Subsurface Stresses (Stresses in soils) ENB371: Geotechnical Engineering Chaminda Gallage
Contents Introduction: Subsurface stresses caused by surface loading Subsurface stresses caused by the soil mass (Geostatic Stresses) ENB7 Vertical Stresses Horiontal Stresses Vertical stress increase within a soil mass caused by vertical surface loading (foundation loads, wheel loads,..etc) Infinitely loaded area Point load (concentrated load) Circular loaded area Strip load Rectangular loaded area
Introduction -1 At a point within a soil mass, stresses will be developed as a result of the soil lying above the point (Geostatic stress) and by any structural or other loading imposed onto that soil mass The subsurface stress at a point is affected by ground water table if it extends to an elevation above the point W Ground level Unit weight γ A h Total stressσ γh A Total stressσ γh+ σ σ
Introduction - Geostatic stresses at a point in soil K o 1 sinφ Ko Coefficeint of earth pressure at rest, φfriction angle Koσ h /σ v σ v * γ s σ h K o * σ v Soil unit weight γ s
Introduction -3 Subsurface stresses in road pavements and airport runways are increased by wheel load on the surface
Introduction -4 Subsurface stresses in soils are increased by foundation loads Burj Dubai (<700m) Dubai, U. A. E
Introduction - 5 Embankments and landfills cause to increase subsoil stresses
Subsoil stress increase -1 It is required to estimate the stress increase in the soil due to the applied loads on the surface. The estimated subsoil stress increase is used: - to estimate settlement of foundation - to check the bearing capacity of the foundation Bearing capacity failure of a raft foundation The Tower of Pisa in Italy, a widely-recognied foundation failure -settlement
Subsoil stress increase - The calculation of subsurface stress increase due to the following surface loading conditions will be discussed. Infinitely loaded area Point load (concentrated load) Circular loaded area Strip loading Rectangular loaded area Though the surface loading is caused to increase both vertical and horiontal stresses in soils, only the vertical stress increase is discussed as it is the major stress component for most practical design problems
Infinitely loaded area -1 The surface loading area is much larger than the depth of a point where vertical stress increment ( σ) is calculated (e.g. land fills, preloading by soil deposition)
Infinitely loaded area - The vertical subsurface stress increment at any depth below the infinitely loaded area is considered to be the same as the surface stress due to external load (filling material) Ground Surface h γ q hγ 1 σ 1 1 σ + 1 σ 1 σ σ + σ σ σ 1 σ q
Finitely loaded area If the surface loading area is finite (point, circular, strip, rectangular, square), the vertical stress increment in the subsoil decreases with increase in the depth and the distance form the surface loading area. Methods have been developed to estimate the vertical stress increment in sub-soil considering the shape of the surface loading area.
Subsurface stress increment due to a point load -1 For homogeneous (properties are the same everywhere), and isotropic (properties are the same in all direction) soils Boussinesq (1885) (French mathematician) developed the following equation to calculate vertical stress increment ( σ σ) in soils due to point load on the surface: Then, 3Q 1 5/ σ 5/ π r 3 1 1+ I p σ Q I p Influence factor, I p π 1+ ( r / ) I p can be obtained numerically or using table where I p is given in terms of r/
Subsurface stress increment due to a point load - Influence factor, I P, for vertical stress increment due to point load
Example -1 (1) In road pavement design, the standard vehicle axel is defined as an axel with two single wheels as shown below. P L For a particular vehicle group, the standard axel load (P) is given as 80 kn and the distance between two wheels (L) is 1.8 m. What is the vertical stress increment in the sub-grade at 4 m depth directly under a wheel if this axel is running. (consider wheel load as a point load)
y 40kN A Z4m σ r 1.8 m 40kN B x 5/ Example -1 () Find: σ σ σ Α + σ B σ Q 3 1 3 1 ( I p ) B π 1+ ( r / ) π 1+ (1.8/ 4) I p 5/ 0.301 ( I p ) A 3 1 π 1+ ( r / ) 5/ 3 1 π 1+ (0 / 4) 5/ 0.477 Therefore, σ Q QB 40 40 A ( I p ) A + ( I p ) B 0.477 + 0.301 1. kpa 4 4 945
Subsurface stress increment due to circular loaded area -1
5/ 0 1 ) 3( ) ( + r dr d r q d π θ σ Considering force on the small element (black) And applying Boussinesq equation, the stress Increase at A caused by this load Subsurface stress increment due to circular loaded area - Subsurface stress increment due to circular loaded area - By integration of Boussinesq solution over the By integration of Boussinesq solution over the whole circular area, the vertical whole circular area, the vertical stress stress increment at depth under the centre (at A) increment at depth under the centre (at A) I c q B q 0 3/ 0 1 1 1 + σ The influence factor I c can be calculated numerically or can be obtained from a table 5/ 1 1 3 + r Q π σ
Subsurface stress increment due to circular loaded area -3 Similar integration can be performed to obtain the vertical stress increase at A, located a distance r from the centre of the loaded area at a depth. This table can be used to calculate I c corresponding to A.
