Portal Frame EC3

Job No. Sheet 1 of 23 Rev A Job Title Worked Example

Subject European Portal Frame

Made by AWC Date November 2002 Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET

Client

Checked by JBL Date January 2003

P:\CMP\Cmp657\European Portal Frame Worked Example.Version 3.doc

Worked Example

on the Preliminary Design of a

EUROPEAN PORTAL FRAME to Eurocode 3

by Dr Andrzej Czechowski

Ascot, November 2002

Preliminary Design of a European Portal Frame to Eurocode

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http://sefie.steelbiz.org/DiscussSteelbizContent.aspx?ResourceID=212154




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1. INTRODUCTION

2. DESIGN BRIEF

3. LOADS 3.1 Permanent actions on the roof 3.2 Snow load 3.3 Wind load 3.4 ‘Second order’ sway action 3.5 Combinations of actions

4. STATICS (ELASTIC ANALYSIS) 4.1 UDL over two spans 4.2 Horizontal force case 4.3 Internal forces and reactions

5. MEMBER DESIGN 5.1 Cross-section resistance – definitions 5.2 Columns 5.3 Rafters 5.4 Deformations – serviceability check


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1. INTRODUCTION

• This is a manual example of the preliminary design of a typical two-span portal frame in steel. The limit state design is carried out in accordance with Eurocodes[1,2]. (U.K. standards [3,4] are used for loading, because of the current unavailability of EC1).

• Elastic analysis of the main frame is performed using the traditional but effective, Prof. Kleinlogel’s Rahmenformeln [5]. In the analysis, initial sway imperfection and second order effects are directly taken into account, so stability of the frame is checked in the non-sway buckling mode.

• The constructional system finally adopted is a very simple and economical solution.

Notation (major symbols) - to Eurocodes[1, 2]

Abbreviations used in the Example:

BMD - Bending Moment Diagram

FB - Flexural Buckling

LTB - Lateral Torsional Buckling

LC - Load Combination

RUR - Resistance Utilization Ratio

SLS - Serviceability Limit State

ULS - Ultimate Limit State

UDL - Uniformly Distributed Load

VA - Variable Action


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2. DESIGN BRIEF Basic data and criteria for structural design

Two-bay portal-frame building – a warehouse (located somewhere in the UK)

• Building size

Width : 2 × 20 = 40 m (column centres)

Length : 60 m (frame centres)

Internal clear height : 6,50 from TOF (± 0,00)

• Loads

Wind : - to BS6399-2 (Ve = 100 km/h)

Snow : - to BS6399-3 (so = sb = 0,75 kN/m2)

Service load : 0,10 kN/m2

• Frame geometry

Span : L = 20 m

Spacing : 6 m

Column base level : –0,20 m

Structural height : h = 7,0 m (base to centre line at knee)

Roof pitch : 5% (~3o)

• Cladding – metal sheeting (panels on purlins/girts)

• Deformation limits for the frame (serviceability)

Vertical deflection : L/200

Horizontal displacement (sway) : H/150


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General arrangement of main frames

B C

12

34

5

C L

89

10

11 C L

10 x

6.0

m =

60

m

3° 3° 3° 3°

7.0

m

0.2

0 m

20 m

±0.00 (TOF)

CL

8.0

m

PLAN

A

20 m

CROSS SECTION

NOTE: All column bases are to be nominally pinned bases.


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3. LOADS

3.1 Permanent actions on the roof

• Characteristic values (self-weights):

G1 - rafters & purlins 0,20 kN/m2

G2 - roofing 0,20 kN/m2

G1 - servicess 0,10 kN/m2

Total: Gk = 0,50 kN/m2

• Design values:

Gd,inf (favourable) = γ f,int Gk = 1.00 × 0.50 = 0,50 kN/m2

Gd,sup (unfavourable) = γ f,sup Gk = 1.35 × 0.50 = 0,68 kN/m2

3.2 Snow load

(to BS6399-3, Ref [4])

• Snow load on the roof – characteristic and design values:

Qk = sd = µ1so = 0,8 × 0,75 = 0,60 kN/m2

Qd = γq Qk = 1,50 × 0,60 = 0,90 kN/m2 (leading VA)

• Local drifting in valley (as alternative and exceptional case)

Snow load shape coefficient : µ = 2h/so = 2 × 0,5/0,75 = 1,33

1

1

0.5 L 0.5 L

h=0.5 m3°α≈

' = 1.33

= 0.8 (uniform snow)µ

µ

NOTE: This case will not govern since ½ µ1 < µ.


