Question 15.3: Preloaded Fasteners in Dynamic Loading Problem Repeat Exampl...
Preloaded Fasteners in Dynamic Loading
Problem Repeat Example 15-2 with a fluctuating load applied to the joint. Determine a suitable bolt size and preload for the joint shown in Figure 15-23 (repeated). Find its safety factors against fatigue, yielding, and separation. Determine the optimum preload as a percentage of proof strength to maximize the fatigue, yielding, and separation safety factors.
Given The joint dimensions are D = 1 in and l = 2 in. The applied load fluctuates between P = 0 and P = 1000 lb.
Assumptions The bolt has rolled threads. Both of the clamped parts are steel. The effects of the flanges on the joint stiffness will be ignored. A preload of 90% of the bolt’s proof strength will be applied as a first trial. Use 99% reliability and an operating temperature of 300°F.
See Figures 15-27 to 15-29.
1 Again, there are too many unknown variables to solve the necessary equations in one pass. Trial values must be chosen for various parameters and iteration used to find a good solution. We actually made several iterations to solve this problem but will present only one in the interest of brevity. The trial values used here have already been massaged to reasonable values.
2 The bolt diameter is the principal trial value to be chosen along with a thread series and a bolt class to define the proof strength. We choose a 5/16-18 UNC-2A steel bolt of SAE class 5.2 based on its successful use in the static loading problem of Example 15-2 (p. 887) which is similar. For a clamp length of 2 in, assume a bolt length of 2.5 in to allow sufficient protrusion for the nut. The preload is taken at 90% of proof strength as assumed above.
3 The proof strength, tensile-stress area, preload force, bolt stiffness, material stiffness, and joint constant are all the same as were found in Example 15-2 for the 90% preload factor. See that example for details. In summary they are
\begin{array}{lll} S_p=85 kpsi & A_t=0.052431 in ^2 & F_i=4011 lb \\ k_b=1.059 E 6 lb / in & k_m=1.063 E 7 lb / \text { in } & C=0.09056 \end{array} (a)
4 The portions of the peak fluctuating load P felt by the bolt and the material and the resulting loads in bolt and material after the load is applied are found the same way as in the previous example, but the peak load here is smaller:
\begin{array}{ll} P_b=91 lb & P_m=909 lb \\ F_b=4102 lb & F_m=3102 lb \end{array} (b)
5 Because these loads are fluctuating, we need to calculate the mean and alternating components of the force felt in the bolt. Figure 15-27 shows the load-deflection diagram for this problem, drawn to scale with the above forces applied. The shallow sine wave between the initial force line A and the maximum bolt force line B is the only fluctuating load felt by the bolt. The mean and alternating forces are then
\begin{array}{c} F_{\text {alt }}=\frac{F_b-F_i}{2}=\frac{4102-4011}{2}=45.3 lb \\ F_{\text {mean }}=\frac{F_b+F_i}{2}=\frac{4102+4011}{2}=4056.2 lb \end{array} (c)
Note how little of the 0–1000-lb fluctuating force is felt by the bolt.
6 The nominal mean and alternating stresses in the bolt are:
\begin{aligned} \sigma_{a_{\text {nom }}} &=\frac{F_{\text {alt }}}{A_t}=\frac{45.3}{0.052431}=864 psi \\ \sigma_{m_{\text {nom }}} &=\frac{F_{\text {mean }}}{A_t}=\frac{4056.2}{0.052431}=77364 psi \end{aligned} (d)
7 The fatigue stress-concentration factor for this diameter thread is found from equation 15.15c and the mean stress-concentration factor K_{fm} is found from equation 6.17 (p. 364).
