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Chapter 3 Restrained Beams J Y R Liew 1 27/9/007 August, 2002 R Liew .1 Chapter 2: Member Design Section 3: LATERALLY RESTRAINED BEAMS 2 June 2003 2 Design Criteria i) Adequate lateral restraint ii) Local buckling iii) Shear iv) Bending and combined bending and shear v) Web bearing and buckling vi) Deflection
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Page 1: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 1

27/9/007 August, 2002

R Liew .1

Chapter 2: Member Design

Section 3:

LATERALLY RESTRAINED BEAMS

2 June 2003

2

Design Criteria

i) Adequate lateral restraintii) Local buckling iii) Shear iv) Bending and combined bending and shearv) Web bearing and bucklingvi) Deflection

Page 2: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 2

2 June 2003

3

Lateral Restraint

"Full lateral restraint may be assumed to exist if the frictional or positive connection of a floor (or other) construction to the compression flange of the member is capable of resisting a lateral force of not less than 2.5% of the maximum force in the compression flange of the member, [under factored loading]. This lateral force should be considered as distributed uniformly along the flange.........."

2 June 2003

4

Top flange laterally restrained by slab

Page 3: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 3

2 June 2003

5

Full Restraint Beam Partial or Unrestraint Beam

2 June 2003

6

Check Shear Fv ≤ Pv

Shear capacity Pv = 0.6 py Av

Shear Area, Av= tD for rolled I, H, Channels= td for welded I= tD for rolled T or Single Notched Beam= AD/(B+D) for RHS= 0.6A for CHS= 0.9Ao for others

Check Shear Buckling

If d/t > 70ε for rolled sections

If d/t > 62ε for welded sections

Page 4: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 4

2 June 2003

7

Shear – allowance for holes

If Av.net ≥ 0.85 Av / Ke, or Aholes ≤ Av(1-0.85 / Ke)

Ignore bolt holes: Pv = 0.6 py Av

But If Av.net < 0.85Av / Ke then

Pv = 0.7 py Ke Av.net

Ke = 1.2 for grade S275= 1.1 for grade S355

Aholes = Av - Anet

2 June 2003

8

Moment CapacityM ≤ Mc

Class 1 Mc =pyS

Class 2 Mc=pyS

Class 3 Mc=pySeffMc=pyZ (Conservative)

ME

PM

Class 4 Mc=pyZeff

Rotation

Page 5: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 5

2 June 2003

9

Stress Blockspy py py

0.4beff

0.6beff

Class 1 & 2 Class 3 Class 4Mc =pyS Mc=pySeff

or Mc=pyZMc=pyZeff

2 June 2003

10

Moment capacity with low shear load

• Fv ≤ 0.6 Pv– For Class 1 or Class 2 Mc= py S – For Class 3 Mc= pySeff or Mc = pyZ– For Class 4 Mc= pyZeff

But Mc < 1.2pyZ for simply supported beams

Page 6: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 6

2 June 2003

11

Effective plastic modulus

• Under pure bending, an effective plastic modulus, Seff may be used instead of Z, for Class 3 sections.

• Different formulae for:– I sections: rolled and welded– RHS: hot finished and cold-formed– CHS

2 June 2003

12

Effective Plastic Modulus for I and H sections

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

−⎟⎟⎠

⎞⎜⎜⎝

−⎟⎠⎞

⎜⎝⎛

−+=

1

1/)( 2

2

3

23

,

w

w

w

xxxeffxtdZSZS

ββ

β

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

−⎟⎟⎠

⎞⎜⎜⎝

−⎟⎟⎠

⎞⎜⎜⎝

−+≤

1

1/

)(

2

3

3

,

f

f

f

xxxeffxTb

ZSZS

ββ

β

But <

For semi-compact sections subject to bending.β is the limiting value of relevant b/T or d/t fora class 2 and a class 3 section

