Foundation Analysis LATERAL EARTH PRESSURE

Foundation Analysis LATERAL EARTH PRESSURE

INTRODUCTION Vertical or near-vertical slopes of soil are supported by retaining walls, cantilever sheet-pile walls, sheet-pile bulkheads, braced cuts, and other similar structures. The proper design of these structures requires an estimation of lateral earth pressure, which is a function of several factors, such as a) the type and amount of wall movement, b) the shear strength parameters of the soil, c) the unit weight of the soil, and d) the drainage conditions in the backfill.

INTRODUCTION Lateral earth pressure is a function of wall movement (or relative lateral movement in the backfill soil).

LATERAL EARTH PRESSURE AT REST (No Lateral Movement) Consider a vertical wall of height H, as shown in Figure 7.3, retaining a soil having a unit weight of γ. A uniformly distributed load, q/unit area, is also applied at the ground surface.

LATERAL EARTH PRESSURE AT REST The shear strength of the soil is, (No Lateral Movement) τ = c + σ tan where, c is the cohesion σ is the effective normal stress ϕ is the effective angle of friction At any depth z below the ground surface, the vertical subsurface stress and lateral earth pressure are expressed as, σ 0 = q + γz σ h = K 0 σ 0 + u where, u is the pore water pressure K 0 is the coefficient of at-rest earth pressure

LATERAL EARTH PRESSURE AT REST (No Lateral Movement) For normally consolidated soil (Jaky, 1944) K 0 = 1 sin For overconsolidated soil (Mayne and Kulhawy, 1982) K 0 = (1 sin )OCR sin The total force, P 0, per unit length of the wall can now be obtained from the area of the pressure diagram as, P 0 = P 1 + P 2 = qk 0 H + 1 2 γh2 K 0 The location of the line of action of the resultant force, P 0, can be obtained by taking the moment about the bottom of the wall. Thus, z = P H 1 2 + P H 2 3 Note: If the surcharge q = 0 and the pore water pressure u = 0, the pressure diagram will be a triangle. P 0

LATERAL EARTH PRESSURE AT REST (No Lateral Movement) If the water table is located at a depth z < H, the at-rest pressure diagram will have to be somewhat modified.

PROBLEM SET 10 1. For the retaining wall shown in the figure below, determine the lateral earth force at rest per unit length of the wall. Also determine the location of the resultant force. Assume OCR = 1.

LATERAL EARTH PRESSURE ACTIVE AND PASSIVE Based on assumptions of the intervening forces and the failure mode, different theories have been developed. These theories differ only in terms of the coefficient of lateral earth pressure but operate with similar stress/pressure equations. The three widely-accepted theories are the following: 1. Rankine s 2. Coulomb s 3. Log-spiral

LATERAL EARTH PRESSURE ACTIVE AND PASSIVE Below are the comparison of the three theories and their applicability. Method Failure Mode Wall Friction Active Case Passive Case based on experimentation and actual failure observations Rankine planar no wall friction poor estimate poor estimate Coulomb planar considered good estimate (less) poor estimate Log-spiral curved considered better estimate better estimate

RANKINE S THEORY A Unit weight of soil = γ c' ' tan ' f σ' v σ' h z Assumptions: Vertical frictionless wall Dry homogeneous soil Horizontal backfill B

ACTIVE EARTH PRESSURE A Unit weight of soil = γ c' ' tan ' f σ' v σ' h z If wall AB is allowed to move away from the soil mass gradually, horizontal stress will decrease. Plastic equilibrium in soil refers to the condition where every point in a soil mass is on the verge of failure. B This is represented by Mohr s circle in the subsequent slide.

ACTIVE EARTH PRESSURE c' ' tan ' f ø' Based on the diagram: c' σ'a Koσ o σ o K a ' a ' ' ' ' a 2 0 tan ' 1- sin ' (45 - ) 2 1 sin ' 0 0 tan K a 2 ' ' (45 - ) - 2c' tan (45 - ) 2 2-2c' K a K a is the Rankine active earth pressure coefficient

ACTIVE EARTH PRESSURE

ACTIVE EARTH PRESSURE DISTRIBUTION - 2c' K a - 2c' K a z c 2c' K a z ' 0 K a ' 0 Ka - 2c' Ka

PASSIVE EARTH PRESSURE A B Unit weight of soil = γ c' ' tan ' f σ' v σ' h z If the wall is pushed into the soil mass, the principal stress σ h will increase. On the verge of failure the stress condition on the soil element can be expressed by Mohr s circle b. The lateral earth pressure, σ p, which is the major principal stress, is called Rankine s passive earth pressure.

