Rc19 footing1

1,955 views

Published on

Published in: Design, Technology, Education
1 Comment
4 Likes
Statistics
Notes
No Downloads
Views
Total views
1,955
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
Downloads
115
Comments
1
Likes
4
Embeds 0
No embeds

No notes for slide

Rc19 footing1

  1. 1. 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
  2. 2. Types of Footings Wall Property line
  3. 3. Isolated spread footing Wall footing
  4. 4. 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
  5. 5. Pile cap Piles Weak soil Bearing stratum Mat Footing
  6. 6. 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
  7. 7. 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
  8. 8. 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
  9. 9. 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
  10. 10. 12.1 .. . . . 2522 ( / . . .) F F 2 F F 5 10 20 25 30 F * F F F 100
  11. 11. 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 / . .
  12. 12. Wall Footings Uniformly loaded wall w w Wall Bending deformation Footing 1-m slice on which design is based
  13. 13. 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
  14. 14. 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
  15. 15. 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
  16. 16. 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
  17. 17. 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
  18. 18. 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
  19. 19. 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
  20. 20. 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
  21. 21. 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
  22. 22. 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
  23. 23. 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
  24. 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
  25. 25. 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 #
  26. 26. 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
  27. 27. 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.
  28. 28. 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
  29. 29. 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)
  30. 30. 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

×