Chapter 6, Problem 41P

6-37* to 6-46* For the problem specified in the table, build upon the results of the original problem to determine the minimum factor of safety for fatigue based on infinite life, using the modified Goodman criterion. The shaft rotates at a constant speed, has a constant diameter, and is made from cold-drawn AISI 1018 steel.

Problem Number	Original Problem, Page Number
6-41*	3–72, 152

3-72* to 3-73* A gear reduction unit uses the countershaft shown in the figure. Gear A receives power from another gear with the transmitted force F_A applied at the 20° pressure angle as shown. The power is transmitted through the shaft and delivered through gear B through a transmitted force F_B at the pressure angle shown.

1. (a)   Determine the force F_B assuming the shaft is running at a constant speed.
2. (b)   Find the bearing reaction forces, assuming the bearings act as simple supports.
3. (c)   Draw shear-force and bending-moment diagrams for the shaft. If needed, make one set for the horizontal plane and another set for the vertical plane.
4. (d)   At the point of maximum bending moment, determine the bending stress and the torsional shear stress.
5. (e)   At the point of maximum bending moment, determine the principal stresses and the maximum shear stress.

To determine

The minimum factor of safety for fatigue based on infinite life.

Answer to Problem 41P

The minimum factor of safety for fatigue based on infinite life is $0.62$.

Explanation of Solution

The following figure shows the free body diagram of the gear A.

Figure-(1)

The following figure shows the free body diagram of the gear B.

Figure-(2)

Calculate the force $F_B$, using the net torque equation.

$\begin{array}{l}{\displaystyle \sum T}=0\\ -\left(F_A\times \frac{d_A}{2}\times \cos {\theta }_1\right)+\left(F_B\times \frac{d_B}{2}\times \cos {\theta }_2\right)=0\\ F_B=F_A\frac{d_A\cos {\theta }_1}{d_B\cos {\theta }_2}\end{array}$ (I)

Here, the force acting on pulley $A$ is $F_A$, the diameter of pulley $A$ is $d_A$, the angle at which force acts on pulley $A$ is ${\theta }_1$, the force acting on pulley $B$ is $F_B$, the diameter of pulley $B$ is $d_B$ and the angle at which force acts on pulley $B$ is ${\theta }_2$.

Write the moment about bearing $O$ in $z\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum M_{Oz}}=0\\ \left[\begin{array}{l}\left(F_A\cos {\theta }_1\times l_{OA}\right)\\ -\left(F_B\sin {\theta }_2\times \left(l_{OA}+l_{AC}+l_{CB}\right)\right)\\ +R_{Cy}\times \left(l_{OA}+l_{AC}\right)\end{array}\right]=0\\ R_{Cy}=\frac{\left[\begin{array}{l}\left(F_B\sin {\theta }_2\times \left(l_{OA}+l_{AC}+l_{CB}\right)\right)\\ -\left(F_A\cos {\theta }_1\times l_{OA}\right)\end{array}\right]}{\left(l_{OA}+l_{AC}\right)}\end{array}$ (II)

Here, the reaction force at bearing $C$ in $y\text{-}$ direction is $R_{Cy}$, the distance between $O$ and $A$ is $l_{OA}$, the distance between $A$ and $C$ is $l_{AC}$ and the distance between $C$ and $B$ is $l_{CB}$.

Write the equation to balance the forces in $y\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{Oy}+F_A\cos {\theta }_1+R_{Cy}-F_B\sin {\theta }_2=0\\ R_{Oy}=F_B\sin {\theta }_2-R_{Cy}-F_A\cos {\theta }_1\end{array}$ (III)

Here, the reaction force at bearing $O$ in $y\text{-}$ direction is $R_{Oy}$.

Write the moment about bearing $O$ in $y\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum M_{Oy}}=0\\ \left[\begin{array}{l}\left(F_A\sin {\theta }_1\times l_{OA}\right)-\left(F_B\cos {\theta }_2\times \left(l_{OA}+l_{AC}+l_{CB}\right)\right)\\ -R_{Cz}\times \left(l_{OA}+l_{AC}\right)\end{array}\right]=0\\ R_{Cz}=\frac{\left[\begin{array}{l}\left(F_A\sin {\theta }_1\times l_{OA}\right)\\ -\left(F_B\cos {\theta }_2\times \left(l_{OA}+l_{AC}+l_{CB}\right)\right)\end{array}\right]}{\left(l_{OA}+l_{AC}\right)}\end{array}$ (IV)

Here, the reaction force at bearing $C$ in $z\text{-}$ direction is $R_{Cz}$.

