(Lecture 18) MAT FOUNDATIONS

Transcription

1 Module 5 (Lecture 18) MAT FOUNDATIONS Topics 1.1 FIELD SETTLEMENT OBSERVATIONS FOR MAT FOUNDATIONS 1.2 COMPENSATED FOUNDATIONS 1.3 Example 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 equations (12 and 13). Meyerhof (1965) compiled the observed maximum settlements for mat foundations constructed on sand and gravel, as listed in table 1. In equation (13) if the depth factor, (DD ff /BB), is assumed to be approximately 1, SS ee = qq all (net ) 0.25NN cor [5.19] Table 2 shows a comparison of the observed maximum settlements in table 1 and the settlements obtained from equation (19). For the cases considered the ratio of SS eecalculated /SS eeobserved varies from 0.84 to 3.6. Thus calculation of the net allowable bearing capacity with equation (12 or 13) will yield a safe and conservative value. Stuart and Graham (1975) reported the case history of the 13-story Ashby Institute building of Queens University, Belfast, Ireland, construction of which began in August It was supported by at foundation 180 ft (length) 65 ft (width). Figure 5.5a shows a schematic diagram of the building cross section. The nature of the subsoil along with the field standard penetration resistance values at the south end of the building are shown in figure 5.5b. The base of the mat was constructed about 20 ft below the ground surface.

2 Figure 5.5 Ashby Institute Building of Queens University, as reported by Stuart and Graham (1975); (a) building cross section; (b) subsoil conditions at south end The variation of the corrected standard penetration number with depth is shown in table 3. Note that the average NN cor value between the bottom of the mat and a depth of 30 ft ( BB/2) is about 17. The engineers estimated the average net dead and live load [equation (16)] at the level of the mat foundation to be about 3360 lb/ft 2. From equation (13) SS ee = qq all (net ) 0.25NN cor DD ff BB [5.20] Substituting appropriate values into equation (20) yield the settlement at the south end of the building: SS ee = (3360 /1000 ) (0.25)(17)[1+0.33(20/65)] = 0.72 in. The construction of the building was completed in February Figure 5.6 shows the variation of the mean settlement of the mat at the south end. In 1972 (eight years after completion of the building) the mean settlement was about 0.55 in. Thus the estimated settlement of 0.72 in. is about 30% higher than that actually observed.

3 Table 1 Observed Maximum Settlement of Mat Foundations on Sand and Gravel Case no. Structure Reference BB(ft) NN cor (aaaaaa) qq all (net ) (kip/ ft 2 ) Observed maximum settlement, SS ee (in. ) 1 T. Edison 2 Banco do Brasil 3 Iparanga 4 C. B. I. Esplanada 5 Riscala 6 Thyssen Dusseldorf, Germany 7 Ministry Dusseldorf, Germany 8 Chimney Cologne, Germany Rios and Silva (1948) Rios and Silva (1948); Vargas (1961) Vargas (1948) Vargas (1961) Vargas (1948) Schultze (1962) Schultze (1962) Schultze (1962)

4 Table 2 Comparison of Settlements Observed and Calculated Case 1 Maximum observed settlement, SS ee (in. ) Calculated settlement, SS ee [equation (19)] SS eecalculated SS eeobserved Refer to table 1 Table 3 Determination of Corrected Standard Penetration Resistance Depth ground (ft) below surface Field standard penetration number, NN FF σσ vv aa (ton/ft 2 ) (bb) CC NN = 1 NN cccccc [equation 7 chapter 2)] σσ γγ σσ vv = (depth); γγ = 120 lb/ft 3 (assumed) Table 4 from chapter 2

5 Figure 5.6 Mean settlement at the south end of the mat foundation, as reported by Stuart and Graham (1975) COMPENSATED FOUNDATIONS The settlement of a mat foundation can be reduced by decreasing the net pressure increase on soil, which can be done by increasing the depth of embedment, DD ff. This increase is particularly important for mats on soft clays, where large consolidation settlements are expected. From equation (16), the net average applied pressure on soil is qq = QQ AA γγdd ff For no increase of the net soil pressure on soil below a raft foundation, qq should be zero. Thus DD ff = QQ AAAA [5.21] This relation for DD ff is usually referred to as the depth of a fully compensated foundation. The factor of safety against bearing capacity failure for partially compensated foundations (that is, DD ff < QQ/AAAA) may be given as FFFF = qq net (uu) qq = qq net (uu ) QQ AA γγdd ff [5.22] For saturated clays, the factor of safety against bearing capacity failure can thus be obtained by substituting equation (10) into equation (22): FFFF = BB LL QQ AA γγdd ff DD ff BB [5.23]

6 Example 3 Refer to figure 5.4. The mat has dimensions of 60 ft 100 ft. The total dead and live load on the mat is kip. The mat is placed over a saturated clay having a unit weight of 120 lb/ft 3 and cc uu = 2800 lb/ft 2. Given DD ff = 5 ft, determine the factor of safety against bearing capacity failure. Solution From equation (23), the factor of safety FFFF = 5.14cc uu BB LL QQ AA γγdd ff DD ff BB Given: cc uu = 2800 lb/ft 2, DD ff = 5 ft, BB = 60 ft, LL = 100 ft, and γγ = 120 lb/ft 3. Hence FFFF = (5.14)(2800 ) 1+(0.195 )(60) = lb (120)(5) Example 4 Consider a mat foundation 90 ft 120 ft in plan, as shown in figure 5.7. The total dead load and live load on the raft is kip. Estimate the consolidation settlement at the center of the foundation. Figure 5.7

7 Solution From equation (64 from chapter 1) SS cc = CC cchh cc log pp oo + pp aaaa 1+ee oo pp oo pp oo = (11)(100) + (40)( ) ( ) 3964lb/ft2 HH cc = in. CC cc = 0.28 ee oo = 0.9 For QQ = lb, the net load per unit area qq = QQ γγdd AA ff = (100)(6) 3567lb/ft In order to calculate pp aaaa, we refer to section 5. The loaded area can be divided into four areas, each measuring45 ft 60 ft. Now using equation (19 from chapter 4), we can calculate the average stress increase in the clay layer below the corner of each rectangular area, or pp aaaa(hh2 /HH 1 ) = qq HH 2II aa (HH 2 ) HH 1 II aa (HH 1 ) HH 2 HH 1 = 3567 ( )II aa (HH2 ) (5+40)II aa (HH 1 ) 18 For II aa(hh2 ), mm = BB HH 2 = = 0.71 nn = LL HH 2 = = 0.95 From figure 5.8, for mm = 0.71 and nn = 0.95, the value of II aa(hh2 ) is Again, for II aa(hh1 ),

8 Figure 5.8 Conventional rigid mat foundation design Fig 5.9 (Continued)

9 mm = BB HH 1 = = 1 nn = LL HH 1 = = 1.33 From figure 5.8, II aa(hh1 ) = 0.225, so pp aaaa(hh2 /HH 1 ) = 3567 (63)(0.21) (45)(0.225) = lb/ft 2 18 So, the stress increase below the center of the 90 ft 120 ft area is (4)(615.3) = lb/ft 2. Thus SS cc = (0.28)(18 12) log = 6.68 in.