S-N Diagram for Steels

We commonly used an idealized S-N diagram for steels, in which the median failure curve is used.

Key points:

$S_{u t}$ , which is the ultimate yield strength (one cycle)
In the low cycle region ( $N < 1000$ ), the line has a low slope and runs between $S_{u t}$ to some $f S_{u t}$ , where $0.8 \leq f \leq 0.9$ .
Between $1 0^{3}$ and $1 0^{6}$ cycles, a line of steeper slope represents the failure curve, which is called Basquin’s Equation, $S_{f} = a N^{b}$ .
After $1 0^{6}$ cycles, the endurance limit is reached, and the failure curve becomes horizontal.

Thus, to create an idealized S-N diagram, we generally need to determine $f S_{u t}$ at $1 0^{3}$ cycles and $S_{e}$ at $1 0^{6}$ cycles.

Estimating the Endurance Limit

For steels, generally we have:

S_{e}^{'} = ⎩ ⎨ ⎧ 0.5 S_{u t} 100 kpsi 700 MPa S_{u t} \leq 200 kpsi (1400 Mpa) S_{u t} > 200 kpsi S_{u t} > 1400 MPa

where $S_{u t}$ is the minimum tensile strength.

We want to estimate the fatigue strength $f S_{u t}$ at $1 0^{3}$ cycles. This can be done with the equations:

f f = 1.06 - 2.8 (1 0^{- 3}) S_{u t} + 6.9 (1 0^{- 6}) S_{u t}^{2} for 70 < S_{u t} < 200 kpsi = 1.06 - 4.1 (1 0^{- 4}) S_{u t} + 1.5 (1 0^{- 7}) S_{u t}^{2} for 500 < S_{u t} < 1400 MPa

In general:

The low cycle line between $1$ and $1 0^{3}$ cycles is given by Basquin’s equation:

S_{f} lo g S_{f} = a N^{b} = b lo g N + lo g a

where $a = S_{u t}$ and $b = - \frac{1}{3} lo g f$ .

The low cycle line between $1 0^{3}$ and $1 0^{6}$ cycles is given by Basquin’s equation:

S_{f} lo g S_{f} = a N^{b} = b lo g N + lo g a

where:

a = \frac{( f S _{u t} ) ^{2}}{S _{e}}, b = - \frac{1}{3} lo g (\frac{f S _{u t}}{S _{e}})

We can use this to solve for $N$ :

N = (\frac{σ _{a r}}{a})^{1/ b}

where $σ_{a r}$ is the alternating stress or stress amplitude.