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3.3 Wind load

(Standard Method to BS6399-2, Ref [3])

• Basic data:

Dynamic pressure: qs = 45 kN/m2 (Ve ≅ 100 km/h)

Roof pitch angle α ≅ 3o

Building overall dimensions:

W (width) ≅ 40,0 m

L (length) ≅ 60,0 m

H (height) ≅ h + 1,0m = 7,0 + 1,0 = 8,0 m

• Wind cases – pressures coefficients ([3]/Tables 5 & 8):

a) Transverse wind b) Longitudinal wind

b/2 = 8.0 m

BC

D = L

-0.7

+0.6 -0.5

±0.2

b/2 = 8.0 m

-0.7

+0.6

C

-0.5

±0.2

D = L

D/H 4≥

b = 2H = 16.0 m

(C = -0.3)pi (C = -0.3)pi

B = L

B = W

(-0.8)

(-0.8)

w∆

D

(-)

(-)

NOTE: In case D, with regard to the roof, two subcases are to be considered:

pC′ = – 0,2 (suction, Cpi = 0) or pC ′′ = + 0,2 – (–0,3) = 0,5 (pressure & underpressure)

• Tributary areas and size effect coefficients ([3]/Fig. 4: B-line):

Walls A = 6 × 8 = 48 m m1086a 22 =+= Ca = 0,95

Roof As = 6 × 40 = 240 m m40406a 22 ≅+= Ca = 0,86


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• Wind loads on the roof in Zone B/C:

UDL : ∆w = γ0 γQ qs Cq Cp × 6.0 m [kN/m]

pC′ = + 0,5: ∆w1 = 0,6 × 1,50 × 0,45 × 0,85 × 0,5 × 6,0 ≅ 1,0 kN/m

pC ′′ = – 0,2: ∆w2 = 1,00 × 1,50 × 0,45 × 0,85 × 0,2 × 6,0 ≅ 0,7 kN/m

Coexistent with ∆w2 the extra up-list force in Zone C

Ca = 0,95; Ac = 48m2

Pc = 1,50 × 0,45 × 0,95 × [–0,7 – (–0,2)] × 48m = –15,4 kN

• Horizontal wind forces (for one frame, case a):

Overall surface load on both windward and leeward walls:

P = 0,85 Σqs Cpe Ca A = 0,85 Ca qs A ΣCpe

= 0,85 × 0,95 × 0,45 × 48 × (0,6 + 0,5) = 19,2 kN

NOTE: Factor 0,85 accounts for non-simultaneous action between faces

Frictional force on the roof:

Ps = qs Cf As Ca = 0,45 × 0,01 × 240 × 0,85 = 0,92 kN

Resultant horizontal wind force at rafters (see 4.2):

Hw= kN0,1292,02,190,85,4PP

HH

s ≅+×=+×∆

NOTE: ∆H – upper ‘tributary’ height: ∆ ≅ ½h + 1.0m = 3.5 + 1.0 = 4.5m

Design values of horizontal forces:

Hd1 = ψoγQ Hw = 0,6 × 1,50 × 12,0 ≅ 11,0 kN (accompanying VA)

3.4 ‘Second order’ sway action

(to prEN 1993, Ref [2]/5.2 & 5.3)

• Initial sway imperfection for global analysis:

φ = φo α h αm

φ = 1/200

αh = 75,07/2h/2 ==

αm = 82,0)3/11(5,0)m/11(5,0 =+×=+


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φ = 2001 × 0,75 × 0,82 = 0,005 × 0,61 = 0,0031