Page 7: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 7

2 June 2003

13

Effective section for Zeff

Non effective Zone

0.4b

0.6b

Elastic Neutral axis

eff

eff

f

f

cw

tw

Elastic neutral axisof gross section

of effective section

⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎢⎢⎣

⎡ −+

=

cw

tw

yw

twcwffe

ff

pff

tb

11

120 ε

For sections which have slender webs under bending

2 June 2003

14

Effective section for Zeff

For sections which have slender flanges under pure bending

Centroidal axis of theEffective section

HF RHS

t

20t ε 20t ε

t

20tε 20tε

Major axis bending Minor axis bending

D

Page 8: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 8

2 June 2003

15

Moment capacity with high shear load

Fv > 0.6 Pv

Class 1 or 2 Mc = py (S - ρSv )Class 3 Mc = py (Z – ρSv / 1.5 ) or Mc= py (Seff - ρSv)Class 4 Mc = py (Zeff – ρSv / 1.5 )

But not greater than 1.2pyZ for simply supported beams.

2 June 2003

16

ρ = (2 (Fv/ Pv) - 1)2

Sv is, for a section with equal flanges, the plastic modulus of the shear area Sv for a rolled section is

= 2((D/2 . t) D/4) = D2 t / 4.

D

t

NA

py

py

D/2

Page 9: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 9

2 June 2003

17

Effect of Shear on Mc - S355

0

500

1000

1500

2000

2500

0.00 0.20 0.40 0.60 0.80 1.00 1.20

Fv / Pv

Mc

(kN

m) 762x267x134 Class 3

762x267x173 Class 2

2 June 2003

18

Other effects

Bearing & Buckling of websWhere loads are applied directly through the flange of the section, for example where a load is applied to the top flange from an incoming beam then the web should be checked for buckling and bearing as dealt with in the lecture on web effects.

DeflectionThis is a serviceability limit state and the check is usually carried out under unfactored applied load only

Page 10: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 10

2 June 2003

19

Suggested deflection limits

Maximum Deflection due to unfactored imposed load

Cantilevers length/180Internal beams

Beams carrying brittle finish span/360 or 40mmAll other beams span/200 or 40mm

Edge beam Span/300 to Span / 500 or 20mmGantry Girders

vertical span/500horizontal span/600

2 June 2003

20

Summary of Design Procedure:

1. Select the section and determine the value of py2. Determine the section classification3. For class 3 (semi-compact) sections calculate

effective plastic modulus.For class 4 (slender) sections calculate effective elastic modulus

4. Check the shear capacity5. Check the moment capacity with low shear or

high shear as appropriate6. Web bearing and buckling7. Deflections

Page 11: 4_RestrainedBeam

Chapter 3 Restrained Beams

J Y R Liew 11

27/9/007 August, 2002

R Liew .21

ExampleConsider a simply supported beam 914 x 419 x 388 UB, S275 steel subjected to a factored shear force of 2500kN and moment of 4000kNm. Check the shear and bending resistance of the beam if it is fully restrained against lateral-torsional buckling.

Using Design Table914 x 419 x 388 UB, S275 steel under pure bendingPage 196: Section is plasticMcx = 4680kNmPv = 3130kNNote that the moment capacity given in the table is for low shear.The moment needs to be reduced for high shear case.

2 June 2003

22

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Chapter 3 Restrained Beams

J Y R Liew 12

2 June 2003

23

0.6Pv = 0.6 x 3130 = 1888kNSince Fv = 2500kN > 0.6Pv = 1888kNMoment capacity needs to be reduced due to high shear

Mc = ρy(S - Svρ) < 1.2ρyZ, Sv =

Mc = 265 (17700 – 4554 x 0.347) / 103 = 4191 kNm1.2ρyZ = 4961 kNmMc = 4191 kNm

> 4000 kNm (factored moment) OK!

32

45544

cmtD=

347.013147250021

PF

222

v

v =⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=ρ