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07 Sx007a en Eu Beam
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SX007a-EN-EU
Example: Simply supported beam with lateral restraint at load application point EN 1993-1-1 Valérie LEMAIRE Alain BUREAU
This worked example deals with a simply supported beam with lateral restraints at supports and at load application point. The following distributed loads are applied to the beam: • • • self-weight of the beam concrete slab imposed load
Created on Monday, November 09, 2009 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
5,0 m 1 1
1 : Lateral restraint
The beam is a I-rolled profile in bending about the strong axis. This example includes : the classification of the cross-section, the calculation of bending resistance, including the exact calculation of the elastic critical moment for lateral torsional buckling, the calculation of shear resistance, including shear buckling resistance, the calculation of the deflection at serviceability limit state.
Partial factors • • • •
γG = 1,35 γQ = 1,50 γM0 = 1,0 γM1 = 1,0
(permanent loads) (variable loads)
EN 1990 EN 1993-1-1 § 6.1 (1)
all rights reserved. November 09. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
5m Secondary beam Concrete slab
Restraints to lateral buckling
. Two secondary beams are connected to the calculated one at mid-span. The beam is assumed to be laterally restrained at mid-span and at the ends
Calculated Beam
Created on Monday. 2009 This material is copyright .Example: Simply supported beam with lateral restraint at load application point
Basic data Design a non composite floor beam of a multi-storey building according to the data given below.
y = 3141 cm3
.5 mm r = 24 mm 108 kg/m A = 137 cm2
Second moment of area /yy Iy = 82920 cm4 Second moment of area /zz Iz = 3116 cm4 Torsion constant Warping constant Elastic modulus /yy Plastic modulus /yy It = 118.15 × 24 kN/m3 = 3.Example: Simply supported beam with lateral restraint at load application point
Span length : Secondary beam:
o Span length: 7 m o Bay width : • • • • • •
Created on Monday.all rights reserved.50 kN/m2 24 kN/m3
Slab depth : Secondary beams
Partitions and false ceiling: Imposed load : Concrete density : Steel grade : S355
Weight of the slab : 0.6 kN/m2
IPEA 600 – Steel grade S355 Depth Width Web thickness Flange thickness Fillet Mass Section area h = 597 mm b = 220 mm tw = 9. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
5m 15 cm 0.8 mm tf = 17.8 cm4 Iw = 2607000 cm6 Wel. 2009 This material is copyright .10 kN/m2 0.y = 2778 cm3 Wpl. November 09.50 kN/m2 2.
70 = 172 kN Maximal shear force at mid-span : Vz.Example: Simply supported beam with lateral restraint at load application point
Self weight of the beam : qG = (108 × 9.50 × 87.3.70 kN
Created on Monday.Ed = 0.0 × 7.5 mm < 40 mm.50 × 5. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
842.85 kN
Maximal shear force at supports : Vz.50 × 1.81) × 10-3 =1.10 + 0.
.125 × 1.06 = 1.50 × 329.06 kN/m Permanent load : FG = (3. November 09.5 = 329.4.0 + 0.10)
γG qG = 1.Ed = 0.1 or the values from the product standard.85 kN
Yield strength Steel grade S355 The maximum thickness is 17.Ed = 0.1
The National Annex may impose either the values of fy from the Table 3.50 × 329. 2009 This material is copyright .13 kNm Shear force diagram
172 kN V 164.0 = 87.0 = 147 kN Variable load (Imposed load) : FQ = 2.43 × 10.35 × 147 + 1.50)× 5. so : fy = 355 N/mm Note :
EN 1993-1-1 Table 3.25 × 329.70 = 164.43 kN/m γG FG + γQ FQ = 1.13 kNm
Maximal moment at mid span : My.002 + 0.all rights reserved.70 × 10 = 842.6 + 0.0 × 7.5 kN
ULS Combination :
EN 1990 § 6.35 × 1.2 (6.43 × 10.
y fy / γM0 = (3141 × 355 / 1.755 < 1 OK EN 1993-1-1 § 6. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
EN 1993-1-1 Table 5.2 (sheet 1 of 3) Class 1
The class of the cross-section is the highest class (i.13 / 1115 = 0.45 < 72 ε = 58.1 / 17.Ed / Mc.2 (sheet 2 of 3)
Internal compression part : web under pure bending c = h – 2 tf – 2 r = 597 – 2 × 17.Rd = Mpl.5
. 2009 This material is copyright .8 – 2 × 24)/2 = 81.5 – 2 × 24 = 514 mm c / tw = 514 / 9.Rd = Wpl.all rights reserved.29 Class 1 235 = 0. here : Class 1 So the ULS verifications should be based on the plastic resistance of the cross-section.e the least favourable) between the flange and the web.Rd = 1115 kNm My.63 ≤ 9 ε = 7.
