Made by PTY/AAT Date Jan 2006

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1 Job No. VALCOSS Sheet of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 DESIGN EXAMPLE 3 - HOLLOW SECTION LATTICE GIRDER The lattice girder supports roof glazing and is made of square and rectangular hollow sections of grade.430 stainless steel; a comparison is made between material in two strength levels - the annealed condition (f y 0 N/mm ) and in the cold worked condition (strength level CP460, f y 460 N/mm ). Calculations are performed at the ultimate limit state and then at the fire limit state for a fire duration of 30 minutes. For the CP460 material the reduction factors for the mechanical properties at elevated temperatures are conservatively taken as those for grade.438 C850 (Table 7.). The structural analysis was carried out using the FE-program WINRAMI marketed by Finnish Constructional Steelwork Association (FCSA) ( The WINRAMI design environment includes square, rectangular and circular hollow sections for stainless steel structural analysis. WINRAMI solves the member forces, deflections and member resistances for room temperature and structural fire design and also joint resistance at room temperature (it also checks all the geometrical restraints of truss girder joints). In the example, the chord members are modelled as continuous beams and the diagonal members as hinge jointed. According to EN 993--, the buckling lengths for the chord and diagonal members could be taken as 0,9 times and 0,75 times the distance between nodal points respectivel but in this example conservatively the distance between nodal points has been used as the buckling length. The member forces were calculated by using WINRAMI with profile sizes based on the annealed strength condition. These member forces were used for both the annealed and CP460 girders. This example focuses on checking 3 members: mainly axial tension loaded lower chord (member 0), axial compression loaded diagonal (member 3) and combination of axial compression and bending loaded upper chord member (member 5). The weight of the girders is also compared. The welded joints should be designed according to the Section 6.3, which is not included in this example. lower chord 00x60x4, upper chord 80x80x5, corner vertical 60x60x5 diagonals from left to middle: 50x50x3, 50x50x3, 40x40x3, 40x40x3, 40x40x3,40x40x3, 40x40x3. lower chord 60x40x4, upper chord 70x70x4, corner vertical 60x60x5, all diagonals 40x40x3. Span length 5m, height in the middle 3,3 m, height at the corner 0,5 m. Weight of girders: Annealed: 407 kg, CP kg. The weight is not fully optimized. 9

2 Job No. VALCOSS Sheet of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 Actions Assuming the girder carries equally distributed snow load, glazing and its support structures and weight of girder : Permanent actions (G): Load of glazing and supports kn/m Dead load of girder (WINRAMI calculates the weight) Variable actions (Q): Snow load kn/m Load case to be considered (ultimate limit state): γ Load case to be considered (fire situation): Ultimate limit state (room temperature design) γ G, j,35 (unfavourable effects) γ Q,,5 j γ j G, j G k, j + γ Q, Q k, GA, j G k, j + k, Fire design γ GA, j,0 γ ψ, 0, γ ψ, Q (Recommended partial safety factors for actions shall be used in this example) Factored actions for ultimate limit state: Permanent action: Load on nodal points:,35 x 4, kn Self weight of girder (is included by WINRAMI ) Variable action Load from snow:,5 x 8, kn Eq..3 EN 990 EN 99-- Forces at critical member are: Forces are determined by the model using profiles in the annealed strength condition Lower chord member, member 0 Annealed: 00x60x4 mm, CP460: 60x40x4 mm N t, 4, kn, N t,fi, 46,9 kn M max, 0,67 knm, M max,fire, 0,45 knm Upper chord member, member 5 Annealed: 80x80x5 mm, CP460: 70x70x4 mm N c, -49, kn, N c,fire, -49, kn M max,,49 knm, M max,fire, 0,73 knm Diagonal member, member 3 Annealed: 50x50x3mm, N c, -65,9 kn, CP460: 40x40x3 mm N c,fire, -,7 kn 9

