PART 5a : Worked examples Example to EN 1991 Part 1-2: Compartment fire

Transcription

1 PART 5a : Worked examples Example to EN 1991 Part 1-: Compartment fire P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK The gas temperature of a fully engulfed fire in an office has to be determined. The room of the Simulated Office test of the Cardington building is chosen for this analysis. The measured temperatures during the fully engulfed fire are shown in Figure 3, so the calculation can be compared with these results. A natural fire model is chosen for the calculation of the gas temperature. For fires with a flash-over, the method of the compartment fires can be used. A simple calculation method for a parametric temperature-time curve is given in Annex A of EN Figure 1. Cardington building (left) and the office of the Simulated Office test (right) Floor area: A f = 135 m² Total area of vertical openings: A v = 7 m² Vertical opening factor: α v = 0. Horizontal opening factor: α h = 0.0 Height: H = 4.0 m Average window height: Lightweight concrete: h eq = 1.8 m (assumption) ρ = 1900 kg/m² c = 840 J/kgK λ = 1.0 W/mK 5a-1

2 DETERMINATION OF FIRE LOAD DENSITY EN For the determination of the fire load density the Annex E of EN offers a calculation model. The design value of the load density may either be given from a national fire load classification of occupancies and/or specific for an individual project by performing a fire load evaluation. At this example, the second method is chosen. q = q m δ δ δ Annex E.1 f, d f, k q1 q n where: m the combustion factor δ q1 the factor considering the danger of fire activation by size of the compartment δ q the factor considering the fire activation risk due to the type of occupancy δ n the factor considering the different active fire fighting measures The fire load consisted of 0 % plastics, 11 % paper and 69 % wood, so it consisted mainly of cellulosic material. Therefore the combustion factor is: m = 0.8 The factor δ q1 considers the danger of fire activation by size of the compartment, as given in Table 1. Table 1. Fire activation risk due to the size of the compartment (see EN , Table E.1) Compartment floor area A f [m²] ,000 Danger of fire activation δ q δ q1 = 1.5 A factor δ q considers the fire activation risk due to the type of occupancy, as given in Table. Table. Fire activation risk due to the type of occupancy (see EN , Table E.1) Danger of fire activation δ q Examples of occupancies 0.78 artgallery, museum, swimming pool 1.00 offices, residence, hotel, paper industry 1. manufactory for machinery & engines 1.44 chemical laboratory, painting workshop 1.66 manufactory for fireworks or paints δ q = 1.5 5a-

3 The factor taking the different active fire fighting measures into account is calculated to: δ 10 n = i= 1 The factors δ ni are given in Table 3. δ ni Table 3. Factors δ ni (see EN , Table E.) δ ni function of active fire fighting measures Automatic fire Automatic water extinguishing suppression system Automatic fire detection Manual fire suppression Independent water supplies Automatic fire detection & alarm Automatic alarm transmission to fire brigade δ n δ n δ n3 By heat or 0.87 δ n4 by smoke 0.73 δ n Work Fire Brigade δ n Off Site Fire Brigade δ n or Safe access routes δ n8 1.0 or 1.5 Fire fighting devices δ n9 1.0 or 1.5 Smoke exhaust system δ n or 1.5 δ n = = 0.50 For calculating the characteristic fire load, the characteristic fire load has to be determined. It is defined as: Qfi, k = Mki, Hui ψ i Annex E. where: M k,i the amount of combustible material [kg] H ui the net calorific value [MJ/kg], see EN , Table E.3 ψ i the optional factor for assessing protected fire loads The total fire loading was equivalent to 46 kg wood/m², so the characteristic fire load is: ( ) Q fi, k = = 108, 675 MJ The characteristic fire load density is determined to: qf, k= Qfi, k Af= 108, = 805 MJ/m² The design value of the fire load density is calculated to: q f d, = = MJ/m² 5a-3

4 3 CALCULATION OF THE PARAMETRIC TEMPERATURE-TIME CURVE Annex A It has to be determined if the fully engulfed fire is fuel or ventilation controlled. For this, the opening factor and the design value of the fire load density related to the total surface are needed. and O= heq Av At = = m 0. qtd, = qf, d Af A t= = 137.6MJ m The determination, if the fire is fuel or ventilation controlled is: 3 3 qtd, O tlim = = 0.36 h > = h The fire is ventilation controlled For calculation of the temperature-time curves for the heating and the cooling phase, the b factor is needed. This factor considers the thermal absorptivity for the boundary of enclosure. The density, the specific heat and the thermal conductivity of the boundary may be taken at ambient temperature. The floor, the slab and the walls are made of lightweight concrete b ρ c λ J ms = = = 1 The temperature-time curve in the heating phase is given by: θ g K t* 1.7 t* 19 t* ( ) = e 0.04 e 0.47 e Because the fire is ventilation controlled, the time t* is calculated to: where: t* = t Γ ( Ob) ( ) ( ) ( ) Γ= = = Now the heating phase can be calculated: 0. ( 3.04 t) 1.7 ( 3.04 t) 19 ( 3.04 t) θ g = ( e 0.04 e 0.47 e ) For calculation of the cooling phase, the maximum temperature is needed. where: max max max ( 0. t * 1.7 t * 19 t e e e * ) θmax = t * max t max = Γ The time t max is determined as below, where t lim is given in Table 4. t max qtd, O= = 0.36 h = max tlim = h 5a-4

5 Table 4. Time t lim for different fire growth rates Slow fire growth Medium fire growth Fast fire growth rate rate rate t lim [h] So t* max is calculated to: t * max = = 1.10 h The maximum temperature is calculated to: ( e e 1.10 e ) θmax = = C During the cooling phase, t* and t* max are calculated to: [ ] t* = t Γ= t 3.04 h 3 ( td ) t* = q O Γ= 1.10h max, The temperature-time curve in the cooling phase for t* max 0.5 is given by: θ ( max ) ( t ) g = θmax 65 t* t* x where: t max > t lim x = 1.0 = Combination of the heating and cooling curves leads to the parametric temperaturetime curve shown in Figure. Figure. Gas temperature of the office calculated by using the parametric temperaturetime curve of the office 5a-5

