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States of Earth Pressure

Lateral earth pressure is the horizontal pressure exerted by soil against a retaining structure. The pressure state depends on wall movement.

  1. At-rest pressure: the wall is restrained and does not move enough to mobilize shear strength.
  2. Active pressure: the wall moves away from the soil, allowing the soil mass to expand.
  3. Passive pressure: the wall moves into the soil, compressing the soil mass.
$$\sigma_h=K\sigma_v$$
$$P=\frac{1}{2}K\gamma H^2$$

For a triangular pressure diagram, the resultant acts at $H/3$ above the base of the wall.

At-Rest Earth Pressure

For normally consolidated soil, Jaky's expression is commonly used. For overconsolidated soil, $K_0$ increases with overconsolidation ratio.

$$K_0=1-\sin\phi$$
$$K_0=\frac{\mu}{1-\mu}$$
$$OCR=\frac{P_c}{P_0}$$
$$K_0=(1-\sin\phi)OCR^{\sin\phi}$$
$$P_0=\frac{1}{2}K_0\gamma H^2$$

Surcharge and Groundwater

A uniform surcharge adds a rectangular lateral pressure diagram. Below the water table, use effective soil pressure plus hydrostatic water pressure.

$$\sigma_{h,q}=Kq$$
$$P_q=KqH$$
$$\gamma'=\gamma_{sat}-\gamma_w$$
$$\sigma_h=K\sigma_v'+u$$
$$u=\gamma_w z_w$$
$$P_w=\frac{1}{2}\gamma_w H_w^2$$

Rankine Pressure for Horizontal Backfill

Rankine theory assumes a vertical, smooth wall with no wall friction for the basic horizontal-backfill case.

$$K_a=\frac{1-\sin\phi}{1+\sin\phi}$$
$$K_a=\tan^2\left(45^\circ-\frac{\phi}{2}\right)$$
$$K_p=\frac{1+\sin\phi}{1-\sin\phi}$$
$$K_p=\tan^2\left(45^\circ+\frac{\phi}{2}\right)$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_p=\frac{1}{2}K_p\gamma H^2$$

Rankine Pressure for Inclined Backfill

For a cohesionless backfill inclined at angle $\alpha$ with the horizontal, Rankine coefficients are modified by the backfill slope.

$$K_a=\cos\alpha\left(\frac{\cos\alpha-\sqrt{\cos^2\alpha-\cos^2\phi}}{\cos\alpha+\sqrt{\cos^2\alpha-\cos^2\phi}}\right)$$
$$K_p=\cos\alpha\left(\frac{\cos\alpha+\sqrt{\cos^2\alpha-\cos^2\phi}}{\cos\alpha-\sqrt{\cos^2\alpha-\cos^2\phi}}\right)$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_p=\frac{1}{2}K_p\gamma H^2$$

Cohesive Backfill and Tension Cracks

Cohesion reduces active pressure near the top of the wall. If the theoretical pressure becomes tensile, a tension crack may form.

$$\sigma_a=K_a\gamma z-2c\sqrt{K_a}$$
$$z_c=\frac{2c}{\gamma\sqrt{K_a}}$$
$$P_a=\frac{1}{2}K_a\gamma H^2-2cH\sqrt{K_a}$$
$$P_a=\frac{1}{2}K_a\gamma(H-z_c)^2\quad \text{after tensile crack occurs}$$
$$K_a=1\quad \text{when }\phi=0^\circ$$

Coulomb Earth Pressure

Coulomb theory includes wall friction and a possible inclined wall back face. For a vertical wall with horizontal backfill, it reduces to a wall-friction correction.

$$\delta=\tan^{-1}(\mu_w)$$
$$K_a=\frac{\sin^2(90^\circ+\phi)}{\sin(90^\circ-\delta)\left[1+\sqrt{\frac{\sin(\phi+\delta)\sin\phi}{\sin(90^\circ-\delta)}}\right]^2}$$
$$K_p=\frac{\sin^2(90^\circ-\phi)}{\sin(90^\circ+\delta)\left[1-\sqrt{\frac{\sin(\phi+\delta)\sin\phi}{\sin(90^\circ+\delta)}}\right]^2}$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_p=\frac{1}{2}K_p\gamma H^2$$

Problem 130: At-Rest Pressure at 6 m

Determine the lateral pressure of soil at rest 6 m below the ground surface. The soil has $\phi=30^\circ$, dry unit weight $17.5$ kN/m3, and a deep water table.

$$\sigma_v=6(17.5)=105\text{ kPa}$$
$$K_0=1-\sin30^\circ=0.50$$
$$\sigma_h=K_0\sigma_v=0.50(105)=52.5\text{ kPa}$$

Answer: $\sigma_h=52.5$ kPa.

