VALLOUREC & MANNESMANN TUBES. Designsupport for MSH sections


 Roxanne French
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1 VALLOUREC & MANNESMANN TUBES Designsupport for MSH sections according to Eurocode 3, DIN EN : 2005 and DIN EN : 2005
2 DesignSupport for MSH sections according to Eurocode 3, DIN EN : 2005 and DIN EN : 2005 in cooperation with Prof. Dr.Ing. R. Kindmann, Dr.Ing. M. Kraus and Dipl.Ing. J. Vette, University of Bochum, Dipl.Ing. O. Josat, Dipl.Ing. J. Krampen and Dipl.Ing. C. Remde, Vallourec & Mannesmann Tubes VALLOUREC & MANNESMANN TUBES is world market leader in the manufacture of seamless hot rolled steel tubes for all applications. The company operates 11 stateoftheart pipe mills worldwide, eight located in Europe (four plants at three locations in Germany and four plants in France), two at a facility in Brazil and one in the USA. With an annual output of up to three million tonnes the world s largest and most comprehensive range of seamless steel tubes is supplied. Hot rolled circular, square and rectangular Mannesmann Structural Hollow Sections of VALLOUREC & MANNESMANN TUBES have been used successfully for several decades. Modern steel architecture, with its elegant and transparent forms, would be practically impossible to create without them. 2 VALLOUREC & MANNESMANN TUBES
3 Descriptions and basic informations 4 Classification of hollow cross sections 5 Calculation methods/determination of internal forces 5 Resistance of cross sections 5 Buckling resistance of members 6 Designsupport for members in compression 8 Lattice girders 9 Joints of lattice girders 10 Designsupport for K gap joints with square MSHchords (SHS) 12 Designsupport for K joints with circular MSH sections (CHS) 13 Designsupport for K gap joints with rectangular MSHchords (RHS) 14 Designsupport for K overlap joints with square MSHchords (SHS) 16 Calculation examples 18 Circular MSH sections 22 Square MSH sections 24 Rectangular MSH sections 26
4 1 Descriptions and basic informations Table 1 Descriptions and available dimensions circular (CHS) square (SHS) rectangular (RHS) Cross section Outer measurement 21.3 mm 40 x 40 mm 50 x 30 mm d, b or h to to to 711 mm 400 x 400 mm 500 x 300 mm Wall thickness t 2.3 mm to 100 mm maximum 20 mm Available lengths up to 16 m; standard up to 12 m This brochure for the design only covers hotrolled MSH sections (according to DIN EN 10210). Due to production differences they provide more favourable characteristics than cold formed profiles: higher loadcarrying capacity for columns and members in compression larger cross section areas as a result of smaller corner radiuses substantially better suitability for welding in comparison to cold formed hollow profiles according to DIN EN there are no restrictions for the ability of welding (DIN EN :2005) Table 2 Materials: Yield strength fy, tensile strength fu, impact energy KV and carbon equivalent CEV Steel designation fy in N/mm 2 fu in N/mm 2 CEV* in % for KV* in J at DIN EN / test temp. t 16 mm 16 < t 40 mm EN old n. EN 1993 for t 40 mm S 355 J0H St 523U C: Structural S 355 J2H St 523N C: steels S 355 K2H C: Normalised S 355 NH StE 355 N C: fine grain S 355 NLH TStE 355 N C: structual S 460 NH StE 460 N C: steels S 460 NLH TStE 460 N C: S 690 approval in each individual case * according to DIN EN According to DIN EN the yield and tensile strengths f y and f u are either to be taken out of the product standard (DIN EN ) or simplified from DIN EN The values in table 2 correspond to the simplified specifications according to DIN EN for t 40 mm. The DIN EN demands a reduction of the yield strength for wall thicknesses > 16 mm already as well as different tensile strengths. The yield strength according to EC 3 specifies a nominal value for calculations, not the actual minimum value of the material. Detailed information and brochures are available at: 4 VALLOUREC & MANNESMANN TUBES
