5-UnrestrainedBeam

Chapter 4 Unrestrained Beams 15/08/2007

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Chapter 2: Member Design

Section 4: Unrestrained Beams

Compression on top-half of the beam


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Basic Principal of lateral torsionalbuckling

Dead weightload appliedvertically

Buckledposition

Unloadedposition

Clamp atroot

In the case of a beam bent about its major axis, failure may occur by a form of buckling which involves both lateral deflection and twisting.

Slender structural elements loaded in a stiff plane tend to fail by buckling in a more flexible plane.

Lateral-Torsional BucklingM M

L

Elevation Section

Plan

y

zx

u

φ

Perfectly elastic, initially straight, loaded by equal and opposite end moments about its major axis.

Unrestrained along its length, L.

End Supports …

– Twisting and lateral deflection prevented.

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


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Elastic buckling of beams

Includes:

Lateral flexural stiffness EIy

Torsional and Warping stiffnesses GJ and EIw

Their relative importance depends on the type of cross-section used.

2 2

2 2y w

cry y

EI I L GJML I EI

ππ

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦

Critical Buckling Moment for uniform bending moment diagram is

Behaviour of Beams

ηLT = 0.007 (λLT - λo)

(ME - Mb)(Mp- Mb) = ηLT ME Mb

y2

LT

2p

E pEM

Mλ

π=

35 150

Practical Region Elastic Failure

PlasticFailure

Elasto - PlasticRegion

Elastic CriticalMoment ME

Slenderness = L/ry

ResistanceMoment Mb

Mp

λ


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Factors Influencing Mb

• Beam slenderness (L/ry) • Bending stiffness (E Iy)• Torsional stiffness (G J)• Moment diagram shape• Restraint and end conditions• Whether the load is destabilizing

Destabilizing Load

Stabilizing Load

Neutrale


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Beam with Intermittent restraint.

Intermediate Lateral Restraint

Resistance ≥ 2.5% of max force in the compression flange divided between the intermediate lateral restraintsShould be connected close to the compression flangeor at any level if torsional restraint providedMust be connected to bracing or have similar support from part of the structure


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BAY 1Triangulated

BAY 2Tied to braced bay

Intermediate Restraint

Unbraced length

Braced bay

Beam Stiffness > 25 times of unbraced element

unbraced element


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Torsional Restraint

Both flanges held in position relative to each other by external meansAt supports provided – by lateral restraint to both flanges– by bearing stiffeners– as for intermediate torsional restraint

Intermediate torsional restraint provided by triangulated bracing

Torsional Restraint

(a) Torsional restraint using connections

(b) Torsional restraint using stiffeners

Positive Connection

Bracing for torsional restraint


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LTB check:

mLT Mx ≤ Mb

Cross section moment capacityMx ≤ Mcx

Table 18

Py

λLT = u v λ βw1/2

Table 19λ /x and η

LE/ry

Section table

{Table 16

ExemptionsNo need to check for LTB in the following cases:– Bending about the minor axis– CHS, square RHS or square/circular bars– RHS, unless LE/ry > limits in Table 15– I, H, channel & box sections when λLT ≤ λL0

Table 15: limiting L/ry for RHSD/B = 1.25 L/ry limit = 770 x 275/pyD/B = 1.5 L/ry limit = 515 x 275/pyD/B = 2.0 L/ry limit = 340 x 275/py


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SHS and CHS under bending

No LTB

Buckling Resistance Moment Mb

For class 1 and 2: Mb = pb Sx

For class 3: Mb = pb Zx

or Mb = pb Sx.eff

For class 4: Mb = pb Zx,eff


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Bending Strength pb

• pb depends on λLT and py cl. 4.3.6.5

• If λLT ≤ λL0 then pb = py

• Otherwise pb is obtained from:– Table 16 for rolled sections– Table 17 for welded sections

Bending Strength pb


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Equivalent Slenderness λLT

λLT = u v λ βw1/2 Cl 4.3.6.7

u is the buckling parameterv is the slenderness factor Table19λ is the slenderness = LE/ry

For class 1 and class 2: βw = 1.0For class 3: βw = Zx/ Sx or Sx eff/SxFor class 4: βw = Zx eff/ Sx

Effective Length LE (Table 13, Cl4.3.5.2)

• Without intermediate lateral restraint: use Tables 13 and 14 with LLT = LE

• With intermediate lateral restraint:– Normal loads LE = 1.0 LLT

– Destabilizing loads LE = 1.2 LLT (simple beams only)



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Effective Length

(1)

(3)

(2)

(4)

(5)

Case (1)

Case (2)

Case (3)

Case (4) Case (5)


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Determination of λLT

• v is obtained from Table 19 for values of λ /x and η – x is the torsional index– η = 0.5 for equal flanged sections– η = Iyc/(Iyc+Iyt) for unequal flanges

• Annex B.2.3 gives formulae for u and x


V slenderness factor


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Equivalent Uniform Moment Factor mLT

