6 Mat Foundations 6.1 Introduction Under normal conditions, square and rectangular footings such as those described in Chapters 3 and 4 are economical for supporting columns and walls. However, under certain circumstances, it may be desirable to construct a footing that supports a line of two or more columns. These footings are referred to as combined footings. When more than one line of columns is supported by a concrete slab, it is called a mat foundation. Combined footings can be classified generally under the following categories: a. Rectangular combined footing b. Trapezoidal combined footing c. Strap footing Mat foundations are generally used with soil that has a low bearing capacity. A brief overview of the principles of combined footings is given in Section 6., followed by a more detailed discussion on mat foundations. 6. Combined Footings Rectangular Combined Footing In several instances, the load to be carried by a column and the soil bearing capacity are such that the standard spread footing design will require extension of the column foundation beyond the property line. In such a case, two or more columns can be supported on a single rectangular foundation, as shown in Figure 6.1. If the net allowable soil pressure is known, the size of the foundation (B 3 L) can be determined in the following manner: a. Determine the area of the foundation A 5 Q 1 1 Q q net(all) where Q 1, Q 5 column loads q net(all) 5 net allowable soil bearing capacity (6.1) 91
9 Chapter 6: Mat Foundations L Q 1 X Q 1 Q L 3 Q L 1 Section B q net(all) /unit length L Property line B Plan Figure 6.1 Rectangular combined footing b. Determine the location of the resultant of the column loads. From Figure 6.1, X 5 Q L 3 Q 1 1 Q (6.) c. For a uniform distribution of soil pressure under the foundation, the resultant of the column loads should pass through the centroid of the foundation. Thus, L 5 (L 1 X) where L 5 length of the foundation. d. Once the length L is determined, the value of can be obtained as follows: L 1 5 L L L 3 Note that the magnitude of L will be known and depends on the location of the property line. e. The width of the foundation is then L 1 B 5 A L (6.3) (6.4) (6.5) Trapezoidal Combined Footing Trapezoidal combined footing (see Figure 6.) is sometimes used as an isolated spread foundation of columns carrying large loads where space is tight. The size of the foundation that will uniformly distribute pressure on the soil can be obtained in the following manner: a. If the net allowable soil pressure is known, determine the area of the foundation: A 5 Q 1 1 Q q net(all)
6. Combined Footings 93 L X Q 1 Q L 1 Q 1 L 3 Q B q net(all) /unit length B 1 q net(all) /unit length Section B 1 B Property line L Plan Figure 6. Trapezoidal combined footing From Figure 6., A 5 B 1 1 B L (6.6) b. Determine the location of the resultant for the column loads: c. From the property of a trapezoid, X 5 Q L 3 Q 1 1 Q X 1 L 5 B 1 1 B L B 1 1 B 3 (6.7) With known values of A, L, X, and L, solve Eqs. (6.6) and (6.7) to obtain B 1 and B. Note that, for a trapezoid, L 3, X 1 L, L
94 Chapter 6: Mat Foundations Strap Section Strap Section Strap Plan Strap Plan (a) (b) Wall Section Strap Strap Plan (c) Figure 6.3 Cantilever footing use of strap beam Cantilever Footing Cantilever footing construction uses a strap beam to connect an eccentrically loaded column foundation to the foundation of an interior column. (See Figure 6.3). Cantilever footings may be used in place of trapezoidal or rectangular combined footings when the allowable soil bearing capacity is high and the distances between the columns are large. 6.3 Common Types of Mat Foundations The mat foundation, which is sometimes referred to as a raft foundation, is a combined footing that may cover the entire area under a structure supporting several columns and walls. Mat foundations are sometimes preferred for soils that have low load-bearing capacities, but that will have to support high column or wall loads. Under some conditions, spread footings would have to cover more than half the building area, and mat foundations might be more economical. Several types of mat foundations are used currently. Some of the common ones are shown schematically in Figure 6.4 and include the following: 1. Flat plate (Figure 6.4a). The mat is of uniform thickness.. Flat plate thickened under columns (Figure 6.4b). 3. Beams and slab (Figure 6.4c). The beams run both ways, and the columns are located at the intersection of the beams. 4. Flat plates with pedestals (Figure 6.4d). 5. Slab with basement walls as a part of the mat (Figure 6.4e). The walls act as stiffeners for the mat.
