Reinforced Concrete Design. Serviceability

Transcription

1 21 Reinforced Concrete Design Serviceability Cracking in Flexural Members ACI Provision for Crack Control Deflection of Elastic Sections Creep and Shrinkage Deflections ACI Provision for Deflection Control Mongkol JIRAVACHARADET S U R A N A R E E UNIVERSITY OF TECHNOLOGY INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING

2 What is serviceability? Normal service = load actually expected to act ( load factor = 1.0 ) Serviceability = Satisfactory performance under normal service condition - Adequate strength - Service load deflections - Long-term deflections - Tension crack virtually disturb and corrosion of steel - Vibration - Fatique

3 Strength Design Method - more accurate assessment of capacity - higher strength materials more slender members more service load problems Crackings Deflections

4 Cracking in Flexural Members E Modular ratio n = s 8 10 E c Modulus of rupture f = 2 f 30 kg/cm r c 2 2 = = Stress in steel f 8(30) 240 kg/cm s f y Concrete always cracks at normal service condition

5 Well designed beam: - Fine flexural crack hair line little corrosion - Well distributed When service load increase more than cracking load, crack width becomes wider and number of cracks becomes larger.

6 Gerely-Lutz Equation for Crack Width Effective tension area of concrete 2y y h 1 h 2 d c Neutral axis Steel centroid 3 3 Cracking w= 0.011β dc A 10 mm w where f s = tensile stress under normal service, kg/cm 2 = 0.6 f y (if no data) d c = concrete cover, cm β = distance ratio h 1 /h 2 = 1.20 for beam = 1.35 for one-way slab A = concrete area around one bar, cm 2 total effective area = = number of bars 2 y b w n

7 Tolerable crack widths for reinforced concrete Tolerable crack width Exposure condition in. mm Dry air or protective membrane Humidity, moist air, soil Deicing chemical Seawater and seawater spray; wetting and drying Water-retaining structures, excluding nonpressure pipes

8 ACI Provision for Crack Control For beam, β = 1.20 Define w w = = = z f 3 s dc A Interior beam z 31,000 kg/cm ( w 0.41 mm ) Exterior beam z 26,000 kg/cm ( w 0.34 mm ) For one-way slab, β = 1.35 Interior slab z 31,000(1.2/1.35) = 28,000 kg/cm Exterior slab z 26,000(1.2/1.35) = 23,000 kg/cm

9 T-beam flange in tension: A s b b E L /10 A s Deep beam (h > 90 cm): 90 cm A sk s d/2 d Ask s max ( d 75) cm / m d / 6 30 cm

10 Minimum number of bar in one layer Total tensile area = 2 d c b w Tensile area per bar: A = 2d c b w m d c 2d c m = number of bars in one layer 4 cm cover b w From z 2d 2 3 c bw dc bw z= fs dc A = m= 3 fs m ( z / fs) Example: SD40: f y = 4,000 kg/cm 2, f s = 0.6(4,000) = 2,400 kg/cm 2 covering = 4 cm stirrup 9-10 mm d c = d b ( + db) ( z / 2,400) bw m= max m= 2 3

11 Deflection of Elastic Sections 1) Excessive deflection cracking of partitions Wall 2) Ponding effect of roof rain 3) Misalignment of machine 4) Visually offensive sag Working Stress Design (WSD) Deflection is controlled indirectly by limiting service load stress result in large member. Ultimate Stress Design (USD) Members become more slender and/or smaller sections may result in deflection problems.

12 Intermediate Deflections Simply supported (ideal condition) w = 4 5 wl 384EI L End moments caused by monolithic joints (real condition) M a w M b L max = 5M 0 3( M + M ) 48EI 2 2 a b a ML β EI M = 0 wl 8 2 M a M b

13 Effective moment of of inertia for Continuous T-beam sections A C B A C B Section A-A Section B-B Section C-C

14 Variation of of Flexural Rigidity with with applied bending moment E c I based on gross section plus transformed area of reinforcement E c I E c I based on cracked transformed section 0.2 M u M u I g I e I cr M a /M cr

15 Deflection of of RC Beam Load Service load Computed deflection based on transformed cracked section Computed deflections using gross I Actual deflection Cracking load E c I g E c I cr Deflection M 2 Nonlinear material range M 1 M cr Deflection

16 Effective Moment of Inertia I cr I e I g 3 3 M M I = I + 1 I I cr cr e g cr g Mmax Mmax where M cr fr Ig = = cracking moment h y t y t M max = Maximum service load moment b I g = Gross section moment of inertia = bh 3 /12 I cr = Transformed cracked section moment of inertia f r = Modulus of rupture = 2 f c y t = Distance from N-A to tension face

17 Single Value of Effective Moment of Inertia I e1 I m I e2 1) Midspan value I e = I m 2) Weighted average I e = 0.70 I m ( I e1 + I e2 ) for both ends continuous I e = 0.85 I m I e1 for one end continuous 3) Simple average I e = 0.50 I m ( I e1 + I e2 ) for both ends continuous I e = 0.75 I m I e1 for one end continuous

18 Dead Load and Live Load Deflections M DL+LL M DL M cr (I e ) DL (I e ) DL+LL I g cr DL LL DL+LL Dead load deflection: = DL β a E ML 2 ( I ) c e DL Dead load and live load deflection: = DL+ LL β a E ML ( I ) 2 c e DL+ LL Live load deflection: = - LL DL+ LL DL

