Page 462 - Mechanical Behavior of Materials

462 Chapter 9 Fatigue of Materials: Introduction and Stress-Based Approach

Example 9.6
Man-Ten steel is subjected in service to a stress amplitude of 180 MPa and a mean stress of
100 MPa for 20,000 cycles.

(a) What are the safety factors in stress and in life?
(b) What load factor Y = Y a = Y m corresponds to the 20,000 cycle service life?

Solution (a) Constants for this material are available from Table 9.1, which will be used as
needed. If we choose the Morrow mean stress equation, ˆσ ar is calculated from Eq. 9.21, and
then the corresponding life is available from Eq. 9.22:

ˆ σ a 180 MPa
ˆ σ ar = = = 198.2MPa
ˆ σ m 100 MPa
1 − 1 −
σ 1089 MPa
f
1/b
$ % 1/(−0.115)
1 ˆ σ ar 1 198.2MPa 6
N f 2 = = = 1.359 × 10 cycles
2 σ 2 1089 MPa
f
Hence, the safety factor in life from Eq. 9.25(b) is

1.359 × 10 6
N f 2
X N = = = 67.93 Ans.
N ˆ 20,000
The safety factor in stress can be obtained from this value and Eq. 9.12(a), for which B = b.

X S = X −b = (67.93) −(−0.115) = 1.624 Ans.
N
(b) To obtain the load factor, we employ Eq. 9.26(b) to obtain σ , with an equality
ar1
applying, as we wish to compute the failure point:
b

ˆ
σ = σ (2N) = (1089 MPa)(2 × 20,000) −0.115 = 322.0MPa
ar1 f
Then, substituting the known quantities into Eq. 9.26(a) gives
Y a ˆσ a Y a (180 MPa)
σ = , 322.0MPa =
ar1
Y m ˆσ m Y m (100 MPa)
1 − 1 −
σ 1089 MPa
f
Invoking Y = Y a = Y m as specified and solving gives Y = 1.536 (Ans.).
Discussion As expected, the load factor Y does not have the same value as X S .Ifwe
rework this problem, choosing the SWT mean stress equation, we obtain identical values
X S = Y = 1.434.