6.3 Common Types of Mat Foundations 95 Section Section Section Plan Plan Plan (a) (b) (c) Section Section Plan Plan (d) (e) Figure 6.4 Common types of mat foundations Mats may be supported by piles, which help reduce the settlement of a structure built over highly compressible soil. Where the water table is high, mats are often placed over piles to control buoyancy. Figure 6.5 shows the difference between the depth D f and the width B of isolated foundations and mat foundations. Figure 6.6 shows a flat-plate mat foundation under construction. B D f D f B Figure 6.5 Comparison of isolated foundation and mat foundation (B 5 width, D f 5 depth)
96 Chapter 6: Mat Foundations Figure 6.6 A flat plate mat foundation under construction (Courtesy of Dharma Shakya, Geotechnical Solutions, Inc., Irvine, California) 6.4 Bearing Capacity of Mat Foundations The gross ultimate bearing capacity of a mat foundation can be determined by the same equation used for shallow foundations (see Section 3.6), or q u 5 crn c F cs F cd F ci 1 qn q F qs F qd F qi 1 1 gbn g F gs F gd F gi [Eq. (3.19)] (Chapter 3 gives the proper values of the bearing capacity factors, as well as the shape depth, and load inclination factors.) The term B in Eq. (3.19) is the smallest dimension of the mat. The net ultimate capacity of a mat foundation is q net(u) 5 q u q [Eq. (3.14)] A suitable factor of safety should be used to calculate the net allowable bearing capacity. For mats on clay, the factor of safety should not be less than 3 under dead load or maximum live load. However, under the most extreme conditions, the factor of safety should be at least 1.75 to. For mats constructed over sand, a factor of safety of 3 should normally be used. Under most working conditions, the factor of safety against bearing capacity failure of mats on sand is very large. For saturated clays with f50 and a vertical loading condition, Eq. (3.19) gives q u 5 c u N c F cs F cd 1 q (6.8)
6.4 Bearing Capacity of Mat Foundations 97 where c u 5 undrained cohesion. (Note: N c 5 5.14, N q 5 1, and N g 5 0. ) From Table 3.4, for f50, and F cs 5 1 1 B L N q 5 1 1 B N c L 1 5.14 5 1 1 0.195B L F cd 5 1 1 0.4 D f B Substitution of the preceding shape and depth factors into Eq. (6.8) yields q u 5 5.14c u 1 1 0.195B L 1 1 0.4 D f B 1 q (6.9) Hence, the net ultimate bearing capacity is q net(u) 5 q u q 5 5.14c u 1 1 0.195B L 1 1 0.4 D f B (6.10) For FS 5 3, the net allowable soil bearing capacity becomes (6.11) The net allowable bearing capacity for mats constructed over granular soil deposits can be adequately determined from the standard penetration resistance numbers. From Eq. (5.64), for shallow foundations, where q net(all) 5 q u(net) FS 5 1.713c 0.195B u 1 1 L 1 1 0.4 D f B N 60 5 standard penetration resistance B 5 width (m) F d 5 1 1 0.33(D f >B) < 1.33 S e 5 settlement, (mm) q net (kn>m ) 5 N 60 0.08 B 1 0.3 F B d S e 5 When the width B is large, the preceding equation can be approximated as [Eq. (5.64)] q net (kn>m ) 5 N 60 0.08 F d S e 5 5 N 60 0.08 B1 1 0.33 D f B RBS e(mm) 5 < 16.63N 60 B S e(mm) R 5 R (6.1)
98 Chapter 6: Mat Foundations In English units, Eq. (6.1) may be expressed as q net(all) (kip>ft ) 5 0.5N 60 B1 1 0.33 D f B R3S e(in.)4 < 0.33N 60 3S e (in.)4 (6.13) Generally, shallow foundations are designed for a maximum settlement of 5 mm and a differential settlement of about 19 mm. However, the width of the raft foundations are larger than those of the isolated spread footings. As shown in Table 5.3, the depth of significant stress increase in the soil below a foundation depends on the width of the foundation. Hence, for a raft foundation, the depth of the zone of influence is likely to be much larger than that of a spread footing. Thus, the loose soil pockets under a raft may be more evenly distributed, resulting in a smaller differential settlement. Accordingly, the customary assumption is that, for a maximum raft settlement of 50 mm, the differential settlement would be 19 mm. Using this logic and conservatively assuming that F d 5 1, we can respectively approximate Eqs. (6.1) and (6.13) as q net(all) 5 q net (kn>m ) < 5N 60 (6.14a) The net allowable pressure applied on a foundation (see Figure 6.7) may be expressed as q 5 Q A gd f (6.15) where Q 5 dead weight of the structure and the live load A 5 area of the raft In all cases, q should be less than or equal to allowable q net. D f Unit weight Q Figure 6.7 Definition of net pressure on soil caused by a mat foundation
6.5 Differential Settlement of Mats 99 Example 6.1 Determine the net ultimate bearing capacity of a mat foundation measuring 15 m 3 10 m on a saturated clay with c u 5 95 kn>m, f50, and D f 5 m. Solution From Eq. (6.10), q net(u) 5 5.14c u B1 1 0.195B L RB1 1 0.4 D f B R 5 (5.14)(95) B1 1 5 595.9 kn,m 0.195 3 10 15 RB1 1 0.4 3 R 10 Example 6. What will be the net allowable bearing capacity of a mat foundation with dimensions of 15 m 3 10 m constructed over a sand deposit? Here, D f 5 m, the allowable settlement is 5 mm, and the average penetration number N 60 5 10. Solution From Eq. (6.1), or q net(all) 5 N 60 0.08 B1 1 0.33 D f B R S e 5 < 16.63N 60 S e 5 q net(all) 5 10 0.33 3 B1 1 R 5 0.08 10 5 5 133.5 kn,m 6.5 Differential Settlement of Mats In 1988, the American Concrete Institute Committee 336 suggested a method for calculating the differential settlement of mat foundations. According to this method, the rigidity factor is calculated as where K r Er 5 modulus of elasticity of the material used in the structure E s 5 modulus of elasticity of the soil B 5 width of foundation I b 5 moment of inertia of the structure per unit length at right angles to B The term ErI b can be expressed as K r 5 ErI b E s B 3 ErI b 5 Er I F 1 a Ir b 1 a ah 3 1 (6.16) (6.17)
