Soil Mechanics Review PDF 2020

CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Grain Size Distribution In any soil mass the sizes of the grains vary to a great extent. To classify the soil, we need to know its grain size distributio...

CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Grain Size Distribution In any soil mass the sizes of the grains vary to a great extent. To classify the soil, we need to know its grain size distribution. Generally, soils can be as coarse-grained soils or fine-grained soils. Figure 1 shows relative soil particle sizes. Coarse-grained soils: their particles are large enough to be visible to the naked eye. They include gravels and sands and are often referred to as cohesionless soils. The grain size distribution can be obtained by sieve analysis. (ASTM D 422) Fine-grained soils: their particles are not visible to the naked eye. They include silts and clays. Clays are often referred to as cohesive soils. The grain size distribution can be obtained by hydrometer analysis. (ASTM D 422) Figure 1: relative soil particle sizes. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 1 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Figure 2: Different grading curves. Two parameters can be determined from the grain size distribution curves of coarse-grained soils: 1) The uniformity coefficients (𝑪𝒖 ) 𝑫𝟔𝟎 𝑪𝒖 = 𝑫𝟏𝟎 2) The coefficient of graduation or curvature (𝑪𝒄 ) 𝑫𝟑𝟎 𝟐 𝑪𝒄 = 𝑫𝟔𝟎 𝑫𝟏𝟎 Where, 𝐷10 , 𝐷30 , and 𝐷60 are the diameters corresponding to percent finer than 10, 30, and 60% respectively. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 2 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Weight-Volume Relationships This section summarizes common soil definitions and weight-volume relationships. Figure 3 illustrates and defines a number of terms used in these relationships. Figure 3: Three phase system 𝑉𝑉 𝑉𝑉 Void Ratio, 𝑒 = Porosity, 𝑛(%) = × 100 % 𝑉𝑠 𝑉𝑇 𝑊𝑤 𝑉𝑤 Water Content, 𝑤(%) = × 100 % Degree of saturation, 𝑆(%) = × 100 % 𝑊𝑠 𝑉𝑉 𝑊𝑇 𝑊𝑠 Bulk (or total) unit weight, 𝛾 = Dry unit weight, 𝛾𝑑 = 𝑉𝑇 𝑉𝑇 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 Specific Gravity, 𝐺𝑠 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑛 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 3 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Table 1: Some common weight-volume relationships (After Das, 7th Ed.) Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 4 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Atterberg Limits The liquid and plastic limits are routinely determined for cohesive soils. The Atterberg limits are based on the moisture content of the soil. The plastic limit (PL) is the moisture content that defines where the soil changes from a semi-solid to a plastic (flexible) state. The liquid limit (LL) is the moisture content that defines where the soil changes from a plastic to a viscous fluid state. From these two limits the plasticity index is computed. a) The liquid limit test b) The plastic limit test Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 5 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Plasticity index The plasticity index (PI) is a measure of the plasticity of a soil. The plasticity index is the size of the range of water contents where the soil exhibits plastic properties. The PI is the difference between the liquid limit and the plastic limit (PI = LL-PL). Soils with a high PI tend to be clay; those with a lower PI tend to be silt. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 6 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Stress-Strain Relationship in General Normal stress and strain Normal stress,  n F n = A Normal strain,  n h n = h n Note: E = n Shear stress and strain Shear stress,  F = A Shear strain,  x = h  Note: G =  Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 7 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Stress-Strain Relationship for Soils The stress-strain characteristics of a soil determine the settlement that a given structure will experience under certain loading conditions. In some cases, they may also serve as an indicator of construction difficulties that may arise during excavation. Due to the particulate nature of soils, the standard concepts of tension and compression applied to solids are not applicable to soils. Consequently, the stress-strain behaviour of soil is not elastic. Furthermore, the response of soil to applied stress is influenced by several factors, such as water content, anisotropy, stress-history, etc. Nevertheless, in certain cases the elastic properties of soil are useful, especially over the limited range of stresses that are normally encountered. The elastic parameters of importance are the elastic modulus, E, the shear modulus, G, and the Poisson’s ratio, ν. Always Keep in mind ➔ (1) In general, soil is NOT an elastic material. (2) Elasticity theory can only be applied under low stress and small strain conditions. (3) the shear modulus G can be related to the elastic modulus E by the following formula E G= 2(1 + ) Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 8 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Total Stress in the Ground  The total vertical stress acting at a point below the ground surface is due to the weight of everything lying above that point (i.e., soil, water, and surface surcharge). Total stresses are calculated from the unit weight of the soil.  