Example Simply Supported IPE Profile Purlin

Document Ref: SX021a-EN-EU Sheet 1 of 10 Title

Example: Simply supported IPE profile purlin

Eurocode Ref EN 1993-1-1 Made by Mladen Lukic Date Jan 2006

CALCULATION SHEET

Checked by Alain Bureau Date Jan 2006

Example: Simply supported IPE profile purlin This example gives the details of the verification according to EN 1993-1-1 of a simply supported purlin under a uniform loading. The purlin is an I-section rolled profile which is laterally restrained by the steel sheeting.

Scope The example covers the design of a hot rolled profile beam used as purlin, under bending about the strong axis. The purlin is laterally restrained by the steel sheeting. The example includes:

• the classification of the cross-section,

• the calculation of the bending resistance, including the exact calculation of the elastic critical moment for lateral torsional buckling,

• the calculation of the shear resistance,

• the calculation of the deflections at serviceability limit state.

The verification of the steel sheeting is out of the scope of this worked example.

Partial safety factors

• γGmax = 1,35 (permanent loads)

• γGmin = 1,00 (permanent loads)

• γQ = 1,50 (variable loads)

• γM0 = 1,0

• γM1 = 1,0

EN 1990

EN 1993-1-1 § 6.1 (1)

Loading The uniformly distributed loading includes:

• self-weight of the beam

• roofing 0,240 kN/m2

• Snow load : 0,618 kN/m2

• Wind load (uplift) : 0,730 kN/m2

Example: Simply supported IPE profile purlinC

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CALCULATION SHEET


Basic geometrical data

Purlin

• Span length: L = 7,20 m

• Spacing: s = 3,00 m

Try IPE 180 – Steel grade S275 Euronorm z

z

y y

tf

tw

b

h

Depth h = 180 mm 19-57

Width b = 91 mm

Web thickness tw = 5,3 mm

Flange thickness tf = 8 mm

Fillet r = 9 mm

Mass 18,8 kg/m

Section area A = 23,9 cm2

Second moment of area /yy Iy = 1317 cm4

Second moment of area /zz Iz = 100,9 cm4

Torsion constant I = 4,79 cm4t

Warping constant Iw = 7430 cm6

Elastic modulus /yy Wel,y = 146,3 cm3

Plastic modulus /yy Wpl.y = 166,4 cm3

Sheeting – Steel grade S350

100 50 25

25

1000

40

Thickness of the sheeting t = 0,7 mm


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CALCULATION SHEET


It is possible to calculate the shear stiffness of trapezoidal sheeting connected to the purlin at each rib and connected in every side overlap according to the EN 1993-1-3. The formula is:

EN 1993-1-3 § 10.1.1(10)

( )w

3roof

3 s1050t1000 h

bS +=

The following expressions are calculated: 2333 mm 586,07,0t ==

3233roof mm 243720010501050 =×+=+ b

7540

3000s

w

==h

Then the shear stiffness is:

m / kNm 10680107524358601000 3 =××××= -,S

Capacity of the sheeting to restrain the purlin

Continuous lateral restraint EN 1993-1-1 §BB.2.1If the following requirement is met, the purlin may be regarded as being

laterally restrained in the plane of the sheeting: EN 1993-1-3 § 10.1.1

( )22

2z

2

t2w

2

min702 hL

hEIGILEISS ×⎟⎟

⎠

⎞⎜⎜⎝

⎛++=≥ππ

(6)

One calculates:

2-92

62

2w

2

kNm 9712,0107200

107430 210000 =×

×××=ππ

LIE

2-94t kNm 3,86910104,7980770 =×××=GI

( ) 2-92

242

2

2z

2

kNm 2683,0107200

0910100,9 210000 2 =×

××××=ππ

LhIE

Then the minimum stiffness is:

( ) kNm/m 706918,0702653,03,8690,2971 2min =×++=S


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CALCULATION SHEET


Therefore:

S = 10680 kNm/m > S = 9706 kNm/m min

The condition is thus met and the steel sheeting can be considered as sufficiently stiff to restrain the purlin. Note that requirements regarding the resistance and the stability of the steel sheeting are given in the EN 1993-1-3, these verifications are out of the scope of this worked example.