19 Example 1: Investigate the instantaneous deflection for the simply supported beam over a span of 10 m. f c =280kg/cm 2, f y =4,000kg/cm 2 40 cm 8 ton (LL) 5 m Beam weight 700 kg/m(dl) 60 cm 52 cm 10 m 8DB25, A s = cm 2 Minimum depth from ACI table: L/16 = 10(100)/16 = 62.5 cm > 60 cm NG Deflection must be checked (a) Dead load deflection: I g 1 = (40)(60) = 720,000 cm M max 1 (0.7)(10) t-m = = 8

20 For transformed cracked section x 40 cm N.A. f = c E c 280 kg/cm 2 = 15,100 f = 254,512 kg/cm c 2 na s n= Es / Ec = = 8 254,512 Compute neutral axis location: [ΣM N-A = 0] 2 x 40 = 8(39.27)(52 x) 2 x x = 0 x = 21.8 cm 1 Icr = Iconcrete + Isteel = (39.27) ( )( ) ( ) 3 2 I cr = 424,663 cm 4

21 M f r cr = = 33.5 kg/cm fr Ig ,000 = = = y t 2 8,040 kg-m M M cr max 8,040 Mcr = = 0.92 ; = ,750 Mmax 3 Effective moment of inertia: Dead load deflection: I e = 0.78(720,000) (424,663) = 655,026 cm wL /100 (10 100) DL = = = 384E I , ,026 c e 0.55 cm

22 (b) Dead load and live load delfection: M max = (10)/4 = t-m M M cr max 8,040 Mcr = = 0.28 ; = ,750 Mmax 3 I e = 0.022(720,000) (424,663) = 431,160 cm wL PL 5 7 (10 100) 8000(10 100) DL+ LL = + = + 384E I 48E I c e c e = = 2.37 cm (c) Live load delfection: LL = DL+LL - DL = = 1.82 cm ( ) L Allowable LL = = = 2.78 cm > 1.82 cm OK

23 Long-Term Deflections Creep and and Shrinkage Strain Creep Shrinkage True elastic strain Nominal elastic strain t 0 Time

24 Creep Effect on on Deflections under sustained load f c E c E ct Sustained loading i Instantaneous loading 0.5f c Service load condition ε i ε cp Creep x cp x i C t i Creep effect Strain A s ε s

25 Creep Effect on Deflections under Sustained Load Factors: 1) Constituents 4) Age and duration of loading 2) Curing temp. and Humidity 5) Magnitude of stress 3) Size of concrete member Creep coefficient: Creep deflection: C t cp εcp = ε i ( ) = C t i DL ACI Code: C t 0.6 t = C t u where t = time in days after loading C u = ultimate creep = 2.35 for 40% humidity

26 Standard creep coefficient variation For 10 cm or less slump, 40% humidity, moist cured and loading age of 7 days C u C t = 0.78C u at 1 year C t = 0.90C u at 5 year C t Duration of loading, days

27 Creep Correction Factor (CF) Conditions to use ACI equation: - 40% relative humidity - 7 days loading age (moist cured) - 10 cm or less slump days loading age (stream cured) - 15 cm average thick 1) Age of loading: ( CF) Moist cured = 1.25t ( CF) Stream cured = 1.13t a a a a where t a = age of loading in days 2) Humidity for H 40% : ( CF) = h where H = relative humidity in % 3) Compression steel effect: k r ( ) cp r t i DL H 0.85 A s =, ρ = 1+ 50ρ bd = k C

28 Shrinkage Effect on Deflections Under Sustained Load Shrinkage deflection: =α φ sh 1 2 sh L where α 1 = 0.50 cantilever beam = simply supported beam = one-end continuous beam = both-end continuous beam φ sh = shrinkage curvature L = span length, m

29 Shrinkage strain: Moist cured 7 days: Stream cured 1-3 days: ε ε t = 35+ t ( ε ) sh sh u t = 55+ t ( ε ) sh sh u where (ε sh ) u = ultimate shrinkage strain = cm/cm for 40% humidity Correction Factor (CF): ( ) CF = H 40% H 80% ( ) h CF = H H 80% h

30 Shrinkage Curvature φ sh Singly reinforced beam: ε sh h d ε s φ sh ε sh ε s = d ε sh ε s = 1 d ε sh φ sh where ε s = compressive strain in steel Singly and doubly reinforced beam: ε sh 1/3 ρ ρ φsh = 0.7 ( ρ ρ ) for ( ρ ρ ) 3% h ρ φ sh ε sh = for ( ρ ρ ) > 3% h 1/ 2 where ρ or ρ = 100( A or A ) / bd s s

31 Creep and Shrinkage Deflections ( ) λ( ) = kξ = cp+ sh r i DL i DL λ= kξ = r ξ 1+ 50ρ Duration of sustained load ξ 5 years or more year months months 1.0

32 Example 2: From beam in Ex.1, check the total deflection for sustained load at 5 years or more 52 cm 60 cm 40 cm 2DB25 8DB25 Solution: (1) Intermediate deflection from Ex.1 ( i ) DL = 0.55 cm ( i ) DL+LL = 2.35 cm ( i ) LL = 1.80 cm (2) Compute creep and shrinkage deflection: ρ = A / bd = 2(4.91)/(40)(52) = s 2.0 λ = krξ = = (0.0047) ( ) = λ = 1.62(0.55) = 0.89 cm cp+ sh i DL Since ( ι ) DL can be accommodated by camber L 10(100) Deflection = ( i ) LL + cp+ sh = = 2.69 cm < = = 2.78 cm OK