Question 15.7: Preloaded Fasteners in Fatigue Loading Figure 15.17a illustr...

See Figure 15.17.

From Table 7.3, the reliability factor is C_{r} = 0.89. The temperature factor is C_{t} = 1 (Section 7.7). Also, S_{p} = 85 ksi, S_{y} = 92 ksi, S_{u} = 120 ksi (from Table 15.4), K_{f} = 3 (by Table 15.6), and

A_t=0.226  in.^2    (from Table 15.1)

Equation 15.39 results in

S_e=C_r C_t\left\lgroup \frac{1}{K_f} \right\rgroup \left(0.45 S_u\right)        (15.39)

S_e=(0.89)(1)\left(\frac{1}{3}\right)(0.45 \times 120)=16  ksi

The Soderberg and Goodman fatigue failure lines are shown in Figure 15.17c.

a. For loosely held parts, when F_{i} = 0, load on the bolt equals the load on parts:

\begin{array}{l} P_m=\frac{1}{2}(7+0)=3.5 kips , \quad P_a=\frac{1}{2}(7-0)=3.5  kips \\ \sigma_a=\sigma_m=\frac{3.5}{0.226}=15.5  ksi \end{array}

A plot of the stresses shown in Figure 15.17c indicates that failure will occur.

b. Through the use of 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=0.75 S_p A_t=0.75(85)(0.226)=14.4  kips

The grip is L = 2.5 in. By Equations 15.31a and 15.34 with E_{b} = E_{p} = E, we obtain

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

\begin{array}{l} k_b=\frac{\pi d^2 E}{4 L}=\frac{\pi(0.625)^2 E}{4(2.5)}=0.123 E \\ k_p=\frac{0.58 \pi E(0.625)}{2 \ln \left[5 \frac{0.58(2.5)+0.5(0.625)}{0.58(2.5)+2.5(0.625)}\right]}=0.53 E \end{array}

The joint constant is then

C=\frac{k_b}{k_b+k_p}=\frac{0.123}{0.123+0.53}=0.188

Comment: The foregoing means that only about 20% of the external load fluctuation is felt by the bolt and hence about 80% goes to decrease clamping pressure.

Applying Equations 15.35 and 15.36,

\sigma_{b m}=\frac{C P_m}{A_t}+\frac{F_i}{A_t}     (15.36a)

\sigma_{b a}=\frac{C P_a}{A_t}      (15.36b)

\begin{aligned} F_{b m} &=C P_m+F_i \\ &=0.188(3.5)+14.4=15.1  kips \\ \sigma_{b m} &=\frac{15.1}{0.226}=66.8  ksi \\ F_{b a} &=C P_a=0.188(3.5)=0.66  kips \\ \sigma_{b a} &=\frac{0.66}{0.226}=2.92  ksi \end{aligned}

A plot on the fatigue diagram shows that failure will not occur (Figure 15.17c).

c. Equation 15.37 with P_{a} = P_{m} becomes

n=\frac{S_u A_t-F_i}{C\left[P_a\left\lgroup \frac{S_u}{S_e} \right\rgroup +P_m\right]}      (15.37)

n=\frac{S_u A_t-F_i}{C P_a\left[\left(\frac{S_u}{S_e}\right)+1\right]}       (15.40)

Introducing the given numerical values,

n=\frac{(120)(0.226)-14.4}{(0.188)(3.5)\left[\left\lgroup \frac{120}{16} \right\rgroup +1\right]}

from which n = 2.27.

Comment: This is the factor of safety guarding against the fatigue failure. Observe from Figure 15.17c that the Goodman criteria led to a less conservative (higher) value for n.

d. Substitution of the given data into Equations 15.27 and 15.28 gives

n=\frac{S_p A_t-F_i}{C P}      (15.27)

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} n=\frac{85(0.226)-14.4}{(0.188)(7)}=3.66 \\ n_s=\frac{14.4}{7(1-0.188)}=2.53 \end{array}

Comments: The factor of 3.66 prevents the bolt stress from becoming equal to proof strength. On the other hand, the factor of 2.53 guards against joint separation and the bolt taking the entire load.