Design Examples using midas Gen to Eurocode 3. Integrated Design System for Building and General Structures

Transcription

1 Design Examples using midas Gen to Eurocode 3 Integrated Design System for Building and General Structures

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3 Introduction This design example book provides a comprehensive guide for steel design as per Eurocode3--:005. Specifically, this guide will review the design algorithms implemented in midas Gen, and go through detailed verification examples and design tutorials. This book is helpful in understanding the Eurocode design concept and verifying design results using midas Gen. CHAPTER Why midas Gen This chapter describes the main features and advantages of midas Gen and showcases prominent project applications. CHAPTER Steel Design Algorithms This chapter discusses the general design concept of EN993-- and how it has been implemented in midas Gen. This enables the user to understand the equations, formulas, program limitations and development scope of the midas Gen design features. CHAPTER 3 Verification Examples This chapter provides comparative results between design reports generated from midas Gen and design examples from reference books. Numerous worked examples for EN993--:005 has been used to verify design results from midas Gen. 7 steel examples of beam and column members has been included. CHAPTER 4 Steel Design Tutorial This chapter enables the user to get acquainted with the steel design procedure in midas Gen as per EN993--: 005. It encompasses the overall design procedure, from generating load combinations to checking design results with updated sections.

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5 CHAPTER Why midas Gen Design Examples using midas Gen to Eurocode3

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7 CHAPTER Why midas Gen

8 CHAPTER. Why midas Gen

9 CHAPTER. Why midas Gen 3

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11 CHAPTER Steel Design Algorithm Design Examples using midas Gen to Eurocode3

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13 CHAPTER Steel Design Algorithm as per EN993--:005. Overview () General Material Properties Section table for the application of Ultimate Limit State Check () Ultimate Limit State Check Resistance of cross-sections Buckling resistance of members (3) Serviceability Limit State Check Vertical deflections Horizontal deflections Dynamic effects. General () Material Properties The nominal values of the yield strength (fy) and the ultimate strength (fu) for structural steel Steel Grade f y (N/mm ) t 40mm f u (N/mm ) f y (N/mm ) t > 40mm f u (N/mm ) S S S S Modulus of Elasticity = 0,000 N/mm Poisson s Ratio, ν, = 0.3 Thermal Coefficient = x0-6 / o C Weight Density = kn/m 3

14 CHAPTER. Steel Design Algorithm () Section table for the application of Ultimate Limit State Check Limit States Cross section Yielding FB () SB Strong axis Weak axis LTB I section Doubly Symmetric N/A Singly Symmetric N/A N/A Box () N/A Angle N/A N/A N/A Channel N/A N/A Tee N/A N/A N/A Double Angle N/A N/A N/A Double Channel N/A N/A Pipe N/A N/A N/A Solid Rectangle N/A N/A N/A Solid Round N/A N/A N/A U-Rib N/A N/A N/A N/A N/A Note FB: Flexural Buckling, SB: Shear Buckling, LTB: Lateral-Torsional Buckling () Torsional Buckling and Torsional-Flexural Buckling are not evaluated. () The thickness of two webs should be identical, and the member type should be column for the weak axis shear buckling check.

15 CHAPTER. Steel Design Algorithm.3 Ultimate Limit State Check () Resistance of cross-sections Tension N pl,rd = Af y γ M 0 Design tension resistance - The design ultimate resistance of the net cross-section at holes for fasteners is not considered in midas Gen. Compression - Design compression resistance N c,rd = Af y γ M 0 For class, and 3 cross sections N c,rd = A eff f y γ M 0 For class 4 cross sections - In the case of unsymmetrical Class 4 sections, the additional moment due to the eccentricity of the centroidal axis of the effective section is considered in midas Gen. Bending moment - Design bending resistance M c,rd = M pl,rd = W pl f y γ M 0 For class or cross sections M c,rd = M el,rd = W el,min f y γ M 0 For class 3 cross sections M c,rd = W eff,min f y γ M 0 For class 4 cross sections Shear - Design shear resistance in the absence of torsion V pl,rd = A v f y 3 γ M 0 - The shear area Av is calculated based on the clause 6..6 (3) as per EN Rolled I and H sections, load parallel to web: A bt f + t w + r t f - but not less than Design elastic shear resistance is not applied. Shear Buckling - The shear buckling resistance for webs without intermediate stiffeners is calculated, according to section 5 of EN , if 3

16 CHAPTER. Steel Design Algorithm h w t w > 7 ε η ε = 35 f y N mm - For steel grades up to and including S460: η =.0 - For higher steel grades: η =.00 - Design resistance V b,rd = V bw,rd + V bf,rd η f yw h w t 3 γ M V bw,rd = χ w f yw V bf,rd = b f t f fyf cγ M h w t 3γ M M Ed M f,rd - Stiffener design to resist shear buckling is not provided in midas Gen. - Stiffener type for end supports is assumed as a non-rigid end post. - It is assumed that the length of an unstiffened plate, a is the same as the unbraced length. Torsion - The torsional resistance is not checked. Bending and Shear - The effect of shear force on the moment resistance is considered. - Where the shear force is less than half the plastic shear resistance, its effect on the moment resistance is neglected. - Where V ED 0.5V pl,rd I-cross-sections with equal flanges and bending about the major axis M y,v,rd = W pl,y ρa w 4tw γ M 0 f y but, M y,v,rd M y,c,rd ρ = V ED V pl,rd Torsion is not considered when calculating ρ For the other cases M V,Rd = ρ M c,rd 4

17 CHAPTER. Steel Design Algorithm Bending and Axial Force - The effect of axial force on the moment resistance is considered. - Class and cross sections For doubly symmetrical I- and H-sections, allowance is not made for the effect of the axial force on the plastic resistance moment about the y-y axis when both the following criteria are satisfied: N Ed 0.5 N pl,rd N Ed 0.5 h w t w f y γ M 0 For doubly symmetrical I- and H-sections, allowance is not made for the effect of the axial force on the plastic resistance moment about the z-z axis when: N Ed h w t w f y γ M 0 The following equations are used for standard rolled I or H sections and for welded I or H sections with equal flanges: M N,y,Rd = M pl,y,rd n 0.5a but M N,y,Rd M pl,y,rd for n a : M N,z,Rd = M pl,z,rd for n > a : M N,z,Rd = M pl,z,rd n a a Where n = N Ed N pl,rd a = A bt f A but a 0.5 Bending and Axial Force M y,ed M N,y,Rd α + M z,ed M N,z,Rd β for Class & sections I and H section: α=; β=5n but β N Ed N Rd + M y,ed M y,rd + M z,ed M z,rd for Class,,3 & 4 sections Bending, Shear and Axial Force - Where the shear force exceeds 50% of the plastic shear resistance, its effect on the moment of resistance is reflected in the formula above. - M pl,y,rd and M pl,z,rd are replaced by M y,v,rd and M z,v,rd respectively in the following equations to consider shear effect in the above criterion a). M N,y,Rd = M pl,y,rd n / 0.5 a w M N,z,Rd = M pl,z,rd n / 0.5 a r 5

18 CHAPTER. Steel Design Algorithm - M y, Rd and M z, Rd are replaced by M y,v,rd and M z,v,rd respectively in the above criterion b) to consider shear effect. () Buckling resistance of members Uniform members in compression For slenderness λ 0. or for N Ed N cr 0.04 the buckling effects are ignored. λ = Af y N cr for Class, and 3 cross-sections λ = A eff f y N cr for Class 4 cross-section N cr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties. N cr = π EI L e Flexural buckling is checked for the L, C, I, T, Box, Pipe, Double L, and Double C section. Torsional and torsional-flexural buckling is not checked. Design buckling resistance N b,rd = χa f y γ M for Class, and 3 cross-sections N b,rd = χ A eff f y γ M for Class 4 cross-sections χ = Ф+ Ф λ but χ.0 Ф = α λ 0. + λ Buckling Curve a 0 a b c d Imperfection factor α Uniform members in bending - For the uniform and doubly symmetric I cross-sections only, the lateral torsional buckling check is provided. - It is assumed that the section is loaded through its shear center, and the boundary conditions at each end are both restrained against lateral movement and restrained against rotation about the longitudinal axis. - For slenderness λlt λlt,0 or for ignored. M ED M cr λ LT,0 the lateral torsional buckling effects are 6

19 CHAPTER. Steel Design Algorithm λ LT = W y f y M cr λ LT,0 = 0.4 M cr is the elastic critical moment for lateral-torsional buckling. The value of C l depends on the moment distribution along the member which is calculated based on the table in the following page π M cr = C EI z l L cr,lt I w + L cr,lt GI t I z π EI z G = E (l+v) I w = I z (h t f ) 4 : Warping Constant - If the member type is column, C is calculated based on the table below. EN 993--: 99 Annex. - If the member type is beam, C is calculated based on the table below. Conditions Bending moment diagram k C Case Class Case 3.0 Same as 0.5 Case Case 4.0 Same as 0.5 Case Case5.0 Same as 0.5 Case 7

20 CHAPTER. Steel Design Algorithm Design buckling Resistance M b,rd = χ LT W y W y = W pl,y for Class or cross-section f y γ M W y = W el,y for Class 3 cross-section W y = W eff,y for Class or cross-section χ LT = Ф LT + Ф LT +λ LT but χ LT.0 Ф LT = α LT λ LT 0.) + λ LT Buckling Curve a b c d Imperfection factor α LT The method in the Clause and of EC3 are not considered. Uniform members in bending and axial compression - For members which are subjected to combined bending and axial compression, the resistance to lateral and lateral-torsional buckling is verified by the following criteria. N Ed χ y N Rk γ M + k yy M y,ed + M y,ed χ LT M y,rk γ M + k yz M z,ed + M z,ed χ LT M z,rk γ M N Ed χ zn Rk γ M + k zy M y,ed + M y,ed χ LT M y,rk γ M + k zz M z,ed + M z,ed χ LT M z,rk γ M K yy, k yz, k zy, k zz are the interaction factors. These values are obtained from Annex A in EN 993--: 005. C my, C mz, C mlt in Annex A can be either user defined or auto-calculated. Vales for N Rk = f y A i, M i,rk =f y W i and M i,ed Class 3 4 A j A A A A eff W y W pl,y W pl,y W el,y W eff,y W z W pl,z W pl,z W el,z W eff,z ΔM y, Ed e Ny,N Ed ΔM z, Ed e Nz,N Ed - When the design axial force, NEd is larger than Ncr,z or Ncr,TF, the criteria above are not applied. - General method of the clause is not considered. 8

21 CHAPTER. Steel Design Algorithm.4 Serviceability Limit State Check () Vertical Deflection Vertical deflection can be checked for beam member. Remaining total deflection (wmax) caused by the permanent and variable actions is automatically checked based on the serviceability load combinations. The default limit value is set to L/50 The deflection due to the variable actions can be checked manually by adding load combination consisting of variable actions and changing the limit value () Horizontal Deflection Horizontal deflection can be checked for column members. Horizontal displacement over a story height Hi is automatically checked based on the serviceability load combinations. The default limit value is set to Hi/300. Overall horizontal displacement over the building height H should be checked separately. (3) Dynamic effects The vibration of structures is not checked. 9

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23 CHAPTER 3 Verification Examples Design Examples using midas Gen to Eurocode3

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25 CHAPTER 3 Steel Design Verification Examples 3. Cross-section resistance under combined bending and shear A short-span (.4m), simply supported, laterally restrained beam is to be designed to carry a central point load of 050 KN, as shown in the right figure. The arrangement of the figure results in a maximum design shear force VED of 55 KN and a maximum design bending moment MED of knm. In this example a 406 x 78 x 74 UB in grade S75 steel is assessed for its suitability for this application. 3.. Material Properties 3.. Section Properties Material S75 f y = 75N/mm E s = 0 GPa Section Name 406 x 78 x 74 UB Depth (H) 4.8mm Width (B) 79.5mm Flange Thickness (T f ) 6.0 mm Web Thickness (T w ) 9.5 mm Gross sectional area (A) mm Shear area (A sz ) mm 3..3 Analysis Model Loading condition

