(Lecture 19) MAT FOUNDATIONS

Module 5 (Lecture 19) MAT FOUNDATIONS Topics 1.1 STRUCTURAL DESIGN OF MAT FOUNDATIONS Conventional Rigid Method 1. Approximate Flexible Method Foundations on Sandy Soils Foundations on Clays 1.3 Example 1.4 PROBLEMS 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 The conventional rigid method of mat foundation design can be explained step by step with reference to figure 5.8. 1. Figure 5.8a shows mat dimensions of LL BB and columns loads of QQ 1, QQ, QQ 3, Calculate the total column load as QQ = QQ 1 + QQ + QQ 3 [5.4]

. Determine the pressure on the soil, q, below the mat at points AA, BB, CC, DD,, by using the equation AA = BBBB qq = QQ AA ± MM yy xx II yy ± MM xx yy II xx [5.5] II xx = (1/1)BBBB 3 = moment of inertia about the xx axis II yy = (1/1)LLLL 3 = moment of inertia about the yy axis MM xx = moment of the column loads about the xx axis = QQ eeyy MM yy = moment of the column loads about the yy axis = QQ eexx The load eccentricities, ee xx and ee yy, in the xx and yy directions can be determined by using (xx, yy ) coordinates: xx = QQ 1xx 1 +QQ xx +QQ 3 xx 3 + QQ [5.6] And ee xx = xx BB [5.7] Similarly yy = QQ 1yy 1 +QQ yy +QQ 3 yy 3 + QQ [5.8] And ee yy = yy LL [5.9] 3. Compare the values of the soil pressures determined in step with the net allowable soil pressure to determine whether qq qq all (net ). 4. Divide the mat into several strips in x and y directions (see figure 5.8a). Let the width of any strip be BB 1. 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 5.8a is qq aaaa qq II+qq FF [5.30]

qq II and qq FF = soil pressures at poins II and FF as determined from step. The total soil reaction is equal to qq aaaa BB 1 BB. Now obtain the total column load on the strip as QQ 1 + QQ + QQ 3 + QQ 4. The sum of the column loads on the strip will not equal qq aaaa BB 1 BB 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 = qq aaaa BB 1 BB+(QQ 1 +QQ +QQ 3 +QQ 4 ) [5.31] Now, the modified average soil reaction becomes qq aaaa (modified ) = qq aaaa average load qq aaaa BB 1 BB [5.3] And the column load modification factor is FF = average load QQ 1 +QQ +QQ 3 +QQ 4 [5.33] So, the modified column loads are FFFF 1, FFFF, FFFF 3, and FFFF 4. This modified loading on the strip under consideration is shown in figure 5.8b. The shear and the moment diagram for this strip can now be drawn. This procedure is repeated for all strips in the x and y directions. 6. Determine the effective depth of the mat d by checking for diagonal tension shear near various columns. According to ACI Code 318-95. American Concrete Institute, 1995), for the critical section, UU = bb oo dd[φφ(0.34) ff cc ] [5.34] UU = factored column load (MN), or (column load) (load factor) φφ = reduction factor = 0.85 ff cc = compressive strength of concrete at 8 days (MN/m ) The units of bb oo and dd in equation (34) are in meters. In English units, equation (34) may be expressed as UU = bb oo dd (4φφ ff cc ) [5.35]

UU is in lb, bb oo and dd are in in., and ff cc is in lb/in The expression for bb oo in terms of dd, which depends on the location of the column with respect to the plan of the mat, can be obtained from figure 5.8c. 7. From the moment diagrams of all strips in one direction (x or y), obtain the maximum positive and negative moments per unit width (that is, MM = MM/BB 1 ). 8. Determine the areas of steep per unit width for positive and negative reinforcement in the x and y directions. MM uu = (MM )(load factor) = φφaa ss ff yy dd aa [5.36] And aa = AA ssff yy 0.85ff cc bb [5.37] AA ss = area of steel per unit width ff yy = yield stress of reinforcement in tension MM uu = factored moment φφ = 0.9 = reduction factor Examples 5 and 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 coincidental with the line of action of the resultant column loads (see figure 5.9). In the approximate flexible method of design, the soil is assumed to be equivalent to infinite number of elastic springs, as shown in figure 5.9b. It 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.

