________________________________________________________________________
PROBLEM (11.16) A control linkage consists of a high-strength ASTM-A242 rod AB of
diameter d and a pivot arm CD (Fig. P11.16). The load is transmitted to the rod through a pin
A. What is the largest value of F that can be applied on the basis of the buckling strength of the
rod and a safety factor of ns ?
Given: a = 160 mm, b = 40 mm, L = 0.6 m, d = 10 mm, ns = 1.7
E = 200 GPa, σy = 345 MPa (from Table B.4)
SOLUTION
44 4
(10) 490.874
64 64
d
I
mm
ππ
== =
22 2
(10) 78.54
44
d
Am
ππ
== = m
22 9 12
2.692 kN
22
(200 10 )(490.874 10 )
(0.6)
Acr EI
PP L
ππ
−
××
== =
=
The corresponding stress is
6
2,692 34.4
78.54 10
cr
cr P
M
Pa
A
σ
−
== =
×
which is well below the yield stress of 345 MPa. We have
0: ( )
Ccr
M
abF bP=+=
∑
or
200 40(2,692), 538.4FF==N
Therefore
538.4 317
1.7
all s
F
FN
n
== =
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
1
/
1
100%