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Chapter 3 Restrained Beams
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Chapter 2: Member Design
Section 3:
LATERALLY RESTRAINED BEAMS
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Design Criteria
i) Adequate lateral restraintii) Local buckling iii) Shear iv) Bending and combined bending and shearv) Web bearing and bucklingvi) Deflection
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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.........."
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Top flange laterally restrained by slab
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Full Restraint Beam Partial or Unrestraint Beam
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Check Shear Fv PvShear 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 > 70 for rolled sectionsIf d/t > 62 for welded sections
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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
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Moment CapacityM Mc
Class 1 Mc =pyS
Class 2 Mc=pyS
Class 3 Mc=pySeffMc=pyZ (Conservative)
MEPM
Class 4 Mc=pyZeffRotation
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Stress Blockspy py py
0.4beff
0.6beff
Class 1 & 2 Class 3 Class 4Mc =pyS Mc=pySeff
or Mc=pyZMc=pyZeff
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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
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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
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Effective Plastic Modulus for I and H sections
+=
1
1/)( 2
2
3
23
,
w
w
w
xxxeffxtdZSZS
+1
1/
)(
2
3
3
,
f
f
f
xxxeffxTb
ZSZS
But 0.6 Pv
Class 1 or 2 Mc = py (S - Sv )Class 3 Mc = py (Z Sv / 1.5 ) or Mc= py (Seff - Sv)Class 4 Mc = py (Zeff Sv / 1.5 )
But not greater than 1.2pyZ for simply supported beams.
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= (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
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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
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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
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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
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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
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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.
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0.6Pv = 0.6 x 3130 = 1888kNSince Fv = 2500kN > 0.6Pv = 1888kNMoment capacity needs to be reduced due to high shear
Mc = y(S - Sv) < 1.2yZ, Sv =
Mc = 265 (17700 4554 x 0.347) / 103 = 4191 kNm1.2yZ = 4961 kNmMc = 4191 kNm
> 4000 kNm (factored moment) OK!
32
45544
cmtD =
347.013147250021
PF
222
v
v =
=
=
