To calculate the force required to push the frame forward, we need to consider the shear resistance from the shear plane. The shear resistance (τ) can be calculated using the Mohr-Coulomb failure criterion, which is given by:

τ = c' + σ'tan(φ')

where:
c' is the soil cohesion,
σ' is the effective normal stress, and
φ' is the friction angle.

Given that the cohesion c' = 0 (as it's mentioned in the problem), the shear resistance simplifies to:

τ = σ'tan(φ')

The effective normal stress (σ') is the difference between the total vertical stress and the pore water pressure. In this case, since the excavation is dry, the pore water pressure is zero. Therefore, the effective normal stress is equal to the total vertical stress, which is the product of the depth of the excavation and the dry unit weight of the soil (γdry).

σ' = γdry * h
σ' = 18 kN/m³ * 3 m
σ' = 54 kN/m²

Substituting σ' and φ' into the equation for τ:

τ = 54 kN/m² * tan(35°)
τ ≈ 37.2 kN/m²

The total shear force (F) is the product of the shear resistance and the area of the shear plane. The area of the shear plane is the product of the width of the caisson frame and the depth of the excavation.

A = 10 m * 3 m
A = 30 m²

Therefore, the total shear force is:

F = τ * A
F = 37.2 kN/m² * 30 m²
F ≈ 1116 kN

So, the force required to push the frame forward is approximately 1116 kN.

Question

To calculate the force required to push the frame forward, we need to consider the shear resistance from the shear plane. The shear resistance (τ) can be calculated using the Mohr-Coulomb failure criterion, which is given by:

τ = c' + σ'tan(φ')

where:
c' is the soil cohesion,
σ' is the effective normal stress, and
φ' is the friction angle.

Given that the cohesion c' = 0 (as it's mentioned in the problem), the shear resistance simplifies to:

τ = σ'tan(φ')

The effective normal stress (σ') is the difference between the total vertical stress and the pore water pressure. In this case, since the excavation is dry, the pore water pressure is zero. Therefore, the effective normal stress is equal to the total vertical stress, which is the product of the depth of the excavation and the dry unit weight of the soil (γdry).

σ' = γdry * h
σ' = 18 kN/m³ * 3 m
σ' = 54 kN/m²

Substituting σ' and φ' into the equation for τ:

τ = 54 kN/m² * tan(35°)
τ ≈ 37.2 kN/m²

The total shear force (F) is the product of the shear resistance and the area of the shear plane. The area of the shear plane is the product of the width of the caisson frame and the depth of the excavation.

A = 10 m * 3 m
A = 30 m²

Therefore, the total shear force is:

F = τ * A
F = 37.2 kN/m² * 30 m²
F ≈ 1116 kN

So, the force required to push the frame forward is approximately 1116 kN.

Knowee AI · Accepted Answer