Problem 131: Total Lateral Pressure with Water Table

Calculate total lateral pressure at 6 m depth. $K_0=0.40$, water table is 2 m below ground surface, and the extracted solution uses 18.8 kN/m3 above the water table and effective unit weight 9.4 kN/m3 below it.

$$\sigma_h'=K_0[2(18.8)+4(9.4)]$$
$$\sigma_h'=0.40(75.2)=30.08\text{ kPa}$$
$$u=9.81(4)=39.24\text{ kPa}$$
$$\sigma_h=30.08+39.24=69.32\text{ kPa}$$

Answer: Total lateral pressure is 69.32 kPa.

Problem 132: At-Rest Pressure with Water Table at Surface

For sand with water table at the ground surface, compute total at-rest lateral pressure at 5 m. Saturated unit weight is 20.5 kN/m3 and $K_0=0.507$.

$$\sigma_v'=(20.5-9.81)(5)=53.45\text{ kPa}$$
$$\sigma_h=K_0\sigma_v'+\gamma_w z$$
$$\sigma_h=0.507(53.45)+9.81(5)=76.15\text{ kPa}$$

Answer: $\sigma_h=76.15$ kPa.

Problem 133: At-Rest Force on Wall

A 5.4 m retaining wall supports horizontal cohesionless backfill with $\gamma=17.30$ kN/m3 and $\phi=36^\circ$. Determine at-rest force per unit wall length.

$$K_0=1-\sin36^\circ=0.41$$
$$P_0=\frac{1}{2}K_0\gamma H^2$$
$$P_0=\frac{1}{2}(0.41)(17.3)(5.4)^2$$
$$P_0\approx103.4\text{ kN/m}$$

Answer: $P_0\approx103.4$ kN/m.

Problem 134: At-Rest Force with Surcharge

A 6 m vertical wall supports horizontal cohesionless backfill with $\gamma=16$ kN/m3, $\phi=32^\circ$, and uniform surcharge $q=15$ kPa. Determine at-rest lateral force per unit wall length.

$$K_0=1-\sin32^\circ=0.47$$
$$P_q=K_0qH=0.47(15)(6)=42.30\text{ kN/m}$$
$$P_s=\frac{1}{2}K_0\gamma H^2=\frac{1}{2}(0.47)(16)(6)^2=135.36\text{ kN/m}$$
$$P_0=P_q+P_s=177.66\text{ kN/m}$$

Answer: $P_0=177.66$ kN/m.

Problem 135: Restrained Wall Force

A 6 m wall is prevented from moving and supports horizontal backfill as shown in the PDF. The extracted solution uses $\phi=32^\circ$ and $\gamma=16$ kN/m3. Compute lateral force per unit wall length.

$$K_0=1-\sin32^\circ=0.47$$
$$P_0=\frac{1}{2}(0.47)(16)(6)^2$$
$$P_0=135.36\text{ kN/m}$$

Answer: $P_0=135.36$ kN/m.

Problem 136: Friction Angle from Active Coefficient

Using Rankine theory for horizontal backfill, find $\phi$ if $K_a=0.50$.

$$0.50=\frac{1-\sin\phi}{1+\sin\phi}$$
$$0.50+0.50\sin\phi=1-\sin\phi$$
$$\sin\phi=\frac{1}{3}$$
$$\phi=19.5^\circ$$

Answer: $\phi=19.5^\circ$.

Problem 137: Rankine Active Force

A 6 m retaining wall supports horizontal backfill as shown. The extracted solution uses $K_a=0.307$, consistent with $\phi=32^\circ$, and $\gamma=16$ kN/m3. Compute active force.

$$K_a=\frac{1-\sin32^\circ}{1+\sin32^\circ}=0.307$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_a=\frac{1}{2}(0.307)(16)(6)^2=88.42\text{ kN/m}$$

Answer: $P_a=88.42$ kN/m.

Problem 138: Rankine Passive Force

A 6 m retaining wall supports horizontal backfill as shown. Using $\phi=32^\circ$ and $\gamma=16$ kN/m3, compute Rankine passive force.

$$K_p=\frac{1+\sin32^\circ}{1-\sin32^\circ}=3.25$$
$$P_p=\frac{1}{2}K_p\gamma H^2$$
$$P_p=\frac{1}{2}(3.25)(16)(6)^2=936\text{ kN/m}$$

Answer: $P_p=936$ kN/m.

