15 Footings 15-1 INTRODUCTION 15-2 SOIL PRESSURE UNDER FOOTINGS

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1 15 Footings 15-1 INTRODUCTION Footings and other foundation units transfer loads from the structure to the soil or rock supporting the structure. Because the soil is generally much weaker than the concrete columns and walls that must be supported, the contact area between the soil and the footing is much larger than that between the supported member and the footing. The more common types of footings are illustrated in Fig Strip footings or wall footings display essentially one-dimensional action, cantilevering out on each side of the wall. Spread footings are pads that distribute the column load in two directions to an area of soil around the column. Sometimes spread footings have pedestals, are stepped, or are tapered to save materials. A pile cap transmits the column load to a series of piles, which, in turn, transmit the load to a strong soil layer at some depth below the surface. Combined footings transmit the loads from two or more columns to the soil. Such a footing is often used when one column is close to a property line. A mat or raft foundation transfers the loads from all the columns in a building to the underlying soil. Mat foundations are used when very weak soils are encountered. Caissons 2 to 5 ft in diameter are sometimes used instead of piles to transmit heavy column loads to deep foundation layers. Frequently, these are enlarged at the bottom (belled) to apply load to a larger area. The choice of foundation type is selected in consultation with the geotechnical engineer. Factors to be considered are the soil strength, the soil type, the variability of the soil type over the area and with increasing depth, and the susceptibility of the soil and the building to deflections. Strip, spread, and combined footings are considered in this chapter because these are the most basic and most common types SOIL PRESSURE UNDER FOOTINGS The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad. A concrete footing on sand will have a pressure distribution similar to Fig. 15-2a. The sand near the edges of the footing tends to displace laterally when the footing is loaded, causing a decrease in soil pressure near the 812

2 Fig Types of footings. Section 15-2 Soil Pressure Under Footings 813

3 814 Chapter 15 Footings Fig Pressure distribution under footings. edges. On the other hand, the pressure distribution under a footing on clay is similar to Fig. 15-2b. As the footing is loaded, the soil under the footing deflects in a bowl-shaped depression, relieving the pressure under the middle of the footing. For structural design purposes, it is customary to assume that the soil pressures are linearly distributed in such a way that the resultant vertical soil force is collinear with the resultant downward force. Design Methods Allowable Stress Design There are two different philosophies for the design of footings [15-1], [15-2]. The first is allowable stress design. Almost exclusively, footing design is based on the allowable stresses acting on the soil at unfactored or service loads. For a concentrically loaded spread footing, P s q a A (15-1) where P s is the specified (unfactored) load acting on the footing. Section of ASCE 7 gives an updated set of load combinations for allowable stress design [15-3]. ACI Committee 318 has not considered these for footing design as yet. q a is the allowable stress for the soil given by Eq. (15-3), presented in the next subsection. A is the area of the footing in contact with the soil. Limit-States Design The second design philosophy is a limit-states design based on factored loads and factored resistances, given by fr n Ú ap s (15-2) where f is a resistance factor to account for the variability of the load-resisting mechanism of the soil under the footing. R n is the engineer s best estimate of the resistance of the soil under the footing. a is a load factor. P s is the specified load acting on the soil at the base of the footing. The load factors, a, in Eq. (15-2) are those used in building design. Load factors and load combinations for design are given in ACI Code Sections 9.2 and 9.3. Resistance factors for limit-states design of footings are still being developed. Current estimates of f values for shallow footings are as follows:

4 Section 15-2 Soil Pressure Under Footings 815 Vertical resistance, f = 0.5. Sliding resistance dependent on friction, with cohesion equal to zero, f = 0.8. Sliding resistance dependent on cohesion, with friction equal to zero, f = 0.6. Serviceability limit states should also be checked [15-1], [15-2], [15-3]. At the time of writing, virtually all building footings in North America are designed by using allowable-stress design applied to failures of the concrete foundation element or the soil itself. The rest of this chapter will apply allowable-stress design to the soil and then use strength design for the reinforced concrete foundation structure. Limit States for the Design of Foundations Limit States Governed by the Soil Three primary limit states of the soil supporting an isolated foundation are [15-1], [15-2], and [15-3]: 1. a bearing failure of the soil under the footing (Fig. 15-3), 2. a serviceability failure in which excessive differential settlement between adjacent footings causes architectural or structural damage to the structure, or 3. excessive total settlement. Settlement occurs in two stages: immediate settlement as the loads are applied, and a longterm settlement known as consolidation. Procedures for minimizing differential settlements involve a degree of geotechnical engineering theory outside the scope of this book. Bearing failures are controlled by limiting the service-load stress under the footing to less than an allowable stress, q a, as in Eq. (15-1). Limit States Governed by the Structure Similarly, there are four primary structural limit states for the foundations themselves [15-1], [15-2]: 1. flexural failure of the portions of the footing that project from the column or wall, 2. shear failure of the footing, 3. bearing failure at member interfaces, and 4. inadequate anchorage of the flexural reinforcement in the footing. As stated earlier, bearing failures of the soil supporting the foundation are prevented by limiting the service-load stress under the footing to less than an allowable stress q a = q ult FS (15-3) Fig Bearing failure of a footing.

5 816 Chapter 15 Footings where q ult is the stress corresponding to the failure of the soil under the footing and FS is a factor of safety in the range of 2.5 to 3. Values of q a are obtained from the principles of geotechnical engineering and depend on the shape of the footing, the depth of the footing, the overburden or surcharge on top of the footing, the position of the water table, and the type of soil. When using a value of q a provided by a geotechnical engineer, it is necessary to know what strengths were measured and in what kind of tests, and what assumptions have been made in arriving at this allowable soil pressure, particularly with respect to overburden and depth to the base of the footing. It should be noted that q a in Eq. (15-3) is a service-load stress, whereas the rest of the structure usually is designed by using factored loads corresponding to the ultimate (strength) limit states. The method of accounting for this difference in philosophy is explained later. Elastic Distribution of Soil Pressure under a Footing The soil pressure under a footing is calculated by assuming linearly elastic action in compression, but no tensile strength across the contact between the footing and the soil. If the column load is applied at, or near, the middle of the footing, as shown in Fig. 15-4, the stress, q, under the footing is q = P A ; My I (15-4) Fig Soil pressure under a footing: loads within kern. (c) Plan view showing Kern dimensions.