5 2 Classification of hollow cross sections By the classification of the cross sections the resistance and rotation capacity due to local buckling is supposed to be determined. Table 3 Classification on the bases of c/t and d/tratios of cross section parts subjected to compression Cross section Class Pure Pure compression bending 1 c/t 33 ε c/t 72 ε 2 c/t 38 ε c/t 83 ε Cross sections which do not comply to the conditions of the classes 1, 2 or 3 are classified as class 4. ε = 235/fy fy in N/mm 2 3 c/t 42 ε c/t 124 ε 1 d/t 50 ε 2 2 d/t 70 ε 2 3 d/t 90 ε 2 fy ε ε The tables of the sections 14 to 16 include details on the classification using the steel grade S 355. The first digit describes the classification for pure compression, the second for pure bending. 3 Calculation methods/determination of internal forces Internal forces can be determined using an elastic or a plastic structural analysis. A plastic analysis can only be performed, if the structure provides sufficient rotation capacity at the locations where plastic hinges occur. For the structural analysis the design values of the loading have to be taken into consideration, which means that partial safety factors γ F and combination factors ψ have to be regarded for the actions. As a result the design values of the internal forces N Ed, V Ed und M Ed. 4 Resistance of cross sections Tension: N Ed N pl/ γ 1.0 M0 Class: N Ed Compression: 1, 2 or 3 N pl/ γ 1.0 M0 N Ed A eff f y/ γ M0 M Ed Bending moment: 1 or 2 M pl/ γ 1.0 M0 M Ed W el f y/ γ M all 4 3 Partial safety factors: According to DIN EN γ M0 = γ M1 = 1.00 is recommended. The definition will be stated in the national annex, which is not yet available. Npl, Vpl and Mpl for fy = 35.5 kn/cm 2 : see tables in sections 14 to 16. For a different yield strength the values can be converted using the ratio of the strengths. Shear: V Ed V pl/ γ 1.0 M0 no shear buckling! Designsupport for MSH sections 5
6 Bending moment and shear force: The influence of the shear force on the bending moment resistance has to be taken into consideration if the shear force V Ed exceeds 0.5 V pl /γ M0. In that case a reduced yield strength has to be regarded for the parts of the cross section subjected to shear: 2 V red f y = (1 ρ) f y where ρ = 2 Ed 1 Bending moment and axial force: For classes 1 and 2 cross sections the following criterion should be satisfied: M Ed M N,Rd where M N,Rd is the design plastic moment resistance reduced by the axial force N Ed. For rectangular hollow sections the following approximation may be used: where: n = N Ed N pl/ γ M0 V pl/ γ M0 M M N,Rd = pl 1 n where: M N,Rd M pl γ M a w γ M0 2bt a w = 1 but a w 0.5 A For circular hollow cross sections EC 3 does not provide any specifications. Analogously the following condition is obtained according to Kindmann/Frickel Elastische und plastische Querschnittstragfähigkeit (Ernst & Sohn publishing, Berlin): N Ed 2 + arc sin M Ed 1.0 N pl/ γ M0 π M pl/ γ M0 5 Buckling resistance of members Buckling resistance of members in compression Uniform members with class 1, 2 and 3 sections shall be verified against buckling as follows: N Ed 1.0 χ N pl/ γ M1 1 χ = aber χ 1.0 Φ + Φ 2 λ 2 Φ = 0.5 [1+α (λ 0.2) + λ 2] λ = N pl L = cr f y ; Ncr = N cr i π E π 2 EI L 2 cr α = 0.21 for buckling curve a (S 235 to S 420) α = 0.13 for buckling curve a 0 (S 460) L cr : buckling length _ λ χ for curve a a _ λ χ for curve a a VALLOUREC & MANNESMANN TUBES
7 Buckling resistance of members in bending Procedure for the verification of sufficient carrying capacity: Application of equivalent geometric imperfections Determination of internal bending moments using a second order theory analysis taking the equivalent geometric imperfections into account (approximations see below) Verification of sufficient cross section carrying capacity according to section 4 for bending and axial force Equivalent geometric imperfections: a) Initial sway imperfections φ = 1/200 α h α m Reduction factor for height h [m] applicable to columns: 2 2 α h = but α h 1.0 h 3 Reduction factor for the number of columns in a row: α m = m m is the