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Prof. Dr.-Ing. Dietmar Hosser Dr.-Ing. Ekkehard Richter

Structural fire design according to the Eurocodes Lecture 4 Introduction to Eurocode 2 Part 1-2 Applications: Design of Concrete Members. Prof. Dr.-Ing. Dietmar Hosser Dr.-Ing. Ekkehard Richter Institute of Building Materials, Concrete Construction and Fire Protection (iBMB)

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Prof. Dr.-Ing. Dietmar Hosser Dr.-Ing. Ekkehard Richter

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  1. Structural fire designaccording to the EurocodesLecture 4Introduction to Eurocode 2 Part 1-2 Applications: Design of Concrete Members Prof. Dr.-Ing. Dietmar Hosser Dr.-Ing. Ekkehard Richter Institute of Building Materials, Concrete Constructionand Fire Protection (iBMB) Braunschweig University of Technology

  2. Lecture 4 • Description of the building • Structural fire design • Continuous solid slab with tabulated data, Table 5.8 • Continuous beam • Tabulated data: Table 5.5 • Simplified calculation method of EC 2-1-2, Annex E • Reinforced concrete column • Tabulated data • Method A: Table 5.2a • Method A: Equation (5.7) • Method B: Table 5.2b • Method B: Annex C, Tables C.1 – C.9 • Advanced calculation method • Conclusion

  3. Place of Assembly 36 4,99 30 2 10,60 1 4,59 2 36 59 4,90 5,98 49 3 30 30 36 36 5,75 5,70 5,75 18,52 Pos 1 Continuous solid slab Pos 2 Continuous beam Pos 3 Reinforced concrete column Non-sway building structure (Braced structure)

  4. Material Properties Concrete C 20 / 25 (siliceous aggregates) fck= 20 N/mm² gc = 1,5 and gc,fi = 1,0 Reinforcing steel 500 S (with high ductility) fyk= 500 N/mm² gs = 1,15 and gs,fi = 1,0

  5. Fire Resistance Requirements according to the Building Code: • Continuous solid slab REI 60 • Continuous beam R 60 • Reinforced concrete column R 60 • Definitions R: load bearing E: integrity (e.g. at joints) I: insulation (e.g. m 140 K)

  6. Continuous Solid Slab Reinforcedconcrete beam masonry 7 20 „A“ with Q= 1,5 and Q,fi= - with G= 1,35 and GA= 1,0 Slab thickness: h = 190 mm Concrete cover: nom c = 20 mm Reinforcement:  = 7 mm (mesh reinforcement R 513) Detail „A“ R 513

  7. Design Moments MEd,B = -31,95 kNm/m B MEd,1 = +31,95 kNm/m l1 = 5,20 • Normal temperature design: • Support B: MEd,B = -31,95 kNm/m • Span 1: MEd,1 = +31,95 kNm/m • Moment redistribution < 15% at support B

  8. Fire Design 7 20 • Fire design with EC 2-1-2, Section 5.7.3 (2): • Each span should be assessed as a simply supported span, if redistribution > 15 % is used • Table 5.8, columns 2, 3, 4 or 5 respectively • Provided dimensions • Axis distance of tension reinforcementprov a = nom c + /2 = 20 + 7/2 = 23,5 mm

  9. EC 2-1-2, Table 5.8 • REI 60: req hs= 80 mm < prov hs= 190 mm • req a = 20 mm < prov a = 23,5 mm • classified as REI 60

  10. Continuous Solid Slab (Alternative) • Fire design with EC 2-1-2, Annex ESimplified calculation method for beams and slabs • Boundary conditions • Loading is predominantly uniformly distributed • The method can also be applied for continuous beams and slabs with moment redistribution > 15%, if sufficient rotational capacity is provided at the supports • The minimum cross-section dimensions given in tables should not be reduced

  11. Simplified Calculation Method, Annex E moment- redistribution • Consideration of bending moments in fire situation • Constrained thermal deformations • Increase of the moments at the supports • Decrease of the moments at the span • Fire designResistance Rd,fi Effects of actions Ed,fi MEd,B,fi = MRd,B,fi ?? • Moments at support B MRd,B,fi  MEd,B,fi • Moments at span 1 MRd1,fi MEd1,fi B MEd1,fi= ?? MRd1,fi = ??

  12. Moment Redistribution in Fire Situation MRd,B,fi MRd,B MEd,B = -31,95 kNm/m B MEd1,fi MRd1,fi MEd,1 = +31,95 kNm/m 9,5 kN/m MRd,1 15,6 kN/m aS,prov > aS,req aS,prov > aS,req = MEd,B,fi

  13. Moment Resistance at the Support • Normal temperature: • MRd,B = MEd,B = fyk / s As,req  z • fyk = MEd,B / (As,req  z)  s (A) • Fire situation: • MRd,B,fi = fyk,fi / s,fi As,prov  zfi (B) • fyk,fi fyk ? (must be checked) • Eq. (A) in Eq. (B) • Approximation: zfi / z  ?? (d) Fsd d = 166 mm Fcd

