Civil Engineering Reference
In-Depth Information
1000
N
cr,y
N
cr,z
750
75
6
500
N
cr,T
N
cr
250
100
y
z
0
0
1
2
3
4
5
Length
L
(m)
(a) Cross-section
(b) Elastic buckling loads
Figure 3.25
Elastic buckling of a simply supported angle section column.
coincidewiththeloadingaxisthroughthecentroid,andanytwistingwhichoccurs
causesthecentroidalaxistodeflect.Forsimplysupportedmembers,itisshownin
[5,24,25]thatthe(lowest)elasticbucklingload
N
cr
isthelowestrootofthecubic
equation
N
cr
i
0
−
y
0
−
z
0
−
N
cr
(
N
cr
,
y
+
N
cr
,
z
+
N
cr
,
T
)
i
0
−
N
cr
,
z
y
0
−
N
cr
,
y
z
0
+
N
cr
i
0
{
N
cr
,
y
N
cr
,
z
+
N
cr
,
z
N
cr
,
T
+
N
cr
,
T
N
cr
,
y
}−
N
cr
,
y
N
cr
,
z
N
cr
,
T
i
0
=
0.
(3.57)
Forexample,theelasticbucklingload
N
cr
forapin-endedunequalangleisshown
in Figure 3.25, where it can be seen that
N
cr
is less than any of
N
cr
,
y
,
N
cr
,
z
,or
N
cr
,
T
.
These and other cases of flexural-torsional buckling (referred to as torsional-
flexural buckling in EC3) are treated in a number of textbooks and papers [4-6,
24-26],andtabulationsofsolutionsarealsoavailable[16,27].Oncethebuckling
load
N
cr
hasbeendetermined,thecompressionresistance
N
b
,
Rd
canbefoundfrom
Figure 3.23.
3.8 Appendix - elastic compression members
3.8.1 Buckling of straight members
Theelasticbucklingload
N
cr
ofthecompressionmembershowninFigure3.3can
be determined by finding a deflected position which is one of equilibrium. The
differential equilibrium equation of bending of the member is
EI
d
2
v
d
x
2
=−
N
cr
v
.
(3.58)
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