Chapter 8, Problem 37P

To determine

The yielding factor of safety.

The load factor.

Joint separation factor.

Answer to Problem 37P

The yielding factor of safety is $1.3$.

The load factor is $10.57$.

Joint separation factor is $12.03$.

Explanation of Solution

Write the expression for grip.

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

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

Write the expression for length of bolt.

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

Write the expression for threaded length.

    $L_t=2d+6$                                                       (III)

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

Write the expression for length of unthreaded portion in grip.

    $l_d=L-L_T$                                                         (IV)

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

Write the expression for length of threaded portion in grip

    $l_t=l-l_d$                                                           (V)

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

Write the expression for bolt stiffness.

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

Here, bolt stiffness is $k_b$.

Write the expression for stiffness of top frusta.

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

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

Write the expression for total spring rate of the member.

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

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

Write the expression for joint stiffness constant.

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

Here, joint stiffness constant is $C$.

Write the expression for total external load.

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

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

Write the expression for cross section area of the cylinder.

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

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

Write the expression for load on the bolt.

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

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

Write the expression for proof load.

    $F_p=S_p{A}_t$                                      (XIII)

Here, the proof load is $F_p$, proof strength is $S_p$ and the area of threaded portion is $A_t$.

Write the expression for preload.

    $F_i=0.75F_p$                           (XIV)

Here, preload is $F_i$.

Write the expression for load factor

    $n_L=\frac{S_p{A}_t-F_i}{C\left(P\right)}$                            (XV)

Here, load factor is $n_L$.

Write the expression for yielding factor of safety.

    $n_P=\frac{S_p{A}_t}{C\left(P\right)+F_i}$                             (XVI)

Here, yielding factor of safety is $n_P$.

Write the expression for load factor guarding against joint separation.

    $n_o=\frac{F_i}{\left(P\right)\left(1-C\right)}$                                    (XVII)

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

Conclusion:

Refer to the Table 8.7, obtain the Thickness of the steel cylinder cap.

    $h=t_1=20\text{mm}$

Substitute $20\text{mm}$ for $h$, and $12\text{mm}$ for $d$ in Equation (I).

    $\begin{array}{c}l=\left(20\text{mm}\right)+\frac{\left(12\text{mm}\right)}{2}\\ =26\text{mm}\end{array}$

Substitute $20\text{mm}$ for $h$, and $12\text{mm}$ for $d$ in Equation (II).

    $\begin{array}{l}L\ge \left(20\text{mm}\right)+1.5\left(12\text{mm}\right)\\ L\ge 38\text{mm}\\ L\approx 40\text{mm}\end{array}$

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

    $\begin{array}{c}{L}_t=2\left(12\text{mm}\right)+6\text{mm}\\ =30\text{mm}\end{array}$

Substitute $40\text{mm}$ for $L$ and $30\text{mm}$ for $L_T$ in Equation (IV).

    $\begin{array}{c}{l}_d=40\text{mm}-30\text{mm}\\ =10\text{mm}\end{array}$

Substitute $26\text{mm}$ for $l$ and $10\text{mm}$ for $l_d$ in Equation (V).

    $\begin{array}{c}{l}_t=26\text{mm}-10\text{mm}\\ =16\text{mm}\end{array}$

Refer to table 8-1 “Diameter and areas of coarse pitch and fine pitch metric threads”, obtain tensile stress area for nominal diameter of $12\text{mm}$ as $84.3{\text{mm}}^2$.

Refer to table 8-11 “Metric mechanical-property classes for steel bolts, screws and studs”, to obtain the minimum proof strength as $650\text{MPa}$ for medium carbon steel.

Substitute $12\text{mm}$ for $d$ to calculate area $A_d$.

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

Substitute $113.1{\text{mm}}^2$ for $A_d$, $84.3{\text{mm}}^2$ for $A_t$, $207\text{GPa}$ for $E$, $16\text{mm}$ for $l_t$ and $10\text{mm}$ for $l_d$ in Equation (VI).

    $\begin{array}{c}{k}_b=\frac{\left(113.1{\text{mm}}^2\right)\left(84.3{\text{mm}}^2\right)\left(207\text{GPa}\right)}{\left(113.1{\text{mm}}^2\right)\left(16\text{mm}\right)+\left(84.3{\text{mm}}^2\right)\left(10\text{mm}\right)}\\ =\frac{\left(113.1{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\right)\left(84.3{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\right)\left(207\text{GPa}\left(\frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}\right)\right)}{\left[\begin{array}{l}\left(113.1{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\right)\left(16\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\\ +\left(84.3{\text{mm}}^2\left(\frac{1{\text{m}}^{\text{2}}}{{10}^6{\text{mm}}^2}\right)\right)\left(10\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\end{array}\right]}\\ =\frac{1973.61\text{N}/\text{m}}{2.6526\times {10}^{-6}}\\ =744\text{MN}/\text{m}\end{array}$

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

Figure-(1)

Here, thickness is $t$, diameter of the top frusta is $D$.