26 CHAPTER 3. Steel Design Verification Examples SF Beam Diagram BM 3..4 Comparison of Design Results midas Gen Example book Error (%) 689.5kN 689.kN 0.0% Bending resistance 4.50kNm 4.0kNm 0.% Combined resistance kNm 386.8kNm 0.06% Shear resistance 3..5 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT ( 35/fy ) = b/t = BTR = sigma = 0.78 kn/mm^. -. sigma = 0.78 kn/mm^. -. BTR < 9*e ( Class : Plastic ). ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = KPa. -. sigma = KPa. -. HTR < 7*e ( Class : Plastic ).. Check Bending Moment Resistance ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = 4.50 kn-m. ( ). Check ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = 0.89 < > O.K. Mc_Rdy Shear resistance of cross-section ( ). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. eta =. (Fy < 460 MPa.) Example book. Cross-section classification (clause 5.5.) ε = 35/fy = 35/75 = 0.9 Outstand flange in compression (Table 5., sheet ): C =(b - tw r)/ = 74.8 mm c/tf = 74.8/6.0 =4.68 Limit for Class flange=9ε= >4.68 flange is Class Web internal part in bending (Table 5., sheet ): C = h - tf r = mm c/tw = 360.4/9.5 = Limit for Class web = 7ε = > web is Class Therefore, the overall cross section classification is Class.. Bending resistance of cross section (clause 6..5) Mc,y, Rd = W pl,y f y γm 0 for Class or cross sections The design bending resistance of the cross-section Mc,y, Rd = = 4 0 N mm = 4 KNm 4 KNm > KNm cross-section resistance in bending is acceptable. 3. Shear resistance of cross-section of cross-section (clause 6..6) Vpl, Rd = A v (f y /3) γm 0

27 -. r = mm. -. Avy = Area - hw*tw = mm^. -. Avz = eta*hw*tw = mm^. -. Avz = Area - *B*tf + (tw + *r)*tf = mm^. -. Avz = MAX[ Avz, Avz ] = mm^. ( ). Calculate plastic shear resistance in local-z direction (Vpl_Rdz). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = kn. ( ). Shear Buckling Check. [ Eurocode3: ] -. HTR < 7*e/Eta ---> No need to check! ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = kn. V_Edz = = 0.76 < > O.K. Vpl_Rdz Check Interaction of Combined Resistance ( ). Calculate Major reduced design resistance of bending and shear. [ Eurocode3: (6.30) ] -. In case of V_Edz / Vpl_Rdz > 0.5 (equal flanges) -. Rho = { *(V_Edz/Vpl_Rdz) - }^ = My.V_Rd= [ Wply - {Rho*Aw^/(4*tw)} ]*fy / Gamma_M0 = kn-m. -. My_Rd = MIN [ My.V_Rdy, Mc_Rdy ] = kn-m. ( ). Calculate Minor reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edy / Vpl_Rdy < Mz_Rd = Mc_Rdz = 73.4 kn-m. ( ). Check general interaction ratio. [ Eurocode3: (6.) ] - Class or Class N_Ed M_Edy M_Edz -. Rmax = N_Rd My_Rd Mz_Rd = 0.95 < > O.K. ( ). Check interaction ratio of bending and axial force member. [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. n = N_Ed / Npl_Rd = a = MIN[ (Area-b*tf)/Area, 0.5 ] = Alpha = Beta = MAX[ 5*n,.0 ] =.000 For a rolled I section, loaded parallel to the web, the shear area A v is given b y A v = A bt f + (t w + r) t f ( bur not less than ηh w t w ) η=. (from EN , though the UK National Annex may specify an alternative value). h w = (h t f ) =4.8 ( 6.0) = mm A v = 9450 ( ) +( ) 6.0 = 484 mm (but not less than = 434 mm ) V p,rd = 434 (75/3).00 = N = 689. KN Shear buckling need not be considered, provided h w t w 7 ε η for unstiffened webs 7 ε 0.9 = 7 η. = 55.5 Actual h w /t w = 380.9/9.5 = no shear buckling check required 689. > 55 KN shear resistance is acceptable 4. Resistance of cross-section to combined bending and shear (clause 6..8) The applied shear force is greater than half the plastic shear resistance of the cross-section, therefore a reduced moment resistance M y, V,Rd must be calculated. For an I section (with equal flanges) and bending about the major axis, clause 6..8(5) and equation (6.30) may be utilized. M y, V, Rd = (W pl,y ρa w /4t w )f y γ M 0 ρ = V ED = 55 = 0.7 V pl,rd 689. A w = h w t w = = mm but M y,v, Rd M y,c, Rd M y, V, Rd = /4 9.5 ) 75.0 = KN KNm > KNm cross-section resistance to combined bending and shear is acceptable Conclusion A UB in grade S75 steel is suitable for the arrangement and loading shown by Fig

28 -. N_Ed < 0.5*Npl_Rd = kn. -. N_Ed < 0.5*hw*tw*fy/Gamma_M0 = kn. Therefore, No allowance for the effect of axial force. -. Mny_Rd = Mply_Rd = kn-m. -. Rmaxy = M_Edy / Mny_Rd = 0.95 < > O.K. -. N_Ed < hw*tw*fy/gamma_m0 = kn. Therefore, No allowance for the effect of axial force. -. Mnz_Rd = Mplz_Rd = 73.4 kn-m. -. Rmaxz = M_Edz / Mnz_Rd = < > O.K. -. Rmax = MAX[ Rmaxy, Rmaxz ] = 0.95 < > O.K. [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, (Example 6.5) 4

29 3. Cross-section resistance under combined bending and compression A member is to be designed to carry a combined major axis moment and an axial force. In this example, a cross-sectional check is performed to determine the maximum bending moment that can be carried by a UB in grade S35 steel, in the presence of an axial force of 400 KN. 3.. Material Properties 3.. Section Properties Material S75 f y = 75N/mm E s = 0 GPa Section Name 406 x 78 x 74 UB Depth (H) 4.8mm Width (B) 79.5mm Flange Thickness (T f ) 6.0 mm Web Thickness (T w ) 9.5 mm Gross sectional area (A) mm Shear area (A sz ) mm 3..3 Comparison of Design Results midas Gen Example book Error (%) Shear resistance 689.5kN 689.kN 0.0% Bending resistance 4.50kNm 4.0kNm 0.% Combined resistance kNm 386.8kNm 0.06% 3..4 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = KPa. -. sigma = KPa. -. BTR < 9*e (Class : Plastic). ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) =.00 Example book. Cross-section classification under pure compression (clause 5.5.) ε = 35/f y = 35/35 =.00 Outstand flanges (Table 5., sheet ): C =(b - t w r)/ = 80.5 mm c/t f = 80.5/9.6 =4. Limit for Class flange=9ε= >4. flange is Class Web internal part in bending (Table 5., sheet ): C = h - t f r = mm c/t w = 407.6/.4 =

30 -. d/t = HTR = sigma = KPa. -. sigma = KPa. -. HTR < 38*e ( Class : Compact ).. Check Axial and Bending Resistance ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 3.3 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = < > O.K. Nc_Rd ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = 0.00 m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = kn-m. -. N_Ed > 0.5*Npl_Rd = kn. -. N_Ed > 0.5*hw*tw*fy/Gamma_M0 = kn. Therefore, Allowance for the effect of axial force. 3. Check Interaction of Combined Resistance ( ). Calculate Major reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edz / Vpl_Rdz < My_Rd = Mc_Rdy = kn-m. ( ). Calculate Minor reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edy / Vpl_Rdy < Mz_Rd = Mc_Rdz = kn-m. ( ). Check interaction ratio of bending and axial force member. [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. n = N_Ed / Npl_Rd = a = MIN[ (Area-b*tf)/Area, 0.5 ] = Alpha = Beta = MAX[ 5*n,.0 ] = N_Ed > 0.5*Npl_Rd = kn. -. N_Ed > 0.5*hw*tw*fy/Gamma_M0 = kn. Therefore, Allowance for the effect of axial force. -. Mny_Rd = MIN[ Mply_Rd*(-n)/(-0.5*a), Mply_Rd ] = kn-m. Limit for Class web = 38ε = > web is Class Under pure compression, the overall cross-section classification is therefore Class.. Bending and axial force (clause 6..9.) No reduction to the plastic resistance moment due to the effect of axial force is required when both of the following criteria are satisfied. N Ed 0.5N pi, Rd And N Ed = 0.5h w t w f y γ M 0 N Ed = 400 KN N pl, Rd = Af y = = KN γ M N pl, Rd = KN KN < 400 KN equation (6.33) is not satisfied 0.5h w t w f y = = KN γ M KN < 400 KN equation (6.34) is not satisfied Therefore, allowance for the effect of axial force on the plastic moment resistance of the cross-section must made. 3. Reduced plastic moment resistance (clause 6..9.I(5)) M N,y, Rd = M pl, y, Rd n 0.5a but M N,y, Rd M pl, y, Rd Where n = N Ed / M pl, y, Rd = 400/937.5 = 0.48 a = (A bt f )/A = [500 ( )]/500 = 0.40 M pl, y, Rd = W pl,y f y = = 54.5 KNm γ M 0.0 M N,y, Rd = ( ) = 34. KNm Conclusion In order to satisfy the cross-sectional checks of clause 6..9, the maximum bending moment that can be carried by a UB in grade S35 steel, in the presence of an axial force 400 KN is 34. KNm. [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, (Example 6.6) 6

31 3.3 Buckling resistance of a compression member A circular hollow section (CHS) member is to be used as an internal column in a multi-storey building. The column has pinned boundary conditions at each end, and the inter-storey height is 4m, as shown in the right figure. The critical combination of actions results in a design axial force of 630 KN. Assess the suitability of a hot-rolled 44.5 x 0 CHS in grade S75 steel for this application Material Properties 3.3. Section Properties Material S75 f y = 75N/mm E s = 0 GPa Section Name 44.5 X 0 CHS Thickness (T) 0.0 mm Gross sectional area (A) 7370 mm Modulus of Elasticity (W el,y ) mm 3 Modulus of Elasticity (W pl,y) mm 3 Moment of Inertia (I) mm Analysis Model Loading condition Comparison of Design Results midas Gen Example book Error (%) Shear resistance kn 06.8 kn 0.00% Bending resistance knm knm 0.06% 7

32 3.3.5 Detailed comparison midas Gen. Class of Cross Section ( ). Determine classification of tublar section(hollow pipe). [ Eurocode3:05 Table 5. (Sheet 3 of 3) ] -. e = SQRT( 35/fy ) = d/t = DTR = DTR < 50*e^ ( Class : Plastic ).. Check Axial Resistance ( ). Check slenderness ratio of axial compression member (Kl/i) [ Eurocode3: ] -. Kl/i = 48. < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = < > O.K. Nc_Rd ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^= kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M = kn. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) = Ncrz = Pi^*Es*Rzz / KLz^ = kn. -. Lambda_bz > 0. and N_Ed/Ncrz > > Need to check. -. Alphaz = Phiz = 0.5 * [ + Alphaz*(Lambda_bz-0.) + Lambda_bz^ ] = Xiz = MIN [ / [Phiz + SQRT(Phiz^ - Lambda_bz^)],.0 ] = Nb_Rdz = Xiz*Beta_A*Area*fy / Gamma_M = kn. Example book. Cross-section classification (clause 5.5.) ε = 35/f y = 35/75 = 0.9 Tubular sections (Table 5., sheet 3): d/t = 44.5/0.0 =4.5 Limit for Class section =50 ε = > 4.5 section is Class. Cross Section Compression resistance (clause 6..4) N c, Rd = Af y γ M 0 for Class, or 3 cross-sections N c, Rd = = N = 06.8 KN > 630 KN cross-section resistance is acceptable 3. Member Buckling resistance in compression (clause 6.3.) N b, Rd = χaf y γ M χ = but.0 ϕ+ ϕ λ where Ф= 0.5* + α(λ - 0.) + λ ] and λ = Af y N cr for Class, or 3 cross-sections for Class, or 3 cross-sections Elastic critical force and non-dimensional slenderness for flexural buckling N cr = π EI L cr = π = 657 KN λ = = 0.56 Selection of buckling curve and imperfection factor α For a hot-rolled CHS, use buckling curve a (Table 6.5 (Table 6. of EN 993--)). For curve buckling curve a, α = 0. (Table 6.4 (Table 6. of EN 993--)). Buckling curves ϕ = 0.5[ + 0. ( ) ] = 0.69 χ = N b, Rd = = N = KN > 630 KN buckling resistance is acceptable 8

33 ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). -. Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = kn. N_Ed = = < > O.K. Nb_Rd Conclusion The chosen cross-section, CHS, in grade S75 steel is acceptable. [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, (Example 6.7) 9

34 3.4 I-section beam design under shear force and bending moment A simply supported primary beam is required to span 0.8m and to support two secondary beams as shown in Fig.6.4. The secondary beams are connected through pin plates to the web of the primary beam, and full lateral restraint may be assumed at these points. Select a suitable member for the primary beam assuming grade S75 steel Material Properties 3.4. Section Properties Material S75 f y = 75N/mm E s = 0 GPa Section Name 76 X 67 X 73 UB Depth (H) 76. mm Width (B) 66.7 mm Flange Thickness (T f ).6 mm Web Thickness (T w ) 4.3 mm Gross sectional area (A) 000mm Shear area (A sz ) 500. mm Analysis Model Loading condition Beam Diagram SF BM 0

