Specific Gravity-Soil Mechanics and Foundations-Lecture 04 Slides-Civil and Environmental Engineering, Slides of Soil Mechanics and Foundations

Download Specific Gravity-Soil Mechanics and Foundations-Lecture 04 Slides-Civil and Environmental Engineering and more Slides Soil Mechanics and Foundations in PDF only on Docsity! CE 240 Soil Mechanics & Foundations Lecture 2.2 Weight-Volume Relationships (Das, Ch. 3) Outline of this Lecture 1.Density, Unit weight, and Specific gravity (Gs) 2.Phases in soil (a porous medium) 3.Three phase diagram 4.Weight-volume relationships Specific Gravity (Gs) Specific gravity is defined as the ratio of unit weight (or density) of a given material to the unit weight (or density) of water since s w w w gG g γ ρ ρ γ ρ ρ = = = Specific Gravity • Expected Value for Gs Type of Soil Gs Sand 2.65 - 2.67 Silty sand 2.67 – 2.70 Inorganic clay 2.70 – 2.80 Soils with mica or iron 2.75 – 3.00 Organic soils < 2.00 Three Phases of Soils Naturally occurred soils always consist of solid particles, water, and air, so that soil has three phases: solid, liquid and gas. Fully Saturated Soils (Two phase) Water Solid Mineral Skeleton Fully Saturated Dry Soils (Two phase) [Oven Dried] Air Solid Dry SoilMineral Skeleton Three Phase Diagram Solid Air Water WT Ws Ww Wa~0 Vs Va Vw Vv VT Weight Volume Void ratio: e = Vv/Vs; Porosity n = Vv/Vt / 1 / 1 / 1 / 1 v v v t s t v v t v v v s t s v v s V V V V ne V V V V V n V V V V en V V V V V e = = = = − − − = = = = + + + Apparently, for the same material we always have e > n. For example, when the porosity is 0.5 (50%), the void ratio is 1.0 already. Solid Air Water WT Ws Ww Wa~0 Vs Va Vw Vv VT Degree of saturation: S =Vw/Vv x 100% Saturation is measured by the ratio of volume. Moisture content (Water content): w = Ww/Ws, Ww – weight of water, Ws – weight of solid Water content is measured by the ratio of weight. So that w can be greater than 100%. Solid Air Water WT Ws Ww Wa~0 Vs Va Vw Vv VT Degree of saturation: S =Vw/Vv x 100% Saturation is measured by the ratio of volume. Moisture content: w = Ww/Ws, Ww=Vwγw Ww – weight of water, Ws – weight of solid Water content is measured by the ratio of weight. Thus, the dry unit weight γd can be approximated as: (1 )d wγ γ= − From the original form of the dry unit weight By taking the Taylor expansion and truncated at the first order term: 1d w γγ = + 2 3 4 5(1 ...) (1 ) 1d w w w w w w w γγ γ γ= = − + − + − + ≈ − + Because the moisture content w is a number always smaller than one, i.e., w<1. Relationships among S, e, w, and Gs , , 1w w v s v VS then V SV Se given V V = = = = A simple way to get Das, Equation 3.18 , 1, 1w w w w s v s s s s w s s W V eS eSw V V W V G G thus Se wG γ γ γ γ = = = = = ∴ = = ∵ When the soil is 100% saturated (S=1) we have, Equation 3.20 se wG= Relationships among γ, n, w, and Gs (1 ), 1 (1 ) s s s s w s w s s w W V G n given V n W wW wG n γ γ γ = = − = − = = − So that the dry unit weight γd is And the moist unit weight γ is (1 ) (1 ) 1 s s w d s w t W G n G n V n n γγ γ−= = = − − + (1 ) (1 ) 1 (1 ) (1 ) (1 )(1 ) t s w s w s w t t s w s w W W W G n wG n V V w G n G n w γ γγ γ γ + − + − = = = = + − = − + Example: • Determine moisture content, void ratio, porosity and degree of saturation of a soil core sample. Also determine the dry unit weight, γd Data: • Weight of soil sample = 1013g • Vol. of soil sample = 585.0cm3 • Specific Gravity, Gs = 2.65 • Dry weight of soil = 904.0g Solid Air Water Wa~0 1013.0g585.0cm3 904.0g γs =2.65 109.0g 341.1cm3 109.0cm3 243.9cm3 134.9cm3 γW =1.00 Example Volumes Weights Results • From the three phase diagram we can find: – Moisture content, w – Void ratio, e – Porosity, n – Degree of saturation, S – Dry unit weight, γd 715.0 1.341 9.243 3 3 === cm cm V Ve s v %7.41100 )(0.585 )(9.243 3 3 =×== cm cm V Vn T v %1.12100 )(904 )(109 =×== g g W Ww s w 355.1585 904 cm g V W T s d ===γ %7.44100 9.243 109 =×== v w V VS In a two-phase system, i.e., if S=100%, and we let Vs=1, we then have: eVVeV vwt ==+= since,1 Consider the not-submerged case, i.e., the two- phase system has been just put in the air: wsswssss wwww GVGVM andeVM ρρρ ρρ === == Thus, the saturated density is e Ge e Ge V MM swwsw t sw sat + + = + + = + = 1 )( 1 ρρρρ and the dry density is e G V M ws t s d + == 1 ρρ Recall that the saturated density is e Ge V MM sw t sw sat + + = + = 1 )(ρρ If we do the following wsat wsws wws w ws wsat ei e G e eGe e eGe e Ge ρρρ ρρρ ρρρρρρ −=′ ′= + − = + −−+ = + +−+ =− + + =− .,. 1 )1( 1 )1( 1 )1()( 1 )( If you have got the submerged density ρ’ and sure you know water density ρw e Ge V MM sw t sw sat + + = + = 1 )(ρρ You can calculate the saturated density ρsat.If you know ρw then you can calculate the void ratio e. if you think you can know e from the dry density ρd e G ws d + = 1 ρρ You can also calculate the submerged density ρ’ when the sample is not 100% saturated. e SeG ws + −+− =′ 1 )1()1( ρρ