FOUNDATION ENGINEERING UNIT V

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1 FOUNDATION ENGINEERING UNIT V RETAINING WALLS Plastic equilibrium in soils active and passive states Rankine s theory cohesion less and cohesive soil - Coloumb s wedge theory condition for critical failure plane - Earth pressure on retaining walls of simple configurations Graphical methods (Rebhmann and Culmann) - pressure on the wall due to line load Stability of retaining walls. S.NO Part A PAGE.NO. Define Active Earth pressure. 4 Define Passive Earth pressure. 4 Define coefficient of earth pressure 5 4 Enumerate the assumptions made in Rankine s theory. 5 5 What is the critical height of an unsupported vertical cut in 6 cohesive soil 6 Enumerate the assumptions made in Coulomb s Wedge theory. 6 7 Give the criteria for the design of gravity retaining wall. 6 8 Sketch the variation of earth pressure and coefficient of earth pressure with the movement of the wall 7 9 What are the stability conditions should be checked for the 7 retaining wall 0 Give the minimum factor of safety for the stability of a 7 retainingwall. Draw the various Drainage provisions in Retaining wall 8 If a retaining wall of 5 m high is restrained from yielding, what will be the total earth pressure at rest per metre length of wall? Given: the back fill is cohesion less soil having φ 0 and γ 8 kn/m. A cantilever retaining wall of 7 metre height retains sand. The properties of the sand are γ d 7.66KN/m and γ sat 9.9 KN/m φ 0. using Rankine s theory determine active earth pressure at the base when the backfill is (i) Dry, (ii) Saturated and (iii) Submerged. 4 A rigid retaining wall of 6 m high, has a saturated backfill of soft clay soil. The properties of the clay soil are γsat 7.56 kn/m, unit cohesion cu 8 kn/m. Determine the expected depth of 8 8 9

2 tensile crack in the soil 5 A retaining wall of 6 m high has a saturated backfill of soft clay soil. The properties of the clay soil are γsat 7.56 kn/m, unit cohesion cu 8 kn/m. Determine (a) the expected depth of tensile crack in the soil (b) the active earth pressure before the occurrence of tensile crack, and (c) the active pressure after the occurrence of tensile crack 9 S.NO Part B PAGE.NO. Explain Rankine s Active earth pressure theory for cohesion less soil Explain Rankine s Active earth pressure theory for cohesive soil 4 Explain Rankine s Passive earth pressure theory for cohesion less 6 and cohesive soil 4 Explain coulomb s wedge theory 6 5 Explain Culmann s construction for active pressure of cohesion less soil 6 Explain the Rehbhann s Graphical method for active pressure of cohesion less soil 7 Explain the Effect of line load on retaining wall 8 Determine the coulomb active force on the retaining wall shown in fig. 9 A gravity retaining wall retains m of a back fill, r 7.7 kn/m, r sub 0kN/m. φ 5 with a uniform horizontal surface. Assume the wall interface to be vertical, determine the magnitude and point of application of the total active pressure For an earth retaining structure shown in Fig. Construct earth pressure diagram for active state find the total thrust per unit length of the wall. 4

3 A wall of 8 m height retains sand having a density of.96 Mg/m and angle of internal friction of 4. If the surface of the backfill slopes upwards at 5 to the horizontal, find the active thrust per unit length of the wall. Use Rankine s conditions. A counter fort wall of 0 m height retains non cohesive backfill. The void ratio and angle of internal friction of the backfill respectively are 0.70 and 0 in the loose state and they are 0.40 and 40 in the dense state. Calculate and compare active and passive earth pressures in both the cases. Take specific gravity of soil grains. 5 7 A retaining wall has a vertical back and is 7. m high. The soil is sandy loam of unit weight 7. kn/m. it shows a cohesion of kn/m and φ 0. Neglecting wall friction, determine the thrust on the wall. The upper surface of the fill is horizontal 4 A rigid retaining wall of 6 m height (fig) has two layers of back fill. The top layer up to depth of.5 m is sandy clay having φ 0 C, c 0, and γ7.5 kn/m. A smooth rigid retaining wall of 6 m high carries a uniform 5 surcharge load of kn/m. The backfill is clayey sand possessing the following properties. γ 6.0 kn/m, φ 5, and c 6.5 kn/m 6 for a retaining wall system, the following data were available: (i) Height of wall 7 m. (ii) Properties of backfill: γ d 6 kn/m, φ 5 (iii) Angle of wall friction, δ 0 (iv) Back of wall is inclined at 0 to the vertical (positive batter) (v) Backfill surface is sloping at :