35 3.4.4 Comparison of Design Results midas Gen Example book Error (%) Shear resistance kN kN 0.00% Bending resistance kNm kNm 0.06% Combined resistance 5.4kN kN.8% Detailed comparison midas Gen. Cross-section classification ( ). compression outstand flanges (Flange) -. e = SQRT( 35/fy ) = b/t = BTR = sigma = KPa. -. sigma = KPa. -. BTR < 9*e ( Class : Plastic ). ( ). bending Internal Parts (Web) -. e = SQRT( 35/fy ) = d/t = HTR = sigma = KPa. -. sigma = KPa. -. HTR < 7*e ( Class : Plastic ).. Shear resistance of cross-section ( ). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. eta =. (Fy < 460 MPa.) -. r = m. -. Avy = Area - hw*tw = 0.07 m^. -. Avz = eta*hw*tw = 0.03 m^. -. Avz = Area - *B*tf + (tw + *r)*tf = 0.05 m^. -. Avz = MAX[ Avz, Avz ] = 0.03 m^. ( ). Plastic shear resistance (Vpl_Rdz) [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = kn. -. Avz =.3380e Fy =.75000e Gamma_M0 =.00 ( ). Shear Buckling Check [ Eurocode3: ] -. HTR < 7*e/Eta ---> No need to check! -. e = SQRT( 35/fy ) = d/t = HTR = Example book. Cross-section classification ε = 35/f y = 35/75 = 0.9 Outstand flanges (Table 5., sheet ): C =(b - t w r)/ = 09.7 mm c/t f = 09.7/.6 = 5.08 Limit for Class flange=9ε= > 5.08 flange is Class Web internal part in bending (Table 5., sheet ): C = h - t f r = mm c/t w = 686.0/4.3 = 48.0 Limit for Class web = 7ε = > 48.0 web is Class. Shear resistance of cross-section V p,rd = A v (f y /3) γ M 0 For a rolled I section, loaded parallel to the web, the shear area A v, is given by A v = A bt f + (t w + r) t f (but not less than ηh w t w ) η=. (from Eurocode 3 part.5, though the UK National Annex may specify an alternative value). h w = (h t f ) = 76. (.6) = 79.0 mm A v = 000 ( ) + ( ).6 = 983 mm (but not less than = 338mm ) V p,rd = 338 (75/3).00 = N = 959 KN Shear buckling need not be considered, provided h w t w 7 ε η for unstiffened webs 7 ε η = = 55.5 Actual h w /t w = 79.0/4.3 = no shear buckling check required

36 ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = kn. V_Edz = = 0.5 < > O.K. Vpl_Rdz Bending resistance of cross-section ( ). Plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = kn-m. ( ). Ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = 0.73 < > O.K. Mc_Rdy ( ). Plastic resistance moment about minor axis. [ Eurocode3:05 6., 6..5 ] -. Wplz = m^3. -. Mc_Rdz = Wplz * fy / Gamma_M0 =.9 kn-m. ( ). Ratio of moment resistance (M_Edz/Mc_Rdz). M_Edz = = < > O.K. Mc_Rdz.9 4. Combined bending and shear resistance ( ). Major reduced design resistance of bending and shear [ Eurocode3: (6.30) ] -. In case of V_Edz / Vpl_Rdz < My_Rd = Mc_Rdy = kn-m. ( ). Minor reduced design resistance of bending and shear [ Eurocode3: (6.30) ] -. In case of V_Edy / Vpl_Rdy < Mz_Rd = Mc_Rdz =.9 kn-m. ( ). General interaction ratio [ Eurocode3: (6.) ] - Class or Class N_Ed M_Edy M_Edz -. Rmax = N_Rd My_Rd Mz_Rd = 0.73 < > O.K. 959 > 493. KN shear resistance is acceptable 3. Bending resistance of cross-section M c, y, Rd = W pl,y f y for Class or cross-sections γ M 0 EN recommends a numerical value of γ M0 =.00 (through for buildings to be constructed in the UK, reference should be made to the National Annex). The design bending resistance of the cross-section M c, y, Rd = = 704 knm = Nmm 704 KNm > 36 KNm cross-section resistance in bending is acceptable 4. Combined bending and shear resistance Clause 6..8 states that provided the shear force V Ed is less than half the plastic shear resistance V pl,rd its effect on the moment resistance may be neglected except where shear buckling reduces the section resistance. In this case, there is no reduction for shear buckling (see above), and maximum shear force (V Ed =493.kN) is less than half the plastic shear resistance (V pl,rd =959kN). Therefore, resistance under combined bending and shear is acceptable. ( ). Interaction ratio of bending and axial force member [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. Alpha = Beta = MAX[ 5*n,.0 ] =.000

37 -. n = N_Ed / Npl_Rd = a = MIN[ (Area-b*tf)/Area, 0.5 ] = Mny_Rd = MIN[ Mply_Rd*(-n)/(-0.5*a), Mply_Rd ] = kn-m. -. Rmaxy = M_Edy / Mny_Rd = 0.73 <.0 ---> O.K. -. In case of n < a -. Mnz_Rd = Mplz_Rd =.9 kn-m. -. Rmaxz = M_Edz / Mnz_Rd = 0.0 <.0 ---> O.K. -. Rmax = max[rmaxy, Rmaxz] = 0.73 <.0 --> O.K. -. Rmax = MAX[Rmax, Rmax] = 0.73 <.0 -->O.K [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, (Example 6.8) 3

38 3.5 Member resistance under combined major axis bending and axial compression A rectangular hollow section (RHS) member is to be used as a primary floor beam of 7. m span in a multi-storey building. Two design point loads of 58 KN are applied to the primary beam (at locations B and C) from secondary beams, as shown in the right figure. The secondary beams are connected through fin plates to the webs of the primary beam, and full lateral and torsional restraint may be assumed at these points. The primary beam is also subjected to a design axial force of 90 KN. Assess the suitability of a hot-rolled 00 X 00 X 6 RHS in grade S355 steel for this application. In this example the interaction factors kij (for member checks under combined bending and axial compression) will be determined using alternative method (Annex A) 3.5. Material Properties Material S355 f y = 355 N/mm E s = 0 GPa 3.5. Section Properties Section Name 00 X 00 X 6 RHS Depth (H) 00.0 mm Width (B) 00.0 mm Flange Thickness (T f ) 6.0 mm Web Thickness (T w ) 6.0 mm Gross sectional area (A) mm Shear area (A sz ) mm Analysis Model Loading condition Beam Diagram SF BM 4

39 3.5.4 Comparison of Design Results midas Gen Example book Error (%) Axial resistance kN kn.78% Shear resistance 54.60kN 34.00kN.78% Bending resistance 79.8kNm knm.78% Buckling resistance 47.9kN 09.00kN 3.07% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR < 33*e ( Class : Plastic ). ( ). Determine classification of bending and compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = 0.64 kn/mm^. -. sigma = kn/mm^. -. Psi = [*(Nsd/A)*(/fy)]- = Alpha = 0.54 > HTR < 396*e/(3*Alpha-) ( Class : Plastic ).. Check Axial Resistance. ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 64.3 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn. Example book.cross-section classification (clause 5.5.) ε = 35/f y = 35/355 = 0.8 For a RHS the compression width c may be taken as h (or b) 3t. Flange-internal part in compression (Table 5., sheet ): C = b - 3t = 00 (3 6.0) = 5.0 mm c/t = 5.0/6.0 = 3.5 Limit for Class flange=33ε= > 3.5 flange is Class Web internal part in compression (Table 5., sheet ): C = h - 3t = 00.0 (3 X 6.0) = 5.0 mm c/t = 5.0/6.0 = 9.50 Limit for Class web = 33ε = > 9.50 web is Class The overall cross-section classification is therefore Class (under pure compression)..compression resistance of cross-section (clause 6..4) The design compression resistance of the cross-section N c, Rd N c, Rd = Af y = γ M for class, and 3 cross-sections = N = KN KN > 90 KN acceptable ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = < > O.K. Nc_Rd

40 3. Check Bending Moment Resistance About Major Axis ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = 79.8 kn-m. ( ). Check ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = < > O.K. Mc_Rdy Shear resistance of cross-section ( ). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. Avy = Area * B/(B+h) = m^. -. Avz = Area * h/(b+h) = m^. ( ). Calculate plastic shear resistance in local-z direction (Vpl_Rdz). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = kn ( ). Shear Buckling Check. [ Eurocode3: ] -. HTR < 7*e/Eta ---> No need to check! ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = kn. V_Edz = = < > O.K. Vpl_Rdz Bending resistance of cross-section (clause 6..5) Maximum bending moment M y, Ed =.4 58 = 39. KN The design major axis bending resistance of the crosssection. M c, y, Rd = W pl,y f y = γ M for Class or cross-sections = Nmm = 74.3 KNm 74.3 KNm > 39. KNm acceptable 4.Shear resistance of cross-section (clause 6..6 ) Maximum shear force V ED = 58.0 KN The design plastic shear resistance of the cross-section V p,rd = A v (f y /3) γ M 0 Or a rolled RHS of uniform thickness, loaded parallel to the depth, the shear area A v is given by A v = Ah/(b + h) = /( ) = mm Shear buckling need to not be considered, provided h w 7 ε for unstiffened webs t w η η=. (from EN , though the UK National Annex may specify an alternative value). h w = (h t) = 00 ( 6.0) = 68 mm 7 ε 0.8 = 7 η. = 48.8 Actual h w /t w = 00/6.0 = no shear buckling check required 34 > 58.0 KN shear resistance is acceptable 5. CHECK INTERACTION OF COMBINED RESISTANCE ( ). Calculate Major reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edz / Vpl_Rdz < My_Rd = Mc_Rdy = 79.8 kn-m. ( ). Calculate Minor reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edy / Vpl_Rdy < Mz_Rd = Mc_Rdz = kn-m. ( ). Check general interaction ratio. [ Eurocode3: (6.) ] - Class or Class 5.Cross-section resistance under Bending, Shear and axial force (clause 6..0) Provided the shear force V ED is less than 50% of the design plastic shear resistance V p,rd and provided shear buckling is not a concern, than the cross-section need only satisfy the requirements for bending and axial force (clause 6..9). In this case V ED < 0.5 V p,rd, and shear buckling is not a concern (see above). Therefore, cross-section only need be checked for bending and axial force. No reduction to the major axis plastic resistance moment 6

41 N_Ed M_Edy M_Edz -. Rmax = N_Rd My_Rd Mz_Rd = < > O.K. ( ). Check interaction ratio of bending and axial force member. [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. n = N_Ed / Npl_Rd = Alpha = MIN[.66/(-.3*n^), 6.0 ] = Beta = MIN[.66/(-.3*n^), 6.0 ] = N_Ed < 0.5*Npl_Rd = kn. -. N_Ed < 0.5*hw*tw*fy/Gamma_M0 = 477. kn. Therefore, No allowance for the effect of axial force. -. Mny_Rd = Mply_Rd = kn-mm. -. Rmaxy = M_Edy / Mny_Rd = < > O.K. due to the effect of axial force is required when both of the following criteria are satisfied: N ED 0.5 N p,rd N ED 0.5h w t w f y γ M N p,rd = = KN KN > 90KN equation (6.33) is satisfied 0.5h w t w f y = γ M 0 γ M 0 = 954. KN 954. KN > 90KN equation (6.34) is satisfied Therefore, no allowance for the effect of axial force on the major axis plastic moment resistance of the crosssection need be made. -. N_Ed < hw*tw*fy/gamma_m0 = kn. Therefore, No allowance for the effect of axial force. -. Mnz_Rd = Mplz_Rd = kn-mm. -. Rmaxz = M_Edz / Mnz_Rd = < > O.K. -. Rmax = MAX[ Rmaxy, Rmaxz ] = < > O.K. ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^ = 5.48 kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M = 47.9 kn. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) = Ncrz = Pi^*Es*Rzz / KLz^ = kn. -. Lambda_bz < 0. or N_Ed/Ncrz < > No need to check. 6.Member buckling resistance in compression (clause 6.3.) N b, Rd = χ = χaf y γ M ϕ+ ϕ λ where Ф= 0.5* + α( λ - 0.) + λ ] and λ = Af y N cr for Class, or 3 cross-sections but.0 for Class, or 3 cross-sections Elastic critical force and non-dimensional slenderness for flexural buckling For buckling about the major (y-y) axis, L cr should be taken as the full length of the beam(ad), which is 7. m. For buckling about the minor (z-z)axis, L cr should be taken as the maximum length between points of lateral restraint, which is.4 m. Thus, N cr,y = π EI L cr = π = 470 X 0 3 λ = =.4 = 470 KN N cr,z = π EI L cr = π = 47 X 0 3 λ = = 0.84 = 47 KN Selection of buckling curve and imperfection factor α For a hot-rolled RHS, use buckling curve a (Table 6.5 (Table 6. of EN 993--)). For curve buckling curve a, α = 0. (Table 6.4 (Table 6. 7