\begin{array}{ll} K_f=5.7+0.6812 d & d \text { in inches } \\ K_f=5.7+0.02682 d & d \text { in mm } \end{array} (15.15c)
\begin{array}{ll} \text {if} K_f\left|\sigma_{\max _{\text {nom }}}\right|<S_y \text { then : } & K_{f m}=K_f \\ \text {if} K_f\left|\sigma_{\max _{\text {nom }}}\right|>S_y \text { then : } & K_{f m}=\frac{S_y-K_f \sigma_{a_{\text {nom }}}}{\left|\sigma_{m_{\text {nom }}}\right|} \\ \text {if} K_f\left|\sigma_{\max _{\text {nom }}}-\sigma_{\text {min }_{\text {nom }}}\right|>2 S_y \text { then: } & K_{f m}=0 \end{array} (6.17)
\begin{array}{l} K_f=5.7+0.6812 d=5.7+0.6812(0.3125)=5.9\\ \text { if } K_f\left|\sigma_{\max _{\text {nom }}}\right|>S_y \quad \text { then : } \quad K_{f m}=\frac{S_y-K_f \sigma_{a_{\text {nom }}}}{\left|\sigma_{m_{\text {nom }}}\right|}\\ K_f\left|\sigma_{\max _{\text {nom }}}\right|=K_f\left|\sigma_{a_{\text {nom }}}+\sigma_{m_{\text {nom }}}\right|=5.9|864+77364|=462548 psi\\ 462548 psi >S_y=92000 psi\\ K_{f m}=\frac{S_y-K_f \sigma_{a_{\text {nom }}}}{\left|\sigma_{m_{\text {nom }}}\right|}=\frac{92000-5.9(864)}{77364}=1.12 \end{array} (e)
8 The local mean and alternating stresses in the bolt are then:
\begin{aligned} \sigma_a &=K_f \sigma_{a_{\text {nom }}}=5.9(864)=5107 psi \\ \sigma_m &=K_{f m} \sigma_{m_{n o m}}=1.12(77364)=86893 psi \end{aligned} (f)
9 The stress at the initial preload is
\sigma_i=K_{f m} \frac{F_i}{A_t}=1.12 \frac{4011}{0.052431}=85923 psi (g)
10 An endurance strength must be found for this material. Using the methods of Section 6.6 (p. 327) we find
S_e=0.5 S_{u t}=0.5(120000)=60000 psi (h)
\begin{aligned} S_e &=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_e \\ &=0.70(0.995)(0.76)(1)(0.81)(60000)=25726 psi \end{aligned} (i)
where the strength reduction factors are taken from the tables and formulas in Section 6.6 (p. 327) for, respectively, axial loading, the bolt size, a machined finish, room temperature, and 99% reliability.
11 The corrected endurance strength and the ultimate tensile strength are used in equation 15.16 (p. 891) to find the safety factor from the Goodman line.
N_f=\frac{S_e\left(S_{u t}-\sigma_i\right)}{S_e\left(\sigma_m-\sigma_i\right)+S_{u t} \sigma_a} (15.16)
\begin{aligned} N_f &=\frac{S_e\left(S_{u t}-\sigma_i\right)}{S_e\left(\sigma_m-\sigma_i\right)+S_{u t} \sigma_a} \\ &=\frac{25837(120000-85923)}{25726(86893-85923)+120000(5107)}=1.38 \end{aligned} (j)
The modified-Goodman diagram for this stress state is shown in Figure 15-28.
12 The static bolt stress after initial local yielding and the yielding safety factor are:
\sigma_s=\frac{F_{b o l t}}{A_t}=\frac{4102}{0.04536}=78227 psi \quad N_y=\frac{S_y}{\sigma_b}=\frac{92000}{78227}=1.18 (k)
13 The preload required to obtain these safety factors is
F_i=0.90 S_p A_t=0.90(85000)(0.052431) \cong 4011 lb (l)
14 The safety factor against joint separation is found from equation 15.14d (p. 887).
N_{ separation }=\frac{F_i}{P(1-C)}=\frac{4011}{1000(1-0.09056)}=4.4 (m)
15 The fatigue and separation safety factors are acceptable. The yielding safety factor is low but is nevertheless acceptable, since the bolt is being deliberately preloaded to a level close to its yield strength.
16 The model was solved for the range of possible preloads from zero to 100 percent of proof strength and the safety factors plotted versus percent preload. The results are shown in Figure 15-29. The fatigue and separation safety factors are < 1 below 40% preload, at which point the preload becomes effective at keeping the joint closed. The fatigue safety factor remains essentially constant as preload is increased above the 40% threshold, but the safety factor against joint separation increases linearly with increasing preload. To protect the bolted joint against possible overloads, it is desirable to use the largest preload that will not yield the bolt when tightened. In this example, preloading to 90% of the bolt’s proof strength gives an overload margin of N_{separation } = 4.4 at A, with an 18% reserve against yielding during preloading (N_{y} = 1.18 at C), along with a safety factor against fatigue failure of N_{f} = 1.38 at B.
17 The recommended design is then a 5/16-18 UNC-2A, grade 5.2 bolt, 2.5 in long, preloaded to 90% of proof strength with a force of 4 011 lb. Note that this one, small, preloaded bolt will support a half-ton of fluctuating load! The files EX15-03 can be found on the CD-ROM.