number of columns in a row including only those columns which carry a vertical load N Ed not less than 50% of the average value of the column in the vertical plane considered. Equivalent loading b) Initial bow imperfections According to DIN EN the initial bow imperfections are recommended as stated in the following table. The definition will be stated in the national annex, which is not available yet. e 0,d / L Buckling Elastic Plastic curve analysis analysis a 0 1/350 1/300 a 1/300 1/250 Equivalent loading Approximations for the bending moment according to second order theory: For the approximation the first order theory bending moment is multiplied by an enlargement factor α: Bending moment according to first order theory and correction factors δ for selected cases Hog. moment: M I =  q L 2 /2 δ = Hog. moment: M I = P L δ = Hog. moment: M I =  N Ed φ L δ = M II α M I 1+δ N Ed /N cr,d α= 1 N Ed /N cr,d Sag. moment: M I = q L 2 /8 δ = Sag. moment: M I = P L/4 δ = Sag. moment: M I = N Ed e 0,d δ = 0 N cr,d = N cr /γ Μ1 Condition: α 3 L = beam length Hog. moment: M I =  q L 2 /8 δ = Sag. moment at: 5/8 L: M I = 9 q L 2 /128 δ = Hog. moment: M I =  3 PL/16 δ = Sag. moment: M I = 5 PL/32 δ = Online verification against buckling at (STACOM) Hog. moment: M I  N Ed e 0,d δ = Sag. moment at: 5/8 L: M I 0.6 N Ed e 0,d δ = Designsupport for MSH sections 7
8 6 Designsupport for members in compression Example 1, P.18 Example 5, P VALLOUREC & MANNESMANN TUBES
9 7 Lattice girders Lattice structures are often designed as single span girders with parallel chords, as for example roof girders of long span constructions. Established structures in practice are:  Warren truss  small amount of work due few joints  long diagonals in compression  few points of load application at the upper chord  Warren truss with vertical posts  Pratt truss  many points of load application at the upper chord  many joints  extensive joints at lower chord and large amount of work  short diagonals in compression  many points of load application at upper chord  many joints and therefore large amount of work Designsupport for MSH sections 9
10 Compared to plate girders lattice girders with parallel chords using hollow members provide economic advantages for span lengths greater than about 20 m. The splitup of the internal bending moment into tension and compression forces leads to light roof constructions saving material. Additionally hollow profiles provide an ideal cross section shape for centrical compression loading. Support for the construction:  Lattice girders with parallel chords usually have heights from 1/10 to 1/20 of the system lengths. As a guide value for predesign a girder height of 1/15 of the span length is appropriate.  The angles between the chords and the brace members should be inbetween 45 and 60. In any case an angle greater than 30 has to be chosen.  Joints of lattice structures should be designed in a way that the centrelines of the members intersect in a single point. In the case they show eccentricities chapter of DIN EN :2005 has to be regarded (see section 8).  Loadings, as for instance from purlins, should be introduced at the joints of the structure.  Moments at the joints, caused by the actual rotational stiffness of the connections, may be neglected in the design of the members and joints, provided that the range of validity for the joints are observed and the ratio of the system length to the heights of the members is not less than 6. The cross section carrying capacity has to be verified for each member and for those in compression the stability as well. For connections the design joint resistance according to DIN EN :2005 has to be verified. 8 Joints of lattice girders Each member of a lattice structure is usually loaded by axial forces for which they have to be designed. At joints, several members come together and the directions of forceactions have to be redistributed in order to fulfil the equilibrium. Joints are highly stressed details of the structure for which the design joint resistance has to be determined and verified. Usually hollow profiles are welded at the connections; welds have to be verified separately however. The ends of brace members may not be flattened or pressed together. The following types of joints are often used: K gap joint K overlap joint K joint with vertical post (N joint) KT joint 10 VALLOUREC & MANNESMANN TUBES