Problem 139: Active Force with Water Table at Top

A vertical 6 m wall retains horizontal backfill with $e=0.60$, $G_s=2.60$, $w=24\%$, and $\phi=25^\circ$. Compute active force if the water table is at the ground surface.

$$K_a=\frac{1-\sin25^\circ}{1+\sin25^\circ}=0.406$$
$$\gamma_{sat}=\frac{G_s+e}{1+e}\gamma_w=\frac{2.60+0.60}{1+0.60}(9.81)=19.62\text{ kN/m}^3$$
$$\gamma'=19.62-9.81=9.81\text{ kN/m}^3$$
$$P_a=\frac{1}{2}K_a\gamma'H^2+\frac{1}{2}\gamma_wH^2$$
$$P_a=\frac{1}{2}(0.406)(9.81)(6)^2+\frac{1}{2}(9.81)(6)^2$$
$$P_a=248.27\text{ kN/m}$$

Answer: $P_a=248.27$ kN/m.

Problem 140: Active Force with Water Table at Bottom

Use the same wall and soil properties as Problem 139, but place the water table at the bottom of the wall. Compute active force.

$$\gamma=\frac{G_s(1+w)}{1+e}\gamma_w$$
$$\gamma=\frac{2.60(1+0.24)}{1+0.60}(9.81)=19.77\text{ kN/m}^3$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_a=\frac{1}{2}(0.406)(19.77)(6)^2=144.48\text{ kN/m}$$

Answer: $P_a=144.48$ kN/m.

Problem 141: Maximum Depth of Tensile Crack

A frictionless vertical wall retaining horizontal backfill is 8 m high. Soil has $\gamma=17.17$ kN/m3, $\phi=20^\circ$, and $c=13$ kPa. Compute maximum tensile crack depth.

$$K_a=\frac{1-\sin20^\circ}{1+\sin20^\circ}=0.490$$
$$z_c=\frac{2c}{\gamma\sqrt{K_a}}$$
$$z_c=\frac{2(13)}{17.17\sqrt{0.490}}=2.16\text{ m}$$

Answer: $z_c=2.16$ m.

Problem 142: Active Force Before Tensile Cracks

A 6 m frictionless wall supports soil with $\gamma=17.4$ kN/m3, $c=14$ kPa, and $\phi=26^\circ$. Compute Rankine active force before tensile cracks occur.

$$K_a=\frac{1-\sin26^\circ}{1+\sin26^\circ}=0.391$$
$$P_a=\frac{1}{2}K_a\gamma H^2-2cH\sqrt{K_a}$$
$$P_a=\frac{1}{2}(0.391)(17.4)(6)^2-2(14)(6)\sqrt{0.391}$$
$$P_a=17.31\text{ kN/m}$$

Answer: $P_a=17.31$ kN/m before cracking.

Problem 143: Lateral Force After Tensile Cracks

A 5.5 m vertical frictionless wall supports soft backfill with $\gamma=15.5$ kN/m3, undrained friction angle $\phi=0^\circ$, and $c=16.6$ kPa. Determine lateral force after tensile cracks occur.

$$K_a=1\quad \text{for }\phi=0^\circ$$
$$z_c=\frac{2(16.6)}{15.5}=2.14\text{ m}$$
$$h=5.5-2.14=3.36\text{ m}$$
$$P_a=\frac{1}{2}\gamma h^2$$
$$P_a=\frac{1}{2}(15.5)(3.36)^2=87.39\text{ kN/m}$$

Answer: $P_a=87.39$ kN/m.

Problem 144: Rankine Active Force for Inclined Backfill

A 9 m wall supports inclined cohesionless backfill making $\alpha=15^\circ$ with the horizontal. The unit weight is 18 kN/m3 and $\phi=30^\circ$. Compute total active force using Rankine theory.

$$K_a=\cos15^\circ\left(\frac{\cos15^\circ-\sqrt{\cos^2 15^\circ-\cos^2 30^\circ}}{\cos15^\circ+\sqrt{\cos^2 15^\circ-\cos^2 30^\circ}}\right)$$
$$K_a=0.373$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_a=\frac{1}{2}(0.373)(18)(9)^2=271.88\text{ kN/m}$$

Answer: $P_a=271.88$ kN/m.

Problem 145: Tensile Crack in Inclined Cohesive Backfill

A frictionless wall supports inclined backfill at $\alpha=5^\circ$ with the horizontal. The soil has $c=10$ kPa, $\gamma=16.5$ kN/m3, and $\phi=30^\circ$. Find the tensile crack depth.

$$K_a=\cos5^\circ\left(\frac{\cos5^\circ-\sqrt{\cos^2 5^\circ-\cos^2 30^\circ}}{\cos5^\circ+\sqrt{\cos^2 5^\circ-\cos^2 30^\circ}}\right)$$
$$K_a=0.337$$
$$z_c=\frac{2c}{\gamma\sqrt{K_a}}$$
$$z_c=\frac{2(10)}{16.5\sqrt{0.337}}=2.09\text{ m}$$

Answer: $z_c=2.09$ m.

