Chapter 6, Problem 27P

An M14 × 2 hex-head bolt with a nut is used to clamp together two 20-mm steel plates. Compare the results of finding the overall member stiffness by use of Eqs. (8-20), (8-22), and (8-23).

6-27 Using the modified Goodman criterion for infinite life, repeat Prob. 6-25 for each of the following loading conditions:

1. (a)   0 kN to 28 kN
2. (b)   12 kN to 28 kN
3. (c)   −28 kN to 12 kN

To determine

The yield factor of safety.

The fatigue factor of safety using Goodman criterion.

The number of cycle to failure.

Answer to Problem 27P

The yield factor of safety is $3.32$.

The fatigue factor of safety using Goodman criterion $0.95$.

The number of cycle to failure is $586000\text{cycles}$.

Explanation of Solution

Write the expression of endurance limit.

$S_e{}^{\prime }=0.5S_{ut}$ (I)

Here, endurance limit is $S_e{}^{\prime }$, and the minimum tensile strength of material is $S_{ut}$.

Write the expression for surface modification factor.

$k_a=a{S}_{ut}^b$ (II)

Here, surface modification factor is $k_a$ and the constants are $a$ and $b$.

Write the expression for endurance limit at critical location of a machine part in the geometry.

$S_e=k_a{k}_b{k}_c{S}_e^{\text{'}}$.                               (III)

Here, the endurance limit at critical location is $S_e$, the surface modification factor is $k_a$, the size modification factor is $k_b$, the load modification factor is $k_c$.

Write the expression for fatigue stress concentration factor.

$K_f=1+q\left(K_t-1\right)$ (IV)

Here, the fatigue stress concentration is factor is $K_f$, the notch sensitivity factor is $q$ and theoretical stress concentration factor is the $K_t$.

Write the expression for area of the steel bar at the hole or the minimum cross section area on which load is acting.

$A=\left(w-d\right)h$ (V)

Here, width of the bar is $w$, diameter of hole is $d$ and thickness of bar is $h$.

Write the expression for the maximum stress.

${\sigma }_{\left(\max \right)}=\frac{F_{\max }}{A}$ (VI)

Here, the maximum force acting on the bar is $F_{\max }$ and maximum stress produced in the bar is ${\sigma }_{\max }$.

Write the expression for the minimum stress.

${\sigma }_{\min }=\frac{F_{\min }}{A}$ (VII)

Here, the minimum force acting on the bar is $F_{\min }$ and minimum stress acting on the bar is ${\sigma }_{\min }$.

Write the expression for the nominal amplitude stress.

${\sigma }_{ao}=\left|\frac{{\sigma }_{\max }-{\sigma }_{\min }}{2}\right|$ (VIII)

Here, the nominal amplitude stress is ${\sigma }_{ao}$.

Write the expression for the nominal midrange stress.

${\sigma }_{mo}=\left|\frac{{\sigma }_{\max }+{\sigma }_{\min }}{2}\right|$ (IX)

Here, the nominal amplitude stress for is ${\sigma }_{mo}$.

Write the expression for the amplitude component.

${\sigma }_a=K_f\times {\sigma }_{ao}$ (X)

Here, the amplitude component is ${\sigma }_a$ and the above expression is valid when notch present in the component.

Write the expression for the midrange component.

${\sigma }_m=K_f\times {\sigma }_{mo}$ . (XI)

Here, the midrange component is ${\sigma }_m$.

The Equation (XI) is valid when notch is present in the component.

Write the Expression for the yield factor of safety.

$n_y=\left|\frac{S_y}{{\sigma }_{\max }}\right|$ (XII)

Here, the yield factor of safety is $n_y$ and the yield strength is $S_y$.

Write the expression for the fatigue factor of safety using the modified Goodman criterion.

${\left(n_f\right)}_G=\frac{1}{\frac{{\sigma }_a}{S_e}+\frac{{\sigma }_{\mathrm{m}}}{S_{ut}}}$ (XIII)

Write the expression for the completely reversed stress.

${\sigma }_{rev}=\frac{{\sigma }_a}{1-\frac{{\sigma }_{\mathrm{m}}}{S_{ut}}}$ (XIV)

Here, the fatigue factor of safety using Goodman criterion is ${\left(n_f\right)}_G$.

Write the expression of number of cycle in case of completely reversed stress.