300 Chapter 6: Mat Foundations where ErI b 5 flexural rigidity of the superstructure and foundation per unit length at right angles to B SErIr b 5 flexural rigidity of the framed members at right angles to B S(Erah 3 >1) 5 flexural rigidity of the shear walls a 5 shear wall thickness h 5 shear wall height ErI F 5 flexibility of the foundation Based on the value of K r, the ratio (d) of the differential settlement to the total settlement can be estimated in the following manner: 1. If K r. 0.5, it can be treated as a rigid mat, and d50.. If K r 5 0.5, then d < 0.1. 3. If K r 5 0, then d50.35 for square mats (B>L 5 1) and d50.5 for long foundations (B>L 5 0). 6.6 Field Settlement Observations for Mat Foundations Several field settlement observations for mat foundations are currently available in the literature. In this section, we compare the observed settlements for some mat foundations constructed over granular soil deposits with those obtained from Eqs. (6.1) and (6.13). Meyerhof (1965) compiled the observed maximum settlements for mat foundations constructed on sand and gravel, as listed in Table 6.1. In Eq. (6.1), if the depth factor, 1 1 0.33(D f >B), is assumed to be approximately unity, then S e (mm) < q net(all) N 60 (6.18) q net(all) From the values of and N 60 given in Columns 6 and 5, respectively, of Table 6.1, the magnitudes of S e were calculated and are given in Column 8. Column 9 of Table 6.1 gives the ratios of calculated to measured values of S e. These ratios vary from about 0.79 to 3.39. Thus, calculating the net allowable bearing capacity with the use of Eq. (6.1) or (6.13) will yield safe and conservative values. 6.7 Compensated Foundation Figure 6.7 and Eq. (6.15) indicate that the net pressure increase in the soil under a mat foundation can be reduced by increasing the depth D f of the mat. This approach is generally referred to as the compensated foundation design and is extremely useful when structures are to be built on very soft clays. In this design, a deeper basement is made below the higher portion of the superstructure, so that the net pressure increase in soil at any depth is relatively uniform. (See Figure 6.8.) From Eq. (6.15) and Figure 6.7, the net average applied pressure on soil is q 5 Q A gd f
Table 6.1 Settlement of Mat Foundations on Sand and Gravel (Based on Meyerhof, 1965) (Based on Meyerhof, G. G., (1965). Shallow Foundations, Journal of the Soil Mechanics and Foundation Engineering Divsion, American Society of Civil Engineers, Vol. 91, No., pp. 1 31, Table 1. With permission from ASCE.) Observed calculated S e Calculated maximum maximum Case B Average q net(all) settlement, settlement, No. Structure Reference m N kn,m 60 S e mm S e mm observed S e (1) () (3) (4) (5) (6) (7) (8) (9) 1 T. Edison Rios and Silva (1948) 18.9 9.8 15.4 30.64 São Paulo, Brazil 15.01 Banco do Brazil Rios and Silva (1948);.86 39.4 7.94 6.6 São Paulo, Brazil Vargas (1961) 18 0.95 3 Iparanga Vargas (1948) 9.14 304.4 35.56 67.64 São Paulo, Brazil 9 1.9 4 C.B.I., Esplanda Vargas (1961) 14.63 383.0 7.94 34.8 São Paulo, Brazil 1.5 5 Riscala Vargas (1948) 3.96 9.8 1.7.98 São Paulo, Brazil 0 1.81 6 Thyssen Schultze (196).55 39.4 4.13 19.15 Düsseldorf, Germany 5 0.79 7 Ministry Schultze (196) 15.85 0. 0.3.0 Düsseldorf, Germany 0 1.08 8 Chimney Schultze (196) 0.4 17.4 10.16 34.48 Cologne, Germany 10 3.39 301
30 Chapter 6: Mat Foundations Figure 6.8 Compensated foundation For no increase in the net pressure on soil below a mat foundation, q should be zero. Thus, D f 5 Q Ag (6.19) D f This relation for is usually referred to as the depth of a fully compensated foundation. The factor of safety against bearing capacity failure for partially compensated foundations (i.e., D f, Q>Ag) may be given as FS 5 q net(u) q 5 q net(u) Q A gd f (6.0) where q net(u) 5 net ultimate bearing capacity. For saturated clays, the factor of safety against bearing capacity failure can thus be obtained by substituting Eq. (6.10) into Eq. (6.0): FS 5 5.14c u 1 1 0.195B L 1 1 0.4 D f B Q A gd f (6.1) Example 6.3 The mat shown in Figure 6.7 has dimensions of 18.3 m 3 30.5 m. The total dead and live load on the mat is 111 3 10 3 kn kip. The mat is placed over a saturated clay having a unit weight of 18.87 kn>m 3 and c u 5 134 kn>m. Given that D f 5 1.5 m, determine the factor of safety against bearing capacity failure.
Solution From Eq. (6.1), the factor of safety We are given that c u 5 134 kn>m, 18.87 kn>m 3. Hence, FS 5 FS 5 5.14c u 1 1 0.195B L 1 1 0.4 D f B Q A gd f 6.7 Compensated Foundation 303 D f 5 1.5 m, B 5 18.3 m, L 5 30.5 m, (5.14)(134) B1 1 (0.195)(18.3) RB1 1 0.4 1.5 30.5 18.3 R 111 3 103 kn 18.3 3 30.5 (18.87)(1.5) 5 4.66 and g5 Example 6.4 Consider a mat foundation 30 m 3 40 m in plan, as shown in Figure 6.9. The total dead load and live load on the raft is 00 3 10 3 kn. Estimate the consolidation settlement at the center of the foundation. Solution From Eq. (1.61) S c(p) 5 C ch c log sr o 1Dsr av 1 1 e o sr o sr o 5 (3.67)(15.7) 1 (13.33)(19.1 9.81) 1 6 (18.55 9.81) < 08 kn>m H c 5 6 m C c 5 0.8 e o 5 0.9 For Q 5 00 3 10 3 kn, the net load per unit area is q 5 Q A gd 00 3 103 f 5 (15.7)() < 135. kn>m 30 3 40 In order to calculate Dsr av we refer to Section 5.5. The loaded area can be divided into four areas, each measuring 15 m 3 0 m. Now using Eq. (5.19), we can calculate the average stress increase in the clay layer below the corner of each rectangular area, or Dsr av(h >H 1 ) 5 q o B H I a(h ) H 1 I a(h1 ) R H H 1 5 135.B (1.67 1 13.3 1 6)I a(h ) (1.67 1 13.33)I a(h1 ) R 6
304 Chapter 6: Mat Foundations m 1.67 m 30 m 40 m Q z Sand 15.7 kn/m 3 Groundwater table 13.33 m Sand sat 19.1 kn/m 3 6 m I a(h ) For, Sand m 5 B H 5 From Fig. 5.7, for m 5 0.71 and n 5 0.95, the value of is 0.1. Again, for From Figure 5.7, I a(h1 ) 5 0.5, so Dsr av(h >H 1 ) 5 135.B Normally consolidated clay 18.55 kn/m 3 sat 0.8; e o 0.9 C c n 5 L H 5 0 1 5 0.95 15 1.67 1 13.33 1 6 5 0.71 m 5 B H 1 5 15 15 5 1 n 5 L H 1 5 0 15 5 1.33 Figure 6.9 Consolidation settlement under a mat foundation I a(h ) (1)(0.1) (15)(0.5) R 5 3.3 kn>m 6 I a(h1 ), So, the stress increase below the center of the 30 m 3 40 m area is (4)(3.3) 5 93.8 kn>m. Thus S c(p) 5 (0.8)(6) 1 1 0.9 08 1 93.8 log 5 0.14 m 08 5 14 mm 6.8 Structural Design of Mat Foundations The structural design of mat foundations can be carried out by two conventional methods: the conventional rigid method and the approximate flexible method. Finite-difference and finite-element methods can also be used, but this section covers only the basic concepts of the first two design methods.
