Simply Supported Beam With Lateral Restraint

Document Ref: SX007a-EN-EU Sheet 1 of 10 Title

CALCULATION SHEET

Example: Simply supported beam with lateral restraint at load application point

Eurocode Ref EN 1993-1-1 Made by Valérie LEMAIRE Date April 2005 Checked by Alain BUREAU Date April 2005

Example: Simply supported beam with lateral restraint at load application point 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

1 1 5,0 m 5,0 m

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 (permanent loads)

• γQ = 1,50 (variable loads)

• γM0 = 1,0

• γM1 = 1,0

EN 1990

EN 1993-1-1

§ 6.1 (1)

Example: Simply supported beam with lateral restraint at load application pointC

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CALCULATION SHEET



Basic data Design a non composite floor beam of a multi-storey building according to the data given below. 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

7m

7m

5m 5m

Secondary beam

Calculated Beam

Concrete slab

5m 5m

150 mm

597 mm

Restraints to lateral buckling


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CALCULATION SHEET



• Span length : 10 m

• Secondary beam:

o Span length: 7 m

o Bay width : 5 m

• Slab depth : 15 cm

• Secondary beams 0,10 kN/m2

• Partitions and false ceiling: 0,50 kN/m2

• Imposed load : 2,50 kN/m2

• Concrete density : 24 kN/m3

• Steel grade : S355

Weight of the slab : 0,15 × 24 kN/m3 = 3,6 kN/m2

Try IPEA 600 – Steel grade S355

Depth h = 597 mm

Width b = 220 mm

Web thickness tw = 9,8 mm

Flange thickness tf = 17,5 mm

Fillet r = 24 mm

Mass 108 kg/m

z

z

y y

tf

tw

b

h

Euronorm

19-57

Section area A = 137 cm2

Second moment of area /yy Iy = 82920 cm4

Second moment of area /zz Iz = 3116 cm4

Torsion constant It = 118,8 cm4

Warping constant Iw = 2607000 cm6

Elastic modulus /yy Wel,y = 2778 cm3

Plastic modulus /yy Wpl,y = 3141 cm3


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CALCULATION SHEET



Self weight of the beam : qG = (108 × 9,81) × 10-3 =1,06 kN/m

Permanent load :

FG = (3,6 + 0,10 + 0,50)× 5,0 × 7,0 = 147 kN

Variable load (Imposed load) :

FQ = 2,50 × 5,0 × 7,0 = 87,5 kN

ULS Combination :

γG qG = 1,35 × 1,06 = 1,43 kN/m

γG FG + γQ FQ = 1,35 × 147 + 1,50 × 87,5 = 329,70 kN

EN 1990

§ 6.4.3.2 (6.10)

Moment diagram

M 842,13 kNm Maximal moment at mid span :

My,Ed = 0,125 × 1,43 × 10,002 + 0,25 × 329,70 × 10 = 842,13 kNm

Shear force diagram

V

172 kN 164,85 kN

Maximal shear force at supports :

Vz,Ed = 0,50 × 1,43 × 10,0 + 0,50 × 329,70 = 172 kN

Maximal shear force at mid-span :

Vz,Ed = 0,50 × 329,70 = 164,85 kN

Yield strength Steel grade S355

The maximum thickness is 17,5 mm < 40 mm, so : fy = 355 N/mm2

Note : The National Annex may impose either the values of fy from the Table 3.1 or the values from the product standard.

EN 1993-1-1

Table 3.1


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CALCULATION SHEET



Section classification :

The parameter ε is derived from the yield strength : 0,81 ][N/mm

235 2y

==f

ε

Outstand flange : flange under uniform compression

c = (b – tw – 2 r) / 2 = (220 – 9,8 – 2 × 24)/2 = 81,10 mm

c/tf = 81,1 / 17,5 = 4,63 ≤ 9 ε = 7,29 Class 1

EN 1993-1-1

Table 5.2

(sheet 2 of 3)

Internal compression part : web under pure bending

c = h – 2 tf – 2 r = 597 – 2 × 17,5 – 2 × 24 = 514 mm

c / tw = 514 / 9.8 = 52,45 < 72 ε = 58,32 Class 1

The class of the cross-section is the highest class (i.e the least favourable) between the flange and the web, here : Class 1 So the ULS verifications should be based on the plastic resistance of the cross-section.