Problem 146: Coulomb Active Force with Wall Friction

A 6 m vertical retaining wall has cohesionless horizontal backfill with $\gamma=15.5$ kN/m3, $\phi=30^\circ$, and wall-soil coefficient of friction $\mu_w=0.268$. Compute active static force using Coulomb theory.

$$\delta=\tan^{-1}(0.268)=15.0^\circ$$
$$K_a=\frac{\sin^2(90^\circ+30^\circ)}{\sin(90^\circ-15^\circ)\left[1+\sqrt{\frac{\sin(30^\circ+15^\circ)\sin30^\circ}{\sin(90^\circ-15^\circ)}}\right]^2}$$
$$K_a=0.301$$
$$P_a=\frac{1}{2}K_a\gamma H^2$$
$$P_a=\frac{1}{2}(0.301)(15.5)(6)^2=84.09\text{ kN/m}$$

Answer: $P_a=84.09$ kN/m.

Concept: At-Rest, Active, and Passive Pressures

Three states of lateral earth pressure depend on wall movement. The at-rest condition applies when no lateral strain occurs (e.g., rigid basement walls). Jaky's formula for normally consolidated soil:

$$K_0 = 1 - \sin\phi'$$

When a wall moves away from the soil, stress decreases to the active state (minimum lateral pressure). When a wall is pushed into the soil, stress increases to the passive state (maximum lateral pressure). Rankine's coefficients for a level cohesionless backfill:

$$K_a = \tan^2\!\left(45-\frac{\phi}{2}\right) = \frac{1-\sin\phi}{1+\sin\phi}$$
$$K_p = \tan^2\!\left(45+\frac{\phi}{2}\right) = \frac{1+\sin\phi}{1-\sin\phi}$$

For cohesive backfill ($c$-$\phi$ soil), the active horizontal stress is $\sigma_h = K_a\gamma z - 2c\sqrt{K_a}$, producing a tension zone near the surface to depth $z_c = 2c/(\gamma\sqrt{K_a})$.

Problem: At-Rest Pressure on Rigid Basement Wall

A rigid basement wall retains 3.5 m of backfill. The soil has $\phi=30°$ and $\gamma=18$ kN/m³ and is normally consolidated. Find the at-rest horizontal stress at the base of the wall, total at-rest force per unit length, and its point of application.

$$K_0 = 1 - \sin 30° = 1 - 0.5 = 0.5$$
$$\sigma_{h,\text{base}} = K_0\gamma H = 0.5(18)(3.5) = 31.5\text{ kPa}$$

Total force (triangular pressure diagram):

$$P_0 = \frac{1}{2}\sigma_{h,\text{base}} H = \frac{1}{2}(31.5)(3.5) = 55.1\text{ kN/m}$$

Acts at $H/3$ from the base:

$$\bar{y} = \frac{H}{3} = \frac{3.5}{3} = 1.17\text{ m from base}$$

Answer: $\sigma_{h}=31.5$ kPa at base, $P_0=55.1$ kN/m at 1.17 m above base.

Problem: Rankine Active Force with Uniform Surcharge

A 5.0 m retaining wall has a level backfill with $\phi=32°$, $\gamma=17.5$ kN/m³, and $c=0$. A uniform surcharge $q=20$ kPa acts on the surface. Find total active force per unit length and the location of its resultant from the base.

$$K_a = \frac{1-\sin 32°}{1+\sin 32°} = \frac{0.4701}{1.5299} = 0.3073$$

Surcharge adds a uniform pressure block; self-weight adds a triangular block:

$$P_{a,q} = K_a q H = 0.3073(20)(5) = 30.73\text{ kN/m at }H/2=2.5\text{ m from base}$$
$$P_{a,\gamma} = \frac{1}{2}K_a\gamma H^2 = \frac{1}{2}(0.3073)(17.5)(25) = 67.22\text{ kN/m at }H/3=1.67\text{ m from base}$$
$$P_a = 30.73 + 67.22 = 97.95\text{ kN/m}$$

Moment about base: $30.73(2.5) + 67.22(1.67) = 76.83 + 112.26 = 189.09$ kN·m/m.

$$\bar{y} = \frac{189.09}{97.95} = 1.93\text{ m from base}$$

Answer: $P_a=97.95$ kN/m acting 1.93 m above the base.

