Lateral-torsional buckling:

bending about its major axis, failure mayoccur by a form of buckling whichinvolves lateral deflection and twisting.

y

zE I y⋅

E I z⋅

E: modulus of elasticity

Iz: moment of inertia about minor axis

G: shear modulus of elasticity

It: torsion constant

E: modulus of elasticity

Iz: moment of inertia about minor axis

G: shear modulus of elasticity

It: torsion constant

M crπL

E I z⋅ G⋅ I t⋅⋅

• Perfectly elastic, initially straight, loaded byequal and opposite end moments about itsmajor axis.

• Unrestrained along its length

• End Supports …

– Twisting and lateral deflection prevented.

– Free to rotate both in the plane of the weband on plan.

Iw: Warping constantIw: Warping constant

Conditions to determine McrConditions to determine Mcr

M crπL

E I z⋅ G⋅ I t⋅⋅

M crπL

E I z⋅ G I t⋅ E I w⋅L2

π2

⋅+

⋅⋅

Critical Buckling Moment for uniformbending moment diagram is:

+=

z2

t2

z

w2

z2

cr EIGIL

II

LEIM

ππ

Includes:

• Lateral flexural stiffness EIEIzz

• Torsional and Warping stiffnesses GItand EIEIww

Their relative importance depends on thetype of cross-section used.

• The elastic critical moment for a beamunder uniform bending moment is:

t

wtzcr GIL

EIGIEIL

M 2

2

1 ππ+=

• The elastic critical moment (mid-spanmoment) for a beam with a centralpoint load is

t

wtzcr GIL

EIGIEIL

M 2

2

124.4 π+=

which is increased from the basic(uniform moment) case by a factorC1=4.24/π=1.365

EC3 expresses the elastic critical momentMcr for a particular loading case as:

t

wtzcr GIL

EIGIEIL

CM 2

2

1 1ππ

+=

C1 appears as

1/ C10.5 in expressions for λLT.

L Mmax C1

M

M

M

FL/4

FL/8

FL/4

1,00

1,879

2,752

1,365

1,132

1,046

Loads Bendingmoment

M M

M

M -M

F

F

FF= = = =

• Can include the effect of differentsupport conditions by redefining theunrestrained length as an effectiveeffectivelengthlength– Lateral bending restraint: k– Warping restraint: kw

• Basic case assumes end conditionswhich preventprevent lateral movementlateral movement andtwisttwist but permit rotation on plan.permit rotation on plan.

End support conditionsEnd support conditions

•kw = 1.0 unless special provision forwarping fixing is made.

•EC3 recommends k values of

–0.5 for fully fixed ends,

–0.7 for one free and one fixed end

–1.0 for two free ends.

0.50.5

1.01.0

1.01.0

0.50.5

• Loads applied to top flangetop flange aredestabilizing

•destabilizing increases with depth ofincreases with depth ofsectionsection and/or as span reducesas span reduces

Beams with intermediate lateral supports

the segmentssegments of the beam betweenrestraints may be treated in isolationtreated in isolation

Lengths of beams between restraintsshould use an effective length factork of 1,01,0 not 0,70,7

Design of the beam based onthe most critical segmentmost critical segment

beam continuous over a number of spans

--Treat as individual spansindividual spans--taking into account the shapeshape of thebending momentbending moment diagram within each spaneach span--as a result of continuity using the CC11factorfactor

C1 = 1,88 – 1,40ψ + 0,52ψ2C1 = 1,88 – 1,40ψ + 0,52ψ2

Ψ=−1,C1=2,927

Ψ=0,C1=1,88

Non-dimensional plot permits resultsfrom different test series to becompared

Intermediate slenderness: adverselyaffected by inelasticity and geometricalimperfections

0

Stocky Slender

pl

cr

1,0

1,0 MM

λ =LT

Intermediate

pl

crMM

pl

MM

stocky beams λLT ≤ 0.4. Lateral torsionalbuckling is not considered

slender beam λLT ≥ 1.2. Resistance closeto Mcr.

EC3 uses a reduction factor χLT onplastic resistance moment to coverthe whole slenderness range

The design buckling resistancemoment Mb.Rd of a laterallyunrestrained beam is calculated as:

1Myy.plwLTRd.b /fWM γβχ=

0

Welded beams

Rolled sections

SlendernessλLT

χ LTχ LT

Red

uctio

n fa

ctor

1,0 2,0

1,0

β W 1Class 1 and 2Class 1 and 2

Class 3Class 3

β WW el.yW pl.y

Class 4Class 4

β WW eff.yW pl.y

The non-dimensional slenderness is calculated:

• EITHER by

cryyplw

crRdplLT

MfW

MM

/

/

.

.