Write the expression for thickness of frusta.

    $t=13\text{mm}$

Write the expression for diameter of the top frusta.

    $\begin{array}{c}D=1.5d\\ =1.5\left(12\text{mm}\right)\\ =18\text{mm}\end{array}$

Substitute $207\text{GPa}$ for $E$, $12\text{mm}$ for $d$, $13\text{mm}$ for $t$ and $18\text{mm}$ for $D$ in Equation (VII).

    $\begin{array}{c}{k}_1=\frac{0.5774\pi \left(207\text{GPa}\right)\left(12\text{mm}\right)}{\ln \left(\frac{\left(1.155\left(13\text{mm}\right)+18\text{mm}-12\text{mm}\right)\left(18\text{mm}+12\text{mm}\right)}{\left(1.155\left(13\text{mm}\right)+18\text{mm}+12\text{mm}\right)\left(18\text{mm}-12\text{mm}\right)}\right)}\\ =\frac{0.5774\pi \left(207\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(12\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{\ln \left(\frac{\left(630.45\text{mm}\right)}{\left(270.09\text{mm}\right)}\right)}\\ =\frac{4.5058\times {10}^9\text{N}/\text{m}}{0.847}\\ =5317\text{MN}/\text{m}\end{array}$

Write the expression for thickness of middle frusta.

    $\begin{array}{c}t=20\text{mm}-13\text{mm}\\ =7\text{mm}\end{array}$

Write the expression for diameter of the middle frusta.

    $\begin{array}{c}D=18\text{mm}+2\left(13\text{mm}+7\text{mm}\right)\tan 30°\\ =24.93\text{mm}\end{array}$

Substitute $207\text{GPa}$ for $E$, $12\text{mm}$ for $d$, $7\text{mm}$ for $t$ and $24.93\text{mm}$ for $D$ in Equation (VII).

    $\begin{array}{c}{k}_2=\frac{0.5774\pi \left(207\text{GPa}\right)\left(12\text{mm}\right)}{\ln \left(\frac{\left(1.155\left(7\text{mm}\right)+24.93\text{mm}-12\text{mm}\right)\left(24.93\text{mm}+12\text{mm}\right)}{\left(1.155\left(7\text{mm}\right)+24.93\text{mm}+12\text{mm}\right)\left(24.93\text{mm}-12\text{mm}\right)}\right)}\\ =\frac{0.5774\pi \left(207\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(12\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{\ln \left(\frac{\left(776.08\text{mm}\right)}{\left(582.04\text{mm}\right)}\right)}\\ =\frac{4.5058\times {10}^9\text{N}/\text{m}}{0.2880}\\ =15645\text{MN}/\text{m}\end{array}$

Write the expression for thickness of lower frusta.

    $\begin{array}{c}t=26\text{mm}-20\text{mm}\\ =6\text{mm}\end{array}$

Write the expression for diameter of the lower frusta.

    $\begin{array}{c}D=1.5\left(12\text{mm}\right)\\ =18\text{mm}\end{array}$

Substitute $100\text{GPa}$ for $E$, $12\text{mm}$ for $d$, $6\text{mm}$ for $t$ and $18\text{mm}$ for $D$ in Equation (VII).

    $\begin{array}{c}{k}_3=\frac{0.5774\pi \left(100\text{GPa}\right)\left(12\text{mm}\right)}{\ln \left(\frac{\left(1.155\left(6\text{mm}\right)+18\text{mm}-12\text{mm}\right)\left(18\text{mm}+12\text{mm}\right)}{\left(1.155\left(6\text{mm}\right)+18\text{mm}+12\text{mm}\right)\left(18\text{mm}-12\text{mm}\right)}\right)}\\ =\frac{0.5774\pi \left(100\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(12\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{\ln \left(\frac{\left(776.08\text{mm}\right)}{\left(582.04\text{mm}\right)}\right)}\\ =\frac{2.1767\times {10}^9\text{N}/\text{m}}{0.56}\\ =3887\text{MN}/\text{m}\end{array}$

Substitute $5317\text{MN}/\text{m}$ for $k_1$, $15645\text{MN}/\text{m}$ for $k_2$ and $3887\text{MN}/\text{m}$ for $k_3$ in Equation (VIII).

    $\begin{array}{l}\frac{1}{k}=\frac{1}{\frac{1}{\left(5317\text{MN}/\text{m}\right)}+\frac{1}{\left(15645\text{MN}/\text{m}\right)}+\frac{1}{\left(3887\text{MN}/\text{m}\right)}}\\ \frac{1}{k}=\frac{1}{\left(1.8807\times {10}^{-4}\text{MN}/\text{m}\right)+\left(6.3918\times {10}^{-5}\text{MN}/\text{m}\right)+\left(2.5727\times {10}^{-4}\text{MN}/\text{m}\right)}\\ k_m=1963.64\text{MN}/\text{m}\end{array}$

Substitute $744\text{MN}/\text{m}$ for $k_b$ and $1963.64\text{MN}/\text{m}$ for $k_m$ in Equation (IX).

