To overcome the extensive time and cost required to run fatigue tests, Eq. 16.7 can be rep...

To overcome the extensive time and cost required to run fatigue tests, Eq. 16.7 can be replaced by the following relation proposed by Manson (1965) and Muralidharon and Manson (1988), which is known as Manson’s universal slope equation:

where the slope coefficients b = −0.12 and c = −0.6 are fixed for all materials and where the ultimate tensile strength Su, the modulus of elasticity E, and the true fracture strain ϵf are obtained from a single static tension test. For 4340 annealed steel, Su = 827 MPa, E = 200 GPa, and ϵf = 0.57.

a. Rework Example 16.5 using Eq. (a) and compare the results to those obtained in Example 16.5.

 (16.7)

---

b. Comment on any differences in the results and then significance.

EXAMPLE 16.5: Fatigue Life of 4340 Steel

The strain- life equation for 4340 steel, based upon test results of eleven specimens (Graham, 1968), is

ϵt = ϵp + ϵe = 0.58(2N)−0 57 + 0.0062(2N)−0 09    (a)

(a) Determine the total strain ϵtC for the case ϵp = ϵe, that is, at 2N = 2NC (see Figure 16.10).

---

(b) Determine the total strain for 2N = 106 load reversals.

---

 (c) Determine the number of cycles N expected for a fatigue test with constant total strain amplitude ϵt = 0.01.

FIGURE 16.10 Strain- life curves.

Solution:

(a) From Eq. (a), ϵp = 0.58(2N)−0 57 and ϵe = 0.0062(2N)−0 09. So, for ϵp = ϵe, we obtain NC = 6390 cycles. Then, by Eq. (a),

ϵtC = 0.58[2(6390)]−0 57 + 0.0062[2(6390)]−0 09

or

ϵtC = 0.00529

(b) For 2N = 106 load reversals, Eq. (a) yields

ϵt = 0.58(106)−0 57 + 0.0062(106)−0 09

or

ϵt = 0.00201

(c) Forϵt = 0.01, Eq. (a) yields

0.01 = 0.58(2N)−0 57 + 0.0062(2N)−0 09

The solution of this equation is

N = 1184 cycles

This result agrees with the fact that, for E1 "' 0.01, the fatigue life of many metals is about N = 1000 cycles.