Moment resistance The design resistance for bending of a cross section is given by : Mc.10 mm c/tf = 81.Example: Simply supported beam with lateral restraint at load application point
Section classification : The parameter ε is derived from the yield strength : ε = Outstand flange : flange under uniform compression c = (b – tw – 2 r) / 2 = (220 – 9.2. November 09.5 = 4.Rd = 842.81 f y [N/mm 2 ]
EN 1993-1-1 Table 5.0) / 1000 Mc.8 = 52.32
Created on Monday.
since there is no device to prevent the warping at the ends of the beam.Example: Simply supported beam with lateral restraint at load application point
Reduction factor for lateral torsional buckling To determine the design buckling resistance moment of a beam.77 for k = 1 C2 zg = 0
π 2 E Iz
(k Lc )
π 2 × 210000 × 3116 × 10 4
× 10 −3 = 2583 kN
. the reduction factor for lateral torsional buckling must be determined. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
E = 210000 N/mm2 G = 81000 N/mm2
G is the shear modulus :
Lc is the distance between lateral restraints : Lc = 5.
Critical moment for lateral torsional buckling The critical moment may be calculated from the following expression :
2 π 2 E I z ⎪ ⎛ k ⎞ I w (k Lc ) G I t ⎪ ⎜ ⎟ M cr = C1 + + (C 2 z g ) 2 − C2 z g ⎬ 2 2 ⎨ ⎜k ⎟ I π E Iz (k Lc ) ⎪ ⎝ w ⎠ z ⎪ 2
See NCCI SN005
E is the Young modulus :
The C1 and C2 coefficients depend on the moment diagram along the beam segment between lateral restraints.all rights reserved. November 09. The following calculation determines this factor by calculation of the elastic critical moment for lateral torsional buckling. 2009 This material is copyright . the following assumption should be considered : k=1 kw = 1 since the compression flange is free to rotate about the weak axis of the cross-section. then : C1 = 1. It can be assumed that the diagram is linear.0 m
In the expression of Mcr.
λ LT.10 3116 2583000 ⎭
Mcr = 1590 kNm Non-dimensional slenderness The non-dimensional slenderness is obtained from : EN 1993-1-1 § 6.y f y M cr 3141000 × 355 = 0.3
αLT = 0. the reduction factor for lateral torsional buckling is EN 1993-1-1 calculated by : § 6.837 > λ LT.3(1)
Note : So
The value of λ LT. The recommended value is 0.2 (1) Wpl.71 > 2 Curve c
EN 1993-1-1 Table 6.0 = 0.2.3.2.837 1590 × 10 6
For rolled profiles.3 (1) ⎧ χ LT ≤ 1.0 = 0.3. November 09. When applying the method for rolled
profiles.0
Reduction factor For rolled section.all rights reserved.75
. λ LT.5 1 + α LT λ LT − λ LT. the LTB curve has to be selected from the table 6.
λ LT = 0.3.5 : For h/b = 597 / 220 = 2.4.Example: Simply supported beam with lateral restraint at load application point
⎧ M cr = 1.4 and β = 0.0 1 ⎪ χ LT = but ⎨ χ ≤ 1 2 2 ⎪ LT λ 2 φLT + φLT − β λ LT LT ⎩ where : φLT = 0.4
Created on Monday.0 may be given in the National Annex.77 × 2583 × ⎨ ⎩
2607000 81000 × 1188000 ⎫ − 3 × 100 + ⎬.2.0 + β λ LT
αLT is the imperfection factor for LTB.5 Table 6. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
EN 1993-1-1 § 6. 2009 This material is copyright .
