Chapter 8, Problem 54P

8–51 to

8–54 For the pressure cylinder defined in the problem specified in the table, the gas pressure is cycled between zero and p_g. Determine the fatigue factor of safety for the bolts using the following failure criteria:

(a)    Goodman.

(b)    Gerber.

(c)    ASME-elliptic.

Problem Number	Originating Problem Number
8–54	8–36

To determine

The fatigue factor of safety for the bolts using Goodman criteria.

Answer to Problem 54P

The fatigue factor of safety for the bolts using Goodman criteria is $4.72$.

Explanation of Solution

Write the expression of the length of the material squeeze between the bolt face and washer face.

    $l=A+B$                                 (I)

Here, the length of the material squeeze between the bolt face and washer face is $l$, and thickness of pressure vessel  inside the bolt is $A$, the thickness of the cylinder inside the bolt is $B$.

Write the expression for the length of the bolt.

    $L\ge l+H$                                      (II)

Here, the length of bolt is $L$, length of the material squeeze between the bolt face and washer face is $l$ and the thickness of the bolt is $H$.

Write the expression of the threaded length for hexagonal bolt.

    $L_T=2d+0.25$                               (III)

Here, the threaded length is $L_T$.

Write the expression of the length of the unthreaded portion in grip.

    $l_d=L-L_T$                                          (IV)

Here, the length of the unthreaded portion in the grip is $l_d$.

Write the expression of the length of the threaded portion in grip.

    $l_t=l-l_d$                                                 (V)

Here, the length of threaded portion in the grip is $l_t$.

Write the expression of the major area diameter.

    $A_d=\frac{\pi }{4}{d}^2$                                           (VI)

Here, the nominal diameter of the bolt is $d$.

Write the expression of the stiffness for the bolt.

    $k_b=\frac{A_d{A}_t{E}_s}{A_d{l}_t+A_t{l}_d}$                                (VII)

Here, the bolt stiffness is $k_b$, major area diameter of the fastener is $A_d$, tensile stress area is $A_t$, length of threaded portion is $l_t$, length of unthreaded portion is $l_d$ and the Young’s modulus of elasticity of $E_s$.

Write the expression of stiffness for the steel cylinder.

    $k_1=\frac{0.5774\pi E_{s}d}{\ln \frac{\left(1.155t+D-d\right)\left(D+d\right)}{\left(1.155t+D+d\right)\left(D-d\right)}}$           (VIII)

Here, the stiffness of the steel cylinder is $k_1$, Young’s modulus for steel is $E_s$, effective sealing diameter of the gasket sealing is $D$, thickness of steel cylinder head is $t$.

Write the expression for the midpoint of the complete joint.

    $t_o=\frac{A+B}{2}$                                            (IX)

Here, the midpoint of the joint is $t_o$, the thickness of the vessel inside the bolt if $A$ and thickness of the cylinder inside the bolt is $B$.

Write the expression of the thickness of the upper frustum.

    $t_1=t_o-A$                                             (X)

Here, the thickness of upper frustum of the gasket is $t_1$.

Write the expression for the effective sealing diameter of the gasket sealing in upper frustum.

    $D_1=D+2A\tan \alpha$                                   (XI)

Here, the effective sealing diameter of upper frustum of the gasket sealing is $D_1$ semi-angle frustum of cone is $\alpha$.

Write the expression for the stiffness of the upper frustum of cast iron vessel.

    $k_2=\frac{0.5774\pi E_{c}d}{\ln \frac{\left(1.155t_1+D_1-d\right)\left(D_1+d\right)}{\left(1.155t_1+D_1+d\right)\left(D_1-d\right)}}$            (XII)

Here, the stiffness of the cast-iron pressure vessel in the upper frustum is $k_2$ and the Young’s modulus of cast iron is $E_c$.

Write the expression for the stiffness of the lower frustum of the cast iron vessel.

    $k_3=\frac{0.5774\pi E_{c}d}{\ln \frac{\left(1.155t+D-d\right)\left(D+d\right)}{\left(1.155t+D+d\right)\left(D-d\right)}}$                  (XIII)

Here, the stiffness of the cast-iron pressure vessel in the lower frustum is $k_3$.

Write the expression for the stiffness of the member or assembly.

    $k_m={\left[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}\right]}^{-1}$                                   (XIV)

Here, the stiffness of the member is $k_m$.