Write the equation to balance the forces in $z\text{-}$ direction.

$\begin{array}{l}{\displaystyle \sum F_z}=0\\ R_{Oz}-F_A\sin {\theta }_1+R_{Cz}+F_B\cos {\theta }_2=0\\ R_{Oz}=F_A\sin {\theta }_1-R_{Cz}-F_B\cos {\theta }_2\end{array}$ (V)

Here, the reaction force at bearing $O$ in $z\text{-}$ direction is $R_{Oz}$.

Calculate the reaction forces at bearing $O$.

$R_O=\sqrt{R_{Oy}^2+R_{Oz}^2}$ (VI)

Here, the reaction force at bearing $O$ is $R_O$.

Calculate the reaction forces at bearing $C$.

$R_C=\sqrt{R_{Cy}^2+R_{Cz}^2}$ (VII)

Here, the reaction force at bearing $C$ is $R_C$.

The calculations for shear force and bending moment diagram in $y\text{-}$ direction.

Calculate the shear force at $O$ in $y\text{-}$ direction.

$S{F}_{Oy}=R_{Oy}$ (VIII)

Here, the shear force at $O$ in $y\text{-}$ direction is $S{F}_{Oy}$.

Calculate the shear force at $A$ in $y\text{-}$ direction.

$S{F}_{Ay}=S{F}_{Oy}+F_A\cos {\theta }_1$ (IX)

Here, the shear force at $A$ in $y\text{-}$ direction is $S{F}_{Ay}$.

Calculate the shear force at $C$ in $y\text{-}$ direction.

$S{F}_{Cy}=S{F}_{Ay}+R_{Cy}$ (X)

Here, the shear force at $C$ in $y\text{-}$ direction is $S{F}_{Cy}$.

Calculate the shear force at $B$ in $y\text{-}$ direction.

$S{F}_{By}=S{F}_{Cy}-F_B\sin {\theta }_2$ (XI)

Here, the shear force at $B$ in $y\text{-}$ direction is $S{F}_{By}$.

Calculate the moment at $O$ and $B$.

$M_O=M_B=0$ (XII)

Here, the moment at $O$ is $M_O$ and the moment at $B$ is $M_B$.

Calculate the moment at $A$ in $y\text{-}$ direction.

$M_A=S{F}_{Oy}\times l_{OA}$ (XIII)

Here, the moment at $A$ in $y\text{-}$ direction is $M_A$.

Calculate the moment at $C$ in $y\text{-}$ direction.

$M_{Cy}=F_B\sin {\theta }_2\times l_{CB}$ (XIV)

Here, the moment at $C$ in $y\text{-}$ direction is $M_{Cy}$.

The calculations for shear force and bending moment diagram in $z\text{-}$ direction.

Calculate the shear force at $O$ in $y\text{-}$ direction.

$S{F}_{Oz}=-R_{Oz}$ (XV)

Here, the shear force at $O$ in $z\text{-}$ direction is $S{F}_{Oz}$.

Calculate the shear force at $A$ in $z\text{-}$ direction.

$S{F}_{Az}=S{F}_{Oz}+F_A\sin {\theta }_1$ (XVI)

Here, the shear force at $A$ in $z\text{-}$ direction is $S{F}_{Az}$.

Calculate the shear force at $C$ in $z\text{-}$ direction.

$S{F}_{Cz}=S{F}_{Az}-R_{Cz}$ (XVII)

Here, the shear force at $C$ in $z\text{-}$ direction is $S{F}_{Cz}$.

Calculate the shear force at $B$ in $z\text{-}$ direction.

$S{F}_{Bz}=S{F}_{Cz}-F_B\cos {\theta }_2$ (XVIII)

Here, the shear force at $B$ in $z\text{-}$ direction is $S{F}_{Bz}$.

Calculate the moment at $O$ and $B$.

$M_O=M_B=0$ (XIX)

Here, the moment at $O$ is $M_O$ and the moment at $B$ is $M_B$.

Calculate the moment at $A$ in $z\text{-}$ direction.

$M_{Az}=S{F}_{Oz}\times l_{OA}$ (XX)

Here, the moment at $A$ in $z\text{-}$ direction is $M_{Az}$.

Calculate the moment at $C$ in $z\text{-}$ direction.

