FOOTINGS - imcyc.comimcyc.com/biblioteca/ArchivosPDF/Zapatas/4 Footings.pdf · senting the topic of design of footings in this chapter. ... These are similar to the combined footings,

FOOTINGS

12.1 INTRODUCTION

Cumulative floor loads of a superstructure are supported by foundation substructures in direct contact with the soil. The function of the foundation is to transmit safely the high concentrated column and/or wall reactions or lateral loads from earth-retaining walls to the ground without causing unsafe differential settlement of the supported structural system or soil failure.

If the supporting foundations are not adequately proportioned. one part of a structure can settle more than an adjacent part. Various members of such a system become overstressed at the column-beam joints due to iinev.cn settlement of the supports leading to large deformations. The additional bending and torsional moments in excess of the resisting capacity of the members can lead to excessive cracking due to yielding of the reinforcement and ultimately to failure.

If the total structure undergoes even settlement, little or no overstress occurs. Such behavior is observed when the foundation is excessively rigid and the supporting soil highly yielding such that a structure behaves similar to a floating body that can sink or tilt without breakage. Numerous examples of such structures can be found in such locations

Photo 72 One Shell Plaza. New Orleans. (Courtesy of Portland Cement Association.)

520 520

12.1 lntroduction

1 I

i i

52 1

as Mexico City with buildings on mat foundations or rigid supports that sank several feet over the years due to the high consolidation of the supporting soil. Examples of other fa- mous cases of very slow and relatively uneven consolidation process can be cited. Grad- ual loss of stability of a structure undergoing tilting with time, like the leaning Tower of Pisa, is an exarnple of foundation problems resulting from uneven bearing support.

Layouts of structural supports vary widely and soil conditions differ from site to site and within a site. As a result, the type of foundation to be selected has to be governed by these factors and by optimal cost considerations. In summary, the structural engineer has to acquire the maximum economically feasible soil data on the site before embarking on a study of the various posible alternatives for site layout.

Basic knowledge of soil mechanics and foundation engineering is assumed in pre- senting the topic of design of footings in this chapter. Background knowledge of the rnethodology of deterrnining the resistance of cohesive and noncohesive soils is necessary to select the appropriate bearing capacity value for the particular site and the particular foundation system under consideration.

The bearing capacity of soils is usually deterrnined by borings, test pits, or other soil investigations. If these are not available for the preliminary design, representative values at the footing leve1 can normally be used from Table 12.1.

Table 12.1 Presumptive Bearing Capacity (tons/V)

Type of Soil Bearing Capacity

Massive crystalline bedrock, such as granite, diorite, gneiss. and trap rock 1 O0 40 Foliated rocks, such as schist or da te

Sedimentary rocks, such as hard shales, sandstones. limestones, and siltstones Grave1 and gravel-sand mixtures (GW and G P soils)

Densely compacted Medium compacted Loose, not compacted




Sands and gravely sands. well graded (SW soil)

Sands and gravely sands, poorly graded (SP soil)

Silty gravels and gravel-sand-silt mixtures (GM soil)

Silty sand and silt-sand mixtures (SM soil) Clayey gravels. gravel-sand-clay mixtures, clayey sands, sand-clay mixtures

(GC and SC soils)

15

5 4 3

3: 3 21

3 28 1 3

2b

14 2

2 2

Inorganic silts, and fine sands; silty or clayey fine sands and clayey silts,

Inorganic clays of high plasticity, fat clays; micaceous or diatomaceous fine

1 with slight plasticity; inorganic clays of low to medium plasticity; gravely clays: sandy clays; silty clays; lean clays (ML and CL soils)

sand or silty soils, elastic silts (CH and M H soils)

' 1

522 Chapter 12 Footings

12.2 TYPES OF FOUNDATIONS

There are basically six types of foundation substructures, as shown in Figure 12.1. The foundation area must be adequate to carry the column loads, the footing weight. and any overburden weight within the permissible soil pressure.

Property line * 1.1 n

i -- .

l Figure 12.1 Types of foundations: (a) wall footing; (b) isolated footing; (c) corn- bined footing; (d) strap footing; (e) pile foundation; (f) rafí foundation.

12.3 Shear and Flexura1 Behavior of Footings 523

1. Wull footings. Such footings comprise a continuous slab strip along the length of the wall having a width larger than the wall thickness. The projection of the slab footing is treated as a cantilever loaded up by the distributed soil pressure. The length of the projection is determined by the soil bearing pressure, with the critica1 section for bending being at the face of the wall. The main reinforcement is placed perpendicular to the wall direction.

2. Independent isolated column footings. These consist of rectangular or square slabs of either constant thickness or doping toward the cantilever tip. They are reinforced in both directions and are economical for relatively small loads or for footings on rock.

3. Combined footings. Such footings support two or more column loads. They are necessary when a wall column has to be placed on a property line and the footing slab cannot project outside the property line. In such a cáse, an independent footing would be eccentrically loaded, causing apparent tension on the foundation soil.

In order to achieve a relatively uniform stress distribution, the footing for the exterior wall column can be combined with the footing of the adjoining interior column. Additionally, combined footings are also used when the distance between adjoining columns is relatively small, such as in the case of corridor columns, when it becomes more economical to build a combined footing for the closely spaced columns.

4. Cantilever or strup footings. These are similar to the combined footings, except that the footings for the exterior and interior columns are built independently. They are joined by a strap beam to transmit the effect of the bending moment produced by the eccentric wall column load to the interior column footing area.

5. Pile foundutions. This type of foundation is essential when the supporting ground consists of structurally unsound layers of material to large depths. The piles may be driven either to solid bearing on rocks or hardpan or deep enough into the soil to develop the allowable capacity of the pile through skin frictional resistance or a combination of both. The piles could be either precast, and hence driven into the soil, or cast in place by drilling a caisson and subsequently filling it with concrete. The precast piles could be reinforced or prestressed concrete. Other types of piles are made of steel or treated wood. In al1 types, the piles have to be provided with appropriately designed concrete caps reinforced in both directions.

6. Rufr, mal, or floating foundaiions. Such foundation systems are necessary when the allowable bearing capacity of soil is very low to great depths, making pile foundations uneconomical. In this case it becomes necessary to have a deep enough excavation with sufficient depth of soil removed that the net bearing pressure of the soil on the foundation is almost equivalent to the structure load. It becomes necessary to spread the foundation substructure over the entire area of the building such that the superstructure is considered to be theoretically floating on a raft. Continuously consolidating soils require such a substructure, which is basically an inverted floor system. Otherwise, friction piles or piles driven to rock become mandatory.

12.3 SHEAR AND FLEXURAL BEHAVIOR OF FOOTINGS

To simplify foundation design, footings are assumed to be rigid and the supporting soil layers elastic. Consequently, uniform or uniformly varying soil distribution can be assumed. The net soil pressure is used in the calculation of bending moments and shears by subtracting the footing weight intensity and the surcharge from the total soil pressure. If a column footing is considered as an inverted floor segment where the intensity of net soil

524 Chapter 12 Faotings

Pboto 73 Reinforced concrete footing after excavation. (Tests by F. E. Richari.)

pressure is considered to be acting as a column-supported cantilever slab, the slab would be subjected to both bending and shear in a similar manner to a floor slab subjected to gravity loads.

When heavy concentrated loads are involved, it has been found that shear rather than flexure control5 most foundation designs. The mechanism of shear failure in footing slabs is similar to that in supported floor slabs. However, the shear capacity is considerably higher than that of beams, as will be discussed in the next section. Since the footing in most cases bends in double curvature, shear and bending about both principal axes of fhe footing plan have to be considered.