• Equivalent horizontal force due to imperfection:

Hφ = φVEd;

Total max. gravity load (see 3.6/LC1)

VEd = 2w1L = 2 × 9,5 × 20,0 = 380 kN

Hφ = 0,0031 × 380 = 1,2 kN

• Total initial (first order) horizontal action:

Ho = Hd1 + Hφ = 11,0 + 1,2= 12,2 kN

• Critical load factor:

αcrit =

odh

VH

Ed

o

δo = δ(Ho) = Ho/Kh = 12,2/3,2 = 3,8 cm (see 4.2)

αcrit = 8,3

700380

2,12 × = 5,9

NOTE: Since α crit < 10, second order effects (action) cannot be disregarded.

• Total second order (amplified) horizontal action to be used in the global analysis:

19,5

9,52,121

HHcrit

crito

II1 −

×=−

=α

α = 12,2 × 1,20 = 14,7 kN

3.5 Combinations of actions

(to EN 1990, Ref [1])

• The following actions are considered:

G – permanent load (as unfavourable or favourable action)

Q – snow load (as leading variable action)

W – wind load (as accompanying or leading variable action)

• Loading on the frame is represented by two parameters:

- UDL over two spans: w [kN/m]


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- Horizontal force: H [kN]

• ULS load combinations ([1]/Table A1.2B):

(1)Max. gravity load + wind (1,35G + 1,50Q + 0.6 × 1,50W):

w1 = (Gd,sup + Qd)× 6,0m= (0,68 + 0,90) × 6,0 = 9,5 kN/m

I1w = w1 + ∆w1 = 9,5 + 1,0 = 10,5 kN/m (∆w1 – see 3.3)

II1H = 14,8 kN (see 3.4)

(2)Min. gravity load + wind (1,00G + 1,5W):

w2 = Gd,inf × 6,0m = 0,50 × 6,0 = 3,0 kN/m

12w = w2 - ∆w2= 3,0 – 0,7 = 2,3 kN/m (∆W2 – see 3.3)

H2 = Hd2 = 18,0 kN

• SLS characteristic load combinations ([1]/Table A.1.4):

(3)Max. gravity load (G + Q) – for vertical deflection

w3 = (Gk + Qu) × 6,0m = (0,50 + 0.60) × 6,0 = 6,6 kN/m

(4)Wind load (W) – for horizontal displacement (sway)

H4 = HW = 12,0 kN


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4. STATICS (ELASTIC ANALYSIS) (From SDM Ref [5])

4.1 UDL over two spans

([5]/Frame VIII)

A+ +

B B' A'

D D'E

IR

IC

C

w (kN/m)

L = 2s L = 2s

h

_ _ _ __ _

HD H 'DV 'VD DVE

Frame data: L = 20,0 m h = 7,0 m f = 0.5 m Ic ≅ 2IR

• Constants:

φ = 07,00,75,0

hf == but φ ≅ 0 accepted for simplification

xI = x = 35,0107

21

sh

II

c

R =⋅=

N1 = 8x + 12 = 14,8; 4N1 = 59.2

• Influence coefficients (for φ = 0 and x = 0.35):

n11 = 1N

2 = 0,1351; n12 = n21 = 1N

1 = 0,0676; n22 = 1N2x + = 0,1588

r1 = n11 + n21 = -0,067; r2 = n12 + n22 = 0,0912

L1 = 1N4

2 = 0,0338; L2 = 1N42x2 + = 0,0456

m1 = L1 – 2r1 = 0,0673; m2 =L2 +

2r2 = 0,0912

c1 = 14

6N

= 0,101; c2 = 1N46x6 + = 0,137


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• Bending moments:

MA = 4

wLM21

A −= (m1 + 2c1) = –0,067wL2

Mc =4

wL2− (m1 + 2c2) = – 0,091wL2

MB = 2

MM8

wLM cA21

B++−= = 0,046wL2

• Reactions:

V1 = (–Ma + Mc)/L = – 0,024wL; V2 = – V1 = 0,024wL

Vc1 = – V1 + wL/2 = 0,524wL; Vc2 – V2 + wL/2 = 0.524wL

VE = Vc1 + Vc2 = 1,048wL

VD = 1DV = 0,500wL – 0,024wL = 0,476wL

HD = 1DH = MA/h = 0,191wL

• Mid-span deflection:

u(M) = 21M;

EILM125,0

EI8LM 22

== (MA + Mc) = –0,079wL2

u(w) = EI

LM104,0EI384

wL5 2o

4= ; Mo = 0,125wL2

u = u(M) + u(w) = (–0,125 × 0,79 + 0,104 × 0,125)EIwL0031,0

EIwL 44

=

u = ΦwW; where; W = wL, Φw(flexibility) = EI

L0031,0 3

• For I = IR (IPE 360) = 16270cm4:

Φ W = 162700,2100020000031,0 3

×× = 0,073cm/kN; (Kw =

w

1Φ

= 13,7 kN/cm)


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4.2 Horizontal force case

([5]/Frame II)

B A'

D E

C+

V 'DD

HD D' DH '

VE

B'

__

L LV

h

L

δ

H = 2P A

hI C

IR

w

wh

P

• Bending moments:

MA = Hh21M1

A =− = Ph

• Reactions:

VD = 1DV = Pw/L = 0,35P

HD = H21H1

D = = P

• Horizontal deflection – sway stiffness (to [6]):

δo = δR + δ c = R

2

C

3

r

a

E6HLh

EI3Ph

EI3LhM =+ (1 + k) ⇔ H/Ko

δb = R

2

c

3

c

3

EIkHLh42,1

EI3Ph

EIPh

25 =+ ⇔ H/Kb; k =

LIhI

c

R

NOTE: Kb is additive stiffness component due to the rotational stiffness of the nominally pinned base (= 10% column stiffness)

KH = Ko + Kb =

+

+=+

+ k7,0

k16

LhEI

kLhEI7,0

)k1(LhEI6

2R

2R

2r


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For k = 207

21 × = 0,175 and IR (IPE360) = 16270cm3:

KH = 9,1 × 27002000162700,21000

×× = 3,2 kN/cm; (Φ H =

HK1 = 0,31 cm/kN)


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4.3 Internal forces and reactions

Extreme design values

• Bending moments [kNm] at critical sections:

B.M. definition ? Splice at x1 ? Splice at x2 ?

ξ = x/L = 0,00 0,15 0,50 0,80 1,00

Load combination 1 (see 3.5)

M( I1w = 10,5kN/m) – 281 – 29,4 – 193 – 25,2 – 382

M( II1H = 14,7kN) 51,5 44,2 25,8 10,4 0

1a : M( I1w ) + M( II

1H ) – 229 – 14,8 219 14,8 – 382

1b : M( I1w ) – M( II

1H ) – 333 – 44,2 167 – 35,6 – 382

Load combination 2 (see 3.5)

M( I1w = 2,3kN/m) – 61,6 – 6,4 42,3 – 5,5 – 83,7

M( 2H = 18,0kN) 63,0 53,6 31,5 12,6 0

2a : M( I2w ) + M(H2) 1,4 47,2 73,8 7,1 – 83,7

2b : M( I2w ) – M(H2) – 125 – 60,0 10,8 – 181,1 – 83,7

• M(w,x) = MA + Lx (Mc –MA) + 2wL

Lx1

L2x

−

= [– 0,067 – 0,024 ξ + 21 ξ (1 – ξ )] wL2 = [f2(ξ ) + f3(ξ )] wL2, see 4.1

• M(H,x) = MA(H)

