Chapter 8, Problem 51P

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
8–51	8–33

To determine

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

Answer to Problem 51P

The fatigue factor of safety for the bolts using Goodman criteria is $7.55$.

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.

    $L_T=2d+6$                               (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 young’s modulus of elasticity of $E_s$.

Write the expression of stiffness for the steel cylinder.

    $k_1=\frac{0.5774\pi E_{s}d}{\ln \left[\left(\frac{1.155t+D-d}{1.155t+D+d}\right)\left(\frac{D+d}{D-d}\right)\right]}$     (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 gasket is $t$.

Write the expression for the stiffness of the cast iron pressure vessel.

    $k_2=\frac{E_c}{E_s}\times k_1$                                        (IX)

Here, the stiffness of the cast-iron pressure vessel is $k_2$ and the young’s modulus for pressure vessel is $E_c$.

Write the expression for the stiffness of the member.

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

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

Write the expression of joint constant.

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

Here, the joint constant is $C$.

Write the expression of initial tension in the bolt.

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

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.

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

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

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

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

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.

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

Here, the number of bolt is $N$.

Write the expression for the initial stress in the bolt.

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

Write the expression for the average stress.

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

Write the expression for the mean stress.

    ${\sigma }_m=\frac{CP}{2A_t}+\frac{F_i}{A_t}$                                        (XVIII)

Write the expression for factor of safety by Goodman criteria.

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

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

Conclusion:

Substitute $20\text{mm}$ for $A$ and $20\text{mm}$ for $B$ in Equation (I)

    $\begin{array}{c}l=20\text{mm}+20\text{mm}\\ =\text{40}\text{mm}\end{array}$

Refer to Table $\text{A-31}$ “Dimensions of the hexagonal nut” to obtain $10.8\text{mm}$ for $H$.

Substitute $40\text{mm}$ for $l$ and $10.8\text{mm}$ for $H$ Equation (II).

    $\begin{array}{c}L\ge 40\text{mm}+10.8\text{mm}\\ =\text{50}\text{.8}\text{mm}\\ \approx 55\text{mm}\end{array}$

Substitute $12\text{mm}$ for $d$ in Equation (III).

    $\begin{array}{c}{L}_T=2\times 12\text{mm}+6\\ =24\text{mm}+6\\ =30\text{mm}\end{array}$

Substitute $55\text{mm}$ for $L$ and $30\text{mm}$ for $L_T$ in Equation (IV).

    $\begin{array}{c}{l}_d=55\text{mm}-30\text{mm}\\ =\left(55-30\right)\text{mm}\\ =\text{25}\text{mm}\end{array}$

Substitute $40\text{mm}$ for $l$ and $25\text{mm}$ for $l_d$ in Equation (V).

    $\begin{array}{c}{l}_t=40\text{mm}-25\text{mm}\\ =\left(40-25\right)\text{mm}\\ =\text{15}\text{mm}\end{array}$

Substitute $12\text{mm}$ for $d$ in Equation (VI).

    $\begin{array}{c}{A}_d=\frac{\pi }{4}{\left(12\text{mm}\right)}^2\\ =\left(0.785398\right)\left(144\right){\text{mm}}^2\\ =113.09{\text{mm}}^2\\ =113.1{\text{mm}}^2\end{array}$

Refer to Table $8.1$ “Diameter and Area of Unified Screw Threads UNC and UNF” obtain $84.3{\text{mm}}^2$ for $A_t$.

Substitute $113.1{\text{mm}}^2$ for $A_d$, $84.3{\text{mm}}^2$ for $A_t$, $15\text{mm}$ for $l_t$, $25\text{mm}$ for $l_d$ and $207\text{GPa}$ for $E_s$ in Equation (VII).

    $\begin{array}{c}{k}_b=\frac{\left(113.1{\text{mm}}^2\right)\left(84.3{\text{mm}}^2\right)\left(207\text{GPa}\right)}{\left(113.1{\text{mm}}^2\right)\left(15\text{mm}\right)+\left(84.3{\text{mm}}^2\right)\left(25\text{mm}\right)}\\ =\frac{\left(113.1{\text{mm}}^2\right)\left(84.3{\text{mm}}^2\right)\left(207\text{MPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^2}{1\text{MPa}}\right)}{\left(113.1{\text{mm}}^2\right)\left(15\text{mm}\right)+\left(84.3{\text{mm}}^2\right)\left(25\text{mm}\right)}\\ =\left(\frac{1973606.31\times {10}^3}{3804}\text{N}/\text{mm}\right)\left(\frac{{10}^3\text{N}/\text{m}}{1\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{\text{10}}^6\text{N}/\text{m}}\right)\\ =518.8\text{MN}/\text{m}\end{array}$

Substitute $207\text{GPa}$ for $E_s$, $20\text{mm}$ for $t$, $12\text{mm}$ for $d$ and $18\text{mm}$ for $D$ in Equation (VIII).

