Chapter 8, Problem 40P

To determine

The yielding factor of safety.

The load factor.

The joint separation factor.

Answer to Problem 40P

The yielding factor of safety is $1.28$.

The load factor is $8$.

The joint separation factor is $11.40$.

Explanation of Solution

Write the expression tensile load per bolt.

    $T_l=\frac{p_g\times A_s}{N}$                                                                                                 (I)

Here, tensile load per bolt is $T_l$, pressure is $p_g$, cross section area of sealing is $A$ and number of bolt is $N$.

Write the expression for cross section area of sealing.

    $A_s=\frac{\pi }{4}{d}^2$                                                                                                  (II)

Here, area is $A_s$ and sealing diameter is $d$.

Write the expression for grip.

    $l=h+\frac{d}{2}$                                                                                                  (III)

Here, total thickness of plate and washer is $h$ and fastener diameter is $d$.

Write the expression for length of bolt.

    $L\ge h+1.5d$                                                                                            (IV)

Write the expression for threaded length.

    $L_t=2d+6$                                                                                               (V)

Here threaded length is $L_t$ and fastener diameter is $d$.

Write the expression for area of unthreaded portion.

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

Here, the area of unthreaded portion is $A_d$ and the fastener diameter is $d$.

Write the expression for length of unthreaded portion in grip.

    $l_d=L-L_T$                                                                                           (VII)

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

Write the expression for length of threaded portion in grip

    $l_t=l-l_d$                                                                                             (VIII)

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

Write the expression for bolt stiffness.

    $k_b=\frac{A_d{A}_{t}E}{A_d{l}_t+A_t{l}_d}$                                                                                   (IX)

Here, bolt stiffness is $k_b$.

Write the expression for stiffness of top frusta.

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

Here, the stiffness of top frusta is $k_1$.

Write the expression for total spring rate of the member.

    $k=\frac{1}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}$                                                                                  (XI)

Here, the total spring rate of the member is $k_m$.

Write the expression for joint stiffness constant.

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

Here, joint stiffness constant is $C$.

Write the expression for preload.

    $F_i=0.75S_p{A}_t$                                                                                       (XIII)

Here, preload is $F_i$, minimum proof strength is $S_p$ and tensile stress area is $A_t$.

Write the expression for load factor

    $n_L=\frac{S_p{A}_t-F_i}{C\left(P_{\max }\right)}$                                                                                     (XIV)

Here, load factor is $n_L$.

Write the expression for yielding factor of safety.

    $n_P=\frac{S_p{A}_t}{C\left(P_{\max }\right)+F_i}$                                                                               (XV)

Here, yielding factor of safety is $n_P$.

Write the expression for load factor guarding against joint separation.

    $n_o=\frac{F_i}{\left(P_{\max }\right)\left(1-C\right)}$                                                                               (XVI)

Here, load factor guarding against joint is $n_o$.

Write the expression for total external load.

    $P_{total}=p_g\times A_c$                                                                                     (XVII)

Here, the total load is $P_{total}$ and pressure of the gas inside the cylinder is $p_g$

Write the expression for load on the bolt.

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

Here, the load on the bolt is $P$.

Write the expression for cross section area of the cylinder.

    $A_c=\frac{\pi }{4}{d}_c^2$                                                                                              (XIX)

Here, area of cylinder is $A_c$ and diameter is $d_c$.

Conclusion:

Substitute $3.5\text{in}$ for $d$ in Equation (II).

    $\begin{array}{c}{A}_s=\frac{\pi }{4}{\left(3.5\text{in}\right)}^2\\ =9.621{\text{in}}^2\end{array}$

Substitute $9.621{\text{in}}^2$ for $A_s$, $8$ for $N$ and $1200\text{psi}$ for $p_g$ in Equation (I).    $\begin{array}{c}{T}_l=\frac{\left(1200\text{psi}\right)\times \left(9.621{\text{in}}^2\right)}{\left(8\right)}\\ =1443.1\text{lbf}\end{array}$