$N={\left(\frac{{\sigma }_{rev}}{a}\right)}^{1/b}$ . (XV)

Here, the number of cycle are $N$, completely reversed stress is ${\sigma }_{rev}$, constants are $a$ and $b$.

Write the expression for the constant $a$.

$a=\frac{{\left(f{S}_{ut}\right)}^2}{S_e}$ (XVI)

Here, the strength fraction is $f$.

Write the expression for the constant $b$.

$b=-\frac{1}{3}\log \left(\frac{f{S}_{ut}}{S_e}\right)$ (XVII)

Conclusion:

Substitute $590\text{ MPa}$ for $S_{UT}$ in Equation (I).

$\begin{array}{c}{S}_e{}^{\prime }=0.5\times 590\text{MPa}\\ =295\text{MPa}\end{array}$

Refer to Table 6-2 “Parameter of Marin surface modification factor” to obtain the constant $a$ as $0.451$ and the constant $b$ as $-0.265$.

Substitute $4.51$ for $a$ and $-0.265$ for $b$ in Equation (II).

$\begin{array}{c}{k}_a=4.51\times {\left(590\text{MPa}\right)}^{-0.265}\\ =0.832\end{array}$

Refer to equation $6-21$ to obtain $k_b$ as $1$ and equation $6-26$ to obtain the $k_c$ as $0.85$.

Substitute $0.897$ for $K_a$, $1$ for $k_b$ and $0.85$ for $k_c$ in Equation (III).

$\begin{array}{c}{\text{S}}_{\text{e}}=\text{0}\text{.832×1×0}\text{.85×}\left(\text{295}\text{MPa}\right)\\ =\text{208}\text{.6}\text{MPa}\end{array}$

Refer to fig $6-20$ “Notch sensitivity chart for materials” the value of $q$ is $0.83$ and refer to fig $\text{A-15-1}$ “The bar in simple tension or compression with transverse hole” at $d/w=0.24$ the value of theoretical stress concentration factor is $2.44$.

Substitute $2.44$ for $k_t$ and $0.83$ for $q$ in Equation (IV)

$\begin{array}{c}{K}_f=1+0.83\times \left(2.44-1\right)\\ =2.195\\ \approx 2.20\end{array}$

Substitute $\text{25}\text{mm}$ for $w$ and $\text{10}\text{mm}$ for $h$ in Equation (V).

$\begin{array}{c}A=\left(\text{25}\text{mm}-\text{6}\text{mm}\right)\left(\text{10}\text{mm}\right)\\ =\left(19\text{mm}\right)\left(\text{10}\text{mm}\right)\\ =\text{190}{\text{mm}}^{\text{2}}\end{array}$

Substitute $28\text{kN}$ for $F_{\max }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (VI).

$\begin{array}{c}{\sigma }_{\max }=\frac{\text{28}\text{kN}}{\text{190}{\text{mm}}^{\text{2}}}\\ =0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =147.4\text{MPa}\end{array}$

Substitute $0\text{kN}$ for $F_{\min }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (VII).

$\begin{array}{c}{\sigma }_{\min }=\frac{0\text{kN}}{190{\text{mm}}^2}\\ =0\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =0\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =0\text{MPa}\end{array}$

Substitute $\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\max }$ and $0\text{MPa}$ for ${\sigma }_{\min }$ in Equation (VIII).

$\begin{array}{c}{\sigma }_{ao}=\left|\frac{\left(\text{147}\text{.4}\text{MPa}\right)-\left(0\text{MPa}\right)}{2}\right|\\ =\frac{147.4\text{MPa}}{2}\\ =73.7\text{MPa}\end{array}$

Substitute $\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\max }$ and $-\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\min }$ in Equation (IX).

$\begin{array}{c}{\sigma }_{mo}=\left|\frac{\left(147.4\text{MPa}\right)+\left(0\text{MPa}\right)}{2}\right|\\ =\frac{147.4\text{}\text{MPa}}{2}\\ =73.7\text{MPa}\end{array}$

Substitute $73.7\text{MPa}$ for ${\sigma }_{ao}$ and $2.20$ for $K_f$ in Equation (X).

$\begin{array}{c}{\sigma }_a=2.20\times \left(73.7\text{MPa}\right)\\ =162.1\text{MPa}\end{array}$

Substitute $73.7\text{MPa}$ for ${\sigma }_{mo}$ and $2.20$ for $K_f$ in Equation (XI).