42 of EN 993--)) Buckling curves : major (y-y)axis ϕ y = 0.5 [ + 0. (.4-0.) +.4 ] =.63 χ y = = N b,y, Rd = = =09 KN.0 09 KN > 90 KN major axis flexural buckling resistance is acceptable Buckling curve: minor (z-z) axis ϕ z = 0.5 [ + 0. ( ) ] = 0.9 χ y = = N b,y, Rd = = = 66 KN 66 KN > 90 KN minor axis flexural buckling resistance is acceptable ( ). Calculate equivalent uniform moment factors (Cmy,Cmz,CmLT). [ Eurocode3:05 Annex A. Table A., A. ] -. Cmy,0 = Cmz,0 = Cmy (Default or User Defined Value) = Cmz (Default or User Defined Value) = CmLT (Default or User Defined Value) =.000 ( ). Check interaction ratio of bending and axial compression member. [ Eurocode3: , (6.6, 6.6), Annex A ] -. kyy = kyz = kzy = kzz = Xiy = Xiz = XiLT = N_Rk = A*fy = kn. -. My_Rk = Wply*fy = 79.8 kn-m. -. Mz_Rk = Wplz*fy = kn-m. -. N_Ed*eNy = 0.0 (Not Slender) -. N_Ed*eNZ = 0.0 (Not Slender) 7.Member buckling resistance in combined bending and axial compression (clause 6.3.3) Non-dimensional slenderness From the flexural buckling check: λ y =.4 and λ z = 0.84 λ max =.4 From the lateral torsional buckling check: λ LT = 0.3 and λ 0 = 0.3 Equivalent uniform moment factors C mi C my, 0 = ψ y +0.36(ψ y ) N ED N cr,y (0..0 ) ( ) =.0 C mz, 0 = C mz need not be considered since M z,ed =0. ε y = M y,ed N ED A W el,y for class, and 3 cross-sections = 8300 = α LT = - I T.0 = = 0.89 I y C my = C mz,0 +(-C my,0 ).0 ) +( ) =.0 C mlt = C my εy αlt + εy αlt =.0 + ( αlt (N Ed /N cr,z ) (N Ed /N cr,t ) = (9047 ) ( ).0 (but.0) C mlt =.00 N_Ed -. Rmax_LT = Xiy*N_Rk/Gamma_M M_Edy + N_Ed*eNy Interaction factors k ij k yy = C my C mlt =.0.00 μ y N ED /N cr,y C yy / =.06 8

43 + kyy * XiLT*My_Rk/Gamma_M M_Edz + N_Ed*eNz + kyz * Mz_Rk/Gamma_M = < > O.K N_Ed -. Rmax_LT = Xiz*N_Rk/Gamma_M M_Edy + N_Ed*eNy + kzy * XiLT*My_Rk/Gamma_M M_Edz + N_Ed*eNz + kzz * Mz_Rk/Gamma_M = < > O.K. -. Rmax = MAX[ MAX(Rmax, Rmax), MAX(Rmax_LT, Rmax_LT) ] = < > O.K. k zy = C my C mlt = =0.69 μ y N ED /N cr,y C zy 0.6 W y W z / Check compliance with interaction formulae (equations (6.6) and (6.6)) N ED M y,ed M z,ed + k χ y N RK /γ yy + k M χ LT M y,rk / γ zy M M z,rk / γ M ( )/.0 ( )/.0 = = equations (6.6) is satisfied N ED M y,ed M z,ed k χ z N RK /γ zy + k M χ LT M y,rk / γ zz M M z,rk / γ M ( )/.0 ( )/.0 = = equations (6.6) is satisfied Therefore, a hot-rolled RHS in grade S355 steel is suitable for this application. [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, 8-89 (Example 6.9) 9

44 3.6 Member resistance under combined bi-axial bending and axial compression An H section member of length 4. m is to be designed as a ground floor column in a multi-storey building. The frame is moment resisting in-plane and pinned out-of-plane, with diagonal bracing provided in both directions. The column is subjected to major axis bending die to horizontal forces and minor axis bending due to eccentric loading from the floor beams. From the structural analysis, the design action effects of Fig.6.9 arise in the column. Assess the suitability of a hot-rolled 305 X 305 X 40H section in grade S75 steel for this application Material Properties Material S75 f y = 75N/mm E s = 0 GPa 3.6. Section Properties Section Name 305 X 305 X 40H Depth (H) 35.5 mm Width (B) 38.4 mm Flange Thickness (T f ) mm Web Thickness (T w ) 3.0 mm Gross sectional area (A) mm Shear area (A sz ) mm Analysis Model Beam Diagram Axial force(n Ed ) Bending moment (M y,ed ) Bending moment (M z,ed ) 0

45 3.6.4 Comparison of Design Results midas Gen Example book Error (%) Axial resistance kN kN 0.00% Shear resistance knm 68.00kNm 0.06% Bending resistance kN 75.00kN 6.69% Reduced plastic moment resistance kNm kNm 0.39% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = KPa. -. sigma = KPa. -. BTR < 9*e ( Class : Plastic ). ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = KPa. -. sigma = KPa. -. HTR < 33*e ( Class : Plastic ).. Check Axial Resistance. ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 5.6 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = 0.58 < > O.K. Nc_Rd Check Bending Moment Resistance About Major Axis ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] Example book Cross-section classification (clause 5.5.) ε = 35/f y = 35/75 = 0.9 Outstand flanges (Table 5., sheet ): C =(b - t w r)/ = 3.5 mm c/t f = 3.5/37.7 = 3.5 Limit for Class flange=9ε= > 3.5 flange is Class Web internal part in bending (Table 5., sheet ): C = h - t f r = 46.7 mm c/t w = 46.7/3.0 = 0.73 Limit for Class web = 33ε = > 0.73 web is Class The overall cross-section classification is therefore Class. Compression resistance of cross-section (clause 6..4) The design compression resistance of the cross-section N c, Rd = Af y γ M =.00 for class, and 3 cross-sections = N = 845 KN 845 KN > KN acceptable Bending resistance of cross-section (clause 6..5) major (y-y) axis Maximum bending moment M y, Ed = 40.0 KN The design major axis bending resistance of the crosssection.

46 -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = kn-m. ( ). Check ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = < > O.K. Mc_Rdy ( ). Calculate plastic resistance moment about minor axis. [ Eurocode3:05 6., 6..5 ] -. Wplz = m^3m^3. -. Mc_Rdz = Wplz * fy / Gamma_M0 = 536.5kN-m. ( ). Check ratio of moment resistance (M_Edz/Mc_Rdz). M_Edz = = 0.05 < > O.K. Mc_Rdz Shear resistance of cross-section ( ). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. eta =. (Fy < 460 MPa.) -. r = 0.05 m. -. Avy = Area - hw*tw = 0.04 m^. -. Avz = eta*hw*tw = m^. -. Avz = Area - *B*tf + (tw + *r)*tf = m^. -. Avz = MAX[ Avz, Avz ] = m^. ( ). Calculate plastic shear resistance in local-z direction (Vpl_Rdz). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = kN. ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = kn. V_Edz = = 0.46 < > O.K. Vpl_Rdz ( ). Calculate plastic shear resistance in local-y direction (Vpl_Rdy). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdy = [ Avy*fy/SQRT(3) ] / Gamma_M0 = kn. ( ). Check ratio of shear resistance (V_Edy/Vpl_Rdy). ( LCB =, POS = J ) -. Applied shear force : V_Edy = 6.9 kn. V_Edy = = < > O.K. Vpl_Rdy ( ). Shear Buckling Check. [ Eurocode3: ] -. HTR < 7*e/Eta ---> No need to check! M c, y, Rd = W pl,y f y = γ M for Class or cross-sections = Nmm = 68 KNm 68 KNm > 40.0 KNm acceptable minor (z-z) axis Maximum bending moment M y,ed = 0.0 KN The design minor axis bending resistance of the crosssection M c,z,rd = W pl,y f y = = Nmm γ M 0.00 = KNm KNm > 0.0 KNm acceptable Shear resistance of cross-section (clause 6..6 ) The design plastic shear resistance of the cross-section V p,rd = A v (f y /3) γ M 0 Load parallel to web Maximum shear force V ED = 840/4. = 00.0 KN For a rolled H section, loaded parallel to the web, the shear area A v is given by A v = A bt f + (t w + r) t f (but not less than ηh w t w ) η=. (from Eurocode 3 part.5, though the UK National Annex may specify an alternative value). h w = (h t f ) = 35.5 ( 37.7) = 77. mm A v = ( ) +( ) 37.7 = 8033 mm (but not less than = 7648mm ) V p,rd = 8033 (75/3).00 75KN > 00 KN = N = 75KN acceptable Load parallel to flanges Maximum shear force V ED = 0/3.7 = 6. KN No guidance on the determination o the shear area for a rolled I or H section loaded parallel to the flanges is presented in EN 993--, though it may be assumed that adopting the recommendations provided for a welded I or H section would be acceptable. The shear area A v is therefore taken as A w = A - (h w t w ) = ( ) = 47 mm V p,rd = 47 (75/3) KN > 6. KN = N = 3847KN acceptable Shear buckling Shear buckling need not be considered, provided

47 Note. When calculating shear area for H sections, following equation was applied as per EN993--:005, sub clause 6..6(3) a). Av = A-bt f + (t w +r)t f However, in the example book, following equation was applied. A v = A bt f + (t w + r) t f For this reason, the difference in shear resistance occurred. 5. Check Interaction of Combined Resistance ( ). Check interaction ratio of bending and axial force member. [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. n = N_Ed / Npl_Rd = a = MIN[ (Area-b*tf)/Area, 0.5 ] = Alpha = Beta = MAX[ 5*n,.0 ] = N_Ed > 0.5*Npl_Rd = kn. -. N_Ed > 0.5*hw*tw*fy/Gamma_M0 = kn. Therefore, Allowance for the effect of axial force. -. Mny_Rd = MIN[ Mply_Rd*(-n)/(-0.5*a), Mply_Rd ] = kn-m. -. Rmaxy = M_Edy / Mny_Rd = < >O.K. -. N_Ed > hw*tw*fy/gamma_m0 = kn. Therefore, Allowance for the effect of axial force. -. In case of n > a -. Mnz_Rd = Mplz_Rd * [ - ((n-a)/(-a))^ ] = kn-m. -. Rmaxz = M_Edz / Mnz_Rd = 0.9 < > O.K. [ M_Edy ^(Alpha) M_Edz ^(Beta) ] -. Rmax = [ ] [ Mny_Rd Mnz_Rd ] = 0.34 < > O.K. ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^ = kn. -. Lambda_by < 0. or N_Ed/Ncry < > No need to check. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) = Ncrz = Pi^*Es*Rzz / KLz^ = kn. -. Lambda_bz > 0. and N_Ed/Ncrz > > Need to check. -. Alphaz = h w 7 ε for unstiffened webs t w η η=. (from Eurocode 3 part.5, though the UK National Annex may specify an alternative value). 7 ε 0.9 = 7 η. = 55.5 Actual h w /t w = 77./3.0 = no shear buckling check required Cross-section resistance under Bending, Shear and axial force (clause 6..0) Reduced plastic moment resistances (clause 6..9.(5)) major (y-y)axis: M N,y, Rd = M pl, y, Rd n 0.5a (but M N,y, Rd M pl, y, Rd ) Where n = N Ed / N pl, y, Rd = 34400/845 = 0.4 a = (A bt f )/A = [30600 ( )]/30600 = 0. M n, y, Rd = (0.5 0.) = KNm KNm > 70 KNm acceptable minor (z-z) axis For n > a M N,z, Rd = M pl, y, Rd M N,z, Rd = n a a KNm > 0 KNm acceptable = KNm Cross-section check for bi-axial bending (with reduced moment resistance) M y,ed M N,y,Rd α + M z,ed M N,z,Rd β For I and H sections: α= and β= 5n (but β ) = ( 5 x 0.4) = = acceptable Member buckling resistance in compression (clause 6.3.) N b, Rd = χ = χaf y γ M ϕ+ ϕ λ for Class, or 3 cross-sections but χ.0 where, Ф = 0.5* + α(λ- 0.) + λ ] and λ = Af y N cr for Class, or 3 cross-sections Elastic critical force and non-dimensional slenderness for flexural buckling For buckling about the major (y-y) axis: L cr = 0.7L =0.7 x 4. =.94 m (see Table 6.6) For buckling about the minor (z-z)axis: 3