6 Section 15-2 Soil Pressure Under Footings 817 where P = vertical load, positive in compression A = area of the contact surface between the soil and the footing I = moment of inertia of this area M = moment about the centroidal axis of the footing area y = distance from the centroidal axis to the point where the stresses are being calculated The moment, M, can be expressed as Pe, where e is the eccentricity of the load relative to the centroidal axis of the area A. The maximum eccentricity e for which Eq. (15-4) applies is that which first causes q = 0 at some point. Larger eccentricities will cause a portion of the footing to lift off the soil, because the soil footing interface cannot resist tension. For a rectangular footing, this occurs when the eccentricity exceeds e k = / 6, or e k = b 6 (15-5) This is referred to as the kern distance. Loads applied within the kern, the shaded area in Fig. 15-4c, will cause compression over the entire area of the footing, and Eq. (15-4) can be used to compute q. Various pressure distributions for rectangular footings are shown in Fig If the load is applied concentrically, the soil pressure q is q avg = P>A. If the load acts through the kern point (Fig. 15-5c), q = 0 at one side and q = 2q avg at the other. If the load falls outside the kern point, the resultant upward load is equal and opposite to the resultant downward load, as shown in Fig. 15-5d. Generally, such a pressure distribution would not be acceptable, because it makes inefficient use of the footing concrete, tends to overload the soil, and may cause the structure to tilt. Elastic and Plastic Soil-Pressure Distributions The soil-pressure diagrams in Fig are based on the assumption that the soil pressure is linearly distributed under a footing. This is a satisfactory assumption at service-load levels and for footings on rock or dense glacial till. For yielding soils, the soil-pressure distribution will approach a uniform (plastic) distribution over part of the base of the footing in such a way that the resultant load on the footing and the resultant of the soil pressure coincide as is required for equilibrium. For the design of concentrically loaded footings, the distribution of soil pressures is taken to be uniform over the entire contact area, as shown in Fig. 15-5a. For the structural design of eccentrically loaded footings, such as those for retaining walls or bridge abutments, the pressure distribution is a linearly varying distribution like those in Fig. 15-5b, c and d with the resultant of the soil pressure coincident with the resultant of the applied loads. The examples in this chapter are limited to the predominant case of concentric loads and a uniform or linear soil-pressure distribution over the entire contact area. Load and Resistance Factors for Footing Design ACI Code load and resistance factors are given in ACI Code Sections 9.2 and 9.3. The examples in this chapter are based on those load and resistance factors.

7 818 Chapter 15 Footings Resultant of loads on the footing Fig Pressures under an eccentrically loaded footing. Gross and Net Soil Pressures Figure 15-6a shows a 2-ft-thick spread footing with a column at its center and with its top surface located 2 ft below the ground surface. There is no column load at this stage. The total downward load from the weights of the soil and the footing is 540 psf. This is balanced by an equal, but opposite, upward pressure. As a result, the net effect on the concrete footing is zero. There are no moments or shears in the footing due to this loading. When the column load P c is added, the pressure under the footing increases by q n = P c >A, as shown in Fig. 15-6b. The total soil pressure is q = q n. This is referred to as the gross soil pressure and must not exceed the allowable soil pressure, q a. When moments and shears in the concrete footing are calculated, the upward and

8 Section 15-2 Soil Pressure Under Footings Fig Gross and net soil pressures. downward pressures of 540 psf cancel out, leaving only the net soil pressure, q n, to cause internal forces in the footing, as shown in Fig. 15-6c. In design, the area of the footing is selected so that the gross soil pressure does not exceed the allowable soil pressure. The flexural reinforcement and the shear strength of the footing are then calculated by using the net soil pressure. Thus, the area of the footing is selected to be D1structure, footing, surcharge2 + L A = (15-6) q a where D and L refer to the unfactored service dead and live loads.

9 820 Chapter 15 Footings For service-load combinations including wind, W, most codes allow a 33 percent increase in q a. For such a load combination, the required area would be D1structure, footing, surcharge2 + L + W A = (15-7) 1.33q a but not less than the value given by Eq. (15-6). In Eqs. (15-6) and (15-7), the loads are the unfactored service loads. Once the area of the footing is known, the rest of the design of the footing is based on soil stresses due to the factored loads. Factored Net Soil Pressure, q nu. The factored net soil pressures used to design the footing are: q nu = 1factored load2/a (15-8) where the factored loads come from ACI Code Equations 9-1, 9-2, 9-3, 9-4, and 9-6. The factored net soil pressure, q nu, is based on the factored loads and will exceed q a in most cases. This is acceptable, because the factored loads are roughly 1.5 times the service loads, whereas the factor of safety implicit in q a is 2.5 to 3. Hence, the factored net soil pressure will be less than the pressure that would cause failure of the soil. If both load and moment are transmitted to the footing, it is necessary to use Eq. (15-4) (if the load is within the kern) or other relationships (as illustrated in Fig. 15-5d) to compute q nu. In such calculations, the factored loads would be used STRUCTURAL ACTION OF STRIP AND SPREAD FOOTINGS The behavior of footings has been studied experimentally at various times. Our current design procedures have been strongly affected by the tests reported in [15-4], [15-5], and [15-6]. The design of a footing must consider bending, development of reinforcement, shear, and the transfer of load from the column or wall to the footing. Each of these is considered separately here, followed by a series of examples in the ensuing sections. In this section, only axially loaded footings with uniformly distributed soil pressures, q nu, are considered. Flexure A spread footing is shown in Fig Soil pressures acting under the crosshatched portion of the footing in Fig. 15-7b cause the moments about axis A A at the face of the column. From Fig. 15-7c, we see that these moments are M u = 1q nu bf2 f 2 (15-9) where q nu bf is the resultant of the soil pressure on the crosshatched area and f / 2 is the distance from the resultant to section A A. This moment must be resisted by reinforcement placed as shown in Fig. 15-7c. The maximum moment will occur adjacent to the face of the column on section A A or on a similar section on the other side of the column. In a similar manner, the soil pressures under the portion outside of section B B in Fig. 15-7a will cause a moment about section B B. Again, this must be resisted by flexural reinforcement perpendicular to B B at the bottom of the footing; the result is two layers of steel, one each way, shown in section A A in Fig. 15-7c.