    $\begin{array}{c}C=\frac{\left(744\text{MN}/\text{m}\right)}{\left(744\text{MN}/\text{m}\right)+\left(1963.64\text{MN}/\text{m}\right)}\\ =0.2748\end{array}$

Substitute $100\text{mm}$ for $d_c$ in Equation (XI).

    $\begin{array}{c}{A}_c=\frac{\pi }{4}{\left(100\text{mm}\right)}^2\\ =\left(0.785\right){\left(100\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ =0.007854{\text{m}}^2\end{array}$

Substitute $0.007854{\text{m}}^2$ for $A_c$ and $6\text{MPa}$ for $p_g$ in Equation (X).

    $\begin{array}{c}{P}_{total}=6\text{MPa}\times 0.007854{\text{m}}^2\\ =6\text{MPa}\left(\frac{{10}^6\text{N}/{\text{m}}^{\text{2}}}{1\text{MPa}}\right)\times 0.007854{\text{m}}^2\\ =47124\text{N}\left(\frac{1\text{kN}}{\text{1000}\text{N}}\right)\\ =47.124\text{kN}\end{array}$

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

    $\begin{array}{c}P=\frac{47.124\text{kN}}{10}\\ =4.712\text{kN}\end{array}$

Substitute $650\text{MPa}$ for $S_P$ and $84.3{\text{mm}}^2$ for $A_t$ in Equation (XIII).

    $\begin{array}{c}{F}_p=\left(650\text{MPa}\right)\left(84.3{\text{mm}}^{\text{2}}\right)\\ =\left(650\text{MPa}\left(\frac{{10}^3\text{kPa}}{1\text{MPa}}\right)\right)\left(84.3{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\right)\\ =57.795\text{kN}\end{array}$

Substitute $57.795\text{kN}$ for $F_p$ in Equation (XIV).

    $\begin{array}{c}{F}_i=0.75\left(57.795\text{kN}\right)\\ =41.1\text{kN}\end{array}$

Substitute $650\text{MPa}$ for $S_P$, $41.1\text{kN}$ for $F_i$, $0.2748$ for $C$, $4.712\text{kN}$ for $P$ and $84.3{\text{mm}}^2$ for $A_t$ in Equation (XV).

    $\begin{array}{c}{n}_L=\frac{\left(650\text{MPa}\right)\left(84.3{\text{mm}}^2\right)-\left(41.1\text{kN}\right)}{0.2748\left(4.712\text{kN}\right)}\\ =\frac{\left(54795\text{MPa}\cdot {\text{mm}}^2\left(\frac{{10}^3\text{kPa}}{1\text{MPa}}\right)\left(\frac{\text{1}{\text{m}}^2}{{\text{10}}^6{\text{mm}}^2}\right)\right)-\left(41.1\text{kN}\right)}{\left(0.2748\right)\left(4.712\text{kN}\right)}\\ =\frac{54.795\text{kPa}\cdot {\text{m}}^2\left(\frac{\text{kN}}{1\text{kPa}\cdot {\text{m}}^2}\right)-41.1\text{kN}}{1.2948\text{kN}}\\ =10.57\end{array}$

Thus the load factor is $10.57$.

Substitute $650\text{MPa}$ for $S_P$, $41.1\text{kN}$ for $F_i$, $0.2748$ for $C$, $4.712\text{kN}$ for $P$ and $84.3{\text{mm}}^2$ for $A_t$ in Equation (XVI).

    $\begin{array}{c}{n}_P=\frac{\left(650\text{MPa}\right)\left(84.3{\text{mm}}^2\right)}{0.2748\left(4.712\text{kN}\right)+\left(41.1\text{kN}\right)}\\ =\frac{\left(54795\text{MPa}\cdot {\text{mm}}^2\left(\frac{{10}^3\text{kPa}}{1\text{MPa}}\right)\left(\frac{\text{1}{\text{m}}^2}{{\text{10}}^6{\text{mm}}^2}\right)\right)}{0.2748\left(4.712\text{kN}\right)+41.1\text{kN}}\\ =\frac{54.795\text{kPa}\cdot {\text{m}}^2\left(\frac{\text{kN}}{1\text{kPa}\cdot {\text{m}}^2}\right)}{42.3948\text{kN}}\\ \approx 1.3\end{array}$

Thus, the yielding factor of safety is $1.3$

Substitute $41.1\text{kN}$ for $F_i$, $0.2748$ for $C$, $4.712\text{kN}$ for $P$ in Equation (XVI).

    $\begin{array}{c}{n}_o=\frac{41.1\text{kN}}{\left(4.712\text{kN}\right)\left(1-0.2748\right)}\\ =12.03\end{array}$

Thus, the load factor against joint separation is $12.03$.