EN 1993-1-1

Table 5.2

(sheet 1 of 3)

Moment resistance

The design resistance for bending of a cross section is given by :

Mc,Rd = Mpl,Rd = Wpl,y fy / γM0 = (3141 × 355 / 1,0) / 1000

Mc.Rd = 1115 kNm

My,Ed / Mc,Rd = 842,13 / 1115 = 0,755 < 1 OK

EN 1993-1-1

§ 6.2.5


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CALCULATION SHEET



Reduction factor for lateral torsional buckling To determine the design buckling resistance moment of a beam, the reduction factor for lateral torsional buckling must be determined. The following calculation determines this factor by calculation of the elastic critical moment for lateral torsional buckling.

Critical moment for lateral torsional buckling The critical moment may be calculated from the following expression :

( )( )

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++⎟⎟⎠

⎞⎜⎜⎝

⎛= g2

2g2

z2

t2

z

w

2

w2z

2

1cr ) (

zCzCI E

IGL kII

kk

L kI ECM c

c ππ

E is the Young modulus : E = 210000 N/mm2

G is the shear modulus : G = 81000 N/mm2

Lc is the distance between lateral restraints : Lc = 5,0 m

See NCCI

SN005

In the expression of Mcr, the following assumption should be considered :

k = 1 since the compression flange is free to rotate about the weak axis of the cross-section,

kw = 1 since there is no device to prevent the warping at the ends of the beam.

The C1 and C2 coefficients depend on the moment diagram along the beam segment between lateral restraints. It can be assumed that the diagram is linear, then :

C1 = 1,77 for k = 1

C2 zg = 0

Therefore :

( )

kN 258310(5000)

103116 210000 32

42

2z

2

=××××

= −ππ

cL kIE


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CALCULATION SHEET



××= 2583 1,77 crM 3.10 2583000

1188000 81000 1003116

2607000 −

⎭⎬⎫

⎩⎨⎧ ×

+×

Mcr = 1590 kNm

Non-dimensional slenderness The non-dimensional slenderness is obtained from :

0,837 101590

355 3141000

6cr

yypl,LT =

××

==M

fWλ

EN 1993-1-1

§ 6.3.2.2 (1)

For rolled profiles, 0,4 LT,0 =λ

Note : The value of LT,0λ may be given in the National Annex. The recommended value is 0,4.

So LT,0LT 0,837 λλ >=

EN 1993-1-1

§ 6.3.2.3(1)

Reduction factor

For rolled section, the reduction factor for lateral torsional buckling is calculated by :

⎪⎩

⎪⎨⎧

≤

≤

−+=

2LT

LT

LT

2LT

2LTLT

LT 1

1.0 but

1 λ

χ

χ

λβφφχ

where : ( )[ ] 1 0,5 LTLT,0LTLTLT

2λβλλαφ +−+=

EN 1993-1-1

§ 6.3.2.3 (1)

αLT is the imperfection factor for LTB. When applying the method for rolled profiles, the LTB curve has to be selected from the table 6.5 :

For h/b = 597 / 220 = 2,71 > 2 Curve c αLT = 0,49

0,4 LT,0 =λ and β = 0,75

EN 1993-1-1

Table 6.5

Table 6.3


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CALCULATION SHEET



Note : The values of LT,0λ and β may be given in the National Annex. The recommended values are 0,40 and 0,75 respectively.

