Page 184 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 159
Integrating two more times for slope and deflection gives

dy R 1 2 w 3 w 3
EI = x − x + x − a + C 1 (6)
dx 2 6 6
R 1 3 w 4 w 4
EIy = x − x + x − a + C 1 x + C 2 (7)
6 24 24
The boundary conditions are y = 0 at x = 0 and y = 0 at x = l. Substituting the first
condition in Eq. (7) shows C 2 = 0. For the second condition
w w
R 1 3 4 4
0 = l − l + (l − a) + C 1 l
6 24 24
Solving for C 1 and substituting into Eq. (7) yields

R 1 2 2 w 3 3 w 4 w 4
EIy = x(x − l ) − x(x − l ) − x(l − a) + x − a
6 24 24l 24
Finally, substitution of R 1 from Eq. (2) and simplifying results gives

w
Answer y = [2ax(2l − a)(x − l ) − xl(x − l ) − x(l − a) + l x − a ]
3
3
2
2
4
4
24EIl
As stated earlier, singularity functions are relatively simple to program, as they are
omitted when their arguments are negative, and the brackets are replaced with ( )
parentheses when the arguments are positive.

EXAMPLE 4–7 The steel step shaft shown in Fig. 4–7a is mounted in bearings at A and F. A pulley
is centered at C where a total radial force of 600 lbf is applied. Using singularity
1
functions evaluate the shaft displacements at -in increments. Assume the shaft is
2
simply supported.
Solution The reactions are found to be R 1 = 360 lbf and R 2 = 240 lbf. Ignoring R 2 , using
singularity functions, the moment equation is

1
M = 360x − 600 x − 8 (1)
This is plotted in Fig. 4–7b.
For simplification, we will consider only the step at D. That is, we will assume sec-
tion AB has the same diameter as BC and section EF has the same diameter as DE.
Since these sections are short and at the supports, the size reduction will not add much
to the deformation. We will examine this simplification later. The second area moments
for BC and DE are
π 4 4 π 4 4
I BC = 1.5 = 0.2485 in I DE = 1.75 = 0.4604 in
64 64