DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4TH EDITION
SCI PUBLICATION P413
DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4TH EDITION
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Front cover credits
Top left: Top right: Canopy, Napp Pharmaceutical, Cambridge, UK Skid for offshore regasification plant, Grade 1.4401, Courtesy: m-tec Grade 1.4301, Courtesy: Montanstahl
Bottom left: Bottom right: Dairy Plant at Cornell University, College of Águilas footbridge, Spain Agriculture and Life Sciences, Grade 1.4462, Courtesy Acuamed Grade 1.4301/7, Courtesy: Stainless Structurals
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ii PART II - DESIGN EXAMPLES
169
This part of the Design Manual gives fifteen design examples that illustrate the application of the design rules. The examples are:
Design example 1 A circular hollow section subject to axial compression.
Design example 2 A welded I-beam with a Class 4 cross-section subject to combined axial compression and bending.
Design example 3 Trapeziodal roof sheeting with a Class 4 cross-section subject to bending.
Design example 4 A welded hollow section joint subject to fatigue loading.
Design example 5 A welded joint.
Design example 6 A bolted joint.
Design example 7 A plate girder with a Class 4 cross-section subject to bending. Shear buckling is critical.
Design example 8 A plate girder with a Class 4 cross-section subject to bending. Resistance to transverse loads is critical.
Design example 9 A cold formed channel subject to bending with intermediate lateral restraints to the compression flange. Lateral torsional buckling between intermediate lateral restraints is critical.
Design example 10 A rectangular hollow section subject to combined axial compression and bending with 30 minutes fire resistance.
Design example 11 Trapezoidal roof sheeting with a Class 4 cross-section subject to bending – a comparison of designs with cold worked material and annealed material.
Design example 12 A lipped channel from cold worked material in an exposed floor subject to bending.
Design example 13 A stainless steel lattice girder from cold worked material subject to combined axial compression and bending with 30 minutes fire resistance.
Design example 14 The enhanced average yield strength of a cold-rolled square hollow section is determined in accordance with the method in Annex B.
171 Design examples
Design example 15 The bending resistance of a cold-rolled square hollow section is determined in accordance with the Continuous Strength Method (CSM) given in Annex D.
The sheeting in example 3 is from ferritic stainless steel grade 1.4003. The plate girders in examples 7 and 8 are from duplex stainless steel grade 1.4462. The members in the other examples are from austenitic stainless steel grades 1.4301 or 1.4401.
The references in the margin of the design examples are to text sections and expressions/equations in this publication, unless specifically noted otherwise.
172 Sheet 1 of 2 Promotion of new eurocode rules for Title Design Example 1 – CHS Column structural stainless steels (PureSt) Made by HS Date 07/02 Research Fund for Client Revised by JBL Date 03/06 caLcuLatIon SHeet Coal and Steel Revised by FW Date 05/17 deSIGn exaMPLe 1 – cHS coLuMn The circular hollow section column to be designed is an interior column of a multi-storey building. The column is simply supported at its ends. The inter-storey height is 3,50 m.
NEd Nsd
t l
d
Structure Simply supported column, length between supports: l = 3,50 m actions Permanent and variable actions result in a vertical design compression force equal to:
NEd = 250 kN cross-section properties Try a 159 4 cold-formed CHS, austenitic grade 1.4307.
Geometric properties d = 159 mm t = 4,0 mm A = 19,5 cm² I = 585,3 cm4 3 3 Wel = 73,6 cm Wpl = 96,1 cm
Material properties 2 Take fy = 220 N/mm (for cold-rolled strip). Table 2.2 E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3.1 classification of the cross-section = 1,01 Table 5.2 Section in compression : For Class 1, , therefore the section is Class 1. 𝑑𝑑/𝑡𝑡 = 159/4 = 39,8 2 𝑑𝑑/𝑡𝑡 ≤ 50𝜀𝜀
173 Design Example 1 Sheet 2 of 2
compression resistance of the cross-section Section 5.7.3 For a Class 1 cross-section: Eq. 5.27
𝑁𝑁c,Rd = 𝐴𝐴g𝑓𝑓y/𝛾𝛾M0 −1 19,5 × 220 × 10 𝑁𝑁c,Rd = = 390 kN 1,1 resistance to flexural buckling Section 6.3.3 Eq. 6.2
𝑁𝑁b,Rd = χ𝐴𝐴𝑓𝑓y/𝛾𝛾M1 Eq. 6.4 1 = 2 2 0,5 ≤ 1 𝜙𝜙 + [𝜙𝜙 − 𝜆𝜆̅ ] Eq. 6.5 2 𝜑𝜑 = 0,5(1 + 𝛼𝛼(𝜆𝜆̅ − λ̅0) + 𝜆𝜆̅ ) Calculate the elastic critical buckling load:
2 2 4 π 𝐸𝐸𝐸𝐸 π × 200000 × 585,3 × 10 −3 𝑁𝑁cr = 2 = 3 2 × 10 = 943,1 kN Calculate𝐿𝐿crthe flexural buckling(3,50 × 10slenderness:)
Eq. 6.6 2 19,5 × 10 × 220 𝜆𝜆̅ = √ 3 = 0,67 Using an 943imperfection,1 × 10 factor = 0,49 and = 0,2 for a cold-formed austenitic Table 6.1 stainless steel CHS: λ̅0
2 𝜙𝜙 = 0,5 × (1 + 0,49 × (0,67 − 0,2) + 0,67 ) = 0,84 1 𝜒𝜒 = 2 2 0,5 = 0,74 0,84 + [0,84 − 0,67 ] −1 b,Rd 10 The𝑁𝑁 applied = 0, 74axial× 19load,5 ×is 220NEd ×= 250 kN.= 288,6 kN 1,1
Thus the member has adequate resistance to flexural buckling.
174 Design Example 1 Sheet 2 of 2 Promotion of new Sheet 1 of 4 compression resistance of the cross-section Section 5.7.3 eurocode rules for Design Example 2 – Welded I-section beam-column structural stainless Title For a Class 1 cross-section: with lateral restraints Eq. 5.27 steels (PureSt) Made by HS Date 07/02 𝑁𝑁c,Rd = 𝐴𝐴g𝑓𝑓y/𝛾𝛾M0 Research Fund for −1 caLcuLatIon Client Revised by JBL Date 03/06 19,5 × 220 × 10 Coal and Steel 𝑁𝑁c,Rd = = 390 kN SHeet 1,1 Revised by FW Date 06/17 resistance to flexural buckling Section 6.3.3 deSIGn exaMPLe 2 – weLded I-SectIon BeaM-coLuMn wItH LateraL reStraIntS Eq. 6.2 The beam-column to be designed is a welded I-section, simply supported at its ends. Minor axis buckling is prevented by lateral restraints. The inter-storey height is equal to 3,50 m. 𝑁𝑁b,Rd = χ𝐴𝐴𝑓𝑓y/𝛾𝛾M1 Eq. 6.4 The column is loaded by a vertical single load with an eccentricity to the major axis. 1 = 0,5 ≤ 1 2 2 NEdSd 𝜙𝜙 + [𝜙𝜙 − 𝜆𝜆̅ ] Eq. 6.5 2 e 𝜑𝜑 = 0,5(1 + 𝛼𝛼(𝜆𝜆̅ − λ̅0) + 𝜆𝜆̅ ) Calculate the elastic critical buckling load: e
2 2 4 6 π 𝐸𝐸𝐸𝐸 π × 200000 × 585,3 × 10 −3 Load cr 2 3 2 l Calculate𝑁𝑁 = crthe= flexural buckling slenderness: × 10 = 943,1 kN 200 𝐿𝐿 (3,50 × 10 ) 6
200 Eq. 6.6 2 19,5 × 10 × 220 𝜆𝜆̅ = √ 3 = 0,67 Using an 943imperfection,1 × 10 factor = 0,49 and = 0,2 for a cold-formed austenitic Table 6.1 Structure stainless steel CHS: λ̅0 Simply supported column, length between supports: l = 3,50 m 2 𝜙𝜙 = 0,5 × (1 + 0,49 × (0,67 − 0,2) + 0,67 ) = 0,84 Eccentricity of the load: 1 e = 200 mm 𝜒𝜒 = 2 2 0,5 = 0,74 0,84 + [0,84 − 0,67 ] −1 actions b,Rd 10 The𝑁𝑁 applied = 0, 74axial× 19load,5 ×is 220NEd ×= 250 kN.= 288,6 kN Permanent and variable actions result in a vertical design compression force equal to: 1,1 = 120 kN Thus the member has adequate resistance to flexural buckling. 𝑁𝑁Ed Structural analysis Maximum bending moment occurs at the top of the column: = 120 ,0 20 = 24 kNm
𝑀𝑀y,max Ed cross-section properties Try a doubly-symmetric welded I-section 200 200, thickness = 6,0 mm, austenitic grade 1.4401.
Geometric properties 3 b = 200 mm tf = 6,0 mm Wel,y = 259,1 cm 3 hw = 188 mm tw = 6,0 mm Wpl,y = 285,8 cm 4 a = 3,0 mm (weld thickness) Iy = 2591,1 cm
Ag = 35,3 cm² iy = 8,6 cm
175 Design Example 2 Sheet 2 of 4
Material properties 2 fy = 220 N/mm (for hot-rolled strip). Table 2.2 E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3
Classification of the cross-section Table 5.2 = 1,01 Web subject to compression:
(188 − 3 − 3) 𝑐𝑐For/𝑡𝑡 Class= 1, = 30, therefore,3 web is Class 1. 6 𝑐𝑐/𝑡𝑡 ≤ 33,0ε Outstand flange subject to compression:
(200/2 − 6/2 − 3) For𝑐𝑐/𝑡𝑡 Class= 3, , therefore= 94/6 =outstand15,7 flange is Class 4. 6 Therefore, overall𝑐𝑐/𝑡𝑡 ≤ classification14,0ε of cross-section is Class 4.
effective section properties Web is fully effective; calculate the reduction factor for welded outstand elements:
Eq. 5.2 1 0,188 𝜌𝜌 = − 2 but ≤ 1 𝜆𝜆̅p ̅p 𝜆𝜆 Eq. 5.3 𝑏𝑏̅⁄𝑡𝑡 𝜆𝜆̅p = where 𝑏𝑏̅ = 𝑐𝑐 = 94 mm Assuming28 a,4 uniform𝜀𝜀√𝑘𝑘σ stress distribution within the compression flange:
σ2 𝜓𝜓 = = 0,43 = 1 Table 5.4 σ1 𝑘𝑘σ 94/6 λ̅p = = 0,833 28,4 × 1,01 × √0,43 1 0,188 1 0,188 2 2 𝜌𝜌 = 𝑝𝑝 − = − = 0,93 λ̅ λ̅𝑝𝑝 0,833 0,833 𝑏𝑏eff = 0,93 94 = 87,4 mm Calculate the effective cross-section for compression only: = −2 2 𝐴𝐴eff = 𝐴𝐴g − 4(1 − ρ)𝑐𝑐𝑐𝑐 35,3 − 4 × (1 − 0,93) × 94 × 6 × 10 = 33,7 cm Calculate the effective cross-section for major axis bending: = −2 2 𝐴𝐴eff = 𝐴𝐴g − 2(1 − ρ)𝑐𝑐𝑐𝑐 35,3 − 2 × (1 − 0,93) × 94 × 6 × 10 = 34,5 cm
176 Design Example 2 Sheet 2 of 4 Design Example 2 Sheet 3 of 4
Material properties Taking area moments about the neutral axis of the gross cross-section, calculate the shift in 2 fy = 220 N/mm (for hot-rolled strip). Table 2.2 the position of the neutral axis: E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3 ′ 2(1 − ρ)𝑐𝑐𝑐𝑐(ℎw + 𝑡𝑡f)/2 2 × (1 − 0,93) × 94 × 6 × (188 + 6)/2 𝑧𝑧̅ == shifted in the direction = away from the compression−2 flange Classification of the cross-section Table 5.2 𝐴𝐴eff 34,5 × 10 = 1,01 ʹǡʹ Web subject to compression: Calculate the effective second moment of inertia for major axis bending:
2 2 (188 − 3 − 3) 𝑡𝑡 (ℎw + 𝑡𝑡f) ′2 𝑐𝑐For/𝑡𝑡 Class= 1, = 30, therefore,3 web is Class 1. 𝐼𝐼eff,y = 𝐼𝐼y − 2(1 − ρ)𝑐𝑐𝑐𝑐 [ + ] − 𝑧𝑧̅ 𝐴𝐴eff 6 12 4 2 2 𝑐𝑐/𝑡𝑡 ≤ 33,0ε 6 (188 + 6) −4 2 −2 𝐼𝐼eff,y = 2591,1 − 2 × (1 − 0,93) × 94 × 6 × [ + ] × 10 − (2,2) × 34,5 × 10 Outstand flange subject to compression: 12 4 4 𝐼𝐼eff,y = 2515,1 cm (200/2 − 6/2 − 3) 𝐼𝐼eff,y 2515,1 3 For𝑐𝑐/𝑡𝑡 Class= 3, , therefore= 94/6 =outstand15,7 flange is Class 4. 𝑊𝑊eff,y = = = 246,1 cm 6 ℎw ′ 18,8 + 𝑡𝑡f + 𝑧𝑧̅ + 0,6 + 0,22 Therefore, overall𝑐𝑐/𝑡𝑡 ≤ classification14,0ε of cross-section is Class 4. 2 2 effective section properties resistance to major axis flexural buckling Web is fully effective; calculate the reduction factor for welded outstand elements: Eq. 6.3 for Class 4 cross-section subject to compression 𝑁𝑁b,Rd = χ𝐴𝐴eff𝑓𝑓y/𝛾𝛾M1 Eq. 5.2 2 1 0,188 𝐴𝐴eff = 33,7 cm 2 Eq. 6.4 𝜌𝜌 = p − but ≤ 1 𝜆𝜆̅ 𝜆𝜆̅p 1 Eq. 5.3 = 2 2 0,5 ≤ 1 𝑏𝑏̅⁄𝑡𝑡 𝜑𝜑 + [𝜑𝜑 − 𝜆𝜆̅ ] 𝜆𝜆̅p = where 𝑏𝑏̅ = 𝑐𝑐 = 94 mm Eq. 6.5 Assuming a uniformσ stress distribution within the compression flange: 2 28,4𝜀𝜀√𝑘𝑘 = 0,5(1 + α(𝜆𝜆̅ − 𝜆𝜆̅0) + 𝜆𝜆̅ )
Eq. 6.7 σ2 𝜓𝜓 = = 0,43 = 1 Table 5.4 𝐴𝐴eff𝑓𝑓y σ1 𝜆𝜆̅ = √ 𝑁𝑁cr (buckling length is equal to actual length) 𝑘𝑘σ cr 94/6 𝐿𝐿 = 350 cm λ̅p = = 0,833 2 2 4 28,4 × 1,01 × √0,43 π 𝐸𝐸𝐸𝐸 π × 200000 × 2591,1 × 10 −3 𝑁𝑁cr = 2 = 2 2 × 10 = 4175,2 kN 1 0,188 1 0,188 𝐿𝐿cr 350 × 10 2 2 𝜌𝜌 = 𝑝𝑝 − = − = 0,93 λ̅ λ̅𝑝𝑝 0,833 0,833 2 33,7 × 10 × 220 𝜆𝜆̅ = √ 3 = 0,421 𝑏𝑏eff = 0,93 94 = 87,4 mm 4175,2 × 10 Calculate the effective cross-section for compression only: Using imperfection factor and for welded open sections, buckling about Table 6.1 = the major axis: ̅0 −2 2 α = 0,49 𝜆𝜆 = 0,2 𝐴𝐴eff = 𝐴𝐴g − 4(1 − ρ)𝑐𝑐𝑐𝑐 35,3 − 4 × (1 − 0,93) × 94 × 6 × 10 = 33,7 cm Calculate the effective cross-section for major axis bending: 2 = 0,5 × (1 + 0,49 × (0,421 − 0,2) + 0,421 ) = 0,643 = 1 −2 2 = 2 2 0,5 =0,886 𝐴𝐴eff = 𝐴𝐴g − 2(1 − ρ)𝑐𝑐𝑐𝑐 35,3 − 2 × (1 − 0,93) × 94 × 6 × 10 = 34,5 cm 0,643 + [0,643 − 0,421 ] 2 −3 𝑁𝑁b,Rd,y = 0,886 33,7 10 220 10 / 1,1 = 597,23 kN
177 Design Example 2 Sheet 4 of 4
resistance to axial compression and uniaxial major axis moment
Eq. 6.56 𝑁𝑁Ed 𝑀𝑀y,Ed + 𝑁𝑁Ed𝑒𝑒Ny + 𝑘𝑘y ≤ 1 (𝑁𝑁b,Rd)=min 𝛽𝛽W,y for𝑊𝑊pl a,y Class 𝑓𝑓y/𝛾𝛾M1 4 cross -section = 246,1/285,8 = 0,861 𝛽𝛽W,y 𝑊𝑊eff/𝑊𝑊pl,y is zero, due to the symmetry of the cross-section
𝑒𝑒Ny Eq. 6.61 𝑁𝑁Ed 120,0 𝑘𝑘y = 1,0 + 2(𝜆𝜆̅y − 0,5) = 1,0 + 2 × (0,421 − 0,5) × = 0,968 𝑁𝑁b,Rd,y 597,23 2𝑁𝑁Ed 2 × 120 but1,2 + = 1,2 + = 1,60 𝑁𝑁b,Rd,y 597,23 1,2 ≤ 𝑘𝑘y ≤ 1,60 ∴ 𝑘𝑘y = 1,2 6 120,0 24,0 × 10 + 1,2 × 3 = 0,786 ≤ 1 597,23 0,861 × 285,8 × 10 × 220/1,1 Thus the member has adequate resistance.
178 Design Example 2 Sheet 4 of 4 Sheet 1 of 7 Promotion of new resistance to axial compression and uniaxial major axis moment eurocode rules for Design Example 3 – Design of a two-span Title Eq. 6.56 structural stainless trapezoidal roof sheeting 𝑁𝑁Ed 𝑀𝑀y,Ed + 𝑁𝑁Ed𝑒𝑒Ny steels (PureSt) + 𝑘𝑘y ≤ 1 Made by AAT Date 06/02 (𝑁𝑁b,Rd)=min 𝛽𝛽W,y for𝑊𝑊pl a,y Class 𝑓𝑓y/𝛾𝛾M1 4 cross -section Research Fund for = 246,1/285,8 = 0,861 Client Revised by JBL Date 04/06 𝛽𝛽W,y 𝑊𝑊eff/𝑊𝑊pl,y caLcuLatIon SHeet Coal and Steel is zero, due to the symmetry of the cross-section Revised by SJ Date 04/17
𝑒𝑒Ny deSIGn exaMPLe 3 – deSIGn of a two-SPan traPeZoIdaL roof SHeetInG Eq. 6.61 This example considers the design of a two-span trapezoidal sheeting. The material is 𝑁𝑁Ed 120,0 y ̅y ferritic grade 1.4003 stainless steel and the material thickness is 0,6 mm. The dimensions 𝑘𝑘 = 1,0 + 2(𝜆𝜆 − 0,5) b,Rd,y = 1,0 + 2 × (0,421 − 0,5) × = 0,968 𝑁𝑁 597,23 of the cross-section are shown below. 2𝑁𝑁Ed 2 × 120 4 x 212,5 = 850 but1,2 + b,Rd,y = 1,2 + = 1,60 𝑁𝑁 597,23 57 65 1,2 ≤ 𝑘𝑘y ≤ 1,60 ∴ 𝑘𝑘y = 1,2 70 6 120,0 24,0 × 10 + 1,2 × 3 = 0,786 ≤ 1 597,23 0,861 × 285,8 × 10 × 220/1,1 The example shows the following design tasks: Thus the member has adequate resistance. - determination of effective section properties at the ultimate limit state; - determination of the bending resistance of the section; - determination of the resistance at the intermediate support; - determination of deflections at serviceability limit state.
design data Spans L = 2900 mm
Width supports ss = 100 mm Design load Q = 1,4 kN/m2 Self-weight G = 0,07 kN/m2 Design thickness t = 0,6 mm 2 Yield strength fy = 280 N/mm Table 2.2 Modulus of elasticity E = 200000 N/mm2 Section 2.3.1
Partial safety factor M0 = 1,1 Table 4.1
Partial safety factor M1 = 1,1 Table 4.1
Load factor G = 1,35 Section 4.3
Load factor Q = 1,5 Section 4.3
Symbols and detailed dimensions used in calculations are shown in the figure below. The position of the cross-section is given so that in bending at the support the upper flange is compressed.
179 Design Example 3 Sheet 2 of 7
bu0/2 Mid line dimensions: bsu/2 mm hsu mm 0 ℎu0= 70 65 mm bsu0/2 0 𝑤𝑤 = 212 mm,5 𝑏𝑏su = mm 𝑙𝑙0 𝑏𝑏su0= 57 mm h0 𝑏𝑏su = 20 mm 𝑏𝑏s = 8 mm ℎs 0= 6 mm 𝑙𝑙 𝑏𝑏s = 20 mm 1 bsl0/2 𝑏𝑏 = 8 ℎ 𝑙𝑙 = 6 mm (internal radius of h the corners) sl 𝑟𝑟 = 2 bsl/2 w0/2 bl0/2
Angle of the web:
atan atan 57,1 ℎ0u0 70 𝜃𝜃 = | | = | | = ° 0,5(𝑤𝑤0 − 𝑏𝑏 − 𝑏𝑏𝑙𝑙0) 0,5 × (212,5 − 65 − 57) effective section properties at the ultimate limit state (uLS) Section 5.2 Check on maximum width to the thickness ratios: Table 5.1
ℎ0/𝑡𝑡 = 70/ 0 ,6 = 117 ≤ 400sinθ = 336 Table 5.1 maxAngle(𝑏𝑏 of𝑙𝑙0/ the𝑡𝑡 ; 𝑏𝑏web𝑢𝑢0/ 𝑡𝑡and) = corner𝑏𝑏𝑢𝑢0/𝑡𝑡 radius:= 65/0,6 = 108 ≤ 400
u0 su 45°p ≤ 𝜃𝜃 = 57,1° ≤ 90° 𝑏𝑏 − 𝑏𝑏 65 − 20 The𝑏𝑏 = influence of= rounded corners= 22,5 on mm cross-section resistance may be neglected if the 2 2 internal radius and
𝑟𝑟 ≤ 5𝑡𝑡 𝑟𝑟 ≤ 0,10𝑏𝑏p mm Section 5.6.2 𝑟𝑟The= 2influence mm ≤ minof rounded(5𝑡𝑡 ; 0, 1corners𝑏𝑏𝑝𝑝) = minon cross(5 ×-section0,6; 0,1 resistance× 22,5) = may2,25 be neglected.