Write the expression of joint constant.

    $C=\frac{k_b}{k_b+k_m}$                                            (XV)

Here, the joint constant is $C$.

Write the expression of initial tension in the bolt.

    $F_i=0.75A_t{S}_p$                                          (XVI)

Here, the tensile stress area is $A_t$ and the proof strength of material of the bolt is $S_p$.

Write the expression of the effective area of the cylinder.

    $A_g=\frac{\pi }{4}\left(C^2\right)$                                         (XVII)

Here, the effective area of the cylinder is $A_g$, diameter of cylinder is $C$.

Write the expression for the total force acting on the assembly.

    $P_{total}=A_g{p}_g$                                            (XVIII)

Here, the total load acting on the assembly is $P_{total}$, stress on the cylinder sealing is $S_p$.

Write the expression for the load acting on each bolt.

    $P=\frac{P_{total}}{N}$                                                 (XIX)

Here, the number of bolt is $N$.

Write the expression for the initial stress in the bolt.

    ${\sigma }_i=0.75S_P$                                            (XX)

Write the expression for the average stress.

    ${\sigma }_a=\frac{CP}{2A_t}$                                                 (XXI)

Write the expression for the mean stress.

    ${\sigma }_m=\frac{CP}{2A_t}+{\sigma }_i$                                        (XXII)

Write the expression for factor of safety by Goodman criteria.

    $n_f=\frac{S_e\left(S_{ut}-{\sigma }_i\right)}{{\sigma }_a\left(S_{ut}+S_e\right)}$                                     (XXIII)

Here, the ultimate strength is $S_{ut}$ and the endurance strength is $S_e$.

Conclusion:

Substitute $\frac{3}{8}\text{in}$ for $A$ and $\frac{1}{2}\text{in}$ for $B$ in Equation (I)

    $\begin{array}{c}l=\frac{3}{8}\text{in}+\frac{1}{2}\text{in}\\ \text{=}\left(0.375+0.5\right)\text{in}\\ =\text{0}\text{.875}\text{in}\end{array}$

Refer to Table $\text{A-31}$ “Dimensions of the hexagonal nut” to obtain $\frac{3}{8}\text{in}$ for $H$.

Substitute $0.875\text{in}$ for $l$ and $\frac{3}{8}\text{in}$ for $H$. Equation (II).

    $\begin{array}{c}L\ge 0.875\text{in}+\frac{3}{8}\text{in}\\ L=0.875\text{in}+\text{0}\text{.375}\text{in}\\ =\text{1}\text{.25}\text{in}\end{array}$

Substitute $\frac{7}{16}\text{in}$ for $d$ in Equation (III).

    $\begin{array}{c}{L}_T=2\times \frac{7}{16}\text{in}+0.25\\ =0.875\text{in}+0.25\\ =1.125\text{in}\end{array}$

Substitute $1.25\text{in}$ for $L$ and $1.125\text{in}$ for $L_T$ in Equation (IV).

    $\begin{array}{c}{l}_d=1.25\text{in}-1.125\text{in}\\ \text{=}\left(1.25-1.125\right)\text{in}\\ =\text{0}\text{.125}\text{in}\end{array}$

Substitute $0.875\text{in}$ for $l$ and $0.125\text{in}$ for $l_d$ in Equation (V).

    $\begin{array}{c}{l}_t=0.875\text{in}-0.125\text{in}\\ \text{=}\left(0.875-0.125\right)\text{in}\\ =\text{0}\text{.75}\text{in}\end{array}$

Substitute $0.5\text{in}$ for $d$ in Equation (VI).

    $\begin{array}{c}{A}_d=\frac{\pi }{4}{\left(\frac{7}{16}\text{in}\right)}^2\\ =\left(\frac{3.14}{4}\right)\left(0.1914\right){\text{in}}^2\\ =0.15025{\text{in}}^2\\ \approx 0.1503{\text{in}}^2\end{array}$

Refer to Table $8.2$ “Diameter and Area of Unified Screw Threads UNC and UNF” to obtain the value of $A_t$ as $0.1063{\text{in}}^2$.

Refer to Table $8-8$ “Stiffness Parameter of Member Material” to obtain the value of $E_s$ as $30\text{Mpsi}$.

