DIF SEK. PART 5a WORKED EXAMPLES. DIFSEK Part 5a: Worked examples 0 / 62
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1 DIF SEK PART 5a WORKED EXAMPLES DIFSEK Part 5a: Worked examples 0 / 62
2 Resistance to fire - Chain of events Θ Loads 1: Ignition time Steel columns 2: Thermal action 3: Mechanical actions R 4: Thermal response time 5: Mechanical response 6: Possible collapse DIF SEK Part 5a: Worked examples 1 / 62
3 Used standards Ambient temperature design EN 1990 EN EN Basis of structural design Design of steel structures Design of composite structures Fire design EN 1990 EN EN EN Basis of structural design Thermal actions Fire design of steel structures Fire design of composite structures DIF SEK Part 5a: Worked examples 2 / 62
4 Worked examples Overview EN 1991: Part 1-2: Actions on structures General actions Actions on structures exposed to fire Number of examples 2 EN 1993: Part 1-2: Design of steel structures General rules Structural fire design 3 EN 1994: Part 1-2: Design of composite steel and concrete structures General rules Structural fire design 4 DIFSEK Part 5a: Worked examples 3 / 62
5 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 4 / 62
6 Compartment Fire Task Determination of the gas temperature of a fully engulfed fire Natural fire model for compartment fires Parametric temperature time curve θ g = f (q f,d, O, b) EN : Annex A DIFSEK Part 5a: Worked examples 5 / 62
7 Compartment Fire Parameters Building: Type: Fire load: Floor area: Height: Average window height: Area of vertical openings: Vertical opening factor: Material of boundaries: Cardington test facility Office q f,d = 483 MJ/m² A f = 135 m² H = 4.0 m h eq = 1.8 m A v = 27 m² O = m 1/2 Lightweight concrete b = J/(m 2 s 1/2 K) DIFSEK Part 5a: Worked examples 6 / 62
8 Compartment Fire Fuel or ventilation controlled? q O h 3 t,d = h < t = h fuel controlled lim h > t = h ventilation controlled lim where qt,d = qf,d Af At DIFSEK Part 5a: Worked examples 7 / 62
9 Compartment Fire Heating curve Calculation of the heating curve: 0.2 t* 1.7 t* 19 t* ( ) θ = e e e g where: t* = t Γ Γ = 2 ( O b) ( ) 2 DIFSEK Part 5a: Worked examples 8 / 62
10 Compartment Fire Maximum temperature Equal to the calculation of the heating curve, except: t t max t qt,d O = max = lim The maximum temperature is needed to determine the cooling curve. DIFSEK Part 5a: Worked examples 9 / 62
11 Compartment Fire Cooling curve Calculation of the cooling curve: max t* = t Γ ( ) θ = θ 625 t * t * x g max max where 3 ( t,d ) t * = q O Γ If fire is ventilation controlled: x = 1.0 If fire is fuel controlled: x = t lim Γ / t* max DIFSEK Part 5a: Worked examples 10 / 62
12 Compartment Fire Final curve and comparison Parametric temperature-time curve Comparison calculation measurement (Factors of q fi,d : δ q1 =1.0, δ q2 = 1.0, δ n = 1.0 DIFSEK Part 5a: Worked examples 11 / 62
13 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 12 / 62
14 Localised fire Task Determination of the steel temperatures of a steel beam exposed to fire by a burning car. Natural fire model for localised fires EN : Annex C DIFSEK Part 5a: Worked examples 13 / 62
15 Localised fire Parameters Building: Type: Height: Horizontal distance from flame axis to beam: Diameter of flame: Steel Beam: Car park Auchan, Luxembourg Underground car park H = 2.7 m r = 0.0 m D = 2.0 m IPE 550 DIFSEK Part 5a: Worked examples 14 / 62
16 Localised fire Rate of Heat Release Curve of the rate of heat release of one car From ECSC project:development of design rules for steel structures subjected to natural fires in closed car parks. DIFSEK Part 5a: Worked examples 15 / 62
