sm11 48

________________________________________________________________________
PROBLEM (11.48) A pin-ended rod AB of diameter d carries an eccentrically applied
load of P as shown in Fig. P11.48. If the maximum deflection at the midlength is vmax ,
calculate:
(a) The eccentricity e.
(b) The maximum stress in the rod.
Given: d = 40 mm, P = 60 kN, vmax = 0.6 mm, E = 200 GPa
SOLUTION
P
22
(20) 1256.64Am π == m
43
(20) 125.66 10
44
I
mm
π
==×
2
2
cr EI
PL
π
=
29
2
(200 10 )(125.66 10 )
(0.7)
9−
×× π
=
506.2 kN
=
(a) Using Eq. (11.18):
360
0.6 10 sec 1 sec(31 ) 1
2 506.2 o
ee
π
−⎡⎤
⎛⎞
⎡
⎤
×= −= −
⎢⎥
⎜⎟
⎣
⎦
⎜⎟
⎢⎥
⎝⎠
⎣⎦
, 3.6emm
=
(b) Referring to Fig. (a):
max
( ) 60(0.6 3.6)MPv e=+=+252 Nm
=
⋅
Hence,
ma
σ =+
xPMc
AI
33
69
60 10 252(20)10
1256.64 10 125.66 10
−
−
−
×
=+
××
87.9
M
Pa
=
P
e
A
max
B
v
40 mm
0.7 m
Figure (a)
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