sections12.1-3
TRANSCRIPT
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12.1 Introduction to column buckling
Buckling: Buckling can be defined as the
sudden large deformation of structure due to a
slight increase of an existing load under whichthe structure had exhibited little, if any,
deformation before the load was increased. No
failure implied!!!
Reinforced concrete steel
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Stability of equilibrium condition
Easy to visualize for a ball on a surface
Note that we use curvature to decidestability, but the surface can curve up with
zero curvature
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Euler formula For simply supported column
( )
2 22
22 or using ,
/
/ : Slenderness ratio
cr cr
EI EAP I Ar P
L r
L r
= = =
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Large displacements Slenderness ratio and yield stress govern type of
post-buckling response
Dashed lines represent behavior without yielding
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Governing equation for beam column
+
=
=
+=
+=
=+
==
xEI
PBx
EI
PAxy
EI
P
ODE
xBxAdx
yd
xBxAxyTry
yEI
P
dx
yd
y
EI
P
EI
M
dx
yd
sincos)(
:intoSubstitute
))sin()cos((
)sin()cos()(:
0
2
2
2
2
2
2
2
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Apply boundary conditions
2
2
2
2
22
2
2
2
22
)/(:loadbucklingcriticalatstress(normal)eCompressiv
ratiosSlendernes:/
)/(,usingor
loadbucklingcriticalthedefines1npinned,endsbothhcolumn witaFor
gyration.ofradiustheis
mode.bucklingthedefiningintegertheis
,..3,2,10sin
0)(0)0(
rL
E
A
P
rL
rL
EIPArI
L
EIP
AIrwhereArI
nL
EInP
nnL
EI
PL
EI
PB
yand
cr
cr
crcr
==
===
=
==
=
===
==
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Example 1A 3m column with the cross section shown is constructed from two
pieces of timber, that act as a unit. If the modulus of elasticity of timberis E=13 GPa, determine a) The slenderness ratio b) Critical bucklingload c) Axial stress in the column when the critical load is applied
mmrr
mmA
Imm
A
I
mmI
mmI
mmy
mmA
yx
y
x
c
3.32
3.32rand51.59r
inertiaofRadii
10x625.1512
)50(150
12
)150(50
10x13.53)50)(50(15012
)150(50)50)(50(150
12
)50(150
sectioncrosstheofcentroidabout theinertiaofMoments
bottomfrom75000,15
)150)(50)(7550()150)(50(25
000,15)50)(150(2sectioncrosstheofProperties
min
yx
4633
4623
23
2
==
====
=+=
=+++=
=++
=
==
MParLE
APc
kNrL
EAPb
a
crcr
cr
85.14)/(
StressBucklingCritical)
75.222)/(
LoadBucklingCritical)
933.32
3000
r
LratiosSlendernes)
2
2
2
2
===
==
==
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Example 2
C229 x 30 structural steel channels with E=200 GPa are used for a 12m column. Determine the total compressive load required to buckle
the column if a) One channel is used b) Two channels are laced 150
mm back to back as shown
4646
2
10x01.110x3.25
8.143795
channeloneforpropertiesSection
mmImmI
mmxmmA
ycxc
c
==
==
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Solution for single channel
.
kNrL
EAP
r
L
mmrr
mmA
Irmm
A
Ir
cr
y
yx
x
82.13)/(
:loadBucklingCritical
2.736
3.16
12000:ratiosSlendernes
3.16
3.16and1.81
channelsingleaofinertiaofRadii
2
2
min
==
==
==
====
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Two laced channels.
[ ]
kNrL
EA
P
r
L
mmrrmm
A
Irmm
A
Ir
mmAII
mmII
mmA
cr
y
yx
x
ycy
cxx
3.694)/(:loadBucklingCritical
9.1467.81
12000:ratiosSlendernes
7.813.91and7.81
:inertiaofRadii
10x23.63)8.1475(2
10x6.502
:sectioncrosstheofcentroidabout theinertiaofMoments
7590)3975(2
2
2
min
462
46
3
==
==
======
=++=
==
==
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Higher buckling modes
With appropriate boundary
conditions can get higher
modes
2
2
)2(
2
2
)/(4
44
rL
E
A
P
PL
EI
P
cr
cr
==
==
Buckling modes : n = 1,2,3.
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Imperfection approach
Put in imperfection in the form of eccentricity
Moment equation
L
x
PePyxM =)(
Eccentrically loaded pinned-end columns
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Solution of eccentric load Differential equation of equilibrium
General solution
With boundary conditions
LxforyEI
Pk
L
xekyk
dx
yd
,00
, 22
2
2
2
==
==+
L
exxkBxkAy += cossin
= L
x
Lk
xk
ey sin
sin
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Reading assignment
Sections 12.3-4: Question: Why equation 12.26 does not really
guarantee buckling? What does it guarantee?
Source: www.library.veryhelpful.co.uk/ Page11.htm