Structural Mechanics Problem Solution: Load Calculation

________________________________________________________________________
PROBLEMS (11.23 and 11.24) Given on a factor of safety ns = 2, determine the largest
load F that may be applied to the structure shown in Figs. P11.23 and P11.24.
Assumption: Each column is a 50-mm-diameter steel bar having E = 210 GPa.
SOLUTION (11.23)
Joint B
0: 0.6 5
x
BC AB
FF F==
∑
2
0: 0.8
5
yBCAB
FFF=+
∑F=
F
Solving,
0.5 0.671
AB BC
FFF
=
=
We have
49
(0.05) 306.8 10
64 4
I
m
π
−
==× 4 12.5rI Ad mm===
Bar AB ( 200Lr=)
22 9 9
22
(210 10 )(306.8 10 )
0.5 , 101.7
2(2.5)
cr s
EI
FP F kN
nL
ππ
−
××
== = =
Bar BC ( 178.9Lr=)
29 9
2
(210 10 )(306.8 10 )
0.671 63.59
2( 5)
F
π
−
××
==N
, 94.77Fk
=
Therefore,
94.77
all
Fk=N
FBC
FAB
2
C
4 1
3
F
A
2
41
3
F
B
Continued on next slide
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SOLUTION (11.24)
49
(0.05) 306.8 10
64
I
π
−
==×
50 12.5
44
Id
rm
m
A
==== 160
L
r=
We have
22 9 9
79.5 kN
22
(210 10 )(306.8 10 )
2(2)
cr s
EI
PnL
ππ
−
××
==
=
Thus,
1.2 2(79.5) 110.4
0.6
all cr
FP== =kN
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.