$M_{Cz}=-F_B\cos {\theta }_2\times l_{CB}$ (XXI)

Here, the moment at $C$ in $z\text{-}$ direction is $M_{Cz}$.

Write the net moment at $A$.

$M_A=\sqrt{M_{Ay}^2+M_{Az}^2}$ (XXII)

Here, the net moment at $A$ is $M_A$.

Write the net moment at $C$.

$M_C=\sqrt{M_{Cy}^2+M_{Cz}^2}$ (XXIII)

Here, the net moment at $C$ is $M_C$.

Write the torque transmitted by shaft from $A$ to $B$.

$T=F_A\cos {\theta }_1\times \frac{d_A}{2}$ . (XXIV)

Here, the torque transmitted by shaft from $A$ to $B$ is $T$.

Calculate the bending stress.

$\sigma =\frac{32M_C}{\pi d^3}$ (XXV)

Here, the bending stress is $\sigma$ and diameter of shaft is $d$.

Calculate the shear stress.

$\tau =\frac{16T}{\pi d^3}$ (XVI)

Here, the shear stress is $\tau$.

Calculate the maximum principal stress.

${\sigma }_1=\frac{\sigma }{2}+\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXVII)

Here, the maximum principal stress is ${\sigma }_1$.

Calculate the minimum principal stress.

${\sigma }_2=\frac{\sigma }{2}-\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXVIII)

Here, the minimum principal stress is ${\sigma }_2$.

Calculate the maximum shear stress.

${\tau }_{\max }=\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$ (XXIX)

Here, maximum shear stress is ${\tau }_{\max }$.

Write the expression for von Mises alternating stress.

${{\sigma }^{\prime }}_a={\left({\sigma }_a^2+3{\tau }_a^2\right)}^{\frac{1}{2}}$ (XXX)

Here, the amplitude component of the axial stress is ${\sigma }_a$ and the amplitude component of the torsional stress is ${\tau }_a$.

Write the expression for von Mises mid-range stress.

${{\sigma }^{\prime }}_m={\left({\sigma }_m^2+3{\tau }_m^2\right)}^{\frac{1}{2}}$ (XXXI)

Here, the mid-range component of the axial stress is ${\sigma }_m$ and the mid-range component of the torsional stress is ${\tau }_m$.

Write the expression for von Mises maximum stress.

${{\sigma }^{\prime }}_{\max }={\left({\sigma }_{\max }^2+3{\tau }_{\max }^2\right)}^{\frac{1}{2}}$ (XXXII)

Here, the maximum component of the axial stress is ${\sigma }_m$ and the maximum component of the torsional stress is ${\tau }_m$.

Write the expression for yield factor of safety.

$n_y=\frac{S_y}{\max \left[{{\sigma }^{\prime }}_{\max },{{\sigma }^{\prime }}_m,{{\sigma }^{\prime }}_a\right]}$ (XXXIII)

Here, the yield strength of the material is $S_y$.

Write the expression for endurance limit for test specimen.

${S^{\prime }}_e=0.5\left(S_{ut}\right)$ (XXXIV)

Here, the minimum tensile strength is $S_{ut}$.

Write the surface factor for the countershaft.

$k_a=a\times S_{ut}^b$ (XXXV)

Here, the constants for surface factor are $a$ and $b$.

Write the size factor for the countershaft.

$k_b=0.879d^{-0.107}$ (XXXVI)

Write the endurance limit at the critical location of the machine part.

$S_e={S^{\prime }}_e\cdot k_a\cdot k_b$ (XXXVII)

Write the modified Goodman equation.

$\frac{1}{n_f}=\frac{{{\sigma }^{\prime }}_a}{S_e}+\frac{{{\sigma }^{\prime }}_m}{S_{ut}}$ (XXXVIII)

Here, the fatigue factor of safety is $n_f$.

Conclusion:

Substitute $300\text{lbf}$ for $F_A$, $20\text{in}$ for $d_A$, $20°$ for ${\theta }_1$, $8\text{in}$ for $d_B$ and $20°$ for ${\theta }_2$ in Equation (I).

