Chapter 7: Settlement of Shallow Foundations

Transcription

1 Chapter 7: Settlement of Shallow Foundations Introduction The settlement of a shallow foundation can be divided into two major categories: (a) elastic, or immediate settlement and (b) consolidation settlement. Immediate, or elastic settlement Immediate, or elastic settlement of a foundation takes place during or immediately after the construction of the structure. Consolidation settlement Consolidation settlement occurs over time. Pore water is extruded from the void spaces of saturated clayey soils submerged in water. The total settlement of a foundation is the sum of the elastic settlement and the consolidation settlement. Consolidation settlement comprises two phases: primary and secondary. The fundamentals of primary consolidation settlement were explained in detail in Chapter 2. Secondary consolidation settlement occurs after the completion of primary consolidation caused by slippage and reorientation of soil particles under a sustained load. Primary consolidation settlement is more significant than secondary settlement in inorganic clays and silty soils. However, in organic soils, secondary consolidation settlement is more significant. Elastic Settlement of Shallow Foundation on Saturated Clay (μ s = 0.5) The average settlement of flexible foundations on saturated clay soils (Poisson s ratio, μ s = 0.5). Engr. Yasser M. Almadhoun Page 1

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3 Example 7.1 See example 7.1 in textbook, page 340. Elastic Settlement in Granular Soil Settlement Based on the Theory of Elasticity Engr. Yasser M. Almadhoun Page 3

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5 To calculate settlement at the centre of the foundation, use: Engr. Yasser M. Almadhoun Page 5

6 To calculate settlement at a corner of the foundation, use: The elastic settlement of a rigid foundation can be estimated as: Due to the nonhomogeneous nature of soil deposits, the magnitude of Es may vary with depth. For that reason, Bowles (1987) recommended using a weighted average of Es: Example 7.2 See example 7.1 in textbook, page 309. Engr. Yasser M. Almadhoun Page 6

7 Settlement of Sandy Soil: Use of Strain Influence Factor The recommended variation of the strain influence factor Iz for square (L/B =1) or circular foundations and for foundations with L/B = 10. Engr. Yasser M. Almadhoun Page 7

8 Note that the maximum value of Iz [ that is, Iz(m) ] occurs at z = z 1 and then reduces to zero at z = z 2. The maximum value of Iz can be calculated as: where qc = cone penetration resistance Engr. Yasser M. Almadhoun Page 8

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10 Substitute in the equation of Se in order to get the elastic settlement of sandy soil. Settlement of Foundation on Sand Based on Standard Penetration Resistance (Meyerhof Method) Meyerhof proposed a correlation for the net bearing pressure for foundations with the standard penetration resistance, N 60. The net pressure has been defined as: According to Meyerhof s theory, for 25 mm (1 in.) of estimated maximum settlement: In English units: Engr. Yasser M. Almadhoun Page 10

11 In SI units: where Se = settlement, in mm. N 60 = the standard penetration resistance between the bottom of the foundation and 2B below the bottom. Later, Meyerhof (1965) suggested that the net allowable bearing pressure should be increased by about 50%. Bowles (1977) proposed that the modified form of the bearing equations be expressed as: In SI units: where B is in meters and Se is in mm. Engr. Yasser M. Almadhoun Page 11

12 Example 7.7 See example 7.7 in textbook, page 328. Effect of the Rise of Water Table on Elastic Settlement Terzaghi suggested that the submergence of soil mass reduces the soil stiffness by about half, which in turn doubles the settlement. In most cases of foundation design, it is considered that, if the ground water table is located 1.5B to 2B below the bottom of the foundation, it will not have any effect on the settlement. The total elastic settlement (S e ) due to the rise of the ground water table can be given as: Engr. Yasser M. Almadhoun Page 12

13 The following are some empirical relationships for Cw: Example 7.9 See example 7.9 in textbook, page 335. Consolidation Settlement Primary Consolidation Settlement Consolidation settlement occurs over time in saturated clayey soils subjected to an increased load caused by construction of the foundation. Engr. Yasser M. Almadhoun Page 13

14 On the basis of the one-dimensional consolidation settlement equations are: σ o + σ av > σ c Engr. Yasser M. Almadhoun Page 14

15 Correlations for Cc and Cs C c = σ 1 = σ o e log (σ 2 /σ 1 ) σ 2 = σ o + σ av = σ o ( σ t + 4 σ m + σ b ) C c = 0.009(LL 10) C s = ( ) C c Secondary Consolidation Settlement At the end of primary consolidation (i.e., after the complete dissipation of excess pore water pressure) some settlement is observed that is due to the plastic adjustment of soil fabrics. This stage of consolidation is called secondary consolidation. A plot of deformation against the logarithm of time during secondary consolidation is practically linear as shown in the figure below. The magnitude of the secondary consolidation can be calculated as: Engr. Yasser M. Almadhoun Page 15