Location of the centroidal axis when the web is fully effective Calculate reduction factor for effective width of the compressed flange:
Section 5.4.1 Eq. 5.1 0,772 0,079 𝜌𝜌 = − 2 but ≤ 1 where 𝜆𝜆̅p 𝜆𝜆̅p Eq. 5.3 p 𝑏𝑏̅⁄𝑡𝑡 𝜆𝜆̅ = p28,4𝜀𝜀√𝑘𝑘𝜎𝜎 𝑏𝑏̅ = 𝑏𝑏 =22,5 mm Table 5.3 = 1 𝑘𝑘 = 4
180 Design Example 3 Sheet 2 of 7 Design Example 3 Sheet 3 of 7 bu0/2 Mid line dimensions: b /2 0,5 0,5 Table 5.2 su mm y 000 000 h 235 𝐸𝐸 235 200 000 su mm 𝜀𝜀 = [ ] = [ × ] = 0,894 ℎ0 = 70 𝑓𝑓 210 280 210 b /2 u0 65 mm p Eq. 5.3 su0 𝑤𝑤0 = 212,5 mm 22,5/0,6 ̅ 𝑏𝑏su = mm 𝜆𝜆 = = 0,738 𝑙𝑙0 28,4 × 0,894 × √4 𝑏𝑏su0= 57 mm h0 𝑏𝑏su = 20 mm 0,772 0,079 0,772 0,079 𝑏𝑏 = 8 𝜌𝜌 = − 2 = − 2 = 0,901 ≤ 1 s mm 𝜆𝜆̅p 𝜆𝜆̅p 0,738 mm 0, 738 Table 5.3 ℎs 0= 6 mm 𝑙𝑙 eff,u 𝑏𝑏s = 20 mm 𝑏𝑏 = 𝜌𝜌𝑏𝑏̅ = 0,901 × 22,5 = 20,3 1 bsl0/2 𝑏𝑏 = 8 Effective stiffener properties Section 5.5.3 ℎ 𝑙𝑙 = 6 mm (internal radius of h the corners) sl 𝑟𝑟 = 2 bsl/2 w0/2 bl0/2
Angle of the web: atan atan 57,1 su su u0 2 ℎ0 70 2 𝑏𝑏 − 𝑏𝑏 0 2 𝜃𝜃 = | 0 𝑙𝑙0 | = | | = ° su √ su 2 mm 0,5(𝑤𝑤 − 𝑏𝑏 − 𝑏𝑏 ) 0,5 × (212,5 − 65 − 57) ℎ + ( 2 ) √ 20 − 8 effective section properties at the ultimate limit state (uLS) Section 5.2 6 + ( 2 ) 𝑡𝑡 = su 𝑡𝑡 = × 0,6 = 0,849 Fig. 5.3 Check on maximum width to the thickness ratios: ℎ 6 2 Table 5.1 𝐴𝐴s = (𝑏𝑏eff,u + 𝑏𝑏su0)𝑡𝑡 + 2ℎsu𝑡𝑡su = (20,3 + 8) × 0,6 + 2 × 6 × 0,849 = 27,2 mm 0 Table 5.1 ℎsu 6 ℎ /𝑡𝑡 = 70/0,6 = 117 ≤ 400sinθ = 336 𝑏𝑏su0ℎsu𝑡𝑡 + 2ℎsu 𝑡𝑡su 8 × 6 × 0,6 + 2 × 6 × × 0,849 s 2 2 Angle of𝑙𝑙0 the web𝑢𝑢0 and corner𝑢𝑢0 radius: 𝑒𝑒 = = = 2,18 mm max(𝑏𝑏 /𝑡𝑡 ; 𝑏𝑏 /𝑡𝑡) = 𝑏𝑏 /𝑡𝑡 = 65/0,6 = 108 ≤ 400 𝐴𝐴s 27,2 2 4 3 2 2 2 su su0 s s su0 𝑠𝑠𝑠𝑠 s su su ℎ s 15𝑡𝑡 𝑏𝑏 𝑡𝑡 Fig. 5.3 45° ≤ u0𝜃𝜃 = 57su,1° ≤ 90° 𝐼𝐼 = 2(15𝑡𝑡 𝑒𝑒 ) + 𝑏𝑏 𝑡𝑡(ℎ − 𝑒𝑒 ) + 2ℎ 𝑡𝑡 ( − 𝑒𝑒 ) + 2 ( ) + p 3 2 12 12 𝑏𝑏 − 𝑏𝑏 65 − 20 𝑡𝑡suℎsu The𝑏𝑏 = influence of= rounded corners= 22,5 on mm cross-section resistance may be neglected if the + 2 2 2 12 internal radius and 0 2 2 2 2 6 p mm Section 5.6.2 s 𝑟𝑟 ≤ 5𝑡𝑡 𝑟𝑟 ≤ 0,10𝑏𝑏 𝐼𝐼 = 2 × (15 × 0, 6 × 2, 18 ) + 8 × 0,6 × (6 − 2,18) + 2 × 6 × , 849 × ( mm−42,18) 4 3 3 2 𝑟𝑟The= 2influence mm ≤ minof rounded(5𝑡𝑡 ; 0, 1corners𝑏𝑏𝑝𝑝) = minon cross(5 ×-section0,6; 0,1 resistance× 22,5) = may2,25 be neglected. 15 × 0, 6 8 × 0, 6 0,849 × 6 + 2 × ( ) + + 2 × = 159,25 12 12 12 su su0 Location of the centroidal axis when the web is fully effective s 2 su0 2 mm 2 𝑏𝑏 − 𝑏𝑏 2 20 − 8 Calculate reduction factor for effective width of the compressed flange: 𝑏𝑏 = 2√ℎ𝑠𝑠𝑠𝑠 + ( ) + 𝑏𝑏 = 2 × √6 + ( ) + 8 = 25,0 p2 s 2 Section 5.4.1 b s p 1⁄4 Eq. 5.10 Eq. 5.1 0,772 0,079 2 2𝑏𝑏 + 3𝑏𝑏 𝜌𝜌 = − 2 but ≤ 1 𝑙𝑙 = 3,07 [𝐼𝐼 𝑏𝑏 ( 3 )] where 𝜆𝜆̅p 𝜆𝜆̅p 𝑡𝑡 b 1/4 mm Eq. 5.3 p 2 2 × 22,5 + 3 × 25 𝑙𝑙 = 3,07 × [159,25 × 22, 5 × ( 3 )] = 251 𝑏𝑏̅⁄𝑡𝑡 0, 6 𝜆𝜆̅ = p28,4𝜀𝜀√𝑘𝑘𝜎𝜎 Fig. 5.5 𝑏𝑏̅ = 𝑏𝑏 =22,5 mm Table 5.3 2 2 𝑤𝑤0 − 𝑏𝑏u0 − 𝑏𝑏𝑙𝑙0 2 212,5 − 65 − 57 2 𝑠𝑠w = √( ) + ℎ0 = √( ) + 70 = 83,4 mm = 1 𝑘𝑘 = 4 d 2 mm 2 𝑏𝑏 = 2𝑏𝑏p + 𝑏𝑏s = 2 × 22,5 + 25 = 70 181 Design Example 3 Sheet 4 of 7
Eq. 5.11 𝑠𝑠w + 2𝑏𝑏d 83,4 + 2 × 70 𝑘𝑘w0 = √ = √ = 1,37 𝑠𝑠w + 0,5𝑏𝑏d 83,4 + 0,5 × 70 Eq. 5.8 b 𝑙𝑙 251 w w0 w = = 3,01 ≥ 2 𝑘𝑘 = 𝑘𝑘 = 1,37 𝑠𝑠 83,4 w s
cr,s 3 Eq. 5.4 4,2𝑘𝑘s 𝐸𝐸 p 𝐼𝐼 𝑡𝑡p s 𝜎𝜎 = √ 2 𝐴𝐴 4𝑏𝑏 (2𝑏𝑏 + 3𝑏𝑏 ) N/mm2 cr,s 3 3 4,2 × 1,37 × 200 × 10 159,25 × 0, 6 𝜎𝜎 = × √ 2 = 503,4 27,2 4 × 22, 5 × (2 × 22,5 + 3 × 25) d 𝑓𝑓y 280 ̅ 𝜆𝜆 = √ cr,s = √ = 0,746 𝜎𝜎 d 503 ,4 Eq. 5.17
0,65 < 𝜆𝜆̅ = 0,746 sl s sl 2 2 𝑏𝑏 − 𝑏𝑏 𝑙𝑙0 2 mm sl √ℎ + ( ) 2 20 − 8 sl 2 √6 + ( ) 2 𝑡𝑡 = 𝑡𝑡 = × 0,6 = 0,849 w ℎ mm6 𝑡𝑡 = 𝑡𝑡/sin𝜃𝜃 = 0,6/sin(57,1°) = 0,714 2 𝑖𝑖 𝑖𝑖 𝑒𝑒 [mm] 𝐴𝐴 [mm ] tot 2 Ͳ 0,5𝑏𝑏eff,u 𝑡𝑡 = 6,1 mm 𝐴𝐴 = ∑𝐴𝐴𝑖𝑖 Ͳ 0,5𝑏𝑏eff,u 𝜒𝜒d 𝑡𝑡 = 5,66 = 87,3 0,5ℎsu = 3 ℎsu 𝜒𝜒d𝑡𝑡su = 4,74 ℎsu = 6 0,5𝑏𝑏su0 𝜒𝜒d 𝑡𝑡 = 2,23 c ∑𝐴𝐴tot𝑖𝑖𝑒𝑒𝑖𝑖 0 0 w 𝑒𝑒 = mm 0,5ℎ = 35 ℎ 𝑡𝑡 = 49,98 𝐴𝐴 ℎ0 = 70 0,5(𝑏𝑏𝑙𝑙0 − 𝑏𝑏𝑠𝑠𝑠𝑠) 𝑡𝑡 = 11,1 = 34,9 ℎ0 − 0,5ℎs𝑙𝑙 = 67 ℎs𝑙𝑙𝑡𝑡s𝑙𝑙 = 5,09 ℎ0 − ℎs𝑙𝑙 = 64 0,5𝑏𝑏s𝑙𝑙0𝑡𝑡 = 2,4 Effective cross-section of the compression zone of the web EN 1993-1-3 eff, eff, Clause 𝐸𝐸 200 5.5.3.4.3(4-5) 1 0 −3 𝑠𝑠 = 𝑠𝑠 = 0,76𝑡𝑡√ M0 com,Ed = 0,76 × 0,6 × √ = 11,6 mm eff,n eff, 𝛾𝛾 𝜎𝜎 1,1 × 280 × 10 𝑠𝑠 = 1,5𝑠𝑠 0 = 1,5 × 11,6 = 17,4 mm 182 Design Example 3 Sheet 4 of 7 Design Example 3 Sheet 5 of 7 Effective cross-section properties per half corrugation Eq. 5.11 eff,1 eff,1 w d w0 𝑠𝑠 + 2𝑏𝑏 83,4 + 2 × 70 𝑘𝑘 = √ = √ = 1,37 ℎeff, = 𝑠𝑠 eff,sin𝜃𝜃 = 11 ,6 × sin(57,1°) = 9,74 mm 𝑠𝑠w + 0,5𝑏𝑏d 83,4 + 0,5 × 70 Eq. 5.8 𝑛𝑛 𝑛𝑛 b ℎ = 𝑠𝑠 sin𝜃𝜃 = 17,4 × sin(57,1°) = 14,61 mm 𝑙𝑙 251 w w0 w = = 3,01 ≥ 2 𝑘𝑘 = 𝑘𝑘 = 1,37 𝑠𝑠 83,4 w s 𝟐𝟐 𝟒𝟒 cr,s 3 Eq. 5.4 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 s 𝒆𝒆 [𝐦𝐦𝐦𝐦] 𝑨𝑨 [𝐦𝐦𝐦𝐦 ] 𝑰𝑰 [𝐦𝐦𝐦𝐦 ] 4,2𝑘𝑘 𝐸𝐸 p 𝐼𝐼 𝑡𝑡p s 𝜎𝜎 = √ 2 Ͳ 0,5𝑏𝑏eff,u𝑡𝑡 = 6,1 ≈ 0 𝐴𝐴 4𝑏𝑏 (2𝑏𝑏 + 3𝑏𝑏 ) eff,u 𝑑𝑑 N/mm2 Ͳ 0,5𝑏𝑏 𝜒𝜒 𝑡𝑡 = 5,7 ≈ 0 cr,s 3 3 3 su 4,2 × 1,37 × 200 × 10 159,25 × 0, 6 0,5ℎ = 3 ℎsu 𝜒𝜒𝑑𝑑 𝑡𝑡su = 4,7 𝜒𝜒𝑑𝑑𝑡𝑡suℎsu /12 = 14,2 𝜎𝜎 = × √ 2 = 503,4 27,2 4 × 22, 5 × (2 × 22,5 + 3 × 25) ℎsu = 6 0,5𝑏𝑏su0 𝜒𝜒𝑑𝑑 𝑡𝑡 = 2,2 ≈ 0 d 3 0,5ℎeff,1 = 4,9 ℎeff,1𝑡𝑡w = 7,0 𝑡𝑡wℎeff,1 /12 = 55,0 𝑓𝑓y 280 ̅ 3 𝜆𝜆 = √ cr,s = √ = 0,746 0 c eff,𝑛𝑛 𝜎𝜎 503,4 Eq. 5.17 0 0 c eff,𝑛𝑛 0 c eff,𝑛𝑛 𝑤𝑤 d ℎ − 0,5(ℎ − 𝑒𝑒 + ℎ ) = (ℎ − 𝑒𝑒 + ℎ ) 𝑡𝑡 = w (ℎ − 𝑒𝑒 + ℎ ) 𝑡𝑡 = = 45,1 = 35,5 12 0,65 < 𝜆𝜆̅ = 0,746 eff,i eff,i sl s 𝐴𝐴 = ∑𝐴𝐴 = 79,8 mm sl 2 c tot 2 𝑏𝑏 − 𝑏𝑏 𝑙𝑙0 2 mm ∑𝐴𝐴 𝑒𝑒 sl √ℎ + ( ) 2 20 − 8 𝑒𝑒 = = 36,8 2 sl 2 √6 + ( ) tot eff,i eff,i c eeff,i mm 2 𝐴𝐴 𝑡𝑡 = 𝑡𝑡 = mm × 0,6 = 0,849 2 w ℎ 6 𝐼𝐼 = ∑𝐼𝐼 + ∑𝐴𝐴 (𝑒𝑒 − ) = 7 393,3 + 51 667,2 = 59 060,5 𝑡𝑡 = 𝑡𝑡/sin𝜃𝜃 = 0,6/sin(57,1°) = 0,714 Optionally the effective section properties may also be redefined iteratively based on the EN 1993-1-3 location of the effective centroidal axis. 2 𝑖𝑖 𝑖𝑖 𝑒𝑒 [mm] 𝐴𝐴 [mm ] tot Bending strength per unit width (1 m) Section 5.7.4 2 Ͳ 0,5𝑏𝑏eff,u 𝑡𝑡 = 6,1 mm 𝐴𝐴 = ∑𝐴𝐴𝑖𝑖 4 tot mm Ͳ 0,5𝑏𝑏eff,u 𝜒𝜒d 𝑡𝑡 = 5,66 0,5w 0,5 2 = 87,3 10000 1000 su su d su 𝐼𝐼 = 𝐼𝐼 = × 59 060,5 = 555 863,5 0,5ℎ = 3 ℎ 𝜒𝜒 𝑡𝑡 = 4,74 × 12,5 3 u mm ℎsu = 6 0,5𝑏𝑏su0 𝜒𝜒d 𝑡𝑡 = 2,23 c c ∑𝐴𝐴tot𝑖𝑖𝑒𝑒𝑖𝑖 𝐼𝐼 555 863,5 0 0 w mm 𝑊𝑊 = = = 15 105,0 0,5ℎ = 35 ℎ 𝑡𝑡 = 49,98 𝑒𝑒 = 𝑒𝑒 36,8 3 𝐴𝐴 l mm c ℎ0 = 70 0,5(𝑏𝑏𝑙𝑙0 − 𝑏𝑏𝑠𝑠𝑠𝑠) 𝑡𝑡 = 11,1 = 34,9 𝐼𝐼 555 863,5 Because𝑊𝑊 = 0 = = 16 742,9 ℎ0 − 0,5ℎs𝑙𝑙 = 67 ℎs𝑙𝑙𝑡𝑡s𝑙𝑙 = 5,09 ℎ − 𝑒𝑒 70 − 36,8eff,min 3 u l u ℎ0 − ℎs𝑙𝑙 = 64 0,5𝑏𝑏s𝑙𝑙0𝑡𝑡 = 2,4 𝑊𝑊eff,min < 𝑊𝑊 𝑊𝑊 = 𝑊𝑊 = 15 105,0 mm −6 Eq. 5.31 Effective cross-section of the compression zone of the web 𝑊𝑊 𝑓𝑓y 10 𝑀𝑀c,Rd = = 15 105,0 × 280 × = 3,84 kNm EN 1993-1-3 γM0 1,1 eff, eff, Clause determination of the resistance at the intermediate support Section 6.4.4 𝐸𝐸 200 5.5.3.4.3(4-5) 1 0 −3 Web crippling strength 𝑠𝑠 = 𝑠𝑠 = 0,76𝑡𝑡√ M0 com,Ed = 0,76 × 0,6 × √ = 11,6 mm eff,n eff, 𝛾𝛾 𝜎𝜎 1,1 × 280 × 10 mm 𝑠𝑠 = 1,5𝑠𝑠 0 = 1,5 × 11,6 = 17,4 mm 𝑐𝑐 ≥ 40 EN 1993-1-3 𝑟𝑟/𝑡𝑡 = 2/0,6 = 3,33 ≤ 10 Clause 6.1.7 ℎw/𝑡𝑡 = 70/0,6 = 117 ≤ 200sin𝜃𝜃 = 200sin(57,1°) = 168 183 Design Example 3 Sheet 6 of 7 45° ≤ 𝜃𝜃 = 57,1°≤ 90° βV = 0 ≤(category0,2 2) 𝑙𝑙𝑎𝑎 = 𝑠𝑠𝑠𝑠 = 100 mm α = 0,15 EN 1993-1-3 2 Eq. 6.18 2 𝑟𝑟 𝑙𝑙a 𝜑𝜑 1 1000 𝑅𝑅w,Rd = α 𝑡𝑡 √𝑓𝑓y𝐸𝐸 (1 − 0,1√ ) (0,5 + √0,02 ) [2,4 + ( ) ] 𝑡𝑡 𝑡𝑡 90 γM1 0,5w0 2 2 100 𝑅𝑅w,Rd = 0,15 × 0, 6 √280 × 200 000 × (1 − 0,1√ ) (0,5 + √0,02 × ) 0,6 0,6 2 57,1 1 1000 −3 × [2,4 + ( ) ] × × × 10 = 18,4 kN 90 1,1 0,5 × 212,5 combined bending moment and support reaction Factored actions per unit width (1 m): G Q kN/m 𝑞𝑞 = 𝛾𝛾 𝐺𝐺 + 𝛾𝛾 𝑄𝑄 = 1,35 × 0,07 + 1,5 × 1,4 = 2,19 Ed 2 2 kNm 𝑞𝑞𝐿𝐿 2,19 × 2, 9 𝑀𝑀 = = = 2,30 Ed 8 8 kN 5 5 𝐹𝐹 Ed= 𝑞𝑞𝑞𝑞 = × 2,19 × 2,9 = 7,94 Ed 4 4 𝑀𝑀c,Rd 2,30 𝐹𝐹w,Rd 7,94 Ed = Ed = 0,599 ≤ 1,0 = = 0,432 ≤ 1,0 𝑀𝑀 3,84 𝑅𝑅 18,4 EN 1993-1-3 𝑀𝑀c,Rd 𝐹𝐹w,Rd Eq. 6.28a - c + = 0,599 + 0,432 = 1,031 ≤ 1,25 𝑀𝑀Cross-section𝑅𝑅 resistance satisfies the conditions. determination of deflections at serviceability limit state (SLS) Effective cross-section properties For serviceability verification the effective width of compression elements should be based on the compressive stress in element under the SLS loading. A conservative approximation is made for the maximum compressive stress in the effective EN 1993-1-3 section at SLS based on Wu determined above for ULS. Clause 5.5.1 y,Ed,ser 2 2 1,55 kNm (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 (0,07 + 1,4) × 2, 9 𝑀𝑀 = y,Ed,ser = = 8 8 1 N/mm2 com,Ed,ser 6 𝑀𝑀 1,55 × 10 𝜎𝜎 = = = 02,6 The effective section𝑊𝑊u properties15 105are determined as before at ULS except that fy is replaced by com,Ed,ser and the thickness of the flange stiffener is not reduced. The results of the calculation are: 𝜎𝜎 Effective width of the compressed flange: Location of the centroidal axis when the web is fully The flange is fully effective: effective. Effective cross-section of the compression zone of the web: ec = 34,48 mm Effective cross-section properties per half corrugation: The web is fully effective. 184 Design Example 3 Sheet 6 of 7 Design Example 3 Sheet 7 of 7 2 Effective section properties per unit width (1 m): Atot = 88,41 mm ec = 34,48 mm 45° ≤ 𝜃𝜃 = 57,1° ≤ 90° 4 Itot = 63759,0 mm βV = 0 ≤(category0,2 2) 𝑙𝑙𝑎𝑎 = 𝑠𝑠𝑠𝑠 = 100 mm I = 600084,7 mm4 4 Wu = 17403,8 mm α = 0,15 4 EN 1993-1-3 Wl = 16894,3 mm 2 Eq. 6.18 2 𝑟𝑟 𝑙𝑙a 𝜑𝜑 1 1000 𝑅𝑅w,Rd = α 𝑡𝑡 √𝑓𝑓y𝐸𝐸 (1 − 0,1√ ) (0,5 + √0,02 ) [2,4 + ( ) ] 𝑡𝑡 𝑡𝑡 90 γM1 0,5w0 Determination of deflection Secant modulus of elasticity corresponding to maximum value of the bending moment: 2 y,Ed,ser 2 w,Rd 2 100 N/mm 𝑅𝑅 = 0,15 × 0, 6 √280 × 200 000 × (1 − 0,1√ ) (0,5 + √0,02 × ) 6 0,6 0,6 𝑀𝑀 u 1,55 × 10 2 1,Ed,ser = = = 89,06 −3 𝑊𝑊 17403,8 2 57,1 1 1000 6 N/mm × [2,4 + ( ) ] × × × 10 = 18,4 kN l 90 1,1 0,5 × 212,5 𝑀𝑀y,Ed,ser 1,55 × 10 2,Ed,ser = = = 91,75 combined bending moment and support reaction (for ferritic𝑊𝑊 grade16894 1.4003,3 stainless steel) Table 6.4 Factored actions per unit width (1 m): 2 𝑛𝑛 = 14 kN/mm 1,Ed,ser Eq. 6.53 G Q kN/m 𝐸𝐸 200 𝐸𝐸S,1 = 1,Ed,ser y 𝑛𝑛 = 14 = 200,0 𝐸𝐸 𝜎𝜎 200 0,089 𝑞𝑞 = 𝛾𝛾 𝐺𝐺 + 𝛾𝛾 𝑄𝑄 = 1,35 × 0 ,07 + 1,5 kNm× 1,4 = 2,19 1 + 0,002 ( ) 1 + 0,002 × ( ) Ed 2 2 𝜎𝜎 𝑓𝑓 0,089 0,28 kN/mm2 𝑞𝑞𝐿𝐿 2,19 × 2, 9 2,Ed,ser Eq. 6.53 𝑀𝑀 = = = 2,30 𝐸𝐸 200 8 8 kN 𝐸𝐸S,2 = 2,Ed,ser y 𝑛𝑛 = 14 = 200,0 Ed 𝐸𝐸 𝜎𝜎 1 + 0,002 ( ) 200 0,092 5 5 S,2 𝜎𝜎 𝑓𝑓 1 + 0,0022 × ( ) 𝐹𝐹 Ed= 𝑞𝑞𝑞𝑞 = × 2,19 × 2,9 = 7,94 Ed kN/mm 0,092 0,28 Eq. 6.52 4 4 S,1 c,Rd w,Rd 𝐸𝐸 + 𝐸𝐸 200 + 200 𝑀𝑀 2,30 𝐹𝐹 7,94 There𝐸𝐸S = is no effect= of material non=-linearity200 on deflection for the given steel grade and Ed = Ed = 0,599 ≤ 1,0 = = 0,432 ≤ 1,0 2 2 𝑀𝑀 3,84 𝑅𝑅 18,4 EN 1993-1-3 stress level. 𝑀𝑀c,Rd 𝐹𝐹w,Rd Eq. 6.28a - c + = 0,599 + 0,432 = 1,031 ≤ 1,25 𝑀𝑀Cross-section𝑅𝑅 resistance satisfies the conditions. Check of deflection: For cross-section stiffness properties the influence of rounded corners should be taken into determination of deflections at serviceability limit state (SLS) account. The influence is considered by the following approximation: Effective cross-section properties 𝑛𝑛 For serviceability verification the effective width of compression elements should be 𝑗𝑗 𝑜𝑜 j φ o Eq. 5.22 based on the compressive stress in element under the SLS loading. ∑ 𝑟𝑟 294,𝑜𝑜2 j=1 90 2 × A conservative approximation is made for the maximum compressive stress in the effective EN 1993-1-3 δ = 0,43 𝑚𝑚 = 0,43 90 = 0,019 p,i 149,3 section at SLS based on Wu determined above for ULS. Clause 5.5.1 ∑i=1 𝑏𝑏 Eq. 5.20 4 Fory,r the location of maximum deflection: y,Ed,ser 2 2 1,55 kNm 𝐼𝐼 = 𝐼𝐼 (1 - 2 ) = 600084,7 (1 - 2 × 0,019) = 577281,5 mm (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 (0,07 + 1,4) × 2, 9 𝑀𝑀 = = = y,Ed,ser8 8 2 m com,Ed,ser 6 1 N/mm 1 + √33 1 + √33 𝑀𝑀 1,55 × 10 𝑥𝑥 = × 𝐿𝐿 = × 2,9 = 1,22 The𝜎𝜎 effective= sectionu properties= are determined= 02 ,as6 before at ULS except that fy is replaced 16 16 𝑊𝑊 15 105 4 3 4 by com,Ed,ser and the thickness of the flange stiffener is not reduced. The results of the (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 𝑥𝑥 𝑥𝑥 𝑥𝑥 3 4 calculation are: 𝛿𝛿 = S y,r ( − 3 + 2 ) 𝜎𝜎 48𝐸𝐸 𝐼𝐼 𝐿𝐿 𝐿𝐿 𝐿𝐿 Effective width of the compressed flange: 3 4 3 4 (0,07 + 1,4) × 10 × 2, 9 1,48 1, 48 1, 48 Location of the centroidal axis when the web is fully The flange is fully 𝛿𝛿 = 6 −12 × ( − 3 × 3 + 2 × 4 ) effective: effective. 48 × 200 × 10 × 577281,5 × 10 2,9 2, 9 2, 9 The permissible deflection is L/200 = 2900/200 = 14,5 mm 4,64 mm, hence the Effective cross-section of the compression zone of the web: ec = 34,48 mm calculated𝛿𝛿 = 4,64 mmdeflection is acceptable. Effective cross-section properties per half corrugation: The web is fully effective. > 185 186 Sheet 1 of 2 Promotion of new eurocode rules for Design Example 4 – Fatigue strength of a welded Title structural stainless hollow section joint steels (PureSt) Made by AAAT Date 06/02 Research Fund for Client Revised by MEB Date 04/06 caLcuLatIon SHeet Coal and Steel Revised by UDE Date 01/17 deSIGn exaMPLe 4 – fatIGue StrenGtH of a weLded HoLLow SectIon JoInt This example considers the fatigue strength of the chord of a welded hollow section joint. Section 9 Fatigue should be considered in the design of stainless steel structures which are subjected to repeated fluctuations of stresses, e.g. in oil platforms, masts, chimneys, bridges, cranes and transport equipment. EN 1993-1-9 is applicable for estimating the fatigue strength of austenitic and duplex stainless steel structures. The example shows the following design tasks for fatigue assessment: - determination of the fatigue strength curve - determination of secondary bending moments in the joint - determination of the partial safety factor for fatigue strength - fatigue assessment for variable amplitude loading. The chords of the joint are RHS 50×50×4 and braces are RHS 30×30×2. The material is Table 2.2 austenitic grade 1.4301 stainless steel with 0,2% proof stress of 210 N/mm2. 11 30x30x2 3 30x30x2 50x50x4 actions The fatigue stress spectra for the chord during the required design life is: Nominal stress range: Number of cycles: 2 3 Δ1 = 100 N/mm n1 = 10×10 2 3 Δ2 = 70 N/mm n2 = 100×10 2 3 Δ3 = 40 N/mm n3 = 1000×10 Structural analysis All subsequent The detail category of the joint depends on the dimensions of chord and braces. In this references to example b0 = 50 mm, bi = 30 mm, t0 = 4 mm and ti = 2 mm. EN 1993-1-9 Because t0/ti = 2, the detail category is 71. Because 0,5(b0 - bi) = 10 mm, g = 11 mm, 1,1(b0 - bi) = 22 mm and 2t0 = 8 mm, the joint Table 8.7 also satisfies the conditions 0,5(b0 - bi) g 1,1(b0 - bi) and g 2t0. 187 Design Example 4 Sheet 2 of 2 effect of secondary bending moments in the joint The effects of secondary bending moments are taken into account by multiplying the Clause 4 (2), stress ranges due to axial member forces by coefficient k1 = 1,5. Table 4.2 Partial factors When it is assumed that the structure is damage tolerant and the consequence of failure is Clause 3 (7), low, the recommended value for the partial factor for fatigue strength is Mf = 1,0. Table 3.1 The partial safety factor for loading is Ff = 1,0. fatigue assessment Reference stress range corresponding to 2106 stress fluctuations for detail category 71 is 2 c = 71 N/mm . Figure 7.1 The fatigue strength curve for lattice girders has a constant slope m = 5. The number of stress fluctuations corresponding to the nominal stress range Δi is: m 6 c Ni 2 10 Mf Ffk 1 i 2 3 Δ1 = 100 N/mm N1 = 47,5×10 2 3 Δ2 = 70 N/mm N2 = 283×10 2 3 Δ3 = 40 N/mm N3 = 4640×10 Palmgren-Miner rule of cumulative damage Partial damage because of ni cycles of stress range Δi: Dd,i = nEi / NRi Therefore, for 2 Δ1 = 100 N/mm Dd,1 = 0,21 A.5 (1) 2 Δ2 = 70 N/mm Dd,2 = 0,35 2 Δ3 = 40 N/mm Dd,3 = 0,22 The cumulative damage during the design life is: Eq. A.1 n nEi DDd d,i 0,78 1,0 i NRi Because the cumulative damage is less than unity, the calculated design life of the chord Clause 8 (4) exceeds the required design life. The procedure described above shall also be repeated for the brace. 188 Design Example 4 Sheet 2 of 2 Sheet 1 of 7 Promotion of new effect of secondary bending moments in the joint eurocode rules for Title Design Example 5 – Welded joint The effects of secondary bending moments are taken into account by multiplying the Clause 4 (2), structural stainless stress ranges due to axial member forces by coefficient k1 = 1,5. Table 4.2 steels (PureSt) Made by IR Date 08/02 Partial factors Research Fund for Client Revised by MEB Date 04/06 When it is assumed that the structure is damage tolerant and the consequence of failure is Clause 3 (7), caLcuLatIon SHeet Coal and Steel low, the recommended value for the partial factor for fatigue strength is Mf = 1,0. Table 3.1 Revised by UDE Date 01/17 The partial safety factor for loading is Ff = 1,0. deSIGn exaMPLe 5 – weLded JoInt fatigue assessment The joint configuration and its loading are shown in the figure below. Noting that there Reference stress range corresponding to 2106 stress fluctuations for detail category 71 is are two identical plane fillet weld joints of constant throat thickness sharing the applied 2 loading, the required throat thickness for the welds shall be determined. Right angle c = 71 N/mm . Figure 7.1 The fatigue strength curve for lattice girders has a constant slope m = 5. (equal leg) welds will be used throughout. 100 100 The number of stress fluctuations corresponding to the nominal stress range Δi is: Fillet welds : x axi s nz = 300 kn m throat size a 300 6 c throughout Ni 2 10 n 300 x n = 30 kn Mf Ffk 1 i y a cn°2 y axis for joint n°2 ezc = - 140 nz n = 30 kn 2 3 250 c y Δ1 = 100 N/mm N1 = 47,5×10 y axi s c y axis for joint n°1 2 3 n°1 Δ2 = 70 N/mm N2 = 283×10 175 n x = - 20 kn eyc 2 3 Plan Δ3 = 40 N/mm N3 = 4640×10 c : centre of gravity of a weld joint 175 Palmgren-Miner rule of cumulative damage z axis Elevation Partial damage because of ni cycles of stress range Δi: Dd,i = nEi / NRi Material properties Therefore, for Use austenitic grade 1.4401: Table 2.2 2 2 2 2 2 fy = 220 N/mm , fu = 530 N/mm , E = 200000 N/mm and G = 76900 N/mm . Section 2.3.1 Δ1 = 100 N/mm Dd,1 = 0,21 A.5 (1) It is assumed that the yield and ultimate tensile strength of the weld exceed those of the Section 7.4.1 2 Dd,2 = 0,35 Δ 2 = 70 N/mm parent metal. 2 Δ3 = 40 N/mm Dd,3 = 0,22 Partial factor The cumulative damage during the design life is: The partial factor on weld resistance is M2 = 1,25. Table 4.1 Eq. A.1 n The need to include a reduction factor on the weld resistance to account for its length will nEi DDd d,i 0,78 1,0 also be examined. i NRi Because the cumulative damage is less than unity, the calculated design life of the chord Clause 8 (4) analysis of welded joints exceeds the required design life. An elastic analysis approach is used here for designing the right-angle equal-leg fillet EN 1993-1-8 The procedure described above shall also be repeated for the brace. weld for the load case indicated above which leads to a conservative estimate of the joint clause 2.5 resistance. The coordinates (xc, yc,, zc) of a point on the welded joint are taken with reference to a right hand axis system with an origin at the centre of gravity of the welded joint (In the present case the joint is taken to be in the y-z plane so that xc = 0 throughout.). The main purpose of the elastic analysis is to determine the design forces in the weld at the most severely loaded point or points of the welded joint, often referred to as the “critical” points. For the welded joint being examined the critical point can be taken as being the point furthest from the centre of gravity of the joint. 189 Design Example 5 Sheet 2 of 7 The vectors of the applied force, its eccentricity and the resulting moments acting on a welded joint of general form and centre of gravity C can be expressed as follows: Applied force Nw,Ed NNN x,Ed,, y,Ed z,Ed Eccentricity of the applied force eN eee xc,, yc zc (these are the coordinates of the point of application of the force) Applied moments Mxc,Ed eN yc z,Ed eN zc y,Ed Myc,Ed eN zc x,Ed eN xc z,Ed Mzc,Ed eN xc y,Ed eN yc x,Ed A linear elastic analysis of the joint for a general load case leads to the following force components per unit length of weld at a point with coordinates (xc, yc,, zc), where the throat thickness is denoted by a: Nx,Ed zMc yc,Ed yMc zc,Ed Fawx,Ed AIw yc Izc Ny,Ed xMc zc,Ed zM c xc,Ed Fawy,Ed AIw zc Ixc Nz,Ed yM c xc,Ed xMc yc,Ed Fawz,Ed AIw xc Iyc In the above expressions, the resisting sectional throat area and the second moment of area about the principal axes of the welded joint are: Aw adl aii l for a weld of straight segments of length li and throat thickness ai, 22 Ixc a ycc z dl 22 Iyc a xcc z dl 22 Izc a xcc y dl Assuming that all welds have equal thickness a: Aw dl li a and since xc = 0: I zc y2 dl a c I yc z2 dl a c III xc y22 z dl yc zc a cc aa 190 Design Example 5 Sheet 2 of 7 Design Example 5 Sheet 3 of 7 The vectors of the applied force, its eccentricity and the resulting moments acting on a design of fillet welds welded joint of general form and centre of gravity C can be expressed as follows: Two different design methods are allowed for designing fillet welds and thus to determine Section 7.4.2 Applied force the required weld throat thickness at the critical point: The first procedure is based on the simplified and more conservative design shear N NNN,, w,Ed x,Ed y,Ed z,Ed strength for a fillet weld. The design shear force per unit length of the weld at any point of the joint is defined as the vector sum of the design forces per unit length due to all Eccentricity of the applied force forces and moments transmitted by the welded joint. This design shear force per unit e eee,, length should not exceed the design resistance per unit length which is taken as the N xc yc zc design shear strength multiplied by the throat thickness. This approach ignores the throat (these are the coordinates of the point of application of the force) plane orientation to the direction of resultant weld force per unit length. The second procedure is based on comparing the basic design strength of the weaker part Applied moments joined to the applied design weld stress in the weld throat determined by a von Mises Mxc,Ed eN yc z,Ed eN zc y,Ed type of formula. This approach is the most precise as it allows for the throat plane Myc,Ed eN zc x,Ed eN xc z,Ed orientation to the direction of resultant weld force per unit length. Mzc,Ed eN xc y,Ed eN yc x,Ed 1. Simplified method The design resistance check of the fillet weld is as follows: A linear elastic analysis of the joint for a general load case leads to the following force f 3 EN 1993-1-8 components per unit length of weld at a point with coordinates (xc, yc,, zc), where the throat 222 u Fw.Ed F wx,Ed F wy,Ed F wz,Ed F w,Rd af vw,d a clause 4.5.3.3 thickness is denoted by a: w M2 Nx,Ed zMc yc,Ed yMc zc,Ed where: Fawx,Ed AIw yc Izc fvw,d is the design shear strength of the weld, Fw,Rd is the design (shear) resistance per unit length of weld of throat thickness a. Ny,Ed xMc zc,Ed zM c xc,Ed Fawy,Ed For stainless steel, w is taken as 1,0. Section 7.4.2 AIw zc Ixc When the design procedure requires that a suitable throat thickness be obtained, the N yM xM Faz,Ed c xc,Ed c yc,Ed design expression becomes: wz,Ed AI I w xc yc F a w,Ed f In the above expressions, the resisting sectional throat area and the second moment of area vw,d about the principal axes of the welded joint are: 2. Directional method Aw adl aii l In the directional method, the forces transmitted by a weld are resolved into normal for a weld of straight segments of length li and throat thickness ai, stresses and shear stresses with respect to the throat section (see Fig. 4.5 in EN 1993-1-8): 22 A normal stress perpendicular to the throat section, Ixc a ycc z dl A shear stress acting in the throat section parallel to the axis of the weld, 22 Iyc a xcc z dl A shear stress acting in the throat section transverse to the axis of the weld. 22 The normal stress parallel to the weld axis does not need to be considered. Izc a xcc y dl For the combination of stresses , , and , the design requirement is: Eqs. 7.14 and Assuming that all welds have equal thickness a: ff0,9 2 22 uu 7.15 3( ) and Aw w M2 M2 dl li a and since xc = 0: For the present case of a plane fillet weld joint with right angle (equal leg) welds this latter check is not critical. However, it may be critical for partial penetration welds in Izc 2 yc dl bevelled joints. a Instead of having to calculate the stress components in the weld throat the following design Iyc 2 zc dl check expression may be used for y-z plane joints with right angle (equal leg) welds: a 2F2 2 F 2 2 F 2 F 2 Cos 2 F 22 Sin 2 FF Sin 2 FF Cos III w,x w,y w,z w,y w,z w,x w,y w,x w,z xc 22 yc zc ycc z dl 2 a aa fu 2FFw,y w,z Sin Cos a w M2 Note: The subscripts have been shortened: Fw,x for Fwx,Ed etc. 191 Design Example 5 Sheet 4 of 7 In the above expression, the angle is that between the y-axis and the axis of the weld as shown below: attached 1 2 attached element F attached element w,y element Fw,y 1 y y attached support 2 element F support F Fw,z w,x Fw,x w,z z Section 1‑1 z Fillet weld axis Section 2‑2 The force components at the critical point of the weld are determined in the Appendix to this design example. 