6 4 COMPARISON BETWEEN CALCULATION AND FIRE TEST To compare the calculation with the measured temperatures in the test, the factors δ 1, δ and δ ni for calculation of the fire load density have to be set to 1.0 (see Figure 3). Figure 3. Comparison of measured and calculated temperature-time curves REFERENCES EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 The Behaviour of multi-storey steel framed buildings in fire, Moorgate: British Steel plc, Swinden Technology Centre, 1998 Valorisation Project: Natural Fire Safety Concept, Sponsored by ECSC, June 001 5a-6

7 Example to EN 1991 Part 1-: Localised fire P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK The steel temperature of a beam has to be determined. It is part of an underground car park below the shopping mall Auchan in Luxembourg. The beams of the car park are accomplished without any use of fire protection material. The most severe fire scenario is a burning car in the middle of the beam (see Figure 1). For getting the steel temperature, the natural fire model of a localised fire is used. Figure 1. Underground car park of the shopping mall Auchan Figure. Static system and cross-section of the beam 5a-7

8 Diameter of the fire: D =.0 m Vertical distance between fire source and ceiling: H =.7 m Horizontal distance between beam and flame axis: r = 0.0 m Emissivity of the fire: ε f = 1.0 Configuration factor: Φ = 1.0 Stephan Boltzmann constant: σ = W/m K 4 Coefficient of the heat transfer: α c = 5.0 W/m²K Steel profile: IPE 550 Section factor: A m /V = 140 1/m Unit mass: ρ a = 7850 kg/m³ Surface emissivity: ε m = 0.7 Correction factor: k sh = 1.0 RATE OF HEAT RELEASE ECSC Project The rate of heat release is normally determined by using the EN Section E.4. For dimensioning the beams at this car park, the rate of heat release for one car is taken from an ECSC project called "Development of design rules for steel structures subjected to natural fires in CLOSED CAR PARKS" (see Figure 3). Figure 3. Rate of heat release of one car 3 CALCULATION OF THE STEEL TEMPERATURES EN Calculation of the flame length Annex C First of all, the flame length has to be determined. 5 5 L = 1.0 D Q = Q f 5a-8

9 A plot of this function with the values of Figure 3 is shown in Figure 4. With a ceiling height of.80 m, the flame is impacting the ceiling at a time from 16.9 min to 35.3 min (see Figure 4). Figure 4. Flame length of the localised fire It is important to know, if the flame is impacting the ceiling or not, because different calculation methods for the calculation of the net heat flux are used for these two cases (see Figure 5). Figure 5. Flame models: Flame is not impacting the ceiling (A); Flame is impacting the ceiling (B) 3. Calculation of the net heat flux st case: The flame is not impacting the ceiling The net heat flux is calculated according to Section 3.1 of EN h& net = αc ( θ( ) θm ) +Φ εm ε f σ ( ( ) 73) ( 73 z θ + θ z m + ) 4 8 = 5.0 ( θ( ) θm) ( θ( ) 73) ( θm 73 z + + z ) 4 4 Section 3.1 5a-9

10 The gas temperature is calculated to: θ 3 53 = C ( ) ( Q) ( z z z 0 ) 3 5 ( Q) ( Q ) 53 = C where: z is the height along the flame axis (.7 m) z 0 is the virtual origin of the axis [m] z0 = 1.0 D Q = Q 5 5 Annex C 3.. nd case: The flame is impacting the ceiling The net heat flux, if the flame is impacting the ceiling, is given by: h& ( 4 4 ) ( θm 4 4 ) ( 0) ( 73) ( 93) = h& α θ Φ ε ε σ θ + net c m m f m 8 ( θ ) ( ) ( ) = h& m The heat flux depends on the parameter y. For different dimensions of y, different equations for determination of the heat flux have to be used. if y 0.30: h & =100,000 if 0.30 < y < 1.0: if y 1.0: where: h& = 136,300 11,000 y h& = 15, y r+ H + z'.7 + z' y = = L + H + z' L z' h The horizontal flame length is calculated to: where: h * * ( ( ) ) ( ) ( ) L =.9 H Q H = 7.83 Q.7 h H H ( ) ( ) Q = Q H = Q * H The vertical position of the virtual heat source is determined to: if Q D * < 1.0: if Q D * 1.0: where: * * * * (( D ) ( D ) ) ( D ) ( D ) ( ) z' =.4 D Q Q = 4.8 Q Q * * ( ( D ) ) ( D ) ( ) 5 5 z' =.4 D 1.0 Q = Q ( ) ( ) Q = Q D = Q * D 5a-10

11 3.3 Calculation of the steel temperature-time curve EN The specific heat of the steel c a is needed to calculate the steel temperature. The parameter is given by EN , Section depending on the steel temperature. Section Figure 6. Specific heat of carbon steel (see EN 1993 Part 1-, Figure 3.4) Am V 4 θa, t = θm + ksh h& net t = θm h& net Section c ρ a a The steel temperature-time curve is shown in Figure 6. Additionally, the results of the FEM-analysis done by PROFILARBED are shown for comparison. Figure 7. Comparison of the temperature-time curve of the calculation and the FEManalysis of PROFILARBED REFERENCES EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1993, Eurocode 3: Design of steel structures Part 1-: General rules Structural fire design, Brussels: CEN ECSC Project, Development of design rules for steel structures subjected to natural fires in CLOSED CAR PARKS, CEC agreement 710-SA/11/318/518/60/933, Brussels, June a-11

12 Example to EN 1993 Part 1-: Column with axial loads P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK In the following example, a column of a department store will be dimensioned for fire resistance. The column is part of a braced frame and is connected bending resistant to the upper and lower column. The length is 3.0 m. During fire exposure, the buckling length can be reduced as seen below in Figure 1. The loads are centric axial compression forces. The column is exposed to fire on four sides. A hollow encasement of gypsum is chosen for fire protection. The required standard fire resistance class for the column is R 90. Figure 1. Buckling lengths of columns in braced frames Figure. Cross-section of the column Material properties: Column: Profile: rolled section HE 300 B Steel grade: S 35 Cross-section class: 1 Yield stress: f y = 3.5 kn/cm² Cross-sectional area: A a = 149 cm² 5a-1