10 Section 15-3 Structural Action of Strip and Spread Footings 821 A B B A A B B A A-A. Fig Flexural action of a spread footing. (c) Moment about section A-A. A-A. A The critical sections for moment in the footing are taken as follows (ACI Code Sections 15.3 and ): 1. for footings supporting square or rectangular concrete columns or walls, at the face of the column or wall; 2. for footings supporting circular or regular polygonal columns, at the face of an imaginary square column with the same area;

11 822 Chapter 15 Footings 3. for footings supporting masonry walls, halfway between the middle and the edge of the wall; 4. for footings supporting a column with steel base plates, halfway between the face of the column and the edge of the base plate. The moments per unit length vary along lines A A and B B, with the maximum occurring adjacent to the column. To simplify reinforcement placing, however, ACI Code Section states that for square footings the reinforcement shall be distributed uniformly across the entire width of the footing. A banded arrangement is used in rectangular footings, as will be illustrated in Example Although a footing is not a beam, it is desirable that it be ductile in flexure. This can be done by limiting e t, net tensile strain in the extreme tension reinforcement, to the value Ú0.005 as was done in Chapter 4 for design of beams. When calculating f, ACI Code Section states that f = 0.9 for tension-controlled sections, i. e., sections where e t Ú at nominal flexural-strength conditions. ACI Code Section states that, for footings of uniform thickness, the minimum area of flexural tensile reinforcement shall be the same as that required for shrinkage and temperature reinforcement in ACI Code Section For Grade-40 steel, this requires A s,min = bh; for Grade-60 steel, A s,min = bh is specified. This amount of steel should provide a moment capacity between 1.1 and 1.5 times the flexural cracking moment and hence should be enough to prevent sudden failures at the onset of cracking. ACI Code Section gives the maximum spacing of the reinforcement in a footing as the lesser of three times the thickness or 18 in. If the reinforcement required for flexure exceeds the minimum flexural reinforcement, the author recommends the use of the maximum spacing from ACI Code Section , which calls for a maximum spacing equal to twice the slab thickness, but not greater than 18 in. Development of Reinforcement The footing reinforcement is chosen by assuming that the reinforcement stress reaches f y along the maximum-moment section at the face of the column. The reinforcement must extend far enough on each side of the points of maximum bar stress to develop this stress. In other words, the bars must extend / d from the critical section or be hooked at the outer ends. Shear A footing may fail in shear as a wide beam, as shown in Fig. 15-8a or as a result of punching, as shown in Fig. 15-8b. These are referred to as one-way shear and two-way shear and are discussed more fully in Section Concern has been expressed about the shear strength of deep, lightly reinforced concrete members [15-7], [15-8], and [15-9]. Tests of members similar to one-way footings [15-9] suggest that, if the ratio of the length, a, of the portion of the footing that projects outward from the column or wall, to the depth of the footing, d, does not exceed 3, the crack-restraining effect of the soil pressure under the footing tends to offset the strength reduction due to size. ACI Code Section requires minimum stirrups in all flexural 1 members with V u greater than 2 fv c, except for footings and solid slabs. Stirrups are required in all footings for which exceeds fv c. V u

12 Section 15-3 Structural Action of Strip and Spread Footings 823 Fig Critical sections and tributary areas for shear in a spread footing. One-Way Shear A footing failing through one-way shear is designed as a beam with (ACI Code Section ) V u f1v c + V s 2 (15-10) where V c = 2l2f œ c b w d (15-11) Recall that l is a factor used for lightweight concrete and was defined in Chapter 6. For normal-weight concrete, which is commonly used in footings, l = 1.0. Web reinforcement is very seldom used in strip footings or spread footings, due to the difficulty in placing it, and due to the fact that it is usually cheaper and easier to deepen the footing than it is to provide stirrups. Hence, V s = 0 in most cases. The inclined crack shown in Fig. 15-8a intercepts the bottom of the member about d from the face of the column. As a result, the critical section for one-way shear is located at d away from the face of the column or wall, as shown in plan view in Fig. 15-8a. For footings supporting columns with steel base plates, the critical section is d away from a line halfway between the face of the column and the edge of the base plate. The shear is times the tributary area shown shaded in Fig. 15-8a. Two-Way Shear V u q nu Research [15-6], [15-7] has shown that the critical section for punching shear is at the face of the column, while the critical loaded area is that lying outside the area of the portion punched through the slab. To simplify the design equations, the critical-shear perimeter for design purposes has been defined as lying d/2 from the face of the column, as shown by the