Conventional Rigid Method 6.8 Structural Design of Mat Foundations 305 The conventional rigid method of mat foundation design can be explained step by step with reference to Figure 6.10: Step 1. Figure 6.10a shows mat dimensions of L 3 B and column loads of Q 1, Q, Q 3, c. Calculate the total column load as Q 5 Q 1 1 Q 1 Q 3 1 c (6.) Step. Determine the pressure on the soil, q, below the mat at points A, B, C, D, c, by using the equation q 5 Q A 6 M yx I y 6 M xy I x (6.3) where A 5 BL I x 5 (1>1)BL 3 5 moment of inertia about the x-axis I y 5 (1>1)LB 3 5 moment of inertia about the y-axis M x 5 moment of the column loads about the x-axis 5 Qe y M y 5 moment of the column loads about the y-axis 5 Qe x The load eccentricities, e x and e y, in the x and y directions can be determined by using (xr, yr) coordinates: xr 5 Q 1xr 1 1 Q xr 1 Q 3 xr 3 1 c Q (6.4) and e x 5 xr B (6.5) Similarly, yr 5 Q 1yr 1 1 Q yr 1 Q 3 yr 3 1 c Q (6.6) and e y 5 yr L (6.7) Step 3. Step 4. Compare the values of the soil pressures determined in Step with the net allowable soil pressure to determine whether q < q all(net). Divide the mat into several strips in the x and y directions. (See Figure 6.10). Let the width of any strip be B 1.
y y A B 1 B 1 B 1 B 1 B C Q 9 Q 10 Q 11 D Q 1 B 1 e x e y B 1 L J E x Q 5 Q 6 Q 7 Q 8 B 1 Q 1 Q Q 3 Q 4 I H G F B (a) x FQ 1 FQ FQ 3 I H G FQ 4 F B 1 q av(modified) unit length B (b) Edge of mat Edge of mat L d/ d/ L d/ b o L L Edge of mat d/ d/ L b o L L (c) L L d/ d/ d/ d/ b o (L L ) L 306 Figure 6.10 Conventional rigid mat foundation design
6.8 Structural Design of Mat Foundations 307 Step 5. Draw the shear, V, and the moment, M, diagrams for each individual strip (in the x and y directions). For example, the average soil pressure of the bottom strip in the x direction of Figure 6.10a is q av < q I 1 q F (6.8) where q I and q F 5 soil pressures at points I and F, as determined from Step. The total soil reaction is equal to q av B 1 B. Now obtain the total column load on the strip as Q 1 1 Q 1 Q 3 1 Q 4. The sum of the column loads on the strip will not equal q av B 1 B, because the shear between the adjacent strips has not been taken into account. For this reason, the soil reaction and the column loads need to be adjusted, or Average load 5 q avb 1 B 1 (Q 1 1 Q 1 Q 3 1 Q 4 ) (6.9) Now, the modified average soil reaction becomes q av(modified) 5 q av average load q av B 1 B (6.30) and the column load modification factor is average load F 5 Q 1 1 Q 1 Q 3 1 Q 4 (6.31) Step 6. So the modified column loads are FQ 1, FQ, FQ 3, and FQ 4. This modified loading on the strip under consideration is shown in Figure 6.10b. The shear and the moment diagram for this strip can now be drawn, and the procedure is repeated in the x and y directions for all strips. Determine the effective depth d of the mat by checking for diagonal tension shear near various columns. According to ACI Code 318-95 (Section 11.1..1c, American Concrete Institute, 1995), for the critical section, U 5 b o d3f(0.34)"fr c 4 (6.3) where U 5 factored column load (MN), or (column load) 3 (load factor) f5reduction factor 5 0.85 fr c 5 compressive strength of concrete at 8 days (MN>m ) The units of b o and d in Eq. (6.3a) are in meters. The expression for b o in terms of d, which depends on the location of the column with respect to the plan of the mat, can be obtained from Figure 6.10c.