11 The following supports for design may be applied for mainly stationary loading. Moments resulting from eccentricities of the centrelines have to be regarded in the design of tension chord members and brace members as well as the design of the connections if the following limits are not fulfilled: CHS: 0.55 d 0 e 0.25 d 0 resp. RHS/SHS: 0.55 h 0 e 0.25 h 0 In the design of compression chord members eccentricities usually have to be taken into account, even if they are within the limits specified above. According to DIN EN a partial safety factor of γ M5 = 1.00 is recommended for the design of the joints. The definition will be stated in the national annex, which is not available yet. Definition and Notation Table 4 Range of validity for K and N gap joints according to DIN EN :2005 Rectangular MSH sections (RHS): b i /b 0 Max b 0 /t 0 2. b i /t i 35 und h i /t i b 0 /t 0 35 und h 0 /t h 0 /b h i /b i b g Max 0 (1 (h 1 + b 1 + h 2 + b 2 )/(4 b 0 )) t i + t 2 7. If g 1.5 b 0 1 (h 1 + b 1 + h 2 + b 2 ) the joint has to 4 b 0 be treated as two separate Y and T joints 8. Class 2 sections in terms of pure bending at least (see section 2) 9. Θi 30 Table 5 Range of validity for K and N overlap joints according to DIN EN :2005 Circular MSH sections (CHS): d i /d d 0 /t d i /t i Θi Class 2 sections in terms of pure bending at least (see section 2) 6. g t 1 +t 2 Square MSH sections (SHS): Points 19 see RHS b 0 /t 0 35 b 1 + b b 1 Rectangular and square MSH sections (RHP/QHP): 1. b i /b Chord: 0.5 h 0 /b and class 2 section in terms of pure bending at least 3. Braces (in compression): class 1 sections (pure bending) 4. Braces (in tension): b i /t i 35 and h i /t i % λ 0V 100% and b i /b j Θi 30 Circular MSH sections (CHS): Points 15 see table 4 6. λ 0V 25% Designsupport for MSH sections 11
12 9 Designsupport for K gap joints with square MSHchords (SHS) Preconditions: Joint has to be within the range of validity given in table 4! Same yield strength fy for all members Design resistance of the joint n Rd k n N pl,0 N 1,Rd = sin θ 1 γ M5 N 2,Rd = N 1,Rd sin θ 1 sin θ 2 n Rd see diagram below Note: For circular braces the design resistances N i,rd have to be multiplied by the factor π /4. For that case b i =d i is valid. k n is obtained for compression chords by: N o,d γ M5 2 b 0 k n = N pl,0 b 1 + b 2 for tension chords by: k n = 1.0 sin Θ is given by: θ sin θ Example:  Chord member SHS 150x150x6.3 mm (tension)  Brace members SHS 80x80x5 mm  Gradient of the brace members 45 (e = 0 cm) Check of the validity given in table 4: 1. b i /b 0 = 8/15 = b i /t i =h i /t i = 80/5 = g = 15 = 3.7 cm Max 15 = 3.5 cm sin cm 8 7. g = 15 sin 45 = 3.7 cm (1 8/15) = 10.5 cm 8. Members are at least class 2 sections 9. Θ 1 = Θ 2 = b 0 /t 0 = 150/6.3 = b 1 + b b = = Determination of the design resistance: b 1 8 b 0 15 With b = = and = = 23.8: n Rd t The limit brace force for a tension chord is: N 1,Rd = = kn 12 VALLOUREC & MANNESMANN TUBES
13 10 Designsupport for K joints with circular MSH sections (CHS) Preconditions: Joint has to be within the range of validity given in tables 4 and 5! Same yield strength f y for all members Conditions for design: n Rd k p N pl,0 sin θ N 1,Rd Min i γ M5 d i k Θi N pl,0 (d 0 t 0 ) γ M5 N i,d n Rd see diagram below (Interim values may be interpolated) Notes: g<0: overlap joint (table 5) g>0: gap joint (table 4) where i = 1 Comp. 2 Tension k p is obtained for compression chords by: minn o,d γ M5 minn o,d γ M5 k p = (1+ ) 1.0 N pl,0 N pl,0 for tension chords by: k p = 1.0 k Θi is obtained by: k Θi = 1+sin Θ i 12 sin 2 Θ i Θ i k Θi Example:  Chord member: x 6.3 mm (tension)  Brace members: 60.3 x 5mm  Gradient of the brace members: 45 (e = 0 cm) Check of validity given in table 4: d i /d 0 = 60.3/101.6 = d 0 /t 0 = 101.6/6.3 = d i /t i = 60.3/5 = Θ 1 = Θ 2 = Members are at least class 2 sections g = = 1.6 cm t 1 + t 2 = 1.0 cm sin 45 Designsupport for MSH sections Determination of the design resistance: d 1 d 0 g with = 0.594, = 16.1 and = 2.54 and by d 0 t 0 t 0 interpolation: n Rd 0.32 (g/t 0 4), n Rd 0.26 (g/t 0 8) n Rd ( ) (8 2.54)/(4+8) = The limit brace force for a tension chord is: = N Rd = = 272 kn Min = ( )