  14. Design Moment EC 2-1-1, section 5.3.2: MEd,B = 0,65  MEd,x X G ·Gk + Q · Qk 1,35·6,0 + 1,5·5,0=15,6 X 6 15 4,99 MEd,B= -0,65 · 48,56 = -31,56 kNm/m

  15. Temperature of Reinforcement h = 190 d = 166 • Checking the temperature of the reinforcement • Equation MRd,B,fi is valid only for  350 °C • Distance between the upper reinforcement at support B and the heated surface of the slab:d = h - nom c - /2 = 190 – 20 – 7,0/2 = 166 mm • t = 60 min and d = 166 mm:  = ??Fig. A.2

  16. Fig. A.2 (prov a = prov d =166 mm) 100 < prov d = 166 mm   < 100°C

  17. Moment Resistance at the Support Fsd,fi a of R 60 Fcd,fi  500 °C • Ambient temperature: • MRd,B = MEd,B = fyk / s As,req  z • fyk = MEd,B / (As,req  z)  s (A) • Fire situation: • MRd,B,fi = fyk,fi / s,fi As,prov  zfi (B) • fyk,fi fyk ? (it must be checked) • Eq. (A) in Eq. (B) • Approximation: zfi / z  ?? (is checked) Fsd d = 166 mm = (d - aR60) Fcd

  18. Moment Resistance at the Support As,prov < 1,3 As,req • MRdB,fi  > MEdB,fi = 31,56 kNm/m gs / gs,fi partial material factors (1,15 / 1,0) MEd,B applied moment for cold design to EN 1992-1-1 As,prov area of tensile steel provided (5,13 cm2/m) As,req area of tensile steel required (4,70 cm2/m) d effective depth of section (d = 166 mm) a required average bottom axis distance for REI 60: a = 20 mm

  19. Moment Resistance at the Span Eq. (E.3) Strength reduction factor for the given steel temperature  fromFig. A.2

  20. Fig. A.2 (prov a = 23,5 mm)  460 °C a = 23,5 mm ks(460°C) = ??

  21. Strength Reduction Factor of Steel 1 0,7 reinforcing steel 1  460 °C ks(460°C) = 0,7

  22. Moment Resistance at the Span = ü A 5 , 13 cm ² / m 5 , 13 s , prov ý = < 1 , 09 1 , 30 = A 4 , 70 cm ² / m þ 4 , 70 s , req Eq. (E.3) Strength reduction factor for the given steel temperature  fromFig. A.2 ks(460°C) = 0,7

  23. Bending Moment at the Span M Q Gd,fi + Qd,fi = 1,0  6,0 + 0.7  5,0 = 9,5 kN/m2 MRd,B,fi = -34,8 kNm/m 5,20 VEd,fi x MEd1,fi

  24. Simplified Calculation Method, Annex E moment- redistribution MRd,B,fi = - 34.8 kNm/m MRd,B,fi = - 31.6 kNm/m B MEd1,fi= 17.1 kNm/m Classification REI 60 MRd1,fi = 28.1 kNm/m • Consideration of bending moments in fire situation • constrained thermal deformations • increase of the moments at support • decrease of the sagging moments • Fire designResistance Rd,fi Effects of actions Ed,fi • Moments at support B MRd,B,fi  MEd,B,fi • Moments at span 1 MRd1,fi MEd1,fi

  25. Continuous Beam 190 590 400 300 Cross-section at mid-span Stirrups  10 Concrete cover of stirrups: nom c = 25 mm Reinforcement: Reinforcing steel 500 S, 4  25 Design at normal temperature condition: if moment redistribution > 15%, then for Fire design EC 2-1-2, section 5.6.2 (2): each beam should be assessed as a simply supported beam Table 5.5, columns 2, 3, 4 or 5 respectively

  26. EC 2-1-2, Table 5.5

  27. EC 2-1-2, Table 5.5 12,5 10 25 ad 300 Axis distance of tension reinforcement: prov a = nom c + sti +  /2 = 25 + 10 + 25/2 = 47,5 mm prov ad = (300 - 7 25)/2 = 62,5 mm

  28. EC 2-1-2, Table 5.5 12,5 10 25 ad 300 Axis distance of tension reinforcement: prov a = nom c + sti +  /2 = 25 + 10 + 25/2 = 47,5 mm prov ad = (300 - 7 25)/2 = 62,5 mm • R 60: req b = 300 mm = prov b = 300 mm • req a = 25 mm < prov a = 47,5 mm • req ad = 25 + 10 = 35 mm < prov ad = 62,5 mm • classified as R 60

  29. Reinforced Concrete Column NEd  12 mm 300  8 mm 25 300 Concrete C 20/25 Reinforcing steel 500 S Concrete cover of stirrups nom c = 25 mm Reinforcement stirrups st = 8 mm Corner bars 4 L 12 = 4  113 = 452 mm2 Axis distance: prov a = nom c + st + /2 = 25 + 8 + 12/2 = 39 mm

  30. EC 2-1-2, Table 5.2a • Fire design • Comparison of column dimensions: req b < prov b • Comparison of axis distance: req a < prov a • Load level • Boundary conditions • Normal temperature design: both ends of the column with hinges (effective length of the column at room temperature: l0 = 1.0  l) • Fire design: both supports of the column with rotational restraint (effective length of the column under fire conditions: l0,fi = 0.5  l (EC 2-1-2)

