Civil Engineering Reference
In-Depth Information
stiffnesses are such that
Flange end moment
Flange end rotation
=
−
EI
z
R
1
−
R
,
(6.50)
L
in which case
R
2
=
R
4
=
R
.
(6.51)
ItisshowninSection6.13.1thatthemoment
M
cr
atwhichtherestrainedbeam
buckles elastically is given by equations 6.36 and 6.37 when the effective length
factor
k
cr
is the solution of
R
1
−
R
=
−
π
2
k
cr
cot
π
2
k
cr
.
(6.52)
It can be seen from the solutions of this equation shown in Figure 6.18c that
the effective length factor
k
cr
decreases from 1 to 0.5 as the restraint parameter
R
increasesfrom0to1.Thesesolutionsareexactlythesameasthoseobtainedfrom
the compression member effective length chart of Figure 3.21a when
1
−
R
1
−
0.5
R
,
k
1
=
k
2
=
(6.53)
which suggests that the effective length factors
k
cr
for beams with unequal end
restraints may be approximated by using the values given by Figure 3.21a.
The elastic buckling of symmetrically restrained beams with unequal end
moments has also been analysed [36], while solutions have been obtained for
many other minor axis and end warping restraint conditions [16, 27, 32-34].
6.6.3.4 Torsional end restraints
The end torques
T
0
which resist end twist rotations also remain zero until elastic
bucklingoccurs,andthenincreasewiththeendtwistrotations.Ithasbeenassumed
that the ends of all the beams discussed so far are rigidly restrained against end
twistrotations.Whentheendrestraintsareelasticinsteadofrigid,someendtwist
rotationoccursduringbucklingandtheelasticbucklingloadisreduced.Analytical
studies[21]ofbeamsinuniformbendingwithelastictorsionalendrestraintshave
shown that the reduced buckling moment
M
zx
,
r
can be approximated by
1
(
4.9
+
4.5
K
2
)
R
3
+
1
M
zx
,
r
M
zx
=
(6.54)
inwhich1
/
R
3
isthedimensionlessstiffnessofthetorsionalendrestraintsgivenby
1
R
3
=
α
x
L
(6.55)
GI
t
in which
α
x
is the ratio of the restraining torque
T
0
to the end twist rotation
(φ)
0
. Reductions for other loading conditions can be determined from the elastic
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