Download presentation

1
**Course : S0705 – Soil Mechanic**

Year : 2008 TOPIC 3 SOIL STRESS

2
**CONTENT TOTAL STRESS EFFECTIVE STRESS STRESS DISTRIBUTION**

Bina Nusantara

3
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 Bina Nusantara

4
**·A ·B ·C ·D EXAMPLE A = t,1 x 1 m t,1 = 17 kN/m3 = 17 kN/m2**

d,1 = 13 kN/m3 3 m 4 m B = t,1 x 3 m = 51 kN/m2 t,2 = 18 kN/m3 d,2 = 14 kN/m3 C = t,1 x 3 m + t,2 x 4 m = 123 kN/m2 2 m t,3 = 18 kN/m3 d,3 = 15 kN/m3 D = t,1 x 3 m + t,2 x 4 m + t,3 x 2 m = 159 kN/m2 Bina Nusantara

5
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 Bina Nusantara

6
EFFECTIVE STRESS Bina Nusantara

7
**EXAMPLE Sand h1 = 2 m t = 18.0 kN/m3 d = 13.1 kN/m3 h2 = 2.5 m**

MAT Clay t = kN/m3 x Bina Nusantara

8
**EXAMPLE Total Stress = d,1 . h1 + t,1 . h2 + t,2 . h3**

= = kN/m2 Pore Water Pressure u = w . (h2+h3) u = = 70 kN/m2 Effective Stress ’ = - u = 90.3 kN/m2 ’ = d,1 . h1 + (t,2 - w) . h2 + (t,2 - w) . h3 ’ = (18-10) (19,8-10) . 4.5 = 90.3 kN/m2 Bina Nusantara

9
**Profile of Vertical Stress**

EXAMPLE Total Stress () Pore Water Pressure (u) Effective Stress (’) 26.2 kPa 26.2 kPa -2.0 71.2 kPa 25 kPa 46.2 kPa -4.5 160.3 kPa 70 kPa 90.3 kPa -9.0 Bina Nusantara Profile of Vertical Stress

10
**SOIL STRESS CAUSED BY EXTERNAL LOAD**

External Load Types Point Load Line Load Uniform Load Bina Nusantara

11
**LOAD DISTRIBUTION PATTERN**

Bina Nusantara

12
STRESS CONTOUR Bina Nusantara

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

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

15
BOUSSINESQ METHOD Point Load z P r z Bina Nusantara

16
BOUSSINESQ METHOD ] Bina Nusantara

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

18
**BOUSSINESQ METHOD Uniform Load Square/Rectangular Circular Trapezoidal**

Triangle Bina Nusantara

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

20
BOUSSINESQ METHOD Rectangular Bina Nusantara

21
**BOUSSINESQ METHOD Circular At the center of circle (X = 0)**

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

22
BOUSSINESQ METHOD Circular Bina Nusantara

23
BOUSSINESQ METHOD Trapezoidal Bina Nusantara

24
BOUSSINESQ METHOD Triangle Bina Nusantara

25
**EXAMPLE The 5 m x 10 m area uniformly loaded with 100 kPa Y 5 m**

Question : Find the at a depth of 5 m under point Y 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 m m Bina Nusantara

26
**EXAMPLE Question 1 z total = 23.8 – 20.9 – 20.6 + 18 = 0.3 kPa Item**

Area YABC -YAFD -YEGC YEHD x 15 10 5 y z m = x/z 3 2 1 n = y/z I 0.238 0.209 0.206 0.18 z 23.8 - 20.9 -20.6 18.0 z total = 23.8 – 20.9 – = 0.3 kPa Bina Nusantara

27
**EXAMPLE Question 2 z total = 47.6 – 41.9 – 43.8 + 38.6 = 0.5 kPa Item**

Area YABC -YAFD -YEGC YEHD x 15 10 5 y z m = x/z 3 2 1 n = y/z I 0.238 0.209 0.206 0.18 z 47.6 - 41.9 -43.8 38.6 z total = 47.6 – 41.9 – = 0.5 kPa Bina Nusantara

28
**NEWMARK METHOD Where : qo = Uniform Load I = Influence factor**

N = No. of blocks Bina Nusantara

29
**NEWMARK METHOD Diagram Drawing**

Take z/qo between 0 and 1, with increment 0.1 or other, then find r/z value 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 Bina Nusantara

30
**NEWMARK METHOD Example, the depth of point (z) = 6 m**

z/qo r/z Radius (z=6 m) Radius at drawing Operation 0.1 0.27 1.62 m 0.675 cm 1.62/6 x 2.5 cm 0.2 0.40 2.40 m 1 cm 2.4/6 x 2.5 cm 0.3 0.52 3.12 m 1.3 cm 3.12/6 x 2.5 cm 0.4 0.64 3.84 m 1.6 cm 3.84/6 x 2.5 cm And so on, generally up to z/qo 1 because if z/qo = 1 we get r/z = Bina Nusantara

31
NEWMARK METHOD Bina Nusantara

32
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’ Bina Nusantara

33
**EXAMPLE v = 250 . 0.02 . 8 = 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 = qo . I . N v = = 40 kPa Bina Nusantara

34
WESTERGAARD METHOD Point Load = 0 Bina Nusantara

35
WESTERGAARD METHOD ] Bina Nusantara

36
WESTERGAARD METHOD Circular Uniform Load Bina Nusantara

37
WESTERGAARD METHOD Bina Nusantara

38
**BOUSSINESQ VS WESTERGAARD**

Bina Nusantara

39
**BOUSSINESQ VS WESTERGAARD**

Bina Nusantara

40
**BOUSSINESQ VS WESTERGAARD**

Bina Nusantara

Similar presentations

OK

TOPIC 3: DESIGN AND ANALYSIS OF SHALLOW FOUNDATION WEEK 6

TOPIC 3: DESIGN AND ANALYSIS OF SHALLOW FOUNDATION WEEK 6

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Procedure text how to make ppt online Ppt on barack obama leadership ability Ppt on object-oriented concepts and principles Ppt on management information systems Signal generator and display ppt online Ppt on game theory prisoner's dilemma Ppt on information security using steganography Ppt on education in india during british rule Retinal anatomy and physiology ppt on cells Ppt on working of human eye and defects of vision and their correction