Chapter 8, Problem 35P

8-33 to 8-36 The figure illustrates the non-permanent connection of a steel cylinder head to a grade 30 cast- iron pressure vessel using N bolts. A confined gasket seal has an effective sealing diameter D. The cylinder stores gas at a maximum pressure p_g. For the specifications given in the table for the specific problem assigned, select a suitable bolt length from the preferred sizes in Table A-17, then determine the yielding factor of safety n_p, the load factor n_L, and the joint separation factor n₀.

Problem Number	8-33	8-34	8-35	8-36
A	20 mm	$\frac{1}{2}$ in	20 mm	$\frac{3}{8}$ in
B	20 mm	$\frac{5}{8}$ in	25 mm	$\frac{1}{2}$ in
C	100 mm	3.5 in	0.8 m	3.25 in
D	150 mm	4.25 in	0.9 m	3.5 in
E	200 mm	6 in	1.0 m	5.5 in
F	300 mm	8 in	1.1 m	7 in
N	10	10	36	8
pg	6 MPa	1500 psi	550 kPa	1200 psi
Bolt grade	ISO 9.8	SAE 5	ISO 10.9	SAE 8
Bolt spec.	M12 × 1.75	$\frac{1}{2}$ in-13	M10 × 1.5	$\frac{7}{16}$ in-14

To determine

The yield factor of safety

The load factor of safety.

The joint separation factor.

Answer to Problem 35P

The yield factor of safety is $1.27$.

The load factor of safety is $6.88$.

The joint separation factor is $6.09$.

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 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 gasket 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 yield factor of safety.

$n_p=\frac{S_p{A}_t}{CP+F_i}$ (XX)

Here the overload factor of safety is $n_p$, initial tension on the bolt is $F_i$, joint constant is $C$ and the proof strength is $S_p$.

Write the expression of overload factor of safety.

$n_L=\frac{S_p{A}_t-F_i}{CP}$ (XXI)

Here the overload factor of safety is $n_L$.

Write the expression of joint separation factor of safety.

$n_o=\frac{F_i}{p\left(1-C\right)}$ . (XXII)

Here the factor of safety based on joint separation is $n_o$.

Conclusion:

Substitute $20\text{mm}$ for $A$ and $25\text{mm}$ for $B$ in the Equation (I)

$\begin{array}{c}l=20\text{mm}+25\text{mm}\\ \text{=}45\text{mm}\end{array}$

Refer to Table $\text{A-31}$ “Dimensions of the hexagonal nut” to obtain $8.4\text{mm}$ for $H$.

Substitute $45\text{mm}$ for $l$ and $8.4\text{mm}$ for $H$.Equation (II).

$\begin{array}{c}L\ge 45\text{mm}+8.4\text{mm}\\ =\text{53}\text{.4}\text{mm}\\ \approx 55\text{mm}\end{array}$

Substitute $10\text{mm}$ for $d$ in Equation (III).

$\begin{array}{c}{L}_T=2\times 10\text{mm}+6\\ =20\text{mm}+6\\ =26\text{mm}\end{array}$

Substitute $55\text{mm}$ for $L$ and $26\text{mm}$ for $L_T$ in Equation (IV).

$\begin{array}{c}{l}_d=55\text{mm}-26\text{mm}\\ =\left(55-26\right)\text{mm}\\ =\text{29}\text{mm}\end{array}$

Substitute $45\text{mm}$ for $l$ and $29\text{mm}$ for $l_d$ in Equation (V).

$\begin{array}{c}{l}_t=45\text{mm}-29\text{mm}\\ =\left(45-29\right)\text{mm}\\ =\text{16}\text{mm}\end{array}$

Substitute $10\text{mm}$ for $d$ in Equation (VI).

$\begin{array}{c}{A}_d=\frac{\pi }{4}{\left(10\text{mm}\right)}^2\\ =\left(0.785398\right)\left(100\right){\text{mm}}^2\\ =78.5398{\text{mm}}^2\\ \approx 78.5{\text{mm}}^2\end{array}$

Refer to Table $8.1$ “Diameter and Area of Unified Screw Threads UNC and UNF” obtain $58{\text{mm}}^2$ for $A_t$.

