Lateral Earth pressure
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![Active Earth Pressure
Cohesive Soil
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Lateral Earth pressure
- 1. Active Earth Pressure Cohesive Soil Hc 2C cotα Net pressure γH cot2 α-2C cotα [ ]pressureNetCHHP Cdzdzzp dzCzdzppressureTotalNow CZp KCKp a H O H O a H O H O a a aavaha ααγ ααγ ααγ ααγ σσ cot2cot 2 1 )cot2cot( )cot2cot( cot2cot 2 22 2 2 2 −= −=∴ −= −= −== ∫ ∫ ∫∫
- 2. • Practically crack do occur ,neglect the negative pressure and consider the whole positive pressure ααγ ααγ α γ γ cot)(2cot)( 2 1 )cot2cot( tan 4 2 2 22,0 222 2 ZoHCZoH dzCzP C K C H ZcorZobewillthisP H Zo a a c a −−−= −= = =∴ = ∫
- 4. Lateral Earth Pressure Cohesionless Soil Circle shows the soil at rest position Circle represents the active state Circle represents the passive state O A C σv F σhp φ G D E 1 2 3 σha 1 2 3 B
- 6. Active Earth Pressure Cohesive soil or C-φ Soil Active state At rest τ =c′+σ′tanφ Y φ c σha Z σv σn τ X φ Oc/tanφ Zo Zc Dc -2C cotα γD cot2 -2Ccotα Or γD cot2 -2Ccotα
- 11. Passive Earth Pressure Cohesive soil and C-φ soil Pp 2C tanα γD tan2 α τ =c+σ tanφ Q O σha σv P σhp σn τ CφX
- 15. Coulomb’s Earth Pressure Theory (a) Trial Failure wedge (b)Force A B F C Β-α α 90-θ+α 90+θ-β W D β H θ δ Pa φ 90-θ-δ 90+θ+δ-β+φ Β-φ W Pa F
- 16. Derivation of Coulomb’s Earth Pressure Theory Cohesionless Soil Active case Let AB (Fig.) be the back face of a retaining wall supporting a granular soil, the surface of which is constantly sloping at an angle α with the horizontal. BC is a trial failure surface. In the stability consideration of the probable failure wedge ABC, the following forces are involved(per unit width of the wall):
- 17. 1 W, the weight of the soil wedge. 2 F, the resultant of the shear and normal forces on the surface of failure, BC This is inclined at an angle of φ to the normal drawn to the plane BC. 3 Pa , the active force per unit width of the wall. The direction of Pa is inclined at an angle δ is the angle of friction between the wall and the soil. The force triangle for the wedge is shown in Fig. From the law of sines we have,
- 20. • In the preceding expression for Pa , the values of γ, H, θ, α, φ, and δ are constants, and β is the only variable. To determine the critical value of β for maximum Pa we have )10( )cos(.)cos( )sin(.)sin( 1)cos(.cos )(cos : ' )9( 2 1 : ')7.(int ),8(. )8(0 2 2 2 2 → −+ −+ ++ − = →= →= αθθδ αφφδ θδθ θφ γ β β a a aa a K bygivenistcoefficien pressureearthactivesCoulombisKWhere HKP asobtainedispressureearth activesCoulombEqogsubtitutinis ofiprelationshthewhenEqsolvingAfter d dP
- 21. • Note that when α=0o , θ=0o , and δ=0o • ,Coulomb’s active earth pressure coefficient becomes equal to (1-sinφ)/(1+sinφ) ,horizontal which is the same as the Rankine’s earth pressure coefficient. • The variation of the value of Ka for retaining walls with a vertical back (θ=0o ) and horizontal backfill (α=0o ) is given in table. From this table, not that for a given value ofφ the effect of wall friction is to reduce somewhat the active earth pressure coefficient.
- 22. Table Values of Ka (Eq.10) for θ=0o ,α=0o
- 23. C β-α α 90-θ+α 90+θ-β W D β H θ δ Pp φ A W 90-θ+α β+φ Pp F F [180-(90-θ+δ)-(β+φ)] Coulomb’s passive pressure Trial failure wedge Force polygon
- 24. Passive Case • Fig. shows a retaining wall with a sloping cohesionless backfill. The force polygon for equilibrium of the wedge ABC for passive state is shown in fig.(b). Pp is the notation for the passive force. In a procedure similar to the one we followed in the active case, we get: • Pp = ½ Kp γ H2 --------------------(12) • where Kp = coefficient of passive earth pressure for Coulomb’s case: 2 2 2 )cos()cos( )sin()sin( 1)cos(cos )(cos −− +− −− + = θαθδ αφδφ θδθ θφ pK
- 25. • For a frictionless wall with the vertical back face supporting granular soil backfill with a horizontal surface that is, α = 0o ,θ = 90o and δ = 0o Eq. 12 becomes: • Kp =1+sinφ/1-sinφ = tan2 (45+φ/2) • This is the same relationship that was obtained for the passive earth pressure coefficient in Rankine’s case. • The variation of kp with ф and δ for θ =0 and α = 0 is given in Table. It is also observed from this table that for given values of α and ф the value of Kp increases with the wall friction.