13 Elastic modulus: E a = 1,000 kn/cm² Moment of inertia: I a = 8560 cm 4 (weak axis) Encasement: Material: gypsum Thickness: d p = 3.0 cm (hollow encasement) Thermal conductivity: λ p = 0. W/(m K) Specific heat: c p = 1700 J/(kg K) Density: ρ p = 945 kg/m³ Loads: Permanent actions: G k = 100 kn Variable actions: P k = 600 kn FIRE RESISTANCE OF COLUMN.1 Mechanical actions during fire exposure EN The accidental situation is used for the combination of mechanical actions during fire exposure. ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i The combination factor for department stores is ψ,1 = 0.6. So the axial load is determined to: N fi, d = = 1560 kn. Calculation of the maximum steel temperature EN The analysis of EN is used to calculate the steel temperature of the hollow encased column. For a hollow encased member, the section factor is calculated to: ( ) -1 A V = ( b+ h) A = = 81 m Section p a By using the Euro-Nomogram (ECCS No.89), the maximal temperature θ a,max,90 of the steel bar is: 3 ( Ap V) ( p dp) = = λ W m K ECCS No.89 θ a,max, C.3 Verification in the temperature domain EN Within EN the verification in the temperature domain is not allowed for members in which stability phenomena have to be taken into account. Section Verification in the strength domain The verification in the strength domain during fire exposure is carried out as a plastic ultimate state of the load-carrying capacity. E R Section.4. fi, dt, fidt,, In this example, the verification has to be done with the axial forces. N N fi, d b, fitrd,, The design resistance under high temperature conditions is calculated as: 5a-13

14 f Nb, fi, t, Rd= χ fi Aa ky, θ,max γ y M, fi In dependence of θ a,max,90 the reduction factors k y,θ and k E,θ are given in Table 3.1 of the EN For intermediate values of the steel temperature, linear interpolation may be used. Section k y,445 C = Section 3..1 k E,445 C = The load-carrying capacity is determined in consideration of the non-dimensional slenderness during fire exposure. λfi, θ = λ ky, θ ke, θ = = 0.5 Section where: EN ( i ) ( ) ( ) λ = L λ = = 0.1 Section Kz z a With the non-dimensional slenderness the reduction factor for flexural buckling χ fi,θ can be calculated. EN where: and: χ fi 1 1 = = = 0.86 ϕ + ϕ - λ ϕ = + α λ + λ = + + = α = f y = = 0.65 The design resistance arises to: N Verification: b, fi, t, Rd 3.5 = = 713 kn 1.0 N fi, d N b, fi, t, Rd = = 0.58 < 1 Section REFERENCES ECCS No.89, Euro-Nomogram, Brussels: ECCS Technical Committee 3 Fire Safety of Steel Structures, 1995 EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1993, Eurocode 3: Design of steel structures Part 1-1: General rules, Brussels: CEN, May 00 EN 1993, Eurocode 3: Design of steel structures Part 1-: General rules Structural fire design, Brussels: CEN, November 003 5a-14

15 Example to EN 1993 Part 1-: Beam with bending and compression loads P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK This example deals with a beam subjected to a uniform load, which causes bending moment, and an axial load. Stability phenomena have to be considered. The beam is part of an office building. A hollow encasement of gypsum is chosen for fire protection. Due to a concrete slab the beam is exposed to fire on three sides. There is no shear-connection between the beam and the slab. The required standard fire resistance class for the beam is R 90. Figure 1. Static system Figure. Beam Cross-section Material properties: Beam: Profile: rolled section HE 00 B Steel grade: S 35 Cross-section class: 1 Yield stress: f y = 35 N/mm² Elastic modulus: E = 10,000 N/mm² Shear modulus: G = 81,000 N/mm² Cross-sectional area: A a = 7810 mm² Moment of inertia: I z = 000 cm 4 Torsion constant: I t = 59.3 cm 4 5a-15

16 Warping constant: I w = 171,100 cm 6 Section moduli: W el,y = 570 cm² W pl,y = 64.5 cm³ Encasement: Material: gypsum Thickness: d p = 0 mm (hollow encasement) Thermal conductivity: λ p = 0. W/(m K) Specific heat: c p = 1700 J/(kg K) Density: ρ p = 945 kg/m³ Loads: Permanent Loads: G k = 96.3 kn g k = 1.5 kn/m Variable Loads: p k = 1.5 kn/m FIRE RESISTANCE OF BEAM WITH BENDING AND COMPRESSION LOADS.1 Mechanical actions during fire exposure EN The combination of mechanical actions during fire exposure shall be calculated as an accidental situation: ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i The combination factor for office buildings is ψ,1 = 0.3. The design loads under high temperature conditions are: N fi, d = 96.3 kn 10.0 M fi, d = [ ] = 4.38 knm 8. Calculation of steel temperatures pren The steel temperature is given by the Euro-Nomogram (ECCS No.89). Therefore the section factor A p /V is needed. For a hollow encased member exposed to fire on three sides, the section factor is: Ap h+ b m -1 = = = Section V A 78.1 With A V p a λp 0. W = 77 = 770 d the critical temperature arises to: θ a,max, C p m K, ECCS No.89 5a-16

17 .3 Verification in the temperature domain pren Due to section 4..4 () of pren the verification in the temperature domain may not be accomplished, because of stability problems of the beam. Section Verification in the strength domain Members with a Class 1 cross-section should be analysed for the problem of flexural buckling and of lateral torsional buckling..4.1 Flexural buckling The verification for flexural buckling is: N fi, d ky M y, fi, d + χ A k f γ W k f γ min, fi y, θ y M, fi pl, y y, θ y M, fi 1 Section The reduction factor χ min,fi is the minimum of the two reduction factors for flexural buckling χ y,fi and χ z,fi. The non-dimensional slenderness for the temperature θ a is needed for the calculation of these reduction factors. For calculation of the non-dimensional slenderness in the fire situation, the non-dimensional slenderness at ambient temperatures have to be determined. pren Lcr 1000 λy = 1.5 i λ = = y a Lcr 1000 λz =.10 i λ = = z a Section The needed reduction factors k y,θ and k E,θ are given in pren Table 3.1: pren k y,θ = Section 3..1 k E,θ = With the reduction factors, the non-dimensional slenderness in the fire situation can be determined: λ y, θ k y, θ = λy = 1.5 = 1.46 Section k E, θ With and λ z, θ k y, θ = λz =.1 =.44 k E, θ α = f y = = 0.65 ( ) ( ) 1 1 ϕy, θ = 1+ α λy, θ + λy, θ = =.04, ( ) ( ) 1 1 ϕz, θ = 1+ α λz, θ + λz, θ = = 4.7 5a-17