• For the normal loading condition– mLT is obtained from Table 18

• For the destabilizing loading condition– mLT = 1.0

Mx ≤ Mb/ mLT and Mx ≤ Mcx

Factor mLT


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SummaryLTB check:

mLT Mx ≤ Mb

Cross section moment capacity checkMx ≤ Mcx

Table 18

py


Table 19λ /x and η

LE/ry

Section table

Design Procedure

1.Select section & determine py2.Determine section class3.Check Mx ≤ Mcx4.Determine LE5.Calculate slenderness λ = LE/ry6.Determine u, v and βw


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Design Procedure continued

7. Calculate λLT = u v λ βw1/2

8. Determine pb from table 16 or 179. Calculate Mb

10. Determine mLT from table 1811. Check mLTMx ≤ Mb

The moment at A and C = 0The maximum moment occurs at B

Mmax =

EXAMPLE 1Restrained beam

5 m 5 m

225 kNW = 25 kN/m

A B C

533 x 210 x 101 UB Grade S355The maximum shear occurs at A and C.

Fvmax = 2225

21025

+×

kNm8754

102258

1025 2

=×

+×

The shear at B =2

225 = 113 kN

= 238 kN

Design Using Tables533x210UB 101 S355 steelFrom Design Table Page 281Section is plastic Shear capacity = 1200 kN > 238 The shear is low.Moment capacity = 901 kNm >875

Simple connection


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EXAMPLE 2Unrestrained beam

The maximum shear occurs at A&C

Fvmax= 24.1

2225 wtself×

+

= 112.5 + 0.07 = 113 kN

The maximum moment occurs at B

Mmax = wtselftodueMoment+×

410225

= 575 kNm Design using Tables533x210UB 101 S355 steel, effective length = 5mFrom Design Table Page 327Section is plastic Moment capacity Mb = 446 kNm for compact sectionNeed to calculate mLT to complete the design

Lateral restraint


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The bending moment diagram between A & B is as follows:

M = 575B

AM = 0

The requirement is that:Mx = Mb/mLT andMx < Mcx

Mx= 575 kNm < 446/0.6 = 740 kNmMx= 575 kNm < 901 kNm

Therefore, β = 0 mLT = 0.6 table 18


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Example 3: Unrestrained Propped Cantilever Beam

Check the section classification as Class 1 Table 11Section properties: T = 13.3 ⇒Design strength = ρy = 275 N/mm2 Table 9

ry = 3.23cm x = 37.5Sx = 1290cm3 u = 0.868

The three lengths must be checked separately.

(a) Length AB(1) LE = 4m

Beams with intermediate lateral restraints between the supports clause 4.3.5.2ry = 3.24cm λ = LE/ry = 400/3.23 = 124

(2) x = 37.5, λ/x = 123.5/37.6 = 3.3 η = 0.5 ⇒ v = 0.897 Clause 4.3.6.7 or Table 19

(3) u = 0.868(4) λLT = uvλ wβ

= 0.868 x 0.897 x 124 x1.0 ; 1 0 1w . for class sec tionβ =

= 96.5 (5) β = -150/280 = -0.54

mLT = 0.44; mLT Mx = 0.44(280) = 123 kNm Table 18(6) λLT= 96.5 , ρy = 275 N/mm2

⇒ ρb = 131 Table 16Mb = ρbSx = 131 (10-3/10-6) x 12980 (10-6) = 169 kNmmLT Mx = 123 kNm < Mb = 169kNm i.e. length AB is OK!


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(b) Length BC (1) LE = 3m

λ = LE/ry = 300/3.23 = 93 (2) λ/x = 93/37.5 = 2.48

η= 0.5 ⇒ v = 0.935 Clause 4.3.6.7 or Table 19(3) u = 0.868 (4) β = 190/280 = 0.679

mLT = 0.87 Table 18mLT Mx = 0.87(280) = 244kNm

(5) λLT = uvλ wβ

= 0.868 x 0.935 x 93 x1.0 = 75 ⇒ ρb = 176N/mm2 Table 16

Mb = ρbSx = 176(10-3/10-6) x 1290(10-6) = 227 kNm

mLT Mx = 244kNm > Mb = 227kNm NG!

Try 457 x 191 x 67 and check Segment BC againSection properties: T = 12.7 ⇒ Design strength = ρy = 275N/mm2 Table 9

ry = 4.12cm x = 37.9 Sx = 1470cm3 u = 0.872 λ/x = 72.8/37.9 = 1.92

η = 0.5 ⇒ v = 0.96 Clause 4.3.6.7 or Table 19λLT = uvλ wβ

= 0.872 x 0.96 x 72.8 x1.0 = 60.9 ⇒ ρb = 211.1N/mm2 Table 16Mb = ρbSx

= 211 (10-3/10-6) x 1470 (10-6) = 310kNm > mLT Mx = 244kNm OK!


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Using Design Table Page 245457 x 191 x 67 UB S 275 Steel,

Length = 3mMb = 310kNm

for compact section

Tutorial questionsWhat are the main different behaviour between laterally restrained and un-restrained steel beam?

Unrestrained beam deflects and buckles laterallyWhat 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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Reading assignment:

BS5950:Part1 Cls 4.2 & 4.3Chapter 2 Sections 3 & 4Homework : Beams

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