Design of Laterally Unrestrained Flexural

Members

Flexural members1-D structural members which resist transverse load through bending and shear

Image source: N. Subramanian, Design of Steel Structures

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Image source: Geschwinder, Unified design of steel structures

Terminologies

Joists : These beams support roof.

Girders: Beams that support number of joists.

Spandrels: Exterior beams that carry floor and exterior wall loads.

Purlins: Carries roof loads in in trusses

Rafter: Roof beam to support purlins

Lintels: Carries masonry over an openning

Girt: Horizontal beam in industrial buildings to support wall cladding.

Typical cross-sections

Cross-beam

Main-beam

Single Structural ShapesRolled sections (IS808 & IS1173)

Section ClassificationAbbreviated symbol

Column section ISSC SC

Wide Parallel Flange Beams/Columns ISWPB WPB

Angles ISA

Medium weight channels ISMC MC

Light weight channels ISLC LC

Medium weight channels ISMC MC

Medium weight parallel flange channels ISMCP MCP

Rolled normal T bars ISNT NT

Rolled deep legged T bars ISDT DT

Light weight T bars ISLT LT

Slit medium weight T bars ISMT MT

Slit T bars from H-sections ISHT HT

Parallel flange section (IS12778)

3

IChannel

Angle

Tee

Lateral Torsional Buckling

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Lateral torsional buckling of a cantilever beam, Trahair and Bradford, Design of Steel Structures to EC3

EI yd2w '

dx2=−M zϕ

GI tdϕdx

−E Iwd3ϕdx3

=M zdw 'dx

End moment :M z

y’

z’

x’

Lateral torsional buckling of simply supported beam, Engineering Structures, 187 (2019), 329-340

y’ , v’ x’

z’ , w’EI z

d2 v '

dx2=−M z

I t : St .Venant 's Torsion constant

I w :Warping torsion constant

Torsion in Thin-Walled Member

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Warping of an I beamAlexander Chajes, Principles of Structural

Stability

z

x

ϕ

z

T=GI tdϕdx

−E Iwd3ϕdx3

Warping torsion

St. Venant torsion

Effect of torsion on an I-beam Boresi, Advanced Mechanics of Materials

If warping is prevented warping torsion arises

St. Venant’s Torsion Constants

6

y

z

I t=13(b t 2− t 3

4 )≈ bt2

3

Torsion of a rectangular and open section, Trahair and Bradford, Design of Steel Structures to EC3

I t=∑i

n

I ti=∑

i

n bi t i2

3

Thin-walled open section

Clause 8.2.2.1App E1-2b

Rectangular strip Thin-walled closed section

I t=4 A 2

∮S

dst

Clause 8.2.2.1App E1-2b

S: perimeter of the c/s along the centreline

Warping Torsion Constant

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ω B (s)=∫0

s

rB(s) .TB(s)ds

Warping functionP

B

z

y

rB(s)

TB (s)

sI w=∫

0

A

ω B2 (s)dA

Warping torsion constant

As per IS800, E-1.2b

For I−sectionsmonosymmetric about theweak axis(local axis y ) , Iw=(1−β f )β f I y hy2

For angle, Tee, narrow rectangular sections, and for hollow sections, Iw = 0

hy: distance between shear centre of the two flanges of the cross-section

β f=I fc

I fc+ I ft

where, Ifc and Ift are the moment of inertia of the compression and tension flanges, respectively, about the minor axis of the entire section

Lateral Torsional Buckling (LTB) of I-beam Under Pure Bending

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EI yd2w '

dx2=−M crϕ

GI tdϕdx

−E Iwd3ϕdx3

=M crdw 'dx

Assume ,w '=Δ sinπ xL

,ϕ=θ sinπ xL

BC:w '|x=0=0 ,and w '|x=L=0 ,Atends , no transversedeflection

BC: ϕ|x=0=0 ,and ϕ|x=L=0 ,At ends ,no twisting

BC:d2ϕdx2 |

x=0

=0 ,andd2ϕdx2 |

x=L

=0 ,At ends , sections can warp

EI zd2 v '

dx2=−M cr

L

M cr M cr

I t : St .Venant 's Torsion constant

I t :Warping torsion constant

Governing equation for buckling Boundary conditions

BC: v '|x=0=0 ,and v '|x=L=0 ,At ends , no vertical deflection

BC:M|x=0=M cr ,and M|x=L=M cr ,At ends ,M=M cr

M cr=√ π 2E I y

L2 [G I t+π 2E Iw

L2 ]For different boundary conditions L is replaced by LLT (Clause 8.3)

For different loading conditions refer to Appendix E

Lateral Torsional Buckling (LTB) of a Real Beam

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Image source: Subramaniyan

❑ The critical moment is affected by geometric imperfection

❑ Inelasticity

AISC procedure

Image source: Geschwinder, Unified design of Steel Structures

Eurocode Procedure

Compression

Flexure

IS 800, adopts the Eurocode Procedure

Lateral Torsional Buckling with Imperfection

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EI yd2w '

dx2=−M 0(ϕ +ϕ 0)

