1 Unrestrained Beams 23/8/2012
Jan 21, 2016
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Unrestrained Beams
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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
Eurocode 3: Design of Steel Structures R Liew & S D Pang
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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
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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Eurocode 3: Design of Steel Structures R Liew & S D Pang
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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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Eurocode 3: Design of Steel Structures R Liew & S D Pang
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)
Eurocode 3: Design of Steel Structures R Liew & S D Pang
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
Eurocode 3: Design of Steel Structures R Liew & S D Pang
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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
Eurocode 3: Design of Steel Structures R Liew & S D Pang
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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Eurocode 3: Design of Steel Structures R Liew & S D Pang
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
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.
3131
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
Eurocode 3: Design of Steel Structures
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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
Eurocode 3: Design of Steel Structures
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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
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
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
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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Eurocode 3: Design of Steel Structures
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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
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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