CivilGeotechnicalmediumFE ExamCourseworkPE / Advanced

Active Thrust on a Retaining Wall

A smooth vertical retaining wall retains $H = 5.0\ \text{m}$ of dry cohesionless backfill with a horizontal surface.

Soil properties: $\gamma = 18\ \text{kN/m}^3$ , $\phi = 30^\circ$ , $c = 0$ . The water table is well below the base.

Find the total active thrust per metre of wall using Rankine theory.

Given

H	5 m	Wall height
γ	18 kN/m^3	Unit weight
φ	30 °	Friction angle

Hint 1

Two parts: the earth pressure coefficient, then integrating the pressure over the wall height.

Hint 2

K_a = (1 − sinφ)/(1 + sinφ). Check you have not computed the passive coefficient.

Hint 3

Pressure is triangular, so P_a = ½K_aγH².

Worked solution — try the problem first

Rankine active coefficient:

$K_a = \frac{1 - \sin\phi}{1 + \sin\phi} = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3}$

Total active thrust. Pressure grows linearly with depth, so the resultant is the area of a triangular distribution:

$P_a = \tfrac{1}{2} K_a \gamma H^2 = \tfrac{1}{2}\left(\tfrac{1}{3}\right)(18)(5.0)^2$

$= \tfrac{1}{2}(6)(25) = 75\ \text{kN/m}$

$\boxed{P_a = 75\ \text{kN per metre of wall}}$

Point of application. The thrust acts at the centroid of the triangular distribution, $H/3 = 1.67$ m above the base — not at mid-height. That location governs the overturning moment about the toe.

Common alternative form: $K_a = \tan^2(45° - \phi/2) = \tan^2(30°) = 1/3$ ✓

Concepts:Rankine earth pressureRetaining wallsActive vs passive pressure

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