Concrete Design to Eurocode 2

Transcription

1

2 Concrete Design to Eurocode 2 Jenny Burridge MA CEng MICE MIStructE Head of Structural Engineering

3 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Cover Flexure Shear Deflection Further Information

4 The Eurocodes BS EN 1990 (EC0) : Basis of structural design BS EN 1991 (EC1) : Actions on Structures BS EN 1992 (EC2) : Design of concrete structures BS EN 1993 (EC3) : Design of steel structures BS EN 1994 (EC4) : Design of composite steel and concrete structures BS EN 1995 (EC5) : Design of timber structures BS EN 1996 (EC6) : Design of masonry structures BS EN 1997 (EC7) : Geotechnical design BS EN 1998 (EC8) : Design of structures for earthquake resistance BS EN 1999 (EC9) : Design of aluminium structures

5 The Eurocodes BS EN 1990 (EC0): Basis of structural design BS EN 1991 (EC1): Actions on Structures BS EN 1992 (EC2): Design of concrete structures BS EN 1993 (EC3): Design of steel structures BS EN 1994 (EC4): Design of composite steel and concrete structures BS EN 1995 (EC5): Design of timber structures BS EN 1996 (EC6): Design of masonry structures BS EN 1997 (EC7): Geotechnical design BS EN 1998 (EC8): Design of structures for earthquake resistance BS EN 1999 (EC9): Design of aluminium structures

6 Features of the Eurocodes The Eurocodes contain Principles (P) which comprise: General statements and definitions for which there is no alternative, as well as: Requirements and analytical models for which no alternative is permitted They also contain Application Rules, which are generally rules which comply with the Principles The Eurocodes also use a comma (,) as the decimal marker Each Eurocode part has a National Annex which modifies the main text of the Eurocode

7 National Annex The National Annex provides: Values of Nationally Determined Parameters (NDPs) (NDPs have been allowed for reasons of safety, economy and durability) Example: Min diameter for longitudinal steel in columns φ min = 8 mm in text φ min = 12 mm in N.A. The decision where main text allows alternatives Example: Load arrangements in Cl (1) P The choice to adopt informative annexes Example: Annexes E and J are not used in the UK Non-contradictory complementary information (NCCI) TR 43: Post-tensioned concrete floors design handbook

8 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Cover Flexure Shear Deflection Further Information

9 Eurocode Published 27 July 2002 Structures are to be designed, executed and maintained so that, with appropriate forms of reliability, they will: Perform adequately under all expected actions Withstand all actions and other influences likely to occur during construction and use Have adequate durability in relation to the cost Not be damaged disproportionately by exceptional hazards

10 Eurocode The code sets out the following: Basis for calculating design resistance of materials Combinations of actions for ultimate limit state Persistent Transient Accidental Seismic Combinations of actions for serviceability limit state

11 Eurocode Design values of actions, ultimate limit state persistent and transient design situations (Table A1.2(B) Eurocode) Comb tion expression reference Permanent actions Leading variable action Accompanying variable actions Unfavourable Favourable Main(if any) Others Eqn (6.10) γ1.35 G,j,sup GG k k,j,sup 1.0 γ G,j,inf G k G k,j,inf 1.5 γ Q,1 Q k,1 1.5 γ Q,i Ψ 0,i Q k,i Eqn (6.10a) γ1.35 G,j,sup GG k k,j,sup 1.0 γ G,j,inf G k G k,j,inf 1.5 γ Q,1 Ψ 0,1 Q k,1 k 1.5 γ Q,i Ψ 0,i Q k,i Eqn (6.10b) ξ0.925x1.35g γ G,j,sup G k,j,supk 1.0 γ G,j,inf G k G k,j,inf 1.5 γ Q,1 Q k,1 1.5 γ Q,i Ψ 0,i Q k,i For one variable action: 1.25 G k Q k Provided: 1. Permanent actions < 4.5 x variable actions 2. Excludes storage loads

12 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Cover Flexure Shear Deflection Axial Further Information

13 Eurocode 1 Eurocode 1 has ten parts: Densities, self-weight and imposed loads Actions on structures exposed to fire Snow loads Wind actions Thermal actions Actions during execution Accidental actions due to impact and explosions Traffic loads on bridges Actions induced by cranes and machinery Actions in silos and tanks