48 -. Phiz = 0.5 * [ + Alphaz*(Lambda_bz-0.) + Lambda_bz^ ] = Xiz = MIN [ / [Phiz + SQRT(Phiz^ - Lambda_bz^)],.0 ] = Nb_Rdz = Xiz*Beta_A*Area*fy / Gamma_M = kn. ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). -. Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = kn. N_Ed = = 0.58 < > O.K. Nb_Rd L cr =.0L =.0 x 4. = 4.0 m (see Table 6.6) N cr,y = π EI y L cr = π = x 0 3 N = KN λ = = 0.3 N cr,z = π EI y L cr = π = N = 3863 KN λ = = 0.59 Buckling curves : major (y-y)axis ϕ y = 0.5 x [ x (0.3-0.) ] =0.53 χ y = = N b,y, Rd = = =834 KN 834 KN > 3440 KN major axis flexural buckling resistance is acceptable Buckling curve: minor (z-z) axis ϕ z = 0.5 [ ( ) ] = 0.77 χ y = = N b,y, Rd = = = 6640 KN KN > 3440 KN minor axis flexural buckling resistance is acceptable [Reference] L.Gardner and D.A.Nethercot, Designers Guide to EN 993--, The Steel Construction Institute, Thomas Telford, (Example 6.0) 4

49 3.7 I-section beam design under shear force and bending moment The 457 X 9 UB 8 compression member of S75 steel of Figure 3.8a is simply supported about both principle axes at each end (L cr,y =.0 m), and has a central brace which prevents lateral deflections in the minor principle plane (L cr,z = 6.0 m). Check the adequacy of the member for a factored axial compressive load corresponding to a nominal dead of 60 KN and a nominal imposed load of 30 KN Material Properties 3.7. Section Properties Material S75 f y = 75 N/mm E s = 0 GPa Section Name 457 X 9 UB 8 Depth (H) mm Width (B) 9.3 mm Flange Thickness (T f ) 6.0 mm Web Thickness (T w ) 9.9 mm Gross sectional area (A) mm Effective area (A eff ) mm Analysis Model Loading condition & Beam diagram Loading condition Axial force diagram 5

50 3.7.4 Comparison of Design Results midas Gen Example book Error (%) Compression resistance kn kn 0.% Buckling resistance knm knm 0.% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = kn/mm^. -. sigma = kn/mm^. -. BTR < 9*e ( Class : Plastic ).. ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR > 4*e ( Class 4 : Slender ).. CALCULATE EFFECTIVE AREA. ( ). Calculate effective cross-section properties of web of Class 4 (Internal element). [ Eurocode3 Part , Table 4., 4. ] -. RatT = Lambda_p = RatT / [ 8.4*Eps*SQRT(k_sigma) ] = Rho=MIN[ (Lambda_p *(3+psi))/Lambda_p^,.0 ] = sigma_max = MAX(sigma,sigma) = kn/mm^. -. sigma_min = MIN( sigma,sigma ) = kn/mm^. -. r = 0.00 mm. -. Ar = mm^. -. dc = mm. -. deff = *(Rho*dc) / [ 5 - sigma_min/sigma_max ] + r = mm. Classifying the section. Example book For S75 steel with t f = 6 mm, f y =75 N/mm ( EN 005-) ε = (35/75) 0.5 = 0.94 c f /( t f ε)=*( )/] ( ) =5.44 <.4 c w = ( ) = mm c w /( t w ε) = 407.6/( ) = 44.5 > 4 and so th web is Class 4(slender) Effective area. λ p = f y b/t = = σ cr 8.4ε k σ Ec () 407.6/ = ρ = λ p (3+φ) λ p = (3+) = 0.98 Ec () d- d eff = ( ) = 33.6 mm A eff = = 0067 mm -. Aeff = deff * tw + *Ar = mm^. 6

51 -. zeff = deff/ + tf = mm. -. deff = (Rho*dc) - deff + r = mm. -. Aeff = deff * tw + *Ar = mm^. -. zeff = (h+*r) - deff/ + tf = mm. ( ). Calculated effective cross-section properties of Class4 cross-section. -. Aeff = mm^. (for calculating axial resistance) -. Aeffy = mm^. -. Weffy = mm^3. -. Aeffz = mm^. -. Weffz = mm^3. -. eny = mm. -. enz =.84e-04 mm.. CHECK AXIAL RESISTANCE. ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 4.8 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Aeff / Gamma_M0 = kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = 0.03 < > O.K. Nc_Rd ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^ = kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M = kn. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) = Ncrz = Pi^*Es*Rzz / KLz^ = kn. -. Lambda_bz > 0. and N_Ed/Ncrz > > Need to check. -. Alphaz = Phiz = 0.5 * [ + Alphaz*(Lambda_bz-0.) + Lambda_bz^ ] = Xiz = MIN [ / [Phiz + SQRT(Phiz^ - Lambda_bz^)],.0 ] = Cross-section compression resistance. N c,rd = A eff f y = γ Mo.0 Member buckling resistance. λ y = λ z = A eff f y N cr,y = L cr,y i y = A eff f y N cr,z = L cr,z i z = 000 A eff / A λ (8.8 0) 6000 A eff / A λ (4.3 0) =768 KN >56 KN = N Ed 0067 / / = 0.74 =.608 > 0.74 Buckling will occur about the minor (z) axis. For a rolled UB section (with h/b >. and t f 40 mm), buckling about the z-axis, use buckling curve (b) with α = 0.34 Φ z = 0.5 [ ( ) ]=.03 χ z = = N b,z,rd = χa eff f y = γ M.0 = 844 KN >56 KN = N Ed And so the member is satisfactory. -. Nb_Rdz = Xiz*Beta_A*Area*fy / Gamma_M 7

52 = kn. ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). -. Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = kn. N_Ed = = < > O.K. Nb_Rd [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 89-9 (Example 3..) 8

53 3.8 Designing a UC compression member Design suitable UC of S355 steel to resist a factored axial compressive load corresponding to a nominal dead load of 60kN and a nominal imposed load of 30 kn Material Properties 3.8. Section Properties Material S355 f y = 355 N/mm E s = 0 GPa Section Name 5 x5 UC Thickness (T) 06. mm Width (B) 04.3 mm Flange Thickness (T f ).5 mm Web Thickness (T w ) 7.9mm Gross sectional area (A) mm Shear area (A sz ) mm Analysis Model Loading condition & Beam diagram condition Loading Axial force diagram Comparison of Design Results midas Gen Example book Error (%) Buckling resistance 65.6kN 65.00kN 0.03% 9

54 3.8.5 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = kn/mm^. -. sigma = kn/mm^. -. BTR < 9*e ( Class : Plastic ). ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR < 33*e ( Class : Plastic ).. CHECK AXIAL RESISTANCE. ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 34.7 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = 0.38 < > O.K. Nc_Rd ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^ = kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M = 65.6 kn. Example book Target area and first section choice. Assume f y = 355 N/mm and χ = 0.5 A /( ) =36 mm Try a 5 5 UC 30 with A = 38.3 cm, i y =6.76 cm, i z = 3.83 cm, t f = 9.4 mm. ε = (35/355) 0.5 = 0.84 λ y = λ z = Af y N cr,y = L cr,y i y Af y N cr,z = L cr,z i z 000 = λ ( ) 6000 = λ (3.83 0) ) = =.050 <.3 Buckling will occur about the major (y) axis. For a rolled UC section (with h/b. and t f 00 mm), buckling about the y-axis, use buckling curve (b) with α = 0.34 Φ y = 0.5 [ (.3 0. ) +.3 ] = χ y = = which is much less than the guessed value of 0.5. Second section choice. Guess χ =( )/ = 0.33 A /( ) =4789 mm Try a UC 5 with A = 66.3 cm, i y = 8.9 cm,, t f =.5 mm. For S55 steel with t f =.5 mm, f y = 355 N/mm ε = (35/355) 0.5 = 0.84 c f /( t f ε) = *( ) / ] ( ) = 8.65 < 4 c w /( t w ε) = ( ) / ( ) = 5.0 > 4 and so the cross-section is fully effective. λ y = Af y N cr,y = L cr,y i y 000 = λ (8.9 0 ) ) =.763 For a rolled UC section (with h/b >. and t f 00 mm), buckling about the y-axis, use buckling curve (b) with α = 0.34 Φ y = 0.5 [ ( ) ]=.30 30

55 ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). -. Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = 65.6 kn. N_Ed = = 0.9 < > O.K. Nb_Rd 65.6 χ y = = N b,y,rd = χaf y = γ M.0 = 65 KN > 56 KN = N Ed and so the UC 5 is satisfactory. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 9-9 (Example 3..) 3

56 3.9 Design an RHS compression Design a suitable hot finished RHS of S355 steel to resist a factored axial compressive load corresponding to a nominal dead load of 60kN and a nominal imposed load of 30 kn Material Properties 3.9. Section Properties Material S75 f y = 75 N/mm E s = 0 GPa Section Name 50 X 50 X 8 RHS Depth (H) 50.0 mm Width (B) 50.0 mm Flange Thickness (T f ) 8.0 mm Web Thickness (T w ) 8.0 mm Gross sectional area (A) mm i y 9.7mm Analysis Model Loading condition & Beam diagram Loading condition Axial force diagram Comparison of Design Results midas Gen Example book Error (%) Buckling resistance 60.0kN 640.0kN 4.9% 3

57 3.9.5 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR < 38*e ( Class : Compact ).. Check axial resistance. ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 3. < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = 0.60 kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = 0.67 < > O.K. Nc_Rd 0.60 ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ] -. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Ncry = Pi^*Es*Ryy / KLy^ = kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M = 60.8 kn. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) =.83 ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). -. Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = 60.8 kn. N_Ed = = 0.99 < > O.K. Nb_Rd 60.8 Guess χ = 0.3 Example book A = 568 mm Try a RHS, with A = 60.8cm, i y = 9.7cm, i z = 6.5 cm, t = 8.0mm. For S355 steel with t = 8mm, f y = 355 N mm EN 005- ε = (35 355) 0.5 = 0.84 ( ) c w (tε) = ( ) = 34.7 < 4 T5. And so the cross-section is fully effective. Af y λ y = = L cr,y N cr,y i y λ = 000 (9.8 0) =.70 Af y λ z = = L cr,z N cr,z i z λ 6000 = (6.5 0) =.76 Buckling will occur about the major (ν) axis. For a hot- Finished RHS, use bucking curve (a) with α=0. y = =. 56 χ y = = N b,y,rd = χaf y = γ M.0 = 640kN > 56kN = N Ed and so the > 8 RHS is satisractory 33

58 Note. The difference in buckling resistance occurred since currently midas Gen does not consider r value for rolled box section. This can be improved in the future version. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 9-93 (Example 3..3) 34

59 3.0 Compression resistance of a Class 4 compression member Determine the compression resistance of the cross-section of the member shown in Figure the figure below Material Properties 3.0. Section Properties Material S75 f y = 75 N/mm E s = 0 GPa Section Name 40 X 400 Depth (H) 40.0 mm Width (B) mm Flange Thickness (T f ) 0.0 mm Web Thickness (T w ) 0.0 mm Gross sectional area (A) mm Effective area (A eff ) 96.0mm Comparison of Design Results midas Gen Example book Error (%) Axial resistance 37.45kN 37.00kN 0.0% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges.[eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = kn/mm^. -. sigma = kn/mm^. -. BTR > 4*e ( Class 4 : Slender ). ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR > 4*e ( Class 4 : Slender ).. Calculate Effective Area ( ). Calculate effective cross-section properties of left-top Example book Classifying the section plate elements. t f = 0 mm, t w = 0 mm, f y = 355 N/mm ε = (35/355) 0.5 = 0.84 En 005- c f /( t f ε) = (400/ 0/ 8) (0 0.84) = 3.0 > 4 and so the flange is Class 4. c w /( t w ε) = (40 0-8) / (0 0.84) = 47. > 4 and so the web is Class 4. Effective flange area. k σ f = 0.43 λ p f = ( )/ =.3 > ρ f = ( )/.3 = A eff f = (400/ -0/ - 8) 0 +(0 + 8 ) 0 = 5658 mm 35