The state of stress at any element in the footing is due primanly to the combined effects of shear, flexure, and axial compression. Consequently, a basic understanding of the fundamental behavior of the footing slab and the cracking mechanism involved is essen- tiai. It enables developing a background feeling for the underlying hypothesis used in the analysis and design requirements of footings both in shear and in flexure.

12.3.1 Failure Mechanism

The inclined shear cracks develop in essentially the same manner as in beams, stabilizing at approximately 65% of the ultimate load and extending rapidly toward the neutral axis. Thereafter, the cracks propagate slowly toward the compression zone such that a very shallow depth in compression remains at failure.

The inclined cracks always form close to the concentrated load or column reaction in two-way slabs or footings, as seen in Figure 12.2a. This is due partly to the heavy concentration of bending moments in the region close to the column face, forming a trun- cated pyramid at the foot of the column region. The column can perimetrically punch through the slab in this failure form if the slab is not adequately designed to resist shear failure (also called diagonal fension or punching shear). The action of the confining sur- rounding punched slab on the column base interface punching through the slab in Figure 12.2b can be represented by the resulting shear forces VI and V,, the compressive forces, C, and C, and the tensile forces TI and T2, in addition to the interna1 dowel and mem- brane action of the slab.

Figure 12.2~ shows an infinitesimal element taken from the compression zone above the inclined crack. The element is subjected to the following four stress compo- nents: (1) vertical shear stress v,, (2) direct compressive stressf,, (3) vertical compressive stress f,, and (4) lateral compressive stress f,.

12.3 Shear and Flexura1 Behavior of Footings 525

't,

P"

I '4 I 't,

J

flgure 12.2 Two-way action failure mechanism in slabs and footings: (a) footing elevation; (b) failure pyramid; (c) element A in the comprescion zone.

The vertical shearing stress v, is the result of the total shear that has to be entirely transmitted by the compression zone above the inclined crack. "he direct compressive stressf,, which varies along the length of the critica1 section, results from the bending moments. The vertical compressive force f3 is due to the. heavy concentrated column load. It has a major influence on increasing the shear capacity of the slab, as demonstrated in Ref. 12.2 for pressure in an infinite semielastic solid loaded at the surface. The lateral compressive stress fi is the result of the bending moment about an axis perpendicular to the critical section. It contnbutes further to the increase in the compressive strength of the concrete as a result of the triaxial state of stress. Consequently, the existence of the multiaxial forces and stresses in Figure 12.2~ explains why the shear capacity of a slab subjected to concentrated loads in considerably higher than that of a beam.

In addition, the inclined crack generating close to the critical section in two-way slabs and footings due to the high moment concentration justifies considering the critical section to be at a distance of d/2 from the face of the column in slabs and footings, while in beams and one-way slabs and footings, the ACI Code specifies the critica1 section at a distance d from the face of the column support. The nominal shearing stress at failure varies between 6 a and 9 f i for the footing slabs, whereas it does not exceed 2 f i to 4 f i in beams. The Code, however, allows a maximum nominal resisting shear strength of plain concrete not to exceed v, = 4*, for the supported two-way slab or the footing and v, = 2vf;i for beams and one-way-action footings. For plain concrete footings cast


against soil, the effective thickness used in computing stresses is taken as the overall thickness minus 3 in. Tñe overall thickness should not be less than 8 in.

12.3.2 Loads and Reactlons

Based on the foregoing discussion, it is essential to make the correct assumptions for evaluating al1 the combined forces acting on the foundation. The footing slab has to be proportioned to sustain al1 the applied factored loads and induced reactions, which in- clude axial loads, shears, and moments to be resisted at the base of the footing.

After the permissible soil pressure is determined from the available site data and the principles of soil mechanics and the local codes, the footing area size is computed on the basis of the unfucrored (service) loads, such as dead, live, wind, or earthquake loads in whatever combination governs the design.

The minimum eccentricity requirement for column slenderness considerations is neglected in the design of footings or pile caps and only the computed end moments that exist at the base of a column are considered to have been transferred to the footing. In cases where eccentric loads or moments exist due to any loading combinations. the extreme soil pressure resulting from such loading conditions has to be within such permissible bearing values such as those in Table 12.1 or as determined by actual soil tests.

Once the size of a footing or pile cap for a single pile or a group of piles is determined, the design of the footing geometry becomes possible using the principles and methodologies presented in the preceding chapters for shear and flexure design. The ex- terna1 service loads and moments used to determine the size of the foundation area are converted to their ultimate fucrored values using the appropriate load factors and strength reduction factors 4 for determining the nominal resisting values to be used in the analysis and proportioning the size and reinforcement distribution in the footing.

12.4 SOlL BEARING PRESSURE A l BASE OF FOOTINGS

The distribution of soil bearing pressure on the footing depends on the manner in which the column or wall loads are transmitted to the footing slab and the degree of rigidity of the footing. The soil under the footing is assumed to be a homogeneous elastic material, and the footing is assumed to be rigid as a most common type of foundation. Conse- quently, the soil bearing pressure can be considered uniformly distributed if the reaction load acts through the axis of the footing slab area. If the load is not axial or symmetrically applied, the soil pressure distribution becomes trapezoidal due to the combined effects of axial load and bending.

12.4.1 Eccentric Load Effect on Footings

As indicated in Section 12.2, exterior column footings and combined footings can be subjected to eccentric loading. When the eccentric moment is very large. tensile stress on one side of the footing can result, since the bending stress distribution depends on the magnitude of load eccentricity. It is always advisable to proportion the area of these footings such that the load falls within the middle kern, as shown in Figures 12.3 and 12.4. In such a case, the location of the load is in the middle third of the footing dimension in each direction, thereby avoiding tension in the soil that can theoretically occur prior to stress redistribution.

1. Eccenrricity case e e U6 (Figure 12.3a). In this case, the direct stress P/A, is larger than the bending stress MJI. the stress

12.4 Soil Bearing Pressure At Base of Footings

Q Q Q I

Flgure 12.3 Eccentrically loaded footings.

P Pelc p - = - +- Af 1 P Pelc

pmin=--- Af

2. Eccentricity case e, = L/6 (Fig. 12.3b):

P P direct stress = - = - A, SL

C Pez x - Me bendingstress = - = I I

1 6 =-=- c L/2 I s(L3/12) s(L2/6) sL2 - =

(12.la)

(12.lb)

(12.24

(12.2b)

(12.24

l . Figure 12.4 Biaxial loading of footing.

528

t 3 f t

I

Chapter 12 Footings

14 X 14 in. column

where s and L are the width and the length of the footing, respectively. In order to find the limiting case where no tension exists on the footing, the direct stress P/A, has to be equivalent to the bending stress so that

C Pez- = O

P - _ A, I

Substituting for P/Af and C/I from Eqs. 12.2a and c into Eq. 12.2d.

(12.2d)

Consequently, the eccentric load has to act within the middle third of the footing dimension to avoid tension on the soil.

3. Eccentricity case e3 > L/6 (Figure 12 .3~) . As the load acts outside the middle third. tensile stress results at the left side of the footing, as shown in Figure 12 .3~. If the maximum bearing pressure p,,, due to load P does not exceed the allowable bearing capacity of the soil, no uplift is expected at the left end of the footing. and the center of gravity of the triangular bearing stress distribution coincides with the point of action of load p in Figure 12 .3~.

