Page 425 - Mechanical Behavior of Materials

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

Once B is known, A can be calculated from either point:

σ 1 948 MPa
A = = = 1565 MPa Ans.
N B 222 −0.0928
1
(b) Since the stress is chosen in each test, σ a is the independent variable, and N f is the
dependent one. Hence, to proceed with a least-squares fit, solve Eq. 9.6 for N f , and then take the
logarithm of both sides:

1/B
1 1
σ a
N f = , log N f = log σ a − log A
A B B
This is a straight line on a log–log plot, so
y = mx + c

where y is the dependent variable and x is the independent one. Then
1
y = log N f , x = log σ a , m = 1/B, c =− log A
B
Performing a linear least-squares fit on this basis gives

m =−10.582, c = 33.87
so that

B = 1/m =−0.0945, A = 10 −cB = 1587 Ans.

The resulting line is plotted on Fig. E9.1.

Discussion The values of A and B from Solutions (a) and (b) are seen to agree approximately
with each other and with the values in Table 9.1, which are based on a larger set of data.
From Eq. 9.6, the coefficient A is seen to be the σ a intercept at N f = 1. This is shown in
Fig. E9.1. The exponent B is the slope as directly measured on a log–log plot with the same
scales on both axes. In Fig. E9.1, since the logarithmic decades on the σ a axis are twice as large
as those on the N f axis, D v /D h = 2, the slope v/h measured directly from the graph must be
divided by 2 to graphically determine a value of B.
Also, σ and b for Eq. 9.7 may be obtained directly from A and B. Using Eq. 9.8 with the

f
values from the least-squares fit, we get
A 1587 MPa

b = B =−0.0945, σ = = = 1695 MPa
f b −0.0945
2 2
From Eq. 9.7, the constant σ is noted to be the σ a intercept at one-half cycle, N f = 0.5, which

f
is also shown in Fig. E9.1.