60 flange of Class 4 (Outstand element). [ Eurocode3 Part , Table 4., 4. ] -. RatT = Lambda_p = RatT / [ 8.4*Eps*SQRT(k_sigma) ] = Rho = MIN[ (Lambda_p-0.88) / Lambda_p^,.0 ] = sigma_max = MAX( sigma, sigma ) = kn/mm^. -. sigma_min = MIN( sigma, sigma ) = kn/mm^. -. r = mm. -. bc = mm. -. beff = Rho*bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. Effective flange area -. Aeff * 4 = mm^ Effective web area. k σ,w = 4.0 λ p, w = (40 0 8)/ = 0.83 > ρ w = { (3 +!)} / 0.83 =0.885 A eff, w = ( ) ( ) = 3558 mm compression resistance. A eff = = 96 mm N c,rd = /.0 N =37 KN. ( ). Calculate effective cross-section properties of web of Class 4 (Internal element). [ Eurocode3 Part , Table 4., 4. ] -. RatT = Lambda_p = RatT / [ 8.4*Eps*SQRT(k_sigma) ] = Rho = MIN[ (Lambda_p-0.055*(3+psi)) / Lambda_p^,.0 ] = sigma_max = MAX( sigma, sigma ) = kn/mm^. -. sigma_min = MIN( sigma, sigma ) = kn/mm^. -. r = mm. -. Ar = mm^. -. dc = mm. -. deff = *(Rho*dc) / [ 5 - sigma_min/sigma_max ] + r = mm. -. Aeff = deff * tw + *Ar = mm^. -. zeff = deff/ + tf = mm. -. deff = (Rho*dc) - deff + r = mm. -. Aeff = deff * tw + *Ar = mm^. -. zeff = (h+*r) - deff/ + tf = mm. ( ). Calculated effective cross-section properties of Class4 cross-section. -. Aeff = mm^. (for calculating axial resistance) 3. Check Axial Resistance ( ). Check slenderness ratio of axial compression member (Kl/i). [ Eurocode3: ] -. Kl/i = 53. < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Aeff / Gamma_M0 = kn. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 4-4 (Example 4.9.) 36

61 3. Section moment resistance of a Class 3 I-beam Determine the section moment resistance and examine the suitability for plastic design of the 356 X 7 UB 45 of S355 steel shown in the figure below. 3.. Material Properties Material S355 f y = 355 N/mm E s = 0 GPa 3.. Section Properties Section Name 356 X 7 UB 45 Depth (H) Width (B) Flange Thickness (T f ) Web Thickness (T w ) 35.4 mm 7. mm 9.7 mm 7.0 mm Gross sectional area (A) 573,000 mm Plastic section modulus (W pl ) cm Comparison of Design Results midas Gen Example book Error (%) Moment resistance 75.kNm 75.0kNm 0.0% 3..4 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = KPa. -. sigma = KPa. -. BTR < 0*e ( Class : Compact ). Example book Classifying the section- plate elements. t f = 9.7 mm, t w = 7.0 mm, f y = 355 N/mm ε = (35/355) 0.5 = 0.84 En 005- c f /( t f ε) = (7./ 7.0/ 0.) ( ) = 9. > 9 and so the flange is Class. c w /( t w ε) = ( ) / ( ) = 54.7 > 7 and so the web is Class. ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR =

62 -. sigma = KPa. -. sigma = KPa. -. HTR < 7*e ( Class : Plastic ).. Check Bending Moment Resistance About Major Axis. ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = 75. kn-m. section moment resistance The cross-section is Class and therefore unsuitable for plastic design. M c,rd /.0 N = 75. KNm. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 4-43 (Example 4.9.) 38

63 3. Section moment resistance of a Class 4 box beam Determine the section moment resistance of the welded-box section beam of S355 steel shown in the figure below. The weld size is 6 mm. 3.. Material Properties 3.. Section Properties Material S355 f y = 355 N/mm E s = 0 GPa Section Name RHS 430 X 450 Depth (H) mm Width (B) mm Flange Thickness (T f ) 0.0 mm Web Thickness (T w ) 8.0 mm Gross sectional area (A) mm Shear area (A sz ) mm 3..3 Comparison of Design Results midas Gen Example book Error (%) Moment resistance 79.99kN 79.90kNm 0.0% 3..4 Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = 0.08 kn/mm^. -. sigma = 0.08 kn/mm^. -. HTR > 4*e ( Class 4 : Slender ). ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = kn/mm^. -. sigma = kn/mm^. -. HTR < 7*e ( Class : Plastic ). Example book Classifying the section- plate elements. t f = 0 mm, t w = 8 mm, f y = 355 N/mm En 005- ε = (35/355) 0.5 = 0.84 c f /( t f ε) = 40 (0 X 0.84) = 50.4 > 4 and so the flange is Class 4. c w /( t w ε) = (430 X 0- X 6) / (8 X 0.84) = 6. > 7 and so the web is Class. The cross-section is therefore Class 4 since the flange is Class 4. 39

64 . Calculate Effective Section Modulus About Major Axis. ( ). Calculate buckling factor of internal compression element. [ Eurocode3 Part , Table 4. ] -. In case of Psi =.0 -. k_sigma = ( ). Calculate effective cross-section properties of top flange of Class 4 (Internal element). [ Eurocode3 Part , Table 4., 4. ] -. RatT = Lambda_p = RatT / [ 8.4*Eps*SQRT(k_sigma) ] = Rho = MIN[ (Lambda_p-0.055*(3+psi)) / Lambda_p^,.0 ] = sigma_max = MAX( sigma, sigma ) = 0.08 kn/mm^. -. sigma_min = MIN( sigma, sigma ) = 0.08 kn/mm^. -. r = mm. -. bc = mm. -. beff = Rho*bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. Effective cross-section k σ = 4.0 λ p = = > ψ = ρ = ( (3 + )) / =0.848 b c,eff = X 40 = mm A eff = ( ) X0 + (450 0) + (430 0) X 8 = 4935 mm 4935 X z c = ( ) X0 X(430-0/) / + X(430 X0) X 8 X 430/ z c = 06. mm I eff = ( ) X 0 X (430-0/ -06.) ( /) + X(430 X 0) 3 X 8/ +X(430 X0) X8X(430/ 06.) mm 4 = 460. X 0 6 mm 4 ( ). Calculate cross-section properties of bottom flange. -. r = mm. -. bc = mm. -. beff = bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. ( ). Calculate cross-section properties of left web. -. r = mm. -. Ar = mm^. -. dc = mm. -. deff = dc + r = mm. -. Aeff = deff * tw + 4*Ar = mm^. -. zeff = (h+*r) - deff/ = mm. ( ). Calculate cross-section properties of right web. -. r = mm. -. Ar = mm^. -. dc = mm. -. deff = dc + r = mm. -. Aeff = deff * tw + 4*Ar = mm^. -. zeff = (h+*r) - deff/ = mm. ( ). Calculated effective cross-section properties of Class4 cross-section. -. Aeffy = mm^. -. Weffy = mm^3. 40

65 4. CHECK BENDING MOMENT RESISTANCE ABOUT MAJOR AXIS. ( ). Calculate local buckling resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Weffy = mm^3. -. Mc_Rdy = Weffy * fy / Gamma_M = kn-mm. Section moment resistance W eff, min = / ( ) =.056 X 0 6 mm 3 M c,rd =.056 X /.0 N = 79.9 KNm. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, (Example 4.9.3) 4

66 3.3 Section moment resistance of a slender plate girder Determine the section moment resistance of the welded plate girder of S355 steel shown in the figure below. The weld size is 6 mm Material Properties 3.3. Section Properties Material S355 f y = 355 N/mm E s = 0 GPa Section Name 540x400 Depth (H) mm Width (B) mm Flange Thickness (T f ) 0.0 mm Web Thickness (T w ) 0.0 mm Gross sectional area (A) mm Effective area (A eff ) 4.48 x 0 6 mm Comparison of Design Results midas Gen Example book Error (%) Moment resistance kNm kNm 0.0% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = 0.06 kn/mm^. -. sigma = 0.06 kn/mm^. -. BTR < 4*e ( Class 3 : Semi-compact ). ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = 0.05 kn/mm^. -. sigma = kn/mm^. -. HTR > 4*e ( Class 4 : Slender ) Example book t f = 0 mm, t w = 0 mm, f y = 345 N/mm ε = (35/345) 0.5 = 0.85 c f /( t f ε) = (400/ 0/ 6) (0 0.85) =.5 > 0 but < 4, and so the flange is Class 3. c w /( t w ε) = ( ) / (0 0.85) = 80.3 > 4 and so the web is Class 4. A conservative approximation for the cross-section moment resistance may be obtained by ignoring the web completely, so that M c,rd = M f = (400 0) (540 0) 345/.0 N = 495 KNm. A higher resistance may be calculated by determining the 4

67 effective width of the web. ψ = k σ = 3.9 λ p = / =.99. Calculate Effective Section Modulus About Major Axis. ( ). Calculate cross-section properties of left-top flange. [ Eurocode3 Part , Table 4., 4. ] -. r =.000 mm. -. bc = mm. -. beff = bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. ( ). Calculate cross-section properties of right-top flange. -. beff = bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. ( ). Calculate cross-section properties of left-bottom flange. -. beff = bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. ( ). Calculate cross-section properties of right-bottom flange. -. beff = bc + r = mm. -. Aeff = beff * tf = mm^. -. yeff = beff/ = mm. ( ). Calculate buckling factor of internal compression element. -. In case of Psi = k_sigma = ρ = ( (3 )) /.99 =0.705 b c = ( ) / { (-)} = mm b eff = = 54.4 mm b e = =09.8 mm b e = =34.6 mm and the ineffective width of the web is b c - b e - b e = = 9.6 mm A eff = ( ) = 8804 mm 8804 z c = ( (540 0) 0) (540 /) ( /) z c = mm I eff =(400 0) ( ) + (400 0) ( ) + (540 0) 3 0/ + (540 0) 0 540/ 737.6) / ( / 737/6) mm 4 = mm 4 W eff = / ( ) = mm 3 M c,rd = /.0 N = 4996 KNm. ( ). Calculate effective cross-section properties of web -. RatT = Lambda_p = RatT / [ 8.4*Eps*SQRT(k_sigma) ] = Rho = MIN[ (Lambda_p-0.055*(3+psi)) / Lambda_p^,.0 ] = sigma_max = MAX( sigma, sigma ) = 0.05 kn/mm^. -. sigma_min = MIN( sigma, sigma ) = kn/mm^. -. r = mm. -. Ar = mm^. -. dc=(h*sigma_max) / (sigma_max-sigma_min) =744.0mm. -. deff = 0.4*Rho*dc + r = mm. 43

68 -. Aeff = deff * tw + *Ar = mm^. -. zeff = (h+*r) - deff/ + tf = 4.8 mm. -. deff = 0.6*Rho*dc + (h-dc) + r = mm. -. Aeff = deff * tw + *Ar = mm^. -. zeff = deff/ + tf = mm. ( ). Calculated effective cross-section properties of Class4 cross-section. -. Aeffy = mm^. -. Weffy = mm^3. 4. Check Bending Moment Resistance About Major Axis ( ). Calculate local buckling resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Weffy = mm^3. -. Mc_Rdy = Weffy * fy / Gamma_M = kn-mm. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, (Example 4.9.4) 44

69 3.4 Shear buckling resistance of an unstiffened plate girder web Determine the shear buckling resistance of the unstiffened plate girder web of S355 steel shown in the figure below 3.4. Material Properties 3.4. Section Properties Material S355 f y = 355 N/mm E s = 0 GPa Section Name 540x400 Depth (H) mm Width (B) mm Flange Thickness (T f ) 0.0 mm Web Thickness (T w ) 0.0 mm Gross sectional area (A) mm Effective area (A eff ) 4.48 x 0 6 mm Comparison of Design Results midas Gen Example book Error (%) Shear resistance 95.86kN 96.00kN 0.0% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = 0.06 kn/mm^. -. sigma = 0.06 kn/mm^. -. BTR < 4*e ( Class 3 : Semi-compact ). ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = 0.05 kn/mm^. -. sigma = kn/mm^. Example book t f = 0 mm, t w = 0 mm, f y = 345 N/mm ε = (35/355) 0.5 = 0.84 η =. h w = = 500 mm ηh w /( t w ε) =. 500 / ( ) =. > 7 and so the web is slender α/ h w = / h w =, k τst = 0 k τ = 5.34 τ cr = 5.34 X X (0/500) = 45. N/mm λ w = 0.76 X (355/45.) =.3 >.08 Assuming that there is a non-rigid end post, then χ w = 0.83 /.3 =