- 26. Table values of Kp (Eq,12) for θ = 0 and α =0 δ(degree) ф(deg) 0 5 10 15 20 15 1.695 1.90 2.130 2.405 2.735 20 2.040 2.313 2.636 2.030 3.525 25 2.464 2.83 3.286 3.855 4.597 30 3.00 3.506 4.143 4.977 6.105 35 3.69 4.390 5.310 6.854 8.324 40 4.600 5.59 6.946 8.870 11.772
- 27. Example-1 • A retaining wall with a smooth vertical back retains sand backfill for a depth of 6 m. The backfill has a horizontal surface and has the following properties: c′= 0, φ′ = 28o ; γ=16 kN/m3 ; γsat =20 kN/m3 . Calculate the magnitude of the total thrust against the wall for the conditions given below: (a) back fill fully drained but the top of the wall is restrained against yielding, (b) backfill fully drained and the wall is free to yield and (c) wall free to yield, water table at 3 m depth and there is no drainage. Determine the point of application of the resultant thrust for case (c)
- 28. Solution: • (a) If the wall is restrained against yielding, the lateral pressure that would develop against the wall would be the earth pressure ‘at rest’. Calculate coefficient of earth pressure at rest Ko using the following Eq. • Ko =1-sinφ′ = 1-sin28o =0.53 • Hence 6 m Po Po (a) PA PA (b)
- 29. For this case, Fig. shows the pressure distribution diagram for this case. Lateral pressure, po =Ko γ. z At z = 6 m, po = o.53x16x6 =50.88 kN/m2 Total thrust per meter length of the wall is Po = 1/2x 50.88x6 = 152.64 kN (b) For this case again, γ= 16 kN/m3 , but the lateral pressure is the active earth pressure. γ=16 kN/m3.
- 30. Fig. shows the pressure distribution diagram. Above the water table, γ =16 kN/m3 and below the water table,γ = γ′=γsat –γw =20- 9.8 =10.2 kN/m3 . The lateral thrust due to water is shown as P3. p1= Ka γ H1= 0.36 x 16 x 3 =17.28 kN/m2 2 2 /68.103656.34 2 1 /56.3461636.0,6 36.0 sin1 sin1 ,28 mkNP mkNpmzAt zKp K A A AA a o =××=∴ =××== = = + − ==′ γ φ φ φ
- 31. p2= Ka γ′ H2=0.36x10.2x3=11 kN/m p3= γw H2 9.8 x3 = 29.4 kN/m2 The total thrusts which are shown in Fig. are calculated below: P1= ½ x17.28 x 3 = 25.92 kN acting at (3+1/3x3) = 4 m from base P′ = 17.28x3 51.84 kN acting at 2/3 = 1.5 m from base P2 = 1/2x11x3 =16.5 kN acting at 1/3x3 =1 m from base P3 = 1/2x29.4x3 = 44.1 kN acting at 1/3x3=1 m from base.
- 32. 3 m 3 m P3 P′ P1 P2 P1 P2 P3 • It can be seen that the lateral thrust due to water contributes substantially to the total lateral thrust. Total thrust = P1+P′+P2 +P3 = 25.92+51.84+16.5+44.1 = 138.36 kN/m length. The distance of the resultant P from the base of the wall can be obtained by taking moments about the base. mH 75.1 36.138 242 36.138 11.4415.165.184.51492.25 == ×+×+×+× =
- 33. Example-2
- 34. • ΔpA = KA . q = 0.36 x 25 = 9 kN/m2 (Fig) Hence the increase in total lateral pressure ΔpA =9 x 6 = 54 kN/m length. The point of application of the thrust is at a distance 1/2X6=3 m from the base. ΔpA ΔpA =9
- 35. Example-3 • A retaining wall with a smooth vertical back is 10 m high and retains a 2-layer sand backfill with the following properties 0-5 m depth: c′ = 0, φ′=30o γ = 18 kN/m3 Below 5 m: c′ = 0, φ′=34o γ = 20 kN/m3 Show the active earth pressure distribution, assuming that the water table is well below the base of the wall.