308 Chapter 6: Mat Foundations Step 7. Step 8. From the moment diagrams of all strips in one direction (x or y), obtain the maximum positive and negative moments per unit width (i.e., Mr 5 M>B 1 ). Determine the areas of steel per unit width for positive and negative reinforcement in the x and y directions. We have M u 5 (Mr)(load factor) 5fA s f y d a (6.33) and where a 5 A sf y 0.85fr c b A s 5 area of steel per unit width f y 5 yield stress of reinforcement in tension M u 5 factored moment f50.9 5 reduction factor (6.34) Examples 6.5 and 6.6 illustrate the use of the conventional rigid method of mat foundation design. Approximate Flexible Method In the conventional rigid method of design, the mat is assumed to be infinitely rigid. Also, the soil pressure is distributed in a straight line, and the centroid of the soil pressure is coincident with the line of action of the resultant column loads. (See Figure 6.11a.) In the approximate flexible method of design, the soil is assumed to be equivalent to an infinite number of elastic springs, as shown in Figure 6.11b. This assumption is sometimes referred to as the Winkler foundation. The elastic constant of these assumed springs is referred to as the coefficient of subgrade reaction, k. To understand the fundamental concepts behind flexible foundation design, consider a beam of width B 1 having infinite length, as shown in Figure 6.11c. The beam is subjected to a single concentrated load Q. From the fundamentals of mechanics of materials, d z M 5 E F I F dx (6.35) where M 5 moment at any section E F 5 modulus of elasticity of foundation material I moment of inertia of the cross section of the beam 5 ( 1)B 1 1 h 3 F 5 However, and dm dx dv dx 5 shear force 5 V 5 q 5 soil reaction (see Figure 6.11c).
Q 6.8 Structural Design of Mat Foundations 309 Q 1 Q Q 3 Resultant of soil pressure (a) Q Q 1 (b) A Point load B 1 h x Section at A A A q z (c) Figure 6.11 (a) Principles of design by conventional rigid method; (b) principles of approximate flexible method; (c) derivation of Eq. (6.39) for beams on elastic foundation Hence, d M dx 5 q Combining Eqs. (6.35) and (6.36) yields d 4 z E F I F dx 5 q 4 (6.36) (6.37) However, the soil reaction is q 5zkr
310 Chapter 6: Mat Foundations where z 5 deflection kr 5 kb 1 k 5 coefficient So, of subgrade reaction (kn>m 3 or lb>in 3 ) E F I F d4 z dx 4 5zkB 1 (6.38) Solving Eq. (6.38) yields z 5 e ax (Ar cos bx 1 As sin bx) (6.39) where Ar and As are constants and b5 4 Å B 1 k 4E F I F (6.40) The unit of the term b, as defined by the preceding equation, is (length) 1. This parameter is very important in determining whether a mat foundation should be designed by the conventional rigid method or the approximate flexible method. According to the American Concrete Institute Committee 336 (1988), mats should be designed by the conventional rigid method if the spacing of columns in a strip is less than 1.75>b. If the spacing of columns is larger than 1.75>b, the approximate flexible method may be used. To perform the analysis for the structural design of a flexible mat, one must know the principles involved in evaluating the coefficient of subgrade reaction, k. Before proceeding with the discussion of the approximate flexible design method, let us discuss this coefficient in more detail. If a foundation of width B (see Figure 6.1) is subjected to a load per unit area of q, it will undergo a settlement D. The coefficient of subgrade modulus can be defined as k 5 q D (6.41) B q Figure 6.1 Definition of coefficient of subgrade reaction, k
6.8 Structural Design of Mat Foundations 311 The unit of k is kn/m 3. The value of the coefficient of subgrade reaction is not a constant for a given soil, but rather depends on several factors, such as the length L and width B of the foundation and also the depth of embedment of the foundation. A comprehensive study by Terzaghi (1955) of the parameters affecting the coefficient of subgrade reaction indicated that the value of the coefficient decreases with the width of the foundation. In the field, load tests can be carried out by means of square plates measuring 0.3 m 3 0.3 m, and values of k can be calculated. The value of k can be related to large foundations measuring B 3 B in the following ways: Foundations on Sandy Soils For foundations on sandy soils, k 5 k 0.3 B 1 0.3 B (6.4) where k 0.3 and k 5 coefficients of subgrade reaction of foundations measuring 0.3 m 3 0.3 m and B (m) 3 B (m), respectively (unit is kn>m 3 ). Foundations on Clays For foundations on clays, 0.3 (m) k(kn>m 3 ) 5 k 0.3 (kn>m 3 ) B B(m) R (6.43) The definitions of k and k 0.3 in Eq. (6.43) are the same as in Eq. (6.4). For rectangular foundations having dimensions of B 3 L (for similar soil and q), k 5 B k (B3B) 1 1 0.5 L 1.5 (6.44) where k 5 coefficient of subgrade modulus of the rectangular foundation (L 3 B) k (B3B) 5 coefficient of subgrade modulus of a square foundation having dimension of B 3 B