Substitute $0.1503{\text{in}}^2$ for $A_d$, $0.1063{\text{in}}^2$ for $A_t$, $0.75\text{in}$ for $l_t$, $0.125\text{in}$ for $l_d$ and $30\text{Mpsi}$ for $E_s$ in Equation (VII).

    $\begin{array}{c}{k}_b=\frac{\left(0.1503{\text{in}}^2\right)\left(0.1063{\text{in}}^2\right)\left(30\text{Mpsi}\right)}{\left(0.1503{\text{in}}^2\right)\left(0.75\text{in}\right)+\left(0.1063{\text{in}}^2\right)\left(0.125\text{in}\right)}\\ =\frac{\left(0.1503{\text{in}}^2\right)\left(0.1063{\text{in}}^2\right)\left(30\text{Mpsi}\right)\left(\frac{\text{1}\text{Mlbf}/{\text{in}}^{\text{2}}}{\text{1}\text{Mpsi}}\right)}{\left(0.1503{\text{in}}^2\right)\left(0.75\text{in}\right)+\left(0.1063{\text{in}}^2\right)\left(0.125\text{in}\right)}\\ =\frac{0.47930\text{Mlbf}}{0.12601\text{in}}\\ \approx 3.804\text{Mlbf}/\text{in}\end{array}$

Substitute $30\text{Mpsi}$ for $E_s$, $0.375\text{in}$ for $t$, $\frac{7}{16}\text{in}$ for $d$ and $0.65625\text{in}$ for $D$ in Equation (VIII).

    $\begin{array}{c}{k}_1=\frac{0.5774\pi \left(30\text{Mpsi}\right)\left(\frac{7}{16}\text{in}\right)}{\ln \frac{\left(1.155\left(0.375\text{in}\right)+0.65625\text{in}-\frac{7}{16}\text{in}\right)\left(0.65625\text{in}+\frac{7}{16}\text{in}\right)}{\left(1.155\left(0.375\text{in}\right)+0.65625\text{in}+\frac{7}{16}\text{in}\right)\left(0.65625\text{in}-\frac{7}{16}\text{in}\right)}}\\ =\frac{0.5774\left(3.14\right)\left(30\text{Mpsi}\right)\left(\frac{1\text{Mlbf}/{\text{in}}^{\text{2}}}{1\text{Mpsi}}\right)\left(0.4375\text{in}\right)}{\ln \frac{\left(1.155\left(0.375\text{in}\right)+0.65625\text{in}-0.4375\text{in}\right)\left(0.65625\text{in}+0.4375\text{in}\right)}{\left(1.155\left(0.375\text{in}\right)+0.65625\text{in}+0.4375\text{in}\right)\left(0.65625\text{in}-0.4375\text{in}\right)}}\\ =\frac{23.796\text{Mlbf}/\text{in}}{\ln \left(\frac{0.71298}{0.33400}\right)}\\ =\frac{23.796}{0.75832}\text{Mlbf}/\text{in}\end{array}$

    $\begin{array}{c}{k}_1=\frac{23.796}{0.75832}\text{Mlbf}/\text{in}\\ =31.3797\text{Mlbf}/\text{in}\\ \approx 31.4\text{Mlbf}/\text{in}\end{array}$

Substitute $\frac{3}{8}\text{in}$ for $A$ and $\frac{1}{2}\text{in}$ for $B$ in Equation (IX).

    $\begin{array}{c}{t}_o=\frac{\frac{3}{8}\text{in}+\frac{1}{2}\text{in}}{2}\\ =\frac{0.375+0.5}{2}\text{in}\\ =0.4375\text{in}\end{array}$

Substitute $0.4375\text{in}$ for $t_o$ and $\frac{3}{8}\text{in}$ for $A$ in Equation (X).

    $\begin{array}{c}{t}_1=0.4375\text{in}-\frac{3}{8}\text{in}\\ =0.0625\text{in}\end{array}$

Substitute $0.65625\text{in}$ for $D$, $\frac{3}{8}\text{in}$ for $A$ and $30°$ for $\alpha$ in Equation (XI).

    $\begin{array}{c}{D}_1=0.65625\text{in}+2\left(\frac{3}{8}\text{in}\right)\tan \left(30°\right)\\ =\left(0.65625+\left(0.75\right)\left(0.57735\right)\right)\text{in}\\ \text{=1}\text{.0892}\text{in}\\ \approx \text{1}\text{.089}\text{in}\end{array}$

Refer to Table $8-8$ “Stiffness Parameter Of Member Material” obtaining the value of $E_c$ as $14.5\text{Mpsi}$.

