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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