Question 15.2: Preloaded Fasteners in Static Loading Problem Determine a su...

See Figure 15-25.

1    As with most design problems, 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 only present two in the interest of brevity. Thus 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. (This was actually our third trial choice.) 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    Table 15-6 (p. 882) shows the proof strength of this bolt to be 85 kpsi. The tensile-stress area from equation 15-1a (p. 863) is 0.052 431 in². The preload is then

A_t=\frac{\pi}{4}\left\lgroup \frac{d_p+d_r}{2} \right\rgroup ^2       (15.1a)

F_i=0.9 S_p A_t=0.9(85000)(0.052431)=4011  lb         (a)

4    Find the lengths of thread l_{thd} and shank l_{s} of the bolt as shown in Figure 15-21 (p. 883):

from which we can find the length of thread l_{t} that is in the clamp zone:

l_t=l-l_s=2.0-1.625=0.375  \text { in }         (c)

5    Find the stiffness of the bolt from equation 15.11a (p. 884).

\frac{1}{k_b}=\frac{l_t}{A_t E_b}+\frac{l-l_t}{A_b E_b}=\frac{l_t}{A_t E_b}+\frac{l_s}{A_b E_b} \quad \therefore \quad k_b=\frac{A_t A_b}{A_b l_t+A_t l_s} E_b        (15.11a)

6    The calculation for the stiffness of the clamped material is simplified in this example by its relatively small diameter. We can assume in this case that the entire cylinder of material is compressed by the bolt force. (We will soon address the problem of finding the clamped area in a continuum of material.) The material stiffness from equation 15.11d is

k_m=\frac{\pi D_{e f f}^2}{4} \frac{E_m}{l}         (15.11d)

k_m=\frac{\pi\left(D^2-d^2\right)}{4} \frac{E_m}{l}=\frac{\pi\left(1.0^2-0.312^2\right)}{4} \frac{(30 E 6)}{2.0}=1.063 E 7  lb / in         (e)

7    The joint stiffness factor from equation 15.13c (p. 886) is

P_b=\frac{k_b}{k_m+k_b} P \\ P_b=C P \quad \text { where } \quad C=\frac{k_b}{k_m+k_b}       (15.13c)

C=\frac{k_b}{k_m+k_b}=\frac{1.059 E 6}{1.063 E 7+1.059 E 6}=0.09056       (f)

8    The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13.

\Delta \delta=\frac{P_b}{k_b}=\frac{P_m}{k_m}        (15.13a)

P_b=\frac{k_b}{k_m} P_m          (15.13b)

P_b=\frac{k_b}{k_m+k_b} P \\ P_b=C P \quad \text { where } \quad C=\frac{k_b}{k_m+k_b}       (15.13c)

P_m=\frac{k_m}{k_b+k_m} P=(1-C) P       (15.13d)

9    Find the resulting loads in bolt and material after the load P is applied.

Note how little the applied load adds to the preload in the bolt.

10    The maximum tensile stress in the bolt is

\sigma_b=\frac{F_b}{A_t}=\frac{4192}{0.052431}=79953  psi           (i)

Note that no stress-concentration factor is used because it is static loading.

11    This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the applied tensile stress. The safety factor against yielding is then

N_y=\frac{S_y}{\sigma_b}=\frac{92000}{79953}=1.15       (j)

The yield strength is found from Tables 15-6 and 15-7 (p. 882).

12    The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d.

P_0=\frac{F_i}{(1-C)}      (15.14c)

N_{\text {separation }}=\frac{P_0}{P}=\frac{F_i}{P(1-C)}     (15.14d)

P_0=\frac{F_i}{(1-C)}=\frac{4011}{(1-0.09056)}=4410  lb          (k)

N_{\text {separation }}=\frac{P_0}{P}=\frac{4410}{2000}=2.2         (l)

13    The safety factor against separation is acceptable. The yielding safety factor is low but this is to be expected, since the bolt is deliberately preloaded to a level close to its yield strength.

14    The model was solved for the range of possible preloads from zero to 100 percent of proof strength and the safety factors plotted versus preload percentage. The results are shown in Figure 15-25. The separation safety factor rises linearly with increasing preload, but is < 1 until the preload exceeds about 40% of the proof strength. At least that much preload is needed to keep the joint together under the applied load. The yielding safety factor is high at low preloads and decreases nonlinearly with increasing preload. The two lines cross at a preload of about 65% of proof strength at point A. That preload would balance the safety factors against both modes of failure at a value of 1.6. However, if the goal is to protect the joint against possible overloads, then a larger preload is better. At point B, the safety factor against overloading is 2.2, and there is still a 15% reserve against yielding during preloading as shown in the equations above.

15    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 preload force of

F_i=0.90 S_p A_t=0.90(85000)(0.052431) \cong 4011  lb       (m)

15    The files EX15-02 can be found on on the CD-ROM.