Subsurface stress increment due to strip loading -1
Vertical stress increment due to strip area carrying uniform pressure Soil is homogeneous, isotropic The stress increment at point X ( σ ) due to a uniform pressure q on a strip area of width B and infinite length is given by in terms of α and β α β σ q σ + + π { α sinα cos( α β )}
Vertical stress increment due to strip area carrying increasing pressure Soil is homogeneous, isotropic B X The stress increment at point X ( σ ) due to increasing pressure from ero to q on a strip area of width B and infinite length is given by in terms of α and β the lengths R 1 and R R 1 q R α β σ X q X 1 σ α sin π B β
Example - A strip footing m wide carries a uniform pressure of 50 kpa on the surface of a deposit of sand. Water table is at the surface and the saturated unit weight of sand is 0 kn/m 3. Determine the effective vertical stress at a point 3 m below the centre of the footing before and after the application of the surface pressure Before loading Total vertical stress at 3 m depth, σ v σ v γ sat 0 3 60kPa Pore water pressure at 3 m depth, u u γ w 9.8 3 9. 4 kpa Effective vertical stress at 3 m depth, σ v σ v ' σ v u 60 9.4 30. 6 kpa
Strip area carrying uniform pressure -Example After loading B m, q 50 kpa, 3 m, β α σ q σ + π { α + sinα cos( α β )} At the centre of the strip, β α/ Therefore, cos(α+β)1 α tan 1 3 1 1 α tan 3 sinα 0.600 36 0 5' 0.643 radians The increase in stress at 3 m depth due to applied load q 50 σ 0 π π Hence, total effective vertical stress 30.6+99.0 19.6 kpa ( α + sin α ) (0.643+ 0.600) 99. kpa
Subsurface stress increment due to rectangularly/ squarely loaded area -1
Vertical stress increment due to uniformly loaded rectangular area -1 Considering force on the small element (black) and applying Boussinesq equation, the stress Increase at A caused by this load 3Q 1 σ π 1+ r 5/ d( σ ) π 3q 0 ( dx dy) [ ] x + y + 5/ 3 By integration of Boussinesq solution over complete area, the vertical stress increment at depth under the centre (at A) L B σ L B y 0 x 0 π 3q 0 ( dx dy) 3 [ x y ] + + 5/ q 0 I r
Vertical stress increment due to uniformly loaded rectangular area - The stresses increase under a corner of a uniformly loaded flexible rectangular area: σ q 0 I r Influence factor for rectangular footing, I r I r 1 mn(m +n +1) 1/. m +n + 4π m +n -m n +1 m +n +1 + mn(m +n +1) tan-1 m +n -m n +1 +1) 1/ Define m B/ and n L/ Solution by numerically or tables
Influence factor, I r, for vertical stress increment due to uniformly loaded rectangular area -1
Influence factor, I r, for vertical stress increment due to uniformly loaded rectangular area -
Rectangular area The vertical stress increment at a depth below point O, σ q ( I + I + I + I ) 0 1 3 4
Determine the increase in stress at A and A below a rectangular area Example -3 At the centre of rectangular or square area, σ q ( ) 0 I1 + I + I3 + I 4 I B m m 1 I I3 I 4 A A σ σ n A B or A L 1.5 3 σ 0.5, B or L 1.5 ' na' 0.3, I A' 5 A' I A 4 q 0 ( I ) Then, 0 L 1.5 m 4*100*0.084 A A' 4*100*0.037 33.6 14.8 0.084 kpa kpa 0.037