70 -. HTR > 4*e ( Class 4 : Slender ). Check Shear Resistance. ( ). Calculate shear buckling resistance in local-z direction (Vbl_Rdz). [ Eurocode3:05 6., 6..6, EN : ] -. Eta =.0 -. Lambda_w = hw / (86.4*tw*e) = Chi_w = 0.83/Lambda_w = Vbw_Rdz = Chi_w*fy*hw*tw / [sqrt(3)*gamma_m] = kn. Neglecting any contribution from the flanges, V b,rd = V bw,rd = ( ) 3.0 N = 96 KN [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, (Example 4.9.5) 46

71 3.5 Checking a simply supported beam The simply supported 60 X 9 UB 5 of S75 steel shown in the right figure has a span of 6.0m and is laterally braced at.5m intervals. Check the adequacy of the beam for a nominal uniformly distributed dead load of 60 KNm together with a nominal uniformly distributed imposed load of 70 KNm Material Properties Material S75 f y = 65 N/mm E s = 0 GPa 3.5. Section Properties Section Name UB 60x9x5 Depth (H) 6. mm Width (B) 9.0 mm Flange Thickness (T f ) 9.6 mm Web Thickness (T w ).9 mm Gross sectional area (A) mm Plastic section modulus (W pl,y ) cm Analysis Model Loading condition Beam Diagram SF BM Comparison of Design Results midas Gen Example book Error (%) Shear resistance 5.9kN 7.00kN 6.46% Bending resistance 975.kNm kNm 0.% 47

72 Note. Shear resistance is calculated with an error of 6.46% due to the difference in shear area, Av. In the example book, the minimum value of shear area ηh w t w was not considered whereas it was considered in midas Gen. (6..6 (3) EN 993--:005) Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = 0.60 kn/mm^. -. sigma = 0.60 kn/mm^. -. BTR < 9*e ( Class : Plastic ). ( ). Determine classification of bending Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = 0.6 kn/mm^. -. sigma = -0.6 kn/mm^. -. HTR < 7*e ( Class : Plastic ).. Check Bending Moment Resistance About Major Axis ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = mm^3. -. Mc_Rdy = Wply * fy / Gamma_M0 =975.0 kn-m. ( ). Check ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = < > O.K. Mc_Rdy Check Shear Resistance. (). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. eta =. (Fy < 460 MPa.) -. r =.7000 mm. -. Avy = Area - hw*tw = mm^. -. Avz = eta*hw*tw = mm^. -. Avz = Area - *B*tf + (tw + *r)*tf = mm^. -. Avz = MAX[ Avz, Avz ] = mm^. Classifying the section. Example book t f = 9.6 mm, f y = 65 N/mm En 005- ε = (35/65) 0.5 = 0.94 c f /( t f ε) = (9/.9/.7) ( ) = 5.9 < 9 and the flange is Class. c w /( t w ε) = ( ) / ( ) = 48.9 > 7 and the web is Class. (Note the general use of the minimum f y obtained for the flange.) Checking for moment. q Ed = (.35 60) + (.5 70) = 86 KNm M Ed = 86 6 /8 = 837 KNm M c,rd = /.0 Nmm = 975 KNm > 837 KNm = M Ed Which is satisfactory. Checking for shear. V Ed = 86 6 = 558 KN A v = 59X (.9 +.7) 9.6 = 7654 mm V c,rd = 7654 X (65 / 3) /.0 N = 7 KN > 558 KN = V Ed Which is satisfactory. ( ). Calculate plastic shear resistance in local-z direction (Vpl_Rdz). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = 5.90 Checking for bending and shear. 48

73 kn. ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = kn. V_Edz = = < > O.K. Vpl_Rdz 5.90 The maximum M Ed occurs at mid span where V Ed =0, and the maximum V Ed occurs at the support where M Ed = 0, and so there is no need to check for combined bending and shear. (Note that in any case, 0.5V c,rd = = KN > 558 KN = V Ed and so the combined bending and shear condition does not operate.) Note. The difference in shear resistance occurred since the midas Gen consider the additional condition when calculating shear area as per EN993--:005, sub clause 6..6(3) a). Av = A-bt f + (t w +r)t f but not less than ηh w t w [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 09- (Example 5..5) 49

74 3.6 Serviceability of a simply supported beam Check the imposed load deflection of the 60 X 9 UB 5 of right figure for a serviceability limit of L/360v Material Properties Material S75 f y = 75 N/mm E s = 0 GPa 3.6. Section Properties Section Name UB 60x9x5 Depth (H) 6. mm Width (B) 9.0 mm Flange Thickness (T f ) 9.6 mm Web Thickness (T w ).9 mm Gross sectional area (A) mm Plastic section modulus (W pl,y ) cm Analysis Model Loading condition Beam Diagram SF BM Comparison of Design Results midas Gen Example book Error (%) Deflection (ω c ) 5.705mm 5.700mm 0.09% 50

75 3.6.5 Detailed comparison. Check Deflection. midas Gen ( ). Compute Maximum Deflection. -. LCB = -. DAF =.000 (Deflection Amplification Factor). -. Position = mm From i-end(node ). -. Def = * DAF = mm (Golbal Z) -. Def_Lim = 6.667mm Def < Def_Lim ---> O.K! Example book The central deflection ω c of a simply supported beam with uniformly distributed load q can be calculated using ω c = 5qL 4 384EI y = = 5.7 mm. (The same result can be obtained using Figure 5.3.) L/360 =6000/360 = 6.7 mm > 5.7 mm = ω c and so the beam is satisfactory. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, 3 (Example 5..8) 5

76 3.7 Checking the major axis in-plane resistance The 9 m long simply supported beam-column shown in Figure has a factored design axial compression force of 00 KN and a design concentrated load of 0 KN (which includes an allowance for self-weight) acting in the major principal plane at mid-span. The beam-column is the 54 X 46 UB 37 of S75 steel shown in Figure 7.9a. The become-column is continuously braced against lateral deflections ν and twist rotations φ. Check the adequacy of the beam-column Material Properties Material S75 f y = 75 N/mm E s = 0 GPa 3.7. Section Properties Section Name 54x46 UB 37 Depth (H) 56.0 mm Width (B) 46.4 mm Flange Thickness (T f ) 0.9 mm Web Thickness (T w ) 6.3 mm Gross sectional area (A) mm Shear area (A sz ) 500. mm Analysis Model Loading condition Beam Diagram SF BM 5

77 3.7.4 Comparison of Design Results midas Gen Example book Error (%) Shear resistance 900.3kN kN 0.0% Bending resistance 3.8kNm 3.80kNm 0.0% Detailed comparison midas Gen. Cross-section classification ( ). Determine classification of compression outstand flanges. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = b/t = BTR = sigma = KPa. -. sigma = KPa. -. BTR < 9*e ( Class : Plastic ). ( ). Determine classification of bending and compression Internal Parts. [ Eurocode3:05 Table 5. (Sheet of 3), EN ] -. e = SQRT( 35/fy ) = d/t = HTR = sigma = KPa. -. sigma = KPa. -. Psi = [*(Nsd/A)*(/fy)]- = Alpha = > HTR < 396*e/(3*Alpha-) ( Class : Plastic ).. CHECK AXIAL RESISTANCE ( ). Check slenderness ratio of axial compression member (Kl/i) [ Eurocode3: ] -. Kl/i = 83.3 < > O.K. ( ). Calculate axial compressive resistance (Nc_Rd). [ Eurocode3:05 6., 6..4 ] -. Nc_Rd = fy * Area / Gamma_M0 = kn. ( ). Check ratio of axial resistance (N_Ed/Nc_Rd). N_Ed = = 0.54 < > O.K. Nc_Rd ( ). Calculate buckling resistance of compression member (Nb_Rdy, Nb_Rdz). [ Eurocode3: , ]. Beta_A = Aeff / Area = Lambda = Pi * SQRT(Es/fy) = Lambda_by = {(KLy/iy)/Lambda} * SQRT(Beta_A) = Example book Simplified approach for cross-section resistance. t f = 0.9 mm, f y = 75 N/mm En 005- ε = (35/75) 0.5 = 0.94 c f /( t f ε) = (( X 7.6) / ) ( ) =6.0 <9 and the flange is Class. c w = 56.0 ( X 0.9) ( X 7.6) = 9.0 mm The compression proportion of web is α = h t f + r + 56 N Ed t f f y = = 0.76 > 0.5 c w c w t w = 9.0 / 6.3 = 34.8 < 4.3 = and the web is Class ε 3α 9.0 M c,y,rd = 75 X /.0 Nmm =3.8 KNm M y,ed = 0 X 9/4 =45.0 KNm / = And the cross-section resistance is a adequate. Alternative approach for cross-section resistance. Because the section is Class, Clause can be used. No reduction in plastic moment resistance is required provided both N Ed = 00 KN < 34.5 KN = ( /.0)/ 0 3 = 0.5N pl,rd and N Ed = 00 KN < 0.9 KN 0.5 ( ) = = 0.5h w t w f y γ M 0 And so no reduction in the plastic moment resistance is required. 53

78 -. Ncry = Pi^*Es*Ryy / KLy^ = kn. -. Lambda_by > 0. and N_Ed/Ncry > > Need to check. -. Alphay = Phiy = 0.5 * [ + Alphay*(Lambda_by-0.) + Lambda_by^ ] = Xiy = MIN [ / [Phiy + SQRT(Phiy^ - Lambda_by^)],.0 ] = Nb_Rdy = Xiy*Beta_A*Area*fy / Gamma_M =900.3 kn. -. Lambda_bz = {(KLz/iz)/Lambda} * SQRT(Beta_A) =3.30e Ncrz = Pi^*Es*Rzz / KLz^ = kn. -. Lambda_bz < 0. or N_Ed/Ncrz < > No need to check. ( ). Check ratio of buckling resistance (N_Ed/Nb_Rd). - Nb_Rd = MIN[ Nb_Rdy, Nb_Rdz ] = kn. N_Ed = = 0. < > O.K. Nb_Rd CHECK SHEAR RESISTANCE. ( ). Calculate shear area. [ Eurocode3: , EN993--5:04 5. NOTE ] -. eta =. (Fy < 460 MPa.) -. r = m. -. Avy = Area - hw*tw = m^. -. Avz = eta*hw*tw = m^. -. Avz = Area - *B*tf + (tw + *r)*tf = m^. -. Avz = MAX[ Avz, Avz ] = m^. ( ). Calculate plastic shear resistance in local-z direction (Vpl_Rdz). [ Eurocode3:05 6., 6..6 ] -. Vpl_Rdz = [ Avz*fy/SQRT(3) ] / Gamma_M0 = 8. kn. ( ). Shear Buckling Check. [ Eurocode3: ] -. HTR < 7*e/Eta ---> No need to check! ( ). Check ratio of shear resistance (V_Edz/Vpl_Rdz). ( LCB =, POS = J ) -. Applied shear force : V_Edz = 0.00 kn. V_Edz = = < > O.K. Vpl_Rdz CHECK BENDING MOMENT RESISTANCE ABOUT MAJOR AXIS ( ). Calculate plastic resistance moment about major axis. [ Eurocode3:05 6., 6..5 ] -. Wply = m^3. -. Mc_Rdy = Wply * fy / Gamma_M0 = 3.8 kn-m. Thus M N,y, Rd = M pl, y, Rd = 3.9 KNm > 45.0 KNm = M y,ed And the cross-section resistance is adequate. Compression member buckling resistance. Because the member is continuously braced, beam lateral buckling and column minor axis buckling need not be considered. λ y = Af y N cr,y = L cr,y i y 9000 = λ (0.8 0 ) ) = For a rolled UB section (with h/b >. and t f 40 mm), buckling about the y-axis, use buckling curve (a) with α = 0. Φ y = 0.5 [ + 0. ( ) ]=.04 χ y = = N b,y,rd = χaf y = =900KN > 00KN = γ M.0 N Ed Beam-column member resistance more exact approach(annex A) λ max = λ y = N cr,y = π EI L cr = π = = 470 KN Since there is no lateral buckling, λ 0 = 0, b LT = 0, C mlt =.0 C my = C my,0 = 0.8N Ed /N cr,y = / 47 = Wpl,y W y = = 483 =.5, Wel,y 433 n pl = N ED = N RK /γ M ) = /.0 C yy = + (W y ) 54