    $\begin{array}{c}{k}_1=\frac{0.5774\pi \left(207\text{GPa}\right)\left(12\text{mm}\right)}{\ln \frac{\left[1.155\left(20\text{mm}\right)+18\text{mm}-12\text{mm}\right]\left(18\text{mm+12}\text{mm}\right)}{\left[1.155\left(20\text{mm}\right)+18\text{mm}+12\text{mm}\right]\left(18\text{mm}-12\text{mm}\right)}}\\ =\frac{\left(\text{4505}\text{.8657}\left(\text{mm}\right)\left(\text{GPa}\right)\right)\left(\frac{{10}^3{\text{N}/\text{mm}}^2}{\text{1}\text{GPa}}\right)}{\ln 2.740}\\ =\left(4470.108\times {10}^3\text{N}/\text{mm}\right)\left(\frac{{10}^3\text{N}/\text{m}}{1\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{10}^6\text{N}/\text{m}}\right)\\ \approx 4470\text{MN}/\text{m}\end{array}$

Substitute $4470\text{MN}/\text{m}$ for $k_1$, $207\text{GPa}$ for $E_s$, and $100\text{GPa}$ for $E_c$ in Equation (IX).

    $\begin{array}{c}{k}_2=\frac{100\text{GPa}}{207\text{GPa}}\times \left(4470\text{MN}/\text{m}\right)\\ =2159.4\text{MN}/\text{m}\\ \approx 2159\text{MN}/\text{m}\end{array}$

Substitute $4470\text{MN}/\text{m}$ for $k_1$ and $2159\text{MN}/\text{m}$ for $k_2$ in Equation (X).

    $\begin{array}{c}{k}_m={\left[\frac{1}{4470\text{MN}/\text{m}}+\frac{1}{2159\text{MN}/\text{m}}\right]}^{-1}\\ ={\left[\frac{\left(2159\text{MN}/\text{m}\right)+\left(4470\text{MN}/\text{m}\right)}{\left(4470\text{MN}/\text{m}\right)\left(4470\text{MN}/\text{m}\right)}\right]}^{-1}\\ ={\left[6.86891\times {10}^{-4}\text{MN}/\text{m}\right]}^{-1}\text{MN}/\text{m}\\ \approx 1456\text{MN}/\text{m}\end{array}$

Substitute $\text{518}\text{.8}\text{MN}/\text{m}$ for $k_b$ and $\text{1456}\text{MN}/\text{m}$ for $k_m$ in Equation (XI).

    $\begin{array}{c}C=\frac{\text{518}\text{.8}\text{MN}/\text{m}}{\text{518}\text{.8}\text{MN}/\text{m}+\text{1456}\text{MN}/\text{m}}\\ =\frac{518.8}{1947.8}\\ =0.2627\\ \approx 0.263\end{array}$

Refer to Table $8\text{-}11$ “Metric Mechanical Property Classes for Steel Bolts, Screws and Studs” to obtain $650\text{MPa}$ for $S_p$.

Substitute $650\text{MPa}$ for $S_p$, $84.3{\text{mm}}^2$ for $A_t$ in the Equation (XII).

    $\begin{array}{c}{F}_i=0.75\left(84.3{\text{mm}}^2\right)\left(650\text{MPa}\right)\\ =\left(41096.25\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =41.09\text{kN}\\ \approx \text{41}\text{.1}\text{kN}\end{array}$

Substitute $100\text{mm}$ for $C$ in Equation (XIII).

    $\begin{array}{c}{A}_g=\frac{\pi }{4}{\left(100\text{mm}\right)}^2\\ =\left(0.785398\right)\left(10000\right){\text{mm}}^{\text{2}}\\ =7853.98{\text{mm}}^{\text{2}}\\ \approx 7854{\text{mm}}^{\text{2}}\end{array}$

Substitute $7854{\text{mm}}^2$ for $A_g$ and $6\text{MPa}$ for $p_g$ in Equation (XIV).

    $\begin{array}{c}{P}_{total}=\left(7854{\text{mm}}^2\right)\left(6\text{MPa}\right)\\ =\left(47124\text{N}\right)\left(\frac{\text{1}\text{kN}}{1000\text{N}}\right)\\ =47.124\text{kN}\end{array}$

Substitute $47.124\text{kN}$ for $P_{total}$ and $10$ for $N$ in Equation (XV).

    $\begin{array}{c}P=\frac{47.124\text{kN}}{10}\\ =4.7124\text{kN}/\text{bolt}\end{array}$

Refer to Table $8.17$ “Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Threads” to obtain $140\text{MPa}$ for $S_e$ with respect to $\text{ISO}\text{10}\text{.9}$.

Refer to Table 8.11 “Metric Mechanical-Property Classes for Steel Bolts, Screws, and Studs” to obtain $900\text{MPa}$ for $S_{ut}$ with respect to medium carbon.

Substitute $650\text{MPa}$ for $S_p$ in Equation (XVI).

    $\begin{array}{c}{\sigma }_i=0.75\left(650\text{MPa}\right)\\ =487.5\text{Mpa}\end{array}$

Substitute $4.7124\text{kN}/\text{bolt}$ for $P$, $0.263$ for $C$ and $84.3{\text{mm}}^2$ for $A_t$ in the Equation (XVII).