−

Lx1 =

21 (1 – ξ ) Hh = f1(ξ ) Hh, see 4.2

f1(ξ ) = 21 (1 – ξ ) 0,500 0,425 0,250 0,100 0

f2(ξ ) = ξ f1(ξ) 0 0,064 0,125 0,080 0

f3(ξ ) = – (0,067 + 0,024 ξ ) – 0,067 – 0,071 – 0,079 – 0,086 – 0,091

f2(ξ ) + f3(ξ ) – 0,067 – 0,007 0,046 – 0,06 – 0,091

NOTE: M(w,x) = 0 for x = x I0 = 0.17L and x = x II

0 = 0,78L


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• Bending moment envelope:

A

M(-)

M(+)

-60.0

47.2

219

-35.614.8

-333-382

x

BMD (kNm)

x =0.15L

x =0.80L2

L = 20 m

1

CB

NOTE: x1 and x2 are potential splice locations near contraflexure points.

• Reactions:

A'

D E

C

w (kN/m)

V 'VD DVE

DD'

h

(P )C B B'A

H=2P

L L

w =1.2 kN/mo

H '

Leeward suction (C = -0.5)le

(a)‘Windward’ column (LC2a w2 = 2,3 kN/m; p = 21 H2 = 9,0 kN; Pc = 15,4 kN)

min Vd = 0,476 I1w L – 0,35P – Pc = 0,476 × 2,3 × 20 – 0,35 × 9,0 – 15,4 = 3,3 kN

NOTE: There will be no uplift force on footing (min Vd > 0)

(b)Mid column (LC1: w1 = 10,5 kN/m)

max Ve = 1,048 I1w L= 1,048 × 10,5 × 20 = 220 kN

(c) ‘Leeward’ column (LC1b: I1w = 10,5 kN/m); P = 2

1 HD1 = 5,5 KN)

max 1DV = 0,476 I

1w L + 0,35P = 0,476 × 10,5 × 2070,35 × 5,5 = 102 kN

max 1DH = 0,191 I

1w L + P + 0,5woh = 0,191 × 10,5 × 20 + 5,5 + 4,2 = 49.8 kN

NOTE: In above calculations only horizontal force from external (wind) load is used, as contributing to the net reactions


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5. MEMBER DESIGN (to prEN 1993-1-1, Ref. [2])

5.1 Cross-section resistance – definitions

(Class 1/2 I-sections)

Bending: Mc,Rd = Mpl,Rd = fyWpl/γM1

Wpl – plastic modulus

Shear Vc,Rd = Vpl,Rd = 3yf Av/γM1

Av – shear area; Av = A – 2bftf (for V = Vy)

Compression Nc,Rd = Npl,Rd = fyA/γm1,

where fy = specified yield stress [MPa]

γM1 = partial safety factor: γM1 = 1,00

NOTE 1: In the combined load cases (M, V, N), the design moment resistance need not be reduced, provided VEd ≤ Vo = 0,5Vpl,Rd and NEd ≤ No = min [0,25Npl,Rd, 0,5fyhwtw].

(In the present example, where bending predominates, both above conditions will be satisfied at all sections)

NOTE 2: According to Eurocode convention ‘strong’ axis will be denoted by y – y and ‘weak’ axis by z – z

NOTE 3: Cross-sectional geometrical properties are from Ref [7].

5.2 Columns

5.2.1 External Columns

Try IPE 450 (g = 77,6 kg/m) – steel S235

• Section properties and resistances:

A = 98,8cm2; Av = 98,8 – 2 × 19 × 1,46 = 43,3cm2;

Iy = 33740cm4; Wpl,y = 1702cm3; iy = 18,5 cm; Iz = 1680cm3; iz = 4,12cm

Mpl,Rd = 235 × 1702/103 = 400 kNm

Vpl,Rd = (235/√3) × 43,3/10 = 588 kN

Npl,Rd = 235 × 98,8/10 = 2322 kN


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• Flexural buckling reduction factors ([2]/Fig. 6.3):

FB about y – y (curve ‘a’)

Lcr = 0,9h = 0,9 × 700 = 630cm (0,9 × system length)

λ y = Lcr/iy = 630/18,5 = 34

λ y = λ y/λ1 = 34/93,9 = 0,36 → χ y = 0,97

FB about z – z (curve ‘b’)

