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Page 1: 4 Unrestrained Beams - 2011

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Unrestrained Beams

23/8/2012

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Introduction

Lateral Torsional BucklingLateral deflection and twistingSpecial cases where LTB checks can be ignored

Moment ResistanceMoment checkApproaches for determining the reduction factor for LTBLTB Curves – General Case

LTB Curves & Imperfection FactorsBuckling Curves for LTBElastic Critical Moment for LTBCorrection Factor for Non‐Uniform Moment C1

LTB Curves – Rolled Sections or Equivalent Welded SectionsLTB Curves & Imperfection Factors –Comparison of Buckling Curves Given in Clauses 6.3.2.2 & 6.3.2.3

Simplified method for determination of non‐dimensional slenderness

Design Procedure for LTB

ExamplesExample URB‐1 (Buckling resistance of UB)

Outline23/8/2012

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Lateral torsional buckling (LTB) is a form of buckling that involves both lateral deflection and twisting. It is a member buckling mode associated with slender unrestrained beams loaded about their major axis.

Checks for lateral torsional buckling should be carried out on all unrestrained segments of beams(between points where lateral restraint exists).

If continuous lateral restraint is provided to the beam, then lateral torsional buckling will be prevented and failure will be due to in-plane bending and/or shear (refer to restrained beams).

The load at which LTB occurs may be substantially less than the beam in-plane bending capacity.

Introduction23/8/2012

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Lateral Torsional Buckling

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Lateral Torsional Buckling (LTB)

Demonstration of LTB on a cantilever

Cross-section of the free end of an unloaded cantilever

Cross-section of the free end of the cantilever that undergoes LTB when subjected to MAJOR axis moment

minor axismoment Mz

major axismoment My

Cross-section of the deflected free end of the cantilever IFsubjected to minor axis moment

Δ

LTB involves both a lateral deflection and a torsional twist angle

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When member is subjected to MAJOR axis moment, the upper flange & web are in compression and act as a strut.

Being free to move, the compression elements would tend to move laterally

However, the tension flange and web are reluctant to move, creating resistance to lateral movement. As such, the cross-section twists when it deflects, with the tension flange and web dragging behind.

Weak axis of compressive section

Partial section under compression

Strong axis of compressive section

Member subjected to MAJOR axis moment Elements under compression prone to buckling

Elements under tension resist buckling

T-section would deflect in the same direction as minor axis buckling

Lateral Deflection and Twisting23/8/2012

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Both flange free to rotate on plan

Lateral Torsional Buckling

Both flanges restrained from rotation on plan

End Support Conditions

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Special Cases where LTB checks can be ignoredThe following are cases where LTB checks can be ignored:

SHS, CHS, circular or square bar

Fully laterally restrained beams

Minor axis bending

for welded sections

for hot rolled sections

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Z

Z

Minor axis bending

Y

Y

Major axis bending

major axismoment My

No LTB for minor axis bending

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SHS and CHS under bending

No LTB

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Buckling Resistance

EN 1993-1-1 (Cl 6.3.2.1)

Each segment between intermediate lateral restraints or between the end supports of a member subject to major axis bending should be verified against lateral torsional buckling using the following:

The design buckling resistance moment, Mb,Rd of a laterally unrestrained beam should be taken as :

Moment Check

where Wy is the appropriate section modulus– Wy = Wpl,y for Class 1 and 2 cross-sections– Wy = Wel,y for Class 3 cross-sections– Wy = Weff,y for Class 4 cross-sections

LT is the reduction factor for LTB

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EN 1993-1-1 (Cl 6.3.2.2 & 6.3.2.3)

Cl 6.3.2.2 is for general case adopting the lateral torsional buckling curves given by equations 6.56

Cl 6.3.2.3 is for rolled section or equivalent welded section adopting the lateral torsional buckling curves given by equations 6.57.

Approaches for Determining the Reduction Factor for LTB

General Case/Rolled Sections or Equivalent Welded Sections

EN 1993-1-1 (Cl 6.3.2.4)

This method utilizes a simplified assessment approach for beams with restraints in buildings given by equations 6.59 and 6.60. for info only

Simplified Assessment Methods for Beams with Restraints in Buildings

EN 1993-1-1 (Cl 6.3.4)

This method may be used when the above methods do not apply. for info only

General Method for Structural Components

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LTB Curves – General CaseEN 1993-1-1 (Cl 6.3.2.2) For the general case, the value of LT for the appropriate non-dimensional slenderness is given as follows:

where

LT is an imperfection factor

Mcr is the elastic critical moment for LTB

The general case method is meant to be used for deep slender beams that are outside the range of shapes of rolled sections.

The general case is also applicable to rolled and welded sections but provides a more conservative estimate of the buckling resistance.

