Stress At Soil Mass.pdf

Stress at Soil Mass YUSTIAN HERI SUPRAPTO

STRESS AT SOIL MASS (SOIL MECHANICS)

1

Layout Normal Stress and Shear Stress

STRESS AT SOIL MASS (SOIL MECHANICS)

2

Introduction In a given volume of soil, the solid particles are distributed randomly with void spaces in between. The spaces are occupied by water, air, or both.

To analyze the compressibility, bearing capacity, and other geotechnical problems, engineers need to know the nature of distribution of stress along the soil profile.

STRESS AT SOIL MASS (SOIL MECHANICS)

3

Normal and Shear Stresses on a Plane 2D element that is subjected to normal stress and shear stress. (y > x). From freebody diagram (EBF) :

EB = EF cos  ; and ; FB = EF sin Summing the components of forces that act on the element in the direction of N and T,

n(EF) = x(EF) sin2+ y(EF) cos2+ 2xy(EF) sincos Or (a). Soil element with normal shear stress acting on it (b). Freebody diagram of EFB from (a)

n =

 y + x 2

STRESS AT SOIL MASS (SOIL MECHANICS)

+

 y − x 2

cos 2 +  xy sin 2 4

Normal and Shear Stresses on a Plane n = y sin cos  - x sin cos  - xy (cos2-sin2)

n =

 y − x 2

sin 2 −  xy cos 2

Subsituting n = 0 : tan 2 =

2 xy

 y − x

Major principal stress (a). Soil element with normal shear stress acting on it (b). Freebody diagram of EFB from (a)

 n = 1 =

 y + x

STRESS AT SOIL MASS (SOIL MECHANICS)

2

 y −  x  2 +  +  xy   2  2

5

Normal and Shear Stresses on a Plane Minor principal stress : n = 3 =

 y + x 2

 y −  x  2 −   +  xy  2  2

(a). Soil element with normal shear stress acting on it (b). Freebody diagram of EFB from (a) STRESS AT SOIL MASS (SOIL MECHANICS)

6

Mohr’s Circle Mohr’s circle → compressive normal stresses are taken as positive and shear stress are considered positive if they act on opposite faces of the element. Points R and M → stress conditions Point O → point of intersection normal stress axis with the RM line.

the

The radius of the circle:  y −  x   

2

2

2 +  xy  

Principles of the Mohr’s Circle STRESS AT SOIL MASS (SOIL MECHANICS)

7

Mohr’s Circle The ordinates of points N and S are zero, the abscissa of point N is equal to 1 :  n = 1 =

 y + x 2

 y −  x  2 +  +  xy   2  2

And the abscissa for point S is 3 : n = 3 =

 y + x 2

 y −  x  2 −  +  xy   2  2

Principles of the Mohr’s Circle STRESS AT SOIL MASS (SOIL MECHANICS)

8

Example The magnitudes of stresses are x = 120 kN/m2 ;  = 40 kN/m2 ; y = 300 kN/m2, and  = 20o. Determine: (a). Magnitude of the principal stresses

(b). Normal and shear stresses on plane AB

STRESS AT SOIL MASS (SOIL MECHANICS)

9

Solution (a)  n = 1 =

 y + x 2

 y −  x 

+  

2

2

2 +  xy  

300 + 120  300 − 120  2 2  n = 1 = +  + ( − 40 ) = 308 . 5 kN / m  2 2   2

n = 3 =

 y + x 2

 y −  x 

−  

2

2

2  +  xy 

300 + 120  300 − 120  2 2 n = 3 = −  + ( − 40 ) = 111 . 5 kN / m  2 2   2

STRESS AT SOIL MASS (SOIL MECHANICS)

10

Solution (b) n =

 y + x

+

 y − x

cos 2 +  sin 2

2 2 300 + 120 300 − 120 n = + cos(2 x 20) + (−40) sin( 2 x 20) 2 2  n = 252.23kN / m 2

n =

 y − x

sin 2 −  cos 2

2 300 − 120 n = sin( 2 x 20) − (−40) cos(2 x 20) 2  n = 88.49kN / m 2 STRESS AT SOIL MASS (SOIL MECHANICS)

