMIC DESIGN
given, rather than obtained from the guidelines in Eq. (7.1).
S ut = 105 kpsi 299

S = 65 kpsi
e
K f = 1.15 (due to change in diameter)
solution
Step 1. Using Eq. (7.8) and values for the coefficient (a) and exponent (b) from Table 7.1,
calculate the surface finish factor (k a ) as
b −0.085
k a = aS ut = (1.34 kpsi)(105 kpsi) = (1.34)(0.6733) = 0.90
Step 2. Only the larger diameter region of the stepped rod experiences the fluctuating axial
force (F 1 ), so use diameter (d 1 ) in Eq. (7.10) to calculate the size factor (k b ) as
−0.1133 −0.1133
d 0.1875 −0.1133
∼
k b = = = (0.625) = 1.05 = 1
0.3 0.3
Step 3. The stepped rod is axially loaded, so the load type factor (k c ) from the guidelines
in Eq. (7.13) is
k c = 0.923
Step 4. As the stepped rod is operating at room temperature, the temperature factor (k d )
from Eq. (7.15) and Table 7.2 is

k d = 1
Step 5. Using the given reduced stress concentration factor (K f ), calculate the miscella-
neous effect factor (k e ) using Eq. (7.16) as
1 1
k e = = = 0.87
K f 1.15

Step 6. Using the given test specimen endurance limit (S ) and the modifying factors
e
found in steps 1 through 5, calculate the endurance limit (S e ) for the stepped rod using the
Marin equation in Eq. (7.7) as

S e = k a k b k c k d k e S = (0.90)(1)(0.923)(1)(0.87)(65 kpsi)
e
= (0.723)(65 kpsi) = 47.0 kpsi
Step 7. Calculate the maximum axial force (F max ) and minimum axial force (F min ) as
F max = F max + F 2 = (800 lb) + (500 lb) = 1300 lb
1
F min = F min + F 2 = (−200 lb) + (500 lb) = 300 lb
1
Step 8. Calculate the mean axial force (F m ) and the alternating axial force (F a ) as
F max + F min (1,300 lb) + (300 lb) 1,600 lb
F m = = = = 800 lb
2 2 2
F max − F min (1,300 lb) − (300 lb) 1,000 lb
F a = = = = 500 lb
2 2 2