4 4 PART A. Define Active Earth pressure. If the soil exerts a push against the wall by virtue of its tendency to slip laterally and seek its natural slope (angle of repose) thus making the wall to move slightly away from the back filled soil mass. This kind of pressure is known as AEP.. Define Passive Earth pressure. The pressure or resistance which soil develops in response to movement of the structure towards it is called the Passive Earth Pressure.. Define coefficient of earth pressure From Mohr s coulomb equation, σ c 0 σ σ c N φ + σ N φ φ tan 45 + For active case for passive case σ σ v (major) σ σ h (major) σ σ h (minor) σ σ v (minor) σ σ σ σ v h φ tan 45 + Q K a Kp tan (45+ ) φ tan 45 + sin φ sin φ K a Kp + sin φ + sin φ

5 5 4. Enumerate the assumptions made in Rankine s theory.. Soil mass is semi infinitive, homogenous, and cohesion less.. Ground surface is a plane may be horizontal or inclined.. The back of the wall is vertical and smooth. (I.e. no shearing stresses between the wall and the soil) 4. The wall yields about the base and thus satisfies the deformation condition for plastic equilibrium Defects Back of the wall is never smooth, so frictions force develops. the resultant will be inclined to the normal to the wall. 5. What is the critical height of an unsupported vertical cut in cohesive soil given by, The critical height H c of an unsupported vertical cut in cohesive soil is thus H c z o 4c tan α r Where α 45 + φ 6. Enumerate the assumptions made in Coulomb s Wedge theory.. Soil is dry, cohesion less, homogeneous, isotropic and ideally plastic material. The slip surface is a plane passing through the kneel of the wall (failure in active or passive condition occurs when a body of a soil mass known as slip surface). The wall surface is rough. The resultant earth pressure on the wall is inclined at an angle δ to the normal to the wall (δ - angle of friction between wall & back fill). 4. The sliding wedge itself acts as a rigid body

6 6 (The value of earth pressure is obtained by considering the equilibrium) of the sliding wedge as a whole. Retaining wall with inclined back face and a sloping dry granular back fill is shown in figure. 7. Give the criteria for the design of gravity retaining wall.. Maximum pressure should not exceed the bearing capacity of soi (Base width).. No tension should be developed any where in the wall e < b 6 i.e. x b. The wall must be safe against sliding. 4. The wall must be safe against sliding. 5. The wall must be safe against overturning. 8. Sketch the variation of earth pressure and coefficient of earth pressure with the movement of the wall 9. What are the stability conditions should be checked for the retaining walls The stability of retaining walls should be checked fagainst the following conditions (a) The wall should be stable against sliding (b) The wall should be stable against Overturning (c) The wall should be stable against Bearing capacity failure. 0. Give the minimum factor of safety for the stability of a retaining wall. (a) The wall should be stable against sliding.5 (b) The wall should be stable against Overturning For Granular Backfill.5 For cohesive backfill.0 (d) The wall should be stable against Bearing capacity failure. For Granular Backfill.5