18 the reduction factors χ y,fi and χ z,fi can be calculated: χ χ y, fi z, fi 1 1 = = = 0.9 ϕ ϕ λ y, θ + y, θ y, θ = = = 0.13 ϕ ϕ λ z, θ + z, θ z, θ Verification: = 0.94 < where: ( ) µ = 1. β 3 λ β 0.9 y M, y y, θ M, y ( ) = = 1.8 Section k y µ y N fi, d = 1 = 1 = 1.33 χ A f γ y, fi a y m, fi.4. Lateral torsional buckling The second verification deals with the problem of lateral torsional buckling. N fi, d klt M y, fi, d + 1 χ A k f γ χ W k f γ z, fi y, θ y M, fi LT, fi pl, y y, θ y M, fi For calculation of the non-dimensional slenderness in the fire situation, the nondimensional slenderness at ambient temperatures has to be determined pren where: λ LT Wpl, y fy = = = 1.05 Section M 14,40.4 cr π E I ( ) z k I k L G I w t M cr = C1 + + ( C zg) C z g ( k L) kw Iz π E I z π 1, = 1.1 ( ) ( ) , π 1, = 14,40.4 kncm Section C.. During fire exposure, the non-dimensional slenderness changes to: pren With λ LT, θ k y, θ = λlt = 1.0 = 1.19 Section k E, θ 5a-18

19 ( ) ( ) 1 1 φlt, θ = 1+ α λlt, θ + λlt, θ = = 1.59, the reduction factor χ LT,fi is calculated to: χ LT, fi 1 1 = = = 0.38 φ φ λ LT, θ + LT, θ LT, θ Verification: Section where: = = k LT µ LT N fi, d = = = 0.0 χ A k f γ z, fi y, θ y M, fi µ = 0.15 λ β 0.15 < 0.9 LT z, θ M, LT = = 0.33 < 0.9 REFERENCES ECCS No.89, Euro-Nomogram, Brussels: ECCS Technical Committee 3 Fire Safety of Steel Structures, 1995 EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN pren 1993, Eurocode 3: Design of steel structures Part 1-1: General rules, Brussels: CEN, May 00 pren 1993, Eurocode 3: Design of steel structures Part 1-: General rules Structural fire design, Brussels: CEN 5a-19

20 Example to EN 1993 Part 1-: Beam made of a hollow section P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK At this example, a beam made of a welded hollow section has to be dimensioned. It is part of a hall roof structure. The length of the beam is 35.0 m and the beams are arranged at a distance of 10.0 m. It is charged with uniform loads and is restrained against lateral evasion. The beam is executed without any use of fire protection material. The required standard fire resistance class for the tensile bar is R 30. Figure 1. Static system Figure. Cross-section Material properties: Steel grade: S 355 Yield stress: f y = 355 N/mm² Height: h = 700 mm Height of web: h w = 650 mm Width: b = 450 mm Thickness of flange: t f = 5 mm Thickness of web: t w = 5 mm Cross-sectional area of the flange: A f = 11,50 mm² 5a-0

21 Cross-sectional area of the web: A w = 16,50 mm² Specific heat: c a = 600 J/(kg K) Density: ρ a = 7850 kg/m³ Emissivity of the beam: ε m = 0.7 Emissivity of the fire: ε r = 1.0 Configuration factor Φ = 1.0 Coefficient of the heat transfer: α c = 5.0 W/m²K Stephan Boltzmann constant: σ = W/m²K 4 Loads: Permanent actions: Beam: g a,k = 4.3 kn/m Roof: g r,k = 5.0 kn/m Variable actions: Snow: p s,k = 11.5 kn/m FIRE RESISTANCE OF BEAM MADE OF A HOLLOW SECTION.1 Mechanical actions during fire exposure EN The accidental situation is used for the combination of mechanical actions during fire exposure. ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i The combination factor for snow loads is ψ,1 = 0.0. With this parameter, the design bending load is calculated to: 35.0 M fi, d = ( ) = knm 8. Calculation of the steel temperature EN The temperature increase of the steel section is calculated to: Section A V 40 θ = k h& t = h& = h& m 5 a, t sh net, d net net ca ρa where: k sh correction factor for the shadow effect (k sh = 1.0) t time interval ( t = 5 seconds) A m /V section factor for the unprotected beam A V = 1 t = = 40 1 m m The net heat flux is calculated according to EN 1991 Part 1-. EN ( ) 4 4 ( ) ( ) ( 73) ( 73) h& = α θ θ +Φ ε ε σ θ + θ + net c g m m r g m 8 ( θg θ ) m ( θg ) ( θm ) = The standard temperature-time curve is used for getting the gas temperatures. 10 ( t ) Section 3.1 θ = log Section 3..1 g 5a-1

22 The steel temperature-time curve of the hollow section is shown in Figure 3: Figure 3. Steel temperature-time curve of the hollow section θ a,max,30 = 646 C.3 Verification in the temperature domain EN The design moment resistance during fire exposure at the time t = 0 is needed to get the utilization factor. where: and: M = W f k θ γ Section fi, Rd,0 pl y y,,max M, fi 1.0 = 1,875, = knm 6 k y,θ,max = 1.0 for θ = 0 C at the time t = 0 γ M,fi = 1.0 w w w W = A h pl + A h t f = 16, , = 1,875,000 mm The utilization factor is calculated to: µ 0 = E, R,,0 = M, M,,0 = = 0.31 Section 4..4 fi d fi d fi d fi Rd The critical temperature θ a,cr is given in Table 4.1 of the EN 1993 Part 1-. Verification: θ a,cr = 659 C = < 5a-