1. Simplified method The design shear strength for the simplified design approach is: f 530 EN 1993-1-8: u fvw,d 245 N/mm² Eq. 4.4 w M2 3 1, 0 1, 25 3 The value of the resultant induced force per unit length in a weld throat of 1 mm is : 222 222 Fw,Ed FFF wx,Ed wy,Ed wz,Ed 243 747 966 1245 N/mm The required throat thickness is therefore: F 1245 a w,Ed 5,0 mm fvw,d 245 2. directional method At the point (a), where the angle is 0°, the design check expression becomes: 2 2 2 2 fu 2322Fwx,Ed F wy,Ed F wz,Ed FF wx,Ed wz,Ed a M2 The required throat thickness is therefore: 2 ( 243)2 3 (747)2 2 (966)2 2 ( 243) (966) a 4,8 mm 530 /1,25 Adopt a 5 mm throat thickness and assume that the weld is full size over its entire length. Note: A reduction factor is required for splice joints when the effective length of fillet weld is greater than 150a. The reduction factor would seem to be less relevant for the present type of joint. Nevertheless, by considering safely the full length of the welded joint and a throat thickness of 5 mm one obtains: 0,2Lj 0,2 600 EN 1993-1-8 1, 2 1, 2 1, 04 1, 0 LW.1 150a 150 5 Eq. 4.9 Take LW.1 = 1,0. It is concluded that the use of a reduction factor on the design strength of the weld is not required. 192 Design Example 5 Sheet 4 of 7 Design Example 5 Sheet 5 of 7 In the above expression, the angle is that between the y-axis and the axis of the weld as shown below: Appendix – Calculation of the force components at the critical point of the weld attached 1 2 attached element F attached element Geometric properties of the welded joint w,y element Fw,y 1 There are two similar joints, one on each side of the column, resisting the applied loads. y y attached support 2 Only one of the joints needs to be examined. element support Throat area and the positions of the centre of gravity and the critical point F Fw,x F F w,z w,z w,x The throat area (resisting section) of each of the joints made up of straight segments of z Section 1‑1 z Fillet weld axis Section 2‑2 length Li and constant throat thickness a is for each 1 mm of throat thickness: a ds Aw Aw,ii aLw, The force components at the critical point of the weld are determined in the Li 2 175 250 600 mm²/m Appendix to this design example. aa a a Distance of the centre of gravity from the vertical side (parallel to the z axis) of the joint 1. Simplified method with constant throat thickness a: The design shear strength for the simplified design approach is: Aw,i yi EN 1993-1-8: yLii 2 87,5 175 0 250 fu 530 y a 51 mm fvw,d 245 N/mm² Eq. 4.4 Aw,i L 600 3 1, 0 1, 25 3 i w M2 a The value of the resultant induced force per unit length in a weld throat of 1 mm is : yca = +175 ‑ 51 = +124 222 222 51 Fw,Ed FFF wx,Ed wy,Ed wz,Ed 243 747 966 1245 N/mm e yc Load point The required throat thickness is therefore: a rc,a z ca = ‑125 125 e zc Fw,Ed 1245 a 5,0 mm y-y fvw,d 245 c 125 2. directional method d At the point (a), where the angle is 0°, the design check expression becomes: 175 2 z-z f 2 2 2 u The coordinates of the critical point of the joint (a) relative to the principal axes through 2322Fwx,Ed F wy,Ed F wz,Ed FF wx,Ed wz,Ed a M2 the centre of gravity C are: The required throat thickness is therefore: yzca 175 51 124mmca 125mm 2 ( 243)2 3 (747)2 2 (966)2 2 ( 243) (966) Note: The point (d) might also be chosen as a potential critical point, for which: a 4,8 mm 530 /1,25 yzca 175 51 124mmca 125mm Adopt a 5 mm throat thickness and assume that the weld is full size over its entire length. However, for the load case considered it is evident that the point (a) is the most critical. Note: A reduction factor is required for splice joints when the effective length of fillet Second moment of area of the joint resisting section weld is greater than 150a. The reduction factor would seem to be less relevant for For each of the joints, for each 1 mm of throat thickness: the present type of joint. Nevertheless, by considering safely the full length of the 3 Iyc 250 welded joint and a throat thickness of 5 mm one obtains: z2 ds 2 175 1252 6,77 1064 mm /mm a c 12 0,2L EN 1993-1-8 j 0,2 600 3 LW.1 1, 2 1, 2 1, 04 1, 0 I 175 2 150a 150 5 Eq. 4.9 zc y2 ds 250 512 2 2 175 87,5 51 2,01 1064 mm /mm a c 12 Take LW.1 = 1,0. For the “torsion” moment the relevant inertia per joint is: It is concluded that the use of a reduction factor on the design strength of the weld is not 2 2 2 required. Ixc a r c ds a yc ds a zc ds I zc I yc so that: 193 Design Example 5 Sheet 6 of 7 I xc (6,77 2,01) 106 8,78 1064 mm /mm a Applied forces and moments It is assumed that the applied loads and moments are shared equally by both joints. The applied axial and shear force components per joint are: 20 30 300 N 10 kN N 15 kN N 150 kN x,Ed 2 y,Ed 2 z,Ed 2 The applied moments are calculated by using the applied force components and their eccentricities. The eccentricities, i.e. the coordinates of the effective load point, are: exc = 0 as the effective load point is taken to be in the y-z plane of the joint eyc = 300 100 + 175 51 = + 324 mm ezc = 140 mm As a result the applied moments per joint are: Mxc,Ed eN yc z,Ed eN zc y,Ed 324 150 140 15 50,7 kNm Myc,Ed eN zc x,Ed eN xc z,Ed 140 10 0 150 1,4 kNm Mzc,Ed eN xc y,Ed eN yc x,Ed 0 15 324 10 3,24 kNm force components at the critical point of the weld For the y-z plane joint the force components per unit length of the weld, at the point (a) are as follows: Nx,Ed zMca yc,Ed yMca zc,Ed Fwx,Ed Aaw Iyc a Izc a Ny,Ed zMca xc,Ed Fwy,Ed Aaw Ixc a Nzc,Ed yM ca xc,Ed Fwz,Ed Aaw / Ixc a The contributions to the weld force components (at all points of the welded joint) from the applied force components are: N Nx x,Ed 10 Fw,x 17 N/mm Aaw 600 Ny Ny,Ed 15 Fw,y 25 N/mm Aaw 600 N Nz z,Ed 150 Fwz, 250 N/mm Aaw 600 The various contributions to the weld force components per unit length of weld at the point (a) from the applied moment components are : 194 Design Example 5 Sheet 6 of 7 Design Example 5 Sheet 7 of 7 Ixc 6 64 z 125 (6,77 2,01) 10 8,78 10 mm /mm FMMxc c,a 50,7 106 722 N/mm a w,y xc,Ed 6 Iaxc 8,78 10 Applied forces and moments y 124 FMMxc c,a 50,7 106 716 N/mm It is assumed that the applied loads and moments are shared equally by both joints. w,z xc,Ed Ia 8,78 106 The applied axial and shear force components per joint are: xc M z 125 yc c,a 6 20 30 300 FMw,x yc,Ed 1,41 10 6 26 N/mm Nx,Ed 10 kN Ny,Ed 15 kN Nz,Ed 150 kN Ia 6,77 10 2 2 2 yc y 124 Mzc c,a 6 The applied moments are calculated by using the applied force components and their FMw,x zc,Ed 3,24 10 6 200 N/mm Ia 2,01 10 eccentricities. The eccentricities, i.e. the coordinates of the effective load point, are: zc Combining the contributions from the forces and the moments at the point (a) one obtains: exc = 0 as the effective load point is taken to be in the y-z plane of the joint M F FFNx yc FMzc 17 26 200 243 N/mm eyc = 300 100 + 175 51 = + 324 mm wx,Ed w,x w,x w,x N F FFy Mxc 25 722 747 N/mm ezc = 140 mm wy,Ed w,y w,y F FFNz Mxc 250 716 966 N/mm As a result the applied moments per joint are: wz,Ed w,z w,z These resultant design force components per unit length apply to a welded joint with a Mxc,Ed eN yc z,Ed eN zc y,Ed 324 150 140 15 50,7 kNm weld throat thickness of 1 mm throughout its entire effective length. Myc,Ed eN zc x,Ed eN xc z,Ed 140 10 0 150 1,4 kNm Mzc,Ed eN xc y,Ed eN yc x,Ed 0 15 324 10 3,24 kNm force components at the critical point of the weld For the y-z plane joint the force components per unit length of the weld, at the point (a) are as follows: Nx,Ed zMca yc,Ed yMca zc,Ed Fwx,Ed Aaw Iyc a Izc a Ny,Ed zMca xc,Ed Fwy,Ed Aaw Ixc a Nzc,Ed yM ca xc,Ed Fwz,Ed Aaw / Ixc a The contributions to the weld force components (at all points of the welded joint) from the applied force components are: N Nx x,Ed 10 Fw,x 17 N/mm Aaw 600 Ny Ny,Ed 15 Fw,y 25 N/mm Aaw 600 N Nz z,Ed 150 Fwz, 250 N/mm Aaw 600 The various contributions to the weld force components per unit length of weld at the point (a) from the applied moment components are : 195 196 Sheet 1 of 6 Promotion of new eurocode rules for Title Design Example 6 – Bolted joint structural stainless steels (PureSt) Made by IR Date 10/02 Research Fund for Client Revised by MEB Date 04/06 caLcuLatIon SHeet Coal and Steel Revised by UDE Date 01/17 deSIGn exaMPLe 6 – BoLted JoInt An angle (100×100×10) loaded in tension is to be connected to a gusset plate of 10 mm thickness. Stainless steel austenitic grade 1.4401 is used for the angle and the gusset plate. Eight bolts made of austenitic stainless steel, property class 50 with a diameter of 16 mm are used in a staggered line to connect one leg of the angle to the gusset plate. It is required to determine the design resistance of the joint. 100x100x10 angle 10mm thick gusset plate 100 8 M16 property class 50 bolts 18 mm diameter holes 70 40 35 100 25 70 10 9x30 The connection is a Category A: Bearing Type connection. EN 1993-1-8 The design ultimate shear load should not exceed the design shear resistance nor the Clause 3.4.1 design bearing resistance. Material properties Both the angle and the plate are made of stainless steel austenitic grade 1.4401: Table 2.2 2 2 fy = 220 N/mm and fu = 530 N/mm Section 2.3.1 The bolt material is of property class 50: 2 2 fyb = 210 N/mm and fub = 500 N/mm . Table 2.6 Partial factors Partial factor on gross section resistance: M0 = M1 = 1,1 Table 4.1 Partial factor on net section resistance: M2 = 1,25 Partial factor on bolt resistance in shear and in bearing: M2 = 1,25 Position and size of holes Section 7.2.3 For M16 bolts a hole diameter d0 = 18 mm is required. End distances e1 = 30 mm and edge distance e2 = 25 mm. e1 and e2 < 4t + 40 = 4 × 10 + 40 = 80 mm and > 1,2d0 = 1,2 × 18 = 21,6 mm 197 Design Example 6 Sheet 2 of 6 For the staggered bolt rows: - spacing p1 = 60 mm > 2,2d0 = 39,6 mm - distance between two bolts in staggered row: 22 30 35 46,1 mm 2,4d0 43,2 mm - therefore, spacing for staggered rows p2 = 35 mm > 1,2d0 = 21,6 mm Note: For compression loading e2 and p1 should be checked so that they satisfy local buckling requirements for an outstand element and an internal element respectively. Checks on both the angle and gusset plate are required. design resistance of the angle gross cross-section in tension Section 7.2.3 2 Gross cross-sectional area of the angle Ag = 1915 mm Design plastic resistance: Af gy 1915 220 Eq. 7.6 Npl,Rd 3 383 kN M0 1,1 10 design resistance of the angle net cross-section in tension For staggered holes the net cross-sectional area should be taken as the lesser of the gross Section 5.6.4 area minus the deduction for non-staggered holes or: s2 Ag0 t nd 4 p Deductions for non-staggered holes: 2 Ag0 td 1915 10 18 1735 mm Net cross-sectional area through two staggered holes: n = 2, s = 30 mm and p = 35 mm p= 35 2 x 18 mm holes s = 30 s = 30 s22 30 Anet Ag t nd0 1915 10 (2 18) 4 p 4 35 1915 10 36 6,4 1619 mm² Therefore, Anet = 1619 mm². Conservatively the reduction factor for an angle connected by one leg with a single row Table 7.1 of bolts may be used. By interpolation for more than 3 bolts in one row: 3 = 0,57. 198 Design Example 6 Sheet 2 of 6 Design Example 6 Sheet 3 of 6 Design ultimate resistance of the net cross-section of the angle: Section 7.2.3 For the staggered bolt rows: Af 0,57 1619 530 - spacing p1 = 60 mm > 2,2d0 = 39,6 mm N 3 net u 391 kN Eq. 7.10 u,Rd 1,25 103 - distance between two bolts in staggered row: M2 22 30 35 46,1 mm 2,4d0 43,2 mm design resistance of the angle in block tearing - therefore, spacing for staggered rows p2 = 35 mm > 1,2d0 = 21,6 mm The expressions for block tearing are taken from EN 1993-1-8 (instead of EN 1993-1-1) since EN 1993-1-8 explicitly covers angles. Note: For compression loading e2 and p1 should be checked so that they satisfy local buckling requirements for an outstand element and an internal element respectively. Checks on both the angle and gusset plate are required. 60 design resistance of the angle gross cross-section in tension Section 7.2.3 2 Gross cross-sectional area of the angle Ag = 1915 mm 240 Design plastic resistance: Design resistance in block tearing considering rows as staggered: EN 1993-1-8 0,5 fAu nt fAy nv 0,5 530 (60 18) 10 220 (240 4 18) 10 Afgy 1915 220 V N 383 kN Eq. 7.6 eff,2,Rd 3 3 Clause pl,Rd 3 M2 3 M0 1,25 10 3 1,1 10 M0 1,1 10 3.10.2(3) 89 194 283 kN Eq. 3.10 Design resistance in block tearing considering rows as if non staggered: design resistance of the angle net cross-section in tension 0,5 fA fA 0,5 530 (60 18 9) 10 220 (240 3 18 9) 10 EN 1993-1-8 For staggered holes the net cross-sectional area should be taken as the lesser of the gross Section 5.6.4 u nt y nv Veff,2,Rd 3 3 Clause area minus the deduction for non-staggered holes or: 3 1,25 10 3 1,1 10 M2 M0 3.10.2(3) s2 70 204 274 kN Eq. 3.10 Ag0 t nd 4p Deductions for non-staggered holes: design resistance of the gross cross-section of the gusset plate 2 Gross cross-sectional area towards the end of the angle: Section 5.7.2 Ag0 td 1915 10 18 1735 mm 2 Net cross-sectional area through two staggered holes: Ag = 10 × (100 + 70 + 70) = 2400 mm n = 2, s = 30 mm and p = 35 mm Design plastic resistance Af gy 2400 220 Eq. 5.23 Npl,Rd 3 480 kN M0 1,1 10 2 x 18 mm holes p= 35 design resistance of the net cross-section of the gusset plate Net cross-sectional area towards the end of the angle (where the applied load is Section 5.7.2 greatest) through one hole non symmetrically placed on an element of width: b = 100 + 70 +70 = 240 mm 2 s = 30 s = 30 Anet = Ag – d0t = 2400 – 18 × 10 = 2220 mm Net cross-sectional area towards the end of the angle through two staggered holes with s22 30 s = 30 mm and p = 35 mm: Anet Ag t nd0 1915 10 (2 18) 4 p 4 35 st2230 10 A A 2 dt 2400 2 18 10 1915 10 36 6,4 1619 mm² net g04p 4 35 2400 360 64 2104 mm² Therefore, Anet = 1619 mm². Therefore, Anet = 2104 mm². Conservatively the reduction factor for an angle connected by one leg with a single row Table 7.1 of bolts may be used. By interpolation for more than 3 bolts in one row: 3 = 0,57. 199 Design Example 6 Sheet 4 of 6 Design ultimate resistance of the net cross-section of the gusset plate near the end of the angle: kAnet f u Nu,Rd Eq. 5.24 M2 Take factor k = 1,0 for this example (k = 1,0 for sections with smooth holes) 1,0 2104 530 N 892 kN u,Rd 1,25 103 It is advisable to check the resistance of net cross-sections at intermediate cross-sections along the gusset plate. Cross-section at the 1st bolt hole near the gusset plate edge (Where b = 100 + 30 / 240 × 140 = 117,5 mm) 2 Anet = Ag – d0t = 117,5 × 10 – 18 × 10 = 995 mm This cross-section must be capable of transmitting the load from one bolt. Design ultimate resistance at the section: kAnet f u 1,0 995 530 Nu,Rd 3 421 kN Eq. 5.24 M2 1,25 10 It is obvious that there is no need to check any other cross-sections of the gusset plate as the load applied cannot exceed the design resistance of the angle itself, which has been shown to be smaller than the above value. design resistance of the gusset plate in block tearing 3 5 24 0 Design resistance to block tearing considering rows as staggered: EN 1993-1-8 fA fA Clause u nt y nv Veff,1,Rd 3.10.2(2) 3 M2 M0 Eq. 3.9 530 (35 9) 10 220 (240 4 18 240 3 18 9) 10 1,25 103 3 1,1 10 3 110,2 398,4 508 kN Design resistance to block tearing considering rows as if non staggered: EN 1993-1-8 fA fA Clause u nt y nv Veff,1,Rd 3.10.2(2) 3 M2 M0 Eq. 3.9 530 (35 2 9) 10 220 (2 240 6 18 2 9) 10 1,25 103 3 1,1 10 3 72,1 408,8 480 kN 200 Design Example 6 Sheet 4 of 6 Design Example 6 Sheet 5 of 6 Design ultimate resistance of the net cross-section of the gusset plate near the end of the design resistance of the bolts in shear 2 angle: Design resistance of class 50 and M16 bolt of sectional area A = As = 157 mm : kAnet f u α fAub Nu,Rd Eq. 5.24 Fv,Rd Eq. 7.11 M2 M2 Take factor k = 1,0 for this example (k = 1,0 for sections with smooth holes) The value of may be defined in the National Annex. The recommended value is 0,6, Section 7.2.4 1,0 2104 530 which applies if the shear plane passes through the unthreaded or threaded portions of N 892 kN u,Rd 1,25 103 the bolt. It is advisable to check the resistance of net cross-sections at intermediate cross-sections αfAub 0,6 500 157 Fv,Rd 3 37,7 kN along the gusset plate. M2 1,25 10 Cross-section at the 1st bolt hole near the gusset plate edge Design resistance of the bolt group in shear: nb Fv,Rd = 8 × 37,7 = 302 kN (Where b = 100 + 30 / 240 × 140 = 117,5 mm) 2 Anet = Ag – d0t = 117,5 × 10 – 18 × 10 = 995 mm This cross-section must be capable of transmitting the load from one bolt. design resistance of the bolts/ply in bearing Design ultimate resistance at the section: The design resistance for bolted connections susceptible to bearing failure is given by: Section 7.2.3 kAnet f u 1,0 995 530 2,5αbtk td f u N 421 kN Fb,Rd Eq. 7.1 u,Rd 3 Eq. 5.24 M2 1,25 10 M2 It is obvious that there is no need to check any other cross-sections of the gusset plate as Design resistance in bearing on the ply with t = 10 mm for the M16 bolt at the end. the load applied cannot exceed the design resistance of the angle itself, which has been End distances e1 = 30 mm, edge distances e2 = 25 mm ( > 1,2d0 = 21,6 mm), and bolt shown to be smaller than the above value. spacings p1 = 60 mm and p2 = 35 mm. Bolted connections are classified into two groups, based on the thickness of the Section 7.2.3 connected plates. Thick plate connections are those between plates with thicknesses design resistance of the gusset plate in block tearing greater than 4 mm, while connections between plates with thicknesses less than or equal to 4 mm are defined as thin plate connections. This example is a thick plate connection with tmin = 10 mm and deformation should not be a key design consideration. For the end bolt nearest the ends where e1 = 30 mm and p1 = 60 mm the bearing 3 coefficient b in the load transfer direction is determined as follows: 5 1, 0 b min e1 3d0 24 1, 0 0 Design resistance to block tearing considering rows as staggered: EN 1993-1-8 min 30 0,556 0,556 fA fA Clause 3 18 u nt y nv Veff,1,Rd 3.10.2(2) t M2 3 M0 The bearing coefficient k in the direction perpendicular to load transfer is determined Eq. 3.9 as follows: 530 (35 9) 10 220 (240 4 18 240 3 18 9) 10 3 3 1,25 10 3 1,1 10 e2 1, 0 for 1, 5 110,2 398,4 508 kN d0 kt Design resistance to block tearing considering rows as if non staggered: EN 1993-1-8 e 2 Clause 0,8 for 1,5 fAu nt fAy nv d V 0 eff,1,Rd 3.10.2(2) M2 3 M0 e 25 Eq. 3.9 k 0, 8 for 2 1, 39 1, 5 530 (35 2 9) 10 220 (2 240 6 18 2 9) 10 t d0 18 1,25 103 3 1,1 10 3 72,1 408,8 480 kN 201 Design Example 6 Sheet 6 of 6 The design resistance for this bolted connection susceptible to bearing failure for the end bolt is as follows: 2,5αk td f 2,5 0,556 0,8 10 16 530 F bt u 75,44 kN b,Rd 3 Eq. 7.1 M2 1,25 10 Design resistance of the joint in bearing: nb Fb,Rd = 8 × 75,44 = 604 kN design resistance of the joint at the ultimate Limit State Design resistance of the angle gross cross-section in tension Npl,Rd 383 kN Design resistance of the angle net cross-section in tension Nu,Rd 391 kN Design resistance of the angle in block tearing (for staggered rows) Veff,2,Rd 283 kN Design resistance of the angle in block tearing (for non staggered rows) Veff,2,Rd 274 kN Design resistance of the gusset plate gross cross-section in tension Npl,Rd 480 kN Design resistance of the gusset plate net cross-section in tension Nu,Rd 892 kN Design resistance of the gusset plate net cross-section in tension Nu,Rd 421 kN (at the 1st bolt hole near the gusset plate edge) Design resistance of the gusset plate in block tearing (for staggered rows) Veff,1,Rd 508 kN Design resistance of the gusset plate in block tearing (for non staggered rows) Veff,1,Rd 480 kN Design resistance of the bolts in shear Fv,Rd 302 kN Design resistance of the bolts/ply in bearing Fb,Rd 604 kN The smallest design resistance is for the angle in block tearing (for non staggered rows): Veff,2,Rd = 274 kN Note: The critical mode for all of the bolts in the joint is shear (Fv,Rd = 302 kN). 