CONCRETE FOUNDATION DESIGN EXAMPLE FULL WIDTH OF

All aggressive materials in the soil should be stated as well so that adequate protection will be given to the pile material(s) for durability.Design Example 1 Reinforced Concrete Wall For this design example, it is assumed that the foundation system is rigid, and thus the wall is considered to have a xed base. Depending on the soil properties and building loads, the engineer may choose to support the structure on a shallow or deep foundation system.The soil investigation report handed to the structural engineer for the design of pile foundation purpose should contain the embedment length of the piles, the recommended sizes of the pile, the safe working load of each pile size, and other information that may be necessary for the structural engineer to do his design properly. A walk-through of the calculations required to design an isolated footing (ACI 318-14) The foundation is an essential building system that transfers column and wall forces to the supporting soil.

Concrete Foundation Design Example Full Width Of

A typical construction drawing showing this interaction is given in Figure 7.(2) Contractor should maintain a minimum concrete cover of 75 mm(3) There may be need to be cast the pile cap in two stages to achieve the configuration shown in Figure 7. The ground beams are usually embedded into the pile caps, or may be allowed to sit directly on the pile caps depending on the site level. There are construction scenarios where the ground floor slab is placed directly on the pile caps, but note that this concept is quite different from piled raft foundation. This arrangement can be found in Table 3.16 of Reynolds et al (2008), and it is shown in Figure 6.Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m 3 = 1.4 x 7.29 m 2 x 1.3m x 24 kN/m 3 = 318.43 kNTotal load on pile cap at ULS = 2545 kN + 318.43 kN = 2863.43 kNTensile force to be resisted by the reinforcement in both directions F t = x = 528 kNA st = F t/0.95f y = (528 x 1000)/(0.95 x 460) = 1208 mm 2The shear force carried by the piles V = 2863.43/4 = 715.9 kNThe shear stress ОЅ = V/bd = (715.9 x 1000)/(1000 x 1200) = 0.595 MPaThe structural engineer is expected to provide the following drawings (1) Setting out drawing showing the piling points and layout with a known reference point(2) General column/pile cap layout/arrangement(3) Pile cap/ground beam/ground floor slab layout(3) Columns, piles and pile cap reinforcement drawings (detailing)(5) Ground beams and ground floor slab reinforcement detailing(1) Ground beams are usually provided to chain the pile caps together, and to provide the needed support for the ground floor slab. (2008), the tensile force to be resisted within the pile cap is given by N = Column axial load at ultimate limit stateL = Centre to centre spacing of the pilesA = dimension of the column side parallel to the length of the pile capSelf weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m 3 = 1.4 x 2.7m x 0.9m x 1.3m x 24 kN/m 3 = 106.14 kNF t = x = 364 kNA st = F t/0.95f y = (364 x 1000)/(0.95 x 460) = 833 mm 2Provide 6T20 175 c/c (As prov = 1974 mm 2)Critical position for shear on vertical section across full width of pile-cap occurs at distance from face of column given by:A v = 0.5( l вЂ“ c) вЂ“ 0.3П† = 0.5(1800 вЂ“ 450) вЂ“ (0.3 x 600) = 495 mmFig 5: Structural arrangement of 3 pile capsSelf weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m 3 = 1.4 x 5.166 m 2 x 1.3m x 24 kN/m 3 = 225.61 kNTotal load on pile cap at ULS = 1476 kN + 225.61 kN = 1701.61 kNTensile force to be resisted by the reinforcement in the direction parallel to X-X F t,x = x = 284 kNTensile force to be resisted by the reinforcement in the direction parallel to Y-Y F t,y = x = 275 kNLet us use the highest value for design in anticipation that we will provide the same reinforcement in both directionsA st = F t/0.95f y = (284 x 1000)/(0.95 x 460) = 649 mm 2Provide T20 175 c/c in both directions (As prov = 1974 mm 2)The shear force carried by the piles V = 1701.61/3 = 567.2 kNThe shear stress ОЅ = V/bd = (567.2 x 1000)/(1000 x 1200) = 0.472 MPaV c(2d/a v) = 0.292 x = 1.415 MPa > 0.472 MPa This is okayThe number of П†600 mm piles required = 1825/493.48 = 3.69Service load per pile = 1825/4 = 456.25 kNSafe working load of П†600 mm piles = 493.48 kN > 456.25 kN This is okayLet us adopt a square pile cap arranged in such a way that the column load will be equally distributed to the piles.

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