Question 15.6: Preloaded Bolt Clamping of a Cylinder under External Load A ...

See Figures 15.11 and 15.16.

The cross-sectional area of the parts is equal to A_p=\pi\left(D^2-d^2\right) / 4=\pi\left(20^2-10^2\right) / 4= 235.6mm².

a. Through the use of Equation 15.20, the preload is

F_i=\left\{\begin{array}{cc} 0.75 F_p & \text { (reused connections) } \\ 0.9 F_p & \text { (permanent connections) } \end{array}\right.      (15.20)

F_i=0.75 F_p=0.75 S_p A_t=0.75(380)(58)=16.53  kN

This corresponds to an estimated bolt tightening torque (see Section 15.8) of

T=0.2 F_i d=0.2(16.53)(10)=33.06  N \cdot m

b. From Equation 15.33a, the lengths of thread L_{t} and shank L_{s} of the bolt (Figure 15.12) are

L_t=\left\{\begin{array}{cr} 2 d+6 & L \leq 125 \\ 2 d+12 & 125<L \leq 200 \\ 2 d+25 & L>200 \end{array}\right.        (15.33a)

\begin{array}{l} L_t=2 d+6=2(10)+6=26  mm \\ L_s=L-L_t=65-26=39  mm \end{array}

The stiffness constant for the bolt, by Equation 15.32, is

\frac{1}{k_b}=\frac{L_t}{A_t E_b}+\frac{L_{ s }}{A_b E_b}      (15.32)

\frac{1}{k_b}=\frac{L_t}{A_t E}+\frac{L_s}{A_s E}=\frac{1}{200\left(10^6\right)}\left[\frac{26}{58}+\frac{39(4)}{\pi(10)^2}\right], \quad k_b=2.117\left(10^8\right)  N / m

By Equation 15.31b, the stiffness constant for the parts is

k_p=\frac{A_p E_p}{L}          (15.31b)

k_p=\frac{A_p E}{L}=\frac{235.6 \times 10^{-6}\left(200 \times 10^9\right)}{65 \times 10^{-3}}=7.249\left(10^8\right)  N / m

The joint stiffness factor, using Equation 15.22, is therefore

C=\frac{k_b}{k_p+k_b}=\frac{2.117}{7.249+2.117}=0.226

Comment: The results indicate that k_p \approx 3.4 k_b .

c. From Equations 15.23 and 15.24, the forces on the bolt and parts are

F_b=C P+F_i \quad\left(\text { for } F_p<0\right)       (15.23)

F_p=(1-C) P-F_i \quad\left(\text { for } F_p<0\right)     (15.24)

The largest tensile stress in the bolt equals

\sigma_b=\frac{F_b}{A_t}=\frac{18.34\left(10^3\right)}{58\left(10^{-6}\right)}=316  MPa

Comment: No stress-concentration factor applies for a statically loaded ductile material.

d. The factor of safety with respect to onset of yielding is equal to

n=\frac{S_y}{\sigma_b}=\frac{420}{316}=1.33

Applying Equation 15.28, the load required to separate the joint and factor of
safety against joint separation are

P_s=\frac{F_i}{(1-C)}           (15.28a)

n_s=\frac{P_s}{P}=\frac{F_i}{P(1-C)}      (15.28b)

\begin{array}{l} P_s=\frac{F_i}{(1-C)}=\frac{16.53}{(1-0.226)}=21.36  kN \\ n_s=\frac{P_s}{P}=\frac{21.36}{8}=2.67 \end{array}

Comment: Both safety factors found against yielding and separation are acceptable.