14 11 Designsupport for K gap joints with rectangular MSHchords (RHS) Preconditions: Joint has to be within the range of validity given in table 4! Brace members are equal profiles SHS/CHS Same yield strength f y for all members N o,d The compression force in the chord shall satisfy: γ N M5 0.5 pl,0 (For small b i /b 0 ratios and small compression forces or tension forces in the chord the exact verification according to Eurocode can lead to favourable results) Note: For circular braces the design resistances N i,rd have to be multiplied by the factor π /4. For that case b i =d i is valid. Conditions for design: 1. Condition (Verification of the chord force) V Ed = N 1,d sin Θ 1 V pl,0 γ M5 h 0 k α = 0.2 b n 0,Rd = 1 1 N i,rd = Min 1 v 2 Ed k α n N 0,Rd = 0,Rd N pl,0 >N 0,d γ M5 2. Condition (Verification of the brace force) n 1,Rd N pl,i γ M5 > N i,d n 1,Rd and n 2,Rd see diagrams below Diagram for n 1,Rd : n 2,Rd N pl,0 sin Θ i γ M5 14 VALLOUREC & MANNESMANN TUBES
15 Diagrams for n 2,Rd : Designsupport for MSH sections 15
16 12 Designsupport for K overlap joints with square MSHchords (SHS) Preconditions: Joint has to be within the range of validity given in table 5! Brace members are equal profiles SHS/CHS Same yield strength f y for all members Note: For circular braces the design resistancese N i,rd have to be multiplied by the factor π / 4. For that case b i =d i is valid. Design resistance of the joint: N 1,Rd = n Rd N pl,1 γ M5 The overlapping brace member 2 does not have to be verified. Determination n Rd taking into account: λ 0V = q/p 100% 25% (see table 5) With the overlap ratio λ 0V the following cases can be distinguished: Case 1: λ 0V = 25% Case 2: 25% < λ 0V < 50% linear interpolation of n Rd using case 1 and case 3 Case 3: 50% λ 0V < 80% Case 4: 80% λ 0V 100% Diagrams for n Rd in case 1 (λ 0V = 25%): 16 VALLOUREC & MANNESMANN TUBES
17 Diagrams for n Rd in case 3 (50% < λ 0V < 80%): Diagrams for n Rd in case 4 (80% λ 0V 100%): Designsupport for MSH sections 17
18 13 Calculation examples Example 1: Column with hinged supports (L = 6 m): By considering L cr = L = 6 m the following can directly be taken from the designsupport of section 6: CHS x 16 N Ed 6250 kn > 6000 kn In comparison to that the verification against buckling according to sect. 5 is carried out: I = cm 4, N pl = 6966 kn (see section 14), buckling length: L cr = 600 cm (see section 5) π N cr = 2 EI π = = kn; N pl 6966 λ = = = 0.57 L 2 cr N cr χ 0.9 (buckling curve a, see section 5) N Ed 6000 Verification against buckling: = = 0.96 < 1.0 χ N pl /γ M /1.0 Example 2: Column with fixed support (L = 4 m) and lateral loading: The column shown in the figure has to be verified for stability. Due to the lateral loading the verification will be achieved by a second order theory analysis (l y = 9055 cm 4 ). The buckling length of the column is L cr = 2L = 8 m. π 2 EI π N cr = L 2 = cr γ M = kn 1.0 The bending moment at support due to the uniformly distributed load is (see sect. 5): 4 M 2 q l = 15 = 120 knm 2 With δ = 0.40 from the table of section 5 the enlargement factor α is determined as follows: 1+δ N Ed /N cr.d / α = = = N Ed /N cr,d / The second order theory moment is gained with the multiplication of the moment at support by the enlargement factor α (see section 5). M ll q = Ml q α = ( 120) 1.48 = knm 1 The initial sway imperfection is regarded with φ = (for the application of equiv. geo. imperfections see section 5) H 0 = φ N ED = = 6.5 kn; Mϕ l = H l = = 26 knm With δ = 0,18 (see section 5): / α = = ; M ll ϕ = ( 26) 1.65 = 42.9 knm / The internal forces according to the second order theory analysis may only be superposed, if the axial compression force of the loading conditions is identical (limited superposition). In this example the axial compression force is regarded with N d = 1300 kn in combination with the uniformly distributed load q d as well as the sway imperfection φ. MEd ll = Mll q + Mll ϕ = = knm The bending moment resistance is determined according to section 4 with regard of the axial compression force (since V/V pl is smaller than 0.5, a reduction of the yield strength is not necessary): N Ed b t n = = = 0.386; a w = 1 = 1 = N pl /γ M0 3370/1.0 A M pl 1 n ,386 M N,Rd = = = knm MEd ll = knm γ M a w Condition satisfied! 18 VALLOUREC & MANNESMANN TUBES
19 Example 3: K gap joint with RHSchord and SHSbrace members Check of validity: (see section 8) decisive 1. b i /b 0 = = 0.35 Max = b i /ti = h i /t i = 7/0.5 = b 0 /t 0 = 20/1 = und h 0 /t 0 = 10/1 = h 0 /b 0 = 10/20 = h i /b i = 7/7 = = 6.5 cm decisive g = tan 40 sin 40 = 6.99 cm Max = 1.0 cm 7. g = 6.99 cm = 19.5 cm Members are at least class 2 sections 9. Θ 1 = Θ 2 = Check of eccentricities: e h = = Condition satisfied, which means, that the moments resulting from eccentricities may be neglected in the design of the connection. The tension chord does not have to be verified for the moments of eccentricities as well. Determination of the design resistance: 1. Condition (Verification of the chord force) Determination of the plastic shear resistance according to DIN EN :2005: A h 0 f y,d V pl,0,d = b = = kn 0 + h sin With v ed = = 0.599, k α = = 0.45 and n 0,Rd = = N 0,Rd can be determined according to section 11 as follows: 1950 N 0,Rd = = kn N 0,d = 950 kn Condition satisfied! 2. Condition (Verification of the brace force) The following ratios are necessary for the use of the diagrams: t i 5 b i 7 b 0 20 h 0 10 b i 7 t = = 0.5; = = 0.35; = = 20; = = 0.5; = = b 0 20 t 0 1 b 0 20 t i 0.5 With the diagrams n 1,Rd 1,0 and for n 2,Rd is provided. The maximum brace force can be determined as follows: N i,rd = , sin = kn = kn decisive N i,rd = kn N i,d = 350 kn Condition satisfied! Designsupport for MSH sections 19
20 Example 4: K overlap joint with SHS Check of validity: (see table 5) b 1 b b = = = b 0 14 h 0 2. = 1.0 and class 1 cross section b 0 b i h i 3. and 4. = = and class 1 cross section t i t i 4.7 x % b 5. λ 0V = = 45% and i and b = % j 6. Θ i >30 Determination of the design resistance: For λ 0V = 0.45 case 2 is decisive. n Rd has to be interpolated using the cases 1 and 3. With the ratios t i 4 b 1 80 b t = = , = = 0.57 and = = 22.2 the diagrams provide: b t Check of eccentricities: Using the eccentricity e the ratio e/h n Rd, n Rd, can be determined (see section 8): Interpolation: e = 3.68 cm (45 25) n Rd = ( ) = 0.76 N 1,Rd = = 323 kn Verification of the design joint resistance: N 1,Ed 300 = = 0.93 N 1,Rd 323 e = = h 0 14 Condition satisfied, which means, that the moments resulting from eccentricities may be neglected in the design of the connection. For the design of the compression chord the eccentricities have to be regarded. Example 5: Lattice girder L = 40 m Rectangular hollow sections made of S 355 J2H (hot rolled). The example deals with a roof girder designed as a warren truss. The single span girder is loaded by the selfweight, snow and suction forces due to wind. The loads are combined according to DIN EN 1990:2002. Afterwards the verifications against buckling and for the joint resistances are carried out. Loading table: Load case Dead load G Snow S Wind (suction) Ws P kn 7.8 kn 6.9 kn P kn 15.6 kn 13.8 kn (The dead load includes the selfweight of the girder.) Decisive load combinations according to DIN EN 1990:2002: LCC 1: 1.35 * G * S LCC 2: 0.9 * G * Ws Internal forces LCC 1: LCC 2: Load at each node: P 1,d = = 50 kn P 1,d = 3 kn P 2,d = P 1,d /2 = 25 kn P 2,d = 1.5 kn Bearing reaction: A d = = 200 kn A d = = 12 kn max. force in upper chord: (200 25) ( ) O M = 2.5 = 775 kn O M = 46.5 kn max. force in lower chord: U M = /2 = 800 kn U M = 48 kn max. force in braces: D 1 = ± (200 25) 2 = ± kn D 1 = ± 14.9 kn 20 VALLOUREC & MANNESMANN TUBES
