Page 465 - Shigley's Mechanical Engineering Design

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440 Mechanical Engineering Design
(b) From Eq. (8–27), the torque necessary to achieve the preload is

3
Answer T = KF i d = 0.2(25)(10 )(0.75) = 3750 lbf · in
(c) The minor diameter can be determined from the minor area in Table 8–2. Thus d r =
√ √
4A r /π = 4(0.351)/π = 0.6685 in. Thus, the mean diameter is d m = (0.75 +
0.6685)/2 = 0.7093 in. The lead angle is
l 1 1
λ = tan −1 = tan −1 = tan −1 = 1.6066 ◦
πd m πd m N π(0.7093)(16)
For α = 30 , Eq. (8–26) gives
◦
0.7093 tan 1.6066 + 0.15(sec 30 )
◦

◦
3
T = + 0.625(0.15) 25(10 )(0.75)
2(0.75) 1 − 0.15(tan 1.6066 )(sec 30 )
◦
◦
= 3551 lbf · in
which is 5.3 percent less than the value found in part (b).

8–9 Statically Loaded Tension Joint with Preload
Equations (8–24) and (8–25) represent the forces in a bolted joint with preload. The
tensile stress in the bolt can be found as in Ex. 8–3 as

F b CP + F i
σ b = = (a)
A t A t
Thus, the yielding factor of safety guarding against the static stress exceeding the proof
strength is
S p S p
n p = = (b)
σ b (CP + F i )/A t
or
S p A t
n p = (8–28)
CP + F i
Since it is common to load a bolt close to the proof strength, the yielding factor of
safety is often not much greater than unity. Another indicator of yielding that is some-
times used is a load factor, which is applied only to the load P as a guard against
overloading. Applying such a load factor to the load P in Eq. (a), and equating it to
the proof strength gives

Cn L P + F i
= S p (c)
A t
Solving for the load factor gives

S p A t − F i
n L = (8–29)
CP
It is also essential for a safe joint that the external load be smaller than that needed
to cause the joint to separate. If separation does occur, then the entire external load