202 Design Example 6 Sheet 6 of 6 Sheet 1 of 5 Promotion of new The design resistance for this bolted connection susceptible to bearing failure for the end eurocode rules for bolt is as follows: Title Design Example 7 – Shear resistance of plate girder structural stainless 2,5αk td f 2,5 0,556 0,8 10 16 530 F bt u 75,44 kN steels (PureSt) b,Rd 3 Eq. 7.1 Made by AO Date 06/02 M2 1,25 10 Research Fund for Client Revised by MEB Date 04/06 Design resistance of the joint in bearing: caLcuLatIon SHeet Coal and Steel Revised by ER/IA Date 04/17 nb Fb,Rd = 8 × 75,44 = 604 kN deSIGn exaMPLe 7 – SHear reSIStance of PLate GIrder Design a plate girder with respect to shear resistance. The girder is a simply supported design resistance of the joint at the ultimate Limit State I-section with a span according to the figure below. The top flange is laterally restrained. Design resistance of the angle gross cross-section in tension Npl,Rd 383 kN FEd = 440 kN Design resistance of the angle net cross-section in tension Nu,Rd 391 kN Design resistance of the angle in block tearing (for staggered rows) Veff,2,Rd 283 kN hw Design resistance of the angle in block tearing (for non staggered rows) Veff,2,Rd 274 kN Design resistance of the gusset plate gross cross-section in tension Npl,Rd 480 kN bf Design resistance of the gusset plate net cross-section in tension Nu,Rd 892 kN Design resistance of the gusset plate net cross-section in tension Nu,Rd 421 kN 1250 1250 (at the 1st bolt hole near the gusset plate edge) Design resistance of the gusset plate in block tearing (for staggered rows) Veff,1,Rd 508 kN Use lean duplex grade 1.4162 2 fy = 480 N/mm for hot rolled strip Table 2.2 Design resistance of the gusset plate in block tearing (for non staggered rows) Veff,1,Rd 480 kN E = 200000 N/mm2 Section 2.3.1 Design resistance of the bolts in shear Fv,Rd 302 kN Try a cross section with 2 Design resistance of the bolts/ply in bearing Fb,Rd 604 kN Flanges: 12 200 mm Web: 4 500 mm2 The smallest design resistance is for the angle in block tearing (for non staggered rows): Stiffeners: 12 98 mm2 Weld throat thickness: 4 mm Veff,2,Rd = 274 kN Note: The critical mode for all of the bolts in the joint is shear (Fv,Rd = 302 kN). Structural analysis Maximum shear and bending moment are obtained as F 440 V Ed 220kN Ed 2 2 LF 440 5,2 M Ed 275kNm Ed 4 4 Partial factors M0 = 1,1 Table 4.1 M1 = 1,1 classification of the cross-section Section 5.3 235 200 Table 5.2 = ,0 683 480 210 203 Design Example 7 Sheet 2 of 5 Web, subject to bending Table 5.2 c 500 2 2 4 = 178,9 90 therefore the web is Class 4. t 4 0,683 Flange, subject to compression Table 5.2 c 200 4 2 2 4 = 11,3 14,0 therefore the compression flange is Class 3 t 2 12 0,683 Thus, overall classification of cross-section is Class 4. Shear resistance Section 6.4.3 24,3 The shear buckling resistance requires checking when ht/ kfor vertically ww stiffened webs. a/hw = 1250/500 = 2,5 > 1, and since the web is not stiffened, kst=0. Hence, 22 hw 500 k = 5, 34 4 5, 34 4 5, 98 Eq. 6.26 a 1250 EN 1993-1-4 recommended value for 2,1 Section 6.4.3 500 24,3 hw/tw = 125 0,683 5,98 33,8 4 1, 2 Therefore, the shear buckling resistance has to be checked. It is obtained as fyw ht w w 1,2 480 500 4 3 Vb,Rd = Vbw,Rd + Vbf,Rd 10 604,6 kN Eq. 6.22 3 M1 3 1,1 wf yw ht w w Vbw,Rd = Eq. 6.23 3 M1 For non-rigid end posts: h 500 = w = 2,00 0,65 w Eq. 6.25 37,4tkw 37,4 4 0,683 5,98 1,19 w = for w 0,65 Table 6.3 0,54 w Hence the contribution from the web is obtained as 1,19 w = 0,468 Table 6.3 0,54 2,00 wf yw ht w w 0,468 480 500 4 3 Vbw,Rd = 10 = 235,9 kN Eq. 6.23 3 M1 3 1,1 The contribution from the flanges may be utilised if the flanges are not fully utilised in Section 6.4.3 withstanding the bending moment. The bending resistance of a cross section consisting of the flanges only is obtained as 480 6 Mf,Rd = 12 200 (500 12) 10 536,2 kNm 1,1 Mf,Rd > MEd = 275 kNm, therefore the flanges can contribute to the shear buckling resistance. 204 Design Example 7 Sheet 2 of 5 Design Example 7 Sheet 3 of 5 Web, subject to bending Table 5.2 2 2 ftb ff yf MEd c 500 2 2 4 Vbf,Rd = 1 Eq. 6.29 = 178,9 90 therefore the web is Class 4. c M t 4 0,683 M1 f,Rd Flange, subject to compression Table 5.2 3,5bt2 f c f f yf c 200 4 2 2 4 c = a 0,17 2 but ,0 65 Eq. 6.30 = 11,3 14,0 therefore the compression flange is Class 3 thw w f yw a t 2 12 0,683 5,3 2 48012200 Thus, overall classification of cross-section is Class 4. = 1250 ,0 17 2 3385, mm 4 480500 Shear resistance Section 6.4.3 338,5mm < 0,65 1250 812,5mm 24,3 2 2 The shear buckling resistance requires checking when ht/ kfor vertically 48012200 275 ww Vbf,Rd = 1 27 4, kN Eq. 6.29 338 1,1 5362, stiffened webs. Vb,Rd = Vbw,Rd + Vbf,Rd = 235,8 + 27,4 = 263,2 kN 604,6 kN Eq. 6.22 a/hw = 1250/500 = 2,5 > 1, and since the web is not stiffened, kst=0. Hence, 22 hw 500 k = 5, 34 4 5, 34 4 5, 98 Eq. 6.26 transverse stiffeners Section 6.4.5 a 1250 The transverse stiffeners have to be checked for crushing and flexural buckling using Table 6.1 EN 1993-1-4 recommended value for 2,1 Section 6.4.3 = 0,49 and 0 = 0,2. An effective cross section consisting of the stiffeners and parts of 500 24,3 the web is then used. The part of the web included is 11 tw wide, therefore the cross hw/tw = 125 0,683 5,98 33,8 4 1, 2 section of the transverse stiffener is Class 3. Therefore, the shear buckling resistance has to be checked. It is obtained as / ha w 1250/500 2 , 25 , hence the second moment of area of the intermediate Eq. 6.51 fyw ht w w 1,2 480 500 4 3 stiffener has to fulfil Vb,Rd = Vbw,Rd + Vbf,Rd 10 604,6 kN Eq. 6.22 3 3 4 Eq. 6.51 3 M1 3 1,1 st ,0 75 hI t ww ,0 50075 4 24000 mm f ht 3 3 w yw w w (11 ,0 683 4 4) 20012 6 4 Vbw,Rd = Eq. 6.23 Ist = 2 ,8 1000 mm , hence fulfilled. 3 M1 12 12 For non-rigid end posts: The crushing resistance is obtained as h 500 = w = 2,00 0,65 Eq. 6.25 Nc,Rd = Ag fy/M0 Eq. 5.27 w 37,4tkw 37,4 4 0,683 5,98 2 Ag = ( 1120012 ,0 683 24 ) 24601, mm 1,19 3 w = for w 0,65 Table 6.3 Nc,Rd = 2460,1 480 10 /1,1 1073,5kN 0,54 w The flexural buckling resistance is obtained as Hence the contribution from the web is obtained as Nb,Rd = A fy / M1 Eq. 6.2 1,19 w = 0,468 1 Table 6.3 0,54 2,00 = 5,0 1 Eq. 6.4 22 wf yw ht w w 0,468 480 500 4 3 Vbw,Rd = 10 = 235,9 kN Eq. 6.23 2 Eq. 6.5 = 0 , 15 0 3 M1 3 1,1 L 1 fyw = cr Eq. 6.6 The contribution from the flanges may be utilised if the flanges are not fully utilised in Section 6.4.3 i E withstanding the bending moment. The bending resistance of a cross section consisting of Lcr = 0,75hw = 0,75 500 = 375 mm Section 6.4.5 the flanges only is obtained as 375 1 480 480 6 = 0,103 Mf,Rd = 12 200 (500 12) 10 536,2 kNm 6 200000 Eq. 6.6 1,1 8 10 2460,1 Mf,Rd > MEd = 275 kNm, therefore the flanges can contribute to the shear buckling resistance. = 0 , 015 , 49 ,0 103 ,02,0 1032 ,0 48 Eq. 6.5 205 Design Example 7 Sheet 4 of 5 1 = 5,0 ,1 05 11 , 0 Eq. 6.4 ,0 48 ,0 482 ,0 1032 Since Nb,Rd = Nc,Rd =1073,5 kN > NEd, the transverse stiffeners are sufficient. Interaction shear and bending Section 6.4.3 If the utilization of shear resistance, expressed as the factor 3 , exceeds 0,5, the combined effect of bending and shear has to be checked. VEd 3 = 0,1 Eq. 6.36 Vbw,Rd 220 = ,0 933 5,0 , therefore interaction has to be considered. 3 235,9 The condition is M 2 M f,Rd f, Rd 1 1 2 3 11 , 0 for 1 Eq. 6.34 Mpl, Rd M pl,Rd Where: MEd 1 = Eq. 6.35 M pl, Rd Mf,Rd = 536,2 kNm (Sheet 3) Mpl,Rd is the plastic resistance of the cross-section. 2 2 ht ww fh y 4 480500 Mpl,Rd = M ,f Rd 5362, 6 6453, kNm 4 M0 14 , 1 10 Evaluate conditions MEd = 275 kNm, hence: 275 = ,0 426 0,1 OK Eq. 6.35 1 6453, 1 fulfils its condition. Now it remains to check the interaction. M 2 536 2, f, Rd 2 1 1 3 12 0 , 426 1 02 , 933 01 , 553 0,1 M pl,Rd 645 3, It therefore follows that under the conditions given, the resistance of the plate girder is sufficient with respect to shear, bending as well as interaction between shear and bending. Calculation of effective cross-section properties The flanges are Class 3 and hence fully effective. The depth of the web has to be reduced with the reduction factor , welded web. ,0 772 ,0 079 = 2 1 Eq. 5.1 p p / tb p = where b = d = 500 2 24 488,68 mm Eq. 5.3 28 4, k 206 Design Example 7 Sheet 4 of 5 Design Example 7 Sheet 5 of 5 1 Assuming linearly varying, symmetric stress distribution within the web, = ,1 05 11 , 0 5,0 Eq. 6.4 ,0 48 ,0 482 ,0 1032 = 2 = 1 1 Since Nb,Rd = Nc,Rd =1073,5 kN > NEd, the transverse stiffeners are sufficient. k = 23,9 Table 5.3 488,68 4/ Interaction shear and bending p = ,1 29 Eq. 5.3 Section 6.4.3 28 , 04 , 23683 9, If the utilization of shear resistance, expressed as the factor 3 , exceeds 0,5, the ,0 772 ,0 079 combined effect of bending and shear has to be checked. = ,0 55 1 Eq. 5.1 ,1 29 ,1 292 V = Ed 0,1 3 Eq. 6.36 beff = bc = b / (1-) = ,0 48855 ,68/( (1 1 )) 134,76mm Table 5.3 Vbw,Rd 220 be1 = 0,4beff = 4,0 134, 5376 9, mm Table 5.3 3 = ,0 933 5,0 , therefore interaction has to be considered. 235,9 be2 = 0,6beff = 6,0 134, 8076 9, mm The condition is Calculation of effective section modulus under bending Mf,Rd 2 M f, Rd 1 2 11 , 0 for 1 Eq. 6.34 1 3 ei is taken as positive from the centroid of the upper flange and downwards. Mpl, Rd M pl,Rd Where: be1 MEd 1 = Eq. 6.35 M pl, Rd Mf,Rd = 536,2 kNm (Sheet 3) be2 Mpl,Rd is the plastic resistance of the cross-section. 2 2 ht ww fh y 4 480500 Mpl,Rd = M ,f Rd 5362, 6 6453, kNm 4 M0 14 , 1 10 hw/2 Evaluate conditions MEd = 275 kNm, hence: 2 275 Aeff = Ai btff 2 be1 42 tw b e2 t w h w/ 2 t w 6361,7 mm = ,0 426 0,1 OK Eq. 6.35 i 1 6453, eeff = fulfils its condition. Now it remains to check the interaction. 1 1 1 1 Ae bt0 bt h t b 42 t 0,5 b 42 t ii ff ff w f e1 w e1 f Aeff i Aeff Aeff M f, Rd 2 536 2, 2 1 1 3 12 0 , 426 1 02 , 933 01 , 553 0,1 bte2 w0.5 h w t f b e2 / 2 hww / 2 t 0,75 h w 0,5 t f = 266,4 mm M pl,Rd 645 3, 3 It therefore follows that under the conditions given, the resistance of the plate girder is 33tb 42 3 2 btff w e1 tbw e2 thww/2 sufficient with respect to shear, bending as well as interaction between shear and bending. Ieff = Iii Ae(eff ei )2 ii 12 12 12 12 2 2 2 bt e 0 bt e h t b 42 t e 0,5 b 42 t Calculation of effective cross-section properties f f eff f f eff w f e1 w eff e1 f The flanges are Class 3 and hence fully effective. 22 btee2 w eff 0,5 hw t f b e2 hw/ 2 te w eff 0,75 hwf 0,5 t The depth of the web has to be reduced with the reduction factor , welded web. 8 4 ,0 772 ,0 079 = 3,472 10 mm = 2 1 Eq. 5.1 p p / tb p = where b = d = 500 2 24 488,68 mm Eq. 5.3 28 4, k 207 208 Sheet 1 of 5 Promotion of new eurocode rules for Design Example 8 – Resistance to concentrated Title structural stainless loads steels (PureSt) Made by AO Date 06/02 Research Fund for Client Revised by MEB Date 04/06 caLcuLatIon SHeet Coal and Steel Revised by ER/IA Date 04/17 deSIGn exaMPLe 8 – reSIStance to concentrated LoadS An existing plate girder, previously subjected to an evenly distributed load, will be refurbished and will be subjected to a concentrated load. Check if the girder can resist the new load applied through a 12 mm thick plate. The girder is a simply supported I-section with a span according to the figure below. The top flange is laterally restrained. FEd = 110 kN 500 200 1250 1250 Use duplex grade 1.4462 2 fy = 460 N/mm for hot rolled strip Table 2.2 E = 200000 N/mm2 Section 2.3.1 Flanges: 12 200 mm2 Web: 4 500 mm2 Stiffeners: 12 98 mm2 Weld throat thickness: 4 mm Structural analysis Maximum shear and bending moment are obtained as FEd 110 VEd = 55kN 2 2 Ed LF 110 5,2 MEd = 68,75kNm 4 4 Partial safety factors M0 = 1,1 Table 4.1 M1 = 1,1 classification of the cross-section Section 5.3 235 200 = ,0 698 Table 5.2 460 210 209 Design Example 8 Sheet 2 of 5 Web, subject to bending Table 5.2 c 500 22 4 = 1751, 90, therefore the web is Class 4. t 04 , 698 Flange, subject to compression Table 5.2 c 200 4 22 4 = 110, 14 0, , and the compression flange is Class 3. t 2 12 ,0 698 Thus, overall classification of cross-section is Class 4. resistance to concentrated force Section 6.4.4 The design load should not exceed the design resistance, i.e. FRd = yw Lf eff t w / 1M Eq. 6.37 The effective length Leff is given by Leff = lyF Eq. 6.45 where the reduction function is 5,0 F = 0.1 Eq. 6.46 F with the slenderness given by tl wy ft yw F = Eq. 6.47 Fcr The effective loaded length is given by Eq. 6.41 ly = s s 12 mt 1f m 2 Where ss is the length of the stiff bearing and m1 and m2 are dimensionless parameters: yfbf f m1 = Eq. 6.38 ywtf w 2 h w m2 = ,0 02 o r F 0f , 5 Eq. 6.39 tf m2 = 0 o r F 0f , 5 Eq. 6.40 ss is conservatively taken as twice the thickness of the load bearing plate, i.e. 24 mm. Figure 6.5 200460 m1 = 50 Eq. 6.38 460 4 2 500 Eq. 6.39 m2 = ,0 02 34 7, , assuming F 5,0 12 ly = 24 2 12 1 50 34,7 268,9 mm Eq. 6.41 The critical load is obtained as 3 tw Fcr = 9,0 F Ek Eq. 6.48 hw where the buckling coefficient is given by the load situation, type a. 2 2 hw 500 kF = 26 = 26 ,6 08 Figure 6.4 a 2500 210 Design Example 8 Sheet 2 of 5 Design Example 8 Sheet 3 of 5 Web, subject to bending Table 5.2 43 3 Eq. 6.48 c 500 22 4 Fcr = 0,9 6,08 200000 10 140,1kN = 1751, 90, therefore the web is Class 4. 500 t 04 , 698 268,9 4 460 Flange, subject to compression Table 5.2 F = 3 1, 88 0, 5 , assumption OK Eq. 6.47 140,1 10 c 200 4 22 4 = 110, 14 0, , and the compression flange is Class 3. 5,0 t 2 12 ,0 698 = ,0 27 0,1 , OK Eq. 6.46 F ,1 88 Thus, overall classification of cross-section is Class 4. Leff = 0,27 268,9 72,6 mm resistance to concentrated force Section 6.4.4 3 The design load should not exceed the design resistance, i.e. FEd = 110 460 72,6 4 / (1,1 10 ) 121,4 kN Eq. 6.37 FRd = yw Lf eff t w / 1M Eq. 6.37 Hence the resistance exceeds the load. The effective length Leff is given by Interaction between transverse force, bending moment and axial force Leff = lyF Eq. 6.45 Interaction between concentrated load and bending moment is checked according to where the reduction function is EN1993-1-5:2006. 5,0 EN 1993-1-5, F = 0.1 Eq. 6.46 8,0 21 4,1 F Eq. 7.2 with the slenderness given by Where N NM e EN 1993-1-5, tl ft = Ed EdEd N 1 0, wy yw 1 Eq. 4.14 F = Eq. 6.47 Af e f fy / M0 Wf e f fy / M0 Fcr F EN 1993-1-5, The effective loaded length is given by = Ed 0,1 2 Eq. 6.14 Eq. 6.41 yw Lf eff t w / 1M ly = s s 12 mt 1f m 2 Where ss is the length of the stiff bearing and m1 and m2 are dimensionless parameters: Calculation of effective cross-section properties The flanges are Class 3 and hence fully effective. yfbf f m1 = Eq. 6.38 The depth of the web has to be reduced with the reduction factor , welded web. ywtf w ,0 772 ,0 079 2 = 2 1 Eq. 5.1 hw p p m2 = ,0 02 o r F 0f , 5 Eq. 6.39 t f / tb p = where b = d = 500 2 24 488,68 mm Eq. 5.3 m2 = 0 o r F 0f , 5 Eq. 6.40 28 4, k ss is conservatively taken as twice the thickness of the load bearing plate, i.e. 24 mm. Figure 6.5 Assuming linearly varying symmetric stress distribution within the web, 200460 2 m1 = 50 Eq. 6.38 = = 1 460 4 1 2 500 Eq. 6.39 k = 23,9 Table 5.3 m2 = ,0 02 34 7, , assuming F 5,0 12 488,68 4/ p = ,1 26 28 , 04 , 23698 9, ly = 24 2 12 1 50 34,7 268,9 mm Eq. 6.41 ,0 772 ,0 079 The critical load is obtained as = ,0 562 1 ,1 26 ,1 262 3 tw Table 5.3 Fcr = 9,0 F Ek Eq. 6.48 beff = bc = b / (1- ) = 0,562 488,68 / (1 ( 1)) 137,3mm hw be1 = 0,4beff = 0,4 137,3 54,9 mm where the buckling coefficient is given by the load situation, type a. be2 = 0,6beff = 0,6 137,3 82,4 mm 2 2 hw 500 kF = 26 = 26 ,6 08 Figure 6.4 a 2500 211 Design Example 8 Sheet 4 of 5 Calculate effective section modulus under bending ei is taken as positive from the centroid of the upper flange and downwards. be1 be2 hw/2 2 Aeff = Ai btff 2 be1 42 tw b e2 t w h w/ 2 t w 6372,2 mm i eeff= 1 1 1 Ae bt0 bt h t b 42 t 0,5 b 42 t ii ff ff w f e1 w e1 f Aeff i Aeff Aeff bte2 w0,5 h w t f b e2 / 2 hww / 2 t 0,75 h w 0,5 t f = 266,4 mm 3 33tb 42 3 2 btff w e1 tbw e2 thww/2 Ieff = Iii Ae(eff ei )2 ii 12 12 12 12 2 2 2 bt e 0 bt e h t b 42 t e 0,5 b 42 t f f eff f f eff w f e1 w eff e1 f 2 2 btee2 w eff 0,5 hw t f b e2 hw/ 2 te w eff 0,75 hwf 0,5 t = 3,475 108 mm4 Ieff 6 3 Weff = ,1 10293 mm eff 0e , 5 t f 68, 1075 6 EN 1993-1-5 1 = ,0 127 460 ,1 10293 6 1/ , 1 Eq. 4.14 110 EN 1993-1-5 = ,0 919 2 119,63 Eq. 6.14 0,812 0,8 0,1293 0,919 1,021 1,4 Therefore, the resistance of the girder to interaction between concentrated load and bending moment is adequate. Shear resistance Section 6.4.3 56 2, The shear buckling resistance requires checking when /h t for unstiffened ww Eq. 6.20 webs. 56 2, /h t 500 125 ,0 32698 7, ww 4 2,1 Therefore the shear buckling resistance has to be checked. It is obtained as 212 Design Example 8 Sheet 4 of 5 Design Example 8 Sheet 5 of 5 Calculate effective section modulus under bending fyw ht w w Vb,Rd = VV Eq. 6.22 ei is taken as positive from the centroid of the upper flange and downwards. bw,Rd bf,Rd 3 M1 be1 w yw hf t ww Vbw,Rd = Eq. 6.23 M1 3 For non-rigid end posts Table 6.3 provides h 500 be2 w w = = ,2 07 ,0 65 Eq. 6.24 86 4, tw 86 , 044 , 698 1,19 w = for w 0,65 Table 6.3 hw/2 0,54 w 1,19 = 0,455 Table 6.3 w 0,54 2,07 The contribution from the flanges may be utilised if the flanges are not fully utilised to 2 Aeff = Ai btff 2 be1 42 tw b e2 t w h w/ 2 t w 6372,2 mm withstand the bending moment. However, the contribution is small and is conservatively i not taken into account, i.e. Vbf,Rd 0 . eeff= 1 1 1 The shear buckling resistance can be calculated as: Ae bt0 bt h t b 42 t 0,5 b 42 t ii ff ff w f e1 w e1 f f ht Aeff i Aeff Aeff 0,455 460 500 4 3 yw w w Vb,Rd = Vbw,Rd = 10 219,8 kN < 579,45 kN Eq. 6.23 1,1 3 3 M1 bte2 w0,5 h w t f b e2 / 2 hww / 2 t 0,75 h w 0,5 t f = 266,4 mm 3 Vb,Rd = Vbw,Rd > VEd = 55 kN 33tb 42 3 2 btff w e1 tbw e2 thww/2 The shear resistance of the girder is thus adequate. Ieff = Iii Ae(eff ei )2 ii 12 12 12 12 2 2 2 bt e 0 bt e h t b 42 t e 0,5 b 42 t Interaction between shear and bending f f eff f f eff w f e1 w eff e1 f 2 2 If 3 does not exceed 0,5, the resistance to bending moment and axial force does not btee2 w eff 0,5 hw t f b e2 hw/ 2 te w eff 0,75 hwf 0,5 t need to be reduced to allow for shear. = 3,475 108 mm4 V = Ed 1, 0 Ieff 6 3 3 Eq. 6.36 Weff = ,1 10293 mm Vbw,Rd eff 0e , 5 t f 55 68, 1075 6 EN 1993-1-5 = 0,25 0,5 , therefore interaction need not to be considered. 1 = ,0 127 219,8 460 ,1 10293 6 1/ , 1 Eq. 4.14 concluding remarks 110 EN 1993-1-5 = ,0 919 2 119,63 Eq. 6.14 The resistance of the girder exceeds the load imposed. Note that the vertical stiffeners at supports have not been checked. It should be done according to the procedure used in 0,812 0,8 0,1293 0,919 1,021 1,4 Design Example 7. Therefore, the resistance of the girder to interaction between concentrated load and bending moment is adequate. Shear resistance Section 6.4.3 56 2, The shear buckling resistance requires checking when /h t for unstiffened ww Eq. 6.20 webs. 56 2, /h t 500 125 ,0 32698 7, ww 4 2,1 Therefore the shear buckling resistance has to be checked. It is obtained as 213 214 Sheet 1 of 7 Promotion of new eurocode rules for Design Example 9 - Beam with unrestrained Title structural stainless compression flange steels (PureSt) Made by SMH Date 09/01 Research Fund for Client Revised by NRB Date 04/06 caLcuLatIon SHeet Coal and Steel Revised by SJR Date 04/17 deSIGn exaMPLe 9 - BeaM wItH unreStraIned coMPreSSIon fLanGe Design a staircase support beam. The beam is a channel, simply supported between columns. The flight of stairs between A and C provides restraint to the top flange of this part of the beam. The top flange is unrestrained between B and C. The overall span of the beam is taken as 4,2 m. w 1,5 m 1,2 m 1,2 Down Beam 3 A C B RA RB 1,5 m 1,5 m 2,7 m restrained unrestrained 2,2 m actions Assuming the beam carries the load from the first run of stairs to the landing only: Permanent actions (G): Load on stairs 1,0 kN/m2 = 1,0 2,2 = 2,2 kN/m Self-weight of beam 0,13 kN/m Variable actions (Q): Load on stairs 4 kN/m2 = 4,0 2,2 = 8,8 kN/m Load case to be considered (ultimate limit state): G,jG k,j + Qk,1Q,1 + Q,i 0,iQ k,i j1 i1 As there is only one variable action (Qk,1 the last term in the above expression does not need to be considered in this example. G, j = 1,35 (unfavourable effects) Q,1 = 1,5 Factored actions Permanent action: Load on stairs = 1,35 2,2 = 2,97 kN/m Self-weight of beam = 1,35 0,13 = 0,18 kN/m Variable action: Load on stairs = 1,5 8,8 = 13,2 kN/m Structural analysis Reaction at support points: RA + RB = (2,97 + 13,2) 1,5 + 0,18 4,2 = 25,01 kN 215 Design Example 9 Sheet 2 of 7 Taking moments about A: 5,1 , (5 2 , 1397 , 2 0) , 75 ,0 18 4 , (2 4 , 2 / 2 ) RB = = 4,71 kN 2,4 RA = 25,01– 4,71 = 20,30 kN 1, 5 Maximum bending moment occurs at a distance:1, 5 1 = 1,23 m from A. 2 4,2 ,1 232 ,1 232 MEd,max = 20,30 ,1 23 2( , 1397 )2, ,0 18 = 12,60 kNm 2 2 Maximum shear occurs at A: FEd,max = 20,30 kN Material properties Use austenitic grade 1.4401 0,2% proof stress = 240 N/mm2 (for cold formed steel sheet) Table 2.2 2 fy = 240 N/mm E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3.1 Try a 200 75 channel section, thickness t = 5 mm. cross-section properties 6 4 3 3 Iy = 9,456 10 mm Wel,y = 94,56 10 mm 6 4 3 3 Iz = 0,850 10 mm Wpl,y = 112,9 10 mm 6 4 2 Iw = 5085 10 mm Ag = 1650 mm 4 4 It = 1,372 10 mm classification of the cross-section Section 5.3.2 235 E 235 200 000 = = 0,97 Table 5.2 f y 210 000 240 210 000 Assume conservatively that c = h – 2t = 200 – 2 5 = 190 mm for web c 190 Web subject to bending: 38 t 5 c For Class 1, 72 69,8, therefore web is Class 1. Table 5.2 t c 75 Outstand flange subject to compression: 15 t 5 c For Class 3, 14 13,6 , therefore outstand flange is Class 4. Table 5.2 t Therefore, overall classification of cross-section is class 4. 