Substitute $0.375\text{in}$ for $h$ and $0.4375\text{in}$ for $d$ in Equation (III).    $\begin{array}{c}l=\left(0.375\text{in}\right)+\frac{\left(0.4375\text{in}\right)}{2}\\ =0.5937\text{in}\end{array}$

Substitute $0.375\text{in}$ for $h$ and $0.4375\text{in}$ for $d$ in Equation (IV).    $\begin{array}{l}L\ge \left(0.375\text{in}\right)+1.5\left(0.4375\text{in}\right)\\ L\ge 1.031\text{in}\\ L\approx 1.25\text{in}\end{array}$

Substitute $0.4375\text{in}$ for $d$ and $0.875\text{in}$ for $h$ in Equation (V).    $\begin{array}{c}{L}_t=2\left(0.4375\text{in}\right)+0.875\text{in}\\ =1.125\text{in}\end{array}$

Substitute $0.4375\text{in}$ for $d$ in Equation (VI).

    $\begin{array}{c}{A}_d=\frac{\pi }{4}{\left(0.4375\text{in}\right)}^2\\ =0.1503{\text{in}}^2\end{array}$

Substitute $1.6\text{in}$ for $L$ and $1.25\text{in}$ for $L_T$ in Equation (VII).

    $\begin{array}{c}{l}_d=1.25\text{in}-1.125\text{in}\\ =0.125\text{in}\end{array}$

Substitute $1.093\text{in}$ for $l$ and $0.475\text{in}$ for $l_d$ in Equation (VIII).

    $\begin{array}{c}{l}_t=0.5937\text{in}-0.125\text{in}\\ =0.46875\text{in}\end{array}$

Refer to table 8-1 “Diameter and areas of unified screw threads UNC and UNF”, obtain tensile stress area for nominal diameter of $0.4375\text{in}$ as $0.1063{\text{in}}^2$.

Refer to table 8-9 “SAE specification for steel”, to obtain the minimum proof strength as $120\text{kpsi}$ for non permanent connection (low carbon steel).

Substitute $0.1503{\text{in}}^2$ for $A_d$, $0.1063{\text{in}}^2$ for $A_t$, $30\text{kpsi}$ for $E$, $0.46875\text{in}$ for $l_t$ and $0.125\text{in}$ for $l_d$ in Equation (IX).

    $\begin{array}{c}{k}_b=\frac{\left(0.1503{\text{in}}^2\right)\left(0.1063{\text{in}}^2\right)\left(30\text{Mpsi}\right)}{\left(0.1503{\text{in}}^2\right)\left(0.46875\text{in}\right)+\left(0.1063{\text{in}}^2\right)\left(0.125\text{in}\right)}\\ =\frac{\left(0.4793\text{Mpsi}\cdot {\text{in}}^4\right)}{\left(0.08374{\text{in}}^3\right)}\left(\frac{1\text{Mlbf}/\text{in}}{1\text{Mpsi}\cdot \text{in}}\right)\\ =5.72\text{Mlbf}/\text{in}\end{array}$

Figure (1) shows the frustum of the cone for compressive load on the joint.

Figure-(1)

Write the expression for thickness of frusta.

    $\begin{array}{c}t=\frac{0.59375\text{in}}{2}\\ =0.2968\text{in}\end{array}$

Write the expression for diameter of the top frusta.

    $D=0.65625\text{in}$

Substitute $30\text{Mpsi}$ for $E$, $0.4375\text{in}$ for $d$, $0.2968\text{in}$ for $t$ and $0.65625\text{in}$ for $D$ in Equation (X).

    $\begin{array}{c}{k}_1=\frac{0.5774\pi \left(30\text{Mpsi}\right)\left(0.4375\text{in}\right)}{\ln \left(\frac{\left(1.155\left(0.2968\text{in}\right)+0.65625\text{in}-\left(0.4375\text{in}\right)\right)\left(0.65625\text{in}+0.4375\text{in}\right)}{\left(1.155\left(0.2968\text{in}\right)+0.65625\text{in}+\left(0.4375\text{in}\right)\right)\left(0.65625\text{in}-0.4375\text{in}\right)}\right)}\\ =\frac{\left(23.80\text{Mpsi}\text{.in}\right)\left(\frac{1\text{Mlbf}/\text{in}}{1\text{Mpsi}\cdot \text{in}}\right)}{\ln \left(\frac{\left(0.6141\text{in}\right)}{\left(0.3142\text{in}\right)}\right)}\\ =\frac{23.80\text{Mlbf}/\text{in}}{0.6701}\\ \approx 35.52\text{Mlbf}/\text{in}\end{array}$

Write the expression for thickness of middle frusta.