Problem: Rankine Active Force with Water Table in Backfill

A 5.0 m retaining wall has a backfill with the water table at 2.0 m below the surface. Above the water table: $\gamma_{dry}=16$ kN/m³. Below: $\gamma_{sat}=19$ kN/m³. Soil has $\phi=30°$, $c=0$. Compute total lateral force (effective soil pressure + water pressure) and its resultant height above the base.

$$K_a = \frac{1-\sin30°}{1+\sin30°} = \frac{0.5}{1.5} = 0.333,\quad \gamma' = 19-9.81 = 9.19\text{ kN/m}^3$$

Effective horizontal pressures at key levels:

$$\sigma'_{h}(z=2\text{ m}) = K_a(16)(2) = 10.67\text{ kPa}$$
$$\sigma'_{h}(z=5\text{ m}) = K_a[16(2)+9.19(3)] = 0.333(59.57) = 19.84\text{ kPa}$$

Four force components and their heights from base ($H=5$ m):

$$F_1 = \tfrac{1}{2}(10.67)(2) = 10.67\text{ kN/m at }4.33\text{ m}$$
$$F_2 = (10.67)(3) = 32.01\text{ kN/m at }1.50\text{ m}$$
$$F_3 = \tfrac{1}{2}(19.84-10.67)(3) = 13.76\text{ kN/m at }1.00\text{ m}$$
$$F_4 = \tfrac{1}{2}(9.81\times3)(3) = 44.15\text{ kN/m at }1.00\text{ m (water)}$$
$$P_{\text{total}} = 10.67+32.01+13.76+44.15 = 100.6\text{ kN/m}$$

Moment about base $= 10.67(4.33)+32.01(1.5)+13.76(1.0)+44.15(1.0)=152.2$ kN·m/m.

$$\bar{y} = \frac{152.2}{100.6} = 1.51\text{ m from base}$$

Answer: Total lateral force $= 100.6$ kN/m acting 1.51 m above the base.

Problem: Retaining Wall — Factors of Safety

A gravity retaining wall with height $H=5$ m has a net active force $P_a=65$ kN/m acting at 1.80 m above the base. The total weight of wall plus retained soil is $W=200$ kN/m acting 1.00 m from the toe. The base width is $B=2.8$ m and base friction angle $\delta_b=28°$. Passive resistance is neglected. Find FS against overturning about the toe and FS against sliding.

Overturning about toe:

$$M_O = P_a \times 1.80 = 65(1.80) = 117\text{ kN·m/m}$$
$$M_R = W \times 1.00 = 200(1.00) = 200\text{ kN·m/m}$$
$$FS_{OT} = \frac{M_R}{M_O} = \frac{200}{117} = 1.71$$

Sliding:

$$FS_{sliding} = \frac{W\tan\delta_b}{P_a} = \frac{200\tan28°}{65} = \frac{200(0.5317)}{65} = \frac{106.3}{65} = 1.64$$

Answer: $FS_{OT}=1.71$ and $FS_{\text{sliding}}=1.64$. Both exceed 1.5 — wall is adequate.

Problem: Active Force on Cohesive Backfill with Tension Crack

A vertical wall retains a 6.0 m high cohesive backfill with $c=25$ kPa, $\phi=20°$, and $\gamma=18$ kN/m³. Find: (a) the tension crack depth $z_c$, (b) net active force assuming the crack forms but remains dry, and (c) net active force if the tension crack fills with water.

$$K_a = \tan^2\!\left(45-10°\right) = \tan^2 35° = 0.490,\quad \sqrt{K_a} = 0.700$$

(a) Tension crack depth:

$$z_c = \frac{2c}{\gamma\sqrt{K_a}} = \frac{2(25)}{18(0.700)} = \frac{50}{12.60} = 3.97\text{ m}$$

Active pressure at base: $\sigma_h = K_a\gamma H - 2c\sqrt{K_a} = 0.490(18)(6) - 2(25)(0.700) = 52.92 - 35.00 = 17.92$ kPa.

(b) Net active force (dry crack):

$$P_a = \frac{1}{2}(17.92)(6-3.97) = \frac{1}{2}(17.92)(2.03) = 18.2\text{ kN/m}$$

(c) Water in tension crack (triangular water pressure, depth $z_c$):

$$P_w = \frac{1}{2}\gamma_w z_c^2 = \frac{1}{2}(9.81)(3.97)^2 = 77.3\text{ kN/m}$$
$$P_{\text{total}} = 18.2 + 77.3 = 95.5\text{ kN/m}$$

Answer: $z_c=3.97$ m; dry crack $P_a=18.2$ kN/m; water-filled crack $P_{\text{total}}=95.5$ kN/m.