13 824 Chapter 15 Footings dashed line in Fig. 15-8b (ACI Code Sections and ). For the column shown in Fig. 15-8b, the length, b o, of this perimeter is b o = 21c 1 + d2 + 21c 2 + d2 (15-12) where c 1 and c 2 are the lengths of the sides of the column and d is the average effective depth in the two directions. The tributary area assumed critical for design purposes is shown cross-hatched in Fig. 15-8b. Because web reinforcement is rarely used in a footing, V u fv c, where, from ACI Code Section , V c shall be the smallest of (a) V c = a2 + 4 (15-13) b bl2f c œ b o d (ACI Eq ) where b is the ratio of the long side to the short side of the column ( c 2 >c 1 in Fig. 15-8b) and b o is the perimeter of the critical section, (b) V c = a a sd + 2bl2f œ (15-14) b c b o d o (ACI Eq ) where a s is 40 for columns in the center of footing, 30 for columns at an edge of a footing, and 20 for columns at a corner of a footing, and (c) V c = 4l2f œ c b o d (15-15) (ACI Eq ) Transfer of Load from Column to Footing The column applies a concentrated load on the footing. This load is transmitted by bearing stresses in the concrete and by stresses in the dowels or column bars that cross the joint. The design of such a joint is considered in ACI Code Section The area of the dowels can be less than that of the bars in the column above, provided that the area of the dowels is at least times the column area (ACI Code Section ) and is adequate to transmit the necessary forces. Such a joint is shown in Fig Generally, the column bars stop at the bottom of the column, and dowels are used to transfer forces across the column footing joint. Dowels are used because it is awkward to embed the column steel in the footing, due to its unsupported height above the footing and the difficulty in locating it accurately. Figure 15-9a œ shows an 18 in. * 18 in. column with f c = 5000 psi and eight No. 8 bars. The column is supported on a footing made of 3000-psi concrete. There are four No. 7 Grade-60 dowels in the connection. The dowels extend into the footing a distance equal to the compression development length of a No. 7 bottom bar in 3000-psi concrete (19 in.) and into the column a distance equal to the greater of 1. the compression lap-splice length for a No. 7 bar in 5000-psi concrete (26 in.), and 2. the compression development length of a No. 8 bar in 5000-psi concrete (17 in.). This joint could fail by reaching various limit states, including 1. crushing of the concrete at the bottom of the column, where the column bars are no longer effective, 2. crushing in the footing below the column,

14 Section 15-3 Structural Action of Strip and Spread Footings bond failure of the dowels in the footing, and 4. failure in the column of the lap splice between the dowels and the column bars. Bearing Strength The total capacity of the column for pure axial load is 900 kips, of which 197 kips is carried by the steel and the rest by the concrete, as shown in Fig. 15-9b. At the joint, the area of the dowels is less than that of the column bars, and the force transmitted by the dowels is fa sd f y, where A sd is the area of the dowels and f is that for tied columns. As a result, the load carried by the concrete has increased. In Fig. 15-9, the dowels are hooked so that they can be supported on and tied to the mat of footing reinforcement when the footing concrete is placed. The hooks cannot be used to develop compressive force in the bars (ACI Code Section ). The maximum bearing load on the concrete is defined in ACI Code Section as f10.85f œ c A 1 2. If the load combinations from ACI Code Section 9.2 are used, ACI Code Section gives f = 0.65 for bearing and is the area of the contact surface. A Fig Column footing joint.

15 826 Chapter 15 Footings B B B Fig Definition of and A 2. A 1 When the supporting surface is wider on all sides than the loaded area, the maximum bearing load may be taken as f10.85f c œ A 1 2 A A 2 A 1 (15-16) but not more than f11.7f œ c A 1 2, where A 2 is the area of the lower base of a right pyramid or cone formed by extending lines out from the sides of the bearing area at a slope of 2 horizontal to 1 vertical to the point where the first such line intersects an edge. This is illustrated in Fig The first intersection with an edge occurs at point B, resulting in the area A 2 shown crosshatched in Fig b. Two distinct cases must be considered: (1) joints that do not transmit computed moments to the footing and (2) joints that do. These will be discussed separately. No Moment Transferred to Footing If no moments are transmitted, or if the eccentricity falls within the kern of the column, there will be compression over the full section. The total force transferred by bearing is then calculated as 1A g - A sd 2 times the smaller of the bearing stresses allowed on the column or the footing, where A g is total area of the column and A sd is the area of the bars or dowels crossing the joint. Any additional load must be transferred by dowels. Moments Are Transferred to Footing If moments are transmitted to the footing, bearing stresses will exist over part, but not all, of the column cross section. The number of dowels required can be obtained by considering the area of the joint as an eccentrically loaded column with a maximum concrete stress equal to the

16 Section 15-4 Strip or Wall Footings 827 smaller of the bearing stresses allowed on the column or the footing. Sufficient reinforcement must cross the interface to provide the necessary axial load and moment capacity. Generally, this requires that all the column reinforcement or similar-sized dowel bars must cross the interface. This steel must be spliced in accordance with the requirements for column splices. Practical Aspects 15-4 STRIP OR WALL FOOTINGS Three other aspects warrant discussion prior to the examples. The minimum cover to the reinforcement in footings cast against the soil is 3 in. (ACI Code Section 7.7.1). This allows for small irregularities in the surface of the excavation and for potential contamination of the bottom layer of concrete with soil. Sometimes, the bottom of the excavation for the footing is covered with a lean concrete seal coat, to prevent the bottom from becoming uneven after rainstorms and to give a level surface for placing reinforcement. The minimum depth of the footing above the bottom reinforcement is 6 in. for footings on soil and 12 in. for footings on piles (ACI Code Section 15.7). ACI Code Section , covering the distribution of flexural reinforcement in beams and one-way slabs, does not apply to footings. A wall footing cantilevers out on both sides of the wall as shown in Figs. 15-1a and The soil pressure causes the cantilevers to bend upward, and as a result, reinforcement is required at the bottom of the footing, as shown in Fig The critical sections for design for flexure and anchorage are at the face of the wall (section A A in Fig ). One-way shear is critical at a section a distance d from the face of the wall (section B B in Fig ). The presence of the wall prevents two-way shear. Thicknesses of wall footings are chosen in 1-in. increments, widths in 2- or 3-in. increments. EXAMPLE 15-1 Design of a Wall Footing A 12-in.-thick concrete wall carries a service (unfactored) dead load of 10 kips per foot and a service live load of 12.5 kips per foot. From the geotechnical report, the allowable soil pressure, q a, is 5000 psf for shallow foundations. Design a wall footing to be based œ 5-ft below the final ground surface, using f c = 3000 psi normal-weight concrete and f The density of the soil is 120 lb>ft 3 y = 60,000 psi.. Most strip footings on soil have one mat of reinforcement. 1. Estimate the size of the footing and the factored net pressure. Consider a 1-ft strip of footing and wall. Allowable soil pressure is 5 ksf; allowable net soil pressure is 5 ksf - weight>ft 2 of the footing and of the soil over the footing. Because the thickness of the footing is not known at this stage, it is necessary to guess a thickness for a first trial. A d B Fig Structural action of a strip footing. A B