    $\begin{array}{c}t=0.375\text{in}-0.2968\text{in}\\ =0.07812\text{in}\end{array}$

Write the expression for diameter of the middle frusta.

    $\begin{array}{c}D=0.65625\text{in}+2\left(0.5937\text{in}-0.375\text{in}\right)\tan 30°\\ =0.90878\text{in}\\ \approx 0.9088\text{in}\end{array}$

Substitute $30\text{Mpsi}$ for $E$, $0.4375\text{in}$ for $d$, $0.07812\text{in}$ for $t$ and $0.9088\text{in}$ for $D$ in Equation (X).

    $\begin{array}{c}{k}_2=\frac{0.5774\pi \left(30\text{Mpsi}\right)\left(0.4375\text{in}\right)}{\ln \left(\frac{\left(1.155\left(0.07812\text{in}\right)+0.9088\text{in}-0.4375\text{in}\right)\left(0.9088\text{in}+0.4375\text{in}\right)}{\left(1.155\left(0.07812\text{in}\right)+0.9088\text{in}+0.4375\text{in}\right)\left(0.9088\text{in}-0.4375\text{in}\right)}\right)}\\ =\frac{\left(23.80\text{Mpsi}\text{.in}\right)\left(\frac{1\text{Mlbf}/\text{in}}{1\text{Mpsi}\cdot \text{in}}\right)}{\ln \left(\frac{\left(0.7559\text{in}\right)}{\left(0.6770\text{in}\right)}\right)}\\ =\frac{23.80\text{Mlbf}/\text{in}}{0.1102}\\ \approx 215.9\text{Mlbf}/\text{in}\end{array}$

Write the expression for thickness of lower frusta.

    $\begin{array}{c}t=0.5937\text{in}-0.375\text{in}\\ =0.2187\text{in}\end{array}$

Write the expression for diameter of the lower frusta.

    $D=0.65625\text{in}$

Substitute $14.5\text{Mpsi}$ for $E$, $0.4375\text{in}$ for $d$, $0.2187\text{in}$ for $t$ and $0.65625\text{in}$ for $D$ in Equation (X).

    $\begin{array}{c}{k}_3=\frac{0.5774\pi \left(14.5\text{Mpsi}\right)\left(0.4375\text{in}\right)}{\ln \left(\frac{\left(1.155\left(0.2187\text{in}\right)+0.65625\text{in}-0.4375\text{in}\right)\left(0.65625\text{in}+0.4375\text{in}\right)}{\left(1.155\left(0.2187\text{in}\right)+0.65625\text{in}+0.4375\text{in}\right)\left(0.65625\text{in}-0.4375\text{in}\right)}\right)}\\ =\frac{\left(11.50\text{Mpsi}\text{.in}\right)\left(\frac{1\text{Mlbf}/\text{in}}{1\text{Mpsi}\cdot \text{in}}\right)}{\ln \left(\frac{\left(0.5155\text{in}\right)}{\left(0.2945\text{in}\right)}\right)}\\ =\frac{11.50\text{Mlbf}/\text{in}}{0.5598}\\ \approx 20.55\text{Mlbf}/\text{in}\end{array}$

Substitute $35.52\text{Mlbf}/\text{in}$ for $k_1$, $215.9\text{Mlbf}/\text{in}$ for $k_2$ and $20.55\text{Mlbf}/\text{in}$ for $k_3$ in Equation (XI).