Lcr = l1 = 300cm (wall girt spacing)

λ2 = Lcr/i2 = 300/4,12 = 72,8

λ z = λ z/λ1 = 72,8/93,9 = 0,78 → χ z = 0,73

• Lateral torsional buckling assessment:

l =6.0 mo

MA

N

V

Mo L

a) L - l o o

Lc

c

Lc

xxψ

MLT

c

ψ

MLT

c

c

c

b) L - 0.5l

= 0.5

C =0.8

k =0.86 = 0

C =0.6

k =0.75

x

NOTE: Two cases are assessed: (a) without and (b) with one intermediate lateral restraint to compression flange. (Small local bending Mo = 1,2 × 6,02/8 = 5,4 kNm may be ignored)

Simplified approach ([2]/6.3.2.4)

Af = b × tf = 19 × 1,46 = 27,7cm2 (flange area)

Awc ≅ 21 Av = 2

1 × 43,3 = 21,7cm2 (web compression area)

If,z ≅ 21 Iz = 2

1 × 1680 = 840cm4

if,z = 1,249,34

840AA

I

wc31

f

f ==+

≅ 4,9

Case (a)

λ f = 9,939,4

60075,0xiLk

1z,f

cc

××= = 0,98 → χ ≅ 0,55 (curve ‘c’)


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χLT ≅ kfL χ (λ f) = 1,1 × 0.55 = 0,61

Mb,Rd = χLTMpl,Rd =0,61 × 400 =244 kNm

MEd = max/MA/ = 333 kNm (see 4.3); MEd > Mb,Rd

NOTE: In this case the buckling moment resistance is not sufficient.

Case (b)

λ f = 9,935

3008,0×× = 0,51 ≅ λ co = λLT,O + 0,1 = 0,5 (Slenderness limit)

NOTE: In this case the moment resistance will not be affected by LTB.

• Interaction factor – case (b) ([2]/Table B.1):

Cmy (x = 0) = 0,6; ny = NEd/χ yNpl,Rd = 0,45 (see below)

kyy = Cmy [1 + (λ y – 0,2)ny] = 0,6 [1 + (0,37 – 0,2) × 0,045] =0,60

NOTE: Since second-order effects have been accounted for on the global analysis, Cmy as for non-sway mode is used.

• ULS/Bending + compression:

VEd = maxH1d = 49,8 kN; (VEd < Vo = 294 kN)

NEd = maxV1d = 102 kN; (NEd < No = 483 kN)

NOTE: For limiting values Vo and No see 5.1

RUR: ]OK[183,0400333

MN

Rd,pl

Ed <==

Member stability check ([2]/6.3.3 – formula 6.6.1

RUR: ]OK[155,040033360,0

232297,0102

MMk

NN

Rd,pl

Edyy

Rd,ply

El <=+×

=+χ

IPE450 can be adopted for external columns, if at least one mid-height lateral restrain to both flanges is provided – see case (b).

5.2.2 Internal column (pin-ended prop)

Try CHS 193,7 × 5,4 (g = 25,1 kg/m) – steel S235

NEd = max 1EV = 220 kN (See 4.3)


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A =31,9 cm2; i = 6,77

Lcr = h1 = 650 cm; λ = 650/6,66 = 97,6

λ =λ/λ1 = 97,6/93,9 1,04 → χ = 0,65

Nb,Rd = χAfy = 0,65 × 31,9 × 235.10 = 487 kN

ULS/Member stability check

RUR: ]OK[145,0487220

MN

Rd,B

Ed <==

NOTE: This size is a practical minimum for constructional reasons.

5.3 Rafters

5.3.1 Rafter valley segment

Try IPE 450 (g = 77 kg/m) – steel S235

Section properties and resistances – see 5.2

ULS/Section capacity check

VEd = 21 maxVE = 2

1 220 = 110 kN; (VEd < Vo = 294 kN)

MEd = max/Mc/ = 382 kNm (see 4.3)

RUR: 400382

MM

Rd

Ed = = 0,96 < 1 [OK]

NOTE: To prevent this segment from lateral-torsional buckling, adequately located restraints, as in columns (see 5.2), must be provided (Lc ≤ 3,0 m).