Loads

• Purlins: G = (18,8 × 9,81) × 10-3 = 0,184 kN/m 1

• Roofing: G = 0,240 × 3,00 = 0,720 kN/m 2

• Permanent load: G = G + G = 0,904 kN/m 1 2

• Snow: Q = 0,618 × 3,00 = 1,854 kN/m S

• Wind (uplift): QW = 0,730 × 3,00 = 2,190 kN/m

Note: It is assumed that the roof slope is so low that the decomposition of the vertical loads into two components, one parallel to the web and the other one parallel to the plane of the roof, is not needed here.

ULS Combinations:

• Sagging: EN 1990γGmax G + γQ QS = 1,35 × 0,904 + 1,50 × 1,854 = 4,00 kN/m § 6.4.3.2

• Hogging:

γGmin G + γQ Q = 1,00 × 0,904 – 1,50 × 2,190 = -2,38 kN/m S

Moment diagram

• Maximum sagging moment at mid span:

M = 0,125 × 4,00 × 7,202 = 25,92 kNm y,Ed

M

My,Ed

• Maximum hogging moment at mid span: 2M = 0,125 × (–2,38) × 7,20 = –15,42 kNm y,Ed

M

My,Ed


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CALCULATION SHEET


Shear force diagram

• Maximum shear force at supports for sagging:

V = 0,5 × 4,00 × 7,20 = 14,4 kN z,Ed

• Maximum shear force at supports for hogging:

V = 0,5 × 2,38 × 7,20 = 8,57 kN z,Ed

SLS Combination EN 1990 § 6.5.3• Sagging:

G + Q = 0,905 + 1,854 = 2,759 kN/m

• Hogging:

G + Q = 0,905 – 2,190 = –1,285 kN/m

Yield strength

Steel grade S275

The maximum thickness is 8 mm < 40 mm, so: fy = 275 N/mm2 EN 1993-1-1 Table 3.1Note: The National Annex may impose either the values of fy from the

Table 3.1 or the values from the product standard.

Section classification :

EN 1993-1-1The parameter ε is derived from the yield strength: 0,92

][N/mm 235 2

y

==f

ε

Table 5.2

(sheet 2 of 3)Outstand flange: flange under uniform compression

c = (b – tw – 2 r) / 2 = (91 – 5,3 – 2 × 9)/2 = 33,85 mm

c/tf = 33,85 / 8,0 = 4,23 ≤ 9 ε = 8,28 Class 1

Internal compression part: web under pure bending EN 1993-1-1 Table 5.2 c = h – 2 tf – 2 r = 180 – 2 × 8 – 2 × 9 = 146 mm (sheet 1 of 3) c / tw = 146 / 5,3 = 27,5 < 72 ε = 66,24 Class 1

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


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CALCULATION SHEET


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

Mc,Rd = Mpl,Rd = Wpl,y fy / γM0 = (166,4 × 275 / 1,0) × 10-3

Mc.Rd = 45,76 kNm

My,Ed / Mc,Rd = 25,92 / 45,76 = 0,566 < 1 OK

EN 1993-1-1 § 6.2.5

No verification against lateral torsional buckling is needed since the compressed flange is considered as fully restrained laterally.

Moment resistance – Hogging

The bending resistance criterion is:

My,Ed / Mc,Rd = 15,42 / 45,76 = 0,337 < 1 OK

EN 1993-1-1 § 6.2.5

However, resistance to lateral torsional buckling should be checked since the compressed flange is now unrestrained laterally.

Reduction factor for lateral torsional buckling

To determine the design buckling resistance moment of the purlin, the reduction factor for lateral torsional buckling is determined on the basis of the elastic critical moment for lateral torsional buckling taking into account the lateral restraint of the tension flange.

Note: The torsional restraint due to sheeting is also physically present and could be taken into account. In this spreadsheet, it will be supposed as insignificant and won’t be considered.

Critical moment for lateral torsional buckling The critical moment may be calculated using the software LTBeam developed by the CTICM. This software allows the designer to take into account special restraint conditions such as a continuous lateral restraint along the tension flange and its position from the shear centre.

Note : LTBeam is free and can be downloaded on the web site

www.cticm.com

SN011


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CALCULATION SHEET


The continuous lateral restraint is assumed to be located at 90 mm above the shear centre.

The critical moment calculated by LTBeam is:

Mcr = 27,20 kNm

Non-dimensional slenderness

The non-dimensional slenderness is obtained from: EN 1993-1-1 §6.3.2.2

1,297 27,20

10 275 166400

-6

cr

yypl,LT =

××==

MfW

λ (1)

EN 1993-1-1For 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.

