LECTURE-3-BEC304 .ppt

7/27/2019 LECTURE-3-BEC304 .ppt

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LESSON 3Laterally Unrestrained Beam



OUTLINES

Lateral torsional buckling

Factors influencing LTBEffect of non-uniform moment

Buckling resistance moment

Design procedure

Design example



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.



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.)



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.




Dead Weight load

applied vertically

Buckled

positionLateral

deflection




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)



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.



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.



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.



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.



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.



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



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.



a era ors ona uc ngResistance

Verification should be carried out on

all unrestrained of beams (between

the points where lateral restraints

exist).



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.



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



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



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. :




G shear modulus

IT torsion constant

Iwwarping constantIz second moment of area about minor

axis.

Lcr buckling length of the beam




Simplified method of obtaining

slenderness ratio:



Example 1



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.



Solution to Example 1(cont.)

By inspection, segment AB is not critical. LTB

will be investigated along spans BC and CDonly.



X-Section properties:




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.




From Table 5.2: out-stand flanges

=0.92

cf /tf = 5.08 < 9

Flanges are class 1




•Table 5.2: compression web

=0.92

cw/tw= 48.0 < 72

Web is class 1

Therefore, overall section is class 1




2. Bending resistance of cross-section (clause

6.2.5)




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





0.84



therefore, Mb,Rd = LTWyf y/M1

= 0.84x6200x10

3

x275/1.0= 1432.2 kNm > 1362kNm

Span BC is satisfactory against LTB.




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.





0.62



therefore, Mb,Rd = LTWyf y/M1

= 0.67x6200x103x275/1.0

= 1142.4 kNm < 1362kNm

Span BC is not satisfactory against LTB!!

Provide props at midspan of CD.


LECTURE-3-BEC304 .ppt

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