⋅⋅=

=

β

λ

• OR using

5.0w

1

LTLT β

λλλ

=

5.0

1 fyE

= πλ

[ ] 5,022

1

LTLTLT

LTλφφ

χ−+

=

+−∝+= 2)2.0(15,0 LTLTLTLT λλφ

αLT = 0,21 for rolled sections

αLT = 0,49 for welded sections

w

y

cr

Ypl

wy

cr

Ypl

cryyplw

crRdplLT

fE

MWE

Ef

MWE

MfW

MM

βπ

π

βπ

π

β

λ

⋅⋅

⋅⋅

=

⋅⋅

⋅⋅⋅

=

⋅⋅=

=

2

.2

2.

2

.

.

/

/

5.0w

1

LTLT β

λλλ

=

ε

πλ

⋅=

=

93.9

5.0

1 fyE

A main girder is simply supported. At midspan a side girder is connected by anominally pinned connection. Lateralrestraints are assumed at the supports. Thestructural details and the loading are givenin the following figures.

Main girder: IPE 500

Steel grade S275

Span: L = 7 m

Dead load: g = 3.0 kN/m

Imposed load: q = 5.0 kN/m; P = 140 kN

Load combination:

1.35g + 1.5 ( q + P) or 1.35 (g + q +P)

MSd1 = (1.35g)L2/8 + (1.5q)L2/8 + (1.5P)L/4

= 1.35x3.0x72/8 + 1.5x5.0x72/8 + 1.5x140x7/4

= 438.2 kNm

MSd2 = (1.35g)L2/8 + (1.35q)L2/8 + (1.35P)L/4

= 1.35x3.0x72/8 + 1.35x5.0x72/8 + 1.35x140x7/4

= 396.9 kNm

MSd = max (MSd1, MSd2) = 438.2 kNm

Internal moment and forces

VSd1 = (1.35g)L/2 + (1.5q)L/2 + (1.5P)/2

= 1.35x3.0x7/2 + 1.5x5.0x7/2 + 1.5x140/2

= 145.4 kN

VSd2 = (1.35g)L/2 + (1.35q)L/2 + (1.35P)/2

= 1.35x3.0x7/2 + 1.35x5.0x7/2 + 1.35x140/2

= 132.3 kN

VSd = max (VSd1, VSd2) = 145.4 kN

Resistance at ULS

S275à fy = 275 N/mm2 ε = √235/275=0.92

Flange: c/t = 100/16 = 6.3 < 10ε = 9.2 Class 1

Web: d/t = 426/10.2 = 41.8 < 72 ε = 66.24 Class 1

Classification of the cross-section

Shear resistance verification:

VRd = Av( fy/√3)/γM0

= (5.10x102x275/ √3/1.1)x10-3

= 736.1 kN > 145.4 kN OK!

At mid span:

0.5VRd = 0.5x736.1 = 368.1 kN > 105 kN

-- no interaction bending-shear is required

d/t = 41.8 < 69 ε = 63.5

-- no shear buckling verification is required

Moment resistance verification:

Moment resistance for Class 1 or 2 cross-section

Mc.Rd = Wpl fy / γM0

Wpl = 2194 x 103 mm3

Mc.Rd = 2194 x 103 x 275 / 1.1 = 548.5 kNm

Critical bending moment for doubly symmetricand transverse load applied on shear center

kw = 1.0, k = 1.0, L = 3.5 m,

C1 either from Table F.1.1 C1 = 1.88

kNm1905.8102140

107.89105.3039.0101249000105.3

10214021000088.1

4

4626

122

42

=⋅

⋅⋅⋅⋅+⋅

⋅⋅

⋅⋅⋅=

πcrM

or from C1 = 1.88-1.40Ψ + 0.52 Ψ2

Ψ = 0à C1 = 1.88

Lateral torsional buckling verification

0.40.56

/

/

.

.

>=

⋅⋅=

=

cryyplw

crRdplLT

MfW

MM

β

λ

Lateral torsional buckling verification is required

1Myy.plwLTRd.b /fWM γβχ=

β W 1Class 1 and 2Class 1 and 2 β W 1Class 1 and 2Class 1 and 2

αLT = 0,21 for rolled sections

6946.0]0.560.2)-0.21(0.560.5[1

)2.0(15.0

2

2

=++=

+−∝+= LTLTLTLT λλφ

[ ]9045.0

15,022

=−+

=LTLTLT

LTλφφ

χ

kNm496.11.1/2751021940.9045

/3

1..

=⋅⋅⋅=

= MyyplwLTRdb fWM γβχ

MRd = min (Mc.Rd, Mb.Rd) = min (548.5, 496.1)

= 496.1 kNm > 438.2 kNm OK!

Serviceability verification

Load combination for serviceability

g + (q+P)

Deflection due to dead load

δ1 = (5/384) (gL4)/(EIy)

= (5/384)(3 x 74x1012)(210000x48200x104)

= 0.93 mm

Deflection due to imposed load

δ2 = (5/384) (qL4)/(EIy)+(1/48)(PL3)/(EIy)

= (5/384)(5 x 74x1012)(210000x48200x104)

+(1/48)(140x73x1012) (210000x48200x104)

= 11.4 mm = l / 614 < l / 300Maximum deflection

δmax = 11.4 + 0.93 = 12.33 mm

= l / 570 < l / 250 OK!