Civil Engineering Reference
In-Depth Information
5
Q
5
Q
All members have
the same
EI
L
Q
=1
L
2
EI
Q
Q
=1
3
Q
2.50
Q
=1
2
3
0.5
L
2
L
3.00
EI
EI
-0.87
0.5
L
1
1
0.87
-2.60
4
-1.73
2
L
1.16
L
1.16
L
1.16
L
(a) Example1
(b) Example 4
Figure 8.13
Worked examples 1 and 4.
obtained after observing that the end 1 will be heavily restrained by the tension
member 14 and that the end 2 will be moderately restrained by the short lightly
loaded compression member 24. Thus the initial estimate of the effective length
factorofmember12inthisbracedframeshouldbeclosertotherigidlyrestrained
value of 0.5 than to the unrestrained value of 1.0.
Accordingly, assume
k
cr
12
=
0.70.
Then, by using equation 8.2,
N
cr
12
=
π
2
EI
/(
0.70
L
)
2
=
2.04,
and
Q
cr
=
N
cr
12
/
3.00
=
0.68
Thus
N
23
=
1.70,
N
24
=
0.59,
N
14
=−
1.77.
Using the form of equation 3.45,
α
12
=
2
EI
/
L
. Using Figure 3.19,
α
23
=
3
EI
L
1.70
π
2
EI
/
L
2
=−
2.10
EI
1
−
L
,
α
24
=
4
EI
0.58
L
0.59
π
2
EI
/(
0.58
L
)
2
=
6.22
EI
1
−
L
,
α
14
=
4
EI
1.16
L
(
−
1.77
)
2
π
2
EI
/(
1.16
L
)
2
=
7.55
EI
1
−
L
.
Using equation 3.45,
k
1
=
2
/(
0.5
×
7.55
+
2
)
=
0.346
k
2
=
2
/
{
0.5
×
(
−
2.10
+
6.22
)
+
2
}=
0.493
Using Figure 3.21a,
k
cr
12
=
0.65
Thecalculationcanberepeatedusing
k
cr
12
=
0.65insteadoftheinitialestimate
of0.70,inwhichcasethesolution
k
cr
12
=
0.66willbeobtained.Thecorresponding
frame buckling loads are
Q
cr
={
π
2
EI
/(
0.66
L
)
2
}
/
3
=
0.77.
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