7 7 For cohesive backfill.0. Draw the various Drainage provisions in Retaining wall. If a retaining wall of 5 m high is restrained from yielding, what will be the total earth pressure at rest per metre length of wall? Given: the back fill is cohesion less soil having φ 0 and γ 8 kn/m. Solution k o sin φ sin P o H o γ K x0.5x8x5.5kn / m length of wall. A cantilever retaining wall of 7 metre height retains sand. The properties of the sand are γ d 7.66KN/m and γ sat 9.9 KN/m φ 0. using Rankine s theory determine active earth pressure at the base when the backfill is (i) Dry, (ii) Saturated and (iii) Submerged. submerged density γb γsat - γw kN/m sin φ sin 0 for φ 0, K A o + sin φ + sin active earth pressure at the base is (i) for dry backfill, P a K A γ d H x7.66x7 4.kN / m (ii) for saturated backfill, (iii) P a K A γ b H P a K A γ sat H o x0.9x7 for submerged backfill, x.x7 5.9kN / m 48.76kN / m

8 8 4. A rigid retaining wall of 6 m high has a saturated backfill of soft clay soil. The properties of the clay soil are γsat 7.56 kn/m, unit cohesion cu 8 kn/m. Determine the expected depth of tensile crack in the soil Solution At z 0, P a -c - x 8-6kN/m since φ 0 At z H, P a γh c 7.56 x 6 x kn/m. The depth of tensile crack z 0 is (for φ 0) z 0 c γ x m 5. A retaining wall of 6 m high has a saturated backfill of soft clay soil. The properties of the clay soil are γsat 7.56 kn/m, unit cohesion cu 8 kn/m. Determine (a) the expected depth of tensile crack in the soil (b) the active earth pressure before the occurrence of tensile crack, and (c) the active pressure after the occurrence of tensile crack Solution At z 0, P a -c - x 8-6kN/m since φ 0 At z H, P a γh c 7.56 x 6 x kn/m. The active earth pressure before crack occurs. P a γ H ch Since KA for φ 0. substituting, we have P a x7.56x6 x8x kN / m The active earth pressure after the occurrence of tensile crack, Substituting P a ( γ H c)(h z 0 ) P a (7.56x6 x8)(6.05) 7kN / m

9 9 PART B. Explain Rankine s Active earth pressure theory for cohesion less soil Active earth pressure of cohession less soils. Applied to uniform cohesion less soil only [extended to cohesive soil by Resal (90 & Bell (95)] Assumptions of Rankine s theory 4. Soil mass is semi infinitive, homogenous, and cohesion less. 5. Ground surface is a plane may be horizontal or inclined. 6. The back of the wall is vertical and smooth. (I.e. no shearing stresses between the wall and the soil) 7. The wall yields about the base and thus satisfies the deformation condition for plastic equilibrium. Defects Back of the wall is never smooth, so frictions force develops. The resultant will be inclined to the normal to the wall. Dry or moist backfill with no surcharge At z 4 from top, Earth Pa K a r H Total active earth pressure Pa (or) resultant pressure per unit length of the wall, Pa k a rh acting at H from base of wall Submerged backfill P a K a r H +r w H Partially Submerged backfill P a Ka r H + Ka r H + rw H Based on the assumption that φ is same for moist as well as submerged soil.

10 0 Backfill with Uniform surcharge. Backfill with sloping surface K a Cos β Cos φ Cosβ Cosβ + Cos β Cos φ the pressure distribution is triangular. The total active pressure Pa for the wall of height H is given by Pa K a rh The resultant acts at H above the base in direction parallel to the surface shown in figure.. Explain Rankine s Active earth pressure theory for cohesive soil Active earth pressure of cohesive soils. According to bells, σ σ tan α + C tanα P a r C cot φ Q (45+ ) - C Cot (45 + ) at z 0, Q P a -C Cot (45+ ) C When P a 0, z z o tan α r This shows that ve pressure (i.e tension is developed at top level of retaining wall. The tension decreases to zero at a depth Z o For depth, z > z o P a is positive. C tan α r P a r H cot α - ch cot α