17 Localised fire Flame Length if L r H Model A has to be used if L r < H Model B has to be used DIFSEK Part 5a: Worked examples 16 / 62
18 Localised fire Steel temperatures Temperature-time curve for the unprotected steel beam: A / V θ = θ + k h& t p a,r m sh net ca ρa at t θ = a,max θ,max = 770 C min DIFSEK Part 5a: Worked examples 17 / 62
19 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 18 / 62
20 Steel column Task Determination of the design axial resistance for a steel column. Simple calculation model for compression members EN : Section DIFSEK Part 5a: Worked examples 19 / 62
21 Steel column Parameters Building: Fire resistance class: Department store R 90 Loads: G k = 1200 kn P k = 600 kn Profile: Fire protection: Steel grade: Rolled section HE 300 B Hollow encasement of gypsum board (d p = 3 cm) S 235 DIFSEK Part 5a: Worked examples 20 / 62
22 Steel column Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for shopping areas: ψ 2,1 = 0.6 Nfi,d = 1560 kn DIFSEK Part 5a: Worked examples 21 / 62
23 Steel column Maximum steel temperature 2 (b + h) λ W p = Aa dp m K A p V λ d p p Euro-Nomogram: θ a,max, C Reduction factors: k y,θ = k E,θ = DIFSEK Part 5a: Worked examples 22 / 62
24 Steel column Reduction factor and verification Reduction factor χ fi : ky, θ λ fi, θ = λ = 0.25 k E, θ for: θ a = 445 C S 235 χ = fi 0.86 Flexural buckling: N = χ A k θ γ b,fi,t,rd fi y,,max N N = 0.58 < 1 fi,d b,fi,t,rd f y M,fi DIFSEK Part 5a: Worked examples 23 / 62
25 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 24 / 62
26 Steel beam (N + M) Task Verification of a steel beam subjected to bending and compression loads. Simple calculation model for members subjected to bending and compression loads EN : Section DIFSEK Part 5a: Worked examples 25 / 62
27 Steel beam (N + M) Parameters Building: Fire resistance class: Office building R 90 Loads: G k = 96.3 kn g k = 1.5 kn/m p k = 1.5 kn/m Profile: Fire protection: Steel grade: Rolled section HE 200 B Hollow encasement of gypsum board (d p = 2 cm) S 235 DIFSEK Part 5a: Worked examples 26 / 62
28 Steel beam (N + M) Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for office areas: ψ 2,1 = 0.3 Nfi,d = 96.3 kn Mfi,d = knm DIFSEK Part 5a: Worked examples 27 / 62
29 Steel beam (N + M) Maximum steel temperature 2 h b λ W A d m K + p = a p A p V λ d p p Euro-Nomogram: θ a,max, C Reduction factors: k y,θ = k E,θ = DIFSEK Part 5a: Worked examples 28 / 62
30 Steel beam (N + M) Reduction factors and verification Reduction factors χ i,fi : Similar to example Steel column Flexural buckling: Nfi,d ky My,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 Lateral torsional buckling: Nfi,d klt My,fi,d + = 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 DIFSEK Part 5a: Worked examples 29 / 62
31 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 30 / 62
32 Steel beam (hollow section) Task Determination of the design bending resistance for the steel beam. Simple calculation model: for members subjected with bending loads without stability problems EN : Section DIFSEK Part 5a: Worked examples 31 / 62
33 Steel beam (hollow section) Parameters Building: Fire resistance class: Hall roof structure R 30 Loads: g k = 9.32 kn/m p k = kn/m Profile: Fire protection: Steel grade: Welded section h / b = 70 cm / 45 cm t w = t f = 25 mm none S 355 DIFSEK Part 5a: Worked examples 32 / 62
34 Steel beam (hollow section) Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for snow loads: ψ 2,1 = 0.0 Mfi,d = kNm DIFSEK Part 5a: Worked examples 33 / 62
