Lecture on Foundation Settlements-immediate settlement

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Lecture on 21-3-15 CE-319 (3 Credit Hours) Geotechnical and Foundation Engineering Foundation Settlements-immediate

1 Lecture on CE-319 (3 Credit Hours) Geotechnical and Foundation Engineering Foundation Settlements-immediate settlement Instructor: Dr Irshad Ahmad Department of Civil Engineering University of Engineering and Technology, Peshawar 1

2 Contents Foundation Settlement- Introduction Some Typical Loads on Building Foundations Number and Depth of boring Immediate settlement calculations using Theory of Elasticity SPT Procedure SPT corrections SPT correlations Immediate settlement using SPT 2

3 Foundation Settlement 3

4 Foundation Settlement 4

5 Some Typical Loads on Building Foundations Settlements are cause by change in stress in the soil mass Change in stress occurs because Of building loads 5

6 Foundation Settlement 6

7 Number of borings Depth of borings 7

8 Depth of boring For Square footing: 10% qo at 2B For Strip footing 10% qo at 6.5B For circular footing 10% qo at 2 Dia 8

9 Immediate Settlement We will find immediate settlement through THEORY OF ELASTICITY USING STANDARD PENETRATION TEST RESULTS 9

10 Immediate Settlement- Theory Of Elasticity 10

11 Immediate Settlement 11

12 Shape and Rigidity factor I 12

13 Example Method-1: Taking weighted average of Es Method-2: Method of superposition Both methods are discussed in class 13

Example-Using Weighted Average of Es Es (average) = (42.5 3 + 60 8)/11=55 Mpa, H=11m, B =B/2=33.5/2=16.75, H/ B =11/16.75=0.66 (use 0.7) L /B = (39.

14 Example-Using Weighted Average of Es Es (average) = ( )/11=55 Mpa, H=11m, B =B/2=33.5/2=16.75, H/ B =11/16.75=0.66 (use 0.7) L /B = (39.5/2)/(33.5/2)= 1.18 Assume, =0.33, I=0.12 (after interpolation) H=qoB (1-2 )/Es m I = ( )/( ) = m = 17.5 mm 14

15 Example- Method of Superposition Methodology: Calculate Hi (1) Consider the 3m Sand layer with Es overlying sand stone. (2) Consider whole 11m layer as Clay with Es=60 MPa overlying Sand stone. (3) Consider upper 3m layer as Clay with Es=60 Mpa overlaying Sandstone. 15

16 Es = 42.5Mpa, H=3m,B =B/2=16.75, H/ B =3/16.75=0.18, L /B = 1.18 I= H=qoB (1-2 )/Es m I ( )/( = 5.96 mm Es = 60 Mpa, Es = 60 Mpa, H=11m,B =B/2=16.75, H=3m, B =B/2=16.75, H/B =3/16.75=0.18, L /B = 1.18 H/B =3/16.75=0.18, L /B = 1.18 I=0.12 I= H=qoB (1-2 )/Es m I H=qoB (1-2 )/Es m I ( )/( ( )/( = 16 mm = 4.2 mm H= = 17.7 mm 16

17 Standard Penetration Test (ASTM D-1586) Practicing Engineers use the SPT widely in estimating the bearing capacity of soils and to assess the in-situ relative density of a sand deposit. The test is performed using a split spoon barrel sampler 50mm external diameter, 35mm internal diameter and about 650mm in length and connected to the end of boring rods. After boring has been advanced to the desired sampling elevation and excessive cutting has been removed, attach the split spoon sampler to the sampling rods and lover into borehole. Drive the sample with blows from 140lb (63.5 kg) hammer falling freely through a height of 760mm (30 inch). The sampler is advanced under the impact of the hammer into three successive (6 inch) increments. (i.e. total 18 inch). 17

18 Standard Penetration Test (ASTM D-1586) Count the number of blows in each 6 inch increment until one of the following occurs A total of 50 blows have been applied during any of the three 6 inch increments. A total of 100 blows have been applied There is no observed advance of the sampler during the application of 10 successive blows of the hammer. The sampler is advanced the complete 18 inch without the limiting blow counts occurring as describe in 1,2 and 3. The 1 st 6 inch is considered to be a seating drive. The sum of the number of blows required for the 2 nd and 3 rd 6-inch penetration is termed the standard penetration resistance or the N-value. 18

19 Disturbance caused by Split spoon sampler 19

20 Standard Penetration Test (ASTM D-1586) 20

21 Thin wall tube sampler (Shelby tubes) 21

22 Hammers 22

23 Standard Penetration Test (ASTM D-1586) 23

24 Corrected N- value 24

25 N-value corrections 25

26 SPT Correlations 26

27 SPT Correlations 27

28 SPT Correlations 28

29 SPT Correlations Undrained shear strength cu(kn/m 2 ) = 29 (N 60 ) 0.72 [Hara, et al. 1971] 29

30 30

31 Immediate Settlement: SPT (Bowels P263) N60 SI Fps F F F F

32 Design N Values 32

33 Problem (Bowel P-266) Fill up the below table using equation for the widths given and prepare a graph 33

34 PROBLEM 34

35 PROBLEM 35

36 Example (Bowels P-543) Find allowable BC qa to ensure FOS=3 against Shear failure Si <= 50 m Solution: (1) Shear Criteria: qu = c Nc s c + D Nq + ½ B N s BC factors: For ⱷ = 0, Nc = 5.14, Nq = 1 and N = 0 Shape factors: sc=1.3, and s =0.8 qu (net)= qu(gross) - D = ( ) = 1000 kpa qsafe = qu/fos = 334 kn/m 2 36

37 Example (Bowels P-543) Es1= 1000 cu = 1000(300/2)= kpa Es2= 500(N )= 500[18(70/55)+15]=18950 kpa Es2= 500(N )= 500[22(70/55)+15]= kpa Es2= 500(N )= 500[40(70/55)+15]= kpa (2) Settlement Criteria Es(average)= 3.4(150000)+3.3(18950)+7(22000) +13.8(32900)]/27.5 = kpa H from base of the mat to rock = ( ) = 27.5m, Estimate the mat will be on the order of 14m, giving B =B/2=14/2=7, H/ B =27.5/7 4, L /B = 1 For =0.33, I=0.42 (after interpolation) H=qoB (1-2 )/Es m I 0.05 = qo 7 ( )/(42930) qo= 204 kpa Allowable Bearing Capacity: qa=200 kpa, settlement controls: As the design proceeds and B is found to be substantially different from 14m, it may be 37 necessary to revise qa.

Typical set up for Plate Load test assembly

Typical set up for Plate Load test assembly Major disadvantages of field tests are Laborious Time consuming Heavy equipment to be carried to field Short duration behavior Plate Load Test Sand Bags Platform for loading Dial Gauge Testing Plate Foundation