23 .4 Verification in the strength domain To calculate the moment resistance the reduction factor k y,θ has to be determined for the temperature θ a,max,30 = 646 C. This factor is given in Table 3.1 of the EN 1993 Part 1-: k y,θ = Section 3..1 Additionally, the adaptation factors κ 1 and κ have to be determined. The adaptation factor κ 1 considers the non-uniform temperature distribution across the cross-section. Table 1. Adaptation factor κ 1 Section κ 1 [-] Beam exposed on all four sides 1.0 Unprotected beam exposed on three sides with a composite or concrete slab on side four 0.7 Protected beam exposed on three sides with a composite or concrete slab on side four 0.85 The beam in this is an unprotected beam exposed to fire on four sides. Therefore κ 1 is set to: κ 1 = 1.0 The adaptation factor κ considers the non-uniform temperature distribution along a beam. Table. Adaptation factor κ κ [-] At the supports of a statically indeterminate beam 0.85 In all other cases 1.0 The verification is done in the middle of the beam and it is statically determinate. So the adaptation factor κ is set to: κ = 1.0 Therefore the design moment resistance is calculated to: γ M,1 1 M fitrd,, = M plrd,,0 C ky, θ γ κ κ Verification: M, fi = ( ) = ,87, knm = < REFERENCES EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1993, Eurocode 3: Design of steel structures Part 1-: General rules Structural fire design, Brussels: CEN 5a-3

24 Example to EN 1994 Part 1-: Composite slab P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK A composite slab has to be dimensioned in the fire situation. It is part of a shopping centre and the span is 4.8 m. The slab will be dimensioned as a series of simply supported beams. The required standard fire resistance class for the slab is R 90. Figure 1. Static system Figure. Steel sheet Material properties Steel sheet: Yield stress: f yp = 350 N/mm² Cross-sectional area: A p = 156 mm²/m Parameters for m+k method: k = N/mm² Concrete: Strength category: C 5/30 Compression strength: f c = 5 N/mm² Height: h t = 140 mm Cross-sectional area: A c = 131,600 mm²/m 5a-4

25 Scope of application for re-entrant profiles [mm] Existing geometrical parameters [mm] 77.0 l l 1 = l l = l l 3 = h h 1 = h 70.0 h = 51.0 Loads: Permanent loads: Steel sheet g p,k = 0.13 kn/m² Concrete: g c,k = 3.9 kn/m² Finishing load: g f,k = 1. kn/m² Variable loads: Live load: p k = 5.0 kn/m² Design sagging moment at ambient temperatures: M s,d = knm FIRE RESISTANCE OF A COMPOSITE SLAB The composite slab has to be verified according to Section 4.3 and Annex D..1 Geometrical parameters and scope of application Figure 3. Geometry of cross-section h 1 = 89 mm h = 51 mm l 1 = 115 mm l = 140 mm l 3 = 38 mm Table 1. Scope of application for slabs made of normal concrete and re-entrant steel sheets Scope of application for re-entrant profiles [mm] Existing geometrical parameters [mm] 77.0 l l 1 = l l = l l 3 = h h 1 = h 70.0 h = Mechanical actions during fire exposure EN The load is determined by the combination rule for accidental situations. ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i According to EN 1994 Part 1-, the load E d may be reduced by the reduction EN a-5

26 factor η fi. It is calculated to: G k Q,1 k,1 ( ) ( ) Gk + ψ,1 Qk, η fi = = = 0.55 γ G + γ Q With η fi, the design bending moment M fi,d can be calculated: M, = η M = = 1.94 knm/m fi d fi sd Section.4..3 Thermal insulation Section D.1 The thermal insulation criteria I has to ensure the limitation of the thermal condition of the member. The temperature on top of the slab should not exceed 140 C in average and 180 C at its maximum. The verification is done in the time domain. The time in which the slab fulfils the criteria I is calculated to: A 1 A 1 t = a + a h + a Φ+ a + a + a i Lr l3 Lr l3 The rib geometry factor A/L r is equivalent to the section factor A p /V for beams. The factor considers that the mass and height have positive effects on the heating of the slab. L r Figure 4. Definition of the rib geometry factor l1+ l h 5 A = = = 7 mm L r l1 l l + h The view factor Φ considers the shadow effect of the rib on the upper flange. Φ= h + l + h + + l l1 l l1 l = = The coefficients a i for normal weight concrete is given in Table : 5a-6

27 Table. Coefficients for determination of the fire resistance with respect to thermal insulation (see EN , Annex D, Table D.1) a 0 [min] a 1 [min/mm] a [min] a 3 [min/mm] a 4 mm min a 5 [min] Normal weight concrete Light weight concrete With these parameters, t i is calculated to: t i ( ) ( ) ( ) = = min > 90 min.4 Verification of the load carrying-capacity Section 4.3. The plastic moment design resistance is calculated to: f yi, f c, j M fi, t, Rd = Ai zi ky, θ, i + αslab Aj zj kc, θ, j γ M, fi γ M, fi, c To get the reduction factors k y,θ for the upper flange, lower flange and the web, the temperatures have to be determined. These are calculated to: θ = 1 A b + b + b + b Φ+ b Φ Section D. a l3 Lr The coefficients b i can be obtained from Table 3: Table 3. Coefficients for the determination of the temperatures of the parts of the steel decking (see EN , Annex D, Table D.) Concrete Normal weight concrete Fire resistanc e [min] Part of steel sheet Lower flange b 0 [ C] b 1 [ C mm] b [ C/mm] b 3 [ C] b 4 [ C] Web Upper flange Lower flange Web Upper flange Lower flange Web Upper flange For the different parts of the steel sheet, the temperatures are: Lower flange: 1 θ al, = = C 5a-7

28 Web: 1 θ aw, = = C Upper flange: 1 θ al, = = C To get the required load carrying-capacity during fire exposure, reinforcing bars have to be installed which normally are neglected for the ambient temperature design. For each rib, one reinforcing bar Ø 10 mm is chosen. The position of the bar can be seen in Figure 5. Figure 5. Arrangement of the reinforcing bar The temperature of the reinforcing bar is calculated to: where: u A 1 θ = c + c + c z+ c + c α + c s h Lr l = + + z u u u = + + l l h = = 0,393 1 mm z =.54 mm (simplified) Figure 6. Definition of the distances u 1, u, u 3 and the angle α The coefficients c i for normal weight concrete is given in Table 4. 5a-8