$\begin{array}{c}{\sigma }_{\text{m}}=2.20\times \left(73.7\text{MPa}\right)\\ =162.1\text{MPa}\end{array}$

Substitute $490\text{MPa}$ for $S_y$ and $147.4\text{MPa}$ for ${\sigma }_{\max }$ in Equation (XII).

$\begin{array}{c}{n}_y=\left|\frac{490\text{MPa}}{147.4\text{MPa}}\right|\\ =3.32\end{array}$

Thus, the yield factor of safety is $3.32$.

Substitute $208.6\text{MPa}$ for $S_e$, $162.1\text{MPa}$ for ${\sigma }_a$, $162.1\text{MPa}$ for ${\sigma }_{\mathrm{m}}$, and $590\text{MPa}$ for $S_{ut}$ in Equation (XIII).

$\begin{array}{c}{\left(n_f\right)}_G=\frac{1}{\frac{\text{162}\text{.1}\text{MPa}}{\text{208}\text{.6}\text{MPa}}\text{+}\frac{\text{162}\text{.1}\text{MPa}}{\text{590}\text{MPa}}}\\ =\frac{1}{0.777+0.2747}\\ =0.95\end{array}$

Thus, the fatigue factor of safety using Goodman criterion is $0.95$.

Substitute $162.1\text{MPa}$ for ${\sigma }_a$, $162.1\text{MPa}$ for ${\sigma }_{\mathrm{m}}$, and $590\text{MPa}$ for $S_{ut}$ in Equation (XIV).

$\begin{array}{c}{\sigma }_{rev}=\frac{162.1\text{MPa}}{1-\frac{162.1\text{MPa}}{590\text{MPa}}}\\ =\frac{162.1\text{MPa}}{0.7252}\\ =223.5\text{MPa}\end{array}$

From fig-$6-18$ “fatigue strength fraction $f$ of $S_{ut}$ at ${10}^3$ cycles for $S_e$ and $S^{\text{'}}{}_e$ at ${10}^6$ cycles” the value of $f$ is $0.87$.

Substitute $0.87$ for $f$, $\text{590}\text{MPa}$ and $208.6\text{MPa}$ for $S_e$ in Equation (XVI).

$\begin{array}{c}a=\frac{{\left(\left(0.87\right)\left(590\text{MPa}\right)\right)}^2}{208.6\text{MPa}}\\ =\frac{263476.89}{208.6}\text{MPa}\\ =1263.07\text{MPa}\\ \approx 1263.07\text{MPa}\end{array}$

Substitute $0.87$ for $f$, $\text{590}\text{MPa}$ for $S_{ut}$ and $208.6\text{MPa}$ for $S_e$ in Equation (XVII).

$\begin{array}{c}b=-\frac{1}{3}\log \left(\frac{\left(0.87\right)\left(590\text{MPa}\right)}{208.6\text{MPa}}\right)\\ =-\frac{1}{3}\log \left(2.4606\right)\\ =-\frac{1}{3}\left(0.39105\right)\\ =-0.1304\end{array}$

Substitute $1263\text{MPa}$ for $a$, $-0.1304$ for $b$, and $223.5$ for ${\sigma }_{rev}$ in Equation (XV).

$\begin{array}{c}N={\left(\frac{223.5\text{MPa}}{1263\text{MPa}}\right)}^{1/-0.1304}\\ ={\left(0.17695\right)}^{1/-0.1304}\\ =585916.9\text{cycles}\\ \approx 586000\text{cycles}\end{array}$

Thus, the number of cycle to failure is $586000\text{cycles}$.

To determine

The Yield factor of safety.

The fatigue factor of safety using Goodman criterion.

Answer to Problem 27P

The yield factor of safety is $3.32$.

The fatigue factor of safety is $1.2$.

Explanation of Solution

Write the expression of maximum stress.

${\sigma }_{\left(\max \right)}=\frac{F_{\max }}{A}$ (XVIII)

Here, the maximum force acting on the bar is $F_{\max }$ and maximum stress produced in the bar is ${\sigma }_{\max }$.

Write the expression of minimum stress.