31 Chapter 6: Mat Foundations Equation (6.44) indicates that the value of k for a very long foundation with a width B is approximately 0.67k (B3B). The modulus of elasticity of granular soils increases with depth. Because the settlement of a foundation depends on the modulus of elasticity, the value of k increases with the depth of the foundation. Table 6. provides typical ranges of values for the coefficient of subgrade reaction, k 0.3 (k 1 ), for sandy and clayey soils. For long beams, Vesic (1961) proposed an equation for estimating subgrade reaction, namely, kr 5 Bk 5 0.65 1 E s B 4 É E F I F E s 1 m s or k 5 0.65 1 E s B 4 É E F I F E s B(1 m s ) (6.45) where E s 5 modulus of elasticity of soil B 5 foundation width E F 5 modulus of elasticity of foundation material I F 5 moment of inertia of the cross section of the foundation m s 5 Poisson s ratio of soil Table 6. Typical Subgrade Reaction Values, k 0.3 (k 1 ) Soil type k 0.3 (k 1 ) MN,m 3 Dry or moist sand: Loose 8 5 Medium 5 15 Dense 15 375 Saturated sand: Loose 10 15 Medium 35 40 Dense 130 150 Clay: Stiff 10 5 Very stiff 5 50 Hard.50
For most practical purposes, Eq. (6.46) can be approximated as 6.8 Structural Design of Mat Foundations 313 k 5 E s B(1 m s ) (6.46) Now that we have discussed the coefficient of subgrade reaction, we will proceed with the discussion of the approximate flexible method of designing mat foundations. This method, as proposed by the American Concrete Institute Committee 336 (1988), is described step by step. The use of the design procedure, which is based primarily on the theory of plates, allows the effects (i.e., moment, shear, and deflection) of a concentrated column load in the area surrounding it to be evaluated. If the zones of influence of two or more columns overlap, superposition can be employed to obtain the net moment, shear, and deflection at any point. The method is as follows: Step 1. Step. Assume a thickness h for the mat, according to Step 6 of the conventional rigid method. (Note: h is the total thickness of the mat.) Determine the flexural ridigity R of the mat as given by the formula E F h 3 R 5 1(1 m F ) (6.47) where E F 5 modulus m F 5 Poisson s of elasticity of foundation material ratio of foundation material Step 3. Determine the radius of effective stiffness that is, Lr 5 4 R Å k (6.48) Step 4. where k 5 coefficient of subgrade reaction. The zone of influence of any column load will be on the order of 3 to 4 Lr. Determine the moment (in polar coordinates at a point) caused by a column load (see Figure 6.13a). The formulas to use are Q M r 5 radial moment 5 4 C A 1 (1 m F)A r S Lr (6.49) and Q M t 5 tangential moment 5 4 C m FA 1 1 (1 m F)A r S Lr (6.50)
314 Chapter 6: Mat Foundations 6 y 5 M r M y M t 4 r (a) M x x r L 3 1 A A4 A 1 A 3 0 0.4 0.3 0. 0.1 0 0.1 0. 0.3 0.4 A 1, A, A 3, A 4 (b) Figure 6.13 Approximate flexible method of mat design Step 5. Step 6. Step 7. where r 5 radial distance from the column load Q 5 column load A 1, A 5 functions of r>lr The variations of A 1 and A with r>lr are shown in Figure 6.13b. (For details see Hetenyi, 1946.) In the Cartesian coordinate system (see Figure 6.13a), and M x 5 M t sin a1m r cos a (6.51) M y 5 M t cos a1m r sin a (6.5) For the unit width of the mat, determine the shear force V caused by a column load: V 5 Q (6.53) 4Lr A 3 The variation of A 3 with r>lr is shown in Figure 6.13b. If the edge of the mat is located in the zone of influence of a column, determine the moment and shear along the edge. (Assume that the mat is continuous.) Moment and shear opposite in sign to those determined are applied at the edges to satisfy the known conditions. The deflection at any point is given by The variation of A 4 d5 QLr 4R A 4 is presented in Figure 6.13b. (6.54)
6.8 Structural Design of Mat Foundations 315 Example 6.5 The plan of a mat foundation is shown in Figure 6.14. Calculate the soil pressure at points A, B, C, D, E, and F. (Note: All column sections are planned to be 0.5 m 0.5 m.) y y A 400 kn G B I C 500 kn 450 kn 0.5 m 7 m 1500 kn 4.5 m 1500 kn 100 kn 8 m 4.5 m 7 m x 1500 kn 1500 kn 100 kn 7 m 400 kn 500 kn 350 kn F H E J D 8 m 8 m 0.5 m 0.5 m Figure 6.14 Plan of a mat foundation 0.5 m x Solution Eq. (6.3): q 5 Q A 6 M y x I y 6 M x y I x A (16.5)(1.5) 354.75 m I x 5 1 1 BL3 5 1 I y 5 1 1 (16.5)(1.5)3 5 13,665 m 4 1 LB3 5 1 1 (1.5)(16.5)3 5 8,050 m 4 Q 350 ()(400) 450 ()(500) ()(100) (4)(1500) 11,000 kn M y 5 Qe x ; e x 5 x 9 B