  31. Boundary Conditions Separate fire compart- ments in each storey l Separate fire compart- ments in each storey Separate fire compart- ments in each storey l Separate fire compart- ments in each storey Deformation mode at normal temperature Section through the building l l l l Bracingsystem

  32. Boundary Conditions l l l l0= 0,5.l l l Basis of Table 5.2a l Deformation mode at normal temperature Bracingsystem Deformation mode at elevated temperature Section through the building

  33. Boundary Conditions l0= 0,7.l l not covered by Table 5.2a l Deformation mode at normal temperature Section through the building l l l l Bracingsystem Deformation mode at elevated temperature

  34. Buckling Length l0, l0,fi Basis of table. 5.2a: Normal temperature Fire situation l l0 = β0. l l0,fi = βfi. l βfi = 0,5 β0 = 1,0 Design value of the resistance of the column NRd for normal temperature design may be calculated using ß0

  35. EC 2-1-2, Table 5.2a

  36. Load Level Fire design: Reduction factor Load level Normal temp. design:

  37. Buckling Length NEd,fi l0,fi = 0,7 . lcol l0 NEd

  38. Design Load for Normal Temp. Design NRd = ? l‘ = 7,28 m l = 5,20 m • Determination of NRd • Determine the cross-sectionload bearing capacity usingstress-strain diagrams based ondesign values of concrete fcdand steel fyd (EC 2-1-1, Eq.(3.17)) • Determine the structuralbehaviour of the column withconsideration of second ordereffects using stress-straindiagrams of concrete and steel suitable for calculation of deformations (EC 2-1-1, Eq.(3.14)) • Compare the cross-sectionload bearing capacity (1) andthe structural behaviourof the column (2)

  39. Design Load for Normal Temp. Design NRd = ? NRd = 810 kN l‘ = 7,28 m l = 5,20 m

  40. Design Load for Normal Temp. Design NRd = ? l‘ = 7,28 m l = 5,20 m Load level:

  41. EC 2-1-2, Table 5.2a Linear interpolation between mfi = 0.5 and mfi = 0.7 • req b = 218 mm < prov b = 300 mm • req a = 39,5 mm  prov a = 39 mm • classified as R 60 Standard fire resistance R 60 and mfi = 0,57

  42. Reinforced Concrete Column (Alternative 1) • Fire design with EC 2-1-2, Method A, Eq. (5.7) • Simple model for assessing the fire resistance time R taking into account • Load level • Dimensions of the cross section • Length of the column • Reinforcement ratio • Axis distance of the longitudinal bars • Range of application • Axis distance: 25 mm  a  80 mm • Effective length in fire situation: 2 m  l0,fi  6 m • Cross section dimension: h  1.5  b 200 mm  b’  450 b’ = 2  Ac / (b + h)

  43. EC 2-1-2, Eq. (5.7) Rhfi effect of load level in fire situation Ra effect of axis distance Rl effect of length of the column Rb effect of cross section dimension Rn effect of reinforcing bars

  44. EC 2-1-2, Eq. (5.7)

  45. EC 2-1-2, Eq. (5.7) for n = 4 (corner bars only) • classified as R 60

  46. Reinforced Concrete Column (Alternative 2) • Fire design with EC 2-1-2, Method B, Table 5.2b • Extended tabulated data taking into account the mechanical reinforcement ratio  at normal temperature conditions

  47. Reinforced Concrete Column (Alternative 2)  • Fire design with EC 2-1-2, Method B, Table 5.2b • Range of application: eccentricity e = M0Ed,fi / NEd  100 mm prov e = 13.5 mm rel. eccentricity e / b  0,25 prov e / b = 13.5 / 300 = 0.045 slenderness lfi = l0,fi / i  30 prov lfi = (0.7  520) / 8.66 = 42   • Table 5.2b is not valid • use tables in Annex C

  48. Reinforced Concrete Column (Alternative 3) • Fire design with EC 2-1-2, Annex C, Tab. C.1-C.9 • Extended tabulated data taking into account • Load level 0.15  n  0.7 • Slenderness 30 l 80 at normal temperature conditions • mechanical reinforcement ratio 0.1   1.0 at normal temperature conditions • First order eccentricity 0.025  b  e  0.5  b and10 mm  e  200 mm

  49. Reinforced Concrete Column (Alternative 3) • Fire design with EC 2-1-2, Annex C, Tab. C.1-C.9 • Extended tabulated data taking into account • Load level 0.15  n  0.7 • Slenderness 30 l 80 at normal temperature conditions • mechanical reinforcement ratio 0.1   1.0 at normal temperature conditions • First order eccentricity 0.025  b  e  0.5  b and10 mm  e  200 mm • Linear interpolation within each table is permitted • Linear interpolation between the different tables is permitted

  50. Reinforced Concrete Column (Alternative 3) • Table C.1: • = 0.1 e = 0.025  b e  10 mm • Fire design with EC 2-1-2, Annex C, Tab. C.1-C.9

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