– conservative method for general use

(6.56)

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Table 6.4: Recommended lateral torsional buckling curves for cross-sections

Table 6.3: Recommended values for imperfection factors for lateral torsional buckling curves

Buckling curve a b c dImperfection factor LT 0.21 0.34 0.49 0.76

Cross-section Limits Buckling curve

Rolled I-sections h/b ≤ 2h/b > 2

ab

Welded I-sections h/b ≤ 2h/b > 2

cd

Other cross-sections – d

LTB Curves & Imperfection Factors – General Case

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LT

Buckling Curves for LTB

LT

=0.21=0.34=0.49=0.76

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For doubly symmetric cross-sections loaded through its shear center, the elastic critical moment is given by:

where C1 is the correction factor for non-uniform bending momentLcr is the buckling length of the beam/segmentG is the shear modulusIT is the torsion constantIW is the warping constantIz is the section second moment of area about minor axis

NCCI (SN003a-EN-EU)

Elastic Critical Moment for LTB – General Case

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For hot-rolled doubly symmetric I and H sections, may be conservatively simplified to:

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As a further simplification, C1 may also be conservatively taken = 1.0.

Simplified Assessment of 

where

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The value of can be conservatively estimated by the following table for hot-rolled doubly symmetric I and H sections with lateral restraints at both ends of the segment only.

Grade

S235

S275

S355

S420

S460

Lcr = kL = effective lengthiz = radius of gyration about the minor axisE = 210kN/mm2

C1 effect can be included by dividing the value by

Simplified        for I and H Sections

where

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C1 values for end moment loading

C1 values for transverse loading

Correction Factor for Non‐Uniform Moment C1 – General Case

C1 1.0(C1 = 1.0 corresponds to the most severe case loading condition of constant bending moment)

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Effective Length for Beams without Intermediate Restraint

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For destabilizing load, Lcr = DL = 1.2 L

1

3

2

Lcr = kL or DL

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Typical Beam Support Conditions in Building Frame

1.Flanges are fully restrained against rotation on plan

2.Flanges are partially restrained against rotation on plan

3.Flanges are free to rotate on plan

1

3

2

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Destabilizing Load Neutral Load Stabilizing Load

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Cantilever

Source: The Institution of Structural Engineers Manual for the design of steelwork building structures to Eurocode 3

C1 should be 1.0 for cantilever

Effective lengthLcr = kL = DLwhere D = parameter for destabilizing load

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LTB Curves – Rolled Sections or Equivalent Welded SectionsEN 1993-1-1 (Cl 6.3.2.3) – less conservative method

For rolled or equivalent welded sections in bending, the value of LT for the appropriate non-dimensional slenderness is given as follows:

where

= 0.4 (rolled sections, hot finished and cold formed hollow sections)= 0.2 (welded sections)

= 0.75 (rolled sections, hot finished and cold formed hollow sections)= 1.00 (welded sections)

(6.57)

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Table 6.5: Recommended lateral torsional buckling curves for cross-sections

Cross-section Limits Buckling curve

Rolled I- and H- sections, and hot-finished hollow sections h/b ≤ 22.0 < h/b < 3.1

bc

Angles (for moments in the major principal plane) and other hot-rolled sections d

Welded sections and cold-formed hollow sections h/b ≤ 22.0 < h/b < 3.1

cd

Table 6.3: Recommended values for imperfection factors for lateral torsional buckling curves

Buckling curve a b c dImperfection factor LT 0.21 0.34 0.49 0.76

LTB Curves & Imperfection Factors – Rolled Sections or Equivalent Welded Sections h

b

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Modifying LT for Moment Gradient EffectEN1993-1-1 Clause 6.3.2.3(2) and the SS NA.2.18.

The reduction factor is modified to take account of the moment distribution between the lateral restraints of members using the reduction factor f :

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LT

Comparison of Buckling Curves Given in Clauses 6.3.2.2 & 6.3.2.3

LT

Buckling curve for general case (Cl 6.3.2.2)

Rolled I‐ Section with h/b < 2

Buckling curve for rolled section (Cl 6.3.2.3)

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Beams with Intermediate Restraint

Where a beam has effective intermediate restraints the moment resistance can be based on the length between restraints. For destabilizing load, Lcr = 1.2 L.

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Secondary beams providing lateral and torsional restraint.

Effective length of compression flange

Lcr

Effective length of compression flange

Lcr

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Eurocode 3: Design of Steel Structures R Liew & S D Pang

Beams with Intermediate RestraintsLateral torsional buckling resistance checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists).The effect of moment distribution between the lateral restraints may be taken into account by modifying LT using Equation 6.58 from EN1993-1-1 Cl 6.3.2.3(2).

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Illustration of flanges being free to rotate on plan along span

Plan

Bottom flange

Top flange

y

zx

u

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Design Procedure for LTB

Determine buckling curve (a, b, c, or d) from Table 6.4 or Table 6.5.

Determine imperfection factor LT from Table 6.3 after identifying the buckling curve.

Determine the elastic critical lateral torsional buckling moment Mcr.

Determine effective buckling length Lcr.

Check MEd / Mb,Rd ≤ 1.0 for each unrestrained segment.

Calculate buckling reduction factor LT .

Determine buckling resistance Mb,Rd .

