Foundations
Load from roof to column
Load from floor to column
(a) Load from floor to column
(b)
Fig. 3.69 Foundation failures: (a) sliding failure; (b) overturning failure.
Load from column to foundation Foundation loads resisted by ground
Fig. 3.68 Loading on foundations.
3.11 Foundations Foundations are required primarily to carry the dead and imposed loads due to the structure’s floors, beams, walls, columns, etc. and transmit and distribute the loads safely to the ground (Fig. 3.68). The purpose of distributing the load is to avoid the safe bearing capacity of the soil being exceeded otherwise excessive settlement of the structure may occur. Foundation failure can produce catastrophic effects on the overall stability of a structure so that it may slide or even overturn (Fig. 3.69). Such failures are likely to have tremendous financial and safety implications. It is essential, therefore, that much attention is paid to the design of this element of a structure.
3.11.1 FOUNDATION TYPES There are many types of foundations which are commonly used, namely strip, pad and raft. The
(a)
(b)
Fig. 3.70 Pad footing: (a) plan; (b) elevation.
foundations may bear directly on the ground or be supported on piles. The choice of foundation type will largely depend upon (1) ground conditions (i.e. strength and type of soil) and (2) type of structure (i.e. layout and level of loading). Pad footings are usually square or rectangular slabs and used to support a single column (Fig. 3.70). The pad may be constructed using mass concrete or reinforced concrete depending on the relative size of the loading. Detailed design of pad footings is discussed in section 3.11.2.1. Continuous strip footings are used to support loadbearing walls or under a line of closely spaced columns (Fig. 3.71). Strip footings are designed as pad footings in the transverse direction and in the
N = N = N = N A
B
C
D
Elevation
Plan
(a)
(b)
Fig. 3.71 Strip footings: (a) footing supporting columns; ( b) footing supporting wall.
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Design in reinforced concrete to BS 8110
(a)
Plan
(b)
(c)
Typical sections through raft foundations
Fig. 3.72 Raft foundations. Typical sections through raft foundation: (a) flat slab; (b) flat slab and downstand; (c) flat slab and upstand.
longitudinal direction as an inverted continuous beam subject to the ground bearing pressure. Where the ground conditions are relatively poor, a raft foundation may be necessary in order to distribute the loads from the walls and columns over a large area. In its simplest form this may consist of a flat slab, possibly strengthened by upstand or downstand beams for the more heavily loaded structures (Fig. 3.72). Where the ground conditions are so poor that it is not practical to use strip or pad footings but better quality soil is present at lower depths, the use of pile foundations should be considered (Fig. 3.73). The piles may be made of precast reinforced concrete, prestressed concrete or in-situ reinforced concrete. Loads are transmitted from the piles to
Soft strata
Hard strata
Fig. 3.73 Piled foundations.
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the surrounding strata by end bearing and/or friction. End bearing piles derive most of their carrying capacity from the penetration resistance of the soil at the toe of the pile, while friction piles rely on the adhesion or friction between the sides of the pile and the soil.
3.11.2 FOUNDATION DESIGN Foundation failure may arise as a result of (a) allowable bearing capacity of the soil being exceeded, or (b) bending and/or shear failure of the base. The first condition allows the plan-area of the base to be calculated, being equal to the design load divided by the bearing capacity of the soil, i.e. design load bearing Ground pressure = plan area < capacity of soil
(3.25)
Since the settlement of the structure occurs during its working life, the design loadings to be considered when calculating the size of the base should be taken as those for the serviceability limit state (i.e. 1.0Gk + 1.0Qk). The calculations to determine the thickness of the base and the bending and shear reinforcement should, however, be based on ultimate loads (i.e. l.4Gk + 1.6Qk). The design of a pad footing only will be considered here. The reader is referred to more specialised books on this subject for the design of the other foundation types discussed above. However, it should be borne in mind that in most cases the design process would be similar to that for beams and slabs.
3.11.2.1 Pad footing The general procedure to be adopted for the design of pad footings is as follows:
Foundations 1. Calculate the plan area of the footing using serviceability loads. 2. Determine the reinforcement areas required for bending using ultimate loads (Fig. 3.74). 3. Check for punching, face and transverse shear failures (Fig. 3.75).
Load on shaded area to be used in design
Fig. 3.74 Critical section for bending.
1.0d
1.5d
1.5d
Punching shear perimeter = column perimeter + 8 × 1.5d
Face shear
Transverse shear
Fig. 3.75 Critical sections for shear. (Load on shaded areas to be used in design.)
Example 3.15 Design of a pad footing (BS 8110) A 400 mm square column carries a dead load (G k ) of 1050 kN and imposed load (Q k ) of 300 kN. The safe bearing capacity of the soil is 170 kNm−2. Design a square pad footing to resist the loads assuming the following material strengths: fcu = 35 Nmm−2
fy = 500 Nmm−2
Axial loads: dead = 1050 kN imposed = 300 kN
PLAN AREA OF BASE Loading Dead load Assume a footing weight of 130 kN Total dead load (Gk) = 1050 + 130 = 1180 kN Serviceability load Design axial load (N ) = 1.0Gk + 1.0Q k = 1.0 × 1180 + 1.0 × 300 = 1480 kN Plan area Plan area of base =
N 1480 = = 8.70 m2 bearing capacity of soil 170
Hence provide a 3 m square base (plan area = 9 m2) 117
Design in reinforced concrete to BS 8110
Example 3.15 continued Self-weight of footing Assume the overall depth of footing (h) = 600 mm Self weight of footing = area × h × density of concrete = 9 × 0.6 × 24 = 129.6 kN < assumed (130 kN)
BENDING REINFORCEMENT Design moment, M Total ultimate load (W ) = 1.4G k + 1.6Q k = 1.4 × 1050 + 1.6 × 300 = 1950 kN Earth pressure ( ps ) =
W
=
plan area of base
1950 9
= 217 kNm−2
400
1300
1300
217 kN/m2
Maximum design moment occurs at face of column (M ) =
psb 2 217 × 1.300 2 = 2 2
= 183 kNm/m width of slab
Ultimate moment Effective depth Base to be cast against blinding, hence cover (c) to reinforcement = 50 mm (see Table 3.8). Assume 20 mm diameter (Φ) bars will be needed as bending reinforcement in both directions.