3 Job No. VALCOSS Sheet 3 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 Material properties Use material grade.430 Annealed: f y 0 N/mm f u 550 N/mm E N/mm Table 3. CP460: f y 460 N/mm f u 650 N/mm E N/mm Section 3..4 Partial safety factors Table. The following partial safety factors are used throughout the design example: γ M0,, γ M,, γ M,fi,0 Cross section properties Annealed Member 0: A 75 mm W pl,y 37, mm 3 Member 5: A436 mm I y 3, mm 4 i y 30,3 mm W pl,y 39, mm 3 Member 3: A54 mm I y 9, mm 4 i y 9 mm W pl,y 9, mm 3 CP460 Member 0: A 695 mm W pl,y 3,6 000 mm 3 Member 5: A05 mm I y 7, 0 4 mm 4 i y 6,7 mm W pl,y 4, mm 3 Classification of the cross-section of member 5 and member 3 ε,0 ε 0,698 Table 4. Assume conservatively that c h t Annealed 80x80x5 : c mm CP460 70x70x4 : c mm Annealed 50x50x3 : c mm CP460 40x40x3 : c mm Flange/web subject to compression: Table 4. Annealed 80x80x5 : c/t 4 CP460 70x70x4 : c/t 5,5 Annealed 50x50x3 : c/t 4,6 CP460 40x40x3 : c/t,3 For Class, c t 5, 7ε, therefore both profiles are classified as Class Member 3: A4 mm I y 9,3 0 4 mm 4 i y 4,9 mm W pl,y 5,7 0 3 mm 3 LOWER CHORD MEMBER, DESIGN IN ROOM AND FIRE TEMPERATURE (Member 0) A) Room temperature design Tension resistance of cross section Section 4.7. N pl,rd Ag f y γ Eq M N pl,rd 75 mm 0 N/mm /, 35 kn > 4, kn OK. N pl,rd 695 mm 460 N/mm /, 90 kn > 4, kn OK. 93

4 Job No. VALCOSS Sheet 4 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 Moment resistance of cross-section Sec M c,rd Wpl f y γ Eq M M c,rd M c,rd 3 37,93 0 0, 0 3,6 0 3, ,58 knm > 0,67 knm OK. 5,50 knm > 0,67 knm OK. Axial tension and bending moment interaction N M Eq N M Rd Rd 4, kn 0,67 knm + 0,69 35 kn 7,58 knm OK. 4, kn 0,67 knm + 0,6 90 kn 5,50 knm OK. B) Fire temperature design ε res 0, Section Steel temperature after 30 min fire Θ 83 C f, f 0,proof, + g, (f u, f 0,proof, ) Eq. 7. k 0, proof, 0,7-3/00 (0,7-0,4) 0,40 f 0,proof, 0,40 0 N/mm 5,8 N/mm g, (0,35-3/00 (0,35 0,38) 0,357 f u, ( 0,7-3/00 (0,7-0,5)) 550 N/mm 33,3 N/mm f, 5,8 N/mm + 0,357 (33,3 5,8) N/mm 8,5 N/mm k, 8,5/0 0,37 k 0,proof, 0,3-3/00 (0,3-0,) 0,0 f 0,proof, 0,0 460 N/mm 93, N/mm g, 0,5 f u, ( 0,4-3/00 (0,4-0,0)) 650 N/mm 35, N/mm f, 93, N/mm + 0,5 (35, 93,) N/mm 03,6 N/mm k, 03,6 /460 0,5 Section 7. Table 7. Eq. 7. Section 7. Table 7. Eq. 7. Tension resistance of cross section N fi,,rd k, N Rd [γ M0 / γ M,fi ] Eq. 7.6 N fi,,rd 0, kn,/,0 95,6 kn > 46,9 kn OK. N fi,,rd 0,5 90 kn,/,0 59,3 kn > 46,9 kn OK. 94