216 Design Example 9 Sheet 2 of 7 Design Example 9 Sheet 3 of 7 Taking moments about A: calculation of effective section properties Section 5.4.1 5,1 , (5 2 , 1397 , 2 0) , 75 ,0 18 4 , (2 4 , 2 / 2 ) Calculate reduction factor for cold formed outstand elements: RB = = 4,71 kN 2,4 1 0,188 2 but 1 Eq. 5.2 RA = 25,01– 4,71 = 20,30 kN pp ≤ bt/ p where b = c = 75 mm Eq. 5.3 1, 5 28,4 k Maximum bending moment occurs at a distance:1, 5 1 = 1,23 m from A. 2 4,2 Assuming uniform stress distribution within the compression flange ,1 232 ,1 232 MEd,max = 20,30 ,1 23 2( , 1397 )2, ,0 18 = 12,60 kNm = 2 = 1 k = 0,43 Table 5.4 2 2 1 Maximum shear occurs at A: 75 / 5 p 0,830 FEd,max = 20,30 kN 28,4 0,97 0,43 1 0,188 Material properties 2 0,932 0,830 0,830 Use austenitic grade 1.4401 Table 5.4 0,2% proof stress = 240 N/mm2 (for cold formed steel sheet) Table 2.2 ceff = c = 0,932 75 = 69,9 2 fy = 240 N/mm 2 Aeff = Ag 1 ct = 1650 1 0,932 75 5 1 625 mm E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3.1 Calculate shift of neutral axis of section under bending: Try a 200 75 channel section, thickness t = 5 mm. Non-effective zone cross-section properties 6 4 3 3 Iy = 9,456 10 mm Wel,y = 94,56 10 mm Centroidal axis of 6 4 3 3 gross cross-section Iz = 0,850 10 mm Wpl,y = 112,9 10 mm y - y 6 4 2 Iw = 5085 10 mm Ag = 1650 mm 4 4 It = 1,372 10 mm Centroidal axis of y effective cross-section classification of the cross-section Section 5.3.2 235 E 235 200 000 = = 0,97 Table 5.2 ht 200 5 f y 210 000 240 210 000 Ag 1 ct h 1650 1 0,932 75 5 200 2 2 2 2 y Assume conservatively that c = h – 2t = 200 – 2 5 = 190 mm for web Aeff 1625 c 190 Web subject to bending: 38 t 5 y 98,44 c For Class 1, 72 69,8, therefore web is Class 1. Table 5.2 h 200 t Shift of neutral axis position, y – y = y 98,44 1,56 mm 22 3 2 1 ct ht 2 c 75 Ieff,y = I ct A Outstand flange subject to compression: 15 y 1 eff y-y t 5 12 22 c 3 For Class 3, 14 13,6 , therefore outstand flange is Class 4. 621 0,932 75 5 2 Table 5.2 Ieff,y = 9,456 10 1 0,932 75 5 100 2,5 1625 1,56 t 12 Therefore, overall classification of cross-section is class 4. 6 4 Ieff,y = 9,21 10 mm 6 Ieff,y 9,21 10 3 3 Weff,y = 90,69 10 mm h 200 1, 56 22y-y 217 Design Example 9 Sheet 4 of 7 Shear lag Section 5.4.2 Shear lag may be neglected provided that b0 Le/50 for outstand elements. Le = 4200 mm (distance between points of zero moment) Le/50 = 84 mm, b0 = 75 mm, therefore shear lag can be neglected. flange curling Section 5.4.2 2 24b EN 1993-1-3 u = as Etz22 Clause 5.4 2 Eq. 5.3a a = 240 N/mm (maximum possible value) bs = 75 5 = 70 mm z = 100 2,5 = 97,5 mm 2 24024 70 u = = 0,028 mm 20000022 5 97,5 Flange curling can be neglected if u < 0,05 200 = 10 mm Therefore flange curling is negligible. Partial factors The following partial factors are used throughout the design example: M0 = 1,1 and M1 = 1,1 Table 4.1 Moment resistance of cross-section For a class 4 cross-section: 90,69 103 240 Mc,Rd = Wf = 19,79 kNm Eq. 5.31 eff,min y M0 1,1 10 6 MEd,max = 12,60 kNm Mc,Rd = 19,79 kNm cross-section moment resistance is OK. < cross-section resistance to shear Vpl,Rd = Afvy3 M0 Eq. 5.32 2 Av = h t = 200 5 = 1000 mm 1000 240 Vpl,Rd = = 125,97 kN 3 1,1 1000 FEd,max = 20,30 kNm Vpl,Rd = 125,97 kNm cross-section shear resistance is OK. < Check that shear resistance is not limited by shear buckling: Section 6.4.3 Assume that hw = h 2t = 200 2 5 = 190 mm h 190 h 56,2 w = = 38, shear buckling resistance needs to be checked if w Eq. 6.20 t 5 t = 1,20 h 6 ,5 2 56 , 02 , 97 w = 38 = = 45,4 t ,1 20 shear resistance< is not limited by shear buckling. 218 Design Example 9 Sheet 4 of 7 Design Example 9 Sheet 5 of 7 Shear lag Section 5.4.2 resistance to lateral torsional buckling Section 6.4.2 Shear lag may be neglected provided that b0 Le/50 for outstand elements. Compression flange of beam is laterally unrestrained between B and C. Check this portion of beam for lateral torsional buckling. Le = 4200 mm (distance between points of zero moment) Mb,Rd = Wf for a Class 4 cross-section Eq.6.13 Le/50 = 84 mm, b0 = 75 mm, therefore shear lag can be neglected. LT eff,y y M1 3 3 Weff,y = 90,69 10 mm flange curling Section 5.4.2 1 LT = 1 24 220,5 Eq.6.14 2 asb EN 1993-1-3 u = LT LTλ LT Etz22 Clause 5.4 2 Eq.6.15 2 Eq. 5.3a LT = 0,5 1 LT LT 0,4 LT a = 240 N/mm (maximum possible value) bs = 75 5 = 70 mm Wfyy LT = Eq.6.16 z = 100 2,5 = 97,5 mm M cr 2 24024 70 u = = 0,028 mm 22 200000 5 97,5 Determine the elastic critical moment (Mcr): Appendix E 1/2 Flange curling can be neglected if u < 0,05 200 = 10 mm 2 2 2EI k I k L GI MC zwt Cz)2 Cz Therefore flange curling is negligible. cr 1 2 2 2g 2g Eq. E.1 kL kwz I EIz Partial factors C is simply supported, while B approaches full fixity. Assume most conservative case: The following partial factors are used throughout the design example: k = kw = 1,00 M0 = 1,1 and M1 = 1,1 Table 4.1 C1 and C2 are determined from consideration of bending moment diagram and end E.3 conditions. Table E.1 Moment resistance of cross-section From bending moment diagram, = 0, C1 = 1,77 For a class 4 cross-section: C2 = 0 (no transverse loading) 90,69 103 240 Mc,Rd = Wf = 19,79 kNm Eq. 5.31 26 eff,min y M0 1,1 10 6 200000 0,850 10 M cr 1, 77 2 1,00 2700 MEd,max = 12,60 kNm Mc,Rd = 19,79 kNm 0,5 2 2 4 cross-section moment resistance is OK. 1,00 5085 106 1,00 2700 76900 1,372 10 < 6 2 6 1,00 0,850 10 200000 0,850 10 cross-section resistance to shear Mcr = 41,9 kNm Vpl,Rd = Afvy3 M0 Eq. 5.32 2 3 Av = h t = 200 5 = 1000 mm 90,69 10 240 LT = = 0,721 1000 240 41,9 106 Vpl,Rd = = 125,97 kN 3 1,1 1000 Using imperfection factor LT = 0,34 for cold formed sections: Section 6.4.2 FEd,max = 20,30 kNm Vpl,Rd = 125,97 kNm 2 LT = 0,5 1 0,34 0,721 0,4 0,721 = 0,814 cross-section shear resistance is OK. < 1 LT = 0,5 = 0,839 22 Check that shear resistance is not limited by shear buckling: Section 6.4.3 0,814 0,814 0,721 3 -6 Assume that hw = h 2t = 200 2 5 = 190 mm Mb,Rd = 0,839 90,69 10 240 10 / 1,1 h 190 h 56,2 Mb,Rd = 16,60 kNm MEd = 12,0 kNm (max moment in unrestrained portion of beam) w = = 38, shear buckling resistance needs to be checked if w Eq. 6.20 t 5 t member has adequate resistance to lateral torsional buckling. < = 1,20 h 6 ,5 2 56 , 02 , 97 w = 38 = = 45,4 t ,1 20 shear resistance< is not limited by shear buckling. 219 Design Example 9 Sheet 6 of 7 deflection Section 6.4.6 Load case (serviceability limit state): GQk,j k,1 0,i Q k,i j1i1 As there is only one variable action (Qk,1), the last term in the above expression does not need to be considered in this example. Secant modulus is used for deflection calculations – thus it is necessary to find the maximum stress due to unfactored permanent and variable actions. EES1 S2 The secant modulus ES Eq. 6.52 2 E Where ES,i n and i = 1,2 E Eq. 6.53 1 0,002 i,Ed,ser i,Ed,serf y From structural analysis calculations the following were found: Maximum moment due to permanent actions= 1,90 kNm Maximum moment due to imposed actions = 6,68 kNm Total moment due to unfactored actions = 8,58 kNm Section is Class 4, therefore Weff is used in the calculations for maximum stress in the member. Assume, conservatively that the stress in the tension and compression flange are approximately equal, i.e. ES1 = ES2 For austenitic grade 1.4401 stainless steel, n = 7 Table 6.4 6 M max 8,58 10 2 Serviceability design stress, i,Ed,ser 3 94,6 N/mm Weyff, 90,69 10 200000 2 ES,i 7 198757,6 N/mm 200000 94,6 1 0,002 94,6 240 Maximum deflection due to patch loading occurs at a distance of approximately 1,9 m from support A. Deflection at a distance x from support A due to patch load extending a distance a from support A is given by the following formulae: waL4 Steel When x a: n232 m 6 mm 2 (4 n22 ) n 24aE I Designer’s S Manual Where m = x/L and n = a/L (5th Ed) When x = 1,9 m and a = 1,5 m: m = 1,9/4,2 = 0,452; n = 1,5/4,2 = 0,357 Patch load (permanent + variable unfactored actions): w = 11,0 kN/m Uniform load (permanent action): w = 0,128 kN/m Deflection due to patch loads at a distance of 1,9 m from support A, 1: 11000 1,5 42004 1 24 1500 198757,6 9,06 106 2 3 2 2 2 0,357 2 0,452 6 0,452 0,452 4 0,357 0,357 1 = 7,04 mm 220 Design Example 9 Sheet 6 of 7 Design Example 9 Sheet 7 of 7 deflection Section 6.4.6 Deflection at midspan due to self weight of beam, 2 GQ Q 5 (w LL )3 5 (0,128 103 4,2) 42003 Load case (serviceability limit state): k,j k,1 0,i k,i 2 = = 0,29 mm j1i1 6 384 EIS 384 198757,6 9,06 10 As there is only one variable action (Qk,1), the last term in the above expression does not Total deflection 1 + 2 = 7,04 + 0,29 = 7,33 mm need to be considered in this example. L 4200 Secant modulus is used for deflection calculations – thus it is necessary to find the limiting = 16,8 mm 7,33 mm 250 250 maximum stress due to unfactored permanent and variable actions. deflection is acceptably small.> EES1 S2 The secant modulus ES Eq. 6.52 2 (A finite element analysis was carried out on an identical structural arrangement. The total E beam deflection at mid-point was 7,307 mm – see deformed beam shape with deflections Where ES,i n and i = 1,2 E Eq. 6.53 below.) 1 0,002 i,Ed,ser i,Ed,serf y From structural analysis calculations the following were found: Maximum moment due to permanent actions= 1,90 kNm Maximum moment due to imposed actions = 6,68 kNm Total moment due to unfactored actions = 8,58 kNm Section is Class 4, therefore Weff is used in the calculations for maximum stress in the member. Assume, conservatively that the stress in the tension and compression flange are approximately equal, i.e. ES1 = ES2 For austenitic grade 1.4401 stainless steel, n = 7 Table 6.4 6 M max 8,58 10 2 Serviceability design stress, i,Ed,ser 3 94,6 N/mm Weyff, 90,69 10 200000 2 ES,i 7 198757,6 N/mm 200000 94,6 1 0,002 94,6 240 Maximum deflection due to patch loading occurs at a distance of approximately 1,9 m from support A. Deflection at a distance x from support A due to patch load extending a distance a from support A is given by the following formulae: waL4 Steel When x a: n232 m 6 mm 2 (4 n22 ) n 24aE I Designer’s S Manual Where m = x/L and n = a/L (5th Ed) When x = 1,9 m and a = 1,5 m: m = 1,9/4,2 = 0,452; n = 1,5/4,2 = 0,357 Patch load (permanent + variable unfactored actions): w = 11,0 kN/m Uniform load (permanent action): w = 0,128 kN/m Deflection due to patch loads at a distance of 1,9 m from support A, 1: 11000 1,5 42004 1 24 1500 198757,6 9,06 106 2 3 2 2 2 0,357 2 0,452 6 0,452 0,452 4 0,357 0,357 1 = 7,04 mm 221 222 Promotion of new Sheet 1 of 7 eurocode rules for structural stainless Title Design Example 10 – Axially loaded column in fire steels (PureSt) Made by SMH Date 08/01 Research Fund for Client Revised by MEB Date 04/06 caLcuLatIon Coal and Steel SHeet Revised by SA Date 05/17 deSIGn exaMPLe 10 – axIaLLY Loaded coLuMn In fIre Design an unprotected cold-formed rectangular hollow section subject to axial load and bending for 30 minutes fire resistance. The column length is 2,7 m and is subject to axial load from the end reaction of a floor beam at an eccentricity of 90 mm from the narrow face of the column. y h Point of application b of load z z 90 mm y Section A ‑ A A A Floor beam 2,7 m Column actions This eccentricity is taken to be 90 mm + h/2, where h is the depth of the section. Thus the beam introduces a bending moment about the column’s major axis. The unfactored actions are: Permanent action: 6 kN Variable action: 7 kN The column will initially be checked at the ultimate limit state (LC1) and subsequently at the fire limit state (LC2) for fire duration of 30 minutes. The load cases are as follows: 223 Design Example 10 Sheet 2 of 7 LC1 (ultimate limit state) G,jG k,j + Qk,1Q,1 j G, j = 1,35 (unfavourable effects) Q,1 = 1,5 LC2 (fire limit state) GA,jGQ k,j 1,1 k,1 j GA = 1,0 Values for 1,1 are given in EN 1990 and NA for EN 1990, but for this example conservatively assume 1,1 = 1,0. design at the ultimate Limit State (Lc1) Loading on the corner column due to shear force at end of beam (LC1): Axial force NEd = 1,35 6 + 1,5 7 = 18,6 kN Try 100506 cold-formed RHS. Major axis bending moment (due to eccentricity of shear force from centroid of column): My,Ed = 18,6 (0,09 + 0,10/2) = 2,60 kNm Partial factors The following partial factors are used throughout the design example for LC1: Table 4.1 M0 = 1,10 and M1 = 1,10 Material properties Use austenitic grade 1.4401. 2 2 fy = 220 N/mm and fu = 530 N/mm (for hot-rolled strip). Table 2.2 E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3.1 cross-section properties – 100 x 50 x 6 mm rHS 3 3 Wel,y = 32,58 10 mm iy = 32,9 mm 3 3 Wpl,y = 43,75 10 mm iz = 19,1 mm A = 1500 mm2 t = 6,0 mm cross-section classification Section 5.3.2 0,5 0,5 235 E 235 200000 1, 01 Table 5.2 f y 210000 220 210000 For a RHS the compression width c may be taken as h 3t. Table 5.2 For the web, c = 100 3 × 6 = 82 mm Web subject to compression: ct 82 6 = 13,7 Table 5.2 Limit for Class 1 web = 33ε = 33,33 Table 5.2 33,33 > 13,7 Web is Class 1 By inspection, if the web is Class 1 subject to compression, then the flange will also be Class 1. 224 Design Example 10 Sheet 2 of 7 Design Example 10 Sheet 3 of 7 The overall cross-section classification is therefore Class 1 (under pure compression). LC1 (ultimate limit state) G,jG k,j + Qk,1Q,1 j compression resistance of cross-section Section 5.7.3 G, j = 1,35 (unfavourable effects) Afy Q,1 = 1,5 Nc,Rd for Class 1, 2 or 3 cross-sections Eq. 5.27 M0 GQ 1500 220 LC2 (fire limit state) GA,j k,j 1,1 k,1 Nc,Rd 300 kN j 1,1 300 kN > 18,6 kN acceptable GA = 1,0 Values for are given in EN 1990 and NA for EN 1990, but for this example 1,1 Bending resistance of cross-section Section 5.7.4 conservatively assume 1,1 = 1,0. Wfpl,y y M for Class 1, 2 or 3 cross-sections Eq. 5.29 c,y,Rd design at the ultimate Limit State (Lc1) M0 43750 220 Loading on the corner column due to shear force at end of beam (LC1): M 8,75 kNm c,y,Rd 1,1 Axial force NEd = 1,35 6 + 1,5 7 = 18,6 kN 8,75 kNm > 2,60 kNm acceptable Try 100506 cold-formed RHS. Major axis bending moment (due to eccentricity of shear force from centroid of column): axial compression and bending resistance of cross-section Section 5.7.6 My,Ed = 18,6 (0,09 + 0,10/2) = 2,60 kNm Eq. 5.33 The following approximation for may be used for RHS: EN 1993-1- Partial factors 𝑀𝑀y,Ed ≤ 𝑀𝑀N,Rd 1, clause ) but The following partial factors are used throughout the design example for LC1: Table 4.1 𝑀𝑀N,y,Rd 6.2.9.1(5) M0 = 1,10 and M1 = 1,10 𝑀𝑀WhereN,y,Rd = 𝑀𝑀pl,y,Rd (1 − 𝑛𝑛)/(1 − 0,5𝑎𝑎w 𝑀𝑀N,y,Rd ≤ 𝑀𝑀pl,y,Rd Material properties 𝐴𝐴 − 2𝑏𝑏𝑏𝑏 Use austenitic grade 1.4401. 𝑎𝑎w = but 𝑎𝑎w ≤ 0,5 2 2 𝐴𝐴 fy = 220 N/mm and fu = 530 N/mm (for hot-rolled strip). Table 2.2 1500 − 2 × 50 × 6 E = 200000 N/mm2 and G = 76900 N/mm2 Section 2.3.1 𝑎𝑎w = = 0,6 but 𝑎𝑎w ≤ 0,5 , therefore 𝑎𝑎w = 0,5 1500 𝑁𝑁Ed 18,6 cross-section properties – 100 x 50 x 6 mm rHS 𝑛𝑛 = pl,Rd = = 0,062 3 3 𝑁𝑁 300 Wel,y = 32,58 10 mm iy = 32,9 mm 3 3 1 − 0.062 Wpl,y = 43,75 10 mm iz = 19,1 mm 𝑀𝑀N,y,Rd = 8,75 ( ) = 10,94 ≤ 𝑀𝑀pl,y,Rd = 8,75 Therefore 1 − 0,5 × 0,5 , and A = 1500 mm2 t = 6,0 mm 𝑀𝑀N,y,Rd = 8,75 kNm 𝑀𝑀y,Ed ≤ 𝑀𝑀N,Rd Member buckling resistance in compression Section 6.3.3 cross-section classification Section 5.3.2 fA y 0,5 0,5 Nb,Rd for Class 1, 2 or 3 cross-sections Eq. 6.2 235 E 235 200000 M1 1, 01 Table 5.2 f y 210000 220 210000 1 1 220,5 Eq. 6.4 For a RHS the compression width c may be taken as h 3t. Table 5.2 For the web, c = 100 3 × 6 = 82 mm where Web subject to compression: ct 82 6 = 13,7 Table 5.2 2 0,5 1 0 Eq. 6.5 Limit for Class 1 web = 33ε = 33,33 Table 5.2 L 1 f y 33,33 > 13,7 Web is Class 1 cr for Class 1, 2 or 3 cross-sections Eq. 6.6 iE By inspection, if the web is Class 1 subject to compression, then the flange will also be Class 1. Lcr = buckling length of column, taken conservatively as 1,0 column length = 2,7 m 225 Design Example 10 Sheet 4 of 7 2700 1 220 y = 0,866 32,9 200000 2700 1 220 = 1,492 z 19,1 200000 Buckling curves: major (y-y) axis: For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = Table 6.1 0,49 and 0 = 0,30. = 0,5 1 0,49 0,866 0,3 0,8662 = 1,014 1 y = 0,649 220,5 1,014 1,014 0,866 0,649 1500 220 Nb,y,Rd = = 194,70 kN 1,10 194,70 kN > 18,6 kN acceptable Buckling curves: minor (z-z) axis: = 0,5 1 0,49 1,492 0,3 1,4922 = 1,905 1 z = 0,324 220,5 1,905 1,905 1,492 0,324 1500 220 Nb,z,Rd = = 97,20 kN 1,10 97,20 kN > 18,6 kN acceptable (Resistance to torsional buckling will not be critical for a rectangular hollow section with a Section 6.3.1 h/b ratio of 2.) Member buckling resistance in combined bending and axial compression Section 6.5.2 N M Ne Ed k y,Ed Ed Ny 1 Eq. 6.56 y ()Nb,Rd min W,yWf pl,y y/ M1 W,y = 1,0 for Class 1 cross-sections NNEd Ed ky 1, 0 DD1 y 2 1 DD13 D 2 Eq. 6.63 NNb,Rd,y b,Rd,y From Table 6.6, D1 = 2,0 and D2 = 0,3 and D3 = 1,3 Table 6.6 18,6 18,6 k 1, 0 2 0, 866 0, 3 1,108 1 2 1, 3 0, 3 1,191 y 194,7 194,7 ky = 1.108 18,6 2,60 106 0 1,108 3 0,521 1 acceptable 97,20 1,0 43,75 10 220 /1,10 226 Design Example 10 Sheet 4 of 7 Design Example 10 Sheet 5 of 7 2700 1 220 design at the fire Limit State (Lc2) y = 0,866 32,9 200000 For LC2, the column is designed for the following axial loads and moments. Axial compressive force Nfi,Ed = 1,0 6 + 1,0 7 = 13,0 kN 2700 1 220 = 1,492 Maximum bending moment My,fi,Ed = 13,0 (0,09 + 0,05) = 1,82 kNm z 19,1 200000 determine temperature in steel after 30 minutes fire duration Section 8.4.4 Assume that the section is unprotected and that there is a uniform temperature distribution Buckling curves: major (y-y) axis: within the steel section. The increase in temperature during time interval t is found from: For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = AV t m Table 6.1 Δθ = htnet,d Eq. 8.41 0,49 and 0 = 0,30. c = 0,5 1 0,49 0,866 0,3 0,8662 = 1,014 Eq. 8.42 h e t , dn = hh rnet,cnet, 1 Eq. 8.43 y = 0,649 hnet,c = cg 220,5 1,014 1,014 0,866 4 8 4 0,649 1500 220 h e t , rn = res 5,67 10g 273 273 Eq. 8.44 Nb,y,Rd = = 194,70 kN 1,10 where: 194,70 kN > 18,6 kN acceptable g = gas temperature of the environment of the member in fire exposure, given by the nominal temperature time curve: Buckling curves: minor (z-z) axis: g = 20 + 345log10(8t + 1) Eq. 8.45 2 = 0,5 1 0,49 1,492 0,3 1,492 = 1,905 = surface temperature of the member 1 z = 0,324 220,5 Initial input values for determination of final steel temperature are as follows: 1,905 1,905 1,492 -1 Am/V = 200 m 0,324 1500 220 2 Nb,z,Rd = = 97,20 kN c = 25 W/m K Section 8.4.4 1,10 Initial steel temperature: = 20 C 97,20 kN > 18,6 kN acceptable Resultant emissivity: res = 0,4 Section 8.4.4 Density of stainless steel: = 8000 kg/m3 for austenitic grade 1.4401 Table 2.7 (Resistance to torsional buckling will not be critical for a rectangular hollow section with a Section 6.3.1 Configuration factor: = 1,0 EN 1991-1-2 h/b ratio of 2.) cl. 3.1(7) The specific heat is temperature-dependent and is given by the following expression: Member buckling resistance in combined bending and axial compression Section 6.5.2 c = 450 + 0,28 – 2,91 10-4 2 + 1,34 10-7 3 J/kgK Eq. 8.37 N My,Ed Ne Ed Ny Ed k 1 Eq. 6.56 t = 2 seconds y ()Nb,Rd min W,yWf pl,y y/ M1 The above formulae and initial input information were coded in an Excel spreadsheet and the following steel temperature, after a fire duration of 30 minutes, was obtained. W,y = 1,0 for Class 1 cross-sections = 829 C NNEd Ed ky 1, 0 DD1 y 2 1 DD13 D 2 Eq. 6.63 NNb,Rd,y b,Rd,y reduction of mechanical properties at elevated temperature From Table 6.6, D1 = 2,0 and D2 = 0,3 and D3 = 1,3 Table 6.6 The following reduction factors are required for calculation of resistance at elevated Section 8.2 18,6 18,6 temperatures. ky 1, 0 2 0, 866 0, 3 1,108 1 2 1, 3 0, 3 1,191 194,7 194,7 Young’s modulus reduction factor: kE, = E/E Eq. 8.4 ky = 1.108 0,2% proof strength reduction factor: kp0,2, = fp0,2,/fy Eq. 8.1 6 18,6 2,60 10 0 Strength at 2% total strain reduction factor: k2, = f2,/fy but f2, fu, Eq. 8.2 1,108 3 0,521 1 acceptable 97,20 1,0 43,75 10 220 /1,10 The values for the reduction factors at 829 C are obtained by linear interpolation: kE, = 0,578 Table 8.1 kp0,2, = 0,355 Table 8.1 227 Design Example 10 Sheet 6 of 7 k2, = 0,430 ku, = 0,297 f2, = 0,430 220 = 94,6 and fu, =0,297 530 = 157, therefore f2, fu, Partial factor Section 8.1 M,fi = 1,0 cross-section classification Section 8.3.2 Under compression, ky, should be based on fp0,2,, i.e. ky, = kp0,2, Section 8.2 0,5 0,5 k 0,578 E,θ θ 1, 01 1, 29 Eq. 8.6 ky,θ 0,355 Web subject to compression: ct 82 6 = 13,7 Limit for Class 1 web = 33 θ = 42,57 42,57 > 13,7 Web is Class 1 𝜀𝜀 The overall cross-section classification is Class 1 (under pure compression). Member buckling resistance in compression fiAk p0,2,θy f Nb,fi,t,Rd = for Class 1, 2 and 3 cross-sections Eq. 8.10 M,fi 1 fi = 0,5 1 Eq. 8.12 22 θθθ where 2 Eq. 8.13 θ = 0,5 1 θ0 θ 0,5 𝜙𝜙 kp0,2,θ θ = for all classes of cross-section Eq. 8.14 kE,θ 0,5 0,355 y,θ = 0,866= 0,679 0,578 0,5 0,355 z,θ = 1,492= 1,169 0,578 Buckling curves: major (y-y) axis: For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = Table 6.1 0,49 and 0 = 0,30. 