The distance from the load P to the top of footing is r = (L/2) - ei = distance of the centroid of the stress triangle from the base of the triangle. Therefore. the width of the triangle is 3r = 3[(L/2) - e,]. Hence the maximum compressive bearing stress is

4. Eccentricity about two axes, biaxial loading (Fig. 12.4). In the case where a concentrated load has an eccentricity in two directions (both within their respective kern points), the stresses are

(12.3b)

12.4.2 Example 12.1 : Concentrically Loaded Footings

A column support transmits axially a total service load of 400.000 lb (1779 kN) to a square footing at the frost line (3 ft below grade). as shown in Figure 12.5. The frost line is the suh- grade soil leve1 below which the groundwater does not freeze throughout the year. Test bor-

P = 400,000 Ib (1779 kN)

Grade

t t t t t t t I P"

Figure 12.5 Concentrically loaded footing.

12.4 Soil Bearing Pressure At Base of Footings 529

ings indicate a densely compacted gravel-sand soil. Determine the required area of the footing and the net soil pressure intensity p , to which it is subjected. Given:

unit weight of soii y = 135 ib/ft3 (21.1 kN/m3)

footing siab thickness = 2 ft (0.61 m)

Solution: Since the footing is concentrically ioaded, the soil bearing pressure is considered uniformly distributed assuming the footing is rigid. From the soil test borings and Table 12.1. the presumptive bearing capacity of the soii is 5 tonslft’ at the level of the footing, that is, 10,000 Ib/ft2 (478.8 kPa). Assume that the average weight of the soil and concrete above the footing is 135 pcf. Since the top of the footing has to be below the frost line (minimum 3 ft below grade), the net allowable pressure i s

P , ~ = 10,000 - (5 X 135 + 100 psf for surcharge paving) = 9225 psf

400,000 9,225

minimum area of footing Af = ~ - - 43.36 ft2

Use square footing 6 f t 8 in. x 6 f t 8 in. (2.03 m x 2.03 m):

A, = 44.44 ft2 (4.13 mZ) > 43.36 ft2

12.4.3 Example 12.2: Eccentrically Loaded Footings

A reinforced concrete footing supports a 14 in. x 14 in. column reaction P = 400,000 Ib (1779 kN) at the frost iine (3 ft below grade). The load acts at an eccentricity e, = 0.4 ft, e, = 1.3 ft, and e3 = 2.2 ft. Select the necessary area of footing assuming that it is rigid and has a thickness h = 2; ft. Soil test borings indicate that the bearing area is composed of iayers of shale and clay to a considerable depth below the foundation. Use a unit weight y = 140 Ib/ft3.

Solution: From Table 12.1, assume an ailowable bearing capacity p R = 6.5 tons/ft2 (13,000 Ib/ft2) at the footing base level.

Eccentricity e, = 0.4ft

By triai and adjustment, assume a footing 5 ft x 9 ft (1.52 m x 2.74 m), Af= 45 ft2. As- sume that the footing base is 6 ft below grade and that a slab on grade surcharge weighs 120 psf. Assume that the average weight of the soil and footing is = 140 pcf.

net allowable bearing pressure p n = 13,000 - (6 X 140 + 120)

= 12,040 Ib/ft2 (576.5 KPa)

Stress due to the service eccentric column load is

p = - ? - - - P P x e 400,000 400,000 X 0.46 X 6 1 bh2 , where- = - A, I/c 45 5(912 c 6

= 8889 ? 2370 = 11,259 Ib/ft2 (C) and 6519 ib/ft2 (C) < 12,040 ib/ft2

? -

The distribution of the bearing pressure is as shown in Figure 12.6a; therefore, O.K.

Eccentricity e, = 1.3ft

(5.57 m2). The actual service load-bearing pressure is

400,000 400,000 X 1.3 X 6

By triai and adjustment, assume a footing 6 ft x 10 ft (1.83 m x 3.05 m), A,= 60 ft2

+- p=60.0 6( 10)’

= 6667 2 5200 = 11,867 Ib/ft2 (C) and 1467 ib/ft2 (C)

< 12.040 Ib/ft2 therefore, O.K.

530

400,000 Ib

11,259 psf 6519 psf

1 1,067 psf

1467 psf

400,000 Ib

Chapter 12 Footings

(C)

Figure 12.6 Bearing area and bearing stress distribution in Ex. 12.2.

Notice in comparing the two cases that as the moment increases leading to larger eccentrici- ties, the minimum bearing pressure decreases, as seen from Figure 12.6a and b.

Eccentricity e3 = 2.2 f t By trial and adjustment, try a footing 7 ft x 11 ft (2.13 m x 3.35 m), Af= 77 ft2 (7.15 m’).

400,000 400,000 X 2.2 X 6 p = - * 77.0 7(11)’

= 5195 ? 6234 = 11,429 lb/ft’ (C) and - 1039 Ib/ft’ (T)

Check by Eq. 12.3 for e > L/6 > 11.0/6 = 1.83 ft:

Design C

DESIGN

onsiderations in Flexure 53 1

2 X 400,000 3 X 7[(11/2) - 2.21

- - 2P 3S[(L/2) - e3]

= 11,544 lb/ft 2 < 12,040 lb/ft2 O.K.

Figure 12 .6~ shows that as the load acts outside the middle third of the base only part of the footing is subjected to compressive bearing stress.

CONSIDERATIONS IN FLEXURE

The maximum externa1 moment on any section of a footing is determined on the basis of computing the factored moment of the forces acting on the entire area of footing on one side of a vertical plane assumed to pass through the footing. This plane is taken at the following locations:

1. At the face of column, pedestal, or wall for an isolated footing, as in Figure 12.7a 2. Halfway between the middle and edge of wall for footing supporting a masonry

3. Halfway between face of column and edge of steel base for footings supporting a wall, as in Figure 12.7b

column with steel base plates

12.5.1 Reinforcement Distribution

In one-way footings and in two-way square footings, the flexura1 reinforcement should be uniformly distributed across the entire width of the footing. This recommendation is con- servative, particularly if the soil bearing pressure is not uniform. However, no meaningful saving can be accomplished if refinement is made in the bending moment assumptions.

P Plft

a

b

Figure 12.7 Critica1 planes in flexure: (a) concrete column; (b) masonry wall.


In two-way rectangular footings supporting one column, the bending moment in the short direction is taken as equivalent to the bending moment in the long direction. The distribution of reinforcement differs in the long and short directions. The effective depth is assumed without meaningful loss of accuracy to be equal in both the short and long directions, although it differs slightly because of the two-layer reinforcing mats. The following is the recommended reinforcement distribution:

1. Reinforcement in the long direction is to be uniformly distributed across the entire width of the footing.

2. For reinforcement in the short direction, a central band of width equal to the width of footing in the short direction shall contain a major portion of the reinforcement total areas as in Eq. 12.4 uniformly distributed along the band width:

reinforcement in band width total reinforcement in short direction, A,

(12.4)

where p is the ratio of long to short side of footing. The remainder of the reinforcement required in the short direction is uniformly distributed outside the center band of the footing.

2 p + 1

- -

In al1 cases, the depth of the footing above the reinforcement has to be at least 6 in. (152 mm) for footings on soil and at least 12 in. (305 mm) for footings on piles (footings on piles must always be reinforced). A practica1 depth for column footings should not be less than 9 in. (229 mm).