5 Job No. VALCOSS Sheet 5 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 Moment resistance of cross-section M fi,,rd k, M Rd [ γ M 0 / γ M, fi ] Eq. 7.3 M fi,,rd 0,370 7,58 knm,/,0 3,08 knm > 0,45 knm OK. M fi,,rd 0,5 5,50 knm,/,0,36 knm > 0,45 knm OK. Axial tension and bending moment interaction N M + N M Rd Rd 46,9 kn 0,45 knm + 0,57 95,6 kn 3,08 knm OK. 46,9 kn 0,45 knm + 0,97 59,3 kn,36 knm OK. Eq DIAGONAL MEMBER DESIGN IN ROOM AND FIRE TEMPERATURE (Member 3) Buckling length 53 mm A) Room temperature design N b,rd χ Af y / γ M Eq. 5.a Lcr 53 Eq. 5.5a λ ( f y / E) (0/ 00000) 0,696 i π 9 π ϕ 0,5( + α( λ λ 0 ) + λ ) 0,5(+0,49(0,696-0,4)+0,696 ) 0,85 Eq. 5.4 χ ϕ + ( ϕ λ ) 0,85 + (0,85 0,696 ) 0,807 Eq. 5.3 N b,rd 0, mm 0 N/mm /, 87,3 kn > 65,9 kn OK. Lcr 53 λ ( f y / E) (460 / 00000),83 i π 4,9 π ϕ 0,5( + α( λ λ0) + λ ) 0,5(+0,49(,83-0,4)+,83 ),540 Eq. 5.5a Eq. 5.4 χ ϕ + ( ϕ λ ),540 + (,540,83 ) 0,48 Eq. 5.3 N b,rd 0,48 4 mm 460 N/mm /, 73,6 kn > 65,9 kn OK. 95

6 Job No. VALCOSS Sheet 6 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 B) Fire temperature design ε res 0, Steel temperature after 30 min fire Θ 83 C k 0,proof, 0,7-3/00 (0,7-0,4) 0,8 k E, 0,63-3/00 (0,63-0,45) 0,57 k 0,proof, 0,3-3/00 (0,3-0,) 0,9 k E, 0,5-3/00 (0,5-0,35) 0,465 Section Section 7. Table 7. Section 7. Table 7. N b,fi,t,rd χ fi A k0, proof, f y /γ M, fi Eq. 7.8 λ λ k / k ) 0,696 (0,8/ 0,57) 0,439 Eq. 7. ( 0, proof, E, 0 ϕ 0,5( + α( λ λ ) + λ ) 0,5(+0,49(0,439-0,4)+0,439 ) 0,606 χ fi ϕ + ( ϕ λ ) 0,606 + (0,606 0,977 0,439 ) N b,fi,t,rd 0, mm 0,8 0 N/mm /,0 6,5 kn >,7 kn OK. Eq. 7. Eq. 7.0 λ λ k / k ),83 (0,9/ 0,465) 0,8 Eq. 7. ( 0, proof, E, 0 ϕ 0,5( + α( λ λ ) + λ ) 0,5(+0,49(0,8-0,4)+0,8 ) 0,94 χ fi ϕ + ( ϕ λ ) 0,94+ (0,94 0,74 0,8 ) N b,fi,t,rd 0,74 4 mm 0,9 460 N/mm /,0 6,4 kn >,7 kn OK. Eq. 7. Eq. 7.0 UPPER CHORD MEMBER DESIGN IN ROOM AND FIRE TEMPERATURE (Member 5) Buckling length 536 mm A) Room temperature design N M + N e + k y N, ) b Rd min βw, yw pl, y f y / γ Ny ( M Eq. 5.40,0 β W,y,0 class cross section Sec k y,0+(λ y -0,5)N /N b,rd,y, but, k y,+n /N b,rd,y Lcr 536 λ ( f y / E) (0 / 00000) 0,535 i π 30,3 π Eq. 5.5a 96

7 Job No. VALCOSS Sheet 7 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 ϕ 0,5( + α( λ λ 0 ) + λ ) 0,5(+0,49(0,535-0,4)+0,535 ) 0,676 χ ϕ + ( ϕ λ ) 0,676 + (0,676 Eq ,97 Eq ,535 ) N b,rd,y 0, mm 0 N/mm /, 63,3 kn > 49, kn Eq. 5.a k y,0+(0,535-0,5) 49,/63,3,039 because calculated value is less than,, the value of k y,. 49,, , 0,890 <,0 OK. 63,3 3 Eq. 5.40,0 39,74 0 0/, CP460 β W,y,0 class cross section Sec Lcr 536 Eq. 5.5a λ ( f y / E) (460 / 00000) 0,878 i π 6,7 π ϕ 0,5( + α( λ λ 0) + λ ) 0,5(+0,49(0,878-0,4)+0,878 ),00 Eq. 5.4 χ ϕ + ( ϕ λ ),00 + (,00 0,673 0,878 ) Eq. 5.3 N b,rd,y 0, mm 460 N/mm /, 85,6 kn > 49, kn Eq. 5.a k y,0+(0,878-0,5)49,/85,6,394, but, k y, + (49,/85,6),44, thus k y,394 49,, ,394 0,8 <,0 OK. 85,6 3 Eq. 5.40,0 4, /, B) Fire temperature design ε res 0, Steel temperature 80x80x5 mm Θ 80 C Steel temperature 70x70x4 mm Θ 83 C k 0,proof, 0,7-0/00 (0,7-0,4) 0,57 f 0,proof, 0,57 0 N/mm 56,5 N/mm g, (0,35-0/00 (0,35 0,38) 0,353 f u, (0,7-0/00 (0,7-0,5)) 550 N/mm 4,9 N/mm f, 56,5N/mm + 0,353 (4,9 56,5) N/mm 86,6 N/mm k, 86,6 /0 0,394 k E, 0,63 0/00 (0,63-0,45) 0,6 Section Section 7. Table 7. Eq