Shear stress PASSIVE EARTH PRESSURE c' ' tan ' f D b A ' c' O K o σ o a σ o C σ' p Normal stress ' ' p ' p 2 K p tan 0 ' ' 0 0 K tan p 2 ' ' (45 ) 2c' tan (45 ) 2 2 2c' ' 1 sin ' (45 ) 2 1 sin ' K p D K p is the Rankine passive earth pressure coefficient

PASSIVE EARTH PRESSURE

PASSIVE EARTH PRESSURE DISTRIBUTION z For cohesionless soil, p v K p z K p 2c' K p z K p

SPECIAL CASES Submergence: Inclined backfill: Inclined but smooth back face of wall:

SPECIAL CASES Inclined backfill with c -ϕ soil:

ILLUSTRATIVE PROBLEM A frictionless retaining wall is shown in Figure 12.22a. Determine: a) The active force Pa after the tensile crack occurs. b) The passive force Pp. A q = 10 kn/m 2 γ = 15 kn/m 3 ø = 26 o c = 8 kn/m 2 z H = 4 m B Figure 12.22a Frictionless retaining wall AB

SOLUTION a) The active force Pa after the tensile crack occurs. At z = 0 m; At z = 4 m; K a = 1 sinø 1 + sinø σ a = K a σ v 2c σ v = γ z σ v = q = 10 kn/m 2 = 1 si n( 26) 1 + si n( 26) K a = 0. 39 σ a = 0.39(10) 2(8)( 0.39) = 6. 09 kn/m 2 σ v = 10 + 15(4) = 60 kn/m 2 σ a = 0.39(60) 2(8)( 0.39) = 17. 31 kn/m 2

SOLUTION A q = 10 kn/m 2-6.09 kn/m 2 γ = 15 kn/m 3 ø = 26 o y = 1.04 m c = 8 kn/m 2 H = 4 m 4 y = 2.96 m B (a) (b) 17.32 kn/m 2 Figure 12.22 (a) Frictionless retaining wall AB, and (b) active pressure distribution diagram

SOLUTION From Figure 12.22b, -6.09 kn/m 2 6.09 y = 17.31 ; y = 1. 04 m, 4 y = 2. 96 m 4 y The active force Pa after the tensile crack occurs is equal to the area of the active pressure distribution diagram below point C, or P a = 1 2 17.31 2.96 = 25. 62 kn/m 4 y = 2.96 m C y = 1.04 m P a and the location is located at z=0.99 m z = 1 3 2.96 = 0. 99 m 17.32 kn/m 2 (b)

b) The passive force Pp. K p = 1 + sinø 1 sinø SOLUTION = 1 + si n( 26) 1 si n( 26) = 2. 56 σ p = K p σ v + 2c σ v = γ z K p At z = 0 m; σ v = q = 10 kn/m 2 σ p = 2.56(10) + 2(8)( 2.56) = 51. 2 kn/m 2 At z = 4 m; σ v = 10 + 15(4) = 60 kn/m 2 σ p = 2.56(60) + 2(8)( 2.56) = 204. 8 kn/m 2

SOLUTION A q = 10 kn/m 2 51.2 kn/m 2 γ = 15 kn/m 3 ø = 26 o c = 8 kn/m 2 H = 4 m B (a) 51.2 kn/m 2 153.6 kn/m 2 (c) Figure 12.22 (a) Frictionless retaining wall AB, and (c) passive pressure distribution diagram

From Figure 12.22c, SOLUTION The passive force Pp is equal to the area of the passive pressure distribution, or 51.2 kn/m 2 P p = 51.2 4 + 1 2 153.6 4 = 512 kn/m and the location is calculated by taking summation moment at the base, or z=1.6 m P p 512 z = 51.2 4 4 2 + 1 2 153.6 (4) 1 3 (4) z = 1. 6 m 51.2 kn/m 2 153.6 kn/m 2 (c)

FINAL ANSWERS a) The active force Pa after the tensile crack occurs has a magnitude of 25.62 kn per unit length of the frictionless retaining wall and is acting at 0.99 meters above the base. b) The passive force Pp has a magnitude of 512 kn per unit length of the frictionless retaining wall and is acting at 1.6 meters above the base.

PROBLEM SET 10 2. Assume that the retaining wall shown in the figure below can yield sufficiently to develop an active state. Determine the Rankine active force per unit length of the wall and the location of the resultant line of action.

PROBLEM SET 10 3. Assume that the retaining wall shown in the figure below can yield sufficiently to develop passive state. Determine the Rankine passive force per unit length of the wall.

COULOMB S LATERAL EARTH PRESSURE ACTIVE EARTH PRESSURE

COULOMB S LATERAL EARTH PRESSURE PASSIVE EARTH PRESSURE

COULOMB S LATERAL EARTH PRESSURE where, β is the angle the back face is inclined with the horizontal α is the inclination of backfill with the horizontal δ is the wall friction angle ϕ is the angle of internal friction

PROBLEM SET 10 4. A retaining wall shown below has a height of 4.5 m. The unit weight of soil is 16.5 kn/m 3. The angle of internal friction of soil is 36, the wall friction angle is 24, and soil cohesion is 0. The wall is supporting a horizontal backfill. 4.1 Compute the Coulomb s active earth pressure coefficient. 4.2 Compute the Coulomb s active force per unit length of wall. 4.3 Compute the location of the Coulomb s active force from the bottom of the wall.

PROBLEM SET 10 5. A vertical retaining wall has a height of 4 m and is supporting a horizontal backfill. The unit weight of soil is 16.5 kn/m 3. The angle of internal friction of soil is 35, the wall friction angle is 20, and soil cohesion is 0. 5.1 Compute the Coulomb s passive earth pressure coefficient. 5.2 Compute the Coulomb s passive force per unit length of wall perpendicular to the wall. 5.3 Compute the location of the Coulomb s passive force from the bottom of the wall.