GI tdϕdx

−EΓ d3ϕdx3

=M 0(dw 'dx

+dw ' 0dx )

Lateral torsional buckling of simply supported beam, Engineering Structures, 187 (2019), 329-340

y’ , v’ x’

z’ , w’EI z

d2 v '

dx2=−M 0

BC:w '|x=0=0 ,and w '|x=L=0 ,At ends , no transverse deflection

BC: ϕ|x=0=0 ,and ϕ|x=L=0 ,At ends ,no twisting

BC:d2ϕdx2 |

x=0

=0 ,andd2ϕdx2 |

x=L

=0 ,At ends , sections can warp

BC: v '|x=0=0 ,and v '|x=L=0 ,At ends , no vertical deflection

BC:M|x=0=M cr ,and M|x=L=M cr ,At ends ,M=M 0

Assume ,w '=Δ sinπ xL

,ϕ=θ sinπ xL

,w '0=Δ0 sinπ xL

,ϕ 0=θ 0sinπ xL

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−M 0 (Δ+Δ0 )+(G I t+π 2E Iw

L2 )θ =⇒−M 0(Δ+Δ0)+r02Pcrxθ =0

Derivation Continues

−π 2E I y

L2 Δ+M 0(θ +θ 0)=0⇒−PcryΔ+M 0 (θ +θ 0)=0

[ Pcry −M 0

−M 0 r02Pcrx

]{Δθ }=[ 0 M 0

M 0 0 ]{Δ0

θ 0}

Polar radius of gyration , r0=√∫A r2dA

A

{Δθ }=[ Pcry −M 0

−M 0 r02Pcrx

]−1

[ 0 M 0

M 0 0 ]{Δ0

θ 0}

ΔT=Δ+Δ0=M cr

2 Δ0+r02 Pcrx M 0θ 0

M cr2 −M 0

2θ T=θ +θ 0=

M 0PcryΔ0+M cr2 θ 0

M cr2 −M 0

2

limM 0→M cr

ΔT

θ T=

M cr

Pcry

As the moment approaches the elastic critical moment the ratio of total deformation amplitudes remains constant

Derivation follows Journal of Constructional Steel Research 66 (2010) 670–679

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Derivation Continues (Ratio of total deformation with given imperfection)

Let us assume the first buckling mode gives the shape of imperfection for any M0

Δ0

θ 0=

M cr

PcryThen,

ΔT

θ T=

M cr2 Δ0+r0

2PcrxM 0θ 0

M 0Pcry Δ0+M cr2 θ 0

⇒M cr

2 Δ0+r02 Pcrx M 0

Δ0Pcry

M cr

M 0Pcry Δ0+M cr2 Δ0

Pcry

M cr

=M cr

Pcry

If we assume the ratio of imperfection amplitudes follows as the ratio of first buckling modeshapes the ratio of total deformations remains constant (same as what we saw in the previous slide)

ΔT=M cr

2 Δ0+r02PcrxM 0

Pcry

M cr

Δ0

M cr2 −M 0

2=

M cr (M cr+M 0)Δ0

M cr2 −M 0

2=

M cr

M cr−M 0

Δ0

θ T=θ +θ 0=M 0PcryΔ0+M cr

2 θ 0

M cr2 −M 0

2

ΔT=Δ+Δ0=M cr

2 Δ0+r02 Pcrx M 0θ 0

M cr2 −M 0

2

θ T=M 0Pcry

M cr

Pcry

θ 0+M cr2 θ 0

M cr2 −M 0

2=

M cr

M cr−M 0

θ 0

Assumed imperfect shape gives linear and consistent amplification.

Derivation (Manipulation) Continues

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θ =−M 0θ 0

M 0−M cr

=−M 0Δ0Pcry

M cr (M 0−M cr)Δ=

M 0Δ0

M 0−M cr

f max=f y=M 0

Z ez

−E I y

Z eyf( d2

dx2 [v ' ( x)− ymaxϕ (x)])x= L2

M y=M 0−M 0M cr

M 0−M cr

η

M y (M 0−M cr)=M 0(M 0−M cr)−η M 0M cr

M 02−(M y+M cr+η M cr )M 0+M y M cr=0

If the maximum moment is within the elastic limit

Δ0

π 2E I y

L2 =M crθ 0⇒Pcry Δ0=M crθ 0

M 0=(1+η)M cr+M y−√[(1+η)M cr+M y ]

2−4M y M cr

2

M y=M 0+Pcry Z ez

Zeyf

M 0Δ0

M 0−M cr[1+ ymax Pcry

M cr] Let ,η≝−

Pcry Z ez

Z eyf f cr

Δ0[1+ ymax Pcry

M cr]