    $\begin{array}{c}{\sigma }_a=\frac{\left(0.263\right)\left(4.7124\text{kN}/\text{bolt}\right)}{2\times 84.3{\text{mm}}^2}\\ =1.5599\times {10}^{-3}{\text{mm}}^{-2}\times 4.7124\text{kN}/\text{bolt}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =7.350\text{MPa}\end{array}$

Substitute $4.7124\text{kN}/\text{bolt}$ for $P$, $0.263$ for $C$, $84.3{\text{mm}}^2$ for $A_t$ and $\text{41}\text{.1}\text{kN}$ for $F_i$ in Equation (XVIII).

    $\begin{array}{c}{\sigma }_m=\frac{0.263\left(4.7124\text{kN}/\text{bolt}\right)}{2\times 84.3{\text{mm}}^2}+\frac{\text{41}\text{.1}\text{kN}}{84.3{\text{mm}}^2}\\ =1.5599\times {10}^{-3}{\text{mm}}^{-2}\times 4.7124\text{kN}/\text{bolt}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)+\frac{\text{41}\text{.1}\text{kN}}{84.3{\text{mm}}^2}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =7.350\text{MPa}+0.4875\text{MPa}\\ \approx 7.8375\text{MPa}\end{array}$

Substitute $7.350\text{MPa}$ for ${\sigma }_a$, $487.5\text{Mpa}$ for ${\sigma }_i$, $900\text{MPa}$ for $S_{ut}$ and $140\text{MPa}$ for $S_e$ in Equation (XIX).

    $\begin{array}{c}{n}_f=\frac{140\text{MPa}\left(900\text{MPa}-487.5\text{Mpa}\right)}{7.350\text{MPa}\left(900\text{MPa}+140\text{MPa}\right)}\\ =\frac{140\text{MPa}\left(412.5\text{Mpa}\right)}{7.350\text{MPa}\left(1040\text{MPa}\right)}\\ =\frac{57750}{7644}\\ \approx 7.55\end{array}$

Thus, the fatigue factor of safety for the bolts using Goodman criteria is $7.55$.

To determine

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

Answer to Problem 51P

The fatigue factor of safety for the bolts using Gerber criteria is $11.4$.

Explanation of Solution

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

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

Conclusion:

Substitute $7.350\text{MPa}$ for ${\sigma }_a$, $487.5\text{MPa}$ for ${\sigma }_i$, $487.5\text{MPa}$ for ${\sigma }_i$, $900\text{MPa}$ for $S_{ut}$ and $140\text{MPa}$ for $S_e$ in Equation (XX).

    $\begin{array}{c}{n}_f=\frac{1}{2\left(7.350\text{MPa}\right)\left(140\text{MPa}\right)}\left[\begin{array}{l}\left(900\text{MPa}\right)\left[\sqrt{\begin{array}{l}{\left(900\text{MPa}\right)}^2\\ +4\times 140\text{MPa}\left(\begin{array}{l}140\text{MPa}\\ +487.5\text{Mpa}\end{array}\right)\end{array}}\right]\\ -{\left(900\text{MPa}\right)}^2-2\left(487.5\text{MPa}\right)\left(140\text{MPa}\right)\end{array}\right]\\ =\frac{1}{2058{\left(\text{MPa}\right)}^2}\left[\begin{array}{l}969914.4292{\left(\text{MPa}\right)}^2\\ -{\left(900\text{MPa}\right)}^2-136500{\left(\text{MPa}\right)}^2\end{array}\right]\\ =11.377\\ \approx 11.4\end{array}$

Thus, the fatigue factor of safety for the bolts using Gerber criteria is $11.4$.

To determine

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

Answer to Problem 51P

The fatigue factor of safety for the bolts using ASME-elliptic criteria is $9.73$.

Explanation of Solution

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

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

Conclusion:

Substitute $7.350\text{MPa}$ for ${\sigma }_a$, $487.5\text{Mpa}$ for ${\sigma }_i$, $900\text{MPa}$ for $S_{ut}$ and $140\text{MPa}$ for $S_e$ in Equation (XX).

    $\begin{array}{c}{n}_f=\frac{140\text{MPa}}{\left(7.350\text{MPa}\right)\left(\begin{array}{l}{\left(650\text{MPa}\right)}^2\\ +{\left(140\text{MPa}\right)}^2\end{array}\right)}\left[\begin{array}{l}\left(650\text{MPa}\right)\sqrt{\begin{array}{l}{\left(650\text{MPa}\right)}^2\\ +{\left(140\text{MPa}\right)}^2-{\left(487.5\text{Mpa}\right)}^2\end{array}}\\ -\left(487.5\text{Mpa}\right)\left(140\text{MPa}\right)\end{array}\right]\\ =\frac{140\text{MPa}}{3249435{\left(\text{MPa}\right)}^3}\left[225650.4668{\left(\text{MPa}\right)}^2\right]\\ \approx 9.73\end{array}$

Thus, the fatigue factor of safety for the bolts using ASME-elliptic criteria is $9.73$.