Made by SMH Date Aug Checked by NRB Date Dec Revised by MEB Date April 2006

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Edwin Crawford
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1 Job No. OSM 4 Sheet 1 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 DESIGN EXAMPLE 9 - BEAM WITH UNRESTRAINED COMPRESSION FLANGE Design a staircase support beam. The beam is a single section channel, simpl supported between columns. The flight of stairs lands between A and C and provides restraint to the top flange of this part of the beam. The top flange is unrestrained between B and C. The overall span of the beam is taken as 4, m. 1,5 m w 1,5 m 1, m Beam, m Down A C R A 1,5 m,7 m restrained unrestrained R B B Actions Assuming the beam carries the load from the first run of stairs to the landing onl: Permanent actions (G): Load on stairs 1,0 kn/m (1,0, ), kn/m Self weight of beam 0,1 kn/m Variable actions (Q): Load on stairs 4 kn/m (4,0,) 8,8 kn/m Load case to be considered (ultimate limit state): γ G, j G k, j + γ Q,1 Q k,1 + γ Q,i ψ 0, i Q k,i j 1 i>1 As there is onl one variable action (Q k,1 ) the last term in the above expression does not need to be considered in this example. γ G, j 1,5 (unfavourable effects) γ Q,1 1,5 Factored actions: Permanent action: Load on stairs 1,5,,97 kn/m Self weight of beam 1,5 0,1 0,17 kn/m Variable action Load on stairs 1,5 8,8 1, kn/m Eqn.. Section.. Structural analsis Reactions at support points R A + R B (,97 + 1,) 1,5 + 0,17 4, 4,97 kn 11

2 Job No. OSM 4 Sheet of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 Taking moments about A 1,5 1,17 0,75 + 0,17 4, (4,/ ) R B 4,9 kn 4, R A 4,97 4,9 0,8 kn Maximum bending moment occurs at a distance 1,5 1,5 1 1, m from A 4, M max 1, 0,8 1, 1,17 1, 0,17 1,58 knm Maximum shear occurs at A F Sd 0,8 kn Material properties Use material grade ,% proof stress 0 N/mm Table.1 Take f as the 0,% proof stress 0 N/mm Section..4 E N/mm and G N/mm Section..4 Tr a channel section, thickness 5 mm Cross section properties I 9,45 mm 4 W el, 94,5 mm I z 0,850 mm 4 W pl, 11,9 mm I w 5085 mm 4 A g 150 mm I t 1,7 4 mm 4 Classification of the cross-section ε 1,01 Table 4. Assume conservativel that c h t mm for web Web subject to bending: c Table 4. t 5 For Class 1, c t 5ε, therefore web is Class 1 Outstand flange subject to compression: c Table 4. t 5 1

3 Job No. OSM 4 Sheet of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 For Class, c t 11,9ε 1,0, therefore outstand flange is Class 4 Therefore, overall classification of cross-section is Class 4 Calculation of effective section properties Calculate reduction factor ρ for cold formed outstand elements 1 0,1 ρ but 1 Eq. 4.1b λ p λ p b / t λ p where b c 75mm Eq. 4. 8,4ε k σ Assuming uniform stress distribution within the, σ ψ 1 σ 1 Table 4.4 k σ 0,4 Table / 5 λ p 0,797 8,4 1,01 0,4 ρ 1 0,797 0,1 0,797 0,891 c eff 0,891 75,8 mm Table 4.4 A eff A ( 1 ρ) ct , mm g ( ) 5 Calculate shift of neutral axis of section under bending Non-effective zone - Centroidal axis of gross cross-section Centroidal axis of effective cross-section A g h 97,5 mm t ( 1 ρ ) c t h ( ) A eff ,

4 Job No. OSM 4 Sheet 4 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 Shift of neutral axis position, h 00 97,5, 47 mm Calculate I eff, I eff, I eff, I ( 1 ρ ) 1 ct ( 1 ρ) h t ct A eff 9,45 9,47 9,0 mm 4 I W eff, h + eff, - ( 1 0,891) ,0 88,4 mm 00 +,47 ( 1 0,891) 75 5 ( 0,5) Shear lag Section 4.4. Shear lag ma be neglected provided that b 0 L e /50 for outstand elements L e distance between points of zero moment 400 mm L e /50 84 mm, b 0 75 mm, therefore shear lag can be neglected Flange curling Section 4.4. u 4 σ b pren 199- a s 1-:004 E t z Clause 5.4() Eq. 5.a σ a average longitudinal stress in flange 0 N/mm (maximum possible value) b s (75 5) 70 mm z (0,5) 97,5 mm u ,5 4 0,04 mm Flange curling can be neglected if u < 0,05 00 mm Therefore flange curling is negligible pren :004 Clause 5.4(1) Partial safet factors The following partial safet factors are used throughout the design example: γ M0 1,1 and γ M1 1,1 Table.1 14