14 Eurocode 1 Eurocode 1 Part 1-1: Densities, self-weight and imposed loads Bulk density of reinforced concrete is 25 kn/m 3 The UK NA uses the same loads as BS 6399 Plant loading not given

15 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Cover Flexure Shear Deflection Further Information

16 Eurocode 2 Relationships BS EN 1997 GEOTECHNICAL DESIGN BS 8500 Specifying Concrete BS EN Execution of Structures BS EN 1990 BASIS OF STRUCTURAL DESIGN BS EN 1991 ACTIONS ON STRUCTURES BS EN 1992 DESIGN OF CONCRETE STRUCTURES Part 1-1: General Rules for Structures Part 1-2: Structural Fire Design BS EN 1998 SEISMIC DESIGN BS 4449 Reinforcing Steels BS EN Reinforcing Steels BS EN 1994 Design of Comp. Struct. BS EN 1992 Part 2: Bridges BS EN 1992 Part 3: Liquid Ret. Structures BS EN Pre-cast Concrete

17 Eurocode 2/BS 8110 Compared Code deals with phenomena, rather than element types Design is based on characteristic cylinder strength Does not contain derived formulae (e.g. only the details of the stress block is given, not the flexural design formulae) Unit of stress in MPa One thousandth is represented by %o Plain or mild steel not covered Notional horizontal loads considered in addition to lateral loads High strength, up to C90/105 covered

18 Materials

19 Concrete properties (Table 3.1) Strength classes for concrete f ck (MPa) f ck,cube (MPa) f cm (MPa) f ctm (MPa) E cm (GPa) BS 8500 includes C28/35 & C32/40 For shear design, max shear strength as for C50/60 f ck f ck,cube f cm f ctm E cm = Concrete cylinder strength = Concrete cube strength = Mean concrete strength = Mean concrete tensile strength = Mean value of elastic modulus

20 Reinforcement properties (Annex C) Product form Bars and de-coiled rods Wire Fabrics Class A B C A B C Characteristic yield strength f yk or f 0,2k (MPa) 400 to 600 k = (f t /f y ) k 1,05 1,08 1,15 <1,35 1,05 1,08 1,15 <1,35 Characteristic strain at maximum force, ε uk (%) 2,5 5,0 7,5 2,5 5,0 7,5 Fatigue stress range (N = 2 x 10 6 ) (MPa) with an upper limit of 0.6f yk In UK NA max. char yield strength, f yk, = 600 MPa BS 4449 and 4483 have adopted 500 MPa

21 Extract BS 8666

22 Cover

23 BS EN & Cover Nominal cover, c nom Minimum cover, c min c min = max {c min,b ; c min,dur ; 10 mm} Allowance for deviation, c dev Axis distance, a Fire protection

24 BS EN & Cover Minimum cover, c min c min = max {c min,b ; c min,dur ;10 mm} c min,b = min cover due to bond (φ)

25 BS EN Structural fire design Scope Part 1-2 Structural fire design gives several methods for fire engineering Tabulated data for various elements is given in section 5 Reinforcement cover Axis distance, a, to centre of bar a Axis Distance a = c + φ m /2 + φ l

26 µ fi = N Ed,fi / N Rd or conservatively 0.7 Columns: Method A

27 Flexure

28 Simplified Stress Block For grades of concrete up to C50/60, ε cu = η = 1 λ = 0.8 f cd = α cc f ck / γ c = 0.85 f ck /1.5 = 0.57 f ck f yd = f yk /1.15 = 435 MPa

29 Design flowchart The following flowchart outlines a design procedure for rectangular beams with concrete classes up to C50/60 and class 500 reinforcement Carry out analysis to determine design moments (M) Determine K and K from: M K = & K' = 0.6δ 0.18δ b d 2 f ck Note: δ = 0.8 means 20% moment redistribution. Yes Beam is under-reinforced - no compression steel needed Is K K? No Beam is over-reinforced - compression steel needed δ K It is often recommended in the UK that K is limited to to ensure ductile failure

