CE 240 Soil Mechanics & Foundations Lecture 7.1 in situ Stresses I (Das, Ch. 8)
Class Outline Stress tensor, stress units Effective stress, Stresses in saturated soil without seepage Stresses in saturated soil with upward seepage Stresses in saturated soil with downward seepage Seepage force
Stress, Normal Stress, and Shear Stress Stress is the force per unit area on a surface. When the force F and the surface outward normal n are in the same (or opposite) direction, this stress is called the normal stress, when the force and the surface outward normal are perpendicular with each other, the stress is the shear stress. A Fv n A n Fh σ n F = v σ = s A F h A
Fv n n Fh A A The normal stress can be either a positive (F and n are in the same direction) or a negative (F and n are in opposite direction) value, means either a tensile normal stress, or a compressive normal stress. The left sketch in the above figure shows a case of compressive normal stress. General mechanics defines tensile stress as positive, whereas earth science defines the compressive stress positive, since the rocks in depth are constantly experiencing compression.
Stress Tensor But the force has directivity, i.e., force is a vector and can be decomposed into components in 3 orthogonal directions. Meanwhile, the surface on which the force is acting on, also has directivity, it can be facing any direction. If we use a unit vector n to represent a unit area on this surface, it can also be decomposed into components on three orthogonal directions. So, the stress is a tensor and has 9 components.
Actually, since the force is a vector, and can act on any direction, it can be dissolved into 3 orthogonal directions (e.g., x, y, z, in Cartesian coordinate); a surface in or on the solid can face any direction, too. The normal of that surface is also a function of x, y, z. Consequently, a complete stress should be a tensor with 9 elements. σ zz σ xz σ zx σ yx x z σ yz σ zy σ xy σ yy σ σ σ σ xx xy xz = σ σ σ yx yy yz σ σ σ zx zy zz σ xx
Stress Units Not long ago, the unit of pressure and stress is measured in bars, 1 bar = 1 standard atmospheric pressure It is also about equal to the pressure under 10 meters of water. For pressures deep in the earth we use the kilo-bar, equal to 1000 bars. The pressure beneath 10 km of water, or at the bottom of the deepest oceanic trenches, is about 1 kilo-bar. Beneath the Antarctic ice cap (maximum thickness about 5 km) the pressure is about half a kilobar at greatest. Normal stress caused by self-gravity (overburden) Pressure (normal stress) = density x gravity x depth = ρgz
Stress Units (cont.) In the SI System, the fundamental unit of length is in meters, time in seconds, and mass in kilograms (basic units). The unit in SI system for stress/pressure is in Pascals 1 Pascal = 1 Newton/m 2 or 1 kg/(m.sec 2 ). A force of one Newton spread out over a square meter is a pretty feeble force. Atmospheric pressure (1 bar) is about 100,000 Pascals. A manila file folder (35 g, 700 cm 2 area) exerts a pressure of about 5 Pascals. 1 g = 0.001 kg, 1 cm = 0.01 m, 1 cm 2 = 0.0001 m 2 So 35 700 0.001 0.0001 9.81 = 0.5 9.81 = 4.905 Pa
Exchange between SI system and other units By comparison the SI with traditional pressure units, we have 1 bar = 100,000 pascals = 0.1 MPa 1 MPa = 1 megapascal = 10 bars 1 GPa = 1 gigapascal = 10 kilobars The normal stress (pressure) at 1 km depth is about 25 MPa. By comparing the SI unit with the more familiar British psi (pound per square inch) units, we have 1 psi = 6895 Pa = 0.006895 MPa 1 Pa = 0.000145 psi You can find many more useful constants and conversion factors in the first page inside the cover of the textbook.
Effective stress, Stresses in saturated soil without seepage
Let s compare the following 2 cases: A: Soil loaded by an applied weight W B: Soil loaded by water weighing W W W
Soil loaded by an applied weight W Soil loaded by water weighing W W W Compression Little or no compression Why?
Definition of Total and Effective Stress σ = total Stress = v Vertical Force Cross Sectional Area Effective stress σ ' = σ u v v σ v u σ v Case A W A 0 W A Case B W A W A 0
Effective stress σ eff : The normal stress asserted on the solid matrix minus the pressure asserted on the pore fluid. σ or eff = σ u σ = σ + eff u
Figure 8.1 Equation (8.1) leads to σ = Hγ + ( H H) γ w A sat = Hγ + H γ Hγ w A sat sat = Hγ + H ( γ γ ) + H γ Hγ w A sat w A w sat = H ( γ γ ) H( γ γ ) + H γ A sat w sat w A w = ( H H)( γ γ ) + H γ A sat w A w = ( H H) γ + H = σ + u γ A A w where γ γ d = 1 + w γ : submerged unit weight; σ : effective stress; u: pore pressure. Apparently, the effective stress is equal to the product of the soil column and the submerged unit weight.
The effective stress in a soil mass controls its volume change and strength. Increasing the effective stress (equivalent to reduce the pore pressure if the total stress is constant) induces soil to move into a denser state of packing. = + u or = u σ σ σ σ σ σ : total stress; : effective stress; u: pore pressure, or neutral stress. In summary, the total stress is the sum of the effective stress and the pore pressure (neutral stress). The effective stress is carried by the soil skeleton. The pore pressure is carried by pore water.
Principle of Effective Stress Effective stress relations for general stress states σ = σ u ; σ = σ xx xx yz yz = u ; = σ σ σ σ yy yy zx zx = u ; = σ σ σ σ zz zz xy xy
Stresses in saturated soil with upward seepage
When pore pressure is built up at depth, the effective stress is lessened.
γ / γ = 1 γ = γ γ = γ γ γ = 2γ we also learned (Table 3.1) that γ so w sat w sat w = [(1 ng ) + n] γ sat s w [(1 ng ) + n] γ = 2γ [(1 ng ) + n] = 2 G ng + n = s s 2 G 2 = ng n s ng ( 1) = G 2 n = s Gs 2 ( G 1) s w s w w s s s soil's specific gravity is a constant (~2.6) 0.6 n = 1.6 37% When the hydraulic gradient reaches the critical condition (i=i cr ), the effective stress is zero. This means that no force is carried out by the solid skeleton of the soil. What is the consequence? Safety means good compaction!
Stresses in saturated soil with downward seepage
When the pore pressure is reduced at depth, the effective stress is increased.
Example: 1960s Rocky Mountain Arsenals Inject waste water into depth of the ground induced microearthquakes (rock failure by micro-cracks).
Seepage force
Figure 8.8 Hydrostatic (overburden) decreased eff. stress Hydrostatic Seepage increased pore pressure increased eff. stress decreased pore pressure
Solutions are in the next page.
Reading Assignment: Das, Ch. 8 Homework: 8.1-8.3