Figure 5.9 (a) Principles of design conventional rigid method; (b) principles of approximate flexible method; (c) derivation of equation (4) for beams on elastic foundation To understand the fundamental concepts behind flexible foundation design, consider a beam of width BB 1 having infinite length, as shown in figure 5.9c. The beam is subjected to a single concentrated load Q. from the fundamental of mechanics of materials, MM = EE FF II FF dd zz dddd [5.38] MM = moment at any section EE FF = modulus of elasticity of foundation material II FF = moment of inertia of the cross section of the beam = 1 1 BB 1h 3 (see figure 5.9c) However dddd dddd = shear force = VV

And dddd dddd = qq = soil reaction Hence dd MM dddd = qq [5.39] Combining equations (38 and 39) yields dd EE FF II 4 zz FF dddd 4 = qq [5.40] However, the soil reaction is qq = zzzz zz = deflection kk = kkbb 1 kk = coefficient of subgrade reaction (kn/m 3 or lb/in 3 ) So EE FF II FF dd 4 zz dddd 4 = zzzzbb 1 [5.41] Solution of equation (41) yields zz = ee αααα (AA cos ββββ + AA" sin ββββ) [5.4] AA and AA" are constants and 4 ββ = BB 1kk 4EE FF II FF [5.43] The unit of the term ββ as defined by the preceding equation is (length) 1. This parameter is very important in determining whether a mat foundation should be designed by conventional rigid method or 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/ββ. If the spacing of columns is larger than 1.75/ββ, the approximate flexible method may be used.

To perform the analysis for the structural design of a flexible mat, you must know the principles of evaluating the coefficient of subgrade reaction, k. before proceeding with the discussion of the approximate flexible design method, let as discuss this coefficient in more detail. If a foundation of width B (figure 5.10) is subjected to a load per unit area of q, it will undergo a settlement, Δ. The coefficient of subgrade modulus, k, can be defined as kk = qq Δ [5.44] Figure 5.10 Definition of coefficient of subgrade reaction, kk The unit of kk is kn/m 3 (or lb/in 3 ). The value of the coefficient of subgrade reaction is not a constant for a given soil. T depends on several factors, such as the length, LL, and width, BB, of the foundation and also the depth of embedment of the foundation. Terzaghi (1955) made a comprehensive study of the parameters affecting the coefficient of subgrade reaction. It indicated that the value of the coefficient of subgrade reaction decreases with the width of the foundation. In the field, load tests can be carried out by means of square plate measuring 1 ft 1 ft (0.3 m 0.3 m) and values of k can be calculated. The value of k can be related to large foundations measuring BB BB in the following ways. Foundations on Sandy Soils kk = kk 0.3 BB+0.3 BB [5.45] kk 0.3 and kk = coefficients of subgrade reaction of foundation measuring 0.3m 0.3m and BB(m) BB(m), respectively (units is kn/m 3 In English units, equation (45) may be expressed as kk = kk 1 BB+1 BB [5.46]