30 Flow chart for under-reinforced beam Calculate lever arm Z from: z = [ K ] 0. 95d d 2 Calculate tension steel required from: A s = M f z yd Check minimum reinforcement requirements: 0.26f b d ctm t A b s,min f yk t d Check max reinforcement provided A s,max 0.04A c (Cl ) Check min spacing between bars > φ bar > 20 > A gg + 5 Check max spacing between bars

31 Flow chart for over-reinforced beam Calculate lever arm Z from: d z = [ K' ] Calculate excess moment from: 2 M = bd f ( K ') ck Calculate compression steel required from: M A 2 s2 = f d d yd ( ) 2 2 K 2 Calculate tension steel required from: M M ' A = + s f z yd A s2 f f sc yd Check max reinforcement provided A s,max 0.04A c (Cl ) Check min spacing between bars > φ bar > 20 > A gg + 5

32 Shear

33 Eurocode 2/BS 8110 Compared

34 Strut inclination method V Rd,max = α cw b w z cotθ + ν 1 f tanθ cd V = Rd, s A sw s z fywd cotθ 21.8 < θ < 45

35 Shear We can manipulate the Expression for the concrete strut: When cot θ = 2.5 (θ = 21.8 ) V Rd,max = b w z f ck (1 - f ck /250) Or in terms of stress: v Rd = f ck (1 - f ck /250) where v Rd = V Rd /(b z) = V Rd /(0.9 bd) When v Rd > v Ed cot θ = 2.5 (θ = 21.8 ) When v Rd < v Ed we can rearrange the concrete strut expression: θ = 0,5 sin -1 [v Rd /(0.20 f ck (1 - f ck /250))] We can also manipulate the reinforcement expression to give: A sw /s = v Ed b w /(f ywd cot θ) f ck v Rd (when cot θ = 2.5)

36 Design flow chart for shear Determine v Ed where: v Ed = design shear stress [v Ed = V Ed /(b w z) = V Ed /(b w 0.9d)] Determine the concrete strut capacity v Rd when cot θ = 2.5 v Rd = 0.138f ck (1-f ck /250) Is v RD > v Ed? No Determine θ from: θ = 0.5 sin -1 [(v Ed /(0.20f ck (1-f ck /250))] Yes (cot θ = 2.5) Calculate area of shear reinforcement: A sw /s = v Ed b w /(f ywd cot θ) Check maximum spacing of shear reinforcement : s,max = 0.75 d For vertical shear reinforcement

37 Deflection

38 Deflection The deflection limits are: Span/250 under quasi-permanent loads to avoid impairment of appearance and general utility Span/500 after construction under the quasi-permanent loads to avoid damage to adjacent parts of the structure. Deflection requirements can be satisfied by the following methods: Direct calculation (Eurocode 2 methods considered to be an improvement on BS 8110). Limiting span-to-effective-depth ratios

39 EC2 Span/effective depth ratios l 3 2 ρ0 ρ0 = K 11+ 1,5 f + ck 3,2 fck 1 if ρ ρ (7.16.a) 0 d ρ ρ l d = K ,5 f ck ρ0 + ρ ρ' 1 12 f ck l/d is the span/depth ratio K is the factor to take into account the different structural systems ρ 0 is the reference reinforcement ratio = f ck 10-3 ρ is the required tension reinforcement ratio at mid-span to resist the moment due to the design loads (at support for cantilevers) ρ is the required compression reinforcement ratio at midspan to resist the moment due to design loads (at support for cantilevers) ρ' ρ 0 if ρ > ρ 0 (7.16.b)

40 EC2 Span/effective depth ratios Structural system K Simply supported beam, one- or two-way simply supported slab End span of continuous beam or one-way spanning slab continuous slab or two-way slab over continuous over one long side Interior span of beam or one-way or two-way spanning slab Slab supported without beams (flat slab) (based on longer span) Cantilever 0.4

41 EC2 Span/effective depth ratios Span to depth ratio (l/d) 18.5 Percentage of tension reinforcement (A s,req d /bd)

42 Flow Chart Determine basic l/d Factor F1 for ribbed and waffle slabs only F 1 = ((b f /b w ) 1) 0.8 Factor F2 for spans supporting brittle partitions > 7m F 2 = 7/l eff Factor F3 accounts for stress in the reinforcement F3 = 310/σ s where σ s is tensile stress under quasi-permanent load Note: A s,prov 1.5 No A s,req d (UK NA) Increase A s,prov or f ck Is basic l/d x F1 x F2 x F3 >Actual l/d? No Yes Check complete