17 828 Chapter 15 Footings Generally, the thickness will be 1 to 1.5 times the wall thickness. We shall try a 12-in.-thick footing. Therefore, q n = 5-11 * * = 4.37 ksf, and we have: Try a footing 5 ft 2 in. = 62 in. (5.17 ft) wide. Using the load factors in ACI Code Section 9.2.1: In the design of the concrete and reinforcement, we shall use q nu = 6.19 ksf. 2. Check the shear. Shear usually governs the thickness of footings. Only oneway shear is significant in a wall footing. Check it at d away from the face of the wall (section B B in Fig ) The tributary area for shear is shown crosshatched in Fig a. where, from ACI Code Section , f = 0.75 for shear design when the load factors from ACI Code Section are used. Because V u 7 fv c, the footing depth is too small. If V u is larger or considerably smaller than fv c, choose a new thickness and repeat steps 1 and 2. Try a 13-in.-thick footing 5 ft 2 in. wide. A 13-in.-thick footing has d 9.50 in. and fv c = 9.37 kips>ft. Because fv c exceeds V u = 8.51 kips>ft, a 13-in.-thick footing 5 ft 2 in. wide is adequate for shear. 3. Design the flexural reinforcement. The critical section for moment is at the face of the wall (section A A in Fig ). The tributary area for moment is shown crosshatched in Fig b. The required moment is We want, 10 kips kips Area required = 4.37 ksf = 5.15 ft 2 per foot length of wall Factored net pressure, q nu = d = 12 in. - 3 in cover V u = 6.19a * 1b ft2 = 8.51 kips>ft fv c = f * 2l2f œ cb w d = 0.75 * 2 * * 12 * 8.5 = 8380 lbs>ft = 8.38 kips>ft M u = 6.19 * 125> M u fm n = fa s f y jd Footings are generally very lightly reinforced. Therefore, assume that j = Therefore, A s = From ACI Code Sections and , 13.4 k-ft>ft * 12 in.>ft 0.9 * 60 ksi10.95 * 9.5 in.2 = in.2 >ft Minimum A s = bh 1.2 * * bar diameter M 8.5 in. * 1 kip-ft>ft = 13.4 kip-ft>ft of length = * 12 * 13 = in 2 >ft = 6.19 ksf

18 Section 15-4 Strip or Wall Footings 829 No. 4 at 7 oc Fig Strip footing Example Maximum spacing of bars = 2h, or 18 in. We could use: No. 5 bars at 11 in. o.c., A s = 0.34 in. 2 >ft No. 4 bars at 7 in. o.c., A s = 0.34 in. 2 >ft Try No. 4 bars at 7 in. o.c., A s = 0.34 in. 2 >ft Because the calculation of capacity: a = A s was based on an assumed j-value, recompute the moment 0.34 * * 3 * 12 = in.

19 830 Chapter 15 Footings Clearly, the section is tension-controlled, and f = 0.9. Because this exceeds M u = 13.4 kip-ft>ft, the moment capacity is OK. For completeness, we could compute directly and use it to check f. From similar triangles and fm n = 0.9 * 0.34 * >22 = 168 k-in.>ft = 14.0 kip-ft>ft where d t 9.5 in. and c = a>b 1 = 0.667/0.85 = in. Calculations show e t = This far exceeds e t limit = Therefore, f s = f y and f = e t c = d t - c e t e t = 1d t - c2 c 4. Check the development. The clear spacing of the bars being developed exceeds 2d b and the clear cover exceeds d b. Therefore, this is Case 2 development in Tables 8 1 and A 6. From Table A 6, / d for a No. 4 bottom bar in 3000-psi concrete is 21.9 in. The distance from the point of maximum bar stress (at the face of the wall) to the end of the bar is 25 in. - 3 in. cover on the ends of the bars = 22 in. This is more than / d = Use No. 4 bars at 7 in. on centers. 5. Select the minimum (temperature) reinforcement. By ACI Code Section we require the following reinforcement along the length of the footing. A s = bh = * 62 in. * 13 in. = 1.45 in. 2 * The maximum spacing is 5 * 13 = 65 in. or 18 in. Provide three No. 7 bars (1.80 in. 2 ) for shrinkage reinforcement, placed as shown in Fig c. 6. Design the connection between the wall and the footing. ACI Code Section requires that reinforcement equivalent to the minimum vertical wall reinforcement extend from the wall into the footing. A cross section through the wall footing designed in this example is shown in Fig c. The section through the strip footing shows a shear key in the top surface of the footing. This is formed by a two-by-four pushed down into the surface. The shear key is intended to resist some shear to prevent the wall from being dislocated laterally. Sometimes the shear key is omitted, and shear friction is used to resist dislocation of the wall during construction SPREAD FOOTINGS Spread footings are square or rectangular pads that spread a column load over an area of soil large enough to support the column load. The soil pressure causes the footing to deflect upward, as shown in Fig. 15-7a, causing tension in two directions at the bottom. As a result, reinforcement is placed in two directions at the bottom, as shown in Fig. 15-7c. Two examples will be presented: a square axially loaded footing and a rectangular axially loaded footing.