kk 1 and kk = coefficient of subgrade reaction of foundation measuring 1 ft a ft and BB (ft) BB, respectively (units is lb/in 3 ) Foundations on Clays kk(kn/m 3 ) = kk 0.3 (kn/m 3 ) 0.3 (m) BB (m) [5.47] The definition of k in equation (47) is the same as in equation (45). In English units, kk(lb/in 3 ) = kk 1 (lb/in 3 ) 1 (ft) [5.48] BB (ft) The definitions of kk and kk 1 are the same as in equation (46). For rectangular foundations having dimensions of BB LL (for similar soil and q), kk = kk (BB BB) 1+0.5 BB LL 1.5 [5.49] kk = coefficient of subgrade modulus of the rectangular foundation (LL BB) kk (BB BB) = coefficient of subgrade modulus of a square foundation having dimension of BB BB Equation (49) indicates that the value of k of a very long foundation with a width B is approximately 0.67kk (BB BB). 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 as the depth of the foundation increases. Following are some typical ranges of value for the coefficient of subgrade reaction kk 1 for sandy and clayey soils. Sand (dry or moist) Loose: 9 9lb/in 3 (8 5MN/m 3 ) Medium: 91 460lb/in 3 (5 15MN/m 3 ) Dense: 460 1380lb/in 3 (15 375MN/m 3 )

Sand (saturated) Loose: 38 55lb/in 3 (10 15MN/m 3 ) Medium: 18 147lb/in 3 (35 40MN/m 3 ) Dense: 478 55lb/in 3 (130 150MN/m 3 ) Clay Stiff: 44 9lb/in 3 (1 5MN/m 3 Very stiff: 9 184lb/in 3 (5 50MN/m 3 Hard: > 184lb/in 3 (> 50MMMM/m 3 Scott (1981) proposed that for sandy soils the value of kk 0.3 can be obtained from standard penetration resistance at any given depth, or kk 0.3 (MN/m 3 ) = 18NN cor [5.50] NN cccccc = cccccccccccccccccc standard penetration resistance In English units, kk 1 (U. S. ton/ft 3 ) = 6N cor [5.51] For long beams, Vesic (1961) proposed an equation for estimating subgrade reaction: 1 EE ss kk BBBB = 0.65 EE ssbb 4 EE FF II FF 1 μμ ss [5.5] EE ss = modulus of elasticity of soil BB = foundation width EE FF = modulus of elasticity of foundation material II FF = moment of inertia of the cross section of the foundation μμ ss = Poisson sratio of soil For most practical purposes, equation (5) can be approximated as

kk = EE ss BB(1 μμ ss ) [5.53] The coefficient of subgrade reaction is also very useful parameter in the design of rigid highway and airfield pavements. The pavements with a concrete wearing surface are generally referred to as a rigid pavement, and the pavement with an asphaltic wearing surface is called a flexible pavement. For surface load acting on a rigid pavement, the maximum tensile stress occurs at the base of the slab. For estimating the magnitude of the maximum horizontal tensile stress developed at the base of the rigid pavement, elastic solutions involving slabs on Winkler foundations are extremely useful. Some of the early work in this area was done by Westergaard (196, 1939, and 1947). 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 design procedure is based primarily on the theory of plates. Its use allows the effects (that is, 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 used to obtain the net moment, shear, and deflection at any point. 1. Assume a thickness, h, for the mat, according to step 6 as outlined for the conventional rigid method. (Note: h is the total thickness of the mat).. Determine the flexural ridigity R of the mat: RR = EE FFh 3 1(1 μμ ss ) [5.54] EE FF = modulus of elasticity of foundation material μμ FF = Poisson s ratio of foundation material 3. Determine the radius of effective stiffness: LL 4 = RR kk [5.55] kk = coefficient of subgrade reaction The zone of influence of any column load will be on the order of 3 to 4 L. 4. Determine the moment (in polar coordinates at a point) caused by a column load (figure 5.11a):

MM rr = radial moment = QQ AA 4 1 (1 μμ FF)AA rr [5.56] LL MM tt = tangential moment = QQ μμ 4 FFAA 1 + (1 μμ FF)AA rr [5.57] LL Figure 5.11 Approximate flexible method of mat design rr = radial distance from the column load QQ = column load AA 1, AA = functions of rr/ll The variations of AA 1 and AA with rr/ll are shown in figure 5.11b (for details, see Hetenyi, 1946). In the Cartesian coordinates system (figure 5.11a), MM xx = MM tt sin αα + MM rr cos αα [5.58] MM yy = MM tt cos αα + MM rr sin αα [5.59] 5. For the unit width of the mat, determine the shear force, V, caused by a column load:

