Question 15.5: Preloaded Bolt Connecting the Head and Cylinder of a Pressur...

a. Referring to Figure 15.15, Equation 15.34 gives

k_p=\frac{0.58 \pi E_p d}{2 \ln \left(5 \frac{0.58 L+0.5 d}{0.58 L+2.5 d}\right)}              (15.34)

k_p=\frac{0.58 \pi\left(E_s / 2\right)(0.75)}{2 \ln \left[5 \frac{0.58(2)+0.5(0.75)}{0.58(2)+2.5(0.75)}\right]}=0.368 E_s

Through the use of Equation 15.31a,

k_b=\frac{A_b E_b}{L}           (15.31a)

k_b=\frac{A E_s}{L}=\frac{\pi d^2 E_s}{4 L}=\frac{\pi(0.75)^2 E_s}{4(2)}=0.221 E_s

Equation 15.22 is therefore

C=\frac{k_b}{k_b+k_p}=\frac{0.221}{0.221+0.368}=0.375

b. From Tables 15.1 and 15.4, we have A_{t} = 0.334 in.² and S_{p} = 105 ksi. Applying Equation 15.20,

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=0.9 S_p A_t=0.90(105)(0.334)=31.6  kips

For N_{b} bolts, Equation 15.26 can be written in the form

\frac{C P n}{A_t}+\frac{F_i}{A_t}=S_p                 (15.26)

\frac{C\left(P / N_b\right) n}{A_t}+\frac{F_i}{A_t}=S_p

from which

N_b=\frac{C P n}{S_p A_t-F_i}

Substituting the numerical values, we have

N_b=\frac{(0.375)(55)(1.5)}{105(0.334)-31.6}=8.92

Comment: Nine bolts should be used.

c. By Equation 15.21,

T=K d F_i      (15.21)

T=0.2 d F_i=0.2(0.75)(31.6)=4.74  kips \cdot in .