110
Chapter 2
Find kt for the belt holes [2.9] using the tension curve where h<_20d or
h_<20(0.046)_< 0.920"
d 0.046
0.0015
b - 0.12~ - 0.368 Kt -~ 3.21 t/r - 0.046"/~ - 0.065 << 3
From Fig. 2.16 and Table 2.7 q~0.96 QT steel.
Now find ~f with
gf = 1 + q( Kt-1) The
Cq-- ~q
-- 4-0.0833 Table 2.7 q ~ 0.96
q
~t
Cx, - - - 4-0.109 Kt ~ 3.21 Sect. A.5
Kt
FtgOKf ~ .~2 [’OKf ~ "~211/2
~,
q) "Jr-~k~tt
kt) J X100
Kf
[1 ÷ q(Kt - 1)]
(2.120)
with parameters from Section A.5 Example 2.17
OKf _= (~t - 1) and OKf
Oq
~ = 0
~q = 0.08333
2t = 0.109Kt
We find
~f -4-12.16%
Now to develop the ~r - °’m plot by establishing Eq. (2.55) and
’
(2.121)
ae = kakbkc ¯ ¯ ¯ a
e
ka-ground finish Eq. (2.56)
}a = 4-0.103 Table 2.4
ka = 0.7429
~ 0.103
Cvo - k~ - 0.742~ - 0.1386
kb-size effect in o" e = 128,800 kh = 1
~b = Cvb = 0
kc - is 1 when solving for pf
~-c = Cvc = 0
kd - 1 since temperature near room temperature ~d = C.,, = 0
ke - use Kf with stress k~ = 1 here and
~ = C,,~, = 0
kf - 0.0015" thk tape kf = 1
}f = C,,~. = 0
Chapter 2
Find kt for the belt holes [2.9] using the tension curve where h<_20d or
h_<20(0.046)_< 0.920"
d 0.046
0.0015
b - 0.12~ - 0.368 Kt -~ 3.21 t/r - 0.046"/~ - 0.065 << 3
From Fig. 2.16 and Table 2.7 q~0.96 QT steel.
Now find ~f with
gf = 1 + q( Kt-1) The
Cq-- ~q
-- 4-0.0833 Table 2.7 q ~ 0.96
q
~t
Cx, - - - 4-0.109 Kt ~ 3.21 Sect. A.5
Kt
FtgOKf ~ .~2 [’OKf ~ "~211/2
~,
q) "Jr-~k~tt
kt) J X100
Kf
[1 ÷ q(Kt - 1)]
(2.120)
with parameters from Section A.5 Example 2.17
OKf _= (~t - 1) and OKf
Oq
~ = 0
~q = 0.08333
2t = 0.109Kt
We find
~f -4-12.16%
Now to develop the ~r - °’m plot by establishing Eq. (2.55) and
’
(2.121)
ae = kakbkc ¯ ¯ ¯ a
e
ka-ground finish Eq. (2.56)
}a = 4-0.103 Table 2.4
ka = 0.7429
~ 0.103
Cvo - k~ - 0.742~ - 0.1386
kb-size effect in o" e = 128,800 kh = 1
~b = Cvb = 0
kc - is 1 when solving for pf
~-c = Cvc = 0
kd - 1 since temperature near room temperature ~d = C.,, = 0
ke - use Kf with stress k~ = 1 here and
~ = C,,~, = 0
kf - 0.0015" thk tape kf = 1
}f = C,,~. = 0