VV = QQ 4LL AA 3 [5.60] The variation of AA 3 with rr/ll is shown in figure 5.11b. 6. If the edge of the mat is located in the zone of influence of a column, determine the moment and shear along the wedge (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. 7. Deflection (δδ) at any point is given by δδ = QQQQ 4RR AA 4 [5.61] The variation of AA 4 is given in figure 5.11. Example 5 The plan of a mat foundation with column loads is shown in figure 5.1. Use equation (5) to calculate the soil pressures at points AA, BB, CC, DD, EE, FF, GG, HH, II, JJ, KK, LL, MM, and NN. The size of the mat is 76 ft 96 ft, all columns are 4 in. 4 in. in section, and qq all (net ) = 1.5 kip/ft. Verify that the soil pressures are less than the net allowable bearing capacity.

Figure 5.1 Plan of a mat foundation Solution From figure 5.1, Column dead load (DDDD) = 100 + 180 + 190 + 110 + 180 + 360 + 400 + 00 + 190 + 400 + 440 + 00 + 10 + 180 + 180 + 10 = 3550 kip Column live load (LLLL) = 60 + 10 + 10 + 70 + 10 + 00 + 50 + 10 + 130 + 40 + 300 + 10 + 70 + 10 + 10 + 70 + 30 kip So Service load = 3550 + 30 = 5780 kip

According to ACI 318-95, factored load, UU = (1.4)(Dead load) + (1.7)(Live load). So Factored load = (1.4)(3550) + (1.7)(30) = 8761 kip The moments of inertia of the foundation are II xx = 1 1 (76)(96)3 = 5603 10 3 ft 4 II yy = 1 1 (96)(76)3 = 351 10 3 ft 4 And MM yy = 0 So 5780xx = (4)(300 + 560 + 640 + 300) + (48)(310 + 650 + 740 + 300) + (7)(180 + 30 + 30 + 190) xx = 36.664 ft And ee xx = 36.664 36.0 = 0.664 ft Similarly, MM xx = 0 So 5780yy = (30)(30 + 640 + 740 + 30) + (60)(300 + 560 + 650 + 30) + (90)(160 + 300 + 310 + 180) yy = 44.73 ft And ee yy = 44.73 90 = 0.77 ft The moments caused by eccentricity are MM xx = QQQQ yy = (8761)(0.77) = 6369 kip ft MM yy = QQQQ xx = (8761)(0.664) = 5817 kip ft

From equation (5) qq = QQ AA ± MM yy xx II yy ± MM xx yy II xx = 8761 (5817 )(xx) ± ± (6369)(yy) (76)(96) 351 10 3 5603 10 3 Or qq = 1.0 ± 0.0017xx ± 0.0011yy (kip/ft ) Now the following table can be prepared. Point QQ AA (kip/ ft ) XX (ft) ±0.0017XX (ft) yy (ft) ±0.0011yy (ft) qq(kip/ ft ) A 1. -38-0.065 48-0.053 1.08 B 1. -4-0.041 48-0.053 1.106 C 1. -1-0.00 48-0.053 1.17 D 1. 0 0.0 48-0.053 1.147 E 1. 1 0.00 48-0.053 1.167 F 1. 4 0.041 48-0.053 1.188 G 1. 38 0.065 48-0.053 1.1 H 1. 38 0.065-48 0.053 1.318 I 1. 4 0.041-48 0.053 1.94 J 1. 1 0.00-48 0.053 1.73 K 1. 0 0.0-48 0.053 1.53 L 1. -1-0.00-48 0.053 1.33 M 1. -4-0.041-48 0.053 1.1 N 1. -38-0.065-48 0.053 1.188 The soil pressures at all points are less than the given value of qq all (net ) = 1.5 kip/ft.

