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

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Dwayne Stevens
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1 Job o. OS 466 Sheet of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 DESIG EXAPLE 0 - AXIALLY LOADED COLU I FIRE Design an unprotected rectangular hollow section subject to axial load and bending for 30 minutes fire resistance. The column length is,7 m and is subject to axial load from the end reaction of a floor beam at an eccentricit of 90 mm from the narrow face of the column. h Point of application of load z b z 90 mm Section A - A A A Floor beam,7 m Column Actions This eccentricit is taen to be 90 mm h/, where h is the depth of the section. Thus the beam introduces a bending moment about the column s major axis. The unfactored actions are: Permanent action: 6 Variable action: 7 The column will initiall be checed at the ultimate limit state (LC) and subsequentl at the fire limit state (LC) for a fire duration of 30 minutes. The loadcases are as follows: 69

2 Job o. OS 466 Sheet of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 LC (ultimate limit state) j G, j =,35 (unfavourable effects) Q, =,5 Q, Q Eqn..3, G, j G, LC (fire limit state) ψ j GA, jg, j,q, GA =,0 Values for ψ, are given in E 990 and A for E 990, but for this example conservativel assume ψ, =,0 j Section.3. Design at the Ultimate Limit State (LC) Loading on the corner column due to shear force at end of beam (LC): Axial force =,35 6,5 7 = 8,6 Tr Rectangular Hollow Section ajor axis bending moment (due to eccentricit of shear force from centroid of column),, = 8,6 (0,090,0/) =,60 m aterial properties Use material grade.440 0,% proof stress = 0 /mm and f u = 530 /mm Table 3. Tae f as the 0,% proof stress = 0 /mm Section 3..4 E = /mm and G = /mm Section 3..4 Cross-section properties 00 x 50 x 6 mm RHS W el, = 3, mm 3 i = 3,9 mm W pl, = 43, mm 3 i z = 9, mm A g = 500 mm t = 6 mm Classification of the cross-section ε =,0 Table 4. Assume conservativel that c = h t = 00 = 88 mm for web Webs subject to compression: For Class, c t 5,7ε c t = 88 6 = 4,7 = 5,96 Web is Class Table 4. 70

3 Job o. OS 466 Sheet 3 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 B inspection, if the web is Class subject to compression, then the flange will also be Class. Cross-section is Class Partial safet factors The following partial safet factors are used throughout the design example for LC: Table. 0 =, =, Bucling resistance to axial compression Section Resistance to flexural bucling about the z-z axis: χ z Ag f Eq. 5.a b,z,rd = For Class, and 3 cross-sections χ = reduction factor to account for bucling = ϕ = 0,5 ( α ( λ λ ) λ ) λ z = L cr λ z = L i cr z π 0 [ ϕ ] ϕ λ 0,5 Eq. 5.3 Eq. 5.4 f E = bucling length of column, taen conservativel as,0 column length =,7 m 700 9, π =,49 Eq. 5.5a For hollow sections subject to flexural bucling, α = 0,49 and λ 0 = 0,40 Table 5. (,49 ) ϕ =,5 0,49(,49 0,4) χ z = 0 =,88,88 [,88,49 ] 0,5 χ z = 0,3305 0, b,z,rd = = 99,5, (Resistance to torsional bucling will not be critical for a rectangular hollow section with a h/b ratio of.) = 8,6 Bucling resistance of cross-section is OK Section

4 Job o. OS 466 Sheet 4 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 Axial compression and bending resistance Cross sectional resistance interaction chec Section 4.7.6, e z, ez pre 993- c,rd c,,rd -3, Clause c,z,rd 6..9 A g f Eq. 4.5 c,rd = = = 300, 0 e = e Z = 0 z, = 0 W pl, f c,,rd = = c,rd 0, c,,rd = 43,75 0, Resistance of cross-section is OK 3 0 = 8,75 m 8,6,60 = 0,06 0,97 = 0,359 <, ,75 Eq. 4.7 Bucling resistance interaction chec Section 5.5., e ( b,rd ) min W, pl, / β W f β W, =,0 for Class cross-sections ( 0. ) =,0 λ 5 but b,rd,., b,rd, Determine b,rd, using the same method used to calculate b,rd,z given on sheet 3. Eq For hollow sections subject to flexural bucling, α = 0,49 and λ 0 = 0,40 Table 5. λ = L i f cr = π E 700 3,9 π ϕ =,5 0,49( 0,866 0,4) χ = ( 0,866 ) = 0,866 0 = 0,989 0,989 χ = 0,68 <,0 b,rd, = [ 0,989 0,866 ] 0,5 0, = 04,6, 7

