Rc19 footing1
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Rc19 footing1 Rc19 footing1 Presentation Transcript

  • 19 Reinforced Concrete Design Design of Footings 1 Types of Footings Bearing Pressure under Footing Eccentrically Loaded Footing Wall Footings Column Footings Mongkol JIRAVACHARADET SURANAREE UNIVERSITY OF TECHNOLOGY INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING
  • Types of Footings Wall Property line
  • Isolated spread footing Wall footing
  • Combined Footings Property line A B A Rectangular, PA = PB B Rectangular, PA < PB Property line Property line A B Rectangular, PA < PB A B Strap or Cantilever
  • Pile cap Piles Weak soil Bearing stratum Mat Footing
  • Bearing pressure under footings R Axially Loaded Footings : Assume uniform pressure Actual pressure is not uniform due to: 1) Footing flexibility p, bearing pressure 2) Depth of footing below ground surface 3) Type of soil, e.g., clay or sand R Heave R Heave Cohesionless soil Cohesive soil
  • Eccentrically Loaded Footings e y P e x x b load pmin = y P Mc − A I pmax = P Mc + A I h Tensile stress cannot be transmitted between soil and concrete. For full compression, setting pmin = 0, P Mc Pec = = A I I P e= I Ac emax = h/6 For rectangular footing of length h and width b, I bh 3 / 12 h e= = = Ac bh(h / 2) 6 h/3 h/3 h/3 Middle Third
  • Large eccentricity of load e > h/6 Centroid of soil pressure concurrent with applied load e P a R= 1 (3ab )pmax = P 2 where a = h/2 - e R 3a pmax pmax = 2P 3ab
  • F 12.1 0.15 . F F F 1.8 . x1.2 . F F 80 0.40 . e = 0.15 e F 0.60 m pmax = Load 0.60 m . < [h/6 = 0.30 .] F 80 80 × 0.15 × 0.9 + 1.8 × 1.2 1.2 × 1.83 / 12 = 37.0 + 18.5 = 55.5 t/m2 0.90 m pmin = 37.0 − 18.5 = 18.5 t/m2 0.90 m e = 0.40 . > [h/6 = 0.30 .] F a = 0.90 – 0.40 = 0.50 pmax = F . 2 × 80 = 88.9 t/m2 3 × 0.50 × 1.20
  • 12.1 .. . . . 2522 ( / . . .) F F 2 F F 5 10 20 25 30 F * F F F 100
  • F F 12.2 1.5 . γs = 2.0 γc = 2.4 / . . DL = 80 ton LL = 40 ton Grade 1.0 m = (1.52)(0.5)(2.4) = 2.7 = (.32)(1.0)(2.4) = 0.22 = (1.0)(1.52-0.32)(2.0) = 4.32 = 80+40 = 120 = 127.24 30x30 cm column 0.5 m pgr = 127.24 = 56.55 t/m2 2 1.5 F pn = pgr – 1.5 / . . F F = 56.55 – 1.5(2.0) = 53.55 / . .