Total stress increases with depth and with unit weight.  The total horizontal stress is related to the vertical stress by the coefficient of earth pressure at rest, Ko as follows h = K ov  Homogeneous soil The total vertical and horizontal stresses at depth Z are:  v = z  h = K o  v = K o z  Multi-layered soil The total vertical stress at depth Z is the sum of the weights of soil in each layer thickness above, i.e.,  v = 1d1 +  2 d 2 +  3 (z − d1 − d 2 ) and the total horizontal stress is:  h = K o 3 ( 1d1 +  2 d 2 +  3 ( z − d1 − d 2 )) Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 9 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng.  With a surface surcharge load The addition of a surface surcharge load will increase the total stresses below it. If the surcharge loading is extensively wide, the increase in vertical total stress below it may be considered constant with depth and equal to the magnitude of the surcharge. i.e. the total vertical stress at depth Z is,  v = z + q and the total horizontal stress is:  h = K o  v = K o ( z + q) Pore pressure o The magnitude of pore pressure (u) depends on both the depth below the water table and the seepage flow. o Under hydrostatic conditions (no water flow) the pore pressure at a given point is given by the hydrostatic pressure: u = whw Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 10 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Effective Stress in the Ground Ground movements and instabilities can be caused by changes in total stress (such as loading due to foundations or unloading due to excavations), but they can also be caused by changes in pore pressures (slopes can fail after rainfall increases the pore pressures). In fact, it is the combined effect of total stress and pore pressure that controls soil behaviour such as shear strength, compression and distortion. The difference between the total stress and the pore pressure is called the effective stress: Effective stress = Total stress - Pore pressure ' =  − u Terzaghi's principle and equation Karl Terzaghi was the first person to propose the relationship for effective stress. His principle states: “All measurable effects of a change of stress, such as compression, distortion and a change of shearing resistance are due exclusively to changes in effective stress. The effective stress s´ is related to total stress and pore pressure by ' =  − u.” Note: (1) Changes in water table below ground result in changes in effective stresses below the water table. (2) Changes in water level above ground (e.g. in lakes, rivers, etc.) do not cause changes in effective stresses in the ground below. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 11 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Example 1: For the profile shown in the Figure below, determine the total vertical stress, pore water pressure, and effective vertical stress at: (a) At the top of saturated sand layer; (b) At the top of the clay layer. (Dry sand: γd = 16 kN/m³ and saturated sand: γg = 20 kN/m³, assume γw = 10 kN/m³ for simplicity) Solution: (a) At the top of saturated sand (z = 2.0 m) Vertical total stress v = = 32.0 kPa Pore pressure u=0 Vertical effective stress, ' =  − u = 32.0 kPa (b) At the top of the clay (z = 5.0 m) Vertical total stress v = = kPa Pore pressure u= Vertical effective stress, ' =  − u = kPa Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 12 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Example 2: The figure shows soil layers on a site. The unit weight of the silty sand is 19.0 kN/m³ both above and below the water table (not real assumption). The water level is presently at the surface of the silty sand, it may drop, or it may rise show the Effect of changing water table on the vertical total and effective stress. assume γw = 10 kN/m³ for simplicity. a) v = u = 10 × 3= 30 kPa ' =  − u = kPa b) c) v = v = u = 10 × 6 = kPa u = 10 × 5 = kPa ' =  − u = kPa ' =  − u = kPa d) e) v = v = u= × = kPa u= × = kPa ' =  − u = kPa ' =  − u = kPa Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 13 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Example 3: A 4.0 m deep compacted fill is to be placed over the soil profile shown in the figure. Plot the values of total vertical stress and effective vertical stress against depth. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 14 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Stress from Elastic Theory How surface loading affects effective stress 1) Spread footing for a building (small loaded area) Z A ′ 𝜎𝑍(𝐴) ′ = 𝜎𝑍𝑂 +  𝜎𝑍(𝑓𝑜𝑜𝑡𝑖𝑛𝑔) ′ 2) Roadway embankment h Z A ′ 𝜎𝑍(𝐴) ′ = 𝜎𝑍𝑂 +  𝜎𝑍(𝑒𝑚𝑏) ′ If the embankment is very large, then  𝜎𝑍(𝑒𝑚𝑏) ′ = ℎ × 𝑒𝑚𝑏 , if not see next page. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 15 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. I- Point Load (Craig, p.144) Boussinesq in 1885 developed relationship for the stress distribution within a semi-infinite, homogenous, isotropic mass with a linear stress-strain relationship due to a point load on the surface. Where, Q: is the applied point load Z: is the depth at which the stress is required r: is the horizontal distance from the load to the point of interest. Notes: The stress distribution due to surface loading area can be obtained by integration. The stresses at a point due to several loads (more than one load) are obtained by superposition. Negative value of loading can be used to represent excavation. Soil is not an elastic material, use Boussinesq cautiously. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 16 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. II- Strip Area Carrying Uniform Pressure (Craig, p.148) The vertical and horizontal stresses at Point X due to a uniform pressure q on a strip area and width B and infinite length is given by the following relation: III- Strip Area Carrying Linearly Increasing Pressure (Craig, p.148) The vertical and horizontal stresses at Point X due to a linearly increasing pressure from zero to q on a strip area and width B and infinite length is given by the following relation: Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 17 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. IV- Circular Area Carrying Uniform Pressure (Craig, p.150) The vertical stress at depth z under the centerline of a circular area of D=2R and carrying a uniform pressure q is given by: or  = q  Ic Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 18 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. V- Rectangular Area Carrying Uniform Pressure (Craig, p.151) The vertical stress at depth z under the corner of a rectangular area mz by nz and carrying a uniform pressure q is given by:  = q  Ir Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 19 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Example 4: (Craig, example 5.3) A strip footing 2m wide carries a uniform pressure of 250kN/m2 on the surface of a deposit of sand. The water table is at the surface. The saturated unit weight of the sand is 20kN/m3 and Ko=0.40. Determine the effective vertical and horizontal stresses at a point 3m below the centre of the footing before and after the application of the pressure. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 20 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Example 5: Two and half meters of fill are compacted over a large area. Then, on top of the fill a rectangular foundation 9 m x 4.5 m carries a uniform pressure of 300 kPa is placed. Knowing that the unit weight of the fill, γfill= 20 kN/m3 and γs= 18 kN/m3. (a) What the stress increment due to the fill (before placing the foundation)? (b) What is the stress increment at point A at 3 m depth after placing the foundation? 1m 4.5 m A 2.5 m B 4.5 m B 9 x 4.5 m A Plan Section B-B Solution: Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 21 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Consolidation Settlement For saturated soils ➔ Well-draining soils (e.g. sand/gravel)  σ′Z is reached almost immediately ➔ Poorly draining soils (e.g. silt/clay)  σ′Z gradually increases until it reaches the applied tress Consolidation Analogy The mechanics of a consolidation process is analogous to a cylinder filled with water and a piston with spring attached to it and a valve as shown in the Figure below. If a load is applied to the piston with the valve closed (i.e., no leakage), the piston will not move as water is incompressible. Thus, the entire load will be carried by the water and no load is transferred to the piston. The increase in PWP is called “excess pore water pressure” (ue). If the valve is opened, water will be expulsed out through the valve and the valve will start to move and the spring compresses. (thus settlement occurs) Note: When dealing with saturated silt/clay we have to consider 2 situations: 1- Undrained condition (short-term behaviour) 2- Drained condition (long-term behaviour) Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 22 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. Shear Strength of Soils Mohr-Coulomb failure criterion According to the Mohr-Coulomb failure criterion which generally characterises the shear strength of soil, the limiting shear stress that may be applied to any plane in the soil mass is found to be given by an equation of the form: 𝜏𝑓 = 𝑐 + 𝜎𝑓 𝑡𝑎𝑛𝜑 Where, 𝜏𝑓 = shear strength c= cohesion 𝜎𝑓 = stress acting on the shear plane φ = angle of internal friction In general shear strength is dependent on the drainage conditions: 1- Drained condition (no excess PWP, long-term condition) - This is always the case with well draining soils (e.g. sand/gravel) - This condition applies to long-term behaviour of poorly draining soils (e.g. silt/clay) after excess PWP have dissipated. - For drained conditions always use an effective stress analysis with effective strength parameters. 𝝉𝒇 = 𝒄′ + 𝝈′ 𝒇 𝒕𝒂𝒏𝝋′ Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 23 CIVL 4111 Geotechnical Engineering Soil Mechanics Review Instructor: Dr. Hany El Naggar, PEng. 2- Undrained condition (short-term condition) - This is always the case with poorly draining soils (e.g. silt/clay) before PWP are able to dissipate. - In this case use the total stress analysis. 𝝉𝒇 = 𝒄𝒖 + 𝝈𝒇𝒖 𝒕𝒂𝒏𝝋𝒖 - cu, u are referred to as the undrained (total) strength parameters. These parameters can only be determined from undrained tests - but u=0, ➔ 𝜏𝑓 = 𝑐𝑢 , independent of normal stress. Copyright © 2020, Professor Hany El Naggar Ch. 1 – Page 24