11

Stress Caused by a Point Load Boussinesq (1883) → solved the problem of stresses produced at any point in a homogeneous, elastic, and isotropic medium as the result of a point load applied on the surface of an infinitely large half-space. Boussinesq solution :  x2 − y2 P  3x 2 z y 2 z   x = + 3 2   5 − (1 − 2  ) 2 2  L  Lr (L + z ) L r    y2 − x2 P 2 y2z x2 z   y = + 3 2   5 − (1 − 2  ) 2 2  L  Lr (L + z ) L r   3Pz 3 3P z3  z = = 5 2L 2 r 2 + z 2 5 / 2

(

Where: r= (x2+y2)

L= (x2+y2+z2) = (r2+z2)  = Poisson’s ratio

)

STRESS AT SOIL MASS (SOIL MECHANICS)

12

Stress Caused by a Point Load The relationship for the vertical normal stress, z can be rewritten as: P  z = 2 z

 3  P 1 = 2 I1  5/ 2  2  2 (r / z ) + 1  z





where 3 1 I1 = 2 (r / z )2 + 1 5 / 2





Stresses in an elastic medium caused by point load STRESS AT SOIL MASS (SOIL MECHANICS)

13

Variation of I1 for Various Values of r/z

STRESS AT SOIL MASS (SOIL MECHANICS)

14

Example Consider a point load P = 5 kN, calculate the vertical stress increase (z) at z = 0 m, 2m, 4m, 6m, 10 m, and 20 m. given x = 3 m and y = 4 m.

STRESS AT SOIL MASS (SOIL MECHANICS)

15

Solution R = (x2+y2) = (32+42) = 5 m.

r (m)

z (m)

r/z

I1

z = (P/z2)I1

5

0



0

0

2

2.5

0.0034

0.0043

4

1.25

0.0424

0.0133

6

0.83

0.1295

0.0180

10

0.5

0.2733

0.0137

20

0.25

0.4103

0.0051

STRESS AT SOIL MASS (SOIL MECHANICS)

16

Vertical Stress Caused by a Line Load The vertical stress increase → z can be determined:

 z =

2qz 3

 (x + z 2

)

2 2

STRESS AT SOIL MASS (SOIL MECHANICS)

17

Variation of z/(q/z) with x/z

STRESS AT SOIL MASS (SOIL MECHANICS)

18

Vertical Stress Caused by a Line Load This equation can be rewritten as: 2q  z = 2 2 z ( x / z ) + 1 or  z 2 = (q / z )  (x / z )2 + 1 2









STRESS AT SOIL MASS (SOIL MECHANICS)

19

Example Two line loads and a point load acting at the ground surface. Determine the increase in vertical stress at point A which located at a depth 1.5 m.

STRESS AT SOIL MASS (SOIL MECHANICS)

20

Solution z = z(1) + z(2) + z(3)  z =  z =

2q1 z 3

 (x + z 2 1

)

2 2

+

2(15)(1.5) 3

 (2 + (1.5) 2

)

2q2 z 3

 (x22 + z 2 )

2 2

2

+

3P z3 + 2 r 2 + z 2

(

2(10)(1.5) 3

 (4 + (1.5) 2

)

2 2

)

5/ 2

3(30) (1.5) 3 + 2 (3) 2 + (4) 2 2 + (1.5) 2

(



)

5/ 2

 z = 0.825 + 0.065 + 0.012 = 0.902kN / m 2

STRESS AT SOIL MASS (SOIL MECHANICS)

21

Vertical Stress Caused by a Strip Loading (Finite Width and Infinite Length) z =

Calculate z /q from 2z /B and 2x /B

STRESS AT SOIL MASS (SOIL MECHANICS)

22

Vertical Stress Due to a Vertical Strip Load

STRESS AT SOIL MASS (SOIL MECHANICS)

23

Vertical Stress due to Embankment Loading

I2 is a function of B1/z and B2/z

STRESS AT SOIL MASS (SOIL MECHANICS)

24

Vertical Stress due to Embankment Loading

STRESS AT SOIL MASS (SOIL MECHANICS)