21 Buckling of the upper chord member (SHS 140 x 8): The nodes of the upper chords are fixed laterally. Buckling length for hollow section chord members according to Annex BB.1.3 of DIN EN : L cr = = 450 cm. Using the designsupport of section 6 the required cross section can be chosen. With. N Ed = 775 kn and L cr = 4.5 m follows a hollow profile 140 x 140 x 8: I = 1195 cm 4, N pl = 1475 kn (see section 15). π N cr = 2 EI π = = 1223 kn; N λ = pl 1475 = = 1.1 χ (buckling curve a, see sect. 5) L 2 cr N cr 1223 N Ed 775 Verification against buckling for O M : χ N = /1.0 = 0.9 < 1.0 pl /γ M1 Buckling of the lower chord member (SHS 140 x 6.3) : Perpendicular to the girderplane the lower chord is fixed at the bearings and in midspan by structural elements. Buckling length for hollow section chord members according to Annex BB.1.3 of DIN EN : L cr = = 1800 cm π 2 EI π ,9 N pl 1181 N cr = L 2 = cr = 62.9 kn; λ = = N cr = 4.33 χ (buckling curve a, see sect. 5) 62.9 N Ed 46.5 Verification against buckling for U M : χ N pl /γ = M /1.0 = 0.77 < 1.0 Buckling of the brace members (SHS 80 x 5): The brace members are welded continuously to the chord members. Since b i = 0.8 cm < 0.6 b 0 = = 8.4 cm and the connection to the chords is presumed as continuously welded the buckling length for the brace members according to Annex BB.1.3 of DIN EN is: L cr = = 265 cm This length is valid for inplane as well as outofplane buckling of the brace members. π 2 EI π N pl 523 N cr = L 2 = cr = 403 kn; λ = = N cr = 1.14 χ (buckling curve a, see sect. 5) 403 N Ed Verification against buckling for D 1 : χ N pl /γ = = 0.84 < 1.0 M /1.0 Design joint resistance of node 1: 1. b i /b 0 = 8/14 = b i /t i = h i /t i = 8/0.5 = and b 0 /t 0 = 14/0.63 = and 5. h i /b i = 1 (QHP) = 3.0 cm 6. and 7. g = 3.1 cm 8 where e = 2 mm = 9.0 cm 14 (The eccentricity is within the range according to section 8) 8. Members are at least class 2 sections 9. Θ 1 = Θ 2 = b 1 + b b = = Determination of N i,rd : b 1 + b 2 With k n = 1.0 (tension chord) and 2 b = 0.57 the diagram in section 9 provides: n Rd 0.20 N i,rd = = 334 kn N Verification of the design joint resistance: i,d N = = 0.74 < 1.0 i,rd 334 Designsupport for MSH sections 21
22 14 Circular MSH sections according DIN EN Dimensions Bending and Torsion fy = 35.5 kn/cm 2 d t A G U l = l T /2 W el i max S N pl V pl M pl Class mm mm cm 2 kg/m m 2 /m cm 4 cm 3 cm cm 3 kn kn knm S For further available dimensions see MSH technical information 1 22 VALLOUREC & MANNESMANN TUBES
23 Dimensions Bending and Torsion fy = 35.5 kn/cm 2 d t A G U l = l T /2 W el i max S N pl V pl M pl Class mm mm cm 2 kg/m m 2 /m cm 4 cm 3 cm cm 3 kn kn knm S For further available dimensions see MSH technical information 1 Designsupport for MSH sections 23
24 15 Square MSH sections according DIN EN Dimensions Bending fy = 35.5 kn/cm 2 b x b t A G U ly = lz W el iy = iz l T N pl V pl M pl Class mm mm cm 2 kg/m m 2 /m cm 4 cm 3 cm cm 4 kn kn knm S x x x x x x x x x x x For further available dimensions see MSH technical information 1 24 VALLOUREC & MANNESMANN TUBES
25 Dimensions Bending fy = 35.5 kn/cm 2 b x b t A G U ly = lz W el iy = iz l T N pl V pl M pl Class mm mm cm 2 kg/m m 2 /m cm 4 cm 3 cm cm 4 kn kn knm S x x x x x x x x x For further available dimensions see MSH technical information 1 Designsupport for MSH sections 25
26 16 Rectangular MSH sections according DIN EN Dimensions Bending strong and weak axis fy = 35.5 kn/cm 2 h x b t A G ly W el.y i y lz W el.z i z l T N pl Mpl.y M pl.z Class mm mm cm 2 kg/m cm 4 cm 3 cm cm 4 cm 3 cm cm 4 kn knm knm S x x x x x x x x x x x x x For further available dimensions see MSH technical information 1 26 VALLOUREC & MANNESMANN TUBES
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