12.6 DESIGN CONSIDERATIONS IN SHEAR

As discussed in Section 12.3.1, the behavior of footings in shear is not different from that of beams and supported slabs. Consequently, the same principles and expressions as those used in Chapter 6 on shear and diagonal tension are applicable to the shear design of foundations. The shear strength of slabs and footings in the vicinity of column reactions is governed by the more severe of the following two conditions.

12.6.1 Beam Action

The critical section for shear in slabs and footings is assumed to extend in a plane across the entire width and located at a distance d from the face of the concentrated load or reaction area. In this case, if only shear and flexure act, the nominal shear strength of the section is

V, = 2 e b,d (12.5)

where b, is the footing width. V, must always be larger than the nominal shear force V, = VU/+ unless shear reinforcement is provided.

12.6.2 Two-way Action

The plane of the critical section perpendicular to the plane of the slab is assumed to be so located that it has a minimum perimeter b,. This critical section need not be closer than d/2 to the perimeter of the concentrated load or reaction area. The fundamental shear failure mechanism in two-way action as presented in Section 12.3.1 demonstrates that the critical section occurs at a distance d/2 from the face of the support and not at d as in beam action. Maximum allowable nominal shear strength is the smallest of

12.6 Design Considerations in Shear T

533

(12.6a)

(12.6b)

where a, is 40 for interior columns, 30 for edge columns, and 20 for comer columns.

(iii) V , = 4.\/T;, bod ( 12.6~)

where p, = long side c,/short side c, of the concentrated load or reaction area b, = perimeter of the critica1 section, that is, the length of the idealized failure plane

Figure 12.8 gives the relationship of the column side ratio p, to the shear strength V, of the footing. V, must always be larger than the nominal shear force V,, = Vu/+ unless shear reinforcement is provided.

In cases of both one- and two-way action, if shear reinforcement consisting of bars or wires is used,

V, = V, + V, 5 6fiL bod (12.7)

where V , = 2 f i bod and V, is based on the shear reinforcement size and spacing as described in Chapter 6, unless shear heads made from steel1 or channel shapes are used.

It is worthwhile to keep in mind that in most footing slabs, as in most supported superstructure slabs or plates, the use of shear reinforcement is not popular, due to practi- cal considerations and the difficulty of holding the shear reinforcement in position.

12.6.3 Force and Moment Transfer at Column Base

The forces and moments at the base of a column or wall are transferred to the footing by bearing on the concrete and by reinforcement, dowels, and mechanical connectors. Such reinforcement can transmit the compressive forces that exceed the concrete bearing strength of the footing or the supported column as well as any tensile force across the interface.

The permisible bearing stress on the actual loaded area of the column base or footing top area of contact is

or f b = +(O.SSf:), where + = 0.70

fb = 0.60fc

(12.8a)

(12.8b)

O 0.5 1 .o 1

8, -

Figure 12.8 Shear strength in footings.


Hence the permisible bearing stress on the column can normally be considered 0.60 f i for the column concrete. The compressive force that exceeds that developed by the permissible bearing stress at the base of the column or at the top of the footing has to be car- ried by dowels or extended longitudinal bars.

As the footing supporting surface is wider on al1 sides than the loaded area. the code allows the design bearing strength on the loaded area to be multiplied by G. but the value of cannot exceed 2.0. A , is the loaded area and A? is the maximum area of the supporting surface that is geometrically similar and concentric with the loaded area.

A minimum area of reinforcement of 0.005Ag (but not less than four bars) has to be provided across the interface of the column and the footing even when the concrete bearing strength is not exceeded, A , (in.*) being the gross area of the column cross section.

Lateral forces due to horizontal normal loads, wind, or earthquake can be resisted by shear-friction reinforcement, as described in Section 6.10.

T l

12.7 OPERATIONAL PROCEDURE FOR THE DESIGN OF FOOTINGS

The following sequence of steps can be used for the selection and geometrical proportioning of the size and reinforcement spacing in footings.

1. Determine the allowable bearing capacity of the soil based on site boring test data and soil investigations.

2. Determine the service loads and bending moments acting at the base of the columns supporting the superstructure. Select the controlling service load and moment combinations.

3. Calculate the required area of the footing by dividing the total controlling seryice load by the selected allowable bearing capacity of the soil if the load is concentric or by also taking into account the controlling bending stress if combined load and bending moments exist.

4. Calculate the factored loads and moments for the controlling loading condition and find the required nominal resisting values by dividing the factored loads and moments by the applicable strength reduction factors 4.

5. By trial and adjustment, determine the required effective depth d of the section that has adequate punching shear capacity at a distance d from the support facgfor one- way action and at a distance dl2 for two-way action such that V , = 2 Vf;. h,,d for one-way action and V , = smallest of values from Eqs. 12.6 for two-way action. \\.here b, is the footing width for one-way action and h, is the perimeter of the failure planes in two-way action. Use an average value of d, since there are two reinforcing mats in the footing. If the footing is rectangular, check the beam shear capacity in each direction on planes at a distance d from the face of the column support.

6. Calculate the factored moment of resistance M I , on a plane at the face of the column support due to the controlling factored loads from that plane to the extremity of the footing. Find M,, = MI,/(+ = 0.9). Select a total reinforcement area A , based on M , and the applicable effective depth.

7. Determine the size and spacing of the flexura1 reinforcement in the long and short directions: (a) Distribute the steel uniformly across the width of the footing in the long direc-

(b) Determine the portion A,, of the total steel area A , determined in step 6 for the tion.

short direction to be uniformly distributed over the central band:

12.7 Operational Procedure for the Design of Footings [ ~

l I

l 1

535

Distribute uniformiy the remainder of the reinforcement (A, - AJ outside the center band of the footing. Verify that the area of steel in each principal direction of the footing plan exceeds the minimum value required for temperature and shrinkage: A, = 0.0018bWd for sections reinforced with grade 60 steel and 0.0020bWd with grade 40 steel.

8. Check the deveiopment length and anchorage avaiiable to verify that bond requirements are satisfied (see Chapter 10).

9. Check the bearing stresses on the column and the footing at their area of contact such that the bearing strength Pnb for both is larger than the nominal value of column reaction P, = PJ(+ = 0.70). For footing bearing Pnb = - (0.85f2)), a not to exceed 2.0.

10. Determine the number and size of the dowel bars that transfer the column load to the footing siab.

Figure 12.9 presents a flowchart for the sequence of calculation operations.

12.7.1 S1 Footing Design Expressions

E, = f.^ 0.043 fi MPa

fr = 0.7 fi where fy, is in MPa

Atrfyr K,, = -

260sn

If (Y = p = y = X = 1.0 and f: = 27.6 MPa,

Shear in beam action

V, = 2 fi b,d

Shear in mo- way action The smallest of

V, = 2 + - b,d/12 ( B,) V, = 4 df: bod/12

ci, = 40 for interior coiumns, a, = 30 for edge columns, and ci, = 20 for comer columns.


