Page 413 - Shigley's Mechanical Engineering Design

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388 Mechanical Engineering Design
From Eq. (7–33),

−4
δ 11 2.061(10 )
w 1c = w 1 = 35 −4 = 17.11 lbf
δ cc 4.215(10 )
−4
δ 22 3.534(10 )
w 2c = w 2 = 55 −4 = 46.11 lbf
δ cc 4.215(10 )
'
& g 386.1
Answer ω = % = = 120.4 rad/s, or 1150 rev/min
−4
δ cc w ic 4.215(10 )(17.11 + 46.11)
which, except for rounding, agrees with part d, as expected.
6
2
2
3
( f ) For the shaft, E = 30(10 ) psi, γ = 0.282 lbf/in , and A = π(1 )/4 = 0.7854 in .
Considering the shaft alone, the critical speed, from Eq. (7–22), is
'
'
2 2
6
gE I π 386.1(30)10 (0.049 09)
π
Answer ω s = =
l Aγ 31 0.7854(0.282)
= 520.4 rad/s, or 4970 rev/min
2
We can simply add 1/ω to the right side of Dunkerley’s equation, Eq. (1), to include
s
the shaft’s contribution,
Answer 1 . 1 + 6.905(10 ) = 7.274(10 )
−5
−5
=
ω 2 520.4 2
1
.
ω 1 = 117.3 rad/s, or 1120 rev/min
which is slightly less than part d, as expected.
The shaft’s first critical speed ω s is just one more single effect to add to Dunkerley’s
equation. Since it does not fit into the summation, it is usually written up front.
n
1 . 1 $ 1
Answer = + (7–34)
ω 2 ω 2 ω 2
1 s i=1 ii
Common shafts are complicated by the stepped-cylinder geometry, which makes the
influence-coefficient determination part of a numerical solution.

7–7 Miscellaneous Shaft Components

Setscrews
Unlike bolts and cap screws, which depend on tension to develop a clamping force, the
setscrew depends on compression to develop the clamping force. The resistance to axial
motion of the collar or hub relative to the shaft is called holding power. This holding
power, which is really a force resistance, is due to frictional resistance of the contact-
ing portions of the collar and shaft as well as any slight penetration of the setscrew into
the shaft.