11 by c cotα. The effect of cohesion in soil is to reduce the pressure intensity evry where Because of negative pressure, a tension crack is usually developed in the soil near the top of the wall, up to a depth zo. Also the total net pressure upon a depth z 0 is zero. This means that a cohesive soil should be able to stand with a vertical face up to a depth z o without any lateral support. The critical height H c of an unsupported vertical cut in cohesive soil is thus given by, H c z o 4c tan α r Where α 45 + φ Total lateral thrust is given by, Pa rh cot c α - ch cotα + r As cracks do occur and soil does not necessarily remain adhered to the top portion of wall up to height z o, it is usual to neglect the ve pr and consider whole of the +ve pr below z o.. Explain Rankine s Passive earth pressure theory for cohesion less and cohesive soil (a) Cohesion less backup Pp passive earth pressure intensity σ n Pp σ σ v σ rz σ σ tan α Pp rz tan α k p rz. when kp + sin φ sin φ co.eff. of passive earth pressure for backfill with inclined surface at an angle of β. K p Cos β Cosβ Cos β Cos φ Cos β Cos φ

12 (b) Cohesive back fill. Pp rztanα + Ctanα Pp rz Nφ + C φ N φ Nφ tan (45+ ) At z 0 At z H Pp C tan α Pp rh tan α + C tan α Total Pressure Pp r H tan α +c H tanα 4. Explain coulomb s wedge theory Coulomb s wedge theory:- Assumptions (i) Soil is dry, cohesion less, homogeneous, isotrophic and ideally plastic material (ii) The slip surface is a plane passing through the neel of the wall (failure in active or passive condition occurs when a body of a soil mass known as slip surface) (iii) The wall surface is rough. The resultant earth pressure on the wall is inclined at an angle δ to the normal to the wall (δ - angle of friction between wall & back fill). (iv) The sliding wedge itself acts as a rigid baly (The value of earth pressure is obtained by considering the equilibrium) of the sliding wedge as a whole. Retaining wall with inclined back face and a sloping dry granular back fill is shown in figure. In the active case, Sliding wedge ABD moves downward, The reaction R acts upward and inclined at an angle φ with the normal (due to the frictional force which develops in the opposite. Direction of motion, the reaction is inclined).

13 The sliding wedge ABD is in equilibrium under three forces.. weight of wedge W. reaction R on the slip surface BD. Reaction Pa from the wall. We can draw the force triangle for solving the active earth pressure. Using sine law, Pa sin ( α φ' ) sin w ( 80 -{( β - δ + ( α - φ' )) }) Pa w sin ( α - φ' ) sin (80 -β + δ - α + φ' ) Where Pa total active pressure force Weight of the wedge can be calculated by le W area of wedge ( ) x x unit wt. of soil. W H rsin sin ( β + α) β sin ( sin( β + α β + α) - 90) ( 90 α + i) sin + sin( α i) Substituting the value of W in, we will get, Pa H rsin ( β + α) sin ( α - φ' ) sin( β + α 90) sin(90 α + L) x + β + δ α + φ sin β sin ( ') sin( β + α) sin( α + L) The active pressure Pa is maximum when the failure makes an angle α with the horizontal. Such that, P a 0 δα Pa k a rh

14 4 Where Ka sin sin β sin ( β - δ) + ( β + φ' ) sin ( φ ' +δ) sin ( φ- i) sin( β δ) sin ( β + i) 5. Explain Culmann s construction for active pressure of cohesion less soil Procedure. Draw the retaining wall AB to scale.. Draw theφ line BX.. Choose trial plane BC. Determine the wt. of wedge ABC as (r.abc ) w. Represent the weight w on the φ line to scale w Bd. 4. At point d, draw d C making an angle Ψ β-δ with φ line to intersect the trial plane BC at e. 5. Repeat the steps & 4 by taking several planes Bc, Bc, Bc, Bc 4 to get vectors Bd, Bd,Bd,Bd 4 etc., & points e, e, e 4 6. Join B, e, e, e, e 4 to get a smooth curve. This is the pressure curve. 7. Draw a line parallel to the base line (φ- line) and tangential to the pressure curve at point e. 8. Through a draw a vector ed, to make an angle ψ with φ - line. This vector is equal to the active pressure Pa in magnitude, to the same scale as Bd is drawn to represent W. 9. Join B and extend to intersect the ground surface at C. then BC is the failure plane. 6. Explain the Rehbhann s Graphical method for active pressure of cohesion less soil. Draw the ground line and φ - line at angles I and φ respectively with the horizontal to meet in point D.. Draw a semi circle (with) BD as diameter.. Though B, draw a line BH at an angle ψ with BD. Line BH is called the earth pressure on the ψ line.