43 Axial

44 Column design process Determine the actions on the column Determine the effective length, l 0 Determine the first order moments Determine slenderness, λ Determine slenderness limit, λ lim No Is λ λ lim? Yes Column is slender Column is not slender, M Ed = M 02 Calculate A s (eg using column chart) Check detailing requirements

45 Effective length Actions Effective length, l 0 θ First order moments θ M Slenderness, λ l 0 = l l 0 = 2l l 0 = 0,7l l 0 = l / 2 l 0 = l l /2 <l 0 < l Braced members: l 0 = 0,5l k1 1+ 0,45 + k Unbraced members: 1 k2 1+ 0,45 + k 2 l 0 > 2l Slenderness limit, λ lim Is λ λ lim? Yes Slender No Not slender, M Ed = M 02 l 0 = l max k k + k k k k k k ; 1 2 Calculate A s Detailing

46 Effective length (2) From Eurocode 2: k = (θ / M) (EΙ / l) Alternatively... k = Where: E Ic lc 2E I l b b 0.1 (From PD 6687: Background paper to UK NA) I b,i c are the beam and column uncracked second moments of area l b,l c are the beam and column lengths Actions Effective length, l 0 First order moments Slenderness, λ Slenderness limit, λ lim Is λ λ lim? Yes Slender No Not slender, M Ed = M 02 Calculate A s Detailing

47 Effective length (3) How to Columns has a look up table l o = Fl Actions Effective length, l 0 First order moments Slenderness, λ Slenderness limit, λ lim Is λ λ lim? Yes No Slender Not slender, M Ed = M 02 Calculate A s Detailing

48 Design moment The design moment M Ed is as follows: M 01 = Min { M top, M bottom } + e i N ed Actions Effective length, l 0 First order moments M 02 = Max { M top, M bottom } + e i N ed e i = Max {I o /400, h/30, 20} M 2 = N ed e 2 For stocky columns: M Ed = M 02 There are alternative, methods for calculating eccentricity, e 2, for slender columns Slenderness, λ Slenderness limit, λ lim Is λ λ lim? Yes Slender No Not slender, M Ed = M 02 Calculate A s Detailing

49 Slenderness Actions Second order effects may be ignored if they are less than 10% of the corresponding first order effects Second order effects may be ignored if the slenderness, λ < λ lim Slenderness λ = l 0 /i where i = (I/A) Effective length, l 0 First order moments Slenderness, λ Slenderness limit, λ lim hence for a rectangular section for a circular section λ = 3.46 l 0 / h λ = 4 l 0 / h Is λ λ lim? Yes No Slender With biaxial bending the slenderness should be checked separately for each direction and only need be considered in the directions where λ lim is exceeded Not slender, M Ed = M 02 Calculate A s Detailing

50 Slenderness Limit λ lim = 20 A B C/ n where: A = 1 / (1+0,2ϕ ef ) ϕ ef is the effective creep ratio; (if ϕ ef is not known, A = 0,7 may be used) B = (1 + 2ω) ω = A s f yd / (A c f cd ) (if ω is not known, B = 1,1 may be used) C = r m r m = M 01 /M 02 M 01, M 02 are first order end moments, M 02 M 01 (if r m is not known, C = 0.7 may be used) n = N Ed / (A c f cd ) Actions Effective length, l 0 First order moments Slenderness, λ Slenderness limit, λ lim Is λ λ lim? Yes Slender No Not slender, M Ed = M 02 Calculate A s Detailing

51 Slenderness limit factor C 105 knm 105 knm 105 knm Actions Effective length, l 0 First order moments Slenderness, λ -105 knm 105 knm Slenderness limit, λ lim r m = M 01 / M 02 = 0 / 105 = 0 C = = 1.7 r m = M 01 / M 02 = 105 / -105 = -1 C = = 2.7 r m = M 01 / M 02 = 105 / 105 = 1 C = = 0.7 Is λ λ lim? Yes Slender No Not slender, M Ed = M 02 Calculate A s Detailing