${\sigma }_{\min }=\frac{F_{\min }}{A}$ (XIX)

Here, the minimum force acting on the bar is $F_{\min }$ and minimum stress acting on the bar is ${\sigma }_{\min }$.

Write the expression of nominal amplitude stress.

${\sigma }_{ao}=\left|\frac{{\sigma }_{\max }-{\sigma }_{\min }}{2}\right|$ (XX)

Here, the nominal amplitude stress is ${\sigma }_{ao}$.

Write the expression of nominal midrange stress.

${\sigma }_{mo}=\left|\frac{{\sigma }_{\max }+{\sigma }_{\min }}{2}\right|$ (XXI)

Here, the nominal amplitude stress for is ${\sigma }_{mo}$.

Write the expression of amplitude component.

${\sigma }_a=K_f\times {\sigma }_{ao}$ (XXII)

Here, the amplitude component is ${\sigma }_a$ and the above expression is valid when notch present in the component.

Write the expression of midrange component.

${\sigma }_m=K_f\times {\sigma }_{mo}$ . (XXIII)

Here, the midrange component is ${\sigma }_m$ and the above expression is valid when notch present in the component.

Write the Expression of yield factor of safety.

$n_y=\left|\frac{S_y}{{\sigma }_{\max }}\right|$ (XXIV)

Here, the yield factor of safety is $n_y$ and the yield strength is $S_y$.

Write the expression of fatigue factor of safety using the modified Goodman criterion.

${\left(n_f\right)}_G=\frac{1}{\frac{{\sigma }_a}{S_e}+\frac{{\sigma }_{\mathrm{m}}}{S_{ut}}}$ (XXV)

Conclusion:

Substitute $28\text{kN}$ for $F_{\max }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (XVIII)

$\begin{array}{c}{\sigma }_{\max }=\frac{\text{28}\text{kN}}{\text{190}{\text{mm}}^{\text{2}}}\\ =0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =147.4\text{MPa}\end{array}$

Substitute $12\text{kN}$ for $F_{\min }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (XIX).

$\begin{array}{c}{\sigma }_{\min }=\frac{\text{12}\text{kN}}{\text{190}{\text{mm}}^{\text{2}}}\\ =0.0632\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =0.0632\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =63.2\text{MPa}\end{array}$

Substitute $\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\max }$ and $63.2\text{MPa}$ for ${\sigma }_{\min }$ in Equation (XX).

$\begin{array}{c}{\sigma }_{ao}=\left|\frac{\left(\text{147}\text{.4}\text{MPa}\right)-\left(63.2\text{MPa}\right)}{2}\right|\\ =\frac{84.2\text{MPa}}{2}\\ =42.1\text{MPa}\end{array}$

Substitute $\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\max }$ and $63.2\text{MPa}$ for ${\sigma }_{\min }$ in Equation (XXI).

$\begin{array}{c}{\sigma }_{mo}=\left|\frac{\left(147.4\text{MPa}\right)+\left(63.2\text{MPa}\right)}{2}\right|\\ =\frac{210.6\text{MPa}}{2}\\ =105.3\text{MPa}\end{array}$

Substitute $42.1\text{MPa}$ for ${\sigma }_{ao}$ and $2.20$ for $K_f$ in Equation (XXI).

$\begin{array}{c}{\sigma }_a=2.20\times 42.1\text{MPa}\\ =92.83\text{MPa}\end{array}$

Substitute $105.3\text{MPa}$ for ${\sigma }_{mo}$ and $2.20$ for $K_f$ in Equation (XXIII).

$\begin{array}{c}{\sigma }_{\text{m}}=2.20\times 105.3\text{MPa}\\ =231.7\text{MPa}\end{array}$

Substitute $490\text{MPa}$ for $S_y$ and $147.4\text{MPa}$ for ${\sigma }_{\max }$ in Equation (XXIV).

$\begin{array}{c}{n}_y=\left|\frac{490\text{MPa}}{147.4\text{MPa}}\right|\\ =3.32\end{array}$

Thus, the yield factor of safety is $3.32$.