Substitute $78.5{\text{mm}}^2$ for $A_d$, $58{\text{mm}}^2$ for $A_t$, $16\text{mm}$ for $l_t$, $29\text{mm}$ for $l_d$ and $207\text{GPa}$ for $E_s$ in Equation (VII).

$\begin{array}{c}{k}_b=\frac{\left(78.5{\text{mm}}^2\right)\left(58{\text{mm}}^2\right)\left(207\text{GPa}\right)}{\left(78.5{\text{mm}}^2\right)\left(16\text{mm}\right)+\left(58{\text{mm}}^2\right)\left(29\text{mm}\right)}\\ =\frac{\left(78.5{\text{mm}}^2\right)\left(58{\text{mm}}^2\right)\left(207\text{GPa}\right)\left(\frac{{10}^3\text{N}/{\text{mm}}^2}{1\text{GPa}}\right)}{\left(78.5{\text{mm}}^2\right)\left(16\text{mm}\right)+\left(58{\text{mm}}^2\right)\left(29\text{mm}\right)}\\ =\left(\frac{942471\times {10}^3}{2938}\text{N}/\text{mm}\right)\left(\frac{{10}^3\text{N}/\text{m}}{1\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{\text{10}}^6\text{N}/\text{m}}\right)\\ \approx 320.8\text{MN}/\text{m}\end{array}$

Substitute $207\text{GPa}$ for $E_s$, $20\text{mm}$ for $t$, $10\text{mm}$ for $d$ and $15\text{mm}$ for $D$ in Equation (VIII).

$\begin{array}{c}{k}_1=\frac{0.5774\pi \left(207\text{GPa}\right)\left(10\text{mm}\right)}{\ln \frac{\left[1.155\left(20\text{mm}\right)+15\text{mm}-10\text{mm}\right]\left(15\text{mm+10}\text{mm}\right)}{\left[1.155\left(20\text{mm}\right)+15\text{mm}+10\text{mm}\right]\left(15\text{mm}-10\text{mm}\right)}}\\ =\frac{\left(\text{3752}\text{.98}\left(\text{mm}\right)\left(\text{GPa}\right)\right)\left(\frac{{10}^3{\text{N}/\text{mm}}^2}{\text{1}\text{GPa}}\right)}{\ln 2.9209}\\ =\left(3501.26\times {10}^3\text{N}/\text{mm}\right)\left(\frac{{10}^3\text{N}/\text{m}}{1\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{10}^6\text{N}/\text{m}}\right)\\ \approx 3501\text{MN}/\text{m}\end{array}$

Substitute $20\text{mm}$ for $A$ and $25\text{mm}$ for $B$ in the Equation (IX)

$\begin{array}{c}{t}_o=\frac{20\text{mm}+25\text{mm}}{2}\\ =\frac{45}{2}\text{mm}\\ =22.5\text{mm}\end{array}$

Substitute $22.5\text{mm}$ for $t_o$ and $20\text{mm}$ for $A$ in the Equation (X)

$\begin{array}{c}{t}_1=22.5\text{mm}-20\text{mm}\\ \text{=}\left(22.5-20\right)\text{mm}\\ =2.5\text{mm}\end{array}$

Substitute $15\text{mm}$ for $D$, $20\text{mm}$ for $A$ and $30°$ for $\alpha$ in the Equation (XI)

$\begin{array}{c}{D}_1=15\text{mm}+2\left(20\text{mm}\right)\tan 30°\\ =\left(15+\left(40\right)\left(0.577\right)\right)\text{mm}\\ =\left(15+23.08\right)\text{mm}\\ =\text{38}\text{.08}\text{mm}\end{array}$

Refer to Table $8-8$ “Stiffness Parameter Of Member Material” obtaining the value of $E_c$ as $100\text{GPa}$.

Substitute $100\text{GPa}$ for $E_c$, $2.5\text{mm}$ for $t_1$, $10\text{mm}$ for $d$ and $38.08\text{mm}$ for $D_1$ in Equation (XII).