- 36. Solution: • When the backfill consists of more than one soil layer, the lateral pressure distribution for each of the layers is worked out and a combined diagram drawn. At the interface of two layers, there will be a break in the pressure distribution diagram, since there will be two values of pressure- one value at the base of the upper layer and another at the top of the lower layer. For the lower layer, the upper layer will act as a surcharge.
- 38. • Active pressure distribution for the lower layer: z = 5 m; vertical pressure σv =90 kN/m2 pA = KA2 σv =0.283 x 90 = 25.47 kN/m2 z = 10 m; σv =90 + 20x5 190 kN/m2 pA = 0.283 x 190 = 53.77 kN/m2 The active pressure distribution is shown in Fig. In reality, there can not be a sudden change in lateral pressure since shear stresses which develop along the interface have not been considered. But this does not introduce any serious error in the magnitude and direction of the resultant thrust
- 39. c′=0,φ=30o γ= 18 kN/m3 c′=0,φ=34o γ= 20 kN/m3 5 m 5 m 30 25.47 53.77 Fig. Retaining wall and earth pressure diagram
- 40. Example-4 • A retaining wall, 8 m high, with a smooth vertical back, retains a clay backfill with c ′= 15 kN/m2 ,φ′=15o and γ=18 kN/m3 . Calculate the total active thrust on the wall assuming that tension cracks may develop to the full theoretical depth.
- 41. Solution: • The active pressure at a depth z in a c-φ soil is given by: m K c zatp KczKpmkNpmzAt KcpmkNpzAt K KczKp A oA AAAA AAA o o A AAA 17.2 588.018 1522 0 )2(/67.6123818588.0,8 )2(/23588.0152,0 588.0 15sin1 15sin1 sin1 sin1 2. 2 2 = × = − == −==−××== −=−=×−== = + − = + − = −= γ γ φ φ γ
- 42. 8 m _ 2.17 5.83 23 61.67 • From Fig., it can be seen that the pressure is negative or tensile upto a depth of 2.17 m. The soil tends to break away from the wall and tension cracks develop in the soil. Hence, the resultant active thrust is obtained by determining the area of the hatched portion of the pressure diagram. PA =1/2 x61.67x 5.83 = 179.8 kN/m
- 43. Example-5 An excavation was made in saturated soft clay (φu = 0), with its sides more or less vertical. When the depth of excavation reached 6 m, the sides caved in. What was the approximate value of cohesion of the clay soil? Take unit weight of clay = 20 kN/m3 . Solution: The depth of tension crack zo in a c-φ soil is equal to and the depth for which the total net pressure is zero is equal to 2zo or . AK c γ 2 AK c γ 4
- 44. • An excavation in such a soil should be able to sustain its vertical faces upto a depth of without any lateral support. • Critical depth of cut in a φu= 0 soil is therefore equal to since KA = 1 for φu= 0 . AK c γ 4 γ dc4 2 /30 4 206 4 6 6 4 mkNcor c u u = × == =∴ γ γ
- 45. Example-6 • Fig.(a) shows a three layered backfill behind a 15 m high retaining wall with a smooth vertical back. Draw the active earth pressure distribution. • Layer 1: (c = 0 soil) 41.0 25sin1 25sin1 )(:2 /2727.0100 /100520;5 0. 0;0 27.0 35sin1 35sin1 2 2 = + − =− =×= =×== == == = + − = o o A A v vAA v o o A KsoilcLayer mkNp mkNmz Kp pressureverticalz K φ σ σ σ
- 46. H1 =5m H2 =5m H3 =5m Layer γ c φ (kN/m3 ) (kN/m2 ) 1 20 0 35 2 18 20 25 3 18 35 0 2715.4 52.3 120 200 5x18x0.41=36.9 KA γ.D (a) (b) Fig. (a) Retaining wall with backfill (b) Earth pressure diagram- Example-6
- 47. • z = 5 m; σv = 20 x 5 = 100 kN/m2 σv = KA .σv – 2c √ KA = 0.41 x 100 -2 x 20 √0.41= 15.4 kN/m2 z = 10 m; σv =100 + 18 x 5 =190 kN/m2 σv=0.41 x 190 – 2 x 20 √0.41= 52.3 kN/m2 Or 15.4 + 18 x 5 x 0.41= 15.4+36.9=52.3 Layer -3: KA =1 (φu =0 soil) z=10 m; σv =190 kN/m2 pA =pv -2cu = 190-2 x 35 =120 kN/m2 [:. pA = kA σv -2cu √KA where KA =1]