+ htan?im üm M brring cqnciry, p., b.rd on Yte borinp DR diti and so11 inwnipnion1; thai dsaniw üm e x m l CWiUOllinp mrvb

UUI I d , P, and the mmmt P.

f Ulculrta the requimri arw of tho footifq, A,, uch dnt

p”’P f 3 whmlic.;bP A, Iic

b - width of the footinp I - bm of the fmtinp +

1 Ulwlns üm hnad aimfml hui. P., ind f r t d e x m l m m m t U,

Enimtn the effrtwe dpth d oí the f d n p rsquind to raaR punchinp h r r hl im; ümn mke the criticil -ion at a d h m d f r m the

uppm fud for onmny r t i o n n d di2 for ru~v.y mion

Y, csritv: Ve - 2 <b,d fw onbmy mion and v. - 1MII.R Of

iil V, - (2 + $) 8 b,d

iiil V. - (5 + 2 ) Ji: b,d

whm 4 - u) for inferiw mlumm, 30 for mlumm and 2ü for mmer columm liiil V, - 4Ji: b,d for fmt‘my d a k

whm b, - fmtinp width

bo - psnmna of the fiilum plim in -y vt ion

1

Ukubu thi frtond imnding m w m . U,, at th. f w of th. mlurnn uppwt dw m üm mmrdlinp f m d I d

1

Dimibum tho mal unifcfmly awoa the width oí thi footina in üm low diration

D N r m i n the ponion A,, of the total atni a r n A, fw thehm dirrtion to be uniiwmly dimibuted o w the mnvil bond:

A,, - L A 8 + 1 ’

Dimibum uniiwmlv the rmuinder of the n i n f w m m t IA, -A,?) niside the mnmf band of the footinp; check to verify ümt A, mtisfies the minimum Pmpsnture minforcsmmt requirmmts; the total =ea A, is the fame in

the iow and the shm dirrtions + Chack the reinfwmmat dsvslopment lcnpth to mtisfy bond mimments 1

1 Determine the number and size of the dowsl b r s that sansfer the mlumn lod to the fwtinp dab 1

(- End )

Figure 12.9 Flowchart for footing design.

Exarnples of Footing Decign 537

EXAMPLES OF FOOTING DESIGN

12.8.1 Example 12.3: Design of Two-way lsolated Footing

Design the footing thickness and reinforcement distribution for the isolated square footing in Ex. 12.1 if the total service load P = 400,000 comprises 230,000 Ib (1023 kN) dead load and 170,000 Ib (756 kN) live load. Given:

f : = 3000 psi (20.7 MPa), normal-weight concrete (footing) f : = 5500 psi (37.9 MPa) in column

f , = 60,000 psi (413.7 MPa)

Solution: Factored load intensity (step 4)

column size = 14 in. x 14 in. (356 mm x 356 mm)

footing area = 6 ft 8 in. x 6 ft in 8 in. (2.03 m x 2.03 m), Af= 44.49 ft2

assumed footing slab thickness h = 2 ft

factored load U = 1.4 x 230,000 + 1.7 x 170,000 = 611,000 Ib

Data from Ex. 12.1:

U 611,000 factored load intensity = qs = - = ___ = 13,733 Ib/ft2 (657.6 kPa)

Af 44.49

Shear capacity (step 5)

Assume that the thickness of the footing slab = 2 ft. The average depth d = h - 3 in. minimum clear cover minus steel diameter = 20 in.

Beam action (at d f r o m support face): The area to be considered for factored shear V, is shown as ABCD in Figure 12.10.

factored V,, = 13.733 ft 8 in.) = 99,340 Ib

V,, 99,340 required V,, = - = - = 116,871 Ib + 0.85

b , = 6 ft 8 in. = 80 in. (2.03 m)

available V, = 2 f i b,d = 2- X 80 X 20 = 175,271 lb

Two-way action (at d 2 f r o m support face): The area to be considered for factored shear V,, is equal to the total area of footing less area EFGH of the failure zone.

factored V,, = 13,733 44.49 - ___ = 500,736Ib [ ( 1 4 3 1

vll + required V,, = - = 589,100 Ib (2620 kN)

bo = perimeter of failure zone EFGH = (14 + 20)4 = 136 in.

14 14

p =-=1 .0

The available nominal shear ctrength from Eqs. 12.6 is the smallest of

538

howeis \ , 4

Chapter 12 Footings

ccirc-

A l

Figure 12.1 O Details of footing in Ex. 12.3.

or

V , = 2 + - d%%Ó X 136 X 20 = 893.882 Ib ( 3 = 1,174,317 Ib

where (Y, = 40 used for interior column value

V, = 4v3000 X 136 X 20 = 595,922 Ib controls

Since available V,. = 595,922 Ib, > required V, = 589,100 lb. Therefore. d = 20 in. is adr- quate for shear.

Bending moment capacity (steps 6 and 7)

The critica1 section is at the face of the column

6 f t 8 i n . 14 moment arm = ~ - __ - - 2 ft 9 in.

2 2 x 12

[ (2 ft in.)'] factored moment M,, = 13,733 X 6.67

12.8 Exarnples of Footing Design

I

l

539

= 346,359.1 ft-lb = 4,156,310 in.-lb

M , 4,156,310 4 0.90

M,=-= = 4,618,122 in.-lb

(521.8 kN-m)

M , = ATf,. (d - 4) Assume that (d - d2) = 0.9d. Use average d = 20 in.

4.618,122 = A, X 60,000 X 0.9 X 20

or 4,618,122

60,000 X 0.9 X 20 A , = = 4.28 in.2/80-in. band

4.28 X 60,000 0.85 X 3000 X 80

= 1.26 in. - - ASfi a = - 0.85f:b

4,618.122 = A, X 60,000

A, 3.98 bd 80 X 20

A, = 3.98 in.2 p = - = ~ = 0.0025

Minimum allowable shrinkage steel

pmin = 0.0018 < p O.K.

Use 14 No. 5 bars (A, = 4.27 in.2) each way spaced at = 51 in. (139.7 mm) center to center.

Development of reinforcement (step 8)

The critical section for development-length determination is the same as the critical section in flexure, that is. at the face of the column. From Table 10.2 e, = 24 in. for No. 5 bars (bottom bars).

1 . 2

check = -0 = 54.8 < 100 O.K.; s = 5 - in. > 2dh

Use l,, = 24 in. The projection length of each bar beyond the column face is

1 - (6 ft 8 in. - 14 in.) - 3 in. cover = 30 in. > 24 in. 2

O.K.

Force transfer at interface of colurnn and footing (step 9) Column fi. = 5500 psi. Factored P,, = 611,000 Ib.

(a) Bearing strength on column using Eq. 12.8b:

or +Pnh = 0.70 X 0.85f:.A1 = 0.60fCA, +P,ih = 0.60fi.A, = 0.60 X 5500 X 14 X 14

= 646,800 Ib > 611,000 O.K.

From step 9 of the design operational procedure on bearing strength on footing concrete.

G = J T = 5.714 > 2.0 use 2.0

+Pnh = 2.0(0.60fi.A,) = 2.0 X 0.60 X 3000 X 14 X 14 = 705,600 Ib

> 611,000 O.K.


Dowel bars benveen column and footing (step 10) Even though the bearing strength at the interface between the column and the footing

slab is adequate to transfer the factored P”, a minimum area of reinforcement is necessary across the interface. The minimum A, = 0.005 (14 x 14) = 0.98 in.’, but not less than four bars. Use four No. 5 bars as dowels (A, = 1.22 in.’).

Development of dowel reinforcement in compression: From Eqs. 10.7 a and b for No. 5 bars and Section 10.3.5,

and ldb 2 0.0003ddy, where db is the dowel bar diameter. Within column,

0.02 X 0.625 X 60,000 = 10.11 in.

0.003 x 0.625 X 60,000 = 11.25 in. VZGi ld =

controls

Within footing,

0.02 X 0.625 X 60,000 = 13.69 in.

v5.G Id =

Available length for development above the footing reinforcement assuming column bars size to be the same as the dowel bars size:

1 = 24 - 3 (cover) - 2 X 0.625 (footing bars) - 0.625 (dowels)

= 19.13 > 13.69in. O.K.