15 5 4. Through A, draw line AG parallel to the ψ line. 5. Draw GJ perpendicular to BD, to meet the semi circle in J. 6. with B as centre, and BJ as radius, draw an arc to cut BD in E 7. Through E draw EC parallel to the φ - line. BC represents the slip plane 8. With E as centre, and EC as radius, draw an arc to cut BD in K. Join ck. 9. Total active pressure Pa r ( KCE) 7. Explain the Effect of line load on retaining wall To take account the effect of line load due to railway track or a long wall of a building running parallel to the retaining wall. Let the intensity of line load be q per unit length acting at a point C4 distant a from the top of the wall.. At d4 the weight is increased by q, mark d4 which includes the line load.. draw a line from d4 parallel to 4 line and it meets the line BC 4 at e4. similarly add the weight q for all wedges to the right ie d 5, d 6, etc., & mark the corresponding points d 5, d 6 etc., 4. the modified culmann s is given by Be e e e 4 e 4 e 5 e 6 a. if d 4 e 4 < de Then slip occurs along BC and there is no influence o line load on the total pr Pa b. if d4 e4 > de then slip occurs along BC4 and the pr. on the wall is increased by Pa d 4 e 4 de Influence line for thrust increment due to line load. culumann s line is drawn without considering line load. { q is placed at the position C and} q is added to all wedges to its right.. modified culmann s line is drawn & tangent is drawn to it parallel to φ line (e ) 4. at e a tangent drawn (parallel to CD - live) is extended to meet the modified culmann s line (e 4 say) This shows that de d 4 e 4 5. This gives the information that if the line load is placed beyond c 4, there is no effect on the lateral pressure. 6. for other position of line load between A and C 4, Pa is identified as before (4 (ii)). And ordinates are plotted at the location of line load

16 6 7. Pa is maximum when the load is at the face of wall and remains constant with position up to C & then changes; finally it reduces to zero at c 4. This method is used in locating the position of the railway line at a safe distance without increasing earth pr. on wall. 8. Determine the coulomb active force on the retaining wall shown in fig. φ 0 (angle of internal friction) δ 0 (angle of wall friction) r 7.5 kn/m k a k a sin sin β.sin( β δ) x 0.89 Pa k a rh 0.9 ( β + φ) ( φ i) sin( φ + δ) sin sin( β δ)sin( β + i) x x.0 x x7.5 x5 9.9 kn This act at H/ & inclined at 0 to normal i.e 5 m 9. A gravity retaining wall retains m of a back fill, r 7.7 kn/m, r sub 0kN/m. φ 5 with a uniform horizontal surface. Assume the wall interface to be vertical, determine the magnitude and point of application of the total active pressure. If the water table is at a height of 6m, how far do the magnitude and the point of application of active pressure changed?

17 7 I. Dry condition sin φ K a + sin φ K a sin 5 + sin 5 Active earth pr. at base of wall ka r H x 7.7 x 86. kn/m Total active thrust / m run of wall This total pressure acts at II. Water table is at a height of 6m Assuming same φ value, Ka (K a rh) xh x86.x 57.kN H i.e. 4m above the base of the wall. Active earth pr. at 6 m depth x 5.7 x 6 4. kn/m Active earth pr. at m depth 4.+kar sub x 6 +r w x x 0 x x6 6.4kN/m. Total active thrust / m run { x6x4. + 6} + { x 6 x 4.4 } P ( ) kn + { x 6 x 58.9} The height of its point of application Taking moment about the base, P x y 9. x (6+ x 6) x ( 6 ) + 7. x ( x 6) y.6 m