§ 6.3.2.3(1)

LT,0LT 1,297 λλ >=So

Reduction factor

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

⎪⎩

⎪⎨⎧

≤

≤

2LT

LT

LT1

1.0

λχ

χ

1 2LT

2LTLT

LTλβφφ

χ−+

= but

⎥⎦⎤

⎢⎣⎡ +⎟

⎠⎞⎜

⎝⎛ −+=

2LTLT,0LTLTLT 1 0,5 λβλλαφ where:

EN 1993-1-1αLT is the imperfection factor for lateral torsional buckling. When applying the method for rolled profiles, the LTB curve has to be selected from the table 6.5:

Table 6.5

Table 6.3 For h/b = 180 / 91 = 1,97 ≤ 2 Curve b (αLT = 0,34)

and β = 0,75 0,4 LT,0 =λ

LT,0λNote: the values of and β may be given in the National Annex. The recommended values are 0,4 and 0,75 respectively.


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CALCULATION SHEET


We obtain: ( )[ ] 1,283 1,2970,75 0,41,297 0,34 1 0,5 2LT =×+−+=φ

0,525 1,2970,751,2831,283

1 22LT =

×−+=χ and:

Then, we check: χLT = 0,525 < 1,0 OK 2LTλ / 1 = 0,525 < = 0,594 and: χLT

Design buckling resistance moment

Mb,Rd = χLT Wpl,y fy / γM1

Mb,Rd = (0,525 × 166400 × 275 / 1,0) × 10-6 EN 1993-1-1 = 24,02 kNm § 6.3.2.1 My,Ed / Mb,Rd = 15,42 / 24,02 = 0,642 < 1 OK

Shear Resistance

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

EN 1993-1-1 § 6.2.6 (3)

Av,z = A – 2 b tf + (tw + 2 r) tf

Av,z = 2390 – 2 × 91 × 8 + (5,3 + 2 × 9) × 8 = 1120 mm2

Shear plastic resistance EN 1993-1-1 § 6.2.6

kN 177,8 10 1,0

)3 / (2751120 )3 / (

-3

M0

yzv,Rdz,pl, =×

×==

γfA

V (2)

= 14,4 / 177,8 = 0,081 < 1 OK V / Vz,Ed pl,z,Rd

Note: No M-V interaction has to be considered since the maximum moment is obtained at mid-span and the maximum shear force is obtained at supports.

EN1993-1-1 § 6.2.8

Note that the verification to shear buckling is not required when:

EN 1993-1-1 hw / tw ≤ 72 ε / η § 6.2.6 η may be conservatively taken as 1.0 (6)

hw / tw = (180 – 2 × 8) / 5,3 = 30,9 < 72 × 0,92 / 1,0 = 66,24


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CALCULATION SHEET


Deflection – Sagging

EN 1993-1-1Deflection under G § 7.2.1

mm 11,4 101317210000384

(7200)904,05 384

5 4

4

y

4

=×××

××==

IELGw = L/632

Deflection under Q

mm 23,5 101317210000384

(7200)854,15 384

5 4

4

y

4

=×××

××==

IELQw = L/306

The deflection under (G+Q) is L/206

Deflection – Hogging

Deflection under G EN 1993-1-1 § 7.2.1

mm 11,4 101317210000384

(7200)904,05 384

5 4

4

y

4

=×××

××==

IELGw = L/632

Deflection under Q

( ) mm 27,7 101317210000384

(7200)190,25 384

5 4

4

y

4

−=×××

×−×==

IELQw = L/260

The deflection under (G+Q) is L/442

Note: the limits of deflection should be specified by the client. The National Annex may specify some limits. Here the result may be considered as satisfactory.

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Example: Simply supported IPE profile purlin SX021a-EN-EU

Quality Record

RESOURCE TITLE Example: Simply supported IPE profile purlin

Reference(s)

ORIGINAL DOCUMENT

Name Company Date

Created by Mladen Lukic CTICM 08/12/05

Technical content checked by Alain Bureau CTICM 08/12/05

Editorial content checked by

Technical content endorsed by the following STEEL Partners:

1. UK G W Owens SCI 7/04/06

2. France A Bureau CTICM 7/04/06

3. Sweden B Uppfeldt SBI 7/04/06

4. Germany C Müller RWTH 7/04/06

5. Spain J Chica Labein 7/04/06

Resource approved by Technical Coordinator

G W Owens SCI 17/07/06

TRANSLATED DOCUMENT

This Translation made and checked by:

Translated resource approved by:


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Example Simply Supported IPE Profile Purlin

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