Substitute $14.5\text{Mpsi}$ for $E_c$, $0.0625\text{in}$ for $t_1$, $\frac{7}{16}\text{in}$ for $d$ and $1.089\text{in}$ for $D_1$ in Equation (XII).

    $\begin{array}{c}{k}_2=\frac{0.5774\pi \left(14.5\text{Mpsi}\right)\left(\frac{7}{16}\text{in}\right)}{\ln \frac{\left(1.155\left(0.0625\text{in}\right)+1.089\text{in}-\frac{7}{16}\text{in}\right)\left(1.089\text{in}+\frac{7}{16}\text{in}\right)}{\left(1.155\left(0.0625\text{in}\right)+1.089\text{in}+\frac{7}{16}\text{in}\right)\left(1.089\text{in}-\frac{7}{16}\text{in}\right)}}\\ =\left\{\frac{11.50144}{\ln \left(\frac{1.10470}{1.04154}\right)}\text{Mpsi}\cdot \text{in}\right\}\left(\frac{\text{Mlbf}/{\text{in}}^{\text{2}}}{\text{Mpsi}}\right)\\ =\frac{11.50144}{0.05886}\text{Mlbf}/\text{in}\\ =195.40\text{Mlbf}/\text{in}\end{array}$

Substitute $14.5\text{Mpsi}$ for $E_c$, $0.4375\text{in}$ for $t$, $0.4375\text{in}$ for $d$ and $0.65625\text{in}$ for $D$ in Equation (XIII).

    $\begin{array}{c}{k}_3=\frac{0.5774\pi \left(14.5\text{Mpsi}\right)\left(\frac{7}{16}\text{in}\right)}{\ln \frac{\left(1.155\left(0.4375\text{in}\right)+0.65625\text{in}-\frac{7}{16}\text{in}\right)\left(0.65625\text{in}+\frac{7}{16}\text{in}\right)}{\left(1.155\left(0.4375\text{in}\right)+0.65625\text{in}+\frac{7}{16}\text{in}\right)\left(0.65625\text{in}-\frac{7}{16}\text{in}\right)}}\\ =\frac{0.5774\left(3.14\right)\left(14.5\text{Mpsi}\right)\left(\frac{1\text{Mlbf}/{\text{in}}^{\text{2}}}{1\text{Mpsi}}\right)\left(0.4375\text{in}\right)}{\ln \frac{\left(1.155\left(0.4375\text{in}\right)+0.65625\text{in}-0.4375\text{in}\right)\left(0.65625\text{in}+0.4375\text{in}\right)}{\left(1.155\left(0.4375\text{in}\right)+0.65625\text{in}+0.4375\text{in}\right)\left(0.65625\text{in}-0.4375\text{in}\right)}}\\ =\frac{11.50144\text{Mlbf}/\text{in}}{\ln \frac{0.79194}{0.34979}}\\ =\frac{11.50144}{0.81713}\text{Mlbf}/\text{in}\end{array}$

    $\begin{array}{c}{k}_3=\frac{11.50144}{0.81713}\text{Mlbf}/\text{in}\\ =14.075\text{Mlbf}/\text{in}\\ =14.08\text{Mlbf}/\text{in}\end{array}$

Substitute $31.4\text{Mlbf}/\text{in}$ for $k_1$, $195.4\text{Mlbf}/\text{in}$ for $k_2$ and $14.08\text{Mlbf}/\text{in}$ for $k_3$ in Equation (XIV).

    $\begin{array}{c}{k}_m={\left[\left(\frac{1}{31.4\text{Mlbf}/\text{in}\text{Mlbf}/\text{in}}\right)+\left(\frac{1}{195.4\text{Mlbf}/\text{in}}\right)+\left(\frac{1}{14.08\text{Mlbf}/\text{in}}\right)\right]}^{-1}\\ =\left[0.031847+5.1177\times {10}^{-3}+0.0710\right]\text{Mlbf}/\text{in}\\ ={\left(0.10798\right)}^{-1}\text{Mlbf}/\text{in}\\ \approx 9.2603\text{Mlbf}/\text{in}\end{array}$

Substitute $3.804\text{Mlbf}/\text{in}$ for $k_b$ and $9.2603\text{Mlbf}/\text{in}$ for $k_m$ in Equation (XV).