5.3.2 Rafter ridge segment

Try IPE360 (g = 57,1 kg/m) – steel S235

• Section properties and resistances:

A = 72,7 cm2; Wpl,y = 1019 cm3; iz = 3,79 cm

Iz = 1043 cm4; IT = 37,4 cm4; Iw = 0,314 cm6

Npl.Rd = 235 × 72,7/10 = 1710 kN

Mpl,Rd = 235 × 1019/103 = 239 kNm


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• Lateral-torsional buckling assessment ([2]/6.2.2):

M=MB

L c

x x x≈w ≈w

Ridge

k 1.0 k 0.5

L = Lc = 2,0 m (purlin spacing) C1 = 1,0 (quasi-constant moment) k = kz = 1,0; kw = 1,0 (no warping fixity)

Mcr = Ncr,z

z,cr

T

z

w2

w NGI

II

k1 +

, where Ncr,z = 2c

z2

LEIπ

Ncr,z = kN5400200

10430,210002

2=××π (Euler’s critical load)

1043

10315,0II 6

z

w ×= = 302 cm; GIT = 8100,0 × 37,4 = 303000 kN cm2

Mcr 22

10/5400

3030003027,015400 +

= = 1400 kNm

λLT = 1400239

MM

cr

Rd,pl = = 0,48→ χLT = 0,88 curve ‘b’)

NOTE: Since h/b = 360/170 = 2,1 ≅ 2, curve ‘b’ is applied.

• Flexural buckling reduction factor (FB about z – z):

λ z = 85,056,054001710

NN

z,cr

Rd.,p =→== χ = (curve ‘b’)

• ULS/Bending + compression:

Member stability check ([2]/6.3.3 – formula 6.6.2)

NEd = 0,191 w1L = 0,181 × 10,5 × 20 = 38 kN (see 3.5 and 4.3)

MEd = maxMB = 219 kNm (see 4.3)

Interaction factor: kzy ≅ 1,0 (very small compression component)


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RUR: ,107,123988,0

219171085,0

38

,,

>=×

+×

=+RdplLT

Edzy

Rdplz

Ed

MM

kNN

χχ

NOTE: Either accept as preliminary design or reduce purlin centres.

5.4 Deformations – serviceability check

• Deflections of the rafter (L = 20m):

Φw(IPE360) = 0,073 cm/kN (see 4.1)

W = W3L = 6,6 × 20= 132 kN (see 3.5/LC3)

Deflection due to max. gravity load (Gu + Qk);

max u = u (Gk + Qk) = ΦwW = 0,073 × 132 = cm10200Lcm6,9 =< [OK]

Deflection due to imposed load (Qk)

u(Qk) = cm8250L2,56,9

6.05,06,0 =<=×

+ [OK]

• Horizontal displacement of the frame (h = 7,0m):

Kw = 3,2 kN/cm (see 4.2, as assumed: IR/Ic = 0,48 ≅ 21 )

Hw= 12,0 kN (see 3.5/LC4)

maxδ = cm7,4150hcm7,3

2,30,12 =<= [OK]


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References

1 EN 1990 : 2002 Eurocode – Basis of structural design 2 prEN 1993-1-1: 2002 Eurocode 3: Design of steel structures Part 1-1: General rules and rules for

Buildings (final draft) 3 BS6399-2: 1997 Loading for buildings. Part 2: Code of practice for wind loads 4 BS6399-3: 1988 Loading for buildings. Part 3: Code of practice for imposed roof loads 5 Steel Designers’ Manual 4 th Edition, SCI, Blackwell’s, Oxford 6 In-plane stability of Portal Frames to BS 5950-1: 2000

SCI Publication P292 7 Structural sections in accordance with European specification. British Steel (1995)


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Documents

Portal Frame EC3