Example 4 (1) A rectangular foundation 6 X 3 m carries a uniform pressure of 300 kpa near the surface of a soil mass. Determine the vertical stress at a depth of 3 m below point A on the centre line 1.5 m outside the long edge of the foundation. Hint: use the principle of superposition
Example - 4 () Using the principle of superposition the problem is dealt as shown below For the two rectangles (1) carrying a positive pressure of 300 kpa, B 3.0 m, L 4.5 m, m B/ 3.0/3.0 1.0, n L/ 4.5/3.0 1.5 Therefore, I r 0.193 (from the table) For the two rectangles () carrying a negative pressure of 300 kpa, B 1.5 m, L 3.0 m, m B/ 1.5/3.0 0.5, n L/ 3.0/3.0 1.0 Therefore, I r 0.10 (from the table) Hence, vertical stress increase at A ( a depth of 3 m), σ ( ) A ( q0 I r ) 1 + ( q0i r ) ( 300 0.193) ( 300 0.10) 44kPa
Influence chart for vertical stress increase - 1 Newmark (194) constructed influence chart, based on the Boussinesq solution to determine the vertical stress increase at any point below an area of any shape carrying uniform pressure. Then, vertical stress increase at σ 0.005 q o N Chart consists of influence areas which has a influence value of 0.005 per unit pressure The loaded area is drawn on tracing paper to a scale such that the length of the scale line on the chart is equal to the depth Position the loaded area on the chart such that the point at which the vertical stress required is at the centre of the chart. The count the number of influence areas covered by the scale drawing, N
Influence chart for vertical stress increase - Use of Newmark s chart to calculate stress increment at A, ( at a depth of 3.0 m) Z 3.0 m q 0 300 kpa Then, vertical stress increase at 3 m below point A σ 0.005 q o N 0.005 X 300 X 30 45 kpa The scale line represents 3 m The scale to which the rectangular loading area must be drawn The area is positioned on the chart such that A is at the centre The number of influence areas covered by the rectangle, N 30 (approximately)
Approximate method to determine the stress subsurface stress increment (60 0 approximation) According to this method, the increase in stress at depth is σ ( B q 0 + B )( L L + )
Example 5 (1) 3 m Compare the stress increase occurring m below the centre of a 3m by 3m square foundation imposing a bearing pressure of 145 kn/m when: 145 kn / m PLAN VIEW (i) The Boussinesq stress distribution is assumed (ii) The 60 0 approximation is assumed SECTIONAL VIEW
(i) Example 5 () The Boussinesq stress distribution is assumed q ( ) 0 I1 + I + I3 + I 4 σ 1 L 3 m L 1.5 m When the centre is considered, I 1 I I3 I 4 3 4 σ Then, 4 q 0 ( I ) q 145 m 0 kn / 145kN/ m PLAN VIEW B' L' 1.5 m m n B' or L' 1.5 0.75, I 0.1367 σ m 4*145*0.1367 79. 3 kpa SECTIONAL VIEW
Example 5 () (ii) The 60 0 approximation is assumed B L 3. 0 m L + q 145 m 0 kn / L B q 145 m 0 kn / m σ At m depth, q 0 ( B + B L )( L + ) (B + ) B 145kN/ m σ 145 3 3 (3 + )(3 + ) 145 9 5 5 5. kn / m σ SECTIONAL VIEW
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