Example 6 Use the results of example 5 and the conventional rigid method. a. Determine the thickness of the slab. b. Divide the mat into four strips (that is, AAAAAAAA, BBBBBBBBBBBB, DDDDDDDDDDDD, and FFFFFFFF) and determine the average soil reaction at the ends of each strips. c. Determine the reinforcement requirements in the y direction for ff cc = 3000 lb/in and ff yy = 60,000 lb/in. Solution Part a: Determination of Mat Thickness For the critical perimeter column as shown in figure 5.13 *(ACI 318-95), Figure 5.13 Critical perimeter column UU = 1.4(DDDD) + 1.7(LLLL) = (1.4)(190) + (1.7)(130) = 487 kip bb oo = (36 + dd/) + (4 + dd) = 96 + dd(in). From ACI 318-95 φφvv cc VV uu VV cc = nominal shear strength of concrete

VV uu = factored shear strength φφvv cc = φφ(4) ff cc bb oodd = (0.85)(4)( 3000)(96 + dd)dd So (0.85)(4)( 3000 )(96+dd)dd 1000 487 (96 + dd)dd 615.1 dd 19.4 in. For the critical internal column shown in figure 5.14, Figure 5.14 Critical internal column bb oo = 4(4 + dd) = 96 + 4dd(in. ) UU = (1.4)(440) + (1.7)(300) = 116 kip And (0.85)(4)( 3000 )(96+4dd)dd 1000 116 (96 + 4dd)dd 6046.4 dd 8.7 in.

Use dd = 9 in. With a minimum cover of 3 in. over the steel reinforcement and 1-in. diameter steel bars, the total slab thickness is h = 9 + 3 + 1 = 33 in. Part b: Average Soil Reaction Refer to figure 5.1. For strip ABMN (width = 14 ft) qq 1 = qq (at AA)+qq (at BB) qq = qq (at MM )+qq (at NN ) = 1.08+1.106 = 1.1+1.188 For strip BCDKLM (width = 4 ft) qq 1 = 1.106+1.17+1.147 3 qq = 1.53+1.33+1.1 3 = 1.17 kip/ft = 1.33 kip/ft For strip DEFIJK (width = 4 ft) qq 1 = 1.147+1.167+1.188 3 qq = 1.94+1.73+1.53 3 = 1.167 kip/ft = 1.73 kip/ft For strip FGHI (width = 14 ft) qq 1 = 1.188+1.1 = 1.0 kip/ft = 1.094 kip/ft = 1.0 kip/ft qq = 1.318+1.94 = 1.306 kip/ft Check for Σ FF vv = 0: Soil reaction for strip AAAAAAAA = 1 (1.094 + 1.0)(14)(96) = 1541.6 kip Soil reaction for strip BBBBBBBBBBBB = 1 (1.17 + 1.33)(4)(96) = 718.7 kip Soil reaction for strip DDDDDDDDDDDD = 1 (1.167 + 1.73)(4)(96) = 810.9 kip Soil reaction for strip FFFFFFFF = 1 (1.0 + 1.306)(14)(96) = 1684.0 kip 8755. kip Column load = 8761 kip OK

Part c: Reinforcement Requirements Refer to figure 5.15 for the design of strip BCDKLM. Figure 5.15 shows the load diagram, in which QQ 1 = (1.4)(180) + (1.7)(10) = 456 kip QQ = (1.4)(360) + (1.7)(00) = 844 kip QQ 3 = (1.4)(400) + (1.7)(40) = 968 kip QQ 4 = (1.4)(180) + (1.7)(10) = 456 kip The shear and moment diagrams are shown in figure 5.15b and c, respectively. From figure 5.15c, the maximum positive moment at the bottom of the foundation = 81.1/4 = 95.05 kip ft/ft.