18 8 the total thrust increases by 0.6kN & the pt. of application gets lowered by ( m) 0. For an earth retaining structure shown in Fig. Construct earth pressure diagram for active state find the total thrust per unit length of the wall. Solution sin 0 For φ 0, K A + sin 0 Dry unit weight γ d γb (G ) γ + e o o Gγ w.65 x kN / m + e w.65 x kN / m Assuming the soil above water table be dry, [Refer Fig.] P K A γ d H P K A γ b H P K A x q x5.7x 5.7kN / x9.8x7 x4 m.9kn / m 4.66kN / m P 4 (K A ) wγ w H x9.8x kn/m Total thrust summation of area of different parts of pressure diagram P H + P H + PH + P (H + H) + P4H x x7 + x.9x (7 + ) + x68.7x7 50 kn/m. A wall of 8 m height retains sand having a density of.96 Mg/m and angle of internal friction of 4. If the surface of the backfill slopes upwards at 5 to the horizontal, find the active thrust per unit length of the wall. Use Rankine s conditions.

19 9 Solution There can be two solutions: analytical and graphical. Analytical solution can be obtained from the equations. P a K A H γ Where K A cos β x cos β cos β + cos cos β cos β cos φ φ Where, β 5, cos β and cos β And φ 4 gives cosφ Hence K A x γ.96 x kn/m Hence P A x 0. x 9(8) 89kN/m wall Graphical Solution Vertical stress at a depth z 8 m is γ H cosβ 9 x 8 x cos 5 47 kn/m. Now draw Mohr envelope at an angle of 4 and ground line at an angle of 5 with the horizontal axis. Using a suitable scale plot OP 47 kn/m. I. Centre of circle C lies on the horizontal axis. II. Circle passes through point P, and III. Circle is tangent to the mohr envelope This point P, at which the circle cuts the ground line represents earth pressure. The length OP measures 47.5 kn/m. Hence thrust per unit length, P a x 47.5 x8 90 kn/m. A counter fort wall of 0 m height retains non cohesive backfill. The void ratio and angle of internal friction of the backfill respectively are 0.70

20 0 and 0 in the loose state and they are 0.40 and 40 in the dense state. Calculate and compare active and passive earth pressures in both the cases. Take specific gravity of soil grains. Solution (i) in the loose state, e 0.70 which gives γd Gγ w + e.7 x kN / m sin φ sin 0 For φ 0, K A and K p o + sin φ + sin 0 Max P a K A γ d H x5.6x0 o 5kN / m Max P p K p γ d H x 5.6 x kn/m. (ii) in the dense state, e 0.40, which gives, γ d sin 40 For φ 40, K A + sin 40.7 x kN / o o m 0.7, Kp 4. 6 K A Max P a K A γ d H 0.7 x8.9 x0 4. kn/m and Max. P p 4.6 x 8.9 x kn/m.. A retaining wall has a vertical back and is 7. m high. The soil is sandy loam of unit weight 7. kn/m. it shows a cohesion of kn/m and φ 0. Neglecting wall friction, determine the thrust on the wall. The upper surface of the fill is horizontal.

21 Solution given by When the material exhibits cohesion, the pressure on the wall at depth z is P a γzk A c K A sin φ sin 0 Where K A 0.49, K A sin φ + sin 0 o o When the depth is small the expression for z is negative because of the effect of cohesion as per theory up to a depth z o. The soil is in tension and the soil draws away from the wall. zo c γ N φ c γ K p + sin φ Where K p.04, and K p. 4 sin φ Therefore z o x x.4.98 m 7. Pressure at surface (z 0) is P a -c K A - x x kn/m. The negative sign indicates tension. Pressure at the base of the wall (z 7. m) is P a 7. x 7. x kn/m. As the theory, the area of the upper triangle in fig 4.5 (b) to the left of the pressure axis represents a tensile force which should be subtracted from the compressive force on the lower part of the wall below depth zo. Since tension cannot be applied physically between the soil and the wall, this tensile force is neglected. It is therefore, commonly assumed that the active earth pressure is represented by the shaded area in fig. 4.5 (c). the total pressure on the wall is equal to the area of the triangle in fig. 4.5 (c). P a ( γ HK A c K A )(H z o ) (7.x7.x0.49 xx0.7)(7..98) 0.8 kn/m