2 θ = 0,5 1 0,49 0,679 0,3 0,679 = 0,823 𝜙𝜙 ,y 1 fi,y = 0,776 220,5 0,823 0,823 0,679 0,776 0,355 1500 220 Nb,y,fi,t,Rd = = 90,91 kN 1, 0 90,91 kN > 13,0 kN acceptable 228 Design Example 10 Sheet 6 of 7 Design Example 10 Sheet 7 of 7 k2, = 0,430 Buckling curves: minor (z-z) axis: 2 ku, = 0,297 θ = 0,5 1 0,49 1,169 0,3 1,169 = 1,396 f2, = 0,430 220 = 94,6 and fu, =0,297 530 = 157, therefore f2, fu, ,z 𝜙𝜙 1 fi,z = 0,5 0,463 1,396 1,39622 1,169 Partial factor Section 8.1 0,463 0,355 1500 220 M,fi = 1,0 Nb,z,fi,t,Rd = = 54,24 kN 1, 0 cross-section classification Section 8.3.2 54,24 kN > 18,6 kN acceptable Under compression, ky, should be based on fp0,2,, i.e. ky, = kp0,2, Section 8.2 0,5 Member buckling resistance in combined bending and axial compression 0,5 kE,θ 0,578 1, 01 1, 29 Eq. 8.6 Nfi,Ed kMy y,fi,Ed kMz z,fi,Ed θ 1 ky,θ 0,355 fy MMy,fi,θ,Rd z,fi,θ,Rd Eq. 8.26 min,fiAk p0,2,θ Web subject to compression: ct 82 6 = 13,7 M,fi Limit for Class 1 web = 33 θ = 42,57 Where 42,57 > 13,7 Web is Class 1 N 𝜀𝜀 y fi,Ed ky = 13 The overall cross-section classification is Class 1 (under pure compression). fy Eq. 8.30 y,fiAk p0,2,θ M,fi Member buckling resistance in compression Eq. 8.31 y = 1,2M,y 3y, 0,44 M,y 0,29 0,8 fiAk p0,2,θy f Nb,fi,t,Rd = for Class 1, 2 and 3 cross-sections Eq. 8.10 Assuming the column is pinned at the base, a triangular bending moment distribution occurs Table 8.3 M,fi and M = 1,8: 1 fi = y = (1,2 1,8 3) 0,679 0,44 1,8 0,29 0,5 1 Eq. 8.12 22 θθθ = 0,068 where ( 0,068) 13,0 103 ky = 1 = 1,010 < 3,0 2 Eq. 8.13 220 θ = 0,5 1 θ0 θ 0,776 1500 0,355 1, 0 0,5 𝜙𝜙 k p0,2,θ θ = for all classes of cross-section Eq. 8.14 M0 k M kM for Class 1, 2 or 3 sections Eq. 8.15 E,θ y,fi, ,Rd 2, Rd M,fi 0,5 0,355 1,10 y,θ = 0,866= 0,679 0,578 M y,fi, ,Rd 0,430 8,75 4,14 kNm 1, 0 0,5 0,355 z,θ = 1,492= 1,169 0,578 13, 0 1,010 1,82 0,444 1 220 Eq. 8.26 0,463 1500 0,355 4,14 Buckling curves: major (y-y) axis: 1, 0 For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = Table 6.1 Therefore the section has adequate resistance after 30 minutes in a fire. 0,49 and 0 = 0,30. 2 θ = 0,5 1 0,49 0,679 0,3 0,679 = 0,823 𝜙𝜙 ,y 1 fi,y = 0,776 220,5 0,823 0,823 0,679 0,776 0,355 1500 220 Nb,y,fi,t,Rd = = 90,91 kN 1, 0 90,91 kN > 13,0 kN acceptable 229 230 Sheet 1 of 8 Promotion of new eurocode rules for Design Example 11 – Design of a two-span cold- Title structural stainless worked trapezoidal roof sheeting steels (PureSt) Made by JG/AO Date 02/06 Research Fund for Client Revised by GZ Date 03/06 caLcuLatIon SHeet Coal and Steel Revised by SJ Date 04/17 deSIGn exaMPLe 11 – deSIGn of a two-SPan coLd-worKed traPeZoIdaL roof SHeetInG This example deals with a two-span trapezoidal roof sheeting with a thickness of 0,6 mm 2 from stainless steel austenitic grade 1.4401 CP500, i.e. cold worked with fy = 460 N/mm . Comparisons will be made against the design of identical sheeting of ferritic grade 1.4003 2 in the annealed condition, i.e. fy = 280 N/mm (see Design Example 3). (There are no differences in the design procedure for ferritic and austenitic sheeting.) The dimensions of the roof sheeting are shown below. 4 x 212,5 = 850 57 65 70 The example shows the following design tasks: - determination of effective section properties at the ultimate limit state; - determination of the bending resistance of the section; - determination of the resistance at the intermediate support; - determination of deflections at serviceability limit state. design data Spans L = 3500 mm Width of supports ss = 100 mm Design load Q = 1,4 kN/m2 Self-weight G = 0,07 kN/m2 Design thickness t = 0,6 mm 2 Yield strength fy = 460 N/mm Table 2.3 Modulus of elasticity E = 200000 N/mm2 Section 2.3.1 Partial safety factor M0 = 1,1 Table 4.1 Partial safety factor M1 = 1,1 Table 4.1 Load factor G = 1,35 Section 4.3 Load factor Q = 1,5 Section 4.3 A detailed sketch of the roof sheeting is given in the figure below. The upper flange will be in compression over the mid support and therefore this case will be checked in this example. 231 Design Example 11 Sheet 2 of 8 bu0/2 Mid line dimensions: bsu/2 h0 70 mm hsu w0 212 5, mm bsu0/2 bu0 65 mm bl0 57 mm bsu 20 mm h0 bsu0 8 mm hsu 6 mm bsl 20 mm bs10 8 mm bsl0/2 hsl 6 mm h sl r 2 mm (internal radius of bsl/2 the corners) w0/2 bl0/2 Angle of the web: atan atan 57,1 ℎ0u0 70 𝜃𝜃 = | | = | | = ° 0,5(𝑤𝑤0 − 𝑏𝑏 − 𝑏𝑏𝑙𝑙0) 0,5 × (212,5 − 65 − 57) effective section properties at the ultimate limit state (uLS) Section 5.2 Check on maximum width to the thickness ratios and angle of web: Table 5.1 ℎAngle0/𝑡𝑡 = of70 the/0 web,6 = and117 corner≤ 400 radius:sinθ = 336 Table 5.1 max(𝑏𝑏𝑙𝑙0/𝑡𝑡 ; 𝑏𝑏𝑢𝑢0/𝑡𝑡) = 𝑏𝑏𝑢𝑢0/𝑡𝑡 = 65/0,6 = 108 ≤ 400 u0 su 45°p ≤ 𝜃𝜃 = 57,1° ≤ 90° 𝑏𝑏 − 𝑏𝑏 65 − 20 𝑏𝑏The= influence of= rounded corners= 22,5 on mm cross-section resistance may be neglected if the 2 2 internal radius and 𝑟𝑟 ≤ 5 𝑡𝑡 𝑟𝑟 ≤ 0,10𝑏𝑏p mm Section 5.6.2 𝑟𝑟The= 2influencemm ≤ min of rounded(5𝑡𝑡 ; 0,1 corners𝑏𝑏𝑝𝑝) = min on cross(5 ×-0section,6 ; 0,1 resistance× 22,5) = may2,25 be neglected. Location of the centroidal axis when the web is fully effective Calculate reduction factor for effective width of the compressed flange: Section 5.4.1 Eq. 5.1 0,772 0,079 𝜌𝜌 = − 2 but ≤ 1 where 𝜆𝜆̅p 𝜆𝜆̅p Eq. 5.3 p 𝑏𝑏̅⁄𝑡𝑡 22,5/0,6 𝜆𝜆̅ = = = 0,946 28,4𝜀𝜀√𝑘𝑘𝜎𝜎 28,4 × 0,698 × √4 Table 5.3 u0 su = p1 𝑘𝑘 = 4 𝑏𝑏 − 𝑏𝑏 65 − 20 𝑏𝑏̅ = 𝑏𝑏 = = = 22,5 mm 2 2 232 Design Example 11 Sheet 2 of 8 Design Example 11 Sheet 3 of 8 bu0/2 Mid line dimensions: Table 5.2 bsu/2 h 70 mm 000 0,5 000 0,5 0 235y 𝐸𝐸 235 200 000 hsu w0 212 5, mm 𝜀𝜀 = [ ] = [ × ] = 0,698 𝑓𝑓 210 460 210 bsu0/2 bu0 65 mm b 57 mm 0,772 0,079 0,772 0,079 l0 𝜌𝜌 = − 2 = − 2 = 0,728 ≤ 1 ̅p ̅p 0,946 mm0, 946 Table 5.3 bsu 20 mm eff,u 𝜆𝜆 𝜆𝜆 b 8 mm h0 su0 𝑏𝑏Effective= 𝜌𝜌 𝑏𝑏̅stiffener= 0,728 properties× 22,5 = 16,38 hsu 6 mm bsl 20 mm bs10 8 mm bsl0/2 hsl 6 mm h sl r 2 mm (internal radius of bsl/2 the corners) w0/2 bl0/2 su su 2 2 0 2 Angle of the web: su √ su 𝑏𝑏 − 𝑏𝑏 2 mm ℎ + ( 2 ) √ 20 − 8 6 + ( 2 ) 𝑡𝑡 = 𝑡𝑡 = × 0,6 = 0,849 2 atan atan 57,1 s eff,u suℎsu su su 6 mm Fig. 5.3 ℎ0u0 70 𝜃𝜃 = | | = | | = ° 𝐴𝐴 = (𝑏𝑏 + 𝑏𝑏 0)𝑡𝑡 +su2ℎ 𝑡𝑡 = (16,38 + 8) × 0,6 + 2 × 6 × 0,849 = 24,82 0,5(𝑤𝑤0 − 𝑏𝑏 − 𝑏𝑏𝑙𝑙0) 0,5 × (212,5 − 65 − 57) su su su su mm effective section properties at the ultimate limit state (uLS) Section 5.2 s ℎ 6 𝑏𝑏 0ℎ 𝑡𝑡 + 2s ℎ 𝑡𝑡 8 × 6 × 0,6 + 2 × 6 × × 0,849 Check on maximum width to the thickness ratios and angle of web: 𝑒𝑒 = 2 = 2 = 2,39 𝐴𝐴 24,su82 su Table 5.1 s s su su s su su s 2 4 3 2 2 2 ℎ 15𝑡𝑡 𝑏𝑏 0𝑡𝑡 Angle0 of the web and corner radius: 0 ℎ /𝑡𝑡 = 70/0,6 = 117 ≤ 400sinθ = 336 Table 5.1 𝐼𝐼 = 2(15𝑡𝑡 𝑒𝑒 ) + 𝑏𝑏 su 𝑡𝑡 su(ℎ − 𝑒𝑒 ) + 2ℎ 𝑡𝑡 ( − 𝑒𝑒 ) + 2 ( ) + 3 2 12 12 𝑡𝑡 ℎ 𝑙𝑙0 𝑢𝑢0 𝑢𝑢0 + 2 max(𝑏𝑏 /𝑡𝑡 ; 𝑏𝑏 /𝑡𝑡) = 𝑏𝑏 /𝑡𝑡 = 65/0,6 = 108 ≤ 400 12 s 0 2 u0 su 45°p ≤ 𝜃𝜃 = 57,1° ≤ 90° 2 2 2 6 Fig. 5.3 𝐼𝐼 = 2 × (15 × 0, 6 × 2, 39 ) + 8 × 0,6 × (6 − 2.39) + 2 × 6 × , 849 × ( mm−42,39) 𝑏𝑏 − 𝑏𝑏 65 − 20 4 3 3 2 𝑏𝑏The= influence of= rounded corners= 22,5 on mm cross-section resistance may be neglected if the 2 2 15 × 0, 6 8 × 0, 6 0,849 × 6 internal radius and + 2 × ( ) + + 2 × = 159,07 12 12 12 su su0 𝑟𝑟 ≤ 5 𝑡𝑡 𝑟𝑟 ≤ 0,10𝑏𝑏p mm Section 5.6.2 s 2 su0 2 mm 2 𝑏𝑏 − 𝑏𝑏 2 20 − 8 𝑟𝑟The= 2influencemm ≤ min of rounded(5𝑡𝑡 ; 0,1 corners𝑏𝑏𝑝𝑝) = min on cross(5 ×-0section,6 ; 0,1 resistance× 22,5) = may2,25 be neglected. 𝑏𝑏 = 2√ℎ𝑠𝑠𝑠𝑠 + ( ) + 𝑏𝑏 = 2 × √6 + ( ) + 8 = 25,0 p2 s 2 b s p 1⁄4 Location of the centroidal axis when the web is fully effective 2 2𝑏𝑏 + 3𝑏𝑏 Eq. 5.10 𝑙𝑙 = 3,07 [𝐼𝐼 𝑏𝑏 ( 3 )] Calculate reduction factor for effective width of the compressed flange: 𝑡𝑡 1/4 mm Section 5.4.1 2 2 × 22,5 + 3 × 25 = 3,07 × [159,07 × 22, 5 × ( 3 )] = 251 Eq. 5.1 0, 6 0,772 0,079 Fig. 5.5 𝜌𝜌 = − 2 but ≤ 1 2 2 where p ̅p 𝜆𝜆̅ 𝜆𝜆 0 u0 𝑙𝑙0 2 2 w 𝑤𝑤 − 𝑏𝑏 − 𝑏𝑏 0 212,5 − 65 − 57 Eq. 5.3 𝑠𝑠 = √( ) + ℎ = √( ) + 70 = 83,4 mm p d 2 mm 2 𝑏𝑏̅⁄𝑡𝑡 22,5/0,6 𝜆𝜆̅ = = = 0,946 𝑏𝑏 = 2𝑏𝑏p + 𝑏𝑏s = 2 × 22,5 + 25 = 70 28,4𝜀𝜀√𝑘𝑘𝜎𝜎 28,4 × 0,698 × √4 Table 5.3 Eq. 5.11 w d u0 su w0 𝑠𝑠 + 2𝑏𝑏 83,4 + 2 × 70 = p1 𝑘𝑘 = 4 𝑘𝑘 = √ = √ = 1,37 𝑠𝑠w + 0,5𝑏𝑏d 83,4 + 0,5 × 70 Eq. 5.8 𝑏𝑏 − 𝑏𝑏 65 − 20 𝑏𝑏̅ = 𝑏𝑏 = = = 22,5 mm 2 2 𝑙𝑙b 251 = = 3,01 ≥ 2 𝑘𝑘w = 𝑘𝑘w0 = 1,37 𝑠𝑠w 83,4 233 Design Example 11 Sheet 4 of 8 w s cr,s 3 Eq. 5.4 4,2𝑘𝑘s 𝐸𝐸 p 𝐼𝐼 𝑡𝑡p s 𝜎𝜎 = √ 2 𝐴𝐴 4𝑏𝑏 (2𝑏𝑏 + 3𝑏𝑏 ) N/mm2 cr,s 3 3 4,2 × 1,37 × 200 × 10 159,07 × 0, 6 𝜎𝜎 = × √ 2 = 551,3 24,82 4 × 22, 5 × (2 × 22,5 + 3 × 25) y d 𝑓𝑓cr,s 460 𝜆𝜆̅ = √ = √ = 0,913 𝜎𝜎 d 551, 3 Eq. 5.17 0,65 < 𝜆𝜆̅ = 0,913 d< 1, 38 χred,ud = 1,47 − 0,723𝜆𝜆̅ = 1,47 − 0 mm,723 × 0,913 = 0,81 𝑡𝑡 = χd 𝑡𝑡 = 0,81 × 0,6 = 0,486 The distance of neutral axis from the compressed flange: sl s sl 2 2 𝑏𝑏 − 𝑏𝑏 𝑙𝑙0 2 mm sl √ℎ + ( ) 2 sl 2 √ 20 − 8 6 + ( 2 ) 𝑡𝑡 = 𝑡𝑡 = × 0,6 = 0,849 w ℎ mm6 𝑡𝑡 = 𝑡𝑡/sin𝜃𝜃 = 0,6/sin(57,1°) = 0,714 2 𝑖𝑖 𝑖𝑖 𝑒𝑒 [mm] 𝐴𝐴 [mm ] 2 tot mm Ͳ 0,5𝑏𝑏eff,u 𝑡𝑡 = 4,9 𝑖𝑖 Ͳ 0,5𝑏𝑏eff,u 𝜒𝜒𝑑𝑑 𝑡𝑡 = 3,98 𝐴𝐴 = ∑𝐴𝐴 = 83,52 0,5ℎsu = 3 ℎsu 𝜒𝜒𝑑𝑑𝑡𝑡su = 4,13 su su0 𝑑𝑑 ℎ = 6 0,5𝑏𝑏 𝜒𝜒 𝑡𝑡 = 1,94 c mm tot𝑖𝑖 𝑖𝑖 0 0 w ∑𝐴𝐴 𝑒𝑒 0,5ℎ = 35 ℎ 𝑡𝑡 = 49,98 𝑒𝑒 = = 36,46 𝐴𝐴 ℎ0 = 70 0,5(𝑏𝑏𝑙𝑙0 − 𝑏𝑏𝑠𝑠𝑠𝑠) 𝑡𝑡 = 11,1 ℎ0 − 0,5ℎs𝑙𝑙 = 67 ℎs𝑙𝑙𝑡𝑡s𝑙𝑙 = 5,09 ℎ0 − ℎs𝑙𝑙 = 64 0,5𝑏𝑏s𝑙𝑙0𝑡𝑡 = 2,4 Effective cross-section of the compression zone of the web eff, eff, 𝐸𝐸 200 EN 1993-1-3 𝑠𝑠 1 = 𝑠𝑠 0 = 0,76𝑡𝑡√ = 0,76 × 0,6 × √ −3 5.5.3.4.3(4-5) mm γM0σcom,Ed 1,1 × 460 × 10 eff,n = 9,07eff, mm 𝑠𝑠 = 1,5𝑠𝑠 0 = 1,5 × 9,07 = 13,61 Effective cross-section properties per half corrugation eff,1 eff,1 mm ℎeff, = 𝑠𝑠eff, sin𝜃𝜃 = 9 ,07 × sin(57,1°) = 7,62 mm ℎ 𝑛𝑛 = 𝑠𝑠 𝑛𝑛sin𝜃𝜃 = 13,61 × sin(57,1°) = 11,43 234 Design Example 11 Sheet 4 of 8 Design Example 11 Sheet 5 of 8 w s cr,s Eq. 5.4 𝟐𝟐 𝟒𝟒 3 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 𝐦𝐦𝐦𝐦 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 𝐞𝐞𝐞𝐞𝐞𝐞,𝒊𝒊 4,2𝑘𝑘s 𝐸𝐸 p 𝐼𝐼 𝑡𝑡p s 𝒆𝒆 [ ] 𝑨𝑨 [𝐦𝐦𝐦𝐦 ] 𝑰𝑰 [𝐦𝐦𝐦𝐦 ] 𝜎𝜎 = √ 2 Ͳ 0,5𝑏𝑏eff,u𝑡𝑡 = 4,9 𝐴𝐴 4𝑏𝑏 (2𝑏𝑏 + 3𝑏𝑏 ) ≈ 0 2 Ͳ 0,5𝑏𝑏eff,u 𝜒𝜒𝑑𝑑 𝑡𝑡 = 4,0 cr,s 3 N/mm ≈ 0 3 3 4,2 × 1,37 × 200 × 10 159,07 × 0, 6 0,5ℎsu = 3 ℎsu 𝜒𝜒𝑑𝑑 𝑡𝑡su = 4,1 𝑑𝑑 su su 𝜎𝜎 = × √ 2 = 551,3 𝜒𝜒 𝑡𝑡 ℎ /12 = 12,4 24,82 4 × 22, 5 × (2 × 22,5 + 3 × 25) su su0 𝑑𝑑 y ℎ = 6 0,5𝑏𝑏 𝜒𝜒 𝑡𝑡 = 1,9 ≈ 0 3 d eff,1 eff,1 w cr,s 0,5ℎ = 3,8 ℎ 𝑡𝑡 = 5,4 𝑡𝑡wℎeff,1 /12 = 26,3 𝑓𝑓 460 3 ̅ √ 0 0 c eff,𝑛𝑛 0 c eff,𝑛𝑛 𝑤𝑤 𝜆𝜆 = √ = = 0,913 ℎ − 0,5(ℎ − 𝑒𝑒 + ℎ ) (ℎ − 𝑒𝑒 + ℎ ) 𝑡𝑡 = 0 c eff,𝑛𝑛 𝜎𝜎 d 551, 3 Eq. 5.17 w (ℎ − 𝑒𝑒 + ℎ ) = 47,5 𝑡𝑡 = 5411,1 = 32,1 12 ̅ d 0 𝑙𝑙0 s𝑙𝑙 0,65 < 𝜆𝜆 = 0,913 < 1,38 ℎ = 70 0,5(𝑏𝑏 − 𝑏𝑏 ) 𝑡𝑡 = 11,1 ≈ 0 3 red,ud ̅ mm 0 sl sl sl χ = 1,47 − 0,723𝜆𝜆 = 1,47 − 0,723 × 0,913 = 0,81 ℎ − 0,5ℎ = 67 ℎ 𝑡𝑡 = 5,1 𝑡𝑡slℎsl /12 = 15,3 d 0 sl 2 sl0 𝑡𝑡 = χ 𝑡𝑡 = 0,81 × 0,6 = 0,486 ℎ tot− ℎ = 64eff,i mm 0,5𝑏𝑏 𝑡𝑡 = 2,4 ≈ 0 The distance of neutral axis from the compressed flange: 𝐴𝐴 = ∑𝐴𝐴eff,i eff,i= 71,0 c mm sl s tot sl ∑𝐴𝐴 𝑒𝑒 2 𝑒𝑒 = = 40,0 e mm2 2 𝑏𝑏 − 𝑏𝑏 𝑙𝑙0 2 mm tot 𝐴𝐴eff,i eff,i c eff,i sl √ℎ + ( ) 2 2 sl 2 √ 20 − 8 6 + ( 2 ) 𝐼𝐼 = ∑𝐼𝐼 + ∑𝐴𝐴 (𝑒𝑒 − ) = 5 465,1 + 46 021,6 = 51 486,7 𝑡𝑡 = 𝑡𝑡 = × 0,6 = 0,849 w mm Optionally the effective section properties may also be redefined iteratively based on the ℎ 6 EN 1993-1-3 location of the effective centroidal axis. 𝑡𝑡 = 𝑡𝑡/sin𝜃𝜃 = 0,6/sin(57,1°) = 0,714 2 Bending strength per unit width (1 m) 𝑖𝑖 𝑖𝑖 𝑒𝑒 [mm] 𝐴𝐴 [mm ] 2 tot mm 4 mm4 0,5𝑏𝑏eff,u 𝑡𝑡 = 4,9 tot Ͳ 0,5w0 0,5 2 𝑖𝑖 1000 1000 eff,u 𝑑𝑑 𝐴𝐴 = ∑𝐴𝐴 = 83,52 Ͳ 0,5𝑏𝑏 𝜒𝜒 𝑡𝑡 = 3,98 𝐼𝐼 = 𝐼𝐼 = × 51 486,7 = 84 580,7 × 12 ,5 mm3 su su 𝑑𝑑 su u 0,5ℎ = 3 ℎ 𝜒𝜒 𝑡𝑡 = 4,13 𝐼𝐼c 484 580,7 su su0 𝑑𝑑 𝑊𝑊 = = = 12 114,5 ℎ = 6 0,5𝑏𝑏 𝜒𝜒 𝑡𝑡 = 1,94 c mm 𝑒𝑒 40 3 tot𝑖𝑖 𝑖𝑖 l mm 0 0 w ∑𝐴𝐴 𝑒𝑒 c 0,5ℎ = 35 ℎ 𝑡𝑡 = 49,98 𝑒𝑒 = = 36,46 𝐼𝐼 484 580,7 3 0 𝑙𝑙0 𝑠𝑠𝑠𝑠 𝐴𝐴 Because𝑊𝑊 = 0 = = 16 152,7 mm ℎ = 70 0,5(𝑏𝑏 − 𝑏𝑏 ) 𝑡𝑡 = 11,1 ℎ −u𝑒𝑒 l70 − 40 eff,min ℎ0 − 0,5ℎs𝑙𝑙 = 67 ℎs𝑙𝑙𝑡𝑡s𝑙𝑙 = 5,09 𝑊𝑊eff,min< 𝑊𝑊 𝑊𝑊 = 𝑊𝑊𝑢𝑢 = 12 114,5 −6 Eq. 5.31 ℎ0 − ℎs𝑙𝑙 = 64 0,5𝑏𝑏s𝑙𝑙0𝑡𝑡 = 2,4 𝑊𝑊 𝑓𝑓y 10 𝑀𝑀c,Rd = = 12114,5 × 460 × = 5,07 kNm Effective cross-section of the compression zone of the web γM0 1,1 determination of the resistance at the intermediate support Section 6.4.4 eff, eff, Web crippling strength 𝐸𝐸 200 EN 1993-1-3 mm EN 1993-1-3 𝑠𝑠 1 = 𝑠𝑠 0 = 0,76𝑡𝑡√ = 0,76 × 0,6 × √ −3 5.5.3.4.3(4-5) mm γM0σcom,Ed 1,1 × 460 × 10 𝑐𝑐 ≥ 40 Clause 6.1.7 eff,n = 9,07eff, mm 𝑟𝑟/𝑡𝑡 = 2/0,6 = 3,33 ≤ 10 0 𝑠𝑠 = 1,5𝑠𝑠 = 1,5 × 9,07 = 13,61 𝑤𝑤 Effective cross-section properties per half corrugation ℎ /𝑡𝑡 = 70/0,6 = 117 ≤ 200sin𝜃𝜃 = 200sin(57,1°) = 168 eff,1 eff,1 mm 45° ≤ 𝜃𝜃 = 57,1° ≤ 90° ℎeff, = 𝑠𝑠eff, sin𝜃𝜃 = 9 ,07 × sin(57,1°) = 7,62 mm βV = 0 ≤(category0,2 2) 𝑙𝑙𝑎𝑎 = 𝑠𝑠𝑠𝑠 = 100 mm ℎ 𝑛𝑛 = 𝑠𝑠 𝑛𝑛sin𝜃𝜃 = 13,61 × sin(57,1°) = 11,43 𝛼𝛼 = 0,15 235 Design Example 11 Sheet 6 of 8 EN 1993-1-3 a Eq. 6.18 w,Rd y 2 0,5w 2 𝑟𝑟 𝑙𝑙 𝜑𝜑 1M1 10000 𝑅𝑅 = α 𝑡𝑡 √𝑓𝑓 𝐸𝐸 (1 − 0,1√ ) (0,5 + √0,02 ) [2,4 + ( ) ] 𝑡𝑡 𝑡𝑡 90 γ w,Rd 2 2 100 𝑅𝑅 = 0,15 × 0, 6 √460 × 200 000 × (1 − 0,1√ ) (0,5 + √0,02 × ) × 0,6 0,6 2 0,5 212,5 57,1 1 1000 −3 × [2,4 + ( ) ] × × × 10 = 23,6 kN 90 1,1 × combined bending moment and support reaction Factored actions per unit width (1 m): G Q kN/m 𝑞𝑞 = 𝛾𝛾 𝐺𝐺 + 𝛾𝛾 𝑄𝑄 = 1,35 × 0,07 + 1,5 × 1,4 = 2,19 Ed 2 2 kNm 𝑞𝑞𝐿𝐿 2,19 × 3, 5 𝑀𝑀 = = = 3,35 Ed 8 8 kN 5 5 𝐹𝐹 Ed= 𝑞𝑞𝑞𝑞 = × 2,19 × 3,5 = 9,58 Ed 4 4 c,Rd w,Rd EN 1993-1-3 𝑀𝑀 3,35 𝐹𝐹 9,58 Eq. 6.28a - c Ed = Ed = 0,661 ≤ 1,0 = = 0,406 ≤ 1,0 𝑀𝑀 5,07 𝑅𝑅 23,6 𝑀𝑀c,Rd 𝐹𝐹w,Rd + = 0,661 + 0,406 = 1,067 ≤ 1,25 𝑀𝑀Cross-section𝑅𝑅 resistance satisfies the conditions. determination of deflections at serviceability limit state (SLS) Effective cross-section properties For serviceability verification the effective width of compression elements should be based EN 1993-1-3 on the compressive stress in the element under the serviceability limit state loading. Clause 5.5.1 Maximum compressive stress in the effective section at SLS. A conservative approximation is made based on Wu determined above for ultimate limit state. y,Ed,ser 2 2 2,25 kNm (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 (0,07 + 1,4) × 3, 5 𝑀𝑀 = = = y,Ed,ser8 8 2 com,Ed,ser 6 1 5,7 N/mm 𝑀𝑀 2,25 × 10 𝜎𝜎 = = = 8 The effective section𝑊𝑊𝑢𝑢 properties12114 are, 5determined as before in ultimate limit state except that fy is replaced by serEd,com, and the thickness of the flange stiffener is not reduced. The results of the calculation are: Effective width of the compressed flange: The flange is fully effective. Location of the centroidal axis when the web is fully ec = 34,1 mm effective: Effective cross-section of the compression zone of the The web is fully effective. web: Effective part of the web: seff,1 14,268mm s neff, 214, mm 2 Effective cross-section properties per half corrugation: Atot = 82,44 mm ec = 36,25 mm 4 Itot = 59726,1 mm 236 Design Example 11 Sheet 6 of 8 Design Example 11 Sheet 7 of 8 EN 1993-1-3 Effective section properties per unit width (1 m): I = 562128,0 mm4 a 4 Eq. 6.18 Wu = 15507,0 mm w,Rd y 2 0,5w 4 2 𝑟𝑟 𝑙𝑙 𝜑𝜑 1M1 10000 Wl = 16655,6 mm 𝑅𝑅 = α 𝑡𝑡 √𝑓𝑓 𝐸𝐸 (1 − 0,1√ ) (0,5 + √0,02 ) [2,4 + ( ) ] 𝑡𝑡 𝑡𝑡 90 γ Determination of deflection w,Rd Secant modulus of elasticity corresponding to maximum value of the bending moment: 2 2 100 𝑅𝑅 = 0,15 × 0, 6 √460 × 200 000 × (1 − 0,1√ ) (0,5 + √0,02 × ) × 0,6 0,6 y,Ed,ser 2 6 N/mm 2 0,5 212,5 𝑀𝑀 u 2,25 × 10 57,1 1 1000 −3 1,Ed,ser = = = 145,096 × [2,4 + ( ) ] × × × 10 = 23,6 kN y,Ed,ser 2 90 1,1 × 𝑊𝑊 15 507 N/mm 6 combined bending moment and support reaction 𝑀𝑀 l 2,25 × 10 2,Ed,ser = = = 135,090 Factored actions per unit width (1 m): 𝑊𝑊 16 655,6 (for austenitic grade 1.4401 stainless steel) Table 6.4 G Q kN/m 𝑛𝑛 = 7 2 𝑞𝑞 = 𝛾𝛾 𝐺𝐺 + 𝛾𝛾 𝑄𝑄 = 1,35 × 0,07 + 1,5 × 1,4 = 2,19 kN/mm kNm 1,Ed,ser Eq. 6.53 Ed 2 2 𝐸𝐸 200 𝑞𝑞𝐿𝐿 2,19 × 3, 5 𝐸𝐸S,1 = 1,Ed,ser y 𝑛𝑛 = 7 = 199,83 𝑀𝑀 = = = 3,35 𝐸𝐸 𝜎𝜎 200 0,145 8 8 kN 1 + 0,002 ( ) 1 + 0,002 × ( ) Ed 𝜎𝜎 𝑓𝑓 0,145 0,460 kN/mm2 5 5 2,Ed,ser Eq. 6.53 𝐹𝐹 Ed= 𝑞𝑞𝑞𝑞 = × 2,19 × 3,5 = 9,58 Ed 𝐸𝐸 200 4 4 𝐸𝐸S,2 = 2,Ed,ser y 𝑛𝑛 = 7 = 199,89 c,Rd w,Rd EN 1993-1-3 𝐸𝐸 𝜎𝜎 200 0,135 𝑀𝑀 3,35 𝐹𝐹 9,58 S1 + 0,002S,2 ( ) 1 + 0,002 × ( ) = = 0,661 ≤ 1,0 = = 0,406 ≤ 1,0 Eq. 6.28a - c 𝜎𝜎 𝑓𝑓 kN/mm0,1352 0,460 Eq. 6.52 𝑀𝑀 Ed 5Ed,07 𝑅𝑅 23,6 𝐸𝐸 ,1 + 𝐸𝐸 199,83 + 199,89 𝑀𝑀c,Rd 𝐹𝐹w,Rd 𝐸𝐸S = = = 199,86 + = 0,661 + 0,406 = 1,067 ≤ 1,25 2 2 𝑀𝑀Cross-section𝑅𝑅 resistance satisfies the conditions. Check of deflection As a conservative simplification, the variation of Es,ser along the length of the member is determination of deflections at serviceability limit state (SLS) neglected. Effective cross-section properties For cross-section stiffness properties the influence of rounded corners should be taken into For serviceability verification the effective width of compression elements should be based EN 1993-1-3 account. The influence is considered by the following approximation: on the compressive stress in the element under the serviceability limit state loading. Clause 5.5.1 Eq. 5.22 Maximum compressive stress in the effective section at SLS. A conservative approximation 𝑛𝑛 φj o is made based on Wu determined above for ultimate limit state. j o 294,2 ∑𝑗𝑗=1 𝑟𝑟 90 2 × o δ = 0,43 𝑚𝑚 = 0,43 90 = 0,019 y,Ed,ser 2,25 kNm 2 2 p,i 149,3 Eq. 5.20 (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 (0,07 + 1,4) × 3, 5 ∑𝑖𝑖=1 𝑏𝑏 𝑀𝑀 = = = 4 y,Ed,ser8 8 2 Fory,r the location of maximum deflection: com,Ed,ser 6 1 5,7 N/mm 𝐼𝐼 = 𝐼𝐼 (1 - 2 ) = 562128,0 (1 - 2 × 0,019) = 540767,1 mm 𝑀𝑀 2,25 × 10 m The𝜎𝜎 effective= section𝑢𝑢 properties= are determined= 8 as before in ultimate limit state except that 𝑊𝑊 12114,5 1 + √33 1 + √33 fy is replaced by serEd,com, and the thickness of the flange stiffener is not reduced. The 𝑥𝑥 = × 𝐿𝐿 = × 3,5 = 1,48 16 16 results of the calculation are: 4 3 4 (𝐺𝐺 + 𝑄𝑄)𝐿𝐿 𝑥𝑥 𝑥𝑥 𝑥𝑥 Effective width of the compressed flange: The flange is fully effective. 𝛿𝛿 = ( − 3 3 + 2 4 ) 48𝐸𝐸S𝐼𝐼y,r 𝐿𝐿 𝐿𝐿 𝐿𝐿 Location of the centroidal axis when the web is fully ec = 34,1 mm 3 4 3 4 effective: (0,07 + 1,4) × 10 × 3, 5 1,48 1, 48 1, 48 𝛿𝛿 = 6 −12 × ( − 3 × 3 + 2 × 4 ) Effective cross-section of the compression zone of the The web is fully effective. 48 × 199,86 × 10 × 540767,1 × 10 3,5 3, 5 3, 5 web: 𝛿𝛿The= permissible11,1 mm deflection is L/200 = 3500/200 = 17,5 mm 11,1 mm, hence the calculated Effective part of the web: seff,1 14,268mm deflection is acceptable. > s neff, 214, mm 2 Effective cross-section properties per half corrugation: Atot = 82,44 mm ec = 36,25 mm 4 Itot = 59726,1 mm 237 Design Example 11 Sheet 8 of 8 comparison between sheeting in the annealed and cold worked conditions A comparison of the bending resistance per unit width and resistance to local transverse 2 forces of identical sheeting in the annealed condition (fy = 280 N/mm ) and cold worked 2 condition (fy = 460 N/mm ) is given below: 2 fy = 280 N/mm (Design example 3) Mc,Rd = 3,84 kNm and Rw,Rd = 18,4 kN 2 fy = 460 N/mm (Design example 11) Mc,Rd = 5,07 kNm and Rw,Rd = 23,6 kN With sheeting in the annealed condition, the span must be reduced to 2,9 m compared to 3,5 m for material in the cold worked strength condition. Hence, sheeting made from cold worked material enables the span to be increased, meaning that the number of secondary beams or purlins could be reduced, leading to cost reductions. 238 Design Example 11 Sheet 8 of 8 Sheet 1 of 7 Promotion of new comparison between sheeting in the annealed and cold worked conditions eurocode rules for Design Example 12 – Design of a lipped channel A comparison of the bending resistance per unit width and resistance to local transverse Title 2 structural stainless subject to bending forces of identical sheeting in the annealed condition (fy = 280 N/mm ) and cold worked 2 condition (fy = 460 N/mm ) is given below: steels (PureSt) Made by ER/EM Date 02/06 2 Research Fund for fy = 280 N/mm (Design example 3) Mc,Rd = 3,84 kNm and Rw,Rd = 18,4 kN Client Revised by HB Date 03/06 caLcuLatIon SHeet Coal and Steel 2 fy = 460 N/mm (Design example 11) Mc,Rd = 5,07 kNm and Rw,Rd = 23,6 kN Revised by ER/IA Date 04/17 With sheeting in the annealed condition, the span must be reduced to 2,9 m compared to 3,5 deSIGn exaMPLe 12 – deSIGn of a LIPPed cHanneL SuBJect to BendInG m for material in the cold worked strength condition. Hence, sheeting made from cold Design a lipped channel subject to bending with an unrestrained compression flange from worked material enables the span to be increased, meaning that the number of secondary austenitic grade 1.4401 in the cold worked condition CP500. The beam is simply beams or purlins could be reduced, leading to cost reductions. supported with a span, l = 4,0 m. The distance between adjacent beams is 1,0 m. As the load is not applied through the shear centre of the channel, it is necessary to check the interaction between the torsional resistance of the cross-section and the lateral torsional buckling resistance of the member. However, this example only checks the lateral torsional buckling resistance of the member. factors Partial factor M0 = 1,1 and M1 = 1,1 Table 4.1 Load factor G = 1,35 (permanent loads) and Q = 1,5 (variable loads) EN 1991 actions Permanent actions (G): 2 kN/m2 Variable actions (Q): 3 kN/m2 Since the distance between adjacent beams is 1m, Gk = 2 kN/m Qk = 3 kN/m Load case to be considered at the ultimate limit state: * q G,j GQ k,j Q,1 k,1 7,2 kN/m EN 1991 j Structural analysis Reactions at support points (Design shear force) q* 4 V 14,4 kN Ed 2 Design bending moment q*24 M 14,4 kNm Ed 8 Material Properties 2 fy = 460 N/mm Table 2.3 Modulus of elasticity E = 200000 N/mm2 and shear modulus G = 76900 N/mm2 Section 2.3.1 cross-section Properties The influence of rounded corners on cross-section resistance may be neglected if the Section 5.6.2 internal radius r ≤ 5t and r ≤ 0,10bp and the cross section may be assumed to consist of plane elements with sharp corners. For cross-section stiffness properties the influence of rounded corners should always be taken into account. 239 Design Example 12 Sheet 2 of 7 z b h = 160 mm c b = 125 mm h c = 30 mm y y t = 5 mm r = 5 mm r t z rm rt2 7,5 mm grrm tan 2 sin 2 2,2 mm bpr bt 2 g 115,6 mm Figure 5.5 rt5 mm 5 25 mm rb5 mm 0,10p 11,56 mm The influence of rounded corners on section properties may be taken into account with sufficient accuracy by reducing the properties calculated for an otherwise similar cross- section with sharp corners, using the following approximations: Notional flat width of the flange, bp,f bt 2 gr 115,6 mm Notional flat width of the web, bp,w ht 2 gr 150,6 mm Notional flat width of the lip, bp,l ct/ 2 gr 25, 3 mm 2 Ag,sh = tb22p,f b p,w b p,l 2162 mm 113 2 3 2 Iyg,sh = 2btp,f bt p,f (0,5 h 0,5 t ) 2 btp,l bt p,l 0,5 h ( c bp,l ) 0,5 bp,l 12 12 1 bt3 = 9,376106 mm4 12 p,w nm j Eq. 5.22 = 0,43rbj o /p,i 0,02 ji1190 2 Ag = Ag,sh (1 – ) = 2119 mm Eq. 5.19 6 4 Ig = Ig,sh (1 – 2) = 9,0 10 mm Eq. 5.20 1.1 classification of the cross-section Section 5.3 0,5 Table 5.2 235 E 0,698 f y 210000 Flange: Internal compression parts. Part subjected to compression. cbp,f 115,6 mm and c/t = 23,12 For Class 2, c/t 3524,43, therefore the flanges are Class 2 Web: Internal compression parts. Part subjected to bending. c = bp,w = 150,6 mm and c/t = 30,12 For Class 1, c/t ≤ 7250,26, therefore the web is Class 1. 240 Design Example 12 Sheet 2 of 7 Design Example 12 Sheet 3 of 7 z Lip: Outstand flanges. Part subjected to compression, tip in compression. b cbp,l 25,30 mm and c/t = 5,06 h = 160 mm c For Class 1, c/t ≤ 9 6,28, therefore the lip is Class 1. b = 125 mm h c = 30 mm 1.2 effects of shear lag Section 5.4.2 y y t = 5 mm r = 5 mm Shear lag in flanges may be neglected if b0 < Le/50, where b0 is taken as the flange outstand r or half the width of an internal element and Le is the length between points of zero bending t moment. z For internal elements: bo = (b – t)/2 = 60 mm The length between points of zero bending moment is: Le = 4000 mm, Le /50 = 80 mm rm rt2 7,5 mm Therefore shear lag can be neglected. grrm tan 2 sin 2 2,2 mm Section 5.4.3 bpr bt 2 g 115,6 mm Figure 5.5 1.3 flange curling rt5 mm 5 25 mm Flange curling can be neglected if the curling of the flange towards the neutral axis, u, is less than 5% of the depth of the profile cross-section: rb5 mm 0,10 11,56 mm 24 p b EN 1993-1-3, u 2 as The influence of rounded corners on section properties may be taken into account with Etz22 clause 5.4 sufficient accuracy by reducing the properties calculated for an otherwise similar cross- Eq. 5.3a 2 section with sharp corners, using the following approximations: a is mean stress in the flanges calculated with gross area (fy=460 N/mm is assumed) s p,f p,l Notional flat width of the flange, bp,f bt 2 gr 115,6 mm b = is the distance between webs = b + b = 140,9 mm t = 5 mm Notional flat width of the web, b ht 2 g 150,6 mm p,w r z = is the distance of the flange under consideration from neutral axis = 77,5 mm Notional flat width of the lip, bp,l ct/ 2 gr 25, 3 mm u = 2,15 mm < 0,05h = 8 mm, therefore flange curling can be neglected. 2 Ag,sh = tb22p,f b p,w b p,l 2162 mm 1.4 Stiffened elements. edge stiffeners 113 2 3 2 Iyg,sh = 2btp,f bt p,f (0,5 h 0,5 t ) 2 btp,l bt p,l 0,5 h ( c bp,l ) 0,5 bp,l 1.5 distortional buckling. Plane elements with edge stiffeners 12 12 Section 5.5.1 1 bt3 = 9,376106 mm4 and EN 12 p,w 1993-1-3, nm clause 5.5.3 j Eq. 5.22 = 0,43rbj o /p,i 0,02 ji1190 2 Ag = Ag,sh (1 – ) = 2119 mm Eq. 5.19 6 4 Ig = Ig,sh (1 – 2) = 9,0 10 mm Eq. 5.20 1.1 classification of the cross-section Section 5.3 0,5 Table 5.2 235 E 0,698 f y 210000 Flange: Internal compression parts. Part subjected to compression. Step 1: Initial effective cross-section for the stiffener EN 1993-1-3, cbp,f 115,6 mm and c/t = 23,12 For flanges (as calculated before) clause 5.5.3.2 For Class 2, c/t 3524,43, therefore the flanges are Class 2 b =125 mm and bp = bp,f = 115,6 mm Web: Internal compression parts. Part subjected to bending. For the lip, the effective width ceff should be calculated using the corresponding buckling c = bp,w = 150,6 mm and c/t = 30,12 factor k p and expressions as follows: For Class 1, c/t ≤ 7250,26, therefore the web is Class 1. bp,c = bp,l = 25,30 mm 241 Design Example 12 Sheet 4 of 7 bp,c/bp = 0,22 < 0,35 then k= 0,5 EN 1993-1-3, Eq. 5.13b bt p 0,36 ( b 25, 3 mm ) Eq. 5.3 28,4 kσ 1 0,188 Cold formed outstand elements: 2 1, 33 1 then = 1,0 Eq. 5.2 p p ceff = bp,c = 25,30 mm EN 1993-1-3, Eq. 5.13a Step 2: Reduction factor for distortional buckling Calculation of geometric properties of effective edge stiffener section be2 = bp,f = 115,6 mm In this example, since the compressed flange is Class 2, be2 already considers the whole flange and therefore be1 = 0 is adopted. ceff = bp,l = 25,30 mm 2 As = (be2 + ceff)t = (bb,f+ bb,l)t = 704,5 mm Calculation of linear spring stiffness Et 3 1 EN 1993-1-3, 2 K1 2 23 6,4 N/mm Eq. 5.10b 41 b1 h w b 10,5 bb 12 h w k f b1 = b – yb – t/2 – r = 71,1 mm (the distance from the web-to-flange junction to the gravity centre of the effective area of the edge stiffener, including the effective part of the flange be2). kf = 0 (flange 2 is in tension) hw = h – 2t – 2r = 160 – 25 – 25 = 140 mm Elastic critical buckling stress for the effective stiffener section, adopting K = K1 EN 1993-1-3, 2 KEIs 2 cr,s 565,8 N/mm Eq. 5.15 As Reduction factor d for distortional buckling EN 1993-1-3, d fyb cr,s 0,90 Eq. 5.12d EN 1993-1-3, 0,65< d <1,38 then d 1,47 0,723d 0,82 Eq. 5.12b Reduced area and thickness of effective stiffener section, considering that com,Ed = fyb/M0 f γ EN 1993-1-3, yb M0 2 AAs,red d s 576,4 mm Eq. 5.17 com,Ed tred = tAs,red/As = 4,1 mm Calculation of effective section properties with distortional buckling effect 2 Ag,sh = tbbb p,f p,w p,l tbb red p,f p,l 2034,0 mm nm 0,43rbj / 0,02 = j o p,i Eq. 5.22 ji1190 2 Ag = Ag,sh (1–) = 1993,3 mm Eq. 5.19 The new eeff, adopting distances from the centroid of the web, positive downwards: 242 Design Example 12 Sheet 4 of 7 Design Example 12 Sheet 5 of 7 bp,c/bp = 0,22 < 0,35 then k= 0,5 EN 1993-1-3, bthtbthtbthtgbp,f red 0,5 0,5red p,f 0,5 0,5 p,l red 0,5 0,5 r 0,5 p,l Eq. 5.13b eeff = Ag,sh bt p 0,36 ( b 25, 3 mm ) Eq. 5.3 btp,l 0,5 h 0,5 t gr 0,5 b p,l b p,l 0 28,4 kσ 4,7 mm Ag,sh 1 0,188 Cold formed outstand elements: 2 1, 33 1 then = 1,0 Eq. 5.2 Iy,g,sh= p p 113 23 2 btp,f tred bt p,f red (0,5 h 0,5 tred e eff ) btp,l red bt p,l red 0,5 h 0,5 t gr 0,5 b p,l e eff ceff = bp,c = 25,30 mm EN 1993-1-3, 12 12 Eq. 5.13a 113 23 2 Step 2: Reduction factor for distortional buckling btp,f bt p,f (0,5 h 0,5 t eeff ) btp,l bt p,l 0,5 h 0,5 t gr 0,5 b p,l e eff 12 12 Calculation of geometric properties of effective edge stiffener section 1 btbte3 ( )2 8,64 1064 mm be2 = bp,f = 115,6 mm 12 p,w p,wef f In this example, since the compressed flange is Class 2, be2 already considers the whole 6 4 Iy,g = Iy,g,sh (1–2) = 8,29710 mm Eq. 5.20 flange and therefore be1 = 0 is adopted. zmax = h/2 + eeff = 160/2 + 4,7 = 84,7 mm (distance from the top fibre to the neutral axis) ceff = bp,l = 25,30 mm 3 3 2 Wy,g = Iy,g / zmax = 97,9510 mm As = (be2 + ceff)t = (bb,f+ bb,l)t = 704,5 mm Calculation of linear spring stiffness 1.6 resistance of cross-section Section 5.7 Et 3 1 EN 1993-1-3, Cross-section subject to bending moment Section 5.7.4 2 K1 23 6,4 N/mm Eq. 5.10b 2 Mc,Rd Wf pl y/ γ M0 41, 0 kNm Eq. 5.29 41 b1 h w b 10,5 bb 12 h w k f Design bending moment M 14,4 kNm, therefore cross-section moment resistance is b1 = b – yb – t/2 – r = 71,1 mm (the distance from the web-to-flange junction to the gravity Ed OK. centre of the effective area of the edge stiffener, including the effective part of the flange be2). kf = 0 (flange 2 is in tension) Cross-section subject to shear Section 5.7.5 2 Av = 800 mm hw = h – 2t – 2r = 160 – 25 – 25 = 140 mm Eq. 5.32 Elastic critical buckling stress for the effective stiffener section, adopting K = K1 Vpl,Rd Af v y 3/γM0 193,15 kN 2 KEI EN 1993-1-3, s 2 Design shear force VEd 14,4 kN , therefore cross-section shear resistance is OK cr,s 565,8 N/mm Eq. 5.15 As Reduction factor d for distortional buckling Cross-section subjected to combination of loads Section 5.7.6 EN 1993-1-3, VEd = 14,4 kN > 0,5Vpl,Rd = 96,57 kN d fyb cr,s 0,90 Eq. 5.12d Therefore, there is no need to take into account interaction between bending moment and EN 1993-1-3, shear force. 0,65< d <1,38 then d 1,47 0,723d 0,82 Eq. 5.12b Reduced area and thickness of effective stiffener section, considering that com,Ed = fyb/M0 flexural members Section 6.4 fybγ M0 EN 1993-1-3, Lateral-torsional buckling Section 6.4.2 AA 576,4 mm2 s,red d s Eq. 5.17 com,Ed Mb,Rd LT Wf y y M1 Eq. 6.13 tred = tAs,red/As = 4,1 mm 1 LT 0,5 1 Eq. 6.14 Calculation of effective section properties with distortional buckling effect 22 LT LTλ LT 2 Ag,sh = tbbb p,f p,w p,l tbb red p,f p,l 2034,0 mm 2 Eq. 6.15 LT 0,5 1 LT LT 0,4 LT nm 0,43rbj / 0,02 = j o p,i Wf 90 Eq. 5.22 yy ji11 LT Eq. 6.16 M cr 2 Ag = Ag,sh (1–) = 1993,3 mm Eq. 5.19 LT = 0,34 for cold-formed sections The new eeff, adopting distances from the centroid of the web, positive downwards: 243 Design Example 12 Sheet 6 of 7 Determination of the elastic critical moment for lateral-torsional buckling 2 2 2 EI k I k L GI 2 MC zwt Cz Cz Eq. E.1 cr 1 2 2 2g 2g kL kwz I EIz For simply supported beams with uniform distributed load: C1 = 1,13, and C2 = 0,454 Table E.2 Assuming normal conditions of restraint at each end: k = kw = 1 za is the coordinate of point load application zs is the coordinate of the shear centre zg = za zs= h/2 = 80 mm yG = distance from the central axis of the web to the gravity centre 2btgbbtbtp,f ( r 0,5 p,f ) 2 p,l ( 0,5 ) yG = 46,4 mm As 6 4 Iz,sh = 4,59010 mm 3 4 It,sh = 18,0210 mm 9 6 Iw,sh = 23,1910 mm 6 4 Iz = Iz,sh (1–2) = 4,40610 mm 3 4 It = It,sh (1–2) = 17,3010 mm 9 6 Iw = Iw,sh (1–4) = 21,3310 mm 2 2 2 EI k I k L GI 2 Then, MC zwt Cz Cz 34,76 kNm Eq. E.1 cr 1 2 2 2g 2g kL kwz I EIz Wfy,g y 3 3 LT 1,14 (Wy,g = 97,9510 mm , compression flange) Eq. 6.16 M cr 2 Eq. 6.15 LT 0,5 1 LT LT 0,4 LT 1,27 1 LT 0,5 0,54 Eq. 6.14 22 LT LTλ LT Mb,Rd LT Wf y yγ M1 22,21 kNm Eq. 6.13 Design moment MEd 14,4 kNm , therefore lateral torsional buckling resistance OK. Note: As the load is not applied through the shear centre of the channel, it is also necessary to check the interaction between the torsional resistance of the cross-section and the lateral torsional buckling resistance of the member. 1.7 Shear buckling resistance Section 6.4.3 The shear buckling resistance only requires checking when htw / 56,2εη for an Eq. 6.20 unstiffened web. The recommended value for = 1,20. htw / ( h 2 t 2 rt ) / 140 / 5 28,0, 56,2ε η 32,67 , therefore no further check required. 244 Design Example 12 Sheet 6 of 7 Design Example 12 Sheet 7 of 7 Determination of the elastic critical moment for lateral-torsional buckling deflections Section 6.4.6 2 2 2 Deflections should be determined for the load combination at the relevant Serviceability EI k I k L GI 2 MC zwt Cz Cz Eq. E.1 Limit State, with: cr 1 2 2 2g 2g kL kwz I EIz Load factorsG = 1,00 (permanent loads) and Q = 1,00 (variable loads) EN 1991 Permanent actions (G): 2 kN/m2 and Variable actions (Q): 3 kN/m2 For simply supported beams with uniform distributed load: C1 = 1,13, and C2 = 0,454 Table E.2 Load case to be considered at SLS, assuming distance between adjacent beams is 1,0 m : Assuming normal conditions of restraint at each end: k = kw = 1 EN 1991 za is the coordinate of point load application qG,j GQ k,j Q ,1 k,1 5, 0 kN/m j zs is the coordinate of the shear centre The deflection of elastic beams may be estimated by standard structural theory, except that zg = za zs= h/2 = 80 mm the secant modulus of elasticity should be used instead of the modulus of elasticity: yG = distance from the central axis of the web to the gravity centre EES1 S2 Eq. 6.52 ES 2btgbbtbtp,f ( r 0,5 p,f ) 2 p,l ( 0,5 ) yG = 46,4 mm 2 As where: 6 4 Iz,sh = 4,59010 mm ES1 is the secant modulus corresponding to the stress in the tension flange and 3 4 ES2 is the secant modulus corresponding to the stress in the compression flange It,sh = 18,0210 mm 9 6 Iw,sh = 23,1910 mm ES1 and ES2 for the appropriate serviceability design stress can be estimated as follows: 6 4 Iz = Iz,sh (1–2) = 4,40610 mm E Eq. 6.53 and i = 1,2 3 4 ES,i n It = It,sh (1–2 ) = 17,30 10 mm E i,Ed,ser 9 6 1 0,002 Iw = Iw,sh (1–4) = 21,3310 mm i,Ed,serf y 2 2 2 EI k I k L GI 2 where: Then, MC zwt Cz Cz 34,76 kNm Eq. E.1 cr 1 2 2 2g 2g i,Ed,ser is the serviceability design stress in the tension or compression flange kL kwz I EIz n is the Ramberg Osgood parameter; for austenitic stainless steel 1.4401, n = 7. Table 6.4 Wfy,g y 3 3 LT 1,14 (Wy,g = 97,9510 mm , compression flange) Eq. 6.16 M cr The non-linear stainless steel stress-strain relationship means that the modulus of elasticity varies within the cross-section and along the length of a member. As a simplification, the 2 Eq. 6.15 LT 0,5 1 LT LT 0,4 LT 1,27 variation of ES along the length of the member may be neglected and the minimum value of ES for that member (corresponding to the maximum values of the stresses σ1 and σ2 in 1 the member) may be used throughout its length. LT 0,5 0,54 Eq. 6.14 22 LT LTλ LT The stresses in the tension and compression flanges are the following: Compression flange: Mb,Rd LT Wf y yγ M1 22,21 kNm Eq. 6.13 M Ed,max 102,1MPa and ES1 = 199979,2 MPa Eq. 6.53 Design moment M 14,4 kNm , therefore lateral torsional buckling resistance OK. Ed,ser,1 Ed Wy,sup 3 3 Note: As the load is not applied through the shear centre of the channel, it is also necessary with MEd,max = 10 kNm and Wy = 97,95×10 mm to check the interaction between the torsional resistance of the cross-section and the lateral Tension flange: torsional buckling resistance of the member. M Ed,max 100,8 MPa and ES2 = 199980,8 MPa Ed,ser,2 W Section 6.4.3 y,inf 1.7 Shear buckling resistance 3 3 with MEd,max = 10 kNm and Wy = 99,24×10 mm The shear buckling resistance only requires checking when htw / 56,2εη for an Eq. 6.20 And therefore: ES = 199980,0 MPa Eq. 6.52 unstiffened web. The maximum deflection can be estimated by standard structural theory assuming the The recommended value for = 1,20. secant modulus of elasticity: 4 htw / ( h 2 t 2 rt ) / 140 / 5 28,0, 56,2ε η 32,67 , therefore no further check 5ql required. dmax 384EISy 6 4 Since Iy = 8,297×10 mm , q = 5,0 kN/m and l = 4,0 m Sheets 1 & 5 dmax 10,0 mm 245 246 Sheet 1 of 8 Promotion of new eurocode rules for Title Design Example 13 – Hollow section lattice girder structural stainless steels (PureSt) Made by PTY/AAT Date 01/06 Research Fund Client Checked by MAP Date 02/06 caLcuLatIon SHeet for Coal and Steel Revised by MIG Date 06/17 deSIGn exaMPLe 13 - HoLLow SectIon LattIce GIrder The lattice girder supports roof glazing and is made of square and rectangular hollow sections of grade 1.4301 stainless steel; a comparison is made between material in two 2 strength levels - the annealed condition ( fy =210 N/mm ) and in the cold worked condition 2 (strength level CP500, fy = 460 N/mm ). Calculations are performed at the ultimate limit state and then at the fire limit state for a fire duration of 30 minutes. For the CP500 material, the reduction factors for the mechanical properties at elevated temperatures are calculated according to Section 8.2. The structural analysis was carried out using the FE-program WINRAMI marketed by Finnish Constructional Steelwork Association (FCSA) (www.terasrakenneyhdistys.fi). The WINRAMI design environment includes square, rectangular and circular hollow sections for stainless steel structural analysis. WINRAMI solves the member forces, deflections and member resistances for room temperature and structural fire design and also joint resistance at room temperature (it also checks all the geometrical restraints of truss girder joints). In the example, the chord members are modelled as continuous beams and the diagonal members as hinge jointed. According to EN 1993-1-1, the buckling lengths for the chord and diagonal members could be taken as 0,9 times and 0,75 times the distance between nodal points respectively, but in this example conservatively the distance between nodal points has been used as the buckling length. The member forces were calculated by using WINRAMI with profile sizes based on the annealed strength condition. These member forces were used for both the annealed and CP500 girders. This example focuses on checking 3 members: mainly axial tension loaded lower chord (member 0), axial compression loaded diagonal (member 31) and combination of axial compression and bending loaded upper chord member (member 5). The weight of the girders is also compared. The welded joints should be designed according to the Section 7.4, which is not included in this example. Annealed : lower chord 100x60x4, upper chord 80x80x5, corner vertical 60x60x5 diagonals from left to middle: 50x50x3, 50x50x3, 40x40x3, 40x40x3, 40x40x3,40x40x3, 40x40x3. CP500 : lower chord 60x40x4, upper chord 70x70x4, corner vertical 60x60x5, all diagonals 40x40x3. Span length 15 m, height in the middle 3,13 m, height at the corner 0,5 m. Weight of girders: Annealed: 407 kg, CP500 307 kg. The weight is not fully optimised. 247 Design Example 13 Sheet 2 of 8 actions Assuming the girder carries equally distributed snow load, glazing and its support structures and weight of girder : Permanent actions (G): Load of glazing and supports 1 kN/m2 Dead load of girder (WINRAMI calculates the weight) Variable actions (Q): Snow load 2 kN/m2 EN 1990 Load case 1 to be considered (ultimate limit state): γG,jG k,j + Qk,1Q,1 j Load case 2 to be considered (fire situation): γGA,jG k,j + Qk,11,1 j Ultimate limit state (room temperature design) Fire design EN 1990 G, j = 1,35 (unfavourable effects) GA, j = 1,0 EN 1991-1-2 Q,1 = 1,5 γψ1,1 = 0,2 (Recommended partial factors for actions shall be used in this example) Factored actions for ultimate limit state: Permanent action: Load on nodal points: 1,35 x 4,1 kN Self weight of girder (is included by WINRAMI) Variable action Load from snow: 1,5 x 8,1 kN forces at critical members are: Forces are determined by the model using profiles in the annealed strength condition Lower chord member, member 0 Annealed: 100x60x4 mm, CP500: 60x40x4 mm Nt,Ed = 142,2 kN, Nt,fi,Ed = 46,9 kN Mmax ,Ed = 0,672 kNm, Mmax,fire,Ed = 0,245 kNm Upper chord member, member 5 Annealed: 80x80x5 mm, CP500: 70x70x4 mm Nc,Ed = -149,1 kN, Nc,fire,Ed = -49,2 kN Mmax ,Ed = 2,149 kNm, Mmax,fire,Ed = 0,731 kNm Diagonal member, member 31 Annealed: 50x50x3mm, CP500: 40x40x3 mm Nc,Ed = -65,9 kN, Nc,fire,Ed = -21,7 kN Material properties Use material grade 1.4301. 2 2 2 Annealed: fy = 210 N/mm fu = 520 N/mm E = 200000 N/mm Table 2.2 2 2 2 CP500: fy = 460 N/mm fu = 650 N/mm E = 200000 N/mm Table 2.3 Partial factors Table 4.1 and The following partial factors are used throughout the design example: Section 8.1 M0 = 1,1, M1 = 1,1, M,fi = 1,0 cross-section properties: annealed 2 3 3 Member 0: A = 1175 mm Wpl,y = 37,93×10 mm 2 4 4 3 3 Member 5: A = 1436 mm Iy = 131,44×10 mm iy=30,3 mm Wpl,y = 39,74×10 mm 2 4 4 3 3 Member 31: A = 541 mm Iy = 19,47×10 mm iy = 19 mm Wpl,y = 9,39×10 mm 248 Design Example 13 Sheet 2 of 8 Design Example 13 Sheet 3 of 8 actions cross-section properties: cP500 3 3 Assuming the girder carries equally distributed snow load, glazing and its support Member 0: A = 695 mm Wpl,y = 13,16×10 mm structures and weight of girder : 2 4 4 3 3 Member 5: A = 1015 mm Iy = 72,12×10 mm iy = 26,7 mm Wpl,y = 24,76×10 mm 2 Permanent actions (G): Load of glazing and supports 1 kN/m 2 4 4 3 3 Member 31: A = 421 mm Iy = 9,32×10 mm iy = 14,9 mm Wpl,y = 5,72×10 mm Dead load of girder (WINRAMI calculates the weight) Variable actions (Q): Snow load 2 kN/m2 classification of the cross-section of member 5 and member 31 Load case 1 to be considered (ultimate limit state): γ G + Q EN 1990 G,j k,j k,1Q,1 Annealed : = 1,03 CP500 : = 0,698 Table 5.2 j Annealed 80x80x5 : c = 80 15 = 65 mm CP500 70x70x4 : c = 70 – 12 = 58 mm Load case 2 to be considered (fire situation): γGA,jG k,j + Qk,11,1 j Annealed 50x50x3 : c = 50 – 9 = 41 mm CP500 40x40x3 : c = 40 – 9 = 31 mm Ultimate limit state (room temperature design) Fire design Flange/web subject to compression: Table 5.2 EN 1990 G, j = 1,35 (unfavourable effects) GA, j = 1,0 EN 1991-1-2 Annealed 80x80x5 : c/t = 13 CP500 70x70x4 : c/t = 14,5 γψ1,1 = 0,2 Q,1 = 1,5 Annealed 50x50x3 : c/t = 13,7 CP500 40x40x3 : c/t = 10,3 (Recommended partial factors for actions shall be used in this example) c For Class 1, 33 0, , therefore both profiles are classified as Class 1 Factored actions for ultimate limit state: t Permanent action: Load on nodal points: 1,35 x 4,1 kN Self weight of girder (is included by WINRAMI) Lower cHord MeMBer, deSIGn In rooM and fIre teMPerature (Member 0) Variable action Load from snow: 1,5 x 8,1 kN a) room temperature design tension resistance of cross-section Section 5.7.2 forces at critical members are: Npl,Rd = fA Eq. 5.23 Forces are determined by the model using profiles in the annealed strength condition g M 0y Lower chord member, member 0 Annealed : Npl,Rd = 1175 210 / 1,1 = 224,3 kN > 142,2 kN OK. Annealed: 100x60x4 mm, CP500: 60x40x4 mm CP500 : Npl,Rd = 695 460 / 1,1 = 290,6 kN > 142,2 kN OK. Nt,Ed = 142,2 kN, Nt,fi,Ed = 46,9 kN Mmax ,Ed = 0,672 kNm, Mmax,fire,Ed = 0,245 kNm Moment resistance of cross-section Sec. 5.7.4 Upper chord member, member 5 Mc,Rd = Wfpl yγ M0 Eq. 5.29 Annealed: 80x80x5 mm, CP500: 70x70x4 mm 37,93 103 210 Annealed : Mc,Rd = = 7,24 kNm > 0,672 kNm OK. Nc,Ed = -149,1 kN, Nc,fire,Ed = -49,2 kN 1,1 10 6 Mmax ,Ed = 2,149 kNm, Mmax,fire,Ed = 0,731 kNm 13,16 103 460 Diagonal member, member 31 CP500 : Mc,Rd = = 5,50 kNm > 0,672 kNm OK. 1,1 10 6 Annealed: 50x50x3mm, CP500: 40x40x3 mm Nc,Ed = -65,9 kN, Nc,fire,Ed = -21,7 kN axial tension and bending moment interaction N M y,Ed Material properties Ed 1 Eq. 6.55 Use material grade 1.4301. NMRd y,Rd 2 2 2 Annealed: fy = 210 N/mm fu = 520 N/mm E = 200000 N/mm Table 2.2 142,2 0,672 2 2 2 Annealed : 0,73 1 OK. CP500: fy = 460 N/mm fu = 650 N/mm E = 200000 N/mm Table 2.3 224,3 7,24 142,2 0,672 CP500 : 0,61 1 OK. Partial factors Table 4.1 and 290,6 5,50 The following partial factors are used throughout the design example: Section 8.1 M0 = 1,1, M1 = 1,1, M,fi = 1,0 B) fire temperature design εres = 0,4 Section 8.4.4 -1 cross-section properties: annealed Steel temperature for 100x60x4 after 30 min fire for Am/V = 275 m : θ= 833 °C 2 3 3 -1 Member 0: A = 1175 mm Wpl,y = 37,93×10 mm Steel temperature for 60x40x4 after 30 min fire for Am/V = 290 m : θ= 834 °C 2 4 4 3 3 Member 5: A = 1436 mm Iy = 131,44×10 mm iy=30,3 mm Wpl,y = 39,74×10 mm Conservatively take θ= 834 °C. 2 4 4 3 3 Member 31: A = 541 mm Iy = 19,47×10 mm iy = 19 mm Wpl,y = 9,39×10 mm 249 Design Example 13 Sheet 4 of 8 Annealed : The values for the reduction factors at 834 °C are obtained by linear interpolation: Section 8.2 Table 8.1 k2,θ = f2,/fy = 0,292, but f2, fu, ku,θ = fu,/fu = 0,209 f2, = 0,292 210 = 61,3 and fu, = 0,209 520 = 108,7, therefore f2, fu, CP500 : For material in the cold worked