8 Job No. VALCOSS Sheet 8 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 k 0,proof, 0,3-3/00 (0,3-0,) 0,0 f 0,proof, 0,0 460 N/mm 93, N/mm g, 0,5 f u, (0,4-3/00 (0,4-0,0)) 650 N/mm 35,N/mm f, 93, N/mm + 0,5 (35, 93,) N/mm 03,6 N/mm k, 03,6 /460 0,5 k E, 0,5 3/00 (0,5-0,35) 0,48 RFCS Checked by MAP Date Feb 006 Section 7. Table 7. Eq. 7. χ min, fi A g N k fi, y 0, proof, γ M, fi f k ym + M fi, fi,, Rd,0 Eq. 7.4 λ λ ( k 0, proof, / k E, ) 0,535 (0,57 / 0,6) 0,347 Eq. 7. ϕ 0,5( + α( λ λ 0 ) + λ ) 0,5(+0,49(0,347-0,4)+0,347 ) 0,547 χ fi ϕ + ( ϕ λ ) 0,547 + (0,547,03,0 0,347 ) Eq. 7. Eq. 7.0 µ y N fi, k y 3 Eq. 7.8 χ A k f / γ fi g 0, proof, y M, fi Eq. 7.9 µ,β 3) λ + 0,44β 0,9 0,8 y ( M, y M, y χmin, fi A gk0, proof, f y / γ,0 436 mm 0,57 0 N/mm /,0 8, kn M, fi Eq. 7.8 > 49, kn OK. M fi,,rd k, [γ M0/ γ M,fi ]M Rd 0,394,/,0 39, /000 3,79 knm Eq. 7.3 >0,73 knm OK. ψ -0,487 knm/0,73 knm -0,666 Table 7.3 β M,y,8-0,7 ψ,466 µ y (,,466-3)0, ,44,466 0,9 0,78, Eq. 7.9 because calculated value is less than 0,8, the value of µ y 0,8 k y - 0,80 49, kn/8, kn 0,55 Eq , 0,73 + 0,55 0,70 <,0 8, 3,79 OK. Eq. 7.4 λ λ ( k 0, proof, / k E, ) 0,878 (0,0 / 0,48) 0,569 Eq ϕ 0,5( + α( λ λ ) + λ ) 0,5(+0,49(0,569-0,4)+0,569 ) 0,703 χ fi ϕ + ( ϕ λ ) 0,703 + (0,703 0,896 0,569 ) Eq. 7. Eq

9 Job No. VALCOSS Sheet 9 of 9 Rev A P.O. Box 000, FI-0044 VTT Tel Fax Design Example 3 Stainless steel lattice girder made Made by PTY/AAT Date Jan 006 RFCS Checked by MAP Date Feb 006 χmin, fi A gk0, proof, f y / γ 0, mm 0,0 460 N/mm /,084,5 kn Eq. 7.8 M, fi >49, kn OK. M fi,,rd k, [γ M0/ γ M,fi ]M Rd 0,5,/,0 4, /000,8 knm Eq. 7.3 >0,73 knm OK. ψ -0,487 knm/0,73 knm -0,666 Table 7.3 β M,y,8-0,7 ψ,466 µ y (,,466-3)0, ,44,466 0,9 0,77, Eq. 7.9 because calculated value is less than 0,8, the value of µ y 0,8 k y - 0,80 49, kn/84,5 kn 0,534 Eq , 0,73 + 0,534 0,7 <,0 OK. Eq ,5,8 99

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