316 Chapter 6: Mat Foundations xr 5 Q 1xr 1 1 Q xr 1 Q 3 xr 3 1 c Q (8.5)(500 1 1500 1 1500 1 500) 5 1 11,000 C 1 (16.5)(350 1 100 1 100 1 450) 1 (0.5)(400 1 1500 1 1500 1 400) S 5 7.814 m e x 5 x 9 B 5 7.814 8.5 50.435 m < 0.44 m Hence, the resultant line of action is located to the left of the center of the mat. So M y (11,000)(0.44) 4840 kn-m. Similarly M x 5 Qe y ; e y 5 yr L yr 5 Q 1yr 1 1 Q yr 1 Q 3 yr 3 1 c Q 5 1 (0.5)(400 1 500 1 350) 1 (7.5)(1500 1 1500 1 100) c 11,000 1(14.5)(1500 1 1500 1 100) 1 (1.5)(400 1 500 1 450) d 5 10.85 m e y 5 y 9 L 5 10.85 10.75 5 0.1m The location of the line of action of the resultant column loads is shown in Figure 6.15. M x (11,000)(0.1) 1100 kn-m. So q 5 11,000 354.75 6 4840x 8050 6 1100y 13,665 5 31.0 6 0.6x 6 0.08y( kn>m ) y y A 400 kn G B I C 500 kn 450 kn 0.5 m 7 m 1500 kn 4.5 m 0.1 m 1500 kn 8 m 0.44 m 100 kn 4.5 m 7 m x 1500 kn 1500 kn 100 kn 7 m 400 kn 500 kn F H E J 8 m 8 m 0.5 m 350 kn D 0.5 m 0.5 m x Figure 6.15
6.8 Structural Design of Mat Foundations 317 Therefore, At A: q 31.0 (0.6)(8.5) (0.08)(10.75) 36.81 kn/m At B: q 31.0 (0.6)(0) (0.08)(10.75) 31.86 kn/m At C: q 31.0 (0.6)(8.5) (0.08)(10.75) 6.91 kn/m At D: q 31.0 (0.6)(8.5) (0.08)(10.75) 5.19 kn/m At E: q 31.0 (0.6)(0) (0.08)(10.75) 30.14 kn/m At F: q 31.0 (0.6)(8.5) (0.08)(10.75) 35.09 kn/m Example 6.6 Divide the mat shown in Figure 6.14 into three strips, such as AGHF (B 1 4.5 m), GIJH (B 1 8 m), and ICDJ (B 1 4.5 m). Use the result of Example 6.5, and determine the reinforcement requirements in the y direction. Here, f c 0.7 MN/m, f y 413.7 MN/m, and the load factor is 1.7. Solution Determination of Shear and Moment Diagrams for Strips: Strip AGHF: 36.81 1 35.09 Average soil pressure 5 q av 5 q ( at A) 1 q ( at F) 5 5 35.95 kn>m Total soil reaction q av B 1 L (35.95)(4.5)(1.50) 385 kn load due to soil reaction 1 column loads Average load 5 385 1 3800 5 5 354.5 kn So, modified average soil pressure, q av(modified) 5 q ava 354.5 385 b 5 (35.95) a354.5b 5 38.768 kn>m 385 The column loads can be modified in a similar manner by multiplying factor F 5 354.5 3800 5 0.93 Figure 6.16 shows the loading on the strip and corresponding shear and moment diagrams. Note that the column loads shown in this figure have been multiplied by F 0.93. Also the load per unit length of the beam is equal to B 1 q av(modified) (4.5)(38.768) 164.76 kn/m. Strip GIJH: In a similar manner, q av 5 q ( at B) 1 q ( at E) 31.86 1 30.14 5 5 31.0 kn>m Total soil reaction (31)(8)(1.5) 533 kn Total column load 4000 kn
318 Chapter 6: Mat Foundations 37.89 kn 0.5 m A 1398.36 kn 1398.36 kn 37.89 kn 0.5 m 7 m 7 m 7 m F 164.76 kn/m 81.65 576.7 331.7 41.19 41.19 Shear (unit: kn) 381.7 576.70 81.65 117.57 177.57 5.15 718.35 5.15 Moment (unit: kn-m) 36.55 36.55 Figure 6.16 Load, shear, and moment diagrams for strip AGHF Average load 5 533 1 4000 q av(modified) 5 (31.0) a 4666 b 5 7.1 kn>m 533 F 5 4666 4000 5 1.1665 5 4666 kn The load, shear, and moment diagrams are shown in Figure 6.17. Strip ICDJ: Figure 6.18 shows the load, shear, and moment diagrams for this strip. Determination of the Thickness of the Mat For this problem, the critical section for diagonal tension shear will be at the column carrying 1500 kn of load at the edge of the mat [Figure 6.19]. So b o 5 a0.5 1 d b 1 a0.5 1 d b 1 (0.5 1 d) 5 1.5 1 d or U 5 (b o d)s(f)(0.34)"fc r T U (1.7)(1500) 550 kn.55 MN.55 5 (1.5 1 d)(d)s(0.85)(0.34)!0.7 T (1.5 d)(d) 1.94; d 0.68 m
6.8 Structural Design of Mat Foundations 319 583.5 kn 0.5 m B 1749.75 kn 1749.75 kn 583.5 kn 0.5 m 7 m 7 m 7 m E 17 kn/m 990.17 759.58 59 54.56 Shear (unit: kn) 54.56 59 759.58 160.89 990.17 160.89 6.75 91.6 6.78 Moment (unit: kn-m) 637.94 637.94 Figure 6.17 Load, shear, and moment diagrams for strip GIJH 39.4 kn 0.5 m C 1046.44 kn 1046.44 kn 305. kn 0.5 m 7 m 7 m 7 m D 19.8 kn/m 3.45 359.95 548.65 410.81 497.79 635.63 7.97 Shear (unit: kn) 3.3 664.56 360.15 4.06 4.06 Moment (unit: kn-m) 495.0 89.95 1196.19 Figure 6.18 Load, shear, and moment diagrams for strip ICDJ
30 Chapter 6: Mat Foundations 1500 kn Column load Edge of mat 0.5 + d 0.5 + d/ Figure 6.19 Critical perimeter column Assuming a minimum cover of 76 mm over the steel reinforcement and also assuming that the steel bars to be used are 5 mm in diameter, the total thickness of the slab is h 0.68 0.076 0.05 0.781 m 0.8 m The thickness of this mat will satisfy the wide beam shear condition across the three strips under consideration. Determination of Reinforcement From the moment diagram shown in Figures 6.16, 6.17, and 6.18, it can be seen that the maximum positive moment is located in strip AGHF, and its magnitude is Similarly, the maximum negative moment is located in strip ICDJ and its magnitude is From Eq. (6.33): M u 5 (Mr)( load factor) 5fA s f y ad a. b For the positive moment, 0.9. Also, from Eq. (6.34), M 9 5 177.57 5 177.57 5 406.5 kn-m>m B 1 4.5 Mr 5 1196.19 5 1196.19 5 81.5 kn-m>m B 1 4.5 M a0.68 a u 5 (406.5)(1.7) 5 (f)(a b s ) (413.7 3 1000) a 5 A s f y 0.85 f c 9 b 5 (A s)(413.7) (0.85)(0.7)(1) 5 3.51A s; or A s 5 0.045a 691.05 5 (0.9)(0.045a)(413,700) a0.68 a b; or a < 0.0645 So, A s (0.045)(0.0645) 0.0074 m /m 740 mm /m.
