Page 185 - Shigley's Mechanical Engineering Design

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160 Mechanical Engineering Design
Figure 4–7 y 600 lbf
1.750
1.500
1.000 1.000
Dimensions in inches. B C D E
A F
x
0.5
8
8.5 R 2
R 1
19.5
(a) 20
2880 lbf-in 2760 lbf-in
M
(b)
M/I a
b
c
d
(c)

A plot of M/I is shown in Fig. 4–7c. The values at points b and c, and the step change are

M 2760 3 M 2760 3

= = 11 106.6 lbf/in = = 5 994.8 lbf/in
I 0.2485 I 0.4604
b c
M 3

= 5 994.8 − 11 106.6 =−5 111.8 lbf/in
I
The slopes for ab and cd, and the change are
2760 − 2880 4 −5 994.8 4
m ab = =−965.8 lbf/in m cd = =−521.3 lbf/in
0.2485(0.5) 11.5
m =−521.3 − (−965.8) = 444.5 lbf/in 4
3
Dividing Eq. (1) by I BC and, at x 8.5 in, adding a step of −5 111.8 lbf/in and a ramp
4
of slope 444.5 lbf/in , gives
M 1 0 1
= 1 448.7x − 2 414.5 x − 8 − 5 111.8 x − 8.5 + 444.5 x − 8.5 (2)
I
Integration gives
dy 2 2 1
E = 724.35x − 1207.3 x − 8 − 5 111.8 x − 8.5
dx
2
+ 222.3 x − 8.5 + C 1 (3)

Integrating again yields
3 3 2 3
Ey = 241.5x − 402.4 x − 8 − 2 555.9 x − 8.5 + 74.08 x − 8.5 + C 1 x + C 2
(4)

At x = 0, y = 0. This gives C 2 = 0 (remember, singularity functions do not exist until
the argument is positive). At x = 20 in, y = 0, and

3
3
3
2
0 = 241.5(20) − 402.4(20 − 8) − 2 555.9(20 − 8.5) + 74.08(20 − 8.5) + C 1 (20)