$\begin{array}{c}{F}_B=\left(300\text{lbf}\right)\times \frac{\left(20\text{in}\right)\cos 20°}{\left(8\text{in}\right)\cos 20°}\\ =\left(300\text{lbf}\right)\times \frac{20\times 0.939}{8\times 0.939}\\ =\left(300\text{lbf}\right)\times 2.5\\ =750\text{lbf}\end{array}$

Substitute $750\text{lbf}$ for $F_B$, $20°$ for ${\theta }_1$, $300\text{lbf}$ for $F_A$, $20°$ for ${\theta }_2$, $16\text{in}$ for $l_{OA}$, $14\text{in}$ for $l_{AC}$ and $9\text{in}$ for $l_{CB}$ in Equation (II).

$\begin{array}{c}{R}_{Cy}=\frac{\left[\begin{array}{l}\left(\left(750\text{lbf}\right)\sin 20°\times \left(16\text{in}+14\text{in}+9\text{in}\right)\right)\\ -\left(300\text{lbf}\cos 20°\times 16\text{in}\right)\end{array}\right]}{\left(16\text{in}+14\text{in}\right)}\\ =\frac{\left[\begin{array}{l}\left(\left(750\text{lbf}\right)0.3420\times \left(16\text{in}+14\text{in}+9\text{in}\right)\right)\\ -\left(300\text{lbf}\times 0.939\times 16\text{in}\right)\end{array}\right]}{\left(16\text{in}+14\text{in}\right)}\\ =\frac{\left[5493.56\text{lbf}\cdot \text{in}\right]}{\left(30\text{in}\right)}\\ =183.1\text{lbf}\end{array}$

Substitute $750\text{lbf}$ for $F_B$, $20°$ for ${\theta }_2$, $183.1\text{lbf}$ for $R_{Cy}$, $300\text{lbf}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (III).

$\begin{array}{c}{R}_{Oy}=\left(750\text{lbf}\right)\sin 20°-183.1\text{lbf}-\left(300\text{lbf}\right)\cos 20\\ =\left(256.515-183.1-281.9\right)\text{lbf}\\ =\left(73.415-281.9\right)\text{lbf}\\ \approx -208.5\text{lbf}\end{array}$

Substitute $750\text{lbf}$ for $F_B$, $20°$ for ${\theta }_1$, $300\text{lbf}$ for $F_A$, $20°$ for ${\theta }_2$, $16\text{in}$ for $l_{OA}$, $14\text{in}$ for $l_{AC}$ and $9\text{in}$ for $l_{CB}$ in Equation (IV).

$\begin{array}{c}{R}_{Cz}=\frac{\left[\begin{array}{l}\left(\left(750\text{lbf}\right)\cos 20°\times \left(16\text{in}+14\text{in}+9\text{in}\right)\right)\\ -\left(\left(300\text{lbf}\right)\sin 20°\times 16\text{in}\right)\end{array}\right]}{\left(16\text{in}+14\text{in}\right)}\\ =\frac{\left[\begin{array}{l}\left(\left(704.76\text{lbf}\right)\times \left(39\text{in}\right)\right)\\ -\left(\left(102.606\text{lbf}\right)\times 16\text{in}\right)\end{array}\right]}{\left(30\text{in}\right)}\\ =\frac{25843.944}{30}\text{lbf}\\ \approx -861.5\text{lbf}\end{array}$

Substitute $300\text{lbf}$ for $F_A$, $20°$ for ${\theta }_1$, $-861.5\text{lbf}$ for $R_{Cz}$, $750\text{lbf}$ for $F_B$ and $20°$ for ${\theta }_2$ in Equation (V).

$\begin{array}{c}{R}_{Oz}=\left(300\text{lbf}\right)\sin 20°-\left(-861.5\text{lbf}\right)-750\text{lbf}\cos 20°\\ =102.6\text{lbf}-861.5\text{lbf}-704.769\text{lbf}\\ =259.3\text{lbf}\end{array}$

Substitute $-208.5\text{lbf}$ for $R_{Oy}$ and $259.3\text{lbf}$ for $R_{Oz}$ in Equation (VI).

$\begin{array}{c}{R}_O=\sqrt{{\left(-208.5\text{lbf}\right)}^2+{\left(259.3\text{lbf}\right)}^2}\\ =\sqrt{110708.74{\text{lbf}}^2}\\ =332.729\text{lbf}\\ \approx 332.73\text{lbf}\end{array}$

Substitute $183.1\text{lbf}$ for $R_{Cy}$ and $-861.5\text{lbf}$ for $R_{Cz}$ in Equation (VII).

