Page 412 - Shigley's Mechanical Engineering Design

bud29281_ch07_358-408.qxd 12/9/09 4:29PM Page 387 ntt 203:MHDQ196:bud29281:0073529281:bud29281_pagefiles:

Shafts and Shaft Components 387
From Eq. set (7–24),

2
2
2
24(7)(31 − 24 − 7 ) −4
δ 11 = = 2.061(10 ) in/lbf
9
0.2739(10 )
2
2
2
11(20)(31 − 11 − 20 ) −4
δ 22 = = 3.534(10 ) in/lbf
9
0.2739(10 )
2
2
2
11(7)(31 − 11 − 7 ) −4
δ 12 = δ 21 = = 2.224(10 ) in/lbf
9
0.2739(10 )
Answer
j
i 1 2
4
4
1 2.061(10 ) 2.224(10 )
4
4
2 2.224(10 ) 3.534(10 )
y 1 = w 1 δ 11 + w 2 δ 12 = 35(2.061)10 −4 + 55(2.224)10 −4 = 0.019 45 in
y 2 = w 1 δ 21 + w 2 δ 22 = 35(2.224)10 −4 + 55(3.534)10 −4 = 0.027 22 in

(b) w i y i = 35(0.019 45) + 55(0.027 22) = 2.178 lbf · in
Answer w i y = 35(0.019 45) + 55(0.027 22) = 0.053 99 lbf · in 2
2
2
2
i

386.1(2.178)
Answer ω = = 124.8 rad/s, or 1192 rev/min
0.053 99
(c)
1 w 1
Answer = δ 11
ω 2 g
11

g 386.1
ω 11 = = −4 = 231.4 rad/s, or 2210 rev/min
w 1 δ 11 35(2.061)10

g 386.1
Answer ω 22 = = = 140.9 rad/s, or 1346 rev/min
w 2 δ 22 55(3.534)10 −4
1 . 1 1 1 −5
(d) = = + = 6.905(10 ) (1)
ω 2 ω 2 231.4 2 140.9 2
1 ii

.
Answer ω 1 = 1 = 120.3 rad/s, or 1149 rev/min
−5
6.905(10 )
which is less than part b, as expected.
(e) From Eq. (7–24),
2 2 2 2 2 2
b cc x cc l − b − x cc 15.5(15.5)(31 − 15.5 − 15.5 )
cc
δ cc = =
9
6EIl 0.2739(10 )
−4
= 4.215(10 ) in/lbf