Define

√ M y

M cr

=λ̄ , ϕ̄ =(1+η)M cr+M y

2=1+η+λ̄2

2M cr=ϕ M cr

M 0=ϕ̄−√ϕ̄ 2−M crM y=M cr M y

ϕ̄ +√ϕ̄ 2−M cr M y

=M y

ϕ +√ϕ 2−λ̄2

η is chosen negative as M0 will be lesser than Mcr

Zeyf is the elastic section modulus of the flange about the minor axis of the cross-section

Most important assumption in the derivation

Final Form of Moment Carrying Capacity

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M 0=ϕ̄−√ϕ̄ 2−M crM y=M cr M y

ϕ̄ +√ϕ̄ 2−M cr M y

=M y

ϕ +√ϕ 2−λ̄2

Similar to what we had in Perry Robertson’s formula for column buckling.

M 0=M y

ϕ +√ϕ 2−λ̄2⇒

M 0

Z ez

=

M y

Z ez

ϕ +√ϕ 2−λ̄ 2⇒ f=

f y

ϕ +√ϕ 2−λ̄ 2

√ M y

M cr

=λ̄ ,ϕ =1+η+λ̄ 2

2

Similar to what we had in Perry Robertson’s formula for column buckling. The above formula gives us the permissible stress f so that the member stays within the elastic limit.

M cr=√ π 2E I y

L2 [G I t+π 2E Iw

L2 ]

Derivation (Manipulation) Continues

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θ =−M 0θ 0

M 0−M cr

=−M 0Δ0Pcry

M cr (M 0−M cr)Δ=

M 0Δ0

M 0−M cr

f max=f y=M 0

Z ez

−E I y

Z eyf( d2

dx2 [v ' ( x)− ymaxϕ (x)])x= L2

M y=M 0−M 0M cr

M 0−M cr

η

M y (M 0−M cr)=M 0(M 0−M cr)−η M 0M cr

M 02−(M y+M cr+η M cr )M 0+M y M cr=0

If the maximum moment is within the elastic limit

Δ0

π 2E I y

L2 =M crθ 0⇒Pcry Δ0=M crθ 0

M 0=(1+η)M cr+M y−√[(1+η)M cr+M y ]

2−4M y M cr

2

M y=M 0+Pcry Z ez

Zeyf

M 0Δ0

M 0−M cr[1+ ymax Pcry

M cr] Let ,η≝−

Pcry Z ez

Z eyf f cr

Δ0[1+ ymax Pcry

M cr]

Define

√ M y

M cr

=λ̄ , ϕ̄ =(1+η)M cr+M y

2=1+η+λ̄2

2M cr=ϕ M cr

M 0=ϕ̄−√ϕ̄ 2−M crM y=M cr M y

ϕ̄ +√ϕ̄ 2−M cr M y

=M y

ϕ +√ϕ 2−λ̄2

η is chosen negative as M0 will be lesser than Mcr

Zeyf is the elastic section modulus of the flange about the minor axis of the cross-section

Most important assumption in the derivation

IS800 Procedure (Clause 8.2.2)

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IS800, somehow extends the elastic lateral torsional buckling formula with imperfection to inelastic lateral torsional buckling but the derivation is not clear

IS800 Procedure Contd.. (Clause 8.2.2)

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This equation is only valid for uniform moment please refer to E1 for non-uniform moment

Plot the Md vs λLT curve for an I-section of your choice. (Please see one of the live examples)

We should expect When λLT is high Md = Mcr

When λLT is low Md = Mp (plastic moment not the yield moment)

When to Ignore LTB Check?

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LLT For Simply Supported Beam (Clause 8.3)

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Clause (8.3.1, amended version)

In simply supported beams with intermediate lateral restraint against LTB the effective length for LTB, LLT to be used in 8.2.2.1 shall be taken as the length of relevant segment between lateral restraints. In the case of intermediate partial lateral restraints the effective length shall be 1.2 length of the relevant segment in between the partial lateral restraints

partially

When the load is on the top flange, the compression in flange and the load acting together creates a destabilizing condition. Why?

P

P

Normal

Destabilizing

LLT For Cantilever Beam (Clause 8.3.1)

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For cantilever, LLT = projected length of the cantilever, see Table 16

Design of Intermediate Lateral Restraint

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For beams, where members are provided for effective lateral restraint

❑ Determine the maximum total force FC in the compression flange

❑ Select the number of lateral restraint

❑ Estimate the maximum force in support, F = Max(0.025*FC/n, 0.01*FC)

❑ Design the lateral support to resist F

LTB With Different Force and Displacement Boundary Conditions (E-1.2)

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M cr=c1π 2E I y

LLT2 {[( K

Kw)2 Iw

I y

+G I t (LLT)2

π 2E I y

+(c2 y g−c3 y j)2]

0.5

−(c2 y g−c3 y j)}

LTB With Different Loading Conditions (Table 42, E-1.2)

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LTB With Different Support Conditions (Table 42, E-1.2)

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