20 Section 15-5 Spread Footings 831 EXAMPLE 15-2 Design of a Square Spread Footing A square spread footing supports an 18-in.-square column supporting a service dead load of 400 kips and a service live load of 270 kips. The column is built with 5000-psi concrete and has eight No. 9 longitudinal bars with f y = 60,000 psi. Design a spread footing to be constructed by using 3000-psi normal-weight concrete and Grade-60 bars. It is quite common for the strength of the concrete in the footing to be lower than that in the column. Dowels may be required to carry some of the column load across the column footing interface. The top of the footing will be covered with 6 in. of fill with a density of 120 lb>ft 3 and a 6-in. basement floor (Fig.15-13). The basement floor loading is 100 psf. For shallow foundations, the allowable bearing pressure on the soil is 6000 psf. Use load and resistance factors from ACI Code Sections 9.2 and Compute the factored loads and the resistance factors, f. Because the statement of the problem mentioned only dead and live loads, we will assume these as the only applicable loads. In effect, we are assuming that the wind loads and roof loads are small compared to the dead loads. This reduces the set of load combinations to the following: U = 1.41D2 From these equations, the design loads are U = 1.21D L2 and U = 1.4 * 400 = 560 kips U = 1.2 * * 270 = 912 kips Strength-reduction factors f are given in ACI Code Section 9.3. These were selected on the basis of whether flexure or shear is being considered. For shear, ACI Code Section gives f = For flexure, f will be a function of the strain in the extreme-tension layer of bars, but f will probably be 0.9 for footings. 4 No. 6 dowels Fig Spread footing Example 15-2.

21 832 Chapter 15 Footings 2. Estimate the footing size and the factored net soil pressure. Allowable net soil pressure q n = 6 ksf - 1weight>ft 2 of the footing and the soil and floor over the footing and the floor loading). Estimate the overall thickness of the footing at between one and two times the width of the column, say, 27 in.: q n = a = 5.43 ksf 400 kips kips Area required = = 123 ft ksf L 11.1 ft square. Try a footing 11 ft 2 in. square by 27 in. thick: Factored net soil pressure = 3. Check the thickness for two-way shear. Generally, the thickness of a spread footing is governed by two-way shear. The shear will be checked on the critical perimeter at d/2 from the face of the column and, if necessary, the thickness will be increased or decreased. Because there is reinforcement in both directions, the average d will be used: 23 in. The critical shear perimeter (ACI Code Section ) is shown dashed in Fig a. The tributary area for two-way shear is shown crosshatched. We have Length of critical shear perimeter: * * * b 1.2 * * = 7.31 ksf Average d = 27 in in. cover2-11 bar diameter2 V u = 7.31 ksf c a b 2 d ft 2 = 827 kips b o = in. = 164 in. fv c is the smallest of the values obtained from Eqs. (15-13), (15-14), and (15-15). For Eq. (15-13), and, b = larger column section dimension shorter column section dimension = 1.0 a2 + 4 b b = Thus, Eq. (15-13) will not govern. For Eq. (15-14), a s = 40, because this is considered to be an interior connection (not at an edge or in a corner of the footing). So, a a sd 40 * b = a b o b =

22 Section 15-5 Spread Footings Fig Critical sections Example Thus, Eq. (15-14) will not govern. So, using Eq. (15-15) to find V c = 4l2f œ c b o d = 4 * psi * 164 in. * 23 in. = 826,000 lbs = 826 kips and fv c = 0.75 * 826 = 620 kips. Because fv c = 620 kips is less than V u = 827 kips, the footing is not thick enough. Try h = 32 in., d = 28 in., and b o = 184 in. The footing is thicker, and it weighs more. Hence, a larger area may be required: q n = a V c, = 5.37 ksf Area required = 5.37 = 125 ft 2 = 11.2 ft square * * * b

23 834 Chapter 15 Footings Try an 11-ft-2-in.-square footing, 32 in. thick: 1.2 * * 270 Factored net soil pressure, q nu = = 7.31 ksf The new critical shear perimeter and tributary area for shear are shown in Fig b. We have Equation (15-15) governs again: V u = 7.31c a b 2 d = 805 kips fv c = 0.75 * 4 * * 184 * 28 = 847,000 lbs = 847 kips This is adequate. A check using h = 30 in. shows that a 30-in.-thick footing is not adequate. Use an 11-ft-2-in.-square footing, 32 in. thick. 4. Check the one-way shear. Although one-way shear is seldom critical, we shall check it. The critical section for one-way shear is located at d away from the face of the column, as shown in Fig c. Thus, V u = 7.31 ksf a11.17 ft * 30 ftb = 204 kips 12 fv c = f2l2f c œ b w d = 0.75 * 2 * * 134 * 28 = 308,000 lbs = 308 kips Therefore, o.k. in one-way shear. 5. Design the flexural reinforcement. The critical section for moment and anchorage of the reinforcement is shown in Fig d. The ultimate moment is M u = 7.31c11.17 * 158>1222 d = 954 kip-ft 2 Assuming that j = 0.95 and f = 0.90, the area of steel required is A s = The average value of d was used in this calculation for simplicity. The same reinforcement will be used in both directions: Minimum A s 1ACI Sections and = bh = * 134 * 32 = 7.72 in. 2 1does not govern2 Maximum spacing 1ACI Code Section = 18 in. Try eleven No. 8 bars each way, A s = 8.69 in. 2. Recompute fm n as a check. a = 954 * * * * * 3 * 134 = 7.97 in.2 = 1.53 in.