35 Steel beam (hollow section) Maximum steel temperature Standard temperature time curve: ( ) θ = log 8 t + 1 g 10 Steel temperature time curve: A / V θ = k h& t p a,r sh net ca ρa Section factor with equal thickness of flanges and web. Ap 1 1 = = 40 V t m DIFSEK Part 5a: Worked examples 34 / 62
36 Steel beam (hollow section) Verification Verification in the temperature domain: µ 0 = Efi,d Rfi,d,0 = 0.31 θ a,cr = 659 C θ a,max,30 θ a,cr = 0.98 < 1 Verification in the strength domain: γm,1 1 Mfi,t,Rd = Mpl,Rd,20 C ky, θ = knm γ κ κ where k y,θ = κ 1 = 1.0 κ 2 = 1.0 M,fi 1 2 M M fi,d fi,t,rd = 0.87 < 1 DIFSEK Part 5a: Worked examples 35 / 62
37 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 36 / 62
38 Composite slab Task Determination of the design sagging moment resistance for the composite slab. Simple calculation model for composite slabs exposed to fire EN : Annex D DIFSEK Part 5a: Worked examples 37 / 62
39 Composite slab Parameters Building: Fire resistance class: Loads: Heigth of slab: Strength class: Steel sheet: Yield stress: Shopping centre R 90 g k = 4.62 kn/m² p k = 5.0 kn/m² h = 14.0 cm C 25/30 Re-entrant h 2 = 5.1 cm f y = 350 N/mm² DIFSEK Part 5a: Worked examples 38 / 62
40 Composite slab Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Q G k,1 k,1 k q = = 1.1 g k Bending moment in fire situation: Mfi,d = ηfi Msd = knm/m DIFSEK Part 5a: Worked examples 39 / 62
41 Composite slab Rib geometry factor Rib geometry factor considers positive effects of mass and height of the rib. l + l h = = A 2 Lr 2 l1 l2 l2 2 h2 27 mm DIFSEK Part 5a: Worked examples 40 / 62
42 Composite slab Thermal insulation The temperature on top of the slab should not exceed 140 C in average and 180 C at its maximum. ti = min > 90 min DIFSEK Part 5a: Worked examples 41 / 62
43 Composite slab Maximum steel temperatures and verification Calculation of the steel temperatures: Steel sheet θ = 1 A 2 a,i b + 0,i b 1,i b2,i b3,i b4,i l + 3 L + Φ + Φ r Reinforcement bars u3 A 1 θ s = c0 + c1 + c2 z + c3 + c4 α + c5 h L l 2 r 3 Verification: fy,i fc,j Mfi,t,Rd = Ai zi ky, θ,i + αslab A j zj kc, θ,j γm,fi γm,fi,c = knm/m > knm/m DIFSEK Part 5a: Worked examples 42 / 62
44 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 43 / 62
45 Composite beam (steel beam) Task Determination of the design sagging moment resistance for the composite beam. Simple calculation model for composite beams exposed to fire EN : Annex E DIFSEK Part 5a: Worked examples 44 / 62
46 Composite beam (steel beam) Parameters Building: Fire resistance class: Loads: Heigth of slab: Strength class: Profile: Fire protection: Steel grade: Office building R 60 g k = 28.0 kn/m p k = 15.0 kn/m h c = 16.0 cm C 25/30 Rolled section HE 160 B Contour encasement of plaster (d p = 1.5 cm) S 235 DIFSEK Part 5a: Worked examples 45 / 62
47 Composite beam (steel beam) Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for office areas: ψ 2,1 = 0.3 Mfi,d = knm DIFSEK Part 5a: Worked examples 46 / 62
48 Composite beam (steel beam) Maximum steel temperature Upper flange: θ a,max,u 390 C k y,θ = 1.0 Web: θ a,max,w 650 C k y,θ = A V p λp d p Lower flange: θ a,max,l 550 C k y,θ = DIFSEK Part 5a: Worked examples 47 / 62
49 Composite beam (steel beam) Temperatures of the concrete compression zone Check, if the temperatures in the compression zone are lower than 250 C: ( ) h h = 12.2 cm > x = 5 cm c u Concrete compression strength is not reduced. where x: Concrete zone with temperatures θ c > 250 C h u : Height of the compression zone DIFSEK Part 5a: Worked examples 48 / 62