$\begin{array}{c}{R}_C=\sqrt{{\left(183.1\text{lbf}\right)}^2+{\left(-861.5\text{lbf}\right)}^2}\\ =\sqrt{775707.86{\text{lbf}}^2}\\ =880.74\text{lbf}\end{array}$

Substitute $-208.5\text{lbf}$ for $R_{Oy}$ in Equation (VIII).

$S{F}_{Oy}=-208.5\text{lbf}$

Substitute $-208.5\text{lbf}$ for $S{F}_{Oy}$, $300\text{lbf}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (IX.)

$\begin{array}{c}S{F}_{Ay}=-208.5+300\cos 20°\\ =73.4\text{lbf}\end{array}$

Substitute $73.4\text{lbf}$ for $S{F}_{Ay}$ and $183.1\text{lbf}$ for $R_C$ in Equation (X).

$\begin{array}{c}S{F}_{Cy}=73.4\text{lbf}+183.1\text{lbf}\\ =256.5\text{lbf}\end{array}$

Substitute $256.5\text{lbf}$ for $S{F}_{Cy}$, $750\text{lbf}$ for $F_B$ and $20°$ for ${\theta }_2$ in Equation (XI).

$\begin{array}{c}S{F}_{By}=256.5\text{lbf}-750\text{lbf}\times \sin 20°\\ =256.5\text{lbf}-256.5\text{lbf}\\ =0\end{array}$

Substitute $-208.5\text{lbf}$ for $S{F}_{Oy}$ and $16\text{in}$ for $l_{OA}$ in Equation (XIII).

$\begin{array}{c}{M}_{Ay}=-208.5\text{lbf}\times 16\text{in}\\ =-3336\text{lbf}\cdot \text{in}\end{array}$

Substitute $750\text{lbf}$ for $F_B$, $20°$ for ${\theta }_2$ and $9\text{in}$ for $l_{CB}$ in Equation (XIV).

$\begin{array}{c}{M}_{Cy}=-750\text{lbf}\sin 20°\times 9\text{in}\\ =\text{256}\text{.515}\text{lbf}\times \text{9}\text{in}\\ =-2308.5\text{lbf}\cdot \text{in}\end{array}$

Thus, the shear force and bending moment diagram in $y\text{-}$ direction is as follow.

Figure-(3)

Substitute $259.3\text{lbf}$ for $R_{Oz}$ in Equation (XV).

$S{F}_{Oz}=-259.3\text{lbf}$

Substitute $-259.3\text{lbf}$ for $S{F}_{Oz}$, $300\text{lbf}$ for $F_A$ and $20°$ for ${\theta }_1$ in Equation (XVI).

$\begin{array}{c}S{F}_{Az}=-259.3\text{lbf}+300\text{lbf}\times \sin 20°\\ =-259.3\text{lbf}+102.6\text{lbf}\\ =-156.7\text{lbf}\end{array}$

Substitute $-156.7\text{lbf}$ for $S{F}_{Az}$ $-861.5\text{lbf}$ for $R_{Cz}$ in Equation (XVII).

$\begin{array}{c}S{F}_{Cz}=-156.7\text{lbf}-\left(-861.5\text{lbf}\right)\\ =-156.7\text{lbf}+861.5\text{lbf}\\ =704.8\text{lbf}\end{array}$

Substitute $704.8\text{lbf}$ for $S{F}_{Cz}$, $750\text{lbf}$ for $F_B$ and $20°$ for ${\theta }_2$ in Equation (XVIII).

$\begin{array}{c}S{F}_{Bz}=704.8\text{lbf}-750\cos 20\\ =704.8\text{lbf}-704.8\text{lbf}\\ =0\text{lbf}\end{array}$

Substitute $-259.3\text{lbf}$ for $S{F}_{Oy}$ and $16\text{in}$ for $l_{OA}$ in Equation (XX).

$\begin{array}{c}{M}_{Az}=-259.3\text{lbf}\times 16\text{in}\\ =-4148.8\text{lbf}\cdot \text{in}\\ \approx -4149\text{lbf}\cdot \text{in}\end{array}$

Substitute $750\text{lbf}$ for $F_B$, $20°$ for ${\theta }_2$ and $9\text{in}$ for $l_{CB}$ in Equation (XXI).

$\begin{array}{c}{M}_{Cz}=-750\text{lbf}\cos 20°\times 9\text{in}\\ =\left(704.769\times 9\right)\text{lbf}\cdot \text{in}\\ =\text{6342}\text{.9}\text{lbf}\cdot \text{in}\end{array}$

Thus, the shear force and bending moment diagram in $z\text{-}$ direction is as follow.