79 ( ). Check ratio of moment resistance (M_Edy/Mc_Rdy). M_Edy = = < > O.K. Mc_Rdy 3.8 ( ). Calculate plastic resistance moment about minor axis. [ Eurocode3:05 6., 6..5 ] -. Wplz = m^3. -. Mc_Rdz = Wplz * fy / Gamma_M0 = 3.73 kn-m. ( ). Check ratio of moment resistance (M_Edz/Mc_Rdz). M_Edz = = < > O.K. Mc_Rdz CHECK INTERACTION OF COMBINED RESISTANCE ( ). Calculate Major reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edz / Vpl_Rdz < My_Rd = Mc_Rdy = 3.8 kn-m. b LT μ y =.6 W y C my λ max W pl, y W el, y.6 W y C my λ max n pl = + (.5 ) = > = /.5 N ED /N cr,y χ y N ED /N cr,y = k yy = C my C mlt = μ y N ED /N cr,y C yy 00/ /47 = =.09 00/ N Ed + k M y,ed N yy = b,y,rd M c,y,rd = = 0.59 < And the member resistance is adequate. ( ). Calculate Minor reduced design resistance of benging and shear. [ Eurocode3: (6.30) ] -. In case of V_Edy / Vpl_Rdy < Mz_Rd = Mc_Rdz = 3.73 kn-m. ( ). Check general interaction ratio. [ Eurocode3: (6.) ] - Class or Class N_Ed M_Edy M_Edz -. Rmax = N_Rd My_Rd Mz_Rd = < > O.K. ( ). Check interaction ratio of bending and axial force member. [ Eurocode3: (6.3 ~ 6.4) ] - Class or Class -. n = N_Ed / Npl_Rd = a = MIN[ (Area-b*tf)/Area, 0.5 ] = Alpha = Beta = MAX[ 5*n,.0 ] = N_Ed < 0.5*Npl_Rd = 5.03 kn. -. N_Ed < 0.5*hw*tw*fy/Gamma_M0 = 0.88 kn. Therefore, No allowance for the effect of axial force. -. Mny_Rd = Mply_Rd = 3.8 kn-m. -. Rmaxy = M_Edy / Mny_Rd = < > O.K. -. N_Ed < hw*tw*fy/gamma_m0 = 70.0 kn. Therefore, No allowance for the effect of axial force. -. Mnz_Rd = Mplz_Rd = 3.73 kn-m. -. Rmaxz = M_Edz / Mnz_Rd = < > O.K. -. Rmax = MAX[ Rmaxy, Rmaxz ] = < > O.K. 55

80 ( ). Check interaction ratio of bending and axial compression member. [ Eurocode3: , (6.6, 6.6), Annex A ] -. kyy = kyz = kzy = kzz = Xiy = Xiz = XiLT = N_Rk = A*fy = kn. -. My_Rk = Wply*fy = 3.8 kn-m. -. Mz_Rk = Wplz*fy = 3.73 kn-m. -. N_Ed*eNy = 0.0 (Not Slender) -. N_Ed*eNZ = 0.0 (Not Slender) N_Ed -. Rmax_LT = Xiy*N_Rk/Gamma_M M_Edy + N_Ed*eNy + kyy * XiLT*My_Rk/Gamma_M M_Edz + N_Ed*eNz + kyz * Mz_Rk/Gamma_M = 0.59 < > O.K. N_Ed -. Rmax_LT = Xiz*N_Rk/Gamma_M M_Edy + N_Ed*eNy + kzy * XiLT*My_Rk/Gamma_M M_Edz + N_Ed*eNz + kzz * Mz_Rk/Gamma_M = < > O.K. -. Rmax = MAX[ MAX(Rmax, Rmax), MAX(Rmax_LT, Rmax_LT) ] = 0.59 < > O.K. [Reference] N.S. Trahair, M.A. Bradford, D.A.Nethercot, and L.Gardner, The behavior and Design of Steel Structures to EC3, Taylor & Francis, (Example 7.7.) 56

81 CHAPTER 4 Steel Design Tutorial Design Examples using midas Gen to Eurocode3

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83 CHAPTER 4. Steel Design Tutorial Step 00 Contents Eurocode 3 - Design of Multi-Story Steel Building Step : Analyze the model. Step : Select the design code. Step 3: Generate load combinations. Step 4: Enter design parameters (Unbraced Length, Moment Factor, etc). Step 5: Enter deflection limits. Step 6: Check design results. Step 7: Change and update the designed sections. Step 00 Overview Eurocode 3 Steel Design Methods midas Gen provides the following two methods:. The program finds optimal sections for gravity loads (Design > Steel Optimal Design) and also finds optimal sections for lateral loads (Design> Displacement Optimal Design). With t he combined use of the two, the user should find optimal sections.. The program checks strength and serviceability based on the sections defined by the user and the design code selected by the user (Design > Steel Code Check). Also, the program searches and proposes sections which satisfy the design conditions entered by the user. Then the user can update the sections referring to the sections proposed by the program. In this tutorial, method is presented.

84 CHAPTER 4. Steel Design Tutorial Step 00 Overview - Details of the example building ROOF 5F 4F 3F F 4,000 4,000 4,000 4,000 5,000 F Figure. Elevation (unit: mm) 7,500 7,500 7,500 7,500 7,500 7,500 7,500,500,500,500,500,500,500,500 Figure. Structural Plan (~Roof) (unit: mm) Step 00 Overview Applied Codes Applied Wind Load: Eurocode (005) Applied Seismic Load: Eurocode 8 (004) Steel Design Code: Eurocode 3 (005) Applied Loads Self Weight Floor loads For floor ~5 Superimposed Dead Load: 3.7 kn/m Structural System Bracing system Materials Beam, Column and Brace: S75 Unit Load Cases Load Name Details Self Weight Self Weight Live Load: 4 kn/m For roof Superimposed Dead Load: 5 kn/m Live Load:.5kN/m Wind loads Basic Wind Velocity: 6 m/s Terrain Category: II Seismic loads Ground Type: B Static Load Cases Response Spectrum Load Cases SID Superimposed Dead Load 3 Live Load Live Load 4 Wind X-dir 5 Wind Y-dir 6 RX 7 RY Wind Load (in the global X-direction) Wind Load (in the global Y-direction) Seismic Load (in the global X-direction) Seismic Load (in the global Y-direction)

85 CHAPTER 4. Steel Design Tutorial Step 00 Overview Applied Sections These are the sections assumed before design updates. Beam Section ID DB Section Size UNI IPE 500 UNI IPE UNI IPE 450 Column Section ID DB Section Size 4 UNI HEB 40 5 UNI HEB 300 Brace Section ID DB Section Size 6 UNI HEA 60 Step 0 Step. Open the model file and perform analysis & Steel Design Code Procedure Step. Open the model file and perform analysis Open EC3 design_start model.mgb Analysis > Perform Analysis 4 Step. Steel Design Code 3 Design > Steel Design Parameter > Design Code 5 4 Design Code: Eurocode3:05 5 Click on OK button. 3

86 CHAPTER 4. Steel Design Tutorial Step 0 Step. Generate Load Combinations Procedure Generate Load Combinations The program automatically creates design load combinations which can be also modified or deleted by the user Result > Combinations Click on Steel Design Tab. 3 Click Auto Generation button. 4 Option: Add 5 Code Selection: Steel 6 Design Code: Eurocode3: Gamma G:.35, Gamma Q:.5 8 Click on OK button Click on Close button. Step 03 Step. 3 Enter Unbraced Length Procedure Enter Unbraced Length 5 View > Select > Identity 6 Select Type: Section 4 3 Select : IPE 500 & IPE Click on Add button and Close 7 button. 4 5 Design > General Design Parameter >Unbraced Length 8 6 Option: Add/Replace 7 8 Laterally Unbraced Length, Lb =.5 Click on Apply button and Close button. 4

87 CHAPTER 4. Steel Design Tutorial Step 04 Step. 4 Enter Equivalent Uniform Moment Factor (Cmy, Cmz) Procedure Enter Equivalent Uniform Moment Factor (Cmy, Cmz) 3 View > Select > Select All Design > General Design Parameter >Equivalent 4 Uniform Moment Factor Option: Add/Replace 4 Check on Calculate by Program 5 Click on Apply button 6 Click on Close button. Step 05 Step. 5 Enter Equivalent Moment Factor (CmLT) Procedure Enter Equivalent Moment Factor (CmLT) 5 View > Select > Identity 6 3 Select Type: Element Type Select BEAM Click on Add button and Close button. Design > Steel Design Parameter > Equivalent Moment Factor 4 6 Option: Add/Replace 7 Check on Calculate by Program. 8 Click on Apply button and Close button. 5

88 CHAPTER 4. Steel Design Tutorial Step 06 Step. 6 Assign/Confirm Serviceability Load Combination Type Procedure Assign/Confirm Serviceability Load Combination Type Design > General Design Parameter > Serviceability Load Combination Type Click on Close button. Step 07 Step. 7 Enter Serviceability Parameters Procedure Enter Serviceability Parameters 5 If the element s local x-axis is parallel to the global Z-axis, the View > Select > Identity Select Type: Element Type Select BEAM. Click on Add button and Close button. Design > Steel Design Parameter > Serviceability Parameters element is considered as a column. If the element s localxaxis is parallel to global X-Y plane, the element is considered as beam. All other elements except for columns and beams are considered to be braces. 6 Option: Add/Replace 7 Selection Type: By Selection 8 Deflection Control For Beams: L 0 9 / 50 Deflection Control For Columns: h / Deflection Amplification Factor: Click on Apply button and Close button. 6

89 CHAPTER 4. Steel Design Tutorial Step 08 Step. 8- Steel Code Checking Procedure Steel Code Checking () Design > Steel Code Check Click on button. 3 Select SECT 5 & SECT 6. 4 Click on Graphic button. 3 Ultimate Limit State Check Results COM: Critical ratio by axial force and bending moment (Yielding & Buckling) SHR: Critical ratio by shear force (Yielding & Buckling) Serviceability Limit State Check Results Beam: Vertical deflection Column: Horizontal deflection 4 Step 08 Step. 8- Steel Code Checking Procedure Steel Code Checking () Click on Close button. 7

90 CHAPTER 4. Steel Design Tutorial Step 9 Step. 9- Change the NG sections Procedure Change the NG sections () Change command will verify the strength for the user-selected section and save the design results until re-analysis is performed. Click on button. Click on Change button. 3 Limit Combined Ratio from 0.8 to Click on Search Satisfied Section. 5 5 Select HEB Click on Change button. 6 Step 9 Step. 9- Change the NG sections Procedure Change the NG sections () Update command will allow the user to update the section and reanalyze. Click on update button. Select Property No Limit Combined Ratio from 0.8 to Click on Search Satisfied Section. 5 Select HEA Click on Change & Close button. 6 8

91 CHAPTER 4. Steel Design Tutorial Step 9 Step. 9-3 Change the NG sections Procedure Change the NG sections (3) Only the section for design review has been changed. The section in the model has not been changed as seen in the Works Tree. Select SECT 6. Click on Graphic button. Step 9 Step. 9-4 Change the NG sections Procedure Change the NG sections (4) Click on Close button. 9

92 CHAPTER 4. Steel Design Tutorial Step 0 Step. 0- Updated the Design Sections Procedure Updated the Design Sections() Click on Update button. Properties Before Change represents the sections usedin the analysis. Properties After Change represents the sections used in the Design Change. Click on Select All Changed Properties button. 3 Click on <- button. 4 Click on Yes button. 3 4 Step 0 Step. 0- Updated the Design Sections Procedure Updated the Design Sections() Click on Re-analysis button. Click on Re-check button. Re-analyze the model. Re-do the steel code check. Final design results 0

93 CHAPTER 4. Steel Design Tutorial Step Step. - Change the Section with Low Ratio Procedure Change the Section with Low Ratio () Click on View Result Ratio button. Select ID:. 3 Ratio Limit: From 0, To Click on Show Graph of Result Ratio button. 5 Click on Select Elements button. 6 Click on Close button Step Step. - Change the Section with Low Ratio Procedure Change the Section with Low Ratio () Among the elements assigned with IPE600, the elements whose combined resistance ratio is less than 0.5 are changed into a smaller section IPE500. Drag & Drop : IPE500 into the Model View. Click on Yes button.

94 CHAPTER 4. Steel Design Tutorial Step Step. -3 Change the Section with Low Ratio Procedure Change the Section with Low Ratio (3) 3 Analysis > Perform Analysis Design > Steel Code Check 3 Click on View Result Ratio button. 4 Select ID:. 5 Ratio Limit: From 0, To 6 Click on Show Graph of Result Ratio button. 4 7 Click on Close button. 5 6 Combined resistance ratios of Section ID. are all above 0.5.