Calculate non-dimensional slenderness .

Determine shear and bending moment diagram from design loads.

Select and classify section.

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Design Flow Chart for Beams Subjected to LTB

Compute and draw the SFD and BMD under design actions

Select a trial section for the most critical segment based on Mb,Rd

Determine fy and perform section classification

Ultimate strength check moment and shear at critical locations

Member buckling resistance check for each segment

Serviceability check

Section classification

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Examples

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Example URB-1: Buckling Resistance of UBA beam of span 10 m is simply supported at its ends and unrestrained along its length. It supports a uniformly distributed load across the entire span and a point load at its mid-span. Check and verify if section UB 533×210×101 in S355 steel is suitable for this beam. Assume that the beam carried plaster finish.Unfactored load values:Dead Load UDL 5 kN/m Imposed Load UDL 10 kN/m

Point load 50 kN Point load 100 kN

5m 5m

50 kN + 100 kN5 kN/m + 10 kN/m

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Ultimate Limit StateThe section and loading are the same as Example RB-1.Perform the same section classification, shear check, deflection check as described in Example RB-1.In this example, we will perform check on the lateral torsional buckling for this unrestrained beam.

5m 5m

67.5 kN + 150 kN

6.75 kN/m + 15 kN/m

217.5 kN 217.5 kN

Design Moment

Maximum bending moment at mid-span: MEd = (6.75+15)*102/8 + (67.5+150)*10/4 = 816 kNm.

Design Shear

Maximum shear force at the supports: VEd = 217.5 kN.

UB 533×210×101 in S355 steel

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Buckling LengthLcr = 10m

Elastic Critical MomentAssume C1 = 1.0 (conservative estimate)

Imperfection Factorh/b = 536.7/210.0 = 2.6 > 2Use buckling curve c (refer to Table 6.5)Imperfection factor αLT = 0.49 (refer to Table 6.3)

Assume beam end conditions: Compression flange laterally restrained; Nominal torsional restraint against rotation about longitudinal axis; Both flanges free to rotate on plan; Normal loading condition; k= 1.0 and Lcr = 10m

Comment: for beam subject to UDL C1 = 1.132 for mid-span point load C1 = 1.365You may select C1 = 1.132 instead of 1.0

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Eurocode 3: Design of Steel Structures

Simplified Assessment

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Non-Dimensional Slenderness

Buckling Reduction Factor

Buckling Resistance

Since MEd = 816 kNm > Mb,Rd, resistance to lateral torsional buckling is inadequate.

(more conservative!)

Less conservative method Eq. 6.57

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Eurocode 3: Design of Steel Structures

Re-design the beam – Options?

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5m 5m

67.5 kN + 150 kN

6.75 kN/m + 15 kN/m

217.5 kN 217.5 kN

UB 533×210×101 in S355 steel is inadequate

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Eurocode 3: Design of Steel Structures

Design Table Page D-65Select

533x312x182 UB S355 Steel,Buckling resistance

Mby,Rd = 965kNm > 816kNmOK

UB 533×210×101 in S355 steel is NOT adequate MEd = 816 kNmL = 10 mC1 = 1.0

Design Table Page D-103533x312x182 UB S355 Steel,

Design shear resistance Vc,Rd = 1740 kN > 217.5 kN

OK

Page D-6539

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Eurocode 3: Design of Steel Structures

B C

Example URB2

The simply supported beam shown below is restrained laterally at the ends and at the points of load applications only. For the given loading, design the beam in S275 steel.

3m 3m 3m

Permanent:self-weight: 3 kN/mpoint load, beam 1, Gk1 = 40 kNpoint load, beam 2, Gk2 = 20 kN

Beam 1 Beam 2

Imposed:point load, beam 1, Qk1 = 60 kNpoint load, beam 2, Qk2 = 30 kN

A D

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Eurocode 3: Design of Steel Structures

Design loads:

UDL = 3 × 1.35 = 4.05 kN/mFD1 = 40 × 1.35 + 60 × 1.5 = 144 kN;FD2 = 20 × 1.35 + 30 × 1.5 = 72 kN

144 kN 72 kN4.05 kN/m

3m 3m 3m138.2

126.1

17.9 30.1102.1 114.2

Shear (kN)

Bending (kNm)

396.5324.5

A B C D

For beam segment within intermediate lateral restraints, C1 = 1.0

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MEd = 396.5kNm ; Lcr = 3m

Page C-67

Select 457x 191 x 82 UB S275 steel, Mb,Rd = 421 kNm > 396 kNm

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Eurocode 3: Design of Steel Structures

Tutorial questions• What are the main different behaviour between laterally

restrained and un-restrained steel beam?Unrestrained beam deflects and buckles laterally

• What are the main factors affecting the bending capacity of laterally unrestrained steel beams?Unbraced length, cross sectional shapes, loading, end support conditions etc.

• How do we prevent lateral torsional buckling of beams?Use hollow sections; provide adequate lateral bracing

• How do we ensure lateral restraints are effective?Need to anchor the lateral tie

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