Figure-(4)

Substitute $-3336\text{lbf}\cdot \text{in}$ for $M_{Ay}$ and $-4149\text{lbf}\cdot \text{in}$ for $M_{Az}$ in Equation (XXII).

$\begin{array}{c}{M}_A=\sqrt{{\left(-3336\text{lbf}\cdot \text{in}\right)}^2+{\left(-4149\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{28343097{\left(\text{lbf}\cdot \text{in}\right)}^2}\\ =5323.82\text{lbf}\cdot \text{in}\\ \approx 5324\text{lbf}\cdot \text{in}\end{array}$

Substitute $-2308.5\text{lbf}\cdot \text{in}$ for $M_{Cy}$ and $-6343\text{lbf}\cdot \text{in}$ for $M_{Cz}$ in Equation (XXIII).

$\begin{array}{c}{M}_C=\sqrt{{\left(-2308.5\text{lbf}\cdot \text{in}\right)}^2+{\left(-6343\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{45562821.25{\left(\text{lbf}\cdot \text{in}\right)}^2}\\ \approx 6750\text{lbf}\cdot \text{in}\end{array}$

Since, $M_C>M_A$.

The critical location is at C.

Substitute $300\text{lbf}$ for $F_A$, $20°$ for ${\theta }_1$ and $20\text{in}$ for $d_A$ in Equation (XXIV).

$\begin{array}{c}T=300\text{lbf}\times \cos 20°\times \frac{20\text{in}}{2}\\ =281.9\text{lbf}\times \text{10}\text{in}\\ =2819\text{lbf}\cdot \text{in}\end{array}$

Substitute $6750\text{lbf}\cdot \text{in}$ for $M_C$ and $1.25\text{in}$ for $d$ in Equation (XXV).

$\begin{array}{c}\sigma =\frac{32\times 6750\text{lbf}\cdot \text{in}}{\pi {\left(1.25\text{in}\right)}^3}\\ =\frac{216000\text{lbf}\cdot \text{in}}{\pi \times 1.953{\text{in}}^3}\\ =\frac{216000\text{lbf}\cdot \text{in}}{6.1355{\text{in}}^3}\left(\frac{1\text{kpsi}}{{10}^3\text{lbf}/{\text{in}}^2}\right)\\ =35.2\text{kpsi}\end{array}$

Substitute $2819\text{lbf}\cdot \text{in}$ for $T$ and $1.25\text{in}$ for $d$ in Equation (XXVI).

$\begin{array}{c}\tau =\frac{16\times 2819\text{lbf}\cdot \text{in}}{\pi {\left(1.25\text{in}\right)}^3}\\ =\frac{45104\text{lbf}\cdot \text{in}}{\pi {\left(1.25\text{in}\right)}^3}\\ =\frac{45104\text{lbf}\cdot \text{in}}{6.1355{\text{in}}^3}\left(\frac{1\text{kpsi}}{{10}^3\text{lbf}/{\text{in}}^2}\right)\\ =7.35\text{kpsi}\end{array}$

Substitute $35.2\text{kpsi}$ for $\sigma$ and $7.35\text{kpsi}$ for $\tau$ in Equation (XXVII).

$\begin{array}{c}{\sigma }_1=\frac{35.2\text{kpsi}}{2}+\sqrt{{\left(\frac{35.2\text{kpsi}}{2}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =17.6\text{kpsi}+\sqrt{{\left(17.6\text{kpsi}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =17.6\text{kpsi}+19.07\text{kpsi}\\ \approx 36.7\text{kpsi}\end{array}$

Substitute $35.2\text{kpsi}$ for $\sigma$ and $7.35\text{kpsi}$ for $\tau$ n Equation (XXVIII).

$\begin{array}{c}{\sigma }_2=\frac{35.2\text{kpsi}}{2}-\sqrt{{\left(\frac{35.2\text{kpsi}}{2}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =17.6\text{kpsi}-\sqrt{{\left(17.6\text{kpsi}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =17.6\text{kpsi}-19.07\text{kpsi}\\ =-1.47\text{kpsi}\end{array}$

Substitute $21.5\text{kpsi}$ for $\sigma$ and $5.09\text{kpsi}$ for $\tau$ in Equation (XXIX).