29 Table 4. Coefficients for the determination of the temperatures of the reinforcement bars in rib (see EN , Annex D, Table D.3) Concrete Fire resistanc e [min] c 0 [ C] c 1 [ C] c [ C/mm 0.5 ] c 3 [ C/mm] c 4 [ C/ ] c 5 [ C] Normal weight concrete With these parameters, the temperature of the reinforcing bar is: 61 θ s = ( 56) + ( 35), ( 5,30) 7 + 1, ( 167) 38 = 407,0 C For the steel sheet, the reduction factors k y,i are given in Table 3. of the EN For the reinforcement the reduction factor is given in Table 3.4, because the reinforcement bars are cold worked. The carrying-capacity for each part of the steel sheet and the reinforcing bars can now be calculated. Table 5. Reduction factors and carrying-capacities Reduction Temperature factor θ i [ C] k y,i [-] Partial area A i [cm²] f y,i [kn/cm²] Z i [kn] Lower flange 960,9 0,047 1,04 35,0 1,98 Web 781,60 0,13 0,904 35,0 4,18 Upper flange 580,87 0,59 0,37 35,0 6,05 Reinforcement 407,0 0,91 0,79 50,0 36,38 The plastic neutral axis is calculated as equilibrium of the horizontal forces. The equilibrium is set up for one rib (b = l 1 + l ). z pl i Z 1,98 + 4,18 + 6, ,38 = = = 15,0 mm a l + l f + slab 3 ( ) 0,85 ( ) The plastic moment resistance for one rib is determined to: c Table 6. Calculation of the moment resistance of one rib Z i [kn] z i [cm] M i [kncm] Lower flange 1,98 14,0 7,7 Web 4,18 14,0 5,1 / = 11,45 47,86 Upper flange 6,05 14,0 5,1 = 8,9 53,85 Reinforcement 36,38 14,0 5,1 1,0 = 7,9 87,4 Concrete -48,59 1,50 / = 0,75-36,44 Σ 380,39 With the plastic moment of M pl,rib = 3.80 knm and the width w rib = 0.15 m of one rib, the plastic moment resistance of the composite slab is: M fi, Rd = 3,80 0,15 = 5,00 knm/m Verification: 5a-9

30 1,94 0,88 1 5,00 = < REFERENCES EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1994, Eurocode 4: Design of composite steel and concrete structures Part 1-: General Rules Structural Fire Design, Brussels: CEN, October 003 5a-30

31 Example to EN 1994 Part 1-: Composite beam P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK A fire safety verification has to be done for a composite beam of an office building. It is a simply supported beam and is loaded uniformly. The concrete slab of the composite beam protects the beam from the top in the fire situation, so the steel beam is exposed to fire on three sides. For fire protection of the steel beam a contour encasement of plaster is chosen. The required standard fire resistance class for the beam is R 60. Figure 1. Static system Figure. Cross-section Material properties: Beam: Profile: rolled section HE 160 B Steel grade: S 355 Height: h = 160 mm Height of web: h w = 134 mm Width: b = b 1 = b = 160 mm 5a-31

32 Thickness of web: e w = 8 mm Thickness of flange: e f = e 1 = e = 13 mm Cross-sectional area: A a = 5430 mm² Yield stress: f y,a = 355 N/mm² Slab: Strength category: C 5/30 Height: h c = 160 mm Effective width: b eff = 1400 mm Compression strength: f c = 5 N/mm² Elastic modulus: E cm = 9,000 N/mm² Shear connectors: Quantity: n = 34 (equidistant) Diameter: d = mm Tensile strength: f u = 500 N/mm² Encasement: Material: plaster Thickness: d p = 15 mm (contour encasement) Thermal conductivity: λ p = 0.1 W/(m K) Specific heat: c p = 1100 J/(kg K) Density: ρ p = 550 kg/m³ Loads: Permanent loads: Self weight: Finishing load: Variable loads: Live load: g k = 0.5 kn/m g k = 7.5 kn/m p k = 15.0 kn/m FIRE RESISTANCE OF A COMPOSITE BEAM.1 Mechanical actions during fire exposure EN Actions on structures from fire exposure are classified as accidental situation: ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i The partial safety factor γ GA for the accidental situation is γ GA = 1.0. The combination factor for the leading variable action for office buildings is set to ψ,1 = 0.3. With these parameters, the design bending moment during fire exposure can be calculated: 5.6 M fi, d = ( ) ( 15.0) = 17.4 knm 8. Calculation of the temperatures in the cross-section EN For the calculation of the temperatures, the cross-section is split into different sections. These are the concrete slab, the upper flange, the web and the lower flange. It is done according to Section of EN The temperatures of the upper flange, the web and the lower flange are determined by using the Euro-Nomogram ( Euro-Nomogram, ECCS No.89, 1996). Therefore, the section factors of these parts are required. 5a-3

33 Lower flange: Web: ( ) ( ) Ap b + e = = = m V b1 e l ( hw ) ( ) -1 Ap = = = 50.0 m V hw ew w Upper flange (more than 85% of the upper flange is in contact with the concrete slab): ( + ) ( ) Ap b e = = = 89.4 m V b e u -1 The temperatures are determined to: Table 1. Temperatures of upper flange, web and lower flange Section ECCS No.89 Ap λp W V d p m³k i θ a,max,60 [ C] Upper flange Web Lower flange The temperature of the concrete slab is not constant over its thickness. Therefore the compression strength varies over the thickness. For temperatures lower than 50 C, the compression strength is not reduced. Above 50 C it is reduced by the reduction factor k c,θ. Assessment of the temperatures may be done in layers of 10 mm thickness on basis of Table. EN Section D.3 Table. Temperature distribution in a solid slab of 100 mm thickness composed of normal weight concrete and not insulated (see EN , Annex D.3, Table D.5) Depth Temperature θ c [ C] after a fire duration in min. of x [mm] a-33

34 .3 Verification using simple calculation model The composite beam is verified by the simple calculation model. It is done in the strength domain. The calculation of the moment resistance is accomplished according to Annex E. Figure 3. Calculation of the moment resistance The temperatures of the parts of the steel beam were determined in Section 3.. The reduction factors k y,θ,i for the calculation of the yield stresses at elevated temperatures, are given in Table 3. of EN , Section Table 3. Calculation of the reduced yield stresses θ a,max,60 [ C] k y,θ [-] f ay,θ [kn/cm²] Upper flange Web 650 ( ) = Lower flange 550 ( ) = The next step is the calculation of the tensile force T of the steel beam according to Figure 3. fay, θ1 ( b ef ) + fay, θw ( hw ew ) + fay, θ ( b ef ) T = γ M, fi, a ( ) ( ) ( ) = 1.0 = kn The location of the tensile force is determined to: Section E.1 y T e f h ef ay, θ1 + ay, θw w w f + + ay, θ f w ( ) ( ) f b f h e e f b e h = T γ M, fi, a ( ) ( ) = = 9.53 cm 5a-34