condition for θ ≥ 800 °C: Section 8.2 k2,θ,CF = f2,,CF/fy = 0,9k2,θ = 0,9f2,/fy = 0,9×0,292 = 0,263, but f2,,CF fu,,CF Table 8.1 ku,θ,CF = ku,θ = fu,,CF/fu = 0,209 f2,,CF = 0,263 460 = 121,0 and fu,,CF = 0,209 650 = 135,9, therefore f2,,CF fu,,CF tension resistance of cross-section Nfi,θ,Rd = k2, NRd [M0 /M,fi] Eq. 8.8 Annealed : Nfi,θ,Rd = 0,292×224,31,1/1,0 = 72,0 kN > 46,9 kN OK. CP500 : Nfi,θ,Rd = 0,263×290,61,1/1,0 = 84,1 kN > 46,9 kN OK. Moment resistance of cross-section Eq. 8.15 Mfi,θ,Rd = kM2,θ Rdγ M0 /γ M,fi Annealed : Mfi,θ,Rd = 0,292×7,24 ×1,1/1,0 = 2,33 kNm > 0,245 kNm OK. CP500 : Mfi,θ,Rd = 0,263× 5,50 ×1,1/1,0 = 1,59 kNm > 0,245 kNm OK. axial tension and bending moment interaction N M Ed y,Ed 1 NMRd y,Rd 46,9 0,245 Eq. 6.55 Annealed 0,75 1 OK 72,0 2,33 46,9 0,245 CP500 : 0,71 1 OK. 84,1 1,59 dIaGonaL MeMBer deSIGn In rooM and fIre teMPerature (Member 31) Buckling length = 1253 mm a) room temperature design Nb,Rd = χAfy /γ M1 Eq. 6.2 Annealed : Lcr 1 1253 1 (fEy /)= (210 / 200000) = 0,680 Eq. 6.6 i 19 2 2 Eq. 6.5 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,680 - 0,3)+0,680 ) = 0,824 Table 6.1 1 1 = = 0,776 22 Eq. 6.4 () 22 0,824 (0,824 0,680 ) Nb,Rd = 0,776 × 541 × 210 /1,1 = 80,1 kN > 65,9 kN OK. 250 Design Example 13 Sheet 4 of 8 Design Example 13 Sheet 5 of 8 Annealed : CP500 : The values for the reduction factors at 834 °C are obtained by linear interpolation: Section 8.2 Lcr 1 1253 1 Table 8.1 (fEy /)= (460 / 200000) = 1,284 Eq. 6.6 k2,θ = f2,/fy = 0,292, but f2, fu, i 14,9 ku,θ = fu,/fu = 0,209 2 2 Eq. 6.5 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(1,284-0,3)+1,284 ) = 1,565 f2, = 0,292 210 = 61,3 and fu, = 0,209 520 = 108,7, therefore f2, fu, Table 6.1 CP500 : 1 1 For material in the cold worked condition for θ ≥ 800 °C: Section 8.2 = = 0,407 22 22 Eq. 6.4 k2,θ,CF = f2,,CF/fy = 0,9k2,θ = 0,9f2,/fy = 0,9×0,292 = 0,263, but f2,,CF fu,,CF Table 8.1 () 1,565 (1,565 1,284 ) u,θ,CF u,θ u,,CF u k = k = f /f = 0,209 Nb,Rd = 0,407×421 × 460 /1,1 = 71,7 kN > 65,9 kN OK. f2,,CF = 0,263 460 = 121,0 and fu,,CF = 0,209 650 = 135,9, therefore f2,,CF fu,,CF B) fire temperature design tension resistance of cross-section εres = 0,4 Section 8.4.4 Nfi,θ,Rd = k2, NRd [M0 /M,fi] Eq. 8.8 -1 Steel temperature for 80x80x5 after 30 min fire for Am/V = 220 m : θ = 830 °C -1 Annealed : Nfi,θ,Rd = 0,292×224,3 1,1/1,0 = 72,0 kN > 46,9 kN OK. Steel temperature for 70x70x5 after 30 min fire for Am/V = 225 m : θ = 831 °C CP500 : Nfi,θ,Rd = 0,263×290,61,1/1,0 = 84,1 kN > 46,9 kN OK. Conservatively take θ = 831 °C. Annealed : Moment resistance of cross-section The values for the reduction factors at 831 C are obtained by linear interpolation: Section 8.2 Table 8.1 Eq. 8.15 kp0,2,θ = 0,219 and kE,θ = 0,574 Mfi,θ,Rd = kM2,θ Rdγ M0 /γ M,fi Annealed : Mfi,θ,Rd = 0,292×7,24 ×1,1/1,0 = 2,33 kNm > 0,245 kNm OK. Cross-section classification Section 8.3.2 0,5 0,5 CP500 : Mfi,θ,Rd = 0,263× 5,50 ×1,1/1,0 = 1,59 kNm > 0,245 kNm OK. k 0,574 E,θ θ 1, 03 1, 67 Eq. 8.6 ky,θ 0,219 axial tension and bending moment interaction Class 1 sections: c/t 33,0 = 33,0×1,67 = 55,1 NEd M y,Ed 1 Class 1, c/t = 13, therefore profileθ is classified as Class 1. NMRd y,Rd ε 46,9 0,245 Eq. 6.55 Annealed 0,75 1 OK CP500 : 72,0 2,33 For material in the cold worked condition for θ ≥ 800 C: Section 8.2 46,9 0,245 Table 8.1 CP500 : 0,71 1 OK. kp0,2,θ,CF = 0,8kp0,2,θ = 0,8×0,219 = 0,175 84,1 1,59 kE,θ,CF = kE,θ = 0,574 dIaGonaL MeMBer deSIGn In rooM and fIre teMPerature (Member 31) Cross-section classification Section 8.3.2 Buckling length = 1253 mm 0,5 0,5 k 0,574 E,θ a) room temperature design θ 0,698 1, 26 Eq. 8.6 ky,θ 0,175 Nb,Rd = χAfy /γ M1 Eq. 6.2 Class 1 sections: c/t 33,0 = 33,0×1,26 = 41,6 Annealed : Class 1, c/t = 14,5, therefore θprofile is classified as Class 1. L 1 1253 1 ε cr (fEy /)= (210 / 200000) = 0,680 Eq. 6.6 Eq. 8.10 i 19 Nb,fi,t,Rd = fiAk p0,2,θ f y/ γ M,fi as both profiles are classified as Class 1. 2 2 Eq. 6.5 Annealed : 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,680 - 0,3)+0,680 ) = 0,824 Table 6.1 θ (kkp0,2,θ /) E,θ = 0,680 (0,219 / 0,574) = 0,420 Eq. 8.14 1 1 = = 0,776 22 22 Eq. 6.4 2 2 () 0,824 (0,824 0,680 ) θ 0,5(1 (θ 0 ) θ ) = 0,5×(1+0,49×(0,420-0,3)+0,420 ) = 0,618 Eq. 8.13 Nb,Rd = 0,776 × 541 × 210 /1,1 = 80,1 kN > 65,9 kN OK. 251 Design Example 13 Sheet 6 of 8 1 1 fi = = 0,933 220,618 (0,61822 0,420 ) Eq. 8.12 θ() θθ Nb,fi,t,Rd = 0,933×541 0,219×210 /1,0 = 23,2 kN > 21,7 kN OK. CP500 : θ (kkp0,2,θ,CF /) E,θ,CF =1,284 (0,175 / 0,574) = 0,709 Eq. 8.14 2 2 θ 0,5(1 (θ 0 ) θ ) = 0,5× (1+0,49× (0,709-0,3)+0,709 ) = 0,852 Eq. 8.13 1 1 = = 0,755 fi 220,852 (0,85222 0,709 ) Eq. 8.12 θ() θθ Nb,fi,t,Rd = 0,755×421 ×0,175×460 /1,0 = 25,6 kN > 21,7 kN OK. uPPer cHord MeMBer deSIGn In rooM and fIre teMPerature (Member 5) Buckling length = 1536 mm a) room temperature design N M Ne Ed k y,Ed Ed Ny 1, 0 y Eq. 6.56 ()Nb,Rd min W,yWf pl,y y/ γ M1 Annealed : βW,y = 1,0 Class 1 cross-section Sec. 6.5.2 ky = 1+D1( - D2)NEd /Nb,Rd,y, but ky ≤ 1+ D1(D3 - D2)NEd /Nb,Rd,y Eq. 6.63 Where D1 =𝜆𝜆𝑦𝑦 2,0, D2 = 0,3 and D3 = 1,3 Table 6.6 Lcr 1 1536 1 (fEy /)= (210 / 200000) = 0,523 Eq. 6.6 i 30,3 2 2 Eq. 6.5 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,523-0,3)+0,523 ) = 0,691 1 1 = = 0,875 22 Eq. 6.4 () 220,691 (0,691 0,523 ) Nb,Rd,y = 0,875×1436 ×210 /1,1 = 239,9 kN > 149,1 kN Eq. 6.2 ky = 1,0+2,0×(0,523 - 0,30)×149,1/239,9 = 1,277 Table 6.6 ky ≤ 1,0+2,0×(1,3 -0,30)×149,1/239,9 = 2,243, therefore, ky = 1,277 149,1 2,149 10002 1,277 3 = 0,98 < 1,0 OK. Eq. 6.56 239,9 1,0 39,74 10 210 /1,1 CP500 βW,y = 1,0 Class 1 cross-section Sec. 6.5.2 Lcr 1 1536 1 (fEy /)= (460 / 200000) = 0,878 Eq. 6.6 i 26,7 2 2 Eq. 6.5 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,878-0,3)+0,878 ) = 1,027 252 Design Example 13 Sheet 6 of 8 Design Example 13 Sheet 7 of 8 1 1 1 1 = = 0,933 = = 0,641 fi 22 22 Eq. 6.4 220,618 (0,618 0,420 ) Eq. 8.12 22 θ() θθ () 1,027 (1,027 0,878 ) Nb,fi,t,Rd = 0,933×541 0,219×210 /1,0 = 23,2 kN > 21,7 kN OK. Nb,Rd,y = 0,641×1015 ×460 /1,1 = 272,1 kN > 149,1 kN Eq. 6.2 ky = 1,0+2×(0,878 - 0,30)×149,1/272,1 = 1,633 Table 6.6 CP500 : ky ≤ 1,0+2,0×(1,3 - 0,30)×149,1/272,1 = 2,096, therefore ky = 1,633 (kk /)=1,284 (0,175 / 0,574) = 0,709 2 θ p0,2,θ,CF E,θ,CF Eq. 8.14 149,1 2,149 1000 1,633 = 0,89 < 1,0 OK. 3 Eq. 6.56 2 2 272,1 1,0 24,76 10 460 /1,1 θ 0,5(1 (θ 0 ) θ ) = 0,5× (1+0,49× (0,709-0,3)+0,709 ) = 0,852 Eq. 8.13 1 1 = = 0,755 fi 22 Eq. 8.12 B) fire temperature design () 220,852 (0,852 0,709 ) θ θθ εres = 0,4 Section 8.4.4 -1 Nb,fi,t,Rd = 0,755×421 ×0,175×460 /1,0 = 25,6 kN > 21,7 kN OK. Steel temperature for 50x50x3 after 30 min fire for Am/V = 370 m : θ = 836 °C -1 Steel temperature for 40x40x3 after 30 min fire for Am/V = 380 m : θ = 836 °C uPPer cHord MeMBer deSIGn In rooM and fIre teMPerature (Member 5) Buckling length = 1536 mm Annealed : The values for the reduction factors at 836 °C are obtained by linear interpolation: Section 8.2 a) room temperature design kp0,2,θ = 0,214 Table 8.1 N My,Ed Ne Ed Ny Ed k 1, 0 k2,θ = f2,/fy = 0,289, but f2, fu, y Eq. 6.56 ()Nb,Rd min W,yWf pl,y y/ γ M1 ku,θ = fu,/fu = 0,207 Annealed : f2, = 0,289 210 = 60,7 and fu, = 0,207 520 = 107,6, therefore f2, fu, βW,y = 1,0 Class 1 cross-section Sec. 6.5.2 kE,θ = 0,565 ky = 1+D1( - D2)NEd /Nb,Rd,y, but ky ≤ 1+ D1(D3 - D2)NEd /Nb,Rd,y Eq. 6.63 Cross-section classification Section 8.3.2 Where D1 =𝑦𝑦 2,0, D2 = 0,3 and D3 = 1,3 Table 6.6 𝜆𝜆 0,5 k 0,5 L 1 1536 1 E,θ 0,565 cr θ 1, 03 1, 67 Eq. 8.6 (fEy /)= (210 / 200000) = 0,523 Eq. 6.6 i 30,3 ky,θ 0,214 2 2 Eq. 6.5 Class 1 sections: c/t 33,0 = 33,0×1,67 = 55,1 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,523-0,3)+0,523 ) = 0,691 Class 1, c/t = 13,7, therefore θprofile is classified as Class 1. 1 1 ε = = 0,875 22 Eq. 6.4 () 220,691 (0,691 0,523 ) CP500 : Nb,Rd,y = 0,875×1436 ×210 /1,1 = 239,9 kN > 149,1 kN Eq. 6.2 For material in the cold worked condition for θ ≥ 800 °C: Section 8.2 kp0,2,θ,CF =0,8kp0,2,θ = 0,8×0,214 = 0,171 Table 8.1 ky = 1,0+2,0×(0,523 - 0,30)×149,1/239,9 = 1,277 Table 6.6 k2,θ,CF = f2,,CF/fy = 0,9k2,θ = 0,9f2,/fy = 0,9×0,289 = 0,260, but f2,,CF fu,,CF ky ≤ 1,0+2,0×(1,3 -0,30)×149,1/239,9 = 2,243, therefore, ky = 1,277 ku,θ,CF = ku,θ = fu,,CF/fu = 0,207 149,1 2,149 10002 1,277 3 = 0,98 < 1,0 OK. Eq. 6.56 f2,,CF = 0,260 460 = 94,8 and fu,,CF = 0,207 650 = 134,6, therefore f2,,CF fu,,CF 239,9 1,0 39,74 10 210 /1,1 kE,θ,CF = kE,θ = 0,565 CP500 Cross-section classification Section 8.3.2 βW,y = 1,0 Class 1 cross-section Sec. 6.5.2 0,5 0,5 k 0,565 E,θ θ 0,698 1, 27 Eq. 8.6 Lcr 1 1536 1 ky,θ 0,171 (fEy /)= (460 / 200000) = 0,878 Eq. 6.6 i 26,7 Class 1 sections: c/t 33,0 = 33,0×1,27 = 41,9 2 2 Eq. 6.5 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,878-0,3)+0,878 ) = 1,027 Class 1, c/t = 10,3 < 41,9, thereforeεθ profile is classified as Class 1. 253 Design Example 13 Sheet 8 of 8 N kM fi,Ed y y,fi,Ed 1, 0 as both profiles are classified as Class 1. f y M y,fi,θ,Rd Eq. 8.26 min,fi Ak g p0,2,θ M,fi Annealed : Eq. 8.14 θ (kkp0,2,θ /) E,θ = 0,523 (0,214 / 0,565) = 0,322 2 2 θ 0,5(1 (θ 0 ) θ ) = 0,5×(1+0,49×(0,322-0,3)+0,322 ) = 0,557 Eq. 8.13 1 1 fi = = 0,989 220,557 (0,55722 0,322 ) Eq. 8.12 θ() θθ yN fi,Ed ky 13 Eq. 8.30 y,fiAk g p0,2,θ f y/ γ M,fi Eq. 8.31 y (1,2 M,y 3)y,θ 0,44 M,y 0,29 0,8 Eq. 8.26 min,fiAk g p0,2,θ fy / γ M,fi =0,989×1436 × 0,214×210 /1,0 = 63,8 kN > 49,2 kN OK. 3 2 My,fi,θ,Rd = k2,θ[γM0/γM,fi]MRd = 0,289×1,1/1,0×39,74×10 ×210/1000 = 2,65 kNm Eq. 8.15 >0,731 kNm OK. ψ = -0,487 /0,731 = -0,666 Table 8.3 βM,y = 1,8-0,7ψ = 2,266 μy = (1,2×2,266-3) ×0,322 + 0,44×2,266 – 0,29 = 0,617 < 0,8 ky = 1- 0,617×49,2 kN/63,8 kN = 0,524 < 3 49 2, ,0 731 ,0 524 = 0,92 < 1,0 OK. 63 8, ,2 65 CP500 : θ (kkp0,2,θ /) E,θ = 0,878 (0,171/ 0,565) = 0,483 Eq. 8.14 2 2 θ 0,5(1 (θ 0 ) θ ) = 0,5×(1+0,49×(0,483-0,3)+0,483 ) = 0,661 Eq. 8.13 1 1 fi = = 0,899 220,661 (0,66122 0,483 ) Eq. 8.12 θ() θθ Eq. 8.26 min,fiAk g p0,2,θ f y/ γ M,fi = 0,899 1015 0,171 460 /1,0 = 71,8 kN >49,2 kN OK. 3 2 My,fi,θ,Rd = k2,θ[γM0/γM,fi]MRd = 0,260×1,1/1,0×24,76×10 ×460/1000 = 3,26 kNm Eq. 8.15 >0,731 kNm OK. ψ = -0,487 /0,731 = -0,666 Table 8.3 βM,y = 1,8-0,7ψ = 2,266 μy = (1,2×2,266-3)×0,483 + 0,44×2,266 – 0,29 = 0,571 ≤ 0,8 ky = 1- 0,571×49,2 /71,8 = 0,609 49 2, ,0 731 ,0 609 = 0,82 < 1,0 OK. 718, ,3 26 254 Design Example 13 Sheet 8 of 8 Promotion of new Sheet 1 of 3 N kM eurocode rules for fi,Ed y y,fi,Ed 1, 0 as both profiles are classified as Class 1. Design Example 14 – Determination of enhanced f structural stainless Title y M y,fi,θ,Rd Eq. 8.26 average yield strength for cold-formed sections min,fi Ak g p0,2,θ M,fi steels (PureSt) Made by SA Date 05/17 Annealed : Research Fund for Eq. 8.14 caLcuLatIon Client Revised by FW Date 05/17 θ (kkp0,2,θ /) E,θ = 0,523 (0,214 / 0,565) = 0,322 Coal and Steel SHeet Revised by LG Date 05/17 2 2 θ 0,5(1 (θ 0 ) θ ) = 0,5×(1+0,49×(0,322-0,3)+0,322 ) = 0,557 Eq. 8.13 deSIGn exaMPLe 14 – deterMInatIon of enHanced aVeraGe YIeLd StrenGtH 1 1 for coLd-forMed SectIonS fi = = 0,989 220,557 (0,55722 0,322 ) Eq. 8.12 θ() θθ This worked example illustrates the determination of the enhanced average yield strength of a cold-rolled square hollow section (SHS) in accordance with the method in Annex B. yN fi,Ed The calculations are carried out for an SHS 80×80×4 in austenitic grade 1.4301 stainless k 13 ya y Eq. 8.30 steel.𝑓𝑓 The predicted cross-section bending resistances based on the minimum specified yield y,fiAk g p0,2,θ f y/ γ M,fi strength and the calculated enhanced average yield strength are then compared. Eq. 8.31 y ya y (1,2 M,y 3)y,θ 0,44 M,y 0,29 0,8 𝑓𝑓 𝑓𝑓 enhanced average yield strength =0,989×1436 × 0,214×210 /1,0 = 63,8 kN > 49,2 kN OK. Eq. 8.26 min,fiAk g p0,2,θ fy / γ M,fi For stainless steel cold-rolled box sections (RHS and SHS), the enhanced average yield 3 2 strength is: My,fi,θ,Rd = k2,θ[γM0/γM,fi]MRd = 0,289×1,1/1,0×39,74×10 ×210/1000 = 2,65 kNm Eq. 8.15 >0,731 kNm OK. 𝑓𝑓ya Eq. B.2 𝑦𝑦𝑦𝑦 𝑐𝑐,𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑦𝑦𝑦𝑦 𝑐𝑐,𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ψ = -0,487 /0,731 = -0,666 Table 8.3 𝑓𝑓 𝐴𝐴 + 𝑓𝑓 (𝐴𝐴 − 𝐴𝐴 ) 𝑓𝑓𝑦𝑦𝑦𝑦 = 𝐴𝐴 βM,y = 1,8-0,7ψ = 2,266 cross-section properties μy = (1,2×2,266-3) ×0,322 + 0,44×2,266 – 0,29 = 0,617 < 0,8 Geometric properties of SHS 80×80×4 (measured properties from a test specimen): ky = 1- 0,617×49,2 kN/63,8 kN = 0,524 < 3 h = 79,9 mm b = 79,6 mm 2 49 2, ,0 731 t = 3,75 mm A = 1099 mm ,0 524 = 0,92 < 1,0 OK. 3 3 63 8, ,2 65 Wel = 25967 mm Wpl = 30860 mm CP500 : ri = 4,40 mm (Note that ri may be taken as 2t if not known) Appendix B θ (kkp0,2,θ /) E,θ = 0,878 (0,171/ 0,565) = 0,483 Eq. 8.14 Eq. B.14 t 2 2 2 c,rolled c i c 0,5(1 ( 0 ) ) = 0,5×(1+0,49×(0,483-0,3)+0,483 ) = 0,661 A = (n π ) (2r + t) + 4n t θ θ θ Eq. 8.13 4 1 1 3,75 2 2 = = 0,899 Ac,rolled = (4×π× ) × (2×4,40 + 3,75) + 4×4×3,75 = 373 mm fi 22 Eq. 8.12 () 220,661 (0,661 0,483 ) 4 θ θθ Material properties Eq. 8.26 = 230 N/mm2 and = 540 N/mm2 (for cold-rolled strip with t ≤ 8 mm) Table 2.2 min,fiAk g p0,2,θ f y/ γ M,fi = 0,899 1015 0,171 460 /1,0 = 71,8 kN >49,2 kN OK. y u Section 2.3.1 3 2 𝑓𝑓 𝑓𝑓 My,fi,θ,Rd = k2,θ[γM0/γM,fi]MRd = 0,260×1,1/1,0×24,76×10 ×460/1000 = 3,26 kNm Eq. 8.15 2 Eq. B.10 𝐸𝐸 = 200000 N/mm >0,731 kNm OK. p0,2 ε = 0,002 + 𝑓𝑓y⁄E = 0,00315 Eq. C.6 ψ = -0,487 /0,731 = -0,666 Table 8.3 u ε = 1- 𝑓𝑓y⁄𝑓𝑓u = 0,57 βM,y = 1,8-0,7ψ = 2,266 corner and flat enhanced yield strengths μy = (1,2×2,266-3)×0,483 + 0,44×2,266 – 0,29 = 0,571 ≤ 0,8 Predicted enhanced yield strength of corner regions : ky = 1- 0,571×49,2 /71,8 = 0,609 and 𝑓𝑓yc Eq. B.4 49 2, ,0 731 𝑛𝑛p ,0 609 = 0,82 < 1,0 OK. 718, ,3 26 𝑓𝑓yc = 0,85𝐾𝐾 (εc + εp0,2) 𝑓𝑓y ≤ 𝑓𝑓yc ≤ 𝑓𝑓u 255 Design Example 14 Sheet 2 of 3 Predicted enhanced yield strength of flat faces : 𝑓𝑓yf and Eq. B.5 𝑛𝑛p 𝑓𝑓yf = 0,85𝐾𝐾 (εf + εp0,2) 𝑓𝑓y ≤ 𝑓𝑓yf ≤ 𝑓𝑓u Corner and flat cold-work induced plastic strains Strain induced in the corner regions ε : c Eq. B.7 𝑡𝑡 εc = 2(2𝑟𝑟i + 𝑡𝑡) 3,75 εc = = 0,149 Strain2 induced× (2 × 4 ,in40 the+ 3flat,75 )faces ε : f Eq. B.8 𝑡𝑡 𝜋𝜋𝜋𝜋 𝜀𝜀𝑓𝑓 = [ ] + [ ] 900 2(𝑏𝑏 + ℎ − 2𝑡𝑡) 3,75 𝜋𝜋 × 3,75 𝜀𝜀𝑓𝑓 = [ ] + [ ] = 0,043 900 2 × (79,6 + 79,9 − 2 × 3,75) Material model parameters Eq. B.12 𝑙𝑙𝑙𝑙(𝑓𝑓𝑦𝑦⁄𝑓𝑓𝑢𝑢) 𝑛𝑛𝑝𝑝 = 𝑙𝑙𝑙𝑙(𝜀𝜀𝑝𝑝0,2⁄𝜀𝜀𝑢𝑢) 𝑙𝑙𝑙𝑙(230/540) 𝑛𝑛𝑝𝑝 = = 0,164 𝑙𝑙𝑙𝑙(0,00315/0,57) Eq. B.11 𝑓𝑓𝑦𝑦 𝐾𝐾 = 𝑛𝑛𝑝𝑝 𝜀𝜀𝑝𝑝0,2 230 2 𝐾𝐾 = 0,164 = 591,6 𝑁𝑁/𝑚𝑚𝑚𝑚 (0,00315) Corner and flat enhanced yield strengths Predicted enhanced yield strength of corner regions : Eq. B.4 𝑓𝑓yc 0,164 𝑓𝑓yc = 0,85 × 591and,6 × (0,149 + 0,00315) 2 Predicted = 369 enhancedN/mm yield 230 strength≤ 369 ≤ of540 flat faces : Eq. B.5 𝑓𝑓yf 0,164 𝑓𝑓yf = 0,85 × 591,6and× (0,043 + 0,00315) 2 = 304 N/mm 230 ≤ 304 ≤ 540 Section enhanced average yield strength Eq. B.2 𝑓𝑓yc 𝐴𝐴c,rolled + 𝑓𝑓yf(𝐴𝐴 − 𝐴𝐴c,rolled) 𝑓𝑓ya = 𝐴𝐴 369 × 373 + 304 × (1099 − 373) 2 = = 326 𝑁𝑁/𝑚𝑚𝑚𝑚 1099 256 Design Example 14 Sheet 2 of 3 Design Example 14 Sheet 3 of 3 Predicted enhanced yield strength of flat faces : cross-section classification Cross-section classification based on minimum specified yield strength : 𝑓𝑓yf y and Eq. B.5 𝑛𝑛p 𝑓𝑓 yf f p0,2 y yf u Table 5.2 𝑓𝑓 = 0,85𝐾𝐾 (ε + ε ) 𝑓𝑓 ≤ 𝑓𝑓 ≤ 𝑓𝑓 000 0,5 000 0,5 235y 𝐸𝐸 235 200000 Corner and flat cold-work induced plastic strains 𝜀𝜀 = [ ] = [ × ] = 0,986 Strain induced in the corner regions ε : 𝑓𝑓 210 230 210 𝑐𝑐 (79,9 − 3 × 3,75) c Therefore,= the cross-section= 18 is,3 classified< 32,5 = as33 Class𝜀𝜀 1. Eq. B.7 𝑡𝑡 3,75 𝑡𝑡 Cross-section classification based on average yield strength : εc = y 2(2𝑟𝑟i + 𝑡𝑡) 𝑓𝑓 3,75 Table 5.2 εc = = 0,149 000 0,5 000 0,5 Strain2 induced× (2 × 4 ,in40 the+ 3flat,75 )faces ε : 235y 𝐸𝐸 235 200000 𝜀𝜀 = [ ] = [ × ] = 0,829 f 𝑓𝑓 210 326 210 Eq. B.8 𝑡𝑡 𝜋𝜋𝜋𝜋 𝑐𝑐 (79,9 − 3 × 3,75) 𝜀𝜀𝑓𝑓 = [ ] + [ ] Therefore,= the cross-section= 18 is,3 classified< 27,4 = as33 Class𝜀𝜀 1. 900 2(𝑏𝑏 + ℎ − 2𝑡𝑡) 𝑡𝑡 3,75 3,75 𝜋𝜋 × 3,75 𝜀𝜀𝑓𝑓 = [ ] + [ ] = 0,043 900 2 × (79,6 + 79,9 − 2 × 3,75) cross-sectional bending resistance Material model parameters For a Class 1 or 2 section: Eq. B.12 Eq. 5.29 𝑦𝑦 𝑢𝑢 𝑙𝑙𝑙𝑙(𝑓𝑓 ⁄𝑓𝑓 ) 𝑐𝑐,𝑅𝑅𝑅𝑅 𝑝𝑝𝑝𝑝 𝑦𝑦 𝑛𝑛𝑝𝑝 = 𝑀𝑀Resistance = 𝑊𝑊 based 𝑓𝑓 / on𝑀𝑀 minimum0 specified yield strength : 𝑙𝑙𝑙𝑙(𝜀𝜀𝑝𝑝0,2⁄𝜀𝜀𝑢𝑢) 𝒇𝒇y 𝑝𝑝 𝑙𝑙𝑙𝑙(230/540) 𝑛𝑛 = = 0,164 30860 × 230 𝑙𝑙𝑙𝑙(0,00315/0,57) 𝑐𝑐,𝑅𝑅𝑅𝑅 Eq. B.11 Re𝑀𝑀 sistance = based on enhanced= 6,45 average 𝑘𝑘𝑘𝑘𝑘𝑘 yield strength y : 𝑦𝑦 1,1 𝑛𝑛𝑓𝑓𝑝𝑝 𝐾𝐾 = 𝒇𝒇 a 𝜀𝜀𝑝𝑝0,2 30860 × 326 230 2 𝑀𝑀𝑐𝑐,𝑅𝑅𝑅𝑅 = = 9,15 𝑘𝑘𝑘𝑘𝑘𝑘 𝐾𝐾 = 0,164 = 591,6 𝑁𝑁/𝑚𝑚𝑚𝑚 Taking into account1,1 the increased strength arising from strain hardening during section (0,00315) forming results in a 42% increase in bending resistance. Corner and flat enhanced yield strengths Predicted enhanced yield strength of corner regions : Eq. B.4 Note: Example 15 illustrates the additional enhanced cross-section bending resistance due to the beneficial influence of work hardening in service using the Continuous Strength 𝑓𝑓yc 0,164 Method, as described in Annex D. 𝑓𝑓yc = 0,85 × 591and,6 × (0,149 + 0,00315) 2 Predicted = 369 enhancedN/mm yield 230 strength≤ 369 ≤ of540 flat faces : Eq. B.5 𝑓𝑓yf 0,164 𝑓𝑓yf = 0,85 × 591,6and× (0,043 + 0,00315) 2 = 304 N/mm 230 ≤ 304 ≤ 540 Section enhanced average yield strength Eq. B.2 𝑓𝑓yc 𝐴𝐴c,rolled + 𝑓𝑓yf(𝐴𝐴 − 𝐴𝐴c,rolled) 𝑓𝑓ya = 𝐴𝐴 369 × 373 + 304 × (1099 − 373) 2 = = 326 𝑁𝑁/𝑚𝑚𝑚𝑚 1099 257 258 Promotion of new Sheet 1 of 2 eurocode rules for Design Example 15 – Cross-section design in structural stainless Title bending using the continuous strength method steels (PureSt) (CSM) Made by SA Date 05/17 Research Fund for caLcuLatIon Client Revised by FW Date 05/17 Coal and Steel SHeet Revised by LG Date 05/17 deSIGn exaMPLe 15 – croSS-SectIon deSIGn In BendInG uSInG tHe contInuouS StrenGtH MetHod (cSM) This worked example determines the design value of the in-plane bending resistance of a cold-rolled SHS 80×80×4 beam in austenitic grade 1.4301 stainless steel according to the Continuous Strength Method (CSM) method given in Annex D. cross-section properties The properties are given in Design Example 14. Material properties 2 2 fy = 326 N/mm * and fu = 540 N/mm Table 2.2 E = 200000 N/mm2 and υ = 0,3 Section 2.3.1 εy = 𝑓𝑓y⁄E = 0,0016 Eq. C.6 u ε = 1- 𝑓𝑓y⁄𝑓𝑓u = 0,40 * In order to illustrate the extra bending resistance obtained by using the CSM, in addition to that obtained from employing the enhanced average yield strength of the section due to section forming, the yield strength is taken as the enhanced average yield strength from Design Example 14. The yield strength may alternatively be taken as the minimum specified value. cross-section slenderness D.3.2 𝑓𝑓y λ̅p = √ 𝑓𝑓cr,p Eq. D.4 and 2 2 2 2 σ 2 Table 5.3 cr,p 𝑘𝑘 π 𝐸𝐸𝑡𝑡 4 × π × 200000 × 3,75 𝑓𝑓 = 2 2 = 2 2 = 2530 N/mm 12(1 − υ )𝑏𝑏̅ 12 × (1 − 0,3 ) × (79,7 − 2(3,75 + 4,40)) 326 λ̅p = √ = 0,36 (< 0,68) 2530 cross-section deformation capacity for Eq. D.2 εcsm 0,25 𝐶𝐶1εu = 3,6 ≤ min (15, ) λ̅p ≤ 0,68 y p y Fromε Tableλ̅ D.1, C1 = 0.1 forε austenitic stainless steel. Table D.1 259 Design Example 15 Sheet 2 of 2 εcsm 0,25 0,1 × 0,40 = 3,6 = 9,9 ≤ min (15, = 25) εy 0,36 0,0016 εcsm ∴ = 9,9 εy Strain hardening slope From Table D.1, C2 = 0,16 for austenitic stainless steel. Table D.1 Eq. D.1 𝑓𝑓u − 𝑓𝑓y 540 − 326 2 𝐸𝐸sh = = = 3429 N/mm 𝐶𝐶2εu − εy 0,16 × 0,40 − 0,0016 cross-section in-plane bending resistance α Eq. D.9 pl y sh el csm el csm c,Rd csm,Rd 𝑊𝑊 𝑓𝑓 𝐸𝐸 𝑊𝑊 ε 𝑊𝑊 ε 𝑀𝑀 = 𝑀𝑀 = M0 [1 + pl ( y − 1) − (1 − pl)⁄( y ) ] α = 2,0 for RHS γ 𝐸𝐸 𝑊𝑊 ε 𝑊𝑊 ε Table D.2 Mc,Rd = Mcsm,Rd 30860 × 326 3429 25967 25967 2,0 = m × [1 + × × (9,9 − 1) − (1 − )⁄(9,9) ] 1,1 200000 30860 30860 Mc,Rd = 10,31 kN The bending resistance determined according to Section 5 is 6,45 kNm. Consideration of strain hardening to give an average enhanced yield strength due to section forming in Example 14 resulted in a resistance of 9,15 kNm. With the added consideration of strain hardening in service using the CSM for cross-section design, a bending resistance of 10,31 kNm is achieved. This corresponds to an overall increase in resistance of 60%. 260 Design Example 15 Sheet 2 of 2 εcsm 0,25 0,1 × 0,40 = 3,6 = 9,9 ≤ min (15, = 25) εy 0,36 0,0016 εcsm ∴ = 9,9 εy Strain hardening slope From Table D.1, C2 = 0,16 for austenitic stainless steel. Table D.1 Eq. D.1 𝑓𝑓u − 𝑓𝑓y 540 − 326 2 𝐸𝐸sh = = = 3429 N/mm 𝐶𝐶2εu − εy 0,16 × 0,40 − 0,0016 cross-section in-plane bending resistance α Eq. D.9 pl y sh el csm el csm c,Rd csm,Rd 𝑊𝑊 𝑓𝑓 𝐸𝐸 𝑊𝑊 ε 𝑊𝑊 ε 𝑀𝑀 = 𝑀𝑀 = M0 [1 + pl ( y − 1) − (1 − pl)⁄( y ) ] α = 2,0 for RHS γ 𝐸𝐸 𝑊𝑊 ε 𝑊𝑊 ε Table D.2 Mc,Rd = Mcsm,Rd 30860 × 326 3429 25967 25967 2,0 = m × [1 + × × (9,9 − 1) − (1 − )⁄(9,9) ] 1,1 200000 30860 30860 Mc,Rd = 10,31 kN The bending resistance determined according to Section 5 is 6,45 kNm. Consideration of strain hardening to give an average enhanced yield strength due to section forming in Example 14 resulted in a resistance of 9,15 kNm. With the added consideration of strain hardening in service using the CSM for cross-section design, a bending resistance of 10,31 kNm is achieved. This corresponds to an overall increase in resistance of 60%. 261 DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4TH EDITION Stainless steel is used for a wide range of structural applications in aggressive environments where reliable performance over long periods with little maintenance is required. In addition, stainless steel has an attractive appearance, is strong yet still light, highly ductile and versatile in terms of manufacturing. This Design Manual gives design rules for austenitic, duplex and ferritic stainless steels. The rules are aligned to the 2015 amendment of the Eurocode for structural stainless steel, EN 1993-1-4. They cover the design of cross-sections, members, connections and design at elevated temperatures as well as new design methods which exploit the beneficial strain hardening characteristics of stainless steel. Guidance on grade selection, durability and fabrication is also provided. Fifteen design examples are included which illustrate the application of the design rules. SCI Ref: P413 SCI Silwood Park, Ascot, Berkshire. SL5 7QN UK T: +44 (0)1344 636525 F: +44 (0)1344 636570 E: [email protected] www.steel-sci.com