50 Composite beam (steel beam) Design sagging moment resistance and verification Design sagging moment resistance: ( ) M = T y y = knm fi,rd F T Verification: M fi,d M = < fi,rd DIFSEK Part 5a: Worked examples 49 / 62
51 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 50 / 62
52 Composite beam (partially encased beam) Task Determination of the design sagging moment resistance for the composite beam Simple calculation model for composite beams comprising steel beam with partial concrete encasement exposed to fire. EN : Annex F DIFSEK Part 5a: Worked examples 51 / 62
53 Composite beam (partially encased beam) Parameters Building: Fire resistance class: Loads: Height of slab: Strength category: Width of encasement: Strength class: Profile: Steel grade: Storehouse R 90 g k = 21.0 kn/m p k = 30.0 kn/m h c = 16 cm C 25/30 b c = b = 20 cm C 25/30 Rolled section IPE 500 S 355 DIFSEK Part 5a: Worked examples 52 / 62
54 Composite beam (partially encased beam) Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for storage areas: ψ 2,1 = 0.8 Mfi,d = knm DIFSEK Part 5a: Worked examples 53 / 62
55 Composite beam (partially encased beam) Reduction of the cross-section in the fire situation Concrete slab Height reduction Upper flange Width reduction Web Determination of height without strength reduction Lower flange Strength reduction Reinforcements Strength reduction DIFSEK Part 5a: Worked examples 54 / 62
56 Composite beam (partially encased beam) Design sagging moment resistance and verification Design sagging moment resistance: M = T z = knm fi,rd i i where T i : z i : tension force of part of the cross-section distance from compression force to the tension force Verification: M fi,d M = < fi,rd DIFSEK Part 5a: Worked examples 55 / 62
57 Worked examples - Overview Actions Compartment fire Localised fire Steel Steel column Steel beam (N + M) Steel beam (hollow section) Composite Composite slab Composite beam (steel beam) Composite beam (partially encased beam) Composite column DIFSEK Part 5a: Worked examples 56 / 62
58 Composite column Task Determination of the design axial compression resistance for the composite column. Simple calculation model for composite columns exposed to fire and tabulated data method EN : Annex G EN : Section DIFSEK Part 5a: Worked examples 57 / 62
59 Composite column Parameters Building: Fire resistance class: Loads: Concrete strength class: Profile: Steel grade: Office building R 60 G k = kn P k = kn C 25/30 Rolled section HE 300 B S 235 DIFSEK Part 5a: Worked examples 58 / 62
60 Composite column Mechanical actions during fire exposure Accidental situation: ( ) E = E G + A + ψ Q da k d 2,i k,i Combination factor for office areas: ψ 2,1 = 0.3 Nfi,d = kn DIFSEK Part 5a: Worked examples 59 / 62
61 Composite column Reduction of the cross-section in fire situation Flanges Strength reduction Stiffness reduction Web Heigth reduction Strength reduction Stiffness reduction Reinforcements Strength reduction Stiffness reduction Concrete Thickness reduction Strength reduction Stiffness reduction DIFSEK Part 5a: Worked examples 60 / 62
62 Composite column Design axial resistance and verification Calculation of the axial design resistance: N = N = kn fi,pl,rd fi,pl,rd,i where N fi,pl,rd,i plastic design resistances of the several parts Flexural buckling: χ z,fi N fi,d N fi,pl,rd = χ z,fi is determined similar to example Steel column DIFSEK Part 5a: Worked examples 61 / 62
63 Composite column Tabulated data method Existing parameters: e e = 0.58 w f b = h = 300 mm us = 50 mm η fi,t = 0.28 A s Ac + A = 3% s DIFSEK Part 5a: Worked examples 62 / 62
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