12.8.2 Example 12.4: Design of Two-way Rectangular lsolated Footing

Determine the size and distribution of the bending reinforcement of an isolated rectangular footing subjected to a concentrated concentric factored column load P, = 770,000 lb (3425 kN) and having an area 10 ft x 15 ft (3.05 m x 4.57 m). Given:

f : = 3000 psi (20.7 MPa), footing

f y = 60,000 psi (413.7 MPa)

columnsize = 14in. X 18in.

770,000 10 x 15

Solution: factored load intensity qs = ~ - - 5134 lb/ft2

Shear capacity (step 5) Through trial and adjustment, assume that the footing slab is 2 f t 4 in. thick. Beam action (at distance d from column face): Average effective depth = 2 ft 4 in. - 3

From Figure 12.11, length CD subjected to bearing intensity qJ, in one-way beam ac- in. (cover) - in. (diameter of bars in first layer ) = 24 in.

tion:

15 ft 18in. 24in. 2 2 x 12 12

- 4 ft 9 in. = 57 in.

factored V, = 5134 X 10 ft X 4 f t 9 in. = 243,865 Ib

V, 243,865 required V, = - = ~ - - 286,900 lb + 0.85

available V, = 2 f i b,d = 2- X 120 X 24

= 315,488 Ib > 286,900 O.K.

7

541

P, = 770.000 Ib

l 1 12.8 Examples of Footing Design

\ Tweway action

Beam action shear plane

c r

- c 11 H

I 1 I

L = 15f t

Figure 12.1 1 Beam action and two-way action planes in Ex. 12.4.

Notice that the shorter side length was used for b, to give the lower available V,, value.

Two-way action (at distance d 2 from column face): loaded area outside the failure zone LMNP in Figure 12.11

= 15 X 10 - (c, + d ) ( ~ , + d )

(18 + 24)(14 + 24) 144

= 150 -

= 138.92 ft2

factored V, = 5134 X 138.92 = 713,215 lb

713 215 0.85

required V, = = 839,077 lb (3732 kN)

perimeter of shear failure plane b, = 2[(c, + d ) + (es + d ) ]

= 2[(18 + 24) + (14 + 24)] = 160 in.

! 542 Chapter 12 Footings

From Eqs. 12.6,

V,= 2 + - f i b , d 1 4 f i b , d ( 2 18 14

p, = - = 1.286

or

V, = 2 + ~ V%6 X 160 X 24 = 1,074,851 lb. ( 1 .;86)

(yg + 2 ) m x 160 x 24

= 1.682,604 Ib

and

V, = 4 f i bod = 4- X 160 X 24 = 841,302 lb controls

Design of two-way reinforcernent

arm is in the long direction: The critica1 section for bending is at the face of the column. The controlling moment

15 ft 18in. 2 2 x 1 2

= 6.75 ft (2.06 m)

10(6.75)* factored moment M, = 5134 X ~

2

= 1,169,589 ft-lb = 14.035.073 in.-lb (1586 kN-m)

14,035,073 0.9 M, = = 15,526 in.-lb (1762 kN-m)

Assume that (d - d 2 ) = 0.9d.

M, = A,f, d - - or 15,594.526 = A, x 60.000 x 0.9 x 24 ( 3 15,594,526

60,000 X 0.9 X 24 A, = = 12.03 in.'/lO-ft-wide strip

Check:

12.03 X 60.000 0.85 X 3000 X 120

= 2.36 in - - ATf, a=- 0.85f:b

15,594,526 = A, X 60,000

11.39 10 ft

A , = 11.39 h2 = ~ = 1.14 in.' ft/width

Try No. 8 bars, A, = 0.79 in.2 per bar.

11.39 0.79

number of bars in the short direction = ~ = 14.42

Use 15 bars.

of Footing Design 543

Reinforcement in the short direction

The band width = s = 10 ft (Fig. 12.11). From Eq. 12.4,

15 p = - = 1.5 10

2 -- A,, or -- As1 2 - As P + i 11.39 1 + 1.5

Therefore,

2 X 11.39 2.5

= 9.11 im2 As1 =

to be placed in the central 10-ft-wide band and the balance (11.39 - 9.11 = 2.28 in.*) to be placed in the remainder of the footing. Use twelve No. 8 bars in the central band = 9.48 in.2 and two No. 8 bars at each side of the band, as in Figure 12.12. To complete the design, a check of the development length, bearing stress at the column-footing interface, and dowel action has to be made, as in Ex. 12.3.

12.8.3 Example 12.5: Proportioning of a Combined Footing

A combined footing has the layout shown in Figure 12.13. Column L at the property line is subjected to a total service axial load PL = 200,000 lb (889.6 kN), and the interna1 column R is subjected to a total service load P, = 350,000 lb (1556.8 kN). The live load is 35% of the total

15 No. 8 I 3 in. clear c m r

Plan

Figure 12.12 Footing reinforcement details of Ex. 12.4.


Resultant

T 6 f t 6 in.

_L 6 in.

Figure 12.13 Combined footing plan geometry in Ex. 12.5.

load. The bearing capacity of the soil at the leve1 of the footing base is 4000 Ibift' (191.5 kPa). and the average value of the soil and footing unit weight y = 120 pcf (1922 kgim'). A surcharge of 100 Ib/ft* results from the slab on grade. Proportion the footing size and select the necessary size and distribution of the footing slab reinforcement and verify the development length required. Given:

f : = 3000 psi (20.7 MPa)

f, = 60,000 psi (413.7 MPa)

base of footing at 7 f t below grade

Solution: capacityp, = p K - 120(7 ft height to base of footing) - 100 or

total columns load = 200,000 + 350,000 = 550,000 Ib (2446.4 kN) net allowable soil

pn = 4000 - 120 X 7 - 100 = 3060 Ib/ft'

P 550,000

P,, 3060 minimum footing area Af = - = ~ = 179.8 f t2

Center of gravity of column loads from the property line:

200,000 X 0.5 + 350.000 X 20.5 550.000

y = = 13.23 ft

length of footing L = 2 X 13.23 = 26.46 ft

Use L = 27 ft.

179.8 27.0

width of footing S = ~ = 6.66 f t

Use S = 6 f t 6 in. as shown in Figure 12.13.

Factored shears and rnornents

colunin L: P, = 0.65 X 200.000 = 130,000 Ib

P, = 200,000 - 130,000 = 70,000 Ib

P , = 1.4 X 130.000 + 1.7 X 70.000 = 301.000 Ib

colunin R: P , = 227.500 Ib

PL = 122,500Ib

Pu = 1.4 X 227,500 + 1.7 X 122.500 = 526.750 Ib

The net factored soil bearing pressure for footing structural design is

12.8 Exarnples of Footing Design [ ~ l I

545

LUS.b/l Ib ,

C L

" 4\ 20 f t

Column +

R

327,473 Ib

Figure 12.14 Shear diagrarn of footing in Ex. 12.5.

P, 301,000 + 526,750 = 4716.5 Ib/ft*

6.5 X 27.0 q s = - = AJ

Assume that the column loads are acting through their axes.

factored bearing pressure per foot width = qs X S = 4,716.5 X 6.5 = 30,658 Ib/ft

6 12

V,, at center line of column L = 301,000 - 30,658 X - = 285,671 lb

V, at center h e of column R = 526,750 - 30,658 X 6.5 = 327,473 lb

The maximum moment is at the point C of zero shear in Figure 12.14 x (ft) from the center of the left column L.

