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Permeability and Seepage -2 13 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Permeability and Seepage -2 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Conditions favourable for the formation quick sand Quick sand is not a type of sand but a flow condition occurring within a cohesion-less soil when its effective stress is reduced to zero due to upward flow of water. Quick sand occurs in nature when water is being forced upward under pressurized conditions. In this case, the pressure of the escaping water exceeds the weight of the soil and the sand grains are forced apart. The result is that the soil has no capability to support a load. Why does quick condition or boiling occurs mostly in fine sands or silts? Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some practical examples of quick conditions Excavations in granular materials behind cofferdams alongside rivers Any place where artesian pressures exist (i.e. where head of water is greater than the usual static water pressure). -- When a pervious underground structure is continuous and connected to a place where head is higher. Behind river embankments to protect floods Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some practical examples of quick conditions Contrary to popular belief, it is not possible to drown in quick sand, because the density of quick sand is much greater than that of water. ρquicksand >> ρwater Consequently, it is literally impossible for a person to be sucked into quicksand and disappear. So, a person walking into quicksand would sink to about waist depth and then float. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Boiling condition • Example • The sand layer of the soil profile shown in figure Water level is under artesian pressure. 4m A trench is to be 6m h 5m excavated in the clay up to a depth of 4m. stiff clay γ = 18kN/m3 Determine the depth of P 3m water h to avoid boiling. Sand Rock Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Boiling condition • Solution: • At P: σ = hγw + (6-4)γclay • = hγw + 2γclay Water level • At P: u = 5γw 4m ’ • At P: σ = hγw + 2γclay - 5γw 6m h = 1.33 m 5m • σ’ = 0 (boiling condition) stiff clay γ = 18kN/m3 • hγw + 2γclay - 5γw = 0 P • h = (5γ - 2γ )/γ 3m w clay w Sand • = (5 x 9.81 – 2x18)/9.81 Rock • h = 1.33m Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay B Seepage forces (F ) Water flowing past a h s c soil particle exerts a drag Wd force on the particle in the direction of flow. L The drag force is caused by pressure gradient and by viscous drag. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage forces B (F ) Direction W s c of flow Wd FBD of grain FBD of soil At critical condition; h = hc h G −1 Gs + e c s γ AL = (h + L)Aγ ic = = 1+ e w c w L 1+ e Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Considering FBD of grain in the direction of flow at i = ic Wd = (Fs )c + B G −1 (F ) = W − B (F ) = s γ V s c d s c 1+ e w Gsγ w (F ) = AL −V γ (Fs )c = icγ wV s c 1+ e s w G γ (V −V ) (F ) = s w AL − V Vγ Seepage pressure s c 1+ e V w ps = icγw G γ (F ) = s w AL − (1− n)ALγ s c 1+ e w If h < hc then seepage force Js is iγwV Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage force Area per unit width of section = ∆l x 1 h1 > h2 Volume affected by ∆h seepage force = ∆l2 x 1 h 1 h2 Direction of ∆l flow Force applied to sand particles = γwh1 -γwh2 (∆l x 1) ∆l Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage force Force applied to sand particles = γw (∆h) (∆l x 1) 2 = γw (∆h/∆l) (∆l x 1) J = i γw V Seepage pressure ps = Seepage force per unit volume = i γw Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Critical hydraulic gradient – Quick condition The quick condition occurs at a critical upward hydraulic gradient ic, when the seepage force just balances the buoyant weight of an element of soil. (Shear stresses on the sides of the element are neglected.) G −1 i = s c υ The critical hydraulic gradient is typically around 1.0 for many soils. Fluidized beds in chemical engineering systems rely on deliberate generation of quick conditions to ensure that the chemical process can occur most efficiently. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem Determine and plot total h stress, PWP and effective a stress diagrams if: (i) h = 1 m; (ii) h = 4 m; and (iii) h = 2 m. 2 m h b i = 1 m 3 2 γsat =20 kN/m c Case – I; i = 0.5; 1 m d Case – II; i = 2; Datum Case – III; i = 1 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - I i = 0.5 Point PH [m] EH [m] TH [m] a 0 4 4 b 2 2 4 d 5 0 5 c 3.5 1 4.5 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - I i = 0.5 h = 1 m σ u σ´ a 2 m b 20 1 m 20 γ =20 kN/m3 c sat 1 m 40 35 5 d 60 50 10 Datum Stress units are in kN/m2 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - II i = 2 Point PH [m] EH [m] TH [m] a 0 4 4 b 2 2 4 d 8 0 8 c 5 1 6 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - II Stress units are in kN/m2 h = 4 m σ u σ´ a 2 m b 20 1 m 20 γ =20 kN/m3 c sat -10 1 m 40 50 d -20 60 80 Datum Quick sand condition would have already occurred… Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - III i = 1 Point PH [m] EH [m] TH [m] a 0 4 4 b 2 2 4 d 6 0 6 c 4 1 5 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - III Stress units are in kN/m2 h = 2 m σ u σ´ a 0 2 m b 0 20 1 m 20 γ =20 kN/m3 c sat 0 1 m 40 40 d 0 60 60 Datum Just subjected to Quick sand condition… Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem A large open excavation was made in a stratum of clay with a saturated unit weight of 17.6 kN/m3. When the depth of the excavation reached 7.5 m, the bottom rose, gradually cracked and was flooded from below by a mixture of sand and water. Subsequent borings showed that the clay was underlain by a bed of sand with its surface at a depth of 11 m. Compute the elevation for which the water would have risen from the sand surface into a drill hole before the excavation was started. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution 11m 3 γsat = 17.6 kN/m Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution Elevation for which the water would have risen 7.5 m from the sand surface into a drill hole before the excavation was h started. h = 6.16 m A For σ´= 0 at A: σ´= 17.6 x (11 – 7.5) - 10 x h = 0 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem Determine and plot the total vertical stress, pore water and effective vertical stress distribution at levels A, B and C. 4 m A 6 m B 3 m Silty clay; Gs = 2.70; w = 45.2 % C 3 m Coarse sand; Gs = 2.65; e = 0.833 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem An artesian pressure 1 m exists in the lower 1 m γ = 18.5 kN/m3 sand layer. d Determine σ, u, and 3 2 m γsat = 19 kN/m σ ′ at A, B, and C. A (TH)A = 2 + 3 = 5m 1.33 m 2 m (TH) = 7 + 0 = 7m 3 B C γclay = 17 kN/m 2 m 1 m B DATUM Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution At point B: σ = 18.5 x 1 + 19 x 2 + 17 x 3 = 107.5 kN/m2 u = 10 x 7 = 70 kN/m2 σ′ = 107.5 – 70 = 37.5 kN/m2 At point A: σ = 18.5 x 1 + 19 x 2 = 56.5 kN/m2 u = 10 x 2 = 20 kN/m2 σ′ = 56.5 – 20 = 36.5 kN/m2 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution At point c: σ = 18.5 x 1 + 19 x 2 + 17 x 2 = 90.5 kN/m2 2 u = 10 x (PH)c = 10 x 5.33 = 53.3 kN/m σ′ = 107.5 – 53.3 = 38.2 kN/m2 (TH)C = (TH)B - i Z = 7 – (2/3) x 1 = 6.33 m (PH)C = (TH)C – Zc = 6.33 – 1 = 5.33 m Prof.
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