Page 468 - Shigley's Mechanical Engineering Design

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Screws, Fasteners, and the Design of Nonpermanent Joints 443

in the grip is l d = 2.25 − 1.50 = 0.75 in. The threaded length in the grip is l t = l −
2
l d = 0.75 in. From Table 8–2, A t = 0.226 in . The major-diameter area is A d =
2
2
π(0.625) /4 = 0.3068 in . The bolt stiffness is then
A d A t E 0.3068(0.226)(30)
k b = =
Answer A d l t + A t l d 0.3068(0.75) + 0.226(0.75)
= 5.21 Mlbf/in
From Table A–24, for no. 25 cast iron we will use E = 14 Mpsi. The stiffness of the
members, from Eq. (8–22), is
0.5774π Ed 0.5774π(14)(0.625)
k m = 0.5774l + 0.5d = 0.5774 (1.5) + 0.5 (0.625)
Answer 2ln 5 0.5774l + 2.5d 2ln 5 0.5774 (1.5) + 2.5 (0.625)
= 8.95 Mlbf/in

If you are using Eq. (8–23), from Table 8–8, A = 0.778 71 and B = 0.616 16, and
k m = Ed A exp(Bd/l)
= 14(0.625)(0.778 71) exp[0.616 16(0.625)/1.5]

= 8.81 Mlbf/in
which is only 1.6 percent lower than the previous result.
From the first calculation for k m , the stiffness constant C is
5.21
k b
Answer C = = = 0.368
k b + k m 5.21 + 8.95
(b) From Table 8–9, S p = 85 kpsi. Then, using Eqs. (8–31) and (8–32), we find the
recommended preload to be
F i = 0.75A t S p = 0.75(0.226)(85) = 14.4 kip

For N bolts, Eq. (8–29) can be written
S p A t − F i
n L = (1)
C(P total /N )
or
Cn L P total 0.368(2)(36)
N = = = 5.52
S p A t − F i 85(0.226) − 14.4
Answer Six bolts should be used to provide the specified load factor.
(c) With six bolts, the load factor actually realized is
85(0.226) − 14.4
Answer n L = = 2.18
0.368(36/6)
From Eq. (8–28), the yielding factor of safety is

Answer n p = S p A t = 85(0.226) = 1.16
C(P total /N) + F i 0.368(36/6) + 14.4
From Eq. (8–30), the load factor guarding against joint separation is
14.4
F i
Answer n 0 = = = 3.80
(P total /N)(1 − C) (36/6)(1 − 0.368)