740 < 1.3 (2)
f = 1 − 0.8
where : kc = 1 1.876
We obtain : χLT.mod = χLT / f = 0.4) + 0. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
The influence of the moment distribution on the design buckling resistance moment of the beam is taken into account through the f-factor :
EN 1993-1-1 § 6.75 × (0.all rights reserved.0 and β may be given in the National Annex.870 + (0. The recommended values are 0.33 − 0.427
Created on Monday.75 respectively.876 = 0.5 (1 − k c )1 − 2 λ LT − 0.8)2] = 0.33
f = 1 – 0.Example: Simply supported beam with lateral restraint at load application point
The values of λ LT.837)2 ] = 0.40 and 0.752 1.837) 2 = 0.5 [1 + 0.6
Simplified moment distribution :
Then : kc = So :
1 = 0.752) [1 – 2 (0.33 ×ψ and ψ = 0
but ≤ 1 EN 1993-1-1 Table 6.740
Then.740 < 1 / λ LT = 1.2.870) 2 − 0. November 09.75 × (0. we check : χLT = 0.837 − 0.49 (0. 2009 This material is copyright .0 and :
χLT = 0.3.837 – 0.
We obtain : and :
φLT = 0.870
1 0.740 / 0.845
.5 (1 – 0.
The value η may be conservatively taken as 1.22 kNm My.Ed / Vpl.0) × 10-6 = 942.z = 13700 – 2 × 220 × 17.5 mm2 But not less than η hw tw = 1.z.z ( f y / 3 )
.mod Wpl.22 = 0. 2009 This material is copyright .Rd = χLT.all rights reserved.35 < 72 × 0. which is given by: Av.0.12 < 1
Resistance to shear buckling EN 1993-1-1 Unstiffened webs with hw/tw greater than 72 ε / η should be checked for resistance to shear buckling and should be provided with transverse stiffeners § 6. hw / tw = (597 – 2 × 17.0 = 58.6 (6) at the supports.81 / 1. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
EN 1993-1-1 § 6.3.8 = 6609.6 (2)
Vz.894 < 1 OK EN 1993-1-1 § 6.12 mm2 OK
Created on Monday.0
§ 6.13 / 942.5 × (355 / 3 )/1000 = 1437 kN 1.5) / 9.Rd =
Av.z = A – 2 b tf + (tw + 2 r) tf Av.Rd = 172 / 1437 = 0.5 + (9.6 (3)
Vpl.8 = 57.Example: Simply supported beam with lateral restraint at load application point
Design buckling resistance moment Mb.Ed / Mb.Rd = 842.2.1
Shear Resistance In the absence of torsion.Rd = (0.3 So the shear buckling resistance does not need to be checked.2.2.2.2 × 562 ×9. November 09.845× 3141000 × 355 / 1.z.y fy / γM1 Mb.5 = 7011.8 + 2 × 24) × 17. the shear plastic resistance depends on the shear area.
1(1)B
w = 10.50 kN EN 1990 § 6. The National Annex may specify some limits.47 mm The deflection under Q is L/955 – OK Note 1 : The limits of deflection should be specified by the client.Example: Simply supported beam with lateral restraint at load application point
Serviceability Limit State verification SLS Combination qg = 1.5.85 mm The deflection under G+Q is L/347 – OK
Deflection due to Q :
FQ L3 48E I y
87500 × (10000)3 48 × 210000 × 82920 ×10 4
EN 1993-1-1 § 7. Here the result may be considered as fully satisfactory.50 = 234.all rights reserved.06 × (10000)4 + 48 × 210000× 82920 ×104 384 × 210000× 82920 ×10 4
w = 28. 2009 This material is copyright . the National Annex may specify limits concerning the frequency. November 09.2.06 kN/m FG + FQ = 147 + 87.2.3(1)B
Deflection due to G+Q :
(FG + FQ ) L3 48E I y
5qg L4 384EI y
Created on Monday. Note 2 : Concerning vibrations. EN 1993-1-1 § 7. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
234500× (10000)3 5 × 1.
2005 21.08.05.2005 17. Spain Resource approved by Technical Coordinator TRANSLATED DOCUIMENT This Translation made by: Translated resource approved by:
. November 09. UK 2.08.08.2005 17.08.all rights reserved.Example: Simply supported beam with lateral restraint at load application point
Example: Simply supported beam with lateral restraint at load application point SX007a-EN-EU
RESOURCE TITLE Reference(s) ORIGINAL DOCUMENT Name Created by Technical content checked by Editorial content checked by Technical content endorsed by the following STEEL Partners: 1. Use of this document is subject to the terms and conditions of the Access Steel Licence Agreement
Date 08/04/2005 11/05/2005
Valérie LEMAIRE Alain BUREAU
G W Owens Alain BUREAU A Olsson C Muller J Chica G W Owens
17. 2009 This material is copyright .2005 17.08.2005
4. Sweden 5. France 3.2005 17. Germany
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