5 Job No. OSM 4 Sheet 5 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 Moment resistance of cross-section Section For a class 4 cross section M c,rd Weff, min f γ M 0 Eq. 4.9 M c,rd 88,4 0 17,7 knm 1,1 Design moment 1,58 knm, cross-section moment resistance is OK Cross-section resistance to shear Section V pl,rd v( f ) γ M0 A Eq. 4.0 A v h t mm V Rd ,5 kn 1,1 00 Design shear force 0,8 kn, therefore shear resistance of cross-section is OK Check that shear resistance is not limited b shear buckling Assume that h w h t mm h w ε 8, shear buckling resistance needs to be checked if 4,ε t 5 t η Section 5.4. Shear resistance is not limited b shear buckling. Resistance to lateral torsional buckling Section 5.4. Compression flange of beam is laterall unrestrained between B and C. Check this portion of beam for lateral torsional buckling. M b,rd χ W f γ for a Class 4 cross-section Eq. 5.8 LT eff, W eff, 88.4 mm χ LT ϕ LT 1 M1 1 0,5 [ ϕlt LT ] ( 1 α λ 0, + ) + λ ϕ LT,5 ( ) λ LT LT LT 4 LT 15 Eq λ Eq. 5. W f M cr Determine the elastic critical moment (M cr ) M cr 1/ π EI z k + ( ) z I w C1 g j g j kzl kw Iz π EIz ( kzl) GIt + ( C z C z ) ( C z C z ) Eq Appendix B Section B.1

6 Job No. OSM 4 Sheet of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 C is simpl supported, while B approaches full fixit. Assume most conservative case: k z k w 1,0. C 1, C and C are determined from consideration of bending moment diagram and end conditions. From bending moment diagram, ψ 0 C 1 1,77, C 0 and C 1,00 Table B.1 z j 0 for a cross-section with equal flanges M M cr cr π ,850 1,77 1,00 1,00 41,9 knm 88,4 λ LT ( 1,00 700) ,850 41,9 0 + ( 1,00 700) 0, ,7 π , ,5 Eq Using imperfection factor α LT 0,4 for cold formed sections Section 5.4. ( 0,8 ) ϕ,5 1 0,4( 0,8 0,4) χ LT ,779 0, [ 0,779 0,8 ] 0, 5 0,8 M b,rd 0, / 1,1 15, knm From bending moment diagram, maximum moment in unrestrained portion of beam 1,0 knm Thus member has adequate resistance to lateral torsional buckling. Deflection Section 5.4. Load case (serviceabilit limit state): G + Q + ψ Q Eq..8 k, j k,1 j 1 i 1 As there is onl one variable action (Q k,1 ) the last term in the above expression does not need to be considered in this example. Secant modulus is used for deflection calculations - thus it is necessar to find the maximum stress due to unfactored permanent and variable actions. 0, i k, i 1

7 Job No. OSM 4 Sheet 7 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 E The secant modulus E S S1 + E S, Where E Es, i and i 1, n E σ i, Ed, ser 1+ 0,00 σ i Ed ser f,, From structural analsis calculations the following were found: Maximum moment due to permanent actions Maximum moment due to imposed actions Total moment due to unfactored actions 1,90 knm,8 knm 8,58 knm Section is Class 4, therefore W eff is used in the calculations for maximum stress in the member. 17 Appendix C Assume, conservativel that the stress in the tension and are approximatel equal, i.e. E S1 E S. The following constants are used to determine the secant moduli: For grade stainless steel, n (longitudinal direction) 7,0 Table C.1 M max 8,58 Serviceabilit design stress, σ i, Ed, ser 97, 1 N/mm W 88,4 eff, E s, i N/mm ,1 1+ 0,00 97,1 0 Maximum deflection due to patch loading occurs at a distance of approximatel 1,9 m from support A. Deflection at a distance x from support A due to patch load extending a distance a from support A is given b the following formulae: 4 wal When x a δ n [ m m + m(4 + n ) n ] 4aE I S Where m x/l and n a/l When x 1,9 m, and a 1,5 m: m 1,9/4, 0,45, n 1,5/4, 0,57 Patch load (permanent+variable unfactored actions) w 11,0 kn/m Uniform load (permanent action) w 0,18 kn/m Deflection due to patch loads at a distance of 1,9 m from support A, δ ,5 400 δ ,0 0,57 7,09 mm 4 [ 0,45 0,45 + 0,45( 4 + 0,57 ) 0,57 ] Steel Designer s Manual (5 th Ed)

8 Job No. OSM 4 Sheet 8 of 8 Rev B Telephone: (0144) 45 Fax: (0144) 944 Made b SMH Date Aug 001 Checked b NRB Date Dec 001 Revised b MEB Date April 00 Deflection at midspan due to self weight of beam, δ 5 ( w L) L 5 (0,18 4,) 400 δ 84 E I ,0 S Total deflection δ + δ 7,09 + 0,9 7,8 mm span 400 δ limiting 1, 8 mm Therefore deflection is acceptabl small. 0,9 mm 18

Job No. Sheet 1 of 7 Rev A. Made by ER/EM Date Feb Checked by HB Date March 2006

Job No. Sheet 1 of 7 Rev A. Made by ER/EM Date Feb Checked by HB Date March 2006 Job No. Sheet of 7 Rev A Design Example Design of a lipped channel in a Made by ER/EM Date Feb 006 Checked by HB Date March 006 DESIGN EXAMPLE DESIGN OF A LIPPED CHANNEL IN AN EXPOSED FLOOR Design a simply