    $\begin{array}{c}C=\frac{3.804\text{Mlbf}/\text{in}}{3.804\text{Mlbf}/\text{in}+9.2603\text{Mlbf}/\text{in}}\\ =\frac{3.804\text{Mlbf}/\text{in}}{13.0643\text{Mlbf}/\text{in}}\\ =0.291\end{array}$

Refer to Table $8-9$ “SAE specification for steel bolts” to obtain the value of $S_p$ as $120\text{kpsi}$.

Substitute $120\text{kpsi}$ for $S_p$, $0.1063{\text{in}}^{\text{2}}$ for $A_t$ in Equation (XVI).

    $\begin{array}{c}{F}_i=0.75\left(0.1063{\text{in}}^2\right)\left(120\text{kpsi}\right)\\ =0.75\left(0.1063{\text{in}}^2\right)\left(120\text{kpsi}\right)\left(\frac{1\text{kips}/{\text{in}}^{\text{2}}}{1\text{kpsi}}\right)\\ =9.567\text{kips}\\ \approx \text{9}\text{.57}\text{kips}\end{array}$

Substitute $3.25\text{in}$ for $C$ in Equation (XVII).

    $\begin{array}{c}{A}_g=\frac{\pi }{4}{\left(3.25\text{in}\text{in}\right)}^2\\ =\left(0.785\right)\left(10.5625\right){\text{in}}^{\text{2}}\\ =9.61625{\text{in}}^{\text{2}}\\ \approx 8.29{\text{in}}^{\text{2}}\end{array}$

Substitute $8.29{\text{in}}^{\text{2}}$ for $A_g$ and $1200\text{psi}$ for $p_g$ in Equation (XVIII).

    $\begin{array}{c}{P}_{total}=\left(8.29{\text{in}}^{\text{2}}\right)\left(1200\text{psi}\right)\\ =\left(8.29{\text{in}}^{\text{2}}\right)\left(1200\text{psi}\right)\left(\frac{{\text{10}}^{-\text{3}}\text{kips}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\\ =9.948\text{kips}\end{array}$

Substitute $9.948\text{kips}$ for $P_{total}$ and $8$ for $N$ in Equation (XIX).

    $\begin{array}{c}P=\frac{9.948\text{kips}}{8}\\ =1.2435\text{kips}/\text{bolt}\\ \approx 1.244\text{kips}/\text{bolt}\end{array}$

Substitute $120\text{kpsi}$ for $S_p$ in Equation (XX).

    $\begin{array}{c}{\sigma }_i=0.75\left(120\text{kpsi}\right)\\ =90\text{kpsi}\end{array}$

Substitute $1.244\text{kips}/\text{bolt}$ for $P$, $0.291$ for $C$ and $0.1063{\text{in}}^{\text{2}}$ for $A_t$ in Equation (XXI).

    $\begin{array}{c}{\sigma }_a=\frac{0.291\left(1.244\text{kips}/\text{bolt}\right)}{2\left(0.1063{\text{in}}^{\text{2}}\right)}\\ =\frac{\left(1.244\text{kips}/\text{bolt}\right)}{\left(0.73058{\text{in}}^{\text{2}}\right)}\left(\frac{1\text{kpsi}}{1\text{kips}/{\text{in}}^2}\right)\\ =1.703\text{kpsi}\end{array}$

Substitute $1.244\text{kips}/\text{bolt}$ for $P$, $0.291$ for $C$, $90\text{kpsi}$ for ${\sigma }_i$ and $0.1063{\text{in}}^{\text{2}}$ for $A_t$ in Equation (XXII).

    $\begin{array}{c}{\sigma }_m=\frac{\left(0.291\right)\left(1.244\text{kips}/\text{bolt}\right)}{2\left(0.1063{\text{in}}^{\text{2}}\right)}+90\text{kpsi}\\ =\frac{\left(1.244\text{kips}/\text{bolt}\right)}{\left(0.73058{\text{in}}^{\text{2}}\right)}\left(\frac{1\text{kpsi}}{1\text{kips}/{\text{in}}^2}\right)+90\text{kpsi}\\ \approx 91.7\text{kpsi}\end{array}$

Refer to Table $8.17$ “Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Threads” to obtain $23.2\text{kpsi}$ for $S_e$ with respect to $\text{SAE}\text{8}$.

