Fatigue Strength and S-N Diagram

Equation of line A \bar{B}:
\begin{aligned} y-y_{1} &=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}\left[x-x_{1}\right] \\ \log _{10} 200-\log _{10} 450 &=\frac{\log _{10} 150-\log _{10} 450}{(6-3)}\left[\log _{10} N-3\right] \\ N &=163840.580 \text { cycles } \end{aligned}

\begin{aligned} \text{Fluctuating stress, } \quad \sigma&=\sigma_{o} \cos (8 \pi t) \\ \sigma_{\max } &=\sigma_{o} \\ \sigma_{\min } &=-\sigma_{o} \\ \sigma_{\text {man }} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{v} &=\frac{\sigma_{\max }-\sigma_{\min }}{2}=\frac{2 \sigma_{o}}{2}=\sigma_{o} \\ \sigma_{e} &=350 \mathrm{MPa} \\ \mathrm{FOS} &=3.5 \end{aligned}
From strength criteria, \frac{\sigma_{v}}{\sigma_{e}} \leq \frac{1}{\mathrm{FOS}}
\begin{aligned} \frac{\sigma_{o}}{350} & \leq \frac{1}{3.5} \\ \sigma_{o} & \leq 100 \mathrm{MPa} \end{aligned}

\begin{array}{l} \sigma_{\min }=-70 \mathrm{MPa} \\ \sigma_{\text {mean }}=140 \mathrm{MPa} \\ \Rightarrow \sigma_{\text {mean }}=\frac{\sigma_{\min }+\sigma_{\max }}{2}=140 \\ \Rightarrow \frac{-70+\sigma_{\max }}{2}=140 \\ \Rightarrow \sigma_{\max }=350 \mathrm{MPa} \\ \text { The R-ratio, } \mathrm{R}=\frac{\sigma_{\min }}{\sigma_{\max }}=-\frac{70}{350}=-0.2 \end{array}

\begin{aligned} \frac{\sigma_{\max }-\sigma_{\min }}{2} &=1100 \mathrm{N}^{-0.15} \\ \frac{\sigma_{\max }-\left(-\sigma_{\max }\right)}{2} &=1100 \mathrm{N}^{-0.15} \\ \frac{2 \sigma_{\max }}{2} &=1100 \mathrm{N}^{-0.15} \\ \sigma_{\max } &=1100 \mathrm{N}^{-0.15}=1100 \times\left(10^{5}\right)^{-0.15} \\ &=1100 \times(10)^{-0.75}=\frac{1100}{5.62} \\ \sigma_{\max } &=195.61 \mathrm{MPa} \end{aligned}

From two similar triangles,
(i.e. \Delta A B D \text { and } \Delta E C D)
\begin{aligned} \frac{A B}{E C}&=\frac{B D}{C D} \\ \frac{\left(\log _{10} 480-\log _{10} 250\right)}{\left(\log _{10} S-\log _{10}-250\right)}&=\frac{6-3}{6-\log _{10} 10000} \\ \frac{2.68124-2.39794}{\log _{10} S-2.39794}&=1.5 \\ S&=386.195 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

\begin{aligned} \sigma _{max}&=150MPa\\ \sigma _{min}&=50MPa\\ S_e&=200MPa\\ S_{yt}&=300MPa\\ S_{ut}&=400MPa\\ \sigma _a&=\frac{\sigma _{max}-\sigma _{min}}{2}\\ &=\frac{150-50}{2}\\ &=50 MPa\\ \sigma _m&=\frac{\sigma _{max}+\sigma _{min}}{2}\\ &=\frac{150+50}{2}\\ &=100 MPa\\ &\text{Goodman sequation}\\ \frac{\sigma _a}{S_e}+\frac{\sigma _m}{S_{ut}}&=\frac{1}{FS}\\ \frac{50}{200}+\frac{100}{400}&=\frac{1}{FS}\\ FS&=2 \end{aligned}

When a component is subjected to fluctuating stresses there is mean stress (\sigma _m) as well as stress amplitude (\sigma _a). It has been observed that the mean stress component has an effect on fatigue failure when it is present in combination with an alternating component.
When the component is subjected to both component of stress (\sigma _m) and (\sigma _a) the actual failure occurs at different scattered points as shown in figure. There exist a border which decides sole region from unsole region for various combinations of (\sigma _m) and (\sigma _a) and different criterion are proposed to constant the borderline dividing safe zone and failure zone, they include Gerber line, soderberg line, Goodman line, ASME elliptic.

Gerber Line:
A parabolic curve jointing \sigma _e on the ordinate to \sigma _{ut} on the abscissa is called Gerber line. It is written as
\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{ut}} \right )^2=1

Soderberg Line:
A straight line joining \sigma _e on the ordinate to \sigma _{yt} on the abscissa is called soderberg line. It is written as
\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{yt}} \right )=1

Goodman Line:
A straight line joining \sigma _e on the ordinate to \sigma _{ut} on the abscissa is called the Goodman line. It is written as
\left (\frac{\sigma _a}{\sigma _e} \right )+\left (\frac{\sigma _m}{\sigma _{ut}} \right )=1

ASME Elliptic:
The ASME elliptic is written as
\left (\frac{\sigma _a}{\sigma _e} \right )^2+\left (\frac{\sigma _m}{\sigma _{yt}} \right )^2=1

The soderberg line is a more conservative failure criterion and there is no need to consider even yielding in this case.

Equation of straight line connecting \left(3, \log _{10} 490\right)
and \left(6, \log _{10} 70\right)
\begin{aligned} \frac{y-1.8451}{x-6}&=\frac{1.84-2.6902}{6-3} \\ y-1.8451&=-0.2817(x-6)\\ \text{ At }\quad y&=\log _{10} 100=2 \\ \Rightarrow 2-1.8451 &=-0.2817(x-6) \\ x &=5.4501 \\ \log _{10} N &=5.4501 \\ N&=281914 \end{aligned}

We know that
K_{f}=1+q\left(K_{t}-1\right), \quad q=\frac{K_{f}-1}{K_{t}-1}