22 4. A rigid retaining wall of 6 m height (fig) has two layers of back fill. The top layer up to depth of.5 m is sandy clay having φ 0 C, c 0, and γ7.5 kn/m. Determine the total active earth pressure acting on the wall and draw the pressure distribution diagram. For the top layer o 0 K A tan , K p The depth of tensile zone, z 0 is z 0 c γ K x p.4m Since the depth of the sandy clay layer is.5 m, which is less z 0 the tensile crack develops up to z.5 m only. KA for the sandy layer is K A tan o φ o 0 45 tan 45 At depth z 6.4 x kn/m. σ v γz 6.4 x.5 4.6kN/m. The active pressure is P a K A γz x4.6 8.kN / m At depth 6 m, the effective vertical pressure is σ v.5 x x kn/m. The active pressure distribution diagram is given in Fig 5. A smooth rigid retaining wall of 6 m high carries a uniform surcharge load of kn/m. The backfill is clayey sand possessing the following properties. γ 6.0 kn/m, φ 5, and c 6.5 kn/m. Determine the passive earth pressure and draw the pressure diagram.

23 Solution For φ 5, the value of Kp is + sin φ K p. 47 sin φ , Pp at any depth z is P p γzkp + c K p σ v K At depth z 0, σv kn/m p c K p P p x.47 + x kn/m. at z 6 m, σv + 6 x 6 08 kn/m P p 08 x.47 + x kn/m. The pressure distribution is shown in fig. The total passive pressure Pp acting on the wall is P p 50 x 6 + x6( ) kN / m length of Location of resultant Taking moments about the base P p x h Or h x 6 x 50 x Pp x 6 x 7.95 x.0m wall 6. for a retaining wall system, the following data were available: (i) Height of wall 7 m. (ii) Properties of backfill: γ d 6 kn/m, φ 5 (iii) Angle of wall friction, δ 0 (iv) Back of wall is inclined at 0 to the vertical (positive batter) (v) Backfill surface is sloping at :0 Determine the magnitude of active earth pressure by Culmann s method.

24 4 Solution (a) Fig. shows the wall drawn to a scale of cm m. φ line and pressure lines are also drawn. (a) The trial rupture lines Bc, Bc, Bc, etc., are drawn by making Ac c c c c, etc., cm. (b) The length of perpendicular from B to the backfill surface.6 cm. (c) The area of wedges BAc, BAc, BAc, etc., are respectively equal to ½ (Base lengths Ac, Ac, Ac, etc.,) x perpendicular length. (d) The weights of the wedges in (d) above per metre length of wall may be found out by multiplying the areas by the unit eight of soil and the results are tabulated below. Wedge Weight, kn Wedge Weight, kn BAc.5 BAc4 460 BAc 0 BAc5 575 BAc 45 (e) The weights of the wedges BAc, BAc, etc., are respectively plotted Bd, Bd, etc., on the φ line using, a scale cm 00kN. (f) Lines drawn parallel to the pressure line from points d, d, d etc., meet respectively the trial rupture lines Bc, Bc, Bc etc., at points e, e, e, etc., (g) Pressure locus is drawn passing through points e, e, e, etc., (h) Line zz is drawn tangential to the pressure locus at a point at which this is parallel to the φ line. This point coincides with the point e. (i) e d gives the active earth pressure. Pa 0.9 cm x kn per metre length of wall. (j) Bc is the critical rupture plane