Hi,

A lateral thrust is applied by the soil backfilled behind a wall at rest, this thrust is known as the lateral thrust at rest. Walls which are restricted against any movements or moment come under this category. Example can be a basement wall, which is topped with a RCC slab, and therefore restricts its movement. The amount of thrust on the wall depends upon the properties of the soil.

Density of the soil, angle of friction, degree of the consolidation and the level of the water table are the major factors which govern the magnitude of the lateral thrust.

OCR= 2, π = 35°, πΎπ = 19.7 kN/m3. Find the lateral thrust and its location from base.

πΎπ = 1 - sinπ = 1 - sin35° = 0.426

πΎπ(overconsolidated) = πΎπ(normally consolidated )√OCR

πΎπ(overconsolidated) = 0.426√2 = 0.602

At any depth z

π′π£ = z.πΎπ, [pore water pressure u= 0)

π′β = πΎπ*π′π£

Depth, z= 0

π′π£ = 0, π′β = 0

Depth, z= H=10 m

π′π£ = 10*19.7 = 197 kN/m2,

π′β = 0.602*197 = 118.594 kN/m2

Lateral Thrust = Area of the lateral stress diagram. = 1/2 (10*118.59) = 592.97 kN

Lateral thrust acts at the centroid of the lateral stress diagram, i.e. at a height of H/3 from base.

H/3 = 10/3 = 3.333 m from base.

Thanks!

A lateral thrust is applied by the soil backfilled behind a wall at rest, this thrust is known as the lateral thrust at rest. Walls which are restricted against any movements or moment come under this category. Example can be a basement wall, which is topped with a RCC slab, and therefore restricts its movement. The amount of thrust on the wall depends upon the properties of the soil.

Density of the soil, angle of friction, degree of the consolidation and the level of the water table are the major factors which govern the magnitude of the lateral thrust.

**Example:**A 10 m high basement wall, behind which is filled a soil with density 19.7 kN/m3, c=0,OCR= 2, π = 35°, πΎπ = 19.7 kN/m3. Find the lateral thrust and its location from base.

**Solution.**Fig. Lateral Thrust diagram for wall at rest. |

πΎπ(overconsolidated) = πΎπ(normally consolidated )√OCR

πΎπ(overconsolidated) = 0.426√2 = 0.602

At any depth z

π′π£ = z.πΎπ, [pore water pressure u= 0)

π′β = πΎπ*π′π£

Depth, z= 0

π′π£ = 0, π′β = 0

Depth, z= H=10 m

π′π£ = 10*19.7 = 197 kN/m2,

π′β = 0.602*197 = 118.594 kN/m2

Lateral Thrust = Area of the lateral stress diagram. = 1/2 (10*118.59) = 592.97 kN

Lateral thrust acts at the centroid of the lateral stress diagram, i.e. at a height of H/3 from base.

H/3 = 10/3 = 3.333 m from base.

Thanks!