  • Wall Footings Uniformly loaded wall w w Wall Bending deformation Footing 1-m slice on which design is based
  • Critical Section for Moment in Isolated Footings b/2 b/2 Critical section Critical section s b/4 Concrete column, pedestal or wall Masonry wall s/2 Column with steel base plate
  • Moment and Shear in Wall Footings wu = 1.4wDL+1.7wLL Required L = (wDL+wLL)/qa qa = Allowable soil pressure, t/m2 b Factored wall load = wu t/m Factored soil pressure, qu = (wu )/L d 2 d qu L 1 L−b 1 Mu = qu  = qu (L − b )2  2  2  8 L−b  Vu = qu  −d  2  Min t = 15 cm for footing on soil, 30 cm for footing on piles Min As = (14 / fy ) (100 cm) d
  • EXAMPLE 12.3: Design of a Wall Footing to carries a dead load D of 12 t/m and a live load L of 8 t/m. The max. soil pressure is 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm2, and γs = 2.0 t/m3. D = 12 t/m L = 8 t/m Consider: 1-m strip Assume footing t = 30 cm 25 cm Net soil pressure: Df = 1.50 m 5 cm typical 8 cm clear pn = 10 - [0.3(2.4) + 1.2(2.0)] = 6.88 t/m2 Req’d footing length: L = (DL + LL) / pn = (12+8)/6.88 L = 2.91 m Ultimate soil pressure: pu = (1.4 x 12 + 1.7 x 8) / 3.0 = 10.13 t/m USE 3.0 m
  • Check Shear: d = 22 cm 25 cm 115.5 cm Vu = 10.13(1)(1.155) = 11.70 30 cm φVc = 0.85(0.53)(100)(22)/1000 = 15.35 > Vu 10.13 t/m2 Flexural design: 25 cm 137.5 cm Mu = 0.5 10.13 1.3752 = 9.58 - Mu 9.58(105 ) Rn = = φ bd 2 0.9 × 100 × 222 = 21.98 10.13 t/m2 ./ .2 OK
  • 0.85 f c′  2 Rn  1 − 1 −  = 0.0058 > [ ρ min = 0.0035] ρ=  fy  0.85 f c′   As = 0.0058(100)(22) = 12.82 F .2/ DB16 @ 0.15 (As = 13.40 .2/ F F ) F As = 0.0018(100)(30) = 5.4 F OK .2/ .2/ DB12 @ 0.20 (As = 5.65 25 cm 30 cm 3.00 m ) DB12 DB12@0.20m DB16@0.15m
  • Weight of footing ≈ 4-8 % of column load Column Footings Critical section for shear 2 d/2 d 1 Punching shear 2 Beam-shear short direction 3 Beam-shear long direction 1 d 3 Critical section for moment 2 1 2 1 Moment short direction Moment long direction
  • Two-Way Action Shear (punching-shear) On perimeter around column at distance d/2 from face of column c1 + d P d/2 c2 + d c2 b0 c1
  • Two-Way Action: cracking occur around column with periphery b0 at distance d / 2 outside column. Vn is the smallest of  4  Vn = Vc = 0.27  2 +  fc′ b0 d βc   ACI Formula (11-35)  αsd  Vn = Vc = 0.27  + 2  fc′ b0 d  b0  ACI Formula (11-36) Vn = Vc = 1.06 fc′ b0 d ACI Formula (11-37) where b0 = perimeter of critical section at distance d /2 outside column βc = ratio of long side to short side of column αs = 40 for interior columns, 30 for edge columns and 20 for corner columns
  • Distribution of Flexural Reinforcement Footing Type Square Footing Rectangular Footing s (typ.) One-way B L s (typ.) AsL As2 L s (typ.) Two-way As1 L B/2 B/2 L As2  2  = s1  AsL  β + 1 B As 2 = AsL − As1 2 AsB L β= B A
  • Transfer of Forces at Base of Column For a supported column, bearing capacity is φ Pnb = φ (0.85 fc′ A1 ) A1 where A1 = loaded area (column area) φ = 0.70 450 For a supporting footing, 2 1 A2 measured on this plane φ Pnb = φ (0.85 fc′ A1 ) A2 ≤ 2 φ (0.85 fc′ A1 ) A1 where A2 = area of lower base of the largest pyramid cone contained within footing having side slope 1 vertical to 2 horizontal
