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INTRODUCTION to SOIL MECHANICS a Focused Approach  Introduction to Soil Mechanics

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INTRODUCTION to SOIL MECHANICS a Focused Approach  Introduction to Soil Mechanics AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach Introduction to Soil Mechanics Phase Relationship An element of soil is multiphase. A typical element of soil contains three distinct phases, solid, gas(air), and liquid(usually water). Given figure represents the three phases as they would typically exist in an element of natural soil. (a) Partially saturated soil (b) Saturated soil (c) Dry soil The phases are dimensioned with volumes on the left and weights on the right side of the sketch. There are three important relationships of volume: porosity, void ratio, and degree of saturation. Porosity is the ratio of void volume to total volume and void ratio is the ratio of void volume to solid volume. Porosity is usually multiplied by 100% and thus the values are given in %. Void ratio is expressed in a decimal value , such as a void ratio of 0.55, and can run to values greater than unity. Both porosity and void ratio indicate the relative portion of void volume in a soil sample. This void volume is filled with fluid, either gas(air) or liquid, usually water. So, we have V v void ratio(e) = Vs V v porosity(n) = x 100 V Va air content = axc 100 Vv Two relationships between porosity(n) and void ratio(e) are SECOND FLOOR, SULTAN TOWER, ROORKEE – 247667 UTTARAKHAND PH: (01332) 266328 Web: www.amiestudycircle.com 1/56 WHATSAPP: +91-9412903929 AMIESTUDYCIRCLE.COM [email protected] AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach e n = 1+e Vw Degree of saturation(S) = Vv The degree of saturation indicates the percentage of the void volume which is filled with water. Thus, a value of S = 0 indicates a dry soil, S = 100% indicates a saturate soil, and a value between 0 and 100% indicates a partially saturated soil. The most useful relationship between phase weights is water content, which is the weight of water divided by the weight of solid in a soil element. W w Water content w = Ws Ws is weight of solid in soil content. The bulk unit weight(γ) is the weight of entire soil element divided by the volume of entire element. Bulk density(or bulk unit weight or moist unit weight or total unit weight) W G(1+w) w t = = V 1+e Where G is specific gravity(unit weight per unit weight of water), w is water content (also 0 referred to as “m” in few books) and γw (also written as rw) unit weight of water at 4 C. For 3 all practical purposes γw is taken as 1 g/cc or 9.8 kN/m . The dry unit weight, often called dry density, is the weight of solid matter divided by the volume of the entire element. Dry density(dry unit weight) WGsw γd = as w = 0 V1e wG Void ratio e = S Saturated unit weight w(G + e) γsat = (as e = wG because S = 1) 1+e Submerged(buoyant) unit weight (G e) (G 1) ww sub sat w1e w 1e You should understand the meanings of these relationships, and add these terms to your active vocabulary. These relationships are basic to most computations in soil mechanics. SECOND FLOOR, SULTAN TOWER, ROORKEE – 247667 UTTARAKHAND PH: (01332) 266328 Web: www.amiestudycircle.com 2/56 WHATSAPP: +91-9412903929 AMIESTUDYCIRCLE.COM [email protected] AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach Example (AMIE S07, 10, W10, 11, 5 marks) Prove the relationship Se = wG Solution V We know S w Vv V And e v Vs VVV Hence Sewv x w VVvs V s W Also we know w w Ws W And G s Vsw WWws WV ww Hence wG x WVsswsws V V Hence Se = wG Example (AMIE W08, 7 marks) (G Se) Prove that w t 1e Solution Ww W1s W WWsw Ws t VVVsv Vv V1s Vs W(1s w) = [Ww/Ws = w and Vv/Vs = e] V(1s e) Se G1w G GSe = w Hence Proved 1e 1e SECOND FLOOR, SULTAN TOWER, ROORKEE – 247667 UTTARAKHAND PH: (01332) 266328 Web: www.amiestudycircle.com 3/56 WHATSAPP: +91-9412903929 AMIESTUDYCIRCLE.COM [email protected] AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach Example G Prove that s w d 1 e Solution WWs ss W W ss G w