35 In a simply supported beam, the value of the tensile force T is limited by: where: N P fi,rd T N P fird, Number of shear connectors in one of the critical lengths of the beam Design resistance in the fire situation of a shear connector To get P fi,rd,, the reduction factors k u,θ and k c,θ (Table 5) as well as the design resistances of a shear connector at ambient temperatures P Rd,1 and P Rd, are needed. The temperatures for getting the reduction factors are determined as 80 % (stud connector) and 40 % (concrete) of the steel flange (see EN 1994 Part 1-, Section ()). The reduction factor for the tensile strength of the stud connector is given it Table 3. of EN , Section The reduction factor for the compression strength of the concrete is given in Table 3.3 of EN , Section θ = = 31 C v k u,θ = 1.0 θ = = 156 C c k c,θ = 0.98 The design resistances of the shear connector are calculated according to EN , with the partial safety factor γ M,fi,v replacing γ v EN P P Rd,1 Rd, fu π d 50.0 π. = 0.8 = 0.8 = 15 kn Section γ M, fi, v fc Ecm = 0.9 α d = = 10 kn γ 1.0 M, fi, v The design resistance in the fire situation of a shear connector is: EN Pfi, Rd, 1 = 0.8 ku, θ PRd, 1 = = 11.6 kn Pfi, Rd = min Section Pfi, Rd, = kc, θ PRd, = = kn relevant So, the limitation can be verified: kn < = kn Section E.1 For equilibrium of forces, the compression force has to be equal to the tension force. Therefore the thickness of the compressive zone h u is determined to: h u = T cm b f γ = = eff c M, fi, c Now, two situations may occur. The first one is that the temperature in every layer of the concrete in the compression zone is lower than 50 C. In the second situation the temperature of some layers of the concrete is above 50 C. To check which situation occurs, following calculation has to be done: ( h h ) = = 1. cm c u If the result of this equation is greater than the depth x according to Table, corresponding to a concrete temperature below 50 C, the concrete in the compression zone may not be reduced. hcr = x= 5.0 cm < 1. cm The point of application of the compression force y F is determined to: 5a-35

36 ( ) ( ) y = h+ h h = = 30.1 cm F c u The moment resistance is calculated to: Verification: ( ) ( ) M, = T y y = = 74. knm fi Rd F T = 0.46 < 1 REFERENCES ECCS No.89, Euro-Nomogram, Brussels: ECCS Technical Committee 3 Fire Safety of Steel Structures, 1995 EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1994, Eurocode 4: Design of composite steel and concrete structures Part 1-1: General Rules and rules for buildings, Brussels: CEN EN 1994, Eurocode 4: Design of composite steel and concrete structures Part 1-: General Rules Structural Fire Design, Brussels: CEN 5a-36

37 Example to EN 1994 Part 1-: Composite beam comprising steel beam with partial concrete encasement P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK A fire safety verification has to be done for a composite beam of a storehouse. It is a simply supported beam with a uniform load and has a span of 1.0 m. The steel beam is partially encased and the slab is a composite slab. The required standard fire resistance class for the beam is R 90. Figure 1. Static system Figure. Cross-section 5a-37

38 Material properties: Beam: Profile: rolled section IPE 500 Steel grade: S 355 Height: h = 500 mm Width: b = 00 mm Thickness of web: e w = 10. mm Thickness of flange: e f = 16 mm Cross-sectional area: A a = 11,600 mm² Yield stress: f y,a = 355 N/mm² Slab: Strength category: C 5/30 Height: h c = 160 mm Effective width: b eff = 3000 mm Compression strength: f c = 5 N/mm² Profiled steel sheet: Type: re-entrant Height: h a = 51 mm Reinforcement in partial concrete encasement: Steel grade: S 500 Diameter: Ø 30 Cross-sectional area: A s = 1410 mm² Axis distances: u 1 = 110 mm u s1 = 60 mm Yield stress: f y,s = 500 N/mm² Concrete between flanges: Strength category: C 5/30 Width: b c = 00 mm Compression strength: f c Loads: Permanent loads: Self-weight: Finishing load: Variable loads: Live load: = 5 N/mm² g s,k = 15.0 kn/m g f,k = 6.0 kn/m p k = 30.0 kn/m FIRE RESISTANCE OF A COMPOSITE BEAM COMPRISING STEEL BEAM WITH PARTIAL CONCRETE ENCASEMENT.1 Mechanical actions during fire exposure EN Actions on structures in fire situation are classified as an accidental situation: ( ψ,, ) Section 4.3 E = E G + A + Q da k d i k i The combination factor for the leading variable action and for a storehouse is ψ,1 = 0.8. With these parameters, the design bending moment during fire exposure can be calculated: 1.0 M fi, d = (( ) ( 30.0) ) = knm 8 5a-38

39 . Verification using simple calculation model EN The composite beam is verified by the simple calculation model. It is accomplished according to EN 1994 Part 1-, Section and Annex F. To use this model, the slab should have a minimum thickness h c. Additionally the steel beam should have a minimum height h, a minimum width b c (where b c is the minimum width of steel beam or concrete encasement) and a minimum area h b c (see Table 1). Section Table 1. Minimum dimensions for the use of the simple calculation model for composite beams comprising steel beams with partial concrete encasement (see EN 1994 Part 1-, Section , Tables 4.9 and 4.10) Standard fire resistance Minimum slab thickness h c [mm] Minimum profile height h and minimum width b c [mm] Minimum area h b c [mm²] R ,500 R ,000 R ,000 R ,000 R ,000 h = 160 mm > min h = 100 mm c c h= 500 mm > min h= 170 mm b= b = 00 mm > min b = 170 mm c c ( h b ) h b = 100,000 mm > min = 35,000 mm c c In the calculation model of Annex F, the cross section is divided into different parts. At some parts the yield stress at other parts the cross-sectional area is reduced. Section F.1 Figure 3. Reduced cross-section for the calculation of the plastic moment resistance and stress distributions in steel (A) and concrete (B) 5a-39