285,671 lb 30,658 plf

X = = 9.32 f t

Taking a free-body diagram to the left of a section through C, the factored moment at point C is

w,1' uc 2 PUF M =- -

(9.32 + 0.50)' 2

- 301,000 X 9.32 M , from left side = 30,658

= -1,327,108 ft-lb = -15,925,293 in.-lb (Fig. 12.15)

(27.0 - 9.82)' 2

- 526,750(20.0 - 9.32) M,, from right side = 30,658

= 1,101,299 ft-lb = -13,215,588 in.-lb; less than -15,925,293 in.-lb

Hence M , from the left side controls. Note that M , from the right side differs from M , from the left side because the footing length of 27 f t is used instead of the computed length of 26.46 ft and because x is rounded off. Therefore, the load is not exactiy uniform due to the small eccentricity.

30,658 Ib/ft

0.5 f t 4 k x = 9.32 f t 4

PUL = 301,000 Ib

Figure 12.15 Free-body diagram


Design of the footing in the longitudinal direction (a) Shear: The combined footing is considered as a beam in the shear computations.

Hence the critical section is at a distance d from the face of the support. Controlling V, at the column center line

V, - 327,473 + 0.85

= 385,262 Ib -~ -

Assume that the total footing thickness = 3 ft (0.92 m). The effective footing depth d = 32 in. for minimum steel cover of = 4 in. For the controlling interior column R, the equivalent rectangular column size = 13.29 in.

(13.2912 + d) 30,658 X-

12 + required V, at d section = 385,262 -

38.65 X 30,658 12 X 0.85

= 385,262 - = 269,107 Ib ( 1 197 kN)

V, = 2 a b,d = 2- X 6.5 X 12 X 32

= 273,423 Ib (1216 kN) > 269,107

(b) Moment and reinforcement in the longitudinal direction (step 4): The distnbution of shear and moment in the longitudinal direction is shown in Figure 12.16. The critical section for moment is taken at the face of the columns.

O.K.

M , 15,925,293 + 0.9 controlling moment M , = - = = 17,694,770 in.-lb (2000 kNm)

M , = A 5 f y ( d - y ) Assume that (d - a/2) 0.9d.

17,694,770 = A, X 60,000(0.9 X 32) or

17,694,770 60,000 X 0.9 X 32

A, = = 10.24 in.2

10.24 X 60,000 0.85 X 3,000 X 6.5 X 12

= 3.09 in. A s f y - a = - - 0.85fL b

17,694,770 = A, X 60,000 32 - ~ ( ,,,> A, = 9.68 in.2 (6245 mm’)

Use 22 No. 6 bars at the top for the middle span.

A, = 9.68 in.’ (22 bars, 19.1-mm diameter)

Design of footing in the transverse direction Both columns are treated as isolated columns. The width of the band should not be

larger than the width of the column plus half the effective depth d on each side of the column. This assumption is on the safe side since the actual bending stress distribution is highly inde- terminate. It is, however, possible to assume that the flexura1 reinforcement in the transverse direction can raise the shear punching capacity within the d/2 zone from the face of the rectangular left column L and the equivalent rectangular right column R. Figure 12.17 shows the transverse band widths for both columns L and R determined on the basis of this discussion.

band width bL = 12 + - = 28 in. = 2.33 ft

The rectangular column size equivalent to the circular interior M-in.-diameter column = 13.29 in.

32 2

12.8 Examples of Footing Design r I

1

j 1

L R 1 i t t t t t t t t t t t t t t t t t t t t t t t t l

v, = 285,671 I b L ” 9 . 3 2 f t d 1 0 . 6 8 f t 4

(b)

M, = 15825.293 in.-lb (1799.8 kN-m.)

Figure 12.16 Longitudinal shear and moment distribution: (a) elevation; (b) shear; (c) moment.

band width bR = 13.29 + 2 = 45.3 in. = 3.77 ft

Column L transverse band reinforcement

6 ft 6 in. moment arm = ~ - - - - 2.50ft = 30.0in.

2 2 x 12

The net factored bearing pressure in the transverse direction is

301,000 6.5

45=-- - 46,308 Ib/ft

1 2 (2.50)‘ 2 2

M , = 4, - = 46,308 - = 144,713 ft-lb = 1,736,550 in.-lb

M , 1,736,550 4) 0.90

M , = - = = 1,929,500 in.-lb (218.0 kN-m)

or 1,929,500 = A, X 60,000 X 0.9 X 32

A, = 1.12 in.’


13.3 in.

Colurnn L

NsA\v

18 in.

1-6 f t 6 im-4

13.3 in.

b 3 2 . 3 5 i n . 1 k 3 2 . 3 5 i n . d

45.3 in m1 k 6 ft 6 in.-----+!

Colurnn L band Colurnn R band

Figure 12.17 Footing transverse band widths.

1.12 X 60,000 0.85 X 3000 X 28

- = 0.94 in. A6fY a = - -

0.85fLb

1,929,500 = A, X 60,000 32 - ~ ( "2") A, = 1.02 in.' (658 mm')

min. A , = 0.0018bWd = 0.0018 X 28 X 32 = 1.62 in.'

1 .o2 = 0.00114 p=m

Use six No. 5 bars, A, = 1.86 in.' (six bars, 15.9-mm diameter) equally spaced in the band. which is to be centered under the column.

Column R transverse band reinforcement Equivalent square column size = 13.3 in. x 13.3 in

6 f t 6 in. 13.3 in. moment arm = ~ - ~ - - 2.69 ft = 32.35 in.

2 2 x 12

Net factored bearing pressure in the transverse direction is

526,750 q s = = - - 81,038 Ib/ft

(2.69)' = 293,200 ft-lb = 3.518.400 in.-lb

1 2

2 2 M , 1 qi - = 81,038 ~

M , 3,518,400 M , = - = 4) 0.90

M,, = Asfv ( d - 4) = 3,909,333 in.-lb (441.8 kN-m)

a 2

assume that d - - = 0.90d

of Footii i g Design 549

or

3,909,333 = A, X 60,000 X 0.9 X 32

2.26 X 60,000 0.85 x 3000 x 45.3 - = 1.17 in. a = ~ - Asfy

A, = 2.26in.= 0.85fCb

3,909,333 = A, X 60,000 32 - - ( ‘z) A, = 2.07 in.‘ (3347 mm’)

2.07 = 0.0014 < pmin

’= 45.3 X 32

where pmin = 0.0018 (shrinkage temperature reinforcement).

minimum A, = 0.0018 X 45.3 X 32 = 2.61 in?

Use nine No. 5 bars, A, = 2.79 in.’ (nine bars, 15.9-mm diameter) equally spaced.

Development length check for top bars in tension

f C = 3000 psi (27.6 MPa)

Top reinforcement more than 12411. concrete below bars, a = 1.3. From Eq. 10.6,

e d - 3 fY aByA d, 40 fi ( c +dKtr) - -- -

(a) Longitudinal top bars: Assume transverse reinforcement index K,, = O. Spacing c = 4.91 in. > 2d,. cld, = 4.9110.75 = 6.5 > 2.5; use 2.5. No. 6 bar d, = 0.75 in. (19.1 mm).

a = 1 . 3 , p = l . O , y=O.8 , A=1.0, K , = O

= 25.7 in. 1.3 X 1.0 X 0.8 X 1.0

2.5

> min t d = 12 in.

Use td = 26 in. = 2.2 ft. (660 mm).

column = 9.32 + 0.50 + 9.82 ft. > 2.3, O.K.