Substitute $208.6\text{MPa}$ for $S_e$, $92.83\text{MPa}$ for ${\sigma }_a$, $231.7\text{MPa}$ for ${\sigma }_{\mathrm{m}}$, and $590\text{MPa}$ for $S_{ut}$ in Equation (XXV)

$\begin{array}{c}{\left(n_f\right)}_G=\frac{1}{\frac{92.83\text{MPa}}{\text{208}\text{.6}\text{MPa}}\text{+}\frac{231.5\text{MPa}}{\text{590}\text{MPa}}}\\ =\frac{1}{0.83738}\\ =1.194\\ \approx 1.2\end{array}$

Thus, the fatigue factor of safety using Goodman criterion is $1.2$.

To determine

Yield factor of safety.

The fatigue factor of safety using Goodman criterion.

Number of cycle to failure.

Answer to Problem 27P

The yield factor of safety is $3.32$.

The fatigue factor of safety is $0.905$.

The number of cycle to failure is $446000\text{cycles}$.

Explanation of Solution

Write the expression of maximum stress.

${\sigma }_{\left(\max \right)}=\frac{F_{\max }}{A}$ (XXVI)

Here, the maximum force acting on the bar is $F_{\max }$ and maximum stress produced in the bar is ${\sigma }_{\max }$.

Write the expression of minimum stress acting on the bar.

${\sigma }_{\min }=\frac{F_{\min }}{A}$ (XXVII)

Here, the minimum force acting on the bar is $F_{\min }$ and minimum stress acting on the bar is ${\sigma }_{\min }$.

Write the expression of nominal amplitude stress.

${\sigma }_{ao}=\left|\frac{{\sigma }_{\max }-{\sigma }_{\min }}{2}\right|$ (XXVIII)

Here the nominal amplitude stress is ${\sigma }_{ao}$.

Write the expression of nominal midrange stress.

${\sigma }_{mo}=\left|\frac{{\sigma }_{\max }+{\sigma }_{\min }}{2}\right|$ (XXIX)

Here, the nominal amplitude stress for is ${\sigma }_{mo}$.

Write the expression of amplitude component.

${\sigma }_a=K_f\times {\sigma }_{ao}$ (XXX)

Here, the amplitude component is ${\sigma }_a$ and the above expression is valid when notch present in the component.

Write the expression of midrange component.

${\sigma }_m=K_f\times {\sigma }_{mo}$ . (XXXI)

Here, the midrange component is ${\sigma }_m$ and the above expression is valid when notch present in the component.

Write the Expression of yield factor of safety.

$n_y=\left|\frac{S_y}{{\sigma }_{\min }}\right|$ (XXXII)

Here the yield factor of safety is $n_y$ and the yield strength is $S_y$.

Write the expression of fatigue factor of safety using the modified Goodman criterion.

${\left(n_f\right)}_G=\frac{S_e}{{\sigma }_a}$ (XXXIII)

Here the fatigue factor of safety is $n_f$.

Write the expression of number of cycle in case of completely reversed stress.

$N={\left(\frac{{\sigma }_{rev}}{a}\right)}^{1/b}$

Here, the number of cycle are $N$, completely reversed stress is ${\sigma }_{rev}$, constants are $a$ and $b$.

Simplify the Equation (IX) for completely reversed stress and substitute ${\sigma }^{\text{'}}{}_a$ for ${\sigma }_{rev}$ in above Equation.

$N={\left(\frac{{\sigma }_a}{a}\right)}^{1/b}$ (XXXIV)

Write the expression of constant $a$.

$a=\frac{{\left(f{S}_{ut}\right)}^2}{S_e}$ (XXXV)

Here, the strength fraction is $f$.

Write the expression of the constant $b$.

$b=-\frac{1}{3}\log \left(\frac{f{S}_{ut}}{S_e}\right)$ (XXXVI)

Conclusion:

Substitute $12\text{kN}$ for $F_{\max }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (XXVI).

$\begin{array}{c}{\sigma }_{\max }=\frac{\text{12}\text{kN}}{\text{190}{\text{mm}}^{\text{2}}}\\ =0.0632\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =0.0632\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =63.2\text{MPa}\end{array}$

Substitute $-28\text{kN}$ for $F_{\min }$ and $\text{190}{\text{mm}}^{\text{2}}$ for $A$ in Equation (XXVII).

$\begin{array}{c}{\sigma }_{\min }=\frac{-28\text{kN}}{190{\text{mm}}^2}\\ =-0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\\ =-0.14737\left(\text{kN}/{\text{mm}}^{\text{2}}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =-147.4\text{MPa}\end{array}$

Substitute $\text{63}\text{.2}\text{MPa}$ for ${\sigma }_{\max }$ and $-\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\min }$ in Equation (XXVIII).