Figure 5.15 Figure 5.16 Rectangular section in bending; (a) section, (b) assumed stress distribution across the section For the design concepts of a rectangular section in bending refer to figure 5.16. Compressive force, CC = 0.85ff cc aaaa Tensile force, TT = AA ss ff yy CC = TT Note that for this case bb = 1 ft = 1 in. (0.85)(3)(1)aa = AA ss (60) AA ss = 0.51aa From equation (36), MM uu = φφaa ss ff yy dd aa (95.05)(1) = (0.9)(0.51aa)(60) 9 aa aa = 1.47 in.

Thus AA ss = (0.51)(1.47) = 0.75 in Minimum reinforcement, ss min (AAAAAA 318 95) = 00/ff yy = 00/60,000 = 0.00333 Minimum AA ss = (0.00333)(1)(9) = 1.16 in /ft. Hence use minimum reinforcement with AA ss = 1.16 in /ft. Use no. 9 bars at 10 in. center-to-center (AA ss = 11. iiii /ffff) at the bottom of the foundation. From figure 5.15c, the maximum negative moment = 447.8 kip ft/4 = 10 kip ft/ft. by observation, AA ss AA ss(min ). Use no. 9 bars at 10 in. center-to-center at the top of the foundation. Example 7 From the plate load test (plate dimension1 ft 1 ft) in the field, the coefficient of subgrade reaction of a sandy soil was determined to be 80 lb/in 3. (a) What will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 30 ft 30 ft? (b) if the full-sized foundation has dimension of 45 ft 30 ft, what will be the value of the coefficient of subgrade reaction? Solution Part a From equation (46), kk = kk 1 BB+1 BB kk 1 = 80 lb/in BB = 30 ft So kk = 80 30+1 ()(30) = 1.36 in 3

Part b From equation (49), kk = kk (BB BB) 1+0.5 BB LL 1.5 kk (30 ft 30 ft) = 1.36 lb/in 3 So kk = (1.36)(1+0.530 45 = 19 lb/in 3 1.5 PROBLEMS 1. Determine the net ultimate bearing capacity of mat foundation with the following characteristics: a. cc uu = 10 kn/m, φφ = 0, BB = 8 m, LL = 18 m, DD ff = 3 m b. cc uu = 500 lb/ft, φφ = 0, BB = 0 ft, LL = 30 ft, DD ff = 6. ft. Following are the results of a standard penetration test in the field (sandy soil): Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0 10.5 Field value of NN FF 9 1 11 7 13 11 13 Estimate the net allowable bearing capacity of a mat foundation 6.5 m 5 m in plan. Here, DD ff = 1.5 m, and allowable settlement mm. assume that the unit weight of soil γγ = 16.5 kn/m 3. 3. A mat foundation on a saturated clay soil has dimensions of 0 m 0 m. Given dead and live load = 48 MN, cc uu = 30 kn/m, γγ clay = 18.5 kn/m 3.

a. Find the depth, DD ff of the mat for a fully compensated foundation. b. What will be the depth of the mat (DD ff ) for a factor of safety of against bearing capacity failure? 4. Repeat problem 4 part b for cc uu = 0 kn/m. 5. A mat foundation is shown in figure P-1. The design considerations are LL = 1 m, BB = 10 m, DD ff =. m, QQ = 30 MN, xx 1 = m, xx = m, xx 3 = 5. m, and preconsolidation pressure pp cc = 105 kn/m. Calculate the consolidation settlement under the center of the mat. Figure P-1 6. Refer to figure P-. For the mat, QQ 1, QQ 3 = 40 tons, QQ 4, QQ 5, QQ 6 = 60 tons, QQ, QQ 9 = 45 tons, and QQ 7, QQ 8 = 50 tons. all columns are 0 in. 0 in. in cross section. Use the procedure outlined in section 7 to determine the pressure on the soil at A, B, C, D, E, F, G, and H.

Figure P- 7. The plan of a mat foundation with column loads is shown in figure P-3. Calculate the soil pressure at points A, B, C, D, E, and F. note: all columns are 0.5 m 0.5 m in plan.

Figure P-3