Page 67 - Analysis and Design of Machine Elements

Solution 1: Safety factor is solved by calculation Strength of Machine Elements 45
Steps Computation Results Units

1. Calculate stress amplitudes From Eq. (2.3)
and mean stress min = r max =0.25 × 480 = 120MPa

Therefore
+ 480 + 120
= max min = = 300MPa
m 2 2
max − min 480 − 120
= 2 = 2 = 180 MPa
a
2. Calculate fatigue strength From Eq. (2.28) S = 1.45 < [S]
ca
safety factor by infinite life
S = −1 =
ca K +
a m
480
= 1.45 < [S]
1.5 × 180 + 0.2 × 300
3. Calculate fatigue strength From Eq. (2.16) S = 1.88 > [S]
N
safety factor by finite life √ √
m N 0 9 10 7
−1N = −1 = 480 × =
N 10 6
619.94 MPa
S = −1N =
N
K +
a m
619.94
= 1.88 > [S]
1.5 × 180 + 0.2 × 300
4. Calculate safety factor From Eq. (2.31) S = 1.67 > [S]
s
by the static strength 800
S = s = = 1.67 > [S]
s
+ a 300 + 180
m
Solution 2: Safety factor is solved by using the − diagram
a
m
Steps Computation Results Units
1. Draw − diagram Draw a line AD from (0, ) with a slope −1N = 619.94 MPa
m
a
−1
for the material, as shown of tan =− =−0.2.

in Figure E2.1
Draw a − diagram for the infinite life
a
m
of material ADC by , , and .
−1
s
Since
√ √
m N 0 9 10 7
= = 480 × = 619.94 MPa
−1N −1 N 10 6
Draw a − diagram for the finite life
m a
of material A D C,where A D is parallel
N N N N
to AD from (0, ).
−1N