24 Section 15-5 Spread Footings 835 Clearly, the beam is tension-controlled, and f = fm n = 1070 kip-ft This exceeds M u = 954 kip-ft. s 6. Check the development. The clear spacing of the bars being developed exceeds 2d b and the clear cover exceeds d b. Therefore, this is Case 2 development in Tables 8-1 and A-6. From Table A-6, / d for a No. 8 bottom bar in 3000-psi concrete is 54.8 d b. The development length is / d = 54.8 d b c e >l where c e = 1.0 for uncoated reinforcement and l = 1.0 for normal-weight concrete. Accordingly, we have / d = 54.8 * 1.00 in. = 54.8 in. The bar extension past the point of maximum moment is 158 in. - 3 in.2 = 55 in. This is o.k. Use eleven No. 8 bars each way; A s 8.69 in Design the column footing joint. The column footing joint is shown in Fig The factored load at the base of the column is 1.2 * * 270 = 912 kips The maximum bearing load on the bottom of the column (ACI Code Section ) is f0.85f œ c A 1, where A 1 is the area of the contact surface between the column and the footing œ and f c is for the column. When the contact supporting surface on the footing is wider on all sides than the loaded area, the maximum bearing load on the top of the footing may be taken as 0.85ff c œ A 1A A 2 A 1, but not more than 1.7ff c œ A 1 (15-16) where A 2 is the area of the lower base of a right pyramid or cone as defined in Fig ACI Code Section defines f equal to 0.65 for bearing. By inspection, 2A 2 >A 1 for the footing exceeds 2; hence, the maximum bearing load on the footing is 0.85 * 0.65 * 3 * 18 2 * 2 = 1070 kips. The allowable bearing on the base of the column is f10.85f c œ A 1 2 = 0.65 * 0.85 * 5 * 18 2 = 895 kips Thus, the maximum load that can be transferred by bearing is 895 kips, and dowels are needed to transfer the excess load. Accordingly, we have Area of dowels required = ff y = 0.44 in. 2 where f = 0.65 has been used. This is the f value from ACI Code Sections (b) for compression-controlled tied columns and for bearing. The area of dowels must also satisfy ACI Code Section : Area of dowels Ú 0.005A g = * 18 2 = 1.62 in. 2 Try four No. 6 dowels (A s = 1.76 in. 2 ); dowel each corner bar. The dowels must extend into the footing a distance equal to the compression-development length (Table A-7) for

25 836 Chapter 15 Footings Fig Rectangular footing. a No. 6 bar in 3000-psi concrete, or 16 in. The bars will be extended down to the level of the main footing steel and hooked 90. The hooks will be tied to the main steel to hold the dowels in place. The dowels must extend into the column a distance equal to the greater of a compression splice (Table A-13) for the dowels (23 in.) or the compressiondevelopment length (Table A-7) of the No. 9 column bars (25 in.). Use four No. 6 dowels; dowel each corner bar. Extend dowels 25 in. into column. (See Fig ) Rectangular Footings Rectangular footings may be used when there is inadequate clearance for a square footing. In such a footing, the reinforcement in the short direction is placed in the three bands shown in Fig , with a closer bar spacing in the band under the column than in the two end bands. The band under the column has a width equal to the length of the short side of the footing, but not less than the width of the column (if that is greater) and is centered on the column. The reinforcement in the central band shall be 2>1b + 12 times the total reinforcement in the short direction, where b is the ratio of the long side of the footing to the short side (ACI Code Section ). The reinforcement within each band is distributed evenly, as is the reinforcement in the long direction. EXAMPLE 15-3 Design of a Rectangular Spread Footing Redesign the footing from Example 15-2, assuming that the maximum width of the footing is limited to 9 ft. Steps 1 through 3 of this example would proceed in the same sequence as in Example 15-2, leading to a footing 9 ft wide by 14 ft long by 33 in. thick. Using the load factors from ACI Code Section 9.2, the factored net soil pressure is 7.24 ksf. 1. Check the one-way shear. One-way shear may be critical in a rectangular footing and must be checked. In this case, it required a 1 in. increase in the thickness of the footing. The critical section and tributary area for one-way shear are shown in Fig a. We have V u = 7.24a * 9b = 250 kips fv c = 0.75 * 2 * * 108 * 29 = 257,000 lbs = 257 kips

26 Section 15-5 Spread Footings (b) Critical section for moment long direction. 3 No No. 5 evenly spaced No No. 8 (c) Critical section for moment short Short direction. Fig Rectangular footing Example This is just o.k. in one-way shear. 2. Design the reinforcement in the long direction. The critical section for moment and reinforcement anchorage is shown in Fig b. The ultimate moment is M u = 7.24a9 * 175>1222 b = 1270 kip-ft 2