Refer to Table 8.11 “Metric Mechanical-Property Classes for Steel Bolts, Screws, and Studs” to obtain $150\text{kpsi}$ for $S_{ut}$ with respect to medium carbon alloy.

Substitute $150\text{kpsi}$ for $S_{ut}$, $23.2\text{kpsi}$ for $S_e$, $1.703\text{kpsi}$ for ${\sigma }_a$ and $90\text{kpsi}$ for ${\sigma }_i$ in Equation (XXIII).

    $\begin{array}{c}{n}_f=\frac{23.2\text{kpsi}\left(150\text{kpsi}-90\text{kpsi}\right)}{1.703\text{kpsi}\left(150\text{kpsi}+23.2\text{kpsi}\right)}\\ =\frac{1392}{294.9596}\\ =4.72\end{array}$

Thus, the fatigue factor of safety for the bolts using Goodman criteria is $4.72$.

To determine

The fatigue factor of safety for the bolts using Gerber criteria.

Answer to Problem 54P

The fatigue factor of safety for the bolts using Gerber criteria is $7.28$.

Explanation of Solution

Write the expression for the factor of safety using Gerber criteria.

    $n_f=\frac{1}{2{\sigma }_a{S}_e}\left[S_{ut}\sqrt{S_{ut}^2+4S_e\left(S_e+{\sigma }_i\right)}-S_{ut}^2-2{\sigma }_i{S}_e\right]$     (XXIV).

Conclusion:

Substitute $150\text{kpsi}$ for $S_{ut}$, $23.2\text{kpsi}$ for $S_e$, $1.703\text{kpsi}$ for ${\sigma }_a$ and $90\text{kpsi}$ for ${\sigma }_i$ in Equation (XXIV).

    $\begin{array}{c}{n}_f=\frac{1}{2\left(1.703\text{kpsi}\right)\left(23.2\text{kpsi}\right)}\left[\begin{array}{l}150\text{kpsi}\sqrt{\begin{array}{l}{\left(150\text{kpsi}\right)}^2\\ +4\left(23.2\text{kpsi}\right)\left(\left(23.2\text{kpsi}\right)+90\text{kpsi}\right)\end{array}}\\ -{\left(150\text{kpsi}\right)}^2-2\left(90\text{kpsi}\right)\left(23.2\text{kpsi}\right)\end{array}\right]\\ =\frac{1}{79.0192}\left(574.9\right)\\ =7.28\end{array}$

Thus, the fatigue factor of safety for the bolts using Gerber criteria is $7.28$.

To determine

The fatigue factor of safety for the bolts using ASME-elliptic criteria.

Answer to Problem 54P

The fatigue factor of safety for the bolts using ASME-elliptic criteria is $7.24$.

Explanation of Solution

Write the expression for the factor of safety using ASME-elliptic criteria.

    $n_f=\frac{S_e}{{\sigma }_a\left(S_P^2+S_e^2\right)}\left[S_P\sqrt{S_P^2+S_e^2-{\sigma }_i^2}-{\sigma }_i{S}_e\right]$   (XXV)

Conclusion:

Substitute $120\text{kpsi}$ for $S_p$, $23.2\text{kpsi}$ for $S_e$, $1.703\text{kpsi}$ for ${\sigma }_a$ and $90\text{kpsi}$ for ${\sigma }_i$ in Equation (XXIV).

    $\begin{array}{c}{n}_f=\frac{23.2\text{kpsi}}{\left(1.703\text{kpsi}\right)\left({\left(120\text{kpsi}\right)}^2+{\left(23.2\text{kpsi}\right)}^2\right)}\left[\begin{array}{l}\left(120\text{kpsi}\right)\sqrt{\begin{array}{l}{\left(120\text{kpsi}\right)}^2\\ +{\left(23.2\text{kpsi}\right)}^2\\ -{\left(90\text{kpsi}\right)}^2\end{array}}\\ -\left(90\text{kpsi}\right)\left(23.2\text{kpsi}\right)\end{array}\right]\\ =\frac{23.2}{25439.822}7835.238\\ \approx 7.24\end{array}$

Thus, the fatigue factor of safety for the bolts using ASME-elliptic criteria is $7.24$.