$\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{35.2\text{kpsi}}{2}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =\sqrt{{\left(17.6\text{kpsi}\right)}^2+{\left(7.35\text{kpsi}\right)}^2}\\ =19.07\text{kpsi}\\ \approx 19.1\text{kpsi}\end{array}$

Substitute $35.2\text{kpsi}$ for ${\sigma }_a$, and $0$ for ${\tau }_a$ in Equation (XXX).

$\begin{array}{c}{{\sigma }^{\prime }}_a={\left[{\left(35.2\text{kpsi}\right)}^2+3{\left(0\right)}^2\right]}^{\frac{1}{2}}\\ ={\left[{\left(35.2\text{kpsi}\right)}^2\right]}^{\frac{1}{2}}\\ =35.2\text{kpsi}\end{array}$

Substitute $0$ for ${\sigma }_m$, and $7.35\text{kpsi}$ for ${\tau }_a$ in Equation (XXXI).

$\begin{array}{c}{{\sigma }^{\prime }}_m={\left[{\left(0\right)}^2+3{\left(7.35\text{kpsi}\right)}^2\right]}^{\frac{1}{2}}\\ ={\left[162.067{\text{kpsi}}^2\right]}^{\frac{1}{2}}\\ =12.7\text{kpsi}\end{array}$

Substitute $35.2\text{kpsi}$ for ${\sigma }_{\max }$, and $7.35\text{kpsi}$ for ${\tau }_{\max }$ in Equation (XXXII).

$\begin{array}{c}{{\sigma }^{\prime }}_{\max }={\left[{\left(35.2\text{kpsi}\right)}^2+3{\left(7.35\text{kpsi}\right)}^2\right]}^{\frac{1}{2}}\\ ={\left[{\left(35.2\text{kpsi}\right)}^2+3\times 54.0225{\text{kpsi}}^2\right]}^{\frac{1}{2}}\\ ={\left[1401.1075{\text{kpsi}}^2\right]}^{\frac{1}{2}}\\ =37.4\text{kpsi}\end{array}$

Refer to the Table A-20 “Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels” to obtain the yield strength as $54\text{kpsi}$ and the minimum tensile strength as $64\text{kpsi}$ for AISI 1018 steel.

Substitute $37.4\text{kpsi}$ for ${{\sigma }^{\prime }}_{\max }$ and $54\text{kpsi}$ for $S_y$ in Equation (XXXIII).

$\begin{array}{c}{n}_y=\frac{54\text{kpsi}}{37.4\text{kpsi}}\\ =1.443\end{array}$

Substitute $64\text{kpsi}$ for $S_{ut}$ in Equation (XXXIV).

$\begin{array}{c}{S^{\prime }}_e=0.5\left(64\text{kpsi}\right)\\ =32\text{kpsi}\end{array}$

Refer to Table 6-2 “Parameters for Marin Surface Modification Factor” to obtain $2.70$ as $a$ and $-0.265$ for $b$.

Substitute $2.70$ for $a$ and $-0.265$ for $b$ in Equation (XXXV).

$\begin{array}{c}{k}_a=2.70{\left(64\right)}^{-0.265}\\ =0.9\end{array}$

Substitute $1.25\text{in}$ for $d$ in Equation (XXXVI).

$\begin{array}{c}{k}_b=0.879{\left(1.25\right)}^{-0.107}\\ =0.86\end{array}$

Substitute $0.86$ for $k_b$, $0.9$ for $k_a$ and $32\text{kpsi}$ for ${S^{\prime }}_e$ in Equation (XXXVII).

$\begin{array}{c}{S}_e=0.9\left(0.86\right)\left(32\text{kpsi}\right)\\ =0.774\left(32\text{kpsi}\right)\\ =24.768\text{kpsi}\\ \approx 24.8\text{kpsi}\end{array}$

Substitute $35.2\text{kpsi}$ for ${{\sigma }^{\prime }}_a$, $24.8\text{kpsi}$ for $S_e$, $64\text{kpsi}$ for $S_{ut}$ and $12.7\text{kpsi}$ for ${{\sigma }^{\prime }}_m$ in Equation (XXXVIII).

$\begin{array}{c}\frac{1}{n_f}=\frac{35.2\text{kpsi}}{24.8\text{kpsi}}+\frac{12.7\text{kpsi}}{64\text{kpsi}}\\ =1.617\\ n_f=0.618\\ \approx 0.62\end{array}$

Thus, the minimum factor of safety for fatigue based on infinite life is $0.62$.