27 838 Chapter 15 Footings Assuming that f = 0.9 and j = 0.95, the area of steel required is A s = 1270 * * * 292 We could use thirteen No. 8 bars, A s = 10.3 in. 2 ten No. 9 bars, A s = 10.0 in. 2 eight No. 10 bars, A s = 10.2 in. 2 Try thirteen No. 8 bars fm n = 1290 kip-ft (with f = 0.9). Check development; from Table A-6: The length available is 72 in. therefore, o.k. Use thirteen No. 8 bars in the long direction. 3. Design the reinforcement in the short direction. The critical section for moment and reinforcement anchorage is shown in Fig c. We have Assuming that j = 0.95, A s 5.75 in. 2, A s,min = * 168 * 33 = 9.98 in. 2 1this governs2 Try thirteen No. 8 bars: A s = 10.3 in. 2. Check development. / d = 54.8 in. and the length available is 42 in. therefore, not o.k. We must consider smaller bars. Try 33 No. 5 bars, A s = 10.2 in. 2, / d = 27.4 in. therefore, o.k. Use thirty-three No. 5 bars in the short direction of the footing. For the arrangement of the bars in the transverse direction (ACI Code Section ), b = In the middle strip of width 9 ft, provide / d = 54.8 * 1.0 * 1.0 * 1.0 = 54.8 in. M u = 7.24a14.0 * 145>1222 b = 713 kip-ft = 10.2 in.2 A s,min = * 108 * 33 = 6.42 in. 2 1does not govern2 long side short side = 1.56 * 33 bars = 25.8 bars Provide twenty-seven No. 5 bars in the middle strip, and provide three No. 5 bars in each end strip. The final design is shown in Fig d. Footings Transferring Vertical Load and Moment On rare occasions, footings must transmit both axial load and moment to the soil. The design of such a footing proceeds in the same manner as that for a square or rectangular footing, except for three things. First, a deeper footing will be necessary, because there will be shear stresses developed both by direct shear and by moment. The calculations concerning this are discussed in Chapter 13. Second, the soil pressures will not be uniform, as discussed in Section 15-2 and shown in Fig. 15-5b. Third, the design for two-way shear must consider moment and shear. (See Section )

28 Section 15-5 Spread Footings 839 The uneven soil pressures will lead to a tilting settlement of the footing, which will relieve some of the moment if the moment results from compatibility at the fixed end. The tilting can be reduced by offsetting the footing so that the column load acts through the center of the footing base area. If the moment is necessary for equilibrium, it will not be reduced by rotation of the foundation. EXAMPLE 15-4 Design a Rectangular Footing for a Column Subjected to Axial Load and Bending Assume the footing has a width of 10 ft and a shape similar to that in Fig Also, assume it will be supporting a 16 in. by 16 in. column that is carrying the following loads. P D = 180 kips and P L = 120 kips M D = 80 k-ft and M L = 60 k-ft The length and depth of the footing are to be determined. The geotechnical report indicates that for shallow foundations, the allowable soil bearing pressure is 4000 psf. Design the footing assuming f c = 3500 psi (normal-weight concrete) and f y = 60 ksi. 1. Factored loads and capacity reduction factors. Assume ACI Code Eq. (9-2) governs. P u = 1.2(180 k) + 1.6(120 k) = 408 kips M u = 1.2(80 k-ft) + 1.6(60 k-ft) = 192 k-ft = 2300 k-in. For shear strength, use f = 0.75 and for flexural strength, use f = Estimate footing length and depth. Estimate the overall thickness, t, of the footing to be between 1.5 and 2.0 times the size of the column, so select t = 26 in. Assuming a final design section similar to that shown in Fig , the net permissible bearing pressure is: q n = c = 3.44 ksf * * * d The (nonfactored) loads acting on the footing are: P a = 180 k k = 300 kips M a = 80 k-ft + 60 k-ft = 140 k-ft Assume the soil pressure distribution at the base of the footing is similar to that should in Fig. 15-5b. Assuming a footing width, S, and a footing length, L, similar to the footing in Fig , an expression for the maximum bearing pressure under the footing is: q(max) = P a SL + Setting q(max) equal to q n (3.44 ksf) and using S = 10 ft, we can use the following expression to solve for the required footing length, L ksf = M a a L 2 b = P a 1 12 SL3 SL + 6M a SL k (10 ft) # L + 6(140 k-ft) (10 ft) # L 2

29 840 Chapter 15 Footings Solving for L yields, L = 11.0 ft, or ft. Using the positive solution, select a footing length of 12 ft. 3. Calculate factored soil pressures. (Note: In the prior step the acting loads were used with the allowable soil bearing pressure to select the size of the footing. Now, the factored loads will be used to determine the soil pressures that will subsequently be used to determine the factored moment and shear used for the design of the footing.) Using P u = 408 kips and M u = 192 k-ft, the minimum and maximum factored soil pressures, as shown in Fig (a), are: q u (min, max) = P u SL ; 6M u SL 2 q u (max) = = 3.40 ksf ksf = 4.20 ksf, and q u (min) = 3.40 ksf ksf = 2.60 ksf 4. Check footing thickness for two-way shear. The critical shear perimeter is located d/2 away from each column face, as shown in Fig (b). Assume the average effective depth for the footing is: d t - 4 in. = 26 in. - 4 in. = 22 in. Thus, each side of the critical shear perimeter has a length of 38 in. Using the average factored shear stress inside the critical perimeter (q u (avg) = 3.40 ksf), the net factored shear to be transferred across the critical perimeter is: V u (net) = 408 k ksf ca ftba38 ftb d = 374 kips 12 Because we have a square column, the coefficient b in Eq. (15-13) is equal to 1.0, and thus, that equation does not govern for the value of V c. Also, because this is not a large critical perimeter (b o = 4 38 = 152 in.) compared to d (b o 20d ), Eq. (15-14) will not govern. Thus, Eq. (15-15) governs and V c is: V c = 4l2f c œ bo d = 4 * psi * 152 in. * 22 in. = 791,000 lbs = 791 kips Using the appropriate strength reduction factor for shear, we get: f V c = 0.75 * 791 k = 594 kips 7 V u (net) (o.k.) Thus, the footing depth satisfies the strength requirement for the net shear force. 5. Check for combined transfer of shear and moment. We must now check the maximum shear stress for the combined transfer of shear and moment using Eq. (13-30) of this text. n u = V u (13-30) b o d ; g vm u c J c From the previous step, it can be seen that V u (net) exceeds 40% of f V c. Therefore, the coefficient gf, and thus gv, cannot be adjusted as permitted in ACI Code Section (b). Using Eq. (13-32), with b 1 = b 2 = 38 in., we get: g f = 408 k (10 ft)(12 ft) + 6 * 192 k-ft (10 ft)(12 ft) b = 1 1>b 2 5>3 = 0.60