6.8 Structural Design of Mat Foundations 31 Use 5-mm diameter bars at 175 mm center-to-center: ca s provided 5 (491) a 1000 175 b 5 805.7 mm >md Similarly, for negative reinforcement, M u 5 (81.5)(1.7) 5 (f)(a s )(413.7 3 1000) a0.68 a b So 0.9. A s 0.045a 478.55 5 (0.9)(0.045a)(413.7 3 1000) a0.68 a b; or a < 0.045 So, A s (0.045)(0.045) 0.001913 m /m 1913 mm /m. Use 5-mm diameter bars at 55 mm center-to-center: [A s provided 195 mm ] Because negative moment occurs at midbay of strip ICDJ, reinforcement should be provided. This moment is Hence, M a0.68 a u 5 (68.)(1.7) 5 (0.9)(0.045a)(413.7 3 1000) b; or a < 0.0108 M 9 5 89.95 4.5 5 68. kn- m>m A s (0.0108)(0.045) 0.000459 m /m 459 mm /m Provide 16-mm diameter bars at 400 mm center-to-center: [A s provided 50 mm ] For general arrangement of the reinforcement see Figure 6.0. Top steel Bottom steel Top steel Additional top steel in strip ICDJ Figure 6.0 General arrangement of reinforcement
3 Chapter 6: Mat Foundations Problems 6.1 Determine the net ultimate bearing capacity of mat foundations with the following characteristics: c u 5 10 kn>m, f 50, B 5 8 m, L 5 18 m, D f 5 3 m 6. Following are the results of a standard penetration test in the field (sandy soil): Depth (m) Field value of N 60 1.5 9 3.0 1 4.5 11 6.0 7 7.5 13 9.0 11 10.5 13 Estimate the net allowable bearing capacity of a mat foundation 6.5 m 3 5 m in plan. Here, D f 5 1.5 m and allowable settlement 5 50 mm. Assume that the unit weight of soil, g516.5 kn>m 3. 6.3 Repeat Problem 6. for an allowable settlement of 30 mm. 6.4 A mat foundation on a saturated clay soil has dimensions of 0 m 3 0 m. Given: dead and live load 5 48 MN, c u 5 30 kn>m, and g clay 5 18.5 kn>m 3. a. Find the depth, D f, of the mat for a fully compensated foundation. b. What will be the depth of the mat (D f ) for a factor of safety of against bearing capacity failure? 6.5 Repeat Problem 6.4 part b for c u 5 0 kn>m. 6.6 A mat foundation is shown in Figure P6.6. The design considerations are L 5 1 m, B 5 10 m, D f 5. m, Q 5 30 MN, x 1 5 m, x 5 m, x and preconsolidation pressure sr c < 105 kn>m 3 5 5. m,. Calculate the consolidation settlement under the center of the mat. 6.7 For the mat foundation in Problem 6.6, estimate the consolidation settlement under the corner of the mat. 6.8 From the plate load test (plate dimensions 0.3 m 3 0.3 m) in the field, the coefficient of subgrade reaction of a sandy soil is determined to be 14,900 kn>m 3. What will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 7.5 m 3 7.5 m? 6.9 Refer to Problem 6.18. If the full-sized foundation had dimensions of 1.3 m 3 9.1 m, what will be the value of the coefficient of subgrade reaction? 6.10 The subgrade reaction of a sandy soil obtained from the plate load test (plate dimensions 1 m 3 0.7 m) is 18 MN>m 3. What will be the value of k on the same soil for a foundation measuring 5 m 3 3.5 m?
References 33 D f Size of mat B L Q Sand 16.0 kn/m 3 x 1 z Groundwater table x Sand sat 18.0 kn/m 3 x 3 sat e o C c C s Clay 17.5 kn/m 3 0.88 0.38 0.1 Figure P6.6 References AMERICAN CONCRETE INSTITUTE (1995). ACI Standard Building Code Requirements for Reinforced Concrete, ACI 318 95, Farmington Hills, MI. AMERICAN CONCRETE INSTITUTE COMMITTEE 336 (1988). Suggested Design Procedures for Combined Footings and Mats, Journal of the American Concrete Institute, Vol. 63, No. 10, pp. 1041 1077. HETENYI,M.(1946). Beams of Elastic Foundations, University of Michigan Press, Ann Arbor, MI. MEYERHOF, G. G. (1965). Shallow Foundations, Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 91, No. SM, pp. 1 31. RIOS, L., and SILVA, F. P. (1948). Foundations in Downtown São Paulo (Brazil), Proceedings, Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam, Vol. 4, p. 69. SCHULTZE, E. (196). Probleme bei der Auswertung von Setzungsmessungen, Proceedings, Baugrundtagung, Essen, Germany, p. 343. TERZAGHI, K. (1955). Evaluation of the Coefficient of Subgrade Reactions, Geotechnique, Institute of Engineers, London, Vol. 5, No. 4, pp. 197 6. VARGAS, M.(1948). Building Settlement Observations in São Paulo, Proceedings Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam, Vol. 4, p. 13. VARGAS, M.(1961). Foundations of Tall Buildings on Sand in São Paulo (Brazil), Proceedings, Fifth International Conference on Soil Mechanics and Foundation Engineering, Paris, Vol. 1, p. 841. VESIC, A. S. (1961). Bending of Beams Resting on Isotropic Solid, Journal of the Engineering Mechanics Division, American Society of Civil Engineers, Vol. 87, No. EM, pp. 35 53.