52 Column design (2)

53 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Cover Flexure Shear Deflection Further Information

54 Design aids from the UK concrete sector Concise Eurocode 2 How to compendium RC Spreadsheets Worked Examples ECFE scheme sizing Properties of concrete

55 TCC Courses Eurocode 2 half-day course for building designers Background to Eurocode 2 for building designers (one day) Eurocode 2 with design workshops for building designers (one day) Background to Eurocode 2, including liquid retaining structures (one day) Design of Concrete Bridges to Eurocodes (one day) Two-day course for building designers

56 Other Resources Updated Detailing Manual Updated Green book Text Books Designer s Guides

57 Design Guidance Recent Concrete Industry Design Guidance is written for Eurocode 2 TR 64 Flat Slab TR43 Posttensioned Slabs TR58 Deflections

58 Introduction to the Eurocodes Eurocode Eurocode 1 Eurocode 2 Materials Flexure Shear Deflection Axial Further Information Worked Example

59 Worked Example G k = 75 kn/m, Q k = 50 kn/m 10 m Cover = 40mm to each face f ck = Check the beam for flexure, shear and deflection 600

60 Solution - Flexure Carry out analysis to determine design moments (M) Determine K and K from: M K = bd 2 fck & K' = 0.6δ 0.18δ Is K K? Yes Beam is under-reinforced - no compression steel needed ULS = (75 x x 1.5) = kn/m M ult = x 10 2 /8 = 2109 knm d = = K = = δ K

61 Solution - Flexure Calculate lever arm Z d z = K [ ] d z = = [ x 0.134] 0.95d Calculate tension steel M A = s f z yd A s x 10 = = 6015 mm 435 x 806 Provide 8 H32 (6430 mm 2 ) 2 Check max reinforcement provided Check min reinforcement provided Check min spacing between bars Check max spacing between bars Space between bars = 35mm > φ OK

62 Design flow chart for shear Determine v Ed where: v Ed = V Ed /(b w d) Determine the concrete strut capacity v Rd Shear force: V Ed = x (10/ ) = kn Shear stress:v Ed = V Ed /(b w d) = x 10 3 /(1000 x 600) = 1.14 MPa

63 Solution - Shear f ck v Rd (when cot θ = 2.5)

64 Design flow chart for shear Determine v Ed where: v Ed = V Ed /(b w d) Determine the concrete strut capacity v Rd Is v RD > v Ed? Shear force: V Ed = x (10/ ) = kn Shear stress:v Ed = V Ed /(b w d) = x 103/(1000 x 600) = 1.14 MPa v Rd = 3.27 MPa v Rd > v Ed cot θ = 2.5 Yes (cot θ = 2.5) Area of shear reinforcement: A sw /s = v Ed b w /(0.9 f ywd cot θ) Check maximum spacing of shear reinforcement : s l,max = 0.75 d A sw /s = 1.14 x 600 /(0.9 x 435 x 2.5) A sw /s = 0.70 mm Try H10 links with 2 legs. A sw s = 157 mm2 < 157 /0.70 = 224 mm provide H10 links at 200 mm CRS

65 Solution - Deflection Determine basic l/d Reinforcement ratio: ρ = A s /bd = 6430 x 100/(600 x 934) = 1.15%

66 Basic span-to-depth ratios (for simply supported condition) Span to depth ratio (l/d) fck = 20 fck = 25 fck = 28 fck = 30 fck = 32 fck = 35 fck = 40 fck = 45 fck = % 0.80% 1.30% 1.80% Percentage of tension reinforcement (As/bd)

67 EC2 Span/effective depth ratios

68 Solution - Deflection Determine basic l/d Is b f > 3b w No j 1 = 1.0 Beam > 7m & support brittle partitions? Reinforcement ratio: ρ = A s /bd = 6430 x 100/(600 x 934) = 1.15% Req d l/d = 14.9 x 1.0 = 14.9 Actual l/d = 10000/934 = 10.7 Basic l/d > Actual l/d No j 2 = 1.0 Is actual l/d < (l/d).j 1.j 2? Yes Check complete

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