25

Example Determine the stress increase under the embankment at points A1 and A2

STRESS AT SOIL MASS (SOIL MECHANICS)

26

Solution H = 17.5 kN/m3 x 7 m = 122.5 kN/m2 At A1 B1 = 2.5 m and B2 = 14 m ; B1/z = 2.5 / 5 = 0.5 B2/z = 14 / 5 = 2.8

From the curve, we obtain I2 = 0.445 Due to symmetrical form, the I2 is equal at the left side and the right side. z = z(1) + z(2) = q0 [I2 left + I2 right] = 122.5 [0.445 + 0.445] = 109.03 kN/m2

STRESS AT SOIL MASS (SOIL MECHANICS)

27

Solution

STRESS AT SOIL MASS (SOIL MECHANICS)

28

Solution Stress increase at A2 :

For the right side;

B2 = 5 m and B1 = 0

B2/z = 9/5 = 1.8 ; B1/z = 0/5 = 0 → I2 = 0.335

B2/z = 5/5 = 1

z = 78.75 (0.335) = 26.38 kN/m2

B1/z = 0/5 = 0 From the curve, we obtain I2 = 0.25

Total Stress Increase at Point A2 is:

z = 43.75 (0.25) = 10.94 kN/m2

z = z(1) + z(2)-z(3)

For the middle section;

z = 10.94 + 60.64 – 26.38

B2/z = 14/5 = 2.8 ; B1/z = 14/5 = 2.8 → I2 = 0.495

z = 45.2 kN/m2

z = 122.5 (0.495) = 60.64 kN/m2 STRESS AT SOIL MASS (SOIL MECHANICS)

29

Vertical Stress Below Center of Uniform Loaded Circular Area           

1

z = q 1    

R/z

 

2

   

+1

3/ 2

          

STRESS AT SOIL MASS (SOIL MECHANICS)

30

Vertical Stress at Any Point Below a Uniformly Loaded Circular Area

STRESS AT SOIL MASS (SOIL MECHANICS)

31

Vertical Stress at Any Point Below a Uniformly Loaded Circular Area

STRESS AT SOIL MASS (SOIL MECHANICS)

32

Vertical Stress at Any Point Below a Uniformly Loaded Circular Area

STRESS AT SOIL MASS (SOIL MECHANICS)

33

Vertical Stress Caused by Rectangular Load (Under corner of the loading area)  = q  I 3

B Z L n= Z

m=

I3 is a function of m and n STRESS AT SOIL MASS (SOIL MECHANICS)

34

Vertical Stress Caused by Rectangular Load

STRESS AT SOIL MASS (SOIL MECHANICS)

35

Vertical Stress Caused by Rectangular Load

STRESS AT SOIL MASS (SOIL MECHANICS)

36

Vertical Stress Caused by Rectangular Load z = q  (I + I + I + I ) 2(1)

1

3

2(2)

2(3)

2(4)

2

4

STRESS AT SOIL MASS (SOIL MECHANICS)

37

Influence Chart for Vertical Pressure The procedure for obtaining vertical pressure at any point below a loaded area is follows: 1. determine the depth (z) below the uniformly loaded area at which the stress increase is required. 2. plot the plan of the loaded area with a scale of z equal to the unit length of the chart. 3. place the plan (plotted at 2) on the influence chart in such way that the point below which the stress is to be determined is located at the center of the chart. 4. count the number of elements (M) of the chart enclosed by the plan of the loaded area. STRESS AT SOIL MASS (SOIL MECHANICS)

38

Example 3 m → 25 mm

    

660  2 z = (IV) q M = 0.005 48.5 = 17.78kN / m 3x3  STRESS AT SOIL MASS (SOIL MECHANICS)

39

Newmark Chart Determine the depth, z, where you wish to calculate the stress increase Adopt a scale as shown in the figure Draw the footing to scale and place the point of interest over the center of the chart Count the number of elements that fall inside the footing, N Calculate the stress increase as:

STRESS AT SOIL MASS (SOIL MECHANICS)

40

THANK YOU STRESS AT SOIL MASS (SOIL MECHANICS)

41