16 where C α = Cα / (1+ e p ) (varies between to 0.001) e p = void ratio at the end of primary consolidation H c = thickness of clay layer From the figure, the secondary compression index can be defined as: where Cα = secondary compression index e = change of void ratio t 1, t 2 = time Secondary consolidation settlement is more important in the case of all organic and highly compressible inorganic soils. In overconsolidated inorganic clays, the secondary compression index is very small and of less practical significance. Example 7.11 See example 7.11 in textbook, page 343. Field Load Test (Plate Load Test) The ultimate load-bearing capacity of a foundation, as well as the allowable bearing capacity based on tolerable settlement considerations, can be effectively determined from the field load test, generally referred to as the plate load test. The plates that are used for tests in the field are usually made of steel and are 25 mm (1 in.) thick and occasionally, square plates that are 305 mm X 305 mm (12 in. X 12 in.) are used. Engr. Yasser M. Almadhoun Page 16

17 To conduct a plate load test, a hole is excavated with a minimum diameter of 4B (B is the diameter of the test plate) to a depth of Df, the depth of the proposed foundation. The plate is placed at the centre of the hole, and a load that is about 1/4 to 1/5 of the estimated ultimate load is applied to the plate in steps by means of a jack. During each step of the application of the load, the settlement of the plate is observed on dial gauges. At least one hour is allowed to elapse between each application. The test should be conducted until failure, or at least until the plate has gone through 25 mm (1 in.) of settlement. For tests in clay: where q u(fd) = ultimate bearing capacity of the proposed foundation q u(pd) = ultimate bearing capacity of the test plate For tests in sandy soils: where B F = width of the foundation B P = width of the test plate Engr. Yasser M. Almadhoun Page 17

18 The allowable bearing capacity of a foundation, based on settlement considerations and for a given intensity of load, qo, is: Problems Problem (1) Find the size of the square footing, using Bowles theory, that carry allowable load of 1000 KN, given that: N 60 = 10 Se (all) = 25 mm Solution: The load is relatively large, so assume that B > 1.22 m q net(all) = N q net(all) = N B 2 = (B B ) F d ( S e 25 ) (B ) ( B B ) (S e 25 ) (B B 1.5 ) ( ) (25 B 25 ) Solving the previous equation for B: B = 2.3 m Engr. Yasser M. Almadhoun Page 18

19 Problem (2) Refer to the following figure, determine the average stress increase in the clay layer below the centre of the foundation due to the net foundation load of 50 ton. After that, determine the primary consolidation settlement for the clay layer. Solution: First of all, you have to determine the average stress in the clay layer below the centre of the foundation, and then you can calculate the settlement using the appropriate equation: σ av = 4q o [ H 2I a(h2 ) H 1 I a(h1 ) H 2 H 1 ] q o = Q A = = 2 ton/ft2 = psf m 2 = B H 1 = B/2 H 1 = 5/2 3 = 0.83 n 2 = L H 1 = L/2 H 1 = 5/2 3 = 0.83 I a(h1 ) = 0.21 (Figure 6.11, page 280 in textbook) Engr. Yasser M. Almadhoun Page 19

20 m 2 = B H 2 = B/2 H 2 = 5/2 11 = 0.23 n 2 = L H 2 = L/2 H 2 = 5/2 11 = 0.23 I a(h1 ) = 0.11 (Figure 6.11, page 280 in textbook) (11)(0.21) (3)(0.11) σ av = 4(4409.2) [ ] = psf 11 3 σ c = 2000 psf σ o = ( ) 3 + 4( ) = psf σ o + σ av = psf Note that: σ o < σ c < σ o + σ av So that: S c = C sh c 1 + e o log ( σ c σ o ) + C ch c 1 + e o log ( σ o + σ av σ c ) S c = log ( S c = ft = 1.65 in ) log ( ) Engr. Yasser M. Almadhoun Page 20

21 Problem (3) Refer to the following figure, the net load per unit area at the level foundation is 3200 Ib/ft 2. Assume that the foundation is rigid, determine the elastic settlement that the foundation will undergo based on the theory of elasticity. Solution: Solution: m = L B = = 1.54 n = H B = H B/2 = /2 = 9.85 F 1 = (Table 7.2, page 305 in textbook) F 2 = (Table 7.3, page 307 in textbook) I s = F μ s 1 μ s I s = (0.3) 1 (0.3) = I f = f ( D f, μ B s, L ) = (Table 7.4, page 309 in textbook) B S e(flexible,center) = q ob α E s S e(flexible,center) = q ob α E s (1 μ s 2 )I s I f (1 μ s 2 )I s I f Engr. Yasser M. Almadhoun Page 21

22 S e(flexible,center) = ( ) (6.5 2 ) (4) (1 0.3 (3200) 2 )(0.016)(0.828) = in S e(rigid) = 0.93 S e(flexible,center) S e(rigid) = 0.93 (0.4914) = in Problem (4) For the following figure, determine the settlement of the foundation shown. Solution: Solution: H B = = 2 L B = = 2 A 1 = 0.66 (Figure 7.1, page 300 in textbook) D f B = = 0.8 A 2 = 0.93 (Figure 7.1, page 300 in textbook) S e = A 1 A 2 q o B E s S e = (0.66)(0.93) (150)(1.5) (600) = cm Engr. Yasser M. Almadhoun Page 22