$\begin{array}{c}{k}_2=\frac{0.5774\pi \left(100\text{GPa}\right)\left(10\text{mm}\right)}{\ln \frac{\left(1.155\left(2.5\text{mm}\right)+38.08\text{mm}-10\text{mm}\right)\left(38.08\text{mm}+10\text{mm}\right)}{\left(1.155\left(2.5\text{mm}\right)+38.08\text{mm}+10\text{mm}\right)\left(38.08\text{mm}-10\text{mm}\right)}}\\ =\left\{\frac{1813.036}{\ln \left(\frac{1488.9174}{1431.1674}\right)}\text{GPa}\cdot \text{mm}\right\}\left(\frac{{\text{10}}^3\text{N}/{\text{mm}}^{\text{2}}}{\text{GPa}}\right)\left(\frac{{10}^3\text{N}/\text{m}}{}\right)\\ =\left(45831.14\times {10}^3\text{N}/\text{mm}\right)\left(\frac{\text{1000}\text{N}/\text{m}}{\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{10}^6\text{N}/\text{m}}\right)\\ \approx 45831\text{MN}/\text{m}\end{array}$

Substitute $100\text{GPa}$ for $E_c$, $22.5\text{mm}$ for $t$, $10\text{mm}$ for $d$ and $15\text{mm}$ for $D$ in Equation (XIII).

$\begin{array}{c}{k}_3=\frac{0.5774\pi \left(100\text{GPa}\right)\left(10\text{mm}\right)}{\ln \frac{\left[1.155\left(22.5\text{mm}\right)+15\text{mm}-10\text{mm}\right]\left(15\text{mm+10}\text{mm}\right)}{\left[1.155\left(22.5\text{mm}\right)+15\text{mm}+10\text{mm}\right]\left(15\text{mm}-10\text{mm}\right)}}\\ =\frac{\left(\text{1813}\text{.036}\left(\text{mm}\right)\left(\text{GPa}\right)\right)\left(\frac{{10}^3{\text{N}/\text{mm}}^2}{\text{1}\text{GPa}}\right)}{\ln 3.03873}\\ =\left(1631.24\times {10}^3\text{N}/\text{mm}\right)\left(\frac{{10}^3\text{N}/\text{m}}{1\text{N}/\text{mm}}\right)\left(\frac{1\text{MN}/\text{m}}{{10}^6\text{N}/\text{m}}\right)\\ \approx 1631\text{MN}/\text{m}\end{array}$

Substitute $3501\text{MN}/\text{m}$ for $k_1$, $45831\text{MN}/\text{m}$ for $k_2$ and $1631\text{MN}/\text{m}$ for $k_3$ in the Equation (XIV).

$\begin{array}{c}{k}_m={\left[\left(\frac{1}{3501\text{MN}/\text{m}}\right)+\left(\frac{1}{45831\text{MN}/\text{m}}\right)+\left(\frac{1}{1631\text{MN}/\text{m}}\right)\right]}^{-1}\\ ={\left(9.2057\right)}^{-1}\text{MN}/\text{m}\\ =1086.28\text{MN}/\text{m}\\ =1087\text{MN}/\text{m}\end{array}$

Substitute $320.8\text{MN}/\text{m}$ for $k_b$ and $1087\text{MN}/\text{m}$ for $k_m$ in Equation (XV).

$\begin{array}{c}C=\frac{320.8\text{MN}/\text{m}}{320.8\text{MN}/\text{m}+1087\text{MN}/\text{m}}\\ =\frac{320.8\text{MN}/\text{m}}{1407.8\text{MN}/\text{m}}\\ =0.2278\\ \approx 0.228\end{array}$

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

Substitute $830\text{MPa}$ for $S_p$, $58{\text{mm}}^2$ for $A_t$ in the Equation (XVI).

$\begin{array}{c}{F}_i=0.75\left(58{\text{mm}}^2\right)\left(830\text{MPa}\right)\\ =\left(36105\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =36.105\text{kN}\\ \approx \text{36}\text{.1}\text{kN}\end{array}$

Substitute $0.8\text{m}$ for $C$ in Equation (XVII).