$\begin{array}{c}{\sigma }_{ao}=\left|\frac{\left(\text{63}\text{.2}\text{MPa}\right)-\left(-\text{147}\text{.4}\text{MPa}\right)}{2}\right|\\ =\frac{210.6\text{MPa}}{2}\\ =105.3\text{MPa}\end{array}$

Substitute $\text{63}\text{.2}\text{MPa}$ for ${\sigma }_{\max }$ and $-\text{147}\text{.4}\text{MPa}$ for ${\sigma }_{\min }$ in Equation (XXIX).

$\begin{array}{c}{\sigma }_{mo}=\left|\frac{\left(\text{63}\text{.2}\text{MPa}\right)+\left(-147.4\text{MPa}\right)}{2}\right|\\ =\frac{-84.2\text{MPa}}{2}\\ =-42.1\text{MPa}\end{array}$

Substitute $105.2\text{MPa}$ for ${\sigma }_{ao}$ and $2.20$ for $K_f$ in Equation (XXX).

$\begin{array}{c}{\sigma }_a=2.20\times 105.2\text{MPa}\\ =\text{231}\text{.44}\text{MPa}\\ \approx 231.4\text{MPa}\end{array}$

Substitute $-42.1\text{MPa}$ for ${\sigma }_{mo}$ and $2.20$ for $K_f$ in Equation (XXXI).

$\begin{array}{c}{\sigma }_{\text{m}}=2.20\times \left(-42.1\text{MPa}\right)\\ =-92.63\text{MPa}\\ =-92.62\text{MPa}\end{array}$

Substitute $490\text{MPa}$ for $S_y$ and $-147.4\text{MPa}$ for ${\sigma }_{\min }$ in Equation (XXXII).

$\begin{array}{c}{n}_y=\left|\frac{490\text{MPa}}{-147.4\text{MPa}}\right|\\ =\frac{490}{147.4}\\ =3.32\end{array}$

Thus, the yield factor of safety is $3.32$.

Substitute $208.6\text{MPa}$ for $S_e$, $231.4\text{MPa}$ for ${\sigma }_a$ in Equation (XXXIII).

$\begin{array}{c}{\left(n_f\right)}_G=\frac{208.6\text{MPa}}{231.4\text{MPa}}\\ =\frac{208.6}{231.4}\\ =0.905\end{array}$

Thus, the fatigue factor of safety is $0.905$.

From fig-$6-18$ “Fatigue strength fraction $f$ of $S_{ut}$ at ${10}^3$ cycles for $S_e$ and $S^{\text{'}}{}_e$ at ${10}^6$ cycles” the value of $f$ is $0.87$.

Substitute $0.87$ for $f$, $\text{590}\text{MPa}$ and $208.6\text{MPa}$ for $S_e$ in Equation (XXXV).

$\begin{array}{c}a=\frac{{\left(\left(0.87\right)\left(590\text{MPa}\right)\right)}^2}{208.6\text{MPa}}\\ =\frac{263476.89}{208.6}\text{MPa}\\ =1263.07\text{MPa}\\ \approx 1263\text{MPa}\end{array}$

Substitute $0.87$ for $f$, $\text{590}\text{MPa}$ for $S_{ut}$ and $208.6\text{MPa}$ for $S_e$ in Equation (XXXVI).

$\begin{array}{c}b=-\frac{1}{3}\log \left(\frac{\left(0.87\right)\left(590\text{MPa}\right)}{208.6\text{MPa}}\right)\\ =-\frac{1}{3}\left(0.39105\right)\\ =-0.13035\\ \approx -0.1304\end{array}$

Substitute $1263\text{MPa}$ for $a$, $-0.1304$ for $b$, and $231.4\text{MPa}$ for ${\sigma }_a$ in Equation (XIV).

$\begin{array}{c}N={\left(\frac{231.4\text{MPa}}{1263\text{Mpa}}\right)}^{1/\left(-0.1304\right)}\\ ={\left(0.1832\right)}^{1/\left(-0.1304\right)}\\ =448897.52\text{cycles}\\ \approx 446000\text{cycles}\end{array}$

Thus, the number of cycle to failure is $446000\text{cycles}$.