(b) Transverse bottom steel: No. 5 bars, d, = 0.625

The distance from c at the maximum moment in Figure 12.14 to the center of the left

a = 1.0 (bottom steel), p = 1.0, y = 0.8, A = 1.0, K,, = O

c = 3.8 in., c/db = 3.810.625 = 6.1 > 2.5, use 2.5

= 16.4in. 1.0 X 1.0 X 0.8 X 1.0

2.5

A, required A, provided

A, =

1.62 1.86

2.61 2.79

column L steel modifier: A, = - = 0.87

column R steel modifier: A, = - = 0.94

modified 8, = 16.4 X 0.94 = 15.4 in.

available development length = (32.35 - 3.0) in. > 15.4 in. O.K.

Therefore, adopt reinforcement as in Figure 12.18. Check for dowel steel from the columns to the footing slab.

550

6 No. 5 22 No. 6

Chapter 12 Footings

9 No. 5

/

in-i Y 2 0 f t - /15 in. dia.

, I

/ - -

3 ft 6 in.

L /

--

ff 3 in. clear cover ’ 1-27 ft-4

Figure 12.18 Combined footing reinforcement.

12.9 STRUCTURAL DESIGN OF OTHER TYPES OF FOUNDATIONS

From the discussion and the examples given in the foregoing sections. it is clear that the design of foundation substructures follows al1 the hypotheses and procedures used in proportioning the superstructures once the intensity and distribution of the soil bearing pressure is determined. If a cluster of piles supports a very heavy reaction through a pile cap, the analysis reduces to determining the punching load for each pile and determining the corresponding thickness of the cap. A determination of the center of gravity of the resultant of al1 pile forces has to be made if the system is subjected to bending in addition to axial load in order to choose the appropriate pile cap layout.

When raft foundations are necessary in poor soil conditions and deep excavations. the design of such a substructure is not too different from the design of any heavily loaded floor system. Once the soil pressure distribution is determined, the design becomes that of an inverted floor supported by deep beams longitudinally and transversel!.

Variations are to be expected in the described foundation types. particularly in cases of specialized or unique structures. Through an understanding of the basic principles presented, the student and the designer should have no difficulty in utilizing the soil data developed by the geotechnical engineer in selecting and proportioning the appropriate foundation substructure.

SELECTED REFERENCES

12.1. Richart, F. E., “Reinforced Concrete Walls and Column Footings.“ Joiirnril o,f thr Atnrricrin Coti- crete Institute, Proc. Vol. 45, October and November 1948. pp. 97-127 and 237-243.

U.2. Timoshenko, S., and Woinowsky-Kreiger. S.. Theory of Plates and Shells, 2nd ed.. McGraw Hill. New York, 1968,580 pp.

12.3. Balmer, G. G., Jones, V., and McHenry, D., Shearing Srrengrh of Concrere iinder High Tria-vid Stress, Structural Research Laboratory Report SP-23. U.S. Department of Interior. Bureau of Reclamation, Washington, D.C.. 1949,26 pp.

12.4. Arnerican Insurance Association. The National Biiilding Code, 1976 ed.. AIA. New York. Decsm- ber 1977,767 pp.

12.5. Moe, J., Shearing Strength of Reinforced Concrete Slabs and Footings iinder Concentrared Locid. Bulletin D47, Portland Cement Association. Skokie, 111.. April 1961, 134 pp.

’ U.6. Furlong, R. W., “Design Aids for Square Footings,” Joiirnal of the Anierican Concrete Instiriite. Proc. Vol. 62. March 1965, pp. 363-371.

Problems for Solution 551

12.7. Hawkins, N. M., Chairman, ASCE-ACI Committee 426, “The Shear Strength of Reinforced Con- crete Members-Slabs,” Journal of the Structural Division, ASCE, Proc. Vol. 100, August 1974,

12.8. Sowers, G. B., and Sowers, G. F., Zntroductory Soil Mechanics and Foundations, 3rd ed., Macmil- lan, New York, 556 pp.

12.9. Bowles, J. E., Foundations Analysis and Design, McGraw-Hill, New York, 1982,816 pp.

New York, 1975,751 pp.

don, 1948,141 pp.

1998,1250 pp.

pp. 1543-1591.

12.10. Winterkorn, H. F., and Fang, H. Y., Foundation Engineering Handbook, Van Nostrand Reinhold,

12.11. Baker, A. L. L., Raft Foundations-The Soil Line Method of Design, Concrete Publications, Lon-

12.12. Nawy, E. G., Ed., Concrete Construction Engineering Handbook, CRC Press, Boca Raton, FL..

PROBLEMS FOR SOLUTION

12.1. Design a reinforced concrete, square, isolated footing to support an axial column service live load P , = 300,000 Ib (1334 kN) and service dead load P, = 625,000 Ib (2780 kN). The size of the column is 30 in. x 24 in. (0.76 m x 0.61 m). The soil test borings indicate that it is composed of medium compacted sands and gravely sands, poorly graded. The frost line is assumed to be 3 ft below grade. Given:

average weight of soil and concrete above the footing, y = 130 pcf (20.4 kN/m3) footing fC = 3000 psi (20.7 MPa) column fC = 4000 psi (27.6 MPa)

surcharge = 120 psf (5.7 kPa) fy = 60,000 psi (413.7 MPa)

12.2. Design a reinforced concrete wall footing for (a) a 10-in. (0.25-m) reinforced concrete wall and (b) a 12411. (0.30-m) masonry wall. The intensity of service linear dead load is W, = 20,000 Ib/ft (292.0 kN/m) and a service linear live load W, = 15,000 lblft (219.0 kNlm) of wall length. Assume an evenly distributed soil bearing pressure and that the average soil bearing pressure at the base of the footing is 3 tons/ft2 (87.6 kN/m). The frost line is assumed to be 2 ft below grade. Given:

average weight of soil and footing above base = 125 pcf (19.6 kN/m3) footing fi. = 3000 psi (20.7 MPa) column fC = 5000 psi (34.5 MPa)

f y = 60,000 psi (413.7 MPa)

12.3. A combined footing is subjected to an exterior 16 in. x 16 in. (0.4 m x 0.4 m) column abutting the property line carrying a total service load P , = 300,000 Ib (1334 kN) and an interior column 20 in. x 20 in. (0.5 m x 0.5 m) carrying a total factored load P , = 400,000 Ib (1779 kN). The live load is 30% of the total load. The center-line distance between the two columns is 22’W’ (6.71 m). Design the appropriate reinforced concrete footing on a soil weighing 135 pcf (21.2 kN/m3). The bearing capacity of the soil at the level of the footing base is 6000 Ib/ft*. The frost line is assumed to be at 3 ’ 4 ” (1.07 m) below grade. Assume a surcharge of 125 psf (19.62 kN/m3) at grade level. Given:

footing f:. = 3500 psi (24.1 MPa) column fc = 5500 psi (37.4 MPa)

f, = 60,000 psi (413.7 MPa)

12.4. Redesign the isolated reinforced concrete footing in Problem 12.1 if the load is applied at an eccentricity (a) e = 0.5 ft (0.15 m) and (b) e = 1.8 ft (0.55 m).

12.5. Redesign the combined reinforced concrete footing in Problem 12.3 if the center-line distance between the two columns is 15’4” (4.6 m).

Documents

FOOTINGS - imcyc.comimcyc.com/biblioteca/ArchivosPDF/Zapatas/4 Footings.pdf · senting the topic of design of footings in this chapter. ... These are similar to the combined footings,