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LESSON 3Laterally Unrestrained Beam
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OUTLINES
Lateral torsional buckling
Factors influencing LTBEffect of non-uniform moment
Buckling resistance moment
Design procedure
Design example
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Introduction
in beam design , usual to think first of need toprovide adequate strength and stiffness against
vertical bending.
leads naturally to x-section in which stiffnessis much larger in vertical than in horizontal
planes.
sections normally used as beams have:-majority of material concentrated in flanges.
flanges which are relatively narrow so as to
prevent local buckling.
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for ease of connection with adjacentmembers, beams of open sections (I or
H) are commonly used.
whenever a slender structural elementis loaded in its stiff plane, there is a
tendency for it to fail by buckling in the
more flexible plane.
compression flange of an I-beam acts
like a column, i.e.:
Introduction (cont.)
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buckles sideways if beam not
sufficiently stiff.
or not retrained laterally.
load at which beam buckles can be
much less than that causing full
moment capacity.
Introduction (cont.)
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Dead Weight load
applied vertically
Buckled
positionLateral
deflection
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Introduction (cont.)
The beam is said in satisfactory when
its cross-sectional and bucklingcapacities are not exceeded.
MEd ≤ Mc, Rd (in-plane buckling)
Med ≤ Mb,Rd (out-of plane buckling)
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Lateral Torsional Buckling
LTB is a type of instability.
Similar to flexural buckling of an
axially loaded strut.But more complex as it involves
both lateral deflection u and twist .
Loading beam in stiffer plane (planeof the web) has induced a failure by
buckling in a less stiff direction.
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Lateral Torsional Buckling (cont.)
LTB may be effectively prevented by
many types of construction thereby
enabling more efficient beam design.
Normal beam and slabconstruction is an example where
the member is restrained to
prevent buckling.However during erection, beam
may receive far less lateral support
– thus stability must be verified.
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Lateral Torsional Buckling (cont.)
LTB influences design of laterallyunrestrained beams in same way that
flexural buckling influences column
design.
Due to LTB, bending strength now a
function of beam slenderness.
Design procedure somewhat more
complex and lengthier.
Situations where lateral torsional
buckling has to be taken into account
are less common.
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Factors Influencing LTB
Unbraced span
Distance between points at which
lateral deflection is prevented.Beam weaker as unbraced span
gets longer.
LTB prevented by providing propsat intermediate points.
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Factors Influencing LTB (cont.)
Uniform or Non-uniform Moments
Shape of BMD between restraints
affects stability.If non-uniform, force in
compression flange not constant.
Member expected to be morestable.
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Factors Influencing LTB (cont.)
Section shape
Sections with greater lateral
bending (EIz) and torsional stiffness
(GJ) have greater resistance tobuckling.
Nature of End-Restraint of Beam
End-restraints which inhibitdevelopment of buckling shape
likely to increase stability
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Behaviour of Beam
Short stocky members will attain the
full plastic moment Mp.Long slender members will fail at
moments approximately equal to the
elastic critical moment ME.
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a era ors ona uc ngResistance
Verification should be carried out on
all unrestrained of beams (between
the points where lateral restraints
exist).
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Methods to verify LTB
The primary method adopts the
lateral torsional buckling curves given
in equations 6.56 and 6.57 and are
set out in clauses 6.3.2.2 and 6.3.2.3.
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Buckling Resistance
The buckling resistance, Mb,Rd of alaterally unrestrained beam (or
segment of beam) should be taken
as:
Mb,Rd = LTWyf y/M1
Wy is either Wpl,y or Wel,y
Reduction factor for LTB
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Buckling Curve
LTB curve in Figure 6.4 may be used
to obtain reduction factor LT.
The non-dimensional slenderness isgiven in clause 6.3.2.2
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Buckling Curve(cont.)
Mcr is the elastic critical LTB moment.However EC3 does not provide any
method to obtain the value!!
L. Gardner proposes a simplifiedmethod for non-uniform and uniform
moments, i.e. :
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Buckling Curve(cont.)
G shear modulus
IT torsion constant
Iwwarping constantIz second moment of area about minor
axis.
Lcr buckling length of the beam
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Buckling Curve(cont.)
Simplified method of obtaining
slenderness ratio:
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Example 1
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Example 1(cont.)
The simply supported beam shown in the
figure is restraint at points B and C.
Check the beam for lateral torsionalbuckling. Try 762x267x173 UKB of S275
steel EN10025-2.
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Solution to Example 1(cont.)
By inspection, segment AB is not critical. LTB
will be investigated along spans BC and CDonly.
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X-Section properties:
Solution to Example 1(cont.)
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1. Section classification
• From Table 3.1:
tf =21.6mm and tw=14.3mm; EN10025-2
Therefore, f y=275N/mm2
From clause 3.2.6, E=210000N/mm2 and
G=81000N/mm2.
Solution to Example 1(cont.)
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From Table 5.2: out-stand flanges
=0.92
cf /tf = 5.08 < 9
Flanges are class 1
Solution to Example 1(cont.)
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•Table 5.2: compression web
=0.92
cw/tw= 48.0 < 72
Web is class 1
Therefore, overall section is class 1
Solution to Example 1(cont.)
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2. Bending resistance of cross-section (clause
6.2.5)
Solution to Example 1(cont.)
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3. Buckling resistance, Mb,Rd (clause 6.3.2.1)
Mb,Rd = LTWyf y/M1
For span BC (Lcr =3.1m):
λLT = Lcr /96iz= 3100/96x55.8
= 0.6
From Table 6.4:
I-section, h/b = 762.2/266.7
= 2.9>2.0
Solution to Example 1(cont.)
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Solution to Example 1(cont.)
0.84
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therefore, Mb,Rd = LTWyf y/M1
= 0.84x6200x10
3
x275/1.0= 1432.2 kNm > 1362kNm
Span BC is satisfactory against LTB.
Solution to Example 1(cont.)
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For span CD (Lcr =5.2m):
λLT = Lcr /96iz= 5200/96x55.8
= 0.97
From Table 6.4:
I-section, h/b = 762.2/266.7
= 2.9>2.0
use buckling curve b.
Solution to Example 1(cont.)
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Solution to Example 1(cont.)
0.62