5 Job o. OS 466 Sheet 5 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 ( 0. ),0 5 = λ b,rd, =,0 8,6 04,6 < Therefore, =, ( 0, ) =,07,, e ( b,rd ) min β W,Wpl, f / 8,6 99,5 6,60 0 0, = 0,88 0,97 = 0,485 <,0 3,0 43, /, Thus member is OK for combined axial loading and uniaxial moment under LC. Design at the Fire Limit State (LC) For LC, the column is designed for the following axial loads and moments. Axial compressive force, fi, =,0 6,0 7 = 3,0 aximum bending moment,fi, = 3,0 (0,090,05) =,8 m Determine temperature in steel after 30 minutes fire duration Section Assume that the section is unprotected and that there is a uniform temperature distribution within the steel section. Increase in temperature during time interval t is found from: Am V θ a,t = h& net,d t c Eq a ρa h & net,d = h & net,c h& Eq net, r α θ θ Eq h & = ( ) net,c c g a Where: θ g = gas temperature of the environment of the member in fire exposure, given b the nominal temperature time curve: θ g = 0 345log 0 (8t ) Eq θ a = surface temperature of the member ϕε 8 5,67 0 θ 4 73 θ 4 Eq h & = ( ) ( ) net,r [ ] res g a 73 Initial input values for determination of final steel temperature are as follows: A m /V = 00 m - α c = 5 W/m K 73

6 Job o. OS 466 Sheet 6 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 Initial steel temperature, θ a = 0 C Resultant emissivit, ε res = 0, Unit mass of stainless steel, ρ a = 7850 g/m 3 Configuration factor, ϕ =,0 The specific heat is temperature-dependent and is given b the following expression: c a = 450 0,8θ a,9 0-4 θ a, θ a J/gK Eq. 7.4 t = seconds The above formulae and initial input information were coded in an Excel spreadsheet and the following steel temperature, after a fire duration of 30 minutes, was obtained. θ a = 8 C Reduction of mechanical properties at elevated temperature The following reduction factors are required for calculation of resistance at elevated temperatures. Young s modulus retention factor E, θ = E θ /E 0,% proof strength retention factor 0,proof,θ = f 0,proof,θ /f Ultimate tensile strength retention factor u,θ = f u,θ /f u The value of the % ield strength at elevated temperature is also required for resistance calculations. This is given b the following expression: f g f f Eq. 7. f,θ = ( ) 0,proof, θ, θ u, θ 0,proof, θ The values for the retention factors at 8 C are obtained b linear interpolation. Table 7. 0,proof,θ = 0,377 u,θ = 0,3 E, θ = 0,60 g θ = 0,353 Thus f,θ = 0, ,353 (0, ,377 0) = 3,9 /mm,θ = 3,9/0 = 0,58 Partial safet factor,fi =,0 Section 7. Bucling resistance Section b,fi,t,rd = χ z,fi A g 0,proof,θ f /,fi Eq

7 Job o. OS 466 Sheet 7 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 χfi = θ θ ϕ ϕ λ θ ϕ θ = 0,5 ( ( λ θ λ 0 ) λ θ ) λ = [ ] z 0, 5 z,θ Eq. 7.0 but,0 α Eq. 7. λ =,49 (0,377/0,60)0,5 =,73 Eq. 7. 0,proof, θ E, θ For flexural bucling of a hollow section, α = 0,49 and λ 0 = 0, 4 Table 5. (.73 ) ϕ z,θ =,5 0.49( ) χ z,fi =,377 0 =,377,377,73 = 0,477 b,fi,t,rd = 0, ,377 0/,0 = 59,3 fi, = 3,0, bucling resistance of member is OK Axial compression and bending moment The following expression for a class cross section must be satisfied χ min,fi In which A g fi, 0,proof, θ f,fi,fi,,fi, θ,rd z z,fi, z,fi, θ,rd Eq. 7.4 µ fi, Eq. 7.8 = 3 f χ,fiag 0,,proof,θ,fi µ = (, 3) λ θ 0,44β 0,9 0, 8 β Eq. 7.9,,, λ = 0,866 Sheet 4 λ = [ ] 0, 5,θ λ = 0,866 (0,377/0,60)0,5 = 0,68 Eq. 7. 0,proof, θ E, θ Assume the column is fixed at the base, a triangular bending moment distribution occurs and β =,8 µ = (,,8 3) 0,68 0,44,8 0, 9 = 0,070 ( 0,68 ) ϕ,θ = 0,5 0,49( 0,68 0,4) χ,fi = 0,80 = 0,80 0,80 0,68 = 0,88 Table

8 Job o. OS 466 Sheet 8 of 8 Rev B Silwood Par, Ascot, Bers SL5 7Q Telephone: (0344) Fax: (0344) 6944 CALCULATIO SHEET Design Example 0 Axiall loaded column in fire ade b SH Date Aug 00 Checed b RB Date ov 00 Revised b EB Date April 006 ( 0,07) 3,0 0 = =,009 < 3,0 0 0, ,377,00 Interaction expression: χ min,fi Ag fi, 0,proof, θ f,fi 3,fi,,fi, θ,rd 0, θ Rd = 0,58 8,75 4,99 m, fi,0, fi, θ,rd =, = 3,0 0 0, , ,0,009, = 0,9 0,368 = 0,587 0,587 <,00 Thus section is OK in fire conditions for combined axial load and bending Eq

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

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