    $\begin{array}{c}{k}_m=\frac{1}{\frac{1}{\left(35.52\text{Mlbf}/\text{in}\right)}+\frac{1}{\left(215.9\text{Mlbf}/\text{in}\right)}+\frac{1}{\left(20.55\text{Mlbf}/\text{in}\right)}}\\ =\frac{1}{\left(0.0281\text{Mlbf}/\text{in}\right)+\left(0.00463\text{Mlbf}/\text{in}\right)+\left(0.0486\text{Mlbf}/\text{in}\right)}\\ =\frac{1}{0.08133\text{Mlbf}/\text{in}}\\ =12.29\text{Mlbf}/\text{in}\end{array}$

Substitute $5.72\text{Mlbf}/\text{in}$ for $k_b$ and $12.29\text{Mlbf}/\text{in}$ for $k_m$ in Equation (XII).

    $\begin{array}{c}C=\frac{\left(5.72\text{Mlbf}/\text{in}\right)}{\left(5.72\text{Mlbf}/\text{in}\right)+\left(12.29\text{Mlbf}/\text{in}\right)}\\ =0.3257\end{array}$

Substitute $120\text{kpsi}$ for $S_P$ and $0.1063{\text{in}}^2$ for $A_t$ in Equation (XIII)

    $\begin{array}{c}{F}_i=0.75\left(120\text{kpsi}\right)\left(0.1063{\text{in}}^2\right)\\ =9.567\text{kpsi}\end{array}$

Substitute $\text{3}\text{.25}\text{in}$ for $d_c$ in Equation (XIX).

    $\begin{array}{c}{A}_c=\frac{\pi }{4}{\left(\text{3}\text{.25}\text{in}\right)}^2\\ =\left(0.785\right)10.5625{\text{in}}^2\\ =8.29{\text{in}}^2\end{array}$

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

    $\begin{array}{c}{P}_{total}=1200\text{Psi}\left(8.29{\text{in}}^2\right)\\ =9948\text{Psi}\cdot {\text{in}}^2\end{array}$

Substitute $9948\text{Psi}\cdot {\text{in}}^2$ for $P_{total}$ and $8$ for $N$ in Equation (XVIII).

    $\begin{array}{c}P=\frac{9948\text{Psi}\cdot {\text{in}}^2}{8}\\ =\frac{9948\text{Psi}\cdot {\text{in}}^2}{8}\left(\frac{{\text{10}}^{-3}\text{kpsi}}{1\text{Psi}\cdot {\text{in}}^2}\right)\\ \approx 1.244\text{kpsi}\end{array}$

Substitute $120\text{kpsi}$ for $S_P$, $9.567\text{kpsi}$ for $F_i$, $0.3257$ for $C$, $0.1063{\text{in}}^2$ for $A_t$, $1.244\text{kpsi}$ for $P_{\max }$ in Equation (XIV).

    $\begin{array}{c}{n}_L=\frac{\left(120\text{kpsi}\right)\left(0.1063{\text{in}}^2\right)-\left(9.567\text{kpsi}\right)}{\left(0.3257\right)\left(1.244\text{kpsi}\right)}\\ =\frac{3.189\text{kN}}{0.4051\text{kN}}\\ =7.872\\ \approx 8\end{array}$

Thus the load factor is $8$.

Substitute $120\text{kpsi}$ for $S_P$, $9.567\text{kpsi}$ for $F_i$, $0.3257$ for $C$, $0.1063{\text{in}}^2$ for $A_t$, $1.244\text{kpsi}$ for $P_{\max }$ in Equation (XV).

    $\begin{array}{c}{n}_p=\frac{\left(120\text{kpsi}\right)\left(0.1063{\text{in}}^2\right)}{\left(0.3257\right)\left(1.244\text{kpsi}\right)+9.567\text{kpsi}}\\ =\frac{12.756}{9.972}\\ =1.279\\ \approx 1.28\end{array}$

Thus the yielding factor of safety is $1.28$.

Substitute $9.567\text{kpsi}$ for $F_i$, $0.3257$ for $C$, $1.244\text{kpsi}$ for $P_{\max }$ in Equation (XVI).

    $\begin{array}{c}{n}_o=\frac{9.567\text{kpsi}}{\left(1.244\text{kpsi}\right)\left(1-0.3257\right)}\\ =\frac{9.567}{0.8388}\\ =11.40\end{array}$

Thus the load factor against joint separation is $11.40$.