$\begin{array}{c}{A}_g=\frac{\pi }{4}{\left(0.8\text{m}\right)}^2\\ =\left(0.785\right){\left(0.64\right)}^2{\text{m}}^{\text{2}}\\ =0.5024{\text{m}}^{\text{2}}\end{array}$

Substitute $0.5024{\text{m}}^{\text{2}}$ for $A_g$ and $550\text{kpa}$ for $p_g$ in Equation (XVIII).

$\begin{array}{c}{P}_{total}=\left(0.5024{\text{m}}^{\text{2}}\right)\left(550\text{kpa}\right)\\ =\left(276.32\text{kpa}\cdot {\text{m}}^2\right)\left(\frac{1\text{kN}/{\text{m}}^{\text{2}}}{\text{1}\text{kpa}}\right)\\ \text{=276}\text{.32}\text{kN}\end{array}$

Substitute $276.32\text{kN}$ for $P_{total}$ and $36$ for $N$ in Equation (XIX).

$\begin{array}{c}P=\frac{276.32\text{kN}}{36}\\ =7.6755\text{kN}/\text{bolt}\\ =7.676\text{kN}/\text{bolt}\end{array}$

Substitute $830\text{MPa}$ for $S_p$, $0.228$ for $C$, $58{\text{mm}}^2$ for $A_t$, $7.676\text{kN}/\text{bolt}$ for $P$ and $36.1\text{kN}$ for $F_i$ in Equation (XX).

$\begin{array}{c}{n}_p=\frac{\left(830\text{MPa}\right)\left(58{\text{mm}}^2\right)}{\left(0.228\right)\left(7.676\text{kN}/\text{bolt}\right)+36.1\text{kN}}\\ =\frac{\left(830\text{MPa}\right)\left(\frac{1\text{N}/{\text{mm}}^{\text{2}}}{1\text{MPa}}\right)\left(58{\text{mm}}^2\right)}{\left(\left[\left(0.228\right)\left(7.676\right)+36.1\right]\text{kN}\right)\left(\frac{\text{1000}\text{N}}{\text{1}\text{kN}}\right)}\\ =\frac{48140}{37850}\\ =1.27\end{array}$

Thus, the yield factor of safety is $1.27$.

Substitute $830\text{MPa}$ for $S_p$, $0.228$ for $C$, $58{\text{mm}}^2$ for $A_t$, $7.676\text{kN}/\text{bolt}$ for $P$ and $36.1\text{kN}$ for $F_i$ in Equation (XXI).

$\begin{array}{c}{n}_L=\frac{\left(830\text{MPa}\right)\left(58{\text{mm}}^2\right)-\left(36.1\text{kN}\right)}{\left(0.228\right)\left(7.676\text{kN}/\text{bolt}\right)}\\ =\frac{\left(830\text{MPa}\right)\left(\frac{1\text{N}/{\text{mm}}^{\text{2}}}{1\text{MPa}}\right)\left(58{\text{mm}}^2\right)-\left(36.1\text{kN}\right)}{\left(0.263\right)\left(7.676\text{kN}/\text{bolt}\right)}\\ =\frac{\left(\left(830\times 58\right)\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)-\left(36.1\text{kN}\right)}{\left(0.228\right)\left(7.676\text{kN}/\text{bolt}\right)}\\ =\frac{48.140-36.1}{1.750}\end{array}$

$\begin{array}{c}{n}_L=\frac{48.140-36.1}{1.750}\\ =6.88\end{array}$

Thus, the load factor of safety is $6.88$.

Substitute $830\text{MPa}$ for $S_p$, $0.228$ for $C$, $58{\text{mm}}^2$ for $A_t$, $7.676\text{kN}/\text{bolt}$ for $P$ and $36.1\text{kN}$ for $F_i$ in Equation (XXII).

$\begin{array}{c}{n}_o=\frac{\left(36.1\text{kN}\right)}{\left(7.676\text{kN}/\text{bolt}\right)\left(1-0.228\right)}\\ =\frac{36.1\text{kN}}{5.9258\text{kN}}\\ =6.09\end{array}$

Thus, the joint separation factor is $6.09$.