We obtain : ( )[ ] 0,870 (0,837)0,75 0,40,837 0,49 1 0,5 2LT =×+−+=φ

and : 0,740 (0,837)0,75(0,870) 0,870

1 22LT =

×−+=χ

Then, we check : χLT = 0,740 < 1,0

and : χLT = 0,740 < 2LTλ / 1 = 1,427

The influence of the moment distribution on the design buckling resistance moment of the beam is taken into account through the f-factor :

( ) ([ ]2LTc 0,8 2 1 1 0,5 1 −−−−= λkf ) but ≤ 1

EN 1993-1-1

§ 6.3.2.3 (2)

where : kc = ψ×− 33,033,1

1 and ψ = 0 EN 1993-1-1

Table 6.6

Simplified moment distribution :

Lc

Then : kc = 33,11 = 0,752

So : f = 1 – 0,5 (1 – 0,752) [1 – 2 (0,837 – 0,8)2] = 0,876

We obtain : χLT,mod = χLT / f = 0,740 / 0,876 = 0,845


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CALCULATION SHEET



Design buckling resistance moment

Mb,Rd = χLT,mod Wpl,y fy / γM1

Mb,Rd = (0,845× 3141000 × 355 / 1,0) × 10-6 = 942,22 kNm

My,Ed / Mb,Rd = 842,13 / 942,22 = 0,894 < 1 OK

EN 1993-1-1

§ 6.3.2.1

Shear Resistance In the absence of torsion, the shear plastic resistance depends on the shear area, which is given by:

Av,z = A – 2 b tf + (tw + 2 r) tf

Av,z = 13700 – 2 × 220 × 17,5 + (9,8 + 2 × 24) × 17,5 = 7011,5 mm2

But not less than η hw tw = 1,2 × 562 ×9,8 = 6609,12 mm2 OK

EN 1993-1-1

§ 6.2.6 (3)

Shear plastic resistance

kN 1437 1,0

)/10003 / (3555,0117 )3 / (

M0

yzv,Rdz,pl, =

×==

γfA

V

Vz,Ed / Vpl,z,Rd = 172 / 1437 = 0,12 < 1 OK

EN 1993-1-1

§ 6.2.6 (2)

Resistance to shear buckling

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 at the supports.

The value η may be conservatively taken as 1,0.

hw / tw = (597 – 2 × 17,5) / 9,8 = 57,35 < 72 × 0,81 / 1,0 = 58,3

So the shear buckling resistance does not need to be checked.

EN 1993-1-1

§ 6.2.6 (6)


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CALCULATION SHEET



Serviceability Limit State verification

SLS Combination qg = 1,06 kN/m

FG + FQ = 147 + 87,50 = 234,50 kN

EN 1990

§ 6.5.3

Deflection due to G+Q :

y

gQG

EILq

IELFF

w3845

48

)(

4

y

3

++

=

4

4

4

3

108292021000038410000)(1,06 5

108292021000048(10000)234500

×××××

+×××

×=w

w = 28,85 mm

The deflection under G+Q is L/347 – OK

Deflection due to Q :

108292021000048

(10000)87500 48

4

3

y

3

××××

==IE

LFw Q

w = 10,47 mm

The deflection under Q is L/955 – OK

Note 1 : The limits of deflection should be specified by the client. The

National Annex may specify some limits. Here the result may be considered as fully satisfactory.

EN 1993-1-1

§ 7.2.1(1)B

Note 2 : Concerning vibrations, the National Annex may specify limits concerning the frequency.

EN 1993-1-1

§ 7.2.3(1)B


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Example: Simply supported beam with lateral restraint at load application point SX007a-EN-EU

Quality Record

RESOURCE TITLE Example: Simply supported beam with lateral restraint at load application point

Reference(s)

ORIGINAL DOCUMENT

Name Company Date

Created by Valérie LEMAIRE CTICM 08/04/2005

Technical content checked by Alain BUREAU CTICM 11/05/2005

Editorial content checked by

Technical content endorsed by the following STEEL Partners:

1. UK G W Owens SCI 17.08.2005

2. France Alain BUREAU CTICM 17.08.2005

3. Germany A Olsson SBI 17.08.2005

4. Sweden C Muller RWTH 17.08.2005

5. Spain J Chica Labein 17.08.2005

Resource approved by Technical Coordinator

G W Owens SCI 21.05.2005

TRANSLATED DOCUIMENT

This Translation made by:

Translated resource approved by:


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Documents

Simply Supported Beam With Lateral Restraint