  • EXAMPLE 12.4: Design of a Square Footing to support a 40 cm square column. The column carries a dead load D of 40 ton and a live load L of 30 ton. The allowable soil pressure 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm2. Unit weight of the soil above footing base = 2.0 t/m3. D = 40 t L = 30 t (1) Determination of base area: Assume footing depth = 40 cm 1.50 m 40 cm Soil net pressure: pn = 10 – [0.4(2.4) + 1.1(2.0)] h = 6.84 t/m2 Required area = (40+30)/6.84 = 10.23 m2 b USE 3.2x3.2m square footing (10.24m2) (2) Factored loads and soil reaction: Pu = 1.4(40) + 1.7(30) = 107 tons Ultimate pressure pu = 107 = 10.45 t/m2 10.24
  • Assume footing depth = 40 cm and effective depth d = 32 cm Punching shear: Vu = 10.45(3.22 – 0.722) = 101.6 40 cm d/2=16 cm bo = 4(72) = 288 . φ Vc = 0.85(1.06) 240 (288)(32)/1000 72 cm = 128.6 > Vu OK Beam shear: 40 cm d=32 cm Vu = 10.45(1.08)(3.2) = 36.12 φ Vc = 0.85(0.53) 240 (320)(32)/1000 = 71.47 108 cm > Vu OK
  • Flexural Design: - Mu = (0.5)(10.45)(3.2)(1.4)2 = 32.77 32.77(105 ) Rn = = 11.11 0.9 × 320 × 322 ./ .2 As = 0.0029(320)(32) = 29.70 ρ = 0.0029 .2 As,min = 0.0018(320)(40) = 23.04 .2 < As OK USE 15DB16# (As = 30.15 cm2) 40 cm 4DB25 DB16 Critical section for moment 0.40 m 3.20 m 15DB16 #
  • Check development of reinforcement Critical section for development is the same as that for moment (at face of column) f ld αβγλ = 0.28 y db fc′  c + K tr    db   Edge distance (bottom and side) = 8 cm Center-to-center bar spacing = (320 - 2(8))/14 = 21.7 cm 8 cm (control) c = minimum of 21.7 / 2 = 10.9 cm Ktr = 0 (no transverse reinforcement) c + K tr 8 + 0 = = 5.0 > 2.5 db 1.6 USE 2.5
  • α = 1.0 (bottom bars) β = 1.0 (uncoated reinforcement) αβ = 1.0 < 1.7 γ = 0.8 (DB20 and smaller) λ = 1.0 (normal weight concrete) ld 4,000 1.0 × 1.0 × 0.8 × 1.0 = 0.28 = 23.1 db 2.5 240 ld = 23.1 x 1.6 = 37.0 cm > 30 cm OK Since ld = 37 cm < available embedment length (320/2 - 40/2 - 8 = 132 cm), DB16 bars can be fully developed.
  • Transfer of Force at Base of Column (1) Bearing strength of column column bars footing dowels φPnb = φ (0.85f’c A1) = 0.70(0.85x240x40x40)/1,000 32 cm = 228.5 tons > 107 tons OK 8 cm cover (2) Bearing strength of footing 40 cm Bearing strength of footing increased by factor A2 A1 ≤ 2 where A2 is area of pyramid cone having side slope 1 vertical to 2 horizontal 320 cm A2 200 cm A1 A2 = A1 200 × 200 = 5 > 2, use 2 40 × 40
  • φPnb = 2φ (0.85f’c A1) = 2(0.70)(0.85x240x40x40)/1,000 = 457 tons > 107 tons OK (3) Required dowel bars between column and footing: Even though column and footing have enough bearing strength to transfer load, area of reinforcement across interface ≥ 0.005(gross area of supported member) As (min) = 0.005(40x40) = 8.0 cm2 Provide 4DB16 bars as dowels (As = 8.04 cm2) (4) Development of dowel reinforcement in compression: In column & footing: For DB16 bars: ld = ld = 0.075d bfy fc′ ≥ 0.0043d bfy 0.075 × 1.6 × 4,000 240 = 31.0 cm l d (min) = 0.0043 × 1.6 × 4,000 = 27.5 cm (control)
  • Available length for development in footing = footing thickness - cover - 2(footing bar dia.) - dowel bar dia. = 40 - 8 - 2(1.6) - 1.6 = 27.2 cm ≈ 27.5 cm OK Therefore, the dowels can be fully developed in the footing. Home work: Design a square spread footing with the following design conditions: Service dead load = 150 ton Service live load = 120 ton Unit weight of soil = 2.0 ton/m3 Allowable soil pressure = 20 ton/m2 Column dimensions = 60 x 30 cm P Ground elev. 1.5 m