d
Φ Cover
Hence, average effective depth of reinforcement, d, is d = h − c − Φ = 600 − 50 − 20 = 530 mm Ultimate moment Mu = 0.156fcubd 2 = 0.156 × 35 × 103 × 5302 = 1534 × 106 Nmm = 1534 kNm Since Mu > M no compression reinforcement is required. 118
Foundations
Example 3.15 continued Main steel K =
M 183 × 106 = 0.0186 = fcubd 2 35 × 1000 × 5302
z = d[0.5 + (0.25 − K /0.9)] = d[0.5 + (0.25 − 0.0186/0.9)] = 0.979d ≤ 0.95d = 0.95 × 530 = 504 mm As =
M
183 × 106
=
0.87f y z
0.87 × 500 × 504
= 835 mm2 /m
Minimum steel area is 0.13%bh = 780 mm2/m < A s
OK 2
Hence from Table 3.22, provide H20 at 300 mm centres (A s = 1050 mm /m) distributed uniformly across the full width of the footing parallel to the x–x and y–y axis (see clause 3.11.3.2, BS 8110).
CRITICAL SHEAR STRESSES Punching shear
1.5d
Critical perimeter
Critical perimeter, pcrit, is = column perimeter + 8 × 1.5d = 4 × 400 + 8 × 1.5 × 530 = 7960 mm Area within perimeter is (400 + 3d)2 = (400 + 3 × 530)2 = 3.96 × 106 mm2 Ultimate punching force, V, is V = load on shaded area = 217 × (9 − 3.96) = 1094 kN Design punching shear stress, υ, is υ=
V p critd
=
1094 × 103 7960 × 530
= 0.26 Nmm−2
100 A s 100 × 1050 = = 0.198 bd 103 × 530
Hence from Table 3.11, design concrete shear stress, υc, is υc = (35/25)1/3 × 0.37 = 0.41 Nmm−2 Since υc > υ, punching failure is unlikely and a 600 mm depth of slab is acceptable. 119
Design in reinforced concrete to BS 8110
Example 3.15 continued Face shear Maximum shear stress (υmax) occurs at face of column. Hence
υmax =
W 1950 × 103 = = 2.3 Nmm−2 < permissible (= 0.8 35 = 4.73 Nmm−2) column perimeter × d (4 × 400) × 530
Transverse shear
770 mm 530 mm
d
Ultimate shear force (V ) = load on shaded area = ps × area = 217(3 × 0.770) = 501 kN Design shear stress, υ, is
υ=
V
=
501 × 103 3 × 103 × 530
bd Hence no shear reinforcement is required.
= 0.32 Nmm−2 , υc
REINFORCEMENT DETAILS The sketch below shows the main reinforcement requirements for the pad footing. 01-11H20-300 alternate bars reversed A
A
75 kicker
01-11H20-300 alternate bars reversed
Column starter bars (not designed)
01
01 01
01
01
Section A–A
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01
Retaining walls Natural ground slope Retaining wall necessary to avoid demolition of building Existing ground level Building
Fig. 3.76 Section through road embankment incorporating a retaining wall.
3.12 Retaining walls Sometimes it is necessary to maintain a difference in ground levels between adjacent areas of land. Typical examples of this include road and railway embankments, reservoirs and ramps. A common solution to this problem is to build a natural slope between the two levels. However, this is not always possible because slopes are very demanding of space. An alternative solution which allows an immediate change in ground levels to be effected is to build a vertical wall which is capable of resisting the pressure of the retained material. These structures are commonly referred to as retaining walls (Fig. 3.76). Retaining walls are important elements in many building and civil engineering projects and the purpose of the following sections is to briefly describe the various types of retaining walls available and outline the design procedure associated with one common type, namely cantilever retaining walls.
3.12.1 TYPES OF RETAINING WALLS Retaining walls are designed on the basis that they are capable of withstanding all horizontal pressures and forces without undue movement arising from deflection, sliding or overturning. There are two
Mesh reinforcement
main categories of concrete retaining walls (a) gravity walls and (b) flexible walls.
3.12.1.1 Gravity walls Where walls up to 2 m in height are required, it is generally economical to choose a gravity retaining wall. Such walls are usually constructed of mass concrete with mesh reinforcement in the faces to reduce thermal and shrinkage cracking. Other construction materials for gravity walls include masonry and stone (Fig. 3.77). Gravity walls are designed so that the resultant force on the wall due to the dead weight and the earth pressures is kept within the middle third of the base. A rough guide is that the width of base should be about a third of the height of the retained material. It is usual to include a granular layer behind the wall and weep holes near the base to minimise hydrostatic pressure behind the wall. Gravity walls rely on their dead weight for strength and stability. The main advantages with this type of wall are simplicity of construction and ease of maintenance.
3.12.1.2 Flexible walls These retaining walls may be of two basic types, namely (i) cantilever and (ii) counterfort.
Granular backfill
Drainage layer
Weep holes
Porous pipe
(a)
Mass concrete footing (b)
Fig. 3.77 Gravity retaining walls: (a) mass concrete wall; (b) masonry wall.
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