Let engineering stress and strain be denoted by S and e, respectively. Similarly, let σ an...

Let engineering stress and strain be denoted by S and e, respectively. Similarly, let σ and ϵ denote true stress and true strain. For a uniaxial test specimen subjected to an axial tension load P

where A0 is the original cross-sectional area, A is the instantaneous cross-sectional area, l0 is the original gage length, and ∆l0 is the change in l0.

a. Show for constant volume (Al = A0l0) that

 

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b. Determine the differences between σ and S and between ϵ and e for engineering strains of 1%, 2%, 4%, and 10%.

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c. For 4340 annealed steel (see Example 16.5), determine the range of engineering strain for which the use of engineering stress and strain in Eq. 16.7 is reasonable.

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).

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

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 (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.