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TOPIC 3 SOIL STRESS Course: S0705 – Soil Mechanic Year: 2008.

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Presentation on theme: "TOPIC 3 SOIL STRESS Course: S0705 – Soil Mechanic Year: 2008."— Presentation transcript:

1 TOPIC 3 SOIL STRESS Course: S0705 – Soil Mechanic Year: 2008

2 Bina Nusantara CONTENT TOTAL STRESS EFFECTIVE STRESS STRESS DISTRIBUTION

3 Bina Nusantara TOTAL NORMAL STRESS Generated by the mass in the soil body, calculated by sum up the unit weight of all the material (soil solids + water) multiflied by soil thickness or depth. Denoted as ,  v, Po The unit weight of soil is in natural condition and the water influence is ignored. z = The depth of point

4 Bina Nusantara EXAMPLE 3 m 4 m  t,1 = 17 kN/m 3  d,1 = 13 kN/m 3  t,2 = 18 kN/m 3  d,2 = 14 kN/m 3  t,3 = 18 kN/m 3  d,3 = 15 kN/m 3 ·A·B·C·D·A·B·C·D 1 m 2 m  A =  t,1 x 1 m = 17 kN/m 2  B =  t,1 x 3 m = 51 kN/m 2  C =  t,1 x 3 m +  t,2 x 4 m = 123 kN/m 2  D =  t,1 x 3 m +  t,2 x 4 m +  t,3 x 2 m = 159 kN/m 2

5 Bina Nusantara EFFECTIVE STRESS Defined as soil stress which influenced by water pressure in soil body. Published first time by Terzaghi at 1923 base on the experimental result Applied to saturated soil and has a relationship with two type of stress i.e.: –Total Normal Stress (  ) –Pore Water Pressure (u) Effective stress formula

6 Bina Nusantara EFFECTIVE STRESS

7 Bina Nusantara EXAMPLE MAT Sand  t = 18.0 kN/m 3  d = 13.1 kN/m 3 Clay  t = kN/m 3 h 1 = 2 m h 2 = 2.5 m h 3 = 4.5 m x

8 Bina Nusantara EXAMPLE Total Stress  =  d,1. h 1 +  t,1. h 2 +  t,2. h 3  = = kN/m 2 Pore Water Pressure u =  w. (h 2 +h 3 ) u = = 70 kN/m 2 Effective Stress  ’ =  - u = 90.3 kN/m 2  ’ =  d,1. h 1 + (  t,2 -  w ). h 2 + (  t,2 -  w ). h 3  ’ = (18-10) (19,8-10). 4.5 = 90.3 kN/m 2

9 Bina Nusantara EXAMPLE kPa 71.2 kPa kPa 25 kPa 70 kPa 26.2 kPa 46.2 kPa 90.3 kPa Total Stress (  ) Pore Water Pressure (u) Effective Stress (  ’) Profile of Vertical Stress

10 Bina Nusantara SOIL STRESS CAUSED BY EXTERNAL LOAD External Load Types –Point Load –Line Load –Uniform Load

11 Bina Nusantara LOAD DISTRIBUTION PATTERN

12 Bina Nusantara STRESS CONTOUR

13 Bina Nusantara STRESS DISTRIBUTION Point Load P z 2 1 zz

14 Bina Nusantara STRESS DISTRIBUTION Uniform Load z B L B+z L+z

15 Bina Nusantara BOUSSINESQ METHOD Point Load z P r zz

16 Bina Nusantara BOUSSINESQ METHOD ]

17 Bina Nusantara BOUSSINESQ METHOD Line Load z q r zz x

18 Bina Nusantara Uniform Load –Square/Rectangular –Circular –Trapezoidal –Triangle BOUSSINESQ METHOD

19 Bina Nusantara BOUSSINESQ METHOD Rectangular z x y qoqo m = x/z n = y/z

20 Bina Nusantara BOUSSINESQ METHOD Rectangular

21 Bina Nusantara BOUSSINESQ METHOD Circular z x zz 2r At the center of circle (X = 0) For other positions (X  0), Use chart for finding the influence factor

22 Bina Nusantara BOUSSINESQ METHOD Circular

23 Bina Nusantara BOUSSINESQ METHOD Trapezoidal

24 Bina Nusantara BOUSSINESQ METHOD Triangle

25 Bina Nusantara EXAMPLE The 5 m x 10 m area uniformly loaded with 100 kPa Question : 1.Find the at a depth of 5 m under point Y 2.Repeat question no.1 if the right half of the 5 x 10 m area were loaded with an additional 100 kPa Y A B C D E F G H I J 5 m 5 m 5 m 5 m

26 Bina Nusantara EXAMPLE ItemArea Y ABC- Y AFD- Y EGC Y EHD x y 555 z5555 m = x/z3321 n = y/z2111 I zz  z total = 23.8 – 20.9 – = 0.3 kPa Question 1

27 Bina Nusantara EXAMPLE Question 2 ItemArea Y ABC- Y AFD- Y EGC Y EHD x y 555 z5555 m = x/z3321 n = y/z2111 I zz  z total = 47.6 – 41.9 – = 0.5 kPa

28 Bina Nusantara NEWMARK METHOD Where : q o = Uniform Load I = Influence factor N = No. of blocks

29 Bina Nusantara NEWMARK METHOD Diagram Drawing 1.Take  z /q o between 0 and 1, with increment 0.1 or other, then find r/z value 2.Determine the scale of depth and length Example : 2.5 cm for 6 m 3. Calculate the radius of each circle by r/z value multiplied with depth (z) 4. Draw the circles with radius at step 3 by considering the scale at step 2

30 Bina Nusantara NEWMARK METHOD Example, the depth of point (z) = 6 m  z /q o r/zRadius (z=6 m)Radius at drawingOperation m0.675 cm1.62/6 x 2.5 cm m1 cm2.4/6 x 2.5 cm m1.3 cm3.12/6 x 2.5 cm m1.6 cm3.84/6 x 2.5 cm And so on, generally up to  z /q o  1 because if  z /q o = 1 we get r/z = 

31 Bina Nusantara NEWMARK METHOD

32 Bina Nusantara EXAMPLE A uniform load of 250 kPa is applied to the loaded area shown in next figure : Find the stress at a depth of 80 m below the ground surface due to the loaded area under point O’

33 Bina Nusantara EXAMPLE  v = = 40 kPa Solution : –Draw the loaded area such that the length of the line OQ is scaled to 80 m. –Place point O’, the point where the stress is required, over the center of the influence chart –The number of blocks are counted under the loaded area –The vertical stress at 80 m is then indicated by :  v = q o. I. N

34 Bina Nusantara WESTERGAARD METHOD Point Load = 0

35 Bina Nusantara WESTERGAARD METHOD ]

36 Bina Nusantara WESTERGAARD METHOD Circular Uniform Load

37 Bina Nusantara WESTERGAARD METHOD

38 Bina Nusantara BOUSSINESQ VS WESTERGAARD

39 Bina Nusantara BOUSSINESQ VS WESTERGAARD

40 Bina Nusantara BOUSSINESQ VS WESTERGAARD


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