Page 295 - Marks Calculation for Machine Design

P1: Shashi
15:4
January 4, 2005
Brown.cls
Brown˙C07
FATIGUE AND DYNAMIC DESIGN
Note that the prime on the endurance limit (S ) in Eq. (7.1) differentiates the endurance
e 277
limit obtained from a fatigue test and the actual endurance limit (S e ) for a machine element
that usually differs in surface finish, size, loading, temperature, and other miscellaneous
effects from the test specimen. These factors, which modify the value of the endurance limit
(S ) obtained from Eq. (7.1), will be discussed shortly.

e
3
6
To get a feel of the difference between (N = 10 ) cycles and (N = 10 ) cycles, consider
the following example.
U.S. Customary SI/Metric
Example 1. How far must a car be driven at Example 1. How far must a car be driven at
30 mph at an engine speed of 2,500 rpm for the 50 kph at an engine speed of 2,500 rpm for the
6
6
crankshaft to rotate 10 cycles? How far for 10 3 crankshaft to rotate 10 cycles? How far for 10 3
cycles? How long will each take? cycles? How long will each take?
solution solution
Step 1. Convert mph to mi/rev. Step 1. Convert kph to km/rev.
mi 1h 1 min 1 mi km 1h 1 min 1 km
30 × × = 50 × × =
h 60 min 2,500 rev 5,000 rev h 60 min 2,500 rev 3,000 rev
Step 2. Multiply the mi/rev found in step 1 Step 2. Multiply the km/rev found in step 1
6
6
by 10 cycles or revolutions to find the distance by 10 cycles or revolutions to find the distance
traveled as traveled as
1 mi 1 km
6
6
dist 10 6 = × 10 rev = 200 mi dist 10 6 = × 10 rev = 333 km
5,000 rev 3,000 rev
Step 3. Multiply the mi/rev found in step 1 by Step 3. Multiply the mi/r found in step 1 by
3
3
10 cycles or revolutions to find the distance 10 cycles or revolutions to find the distance
traveled as traveled as
1 mi 1 km
3
3
dist 10 3 = × 10 rev = 0.2mi dist 10 3 = × 10 rev = 0.333 km
5,000 rev 3,000 rev
= 1,056 ft = 333 m
Step 4. Divide the distance found in step 2 by Step 4. Divide the distance found in step 2 by
6
6
30 mph to find the time for 10 cycles 50 kph to find the time for 10 cycles
200 mi 333 km
time 10 6 = = 6.7h time 10 6 = = 6.7h
30 mi/h 50 km/h
Step 5. Divide the distance found in step 3 by Step 5. Divide the distance found in step 3 by
3
3
30 mph to find the time10 cycles 50 kph to find the time10 cycles
0.2mi 0.333 km
time 10 3 = = 0.0067 h time 10 3 = = 0.0067 h
30 mi/h 50 km/h
= 0.4 min = 24 s = 0.4 min = 24 s
3
The distances and the times found in Example 1 show strikingly how different 10 cycles
6
and 10 cycles can be.
6
Finite Life. For cycles less than (N = 10 ) the test specimen has a finite life. For most
3
materials the fatigue strength data falls on two straight lines, one from (N =1) to (N =10 )
3
6
cycles, and one from (N =10 ) cycles to (N =10 ) cycles.