d VVVsvVv Vs(1 e ) 1 e Vs 1 Vs Example At a certain construction site the natural moisture content is 25%, and void ratio is 0.7. If specific gravity is 2.66, calculate the porosity, moist unit weight, dry unit weight and degree of saturation. Solution w = 25% = 0.25; e = 0.7; G = 2.66, n = ?, t = ?, d = ?, S = ? We know wG Se 0.25x2.66 Sx0.7 S 0.95 95% e0.7 n0.41 1e 10.7 G(1w) 2.66x9.81(1 0.25) w 19.19kN / m3 t 1e 10.7 Example A soil is saturated at 52% moisture content and has a unit weight of 16.5 KN/m3. Calculate its soil ratio, specific gravity, dry unit weight and submerged unit weight. Solution w = 0.52 S = 1 3 t = 16.5 kN/m 16.5 We know 1w t 10.85KN/m3 td dlw 10.52 Se= wG 1 x e = 0.52 x G e = 0.52 G (1) GG Also dww 10.85 1e 1e SECOND FLOOR, SULTAN TOWER, ROORKEE – 247667 UTTARAKHAND PH: (01332) 266328 Web: www.amiestudycircle.com 4/56 WHATSAPP: +91-9412903929 AMIESTUDYCIRCLE.COM [email protected] AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach G 9.81G 10.85 w 1 0.52G 1 0.52G 10-85 + 10.85 x 0.52 G = 9.81G 4.168G = 10.85 G = 2.60 from (1) e = 0.52 G = 0.52 x 2.60 = 1.352 3 sub = sat - w = 16.5 – 9.81 = 6.7 KN/m Example The porosity of a sand sample is 0.4. Assuming a specific gravity of 2.68, compute void ratio, dry unit weight and moist unit weight at 60% saturation. Solution 3 Take t as 9.8 kN/m . Given S = 0.6. en0.4 ne 0.667 1110.4en G 2.68x 9.81 sw 15.77kN / m3 d 1e (1 0.667) ()GSe (2.68 0.6x 0.667) s (9.81) 18.13kN / m3 tw1e (1 0.667) Example For a sand soil with soil grains spherical and uniform in size, determine maximum void ratio. Solution The loosest possible arrangement for round particles of uniform diameter d is shown in Fig. (a), which is its plan view. This is known as simple packing. Each sphere makes contact with six adjacent spheres-four from sides, one from bottom and one from top. Each of the spheres may be assumed to fit within a cube of side dimension dt as shown in Fig.(b). (a) (b) 3 3 Thus, the volume V of the cube is d and the volume of the sphere Vs is d /6. Thus, the maximum void ratio (loosest state) is 33 Vv dd (/6) emax 3 0.91 Vds /6 SECOND FLOOR, SULTAN TOWER, ROORKEE – 247667 UTTARAKHAND PH: (01332) 266328 Web: www.amiestudycircle.com 5/56 WHATSAPP: +91-9412903929 AMIESTUDYCIRCLE.COM [email protected] AMIE(I) STUDY CIRCLE(REGD.) GEOTECHNICAL AND FOUNDATION ENGINEERING INTRODUCTION TO SOIL MECHANICS A Focused Approach Problem The porosity of a sand sample is 0.4. Assuming a specific gravity of 2.68, compute void ratio, dry unit weight and moist unit weight at 60% saturation. Unit weight of water is 9.8 kN/m3. 3 3 Answer: e = 0.667, d = 15.77 kN/m , t = 18.13 kN/m at 60% saturation Example The unit weight of a slightly most soil is 17.15 KN/m3 and its moisture content as determined in the laboratory is 8%. On adding some water to 10 m3 of the soil its water content rises to 17%. Let the specific gravity be 2.65. Determine the qty. of water required. Solution Before adding water 3 t1 = 17.15 KN/m t1 17.15 3 d 15.884kN / m 1w1 1.08 After adding water 3 t2(1 w 2 ) d (1 0.17)x15.884 18.57 kN / m 3 Water added t2 t1 18.57 17.15 1.42kN / m Hence water added per 10 m3 = 1.42 x 10 = 14.2 kN Problem1 A soil sample in its natural state, when fully saturated, has a water content of 32.5 percent. Determine the void ratio, dry and total unit weights. Calculate the total weight of water required to saturate a soil mass of volume 10 m3. Assume G = 2.69. 3 3 Answer: e = 0.874, t = 18.7 kN/m , d = 14.08 kN/m , weight of water required = 46.2 kN Example The undisturbed soil at a borrow pit has a water content of 15 %, void ratio of 0.60 and specific gravity of soil 2.70. The soil from the borrow pit is to be used for construction of an embankment with a finished volume of 40,000 cu m. The specifications for the embankment require a water content of 18% and dry unit weight of 1.76 g/cc. Calculate the quantity of soil required to be excavated for the embankment. 1 3 If unit weight of water is not given in the problem, then take it as w = 9.8 kN/m .
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