sm11 25 26

________________________________________________________________________
PROBLEMS (11.25 and 11.26) Knowing that the factor of safety ns = 1.5, calculate the
maximum load F that can be applied to the structure shown in Figs. P11.25 and P11.26.
Assumption: Each column is of 1 5/8 in.-diameter steel bar with E = 10 x 106 psi.
SOLUTION (11.25)
44
(1.625) 0.342 .
64
I
in
π
==
4 1.625 4 0.406rIAd m=== = m
13 5
12 12
AB BC
FFF==F
Bar AB ( 78 0.406 192Lr
=
=)
F
2
2
cr s
EI
PnL
π
=
26
2
(10 10 )(0.342) 3.7
1.5(78) kips
π ×
==
and
13 3.7, 3.41
12 FF==
kips
Bar BC ( 36 0.342 105Lr==)
26
2
(10 10 )(0.342) 17.36
1.5(36)
cr
Pkips
π ×
==
and
517.36, 41.66
12 FF==
kips
ips
Hence
3.41
all
Fk=
Continued on next slide
12
5
F
BC
FAB
B
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SOLUTION (11.26)
22 1.625
4 2 4.472 53.66 . 0.406 .
44
BC d
Lftinr=+= = == = in
444
(1.625) 0.342 .
64 64
π
I
din
π
== = , 22 2 2 2.828 33.94
AB
Lftin=+= = .
Bar BC (( 53.66 0.406) 132.2Lr==. Thus
26
2
(10 10 )(0.342) 7.82
1.5(53.66)
cr
BC all s
P
Fkips
n
π ×
== =
()
Bar AB
26
2
(10 10 )(0.342)
() 19.53
1.5(33.94)
cr
AB all s
P
Fkips
n
π ×
== =
Joint B
12
0: 0, 0.791
25
x
AB BC BC AB
FFFF=−==
∑
F
B F
11
0: 0, 1.061
25
yABBC
FFFFF=+−==
AB
F
∑
2
1
1
Solving
1.061
AB
FF
=
1.341 BC FF
=
The allowable value for F:
1.341(19.53) 26.2Fk<=ips
ips
ips
FAB
1 FBC
1.341(7.82) 9.4Fk<=
Thus
10.5
all
Fk=
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.