40 The heating of the concrete slab is considered by reducing the cross-sectional area. For the different fire resistance classes, the thickness reduction h c,fi is given in Table. For composite slabs made with re-entrant steel sheets, a minimum thickness reduction h c,fi,min has to be considered. This minimum thickness reduction is equal to the height of the steel sheet. h c,fi = 30 mm h c,fi,min = 51 mm For this, the height of the concrete during fire exposure h c,h is: h ch, = = 109 mm Table. Thickness reduction h c,fi of the concrete slab (see EN 1994 Part 1-, Annex F, Table F.1) Standard fire resistance Slab reduction h c,fi [mm] R R 60 0 R R R Figure 4. Minimum thickness h c,fi,min reduction for re-entrant profiles The heating of the upper flange of the steel beam is considered by reducing its crosssectional area. The calculation of the width reduction b fi is shown in Table 3. ( ) ( ) b = = 38.0 mm fi The effective width is calculated to: b fi, u = = 14.0 mm Table 3. Width reduction b fi of the upper flange (see EN 1994 Part 1-, Annex F, Table F.) Standard fire resistance Width reduction b fi [mm] R 30 ( e ) + ( b b ) f R 60 ( e ) + + ( b b ) f 10 R 90 ( e ) + + ( b b ) f 30 R 10 ( e ) + + ( b b ) f 40 R 180 ( e ) + + ( b b ) f 60 c c c c c 5a-40

41 The web of the steel beam is divided into two parts. The upper part of the web possesses the full yield stress, where the yield stress of the lower part has a linear gradient, from the yield stress of the upper part to the reduced yield stress of the lower flange. The height of the lower part of the web h l is calculated to: h a a e 1 w l = + > bc bc h h l,min The parameters a 1 and a, as well as the minimum height h l,min, are given in Table 4 for h/b c >. h l 14,000 75, = + = 77.7 mm > 40 mm Table 4. Parameters a 1, a and minimum height h l,min for h/b c > (see EN 1994 Part 1-, Annex F, Table F.3) Standard fire resistance a 1 a h l,min [mm] [mm²] [mm²] R R R 90 14,000 75, R 10 3, , R ,000 50, The lower flange is not reduced by its cross-sectional area. Here, the yield stress is reduced by the factor k a. This factor is limited by a minimum and maximum value. These limits, as well as the calculation of k a, are given in Table 5. a0 = e f = = k a > 0.06 = = < 0.1 Table 5. Reduction factor k a of the yield stress of the lower flange (see EN 1994 Part 1-, Annex F, Table F.4) Standard fire resistance Reduction factor k a k a,min k a,max R h a0 bc bc R 60 6 h a0 bc 4 bc R h a0 bc 38 bc R h a0 bc 40 bc R h a0 bc 50 bc a-41

42 The heating of the reinforcing bars in the partial concrete encasement is considered by reducing the yield stress. The reduction factor depends on the fire resistance class and the position of the reinforcing bars. Like the reduction factor k a, the reduction factor k r has an upper and lower limit. A = h+ b = = 100 mm m c V = h b c = = 100,000 mm u = k r = 1 ( 1 u ) + ( 1 u ) + 1 ( b e u ) i si c w si 1 ( 1 110) + ( 1 60) + 1 ( ) = 9.88 mm ( u a a ) a ( ) > 0.1 = = = 0.51 A V ,000 < 1.0 m Table 6. Parameters for calculation of k r (see EN 1994 Part 1-, Annex F, Table F.5) Standard fire resistance a 3 a 4 a 5 k r,min k r,max R R R R R To acquire the plastic moment resistance, the axial forces of the different parts should be determined. Concrete: Upper flange: Upper web: where: Lower web: C = b h, α f = = kn c eff c h c c T, = b, e f = = kn f u fi u f y T, = e h f = = kn wu w h y h = h e h = = cm h f l 1+ ka Twl, = ew hl fy = = kn z wl, ka = hl = 7.77 =.8 cm ( k ) ( ) a 5a-4

43 Lower flange: T, = b e k f, = = kn Reinforcement bars: f l f a y a T = A k f, = = kn r s r y s Due to the fact that the compression force C c is larger than the sum of the tension forces T i, the plastic neutral axis is situated in the concrete slab. So the plastic neutral axis is calculated to: z pl Ti = = = 4.31 cm α f b c c eff To get the moment resistance, the lever arms of the forces are needed: Concrete slab (referring to upper edge of slab): z = z = 4.31 =.16 cm c pl Upper flange (referring to centre of gravity of concrete slab): Upper web: Lower web: Lower flange: Reinforcement: z, = h + e z = = cm f u c f c z, = h + e + h z = = cm wu c f h c z, = h + e + h + z, z = = 57.7 cm wl c f h wl c z, = h + h e z = = cm f l c f c z = h + h e u1 z = = 51.4 cm r c f c The plastic moment resistance is determined to: M fird, = Tf, u zf, u + Twu, zwu, + Twl, zwl, + Tf, l zf, l + Tr zr = = 10, , ,46 = 94,167 kncm = 94.7 knm Verification: = < REFERENCES EN 1991, Eurocode 1:Actions on structures Part 1-: General actions Actions on structures exposed to fire, Brussels: CEN, November 00 EN 1994, Eurocode 4: Design of composite steel and concrete structures Part 1-: General Rules Structural Fire Design, Brussels: CEN 5a-43

44 Example to EN 1994 Part 1-: Composite column with partially encased steel sections P. Schaumann, T. Trautmann University of Hannover Institute for Steel Construction, Hannover, Germany 1 TASK The following example deals with a composite column made of partially encased steel sections. It is part of an office building and has a length of L = 4.0 m. In this example, the simple calculation model and the tabulated data method are used. The column is part of a braced frame and is connected bending resistant to the upper and lower column. Therefore the buckling length can be reduced as seen in Figure 1. The required standard fire resistance class for the column is R 60. Figure 1. Buckling lengths of columns in braced frames Figure. Cross-section of the column 5a-44