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Screws, Fasteners, and the Design of Nonpermanent Joints 439
Table 8–14
30.3, 32.5, 32.5, 32.9, 32.9, 33.8, 34.3, 34.7, 37.4, 40.5

Distribution of Preload F i -
Mean value,F i = 34.18 kN. Standard deviation, ˆσ = 2.88 kN.
for 10 Tests of Lubricated
Bolts Torqued to 90 N · m

Table 8–15
Bolt Condition K
Torque Factors K for Use Nonplated, black finish 0.30
with Eq. (8–27) Zinc-plated 0.20
Lubricated 0.18
Cadmium-plated 0.16
With Bowman Anti-Seize 0.12
With Bowman-Grip nuts 0.09

1
EXAMPLE 8–3 A 3 4 in-16 UNF × 2 in SAE grade 5 bolt is subjected to a load P of 6 kip in a ten-
2
sion joint. The initial bolt tension is F i = 25 kip. The bolt and joint stiffnesses are
k b = 6.50 and k m = 13.8 Mlbf/in, respectively.
(a) Determine the preload and service load stresses in the bolt. Compare these to the
SAE minimum proof strength of the bolt.
(b) Specify the torque necessary to develop the preload, using Eq. (8–27).

(c) Specify the torque necessary to develop the preload, using Eq. (8–26) with f =
f c = 0.15.
2
Solution From Table 8–2, A t = 0.373 in .
(a) The preload stress is

25
F i
Answer σ i = = = 67.02 kpsi
A t 0.373
The stiffness constant is
6.5
k b
C = = = 0.320
k b + k m 6.5 + 13.8
From Eq. (8–24), the stress under the service load is

F b CP + F i P
σ b = = = C + σ i
A t A t A t
Answer
6
= 0.320 + 67.02 = 72.17 kpsi
0.373
From Table 8–9, the SAE minimum proof strength of the bolt is S p = 85 kpsi. The
preload and service load stresses are respectively 21 and 15 percent less than the proof
strength.