[Consideration of given Text]
Given data:
- Height of the wall (H) = 5.8m
- Top of the wall = 610mm
- Front face batter = 1H:6V
- Angle of internal friction (φ) = 35°
- Unit weight of sand (γ) = 18.9 kN/m³
- Unit weight of concrete (γ_concrete) = 26.3 kN/m³
- Coefficient of friction (μ) = 0.55
To find:
- Base width of the wall (B) so that the resultant of earth pressure and weight of the wall will pass through the outside third point of the bar
- Check the safety of the foundation against sliding
Solution:
We can start by drawing a free-body diagram of the wall and the soil behind it. Let's assume that the wall has a rectangular cross-section with base width "B" and height "H". The weight of the wall (W_wall) can be calculated as:
W_wall = γ_concrete * B * H
The horizontal and vertical forces acting on the soil are shown in the figure below:
The horizontal force is the earth pressure acting on the wall, given by:
P = K * γ * H^2 / 2
where K is the earth pressure coefficient, which can be calculated using Coulomb's active earth pressure theory:
K = tan^2(45° - φ/2)
Substituting the given values, we get:
K = tan^2(45° - 35°/2) = 0.216
Therefore,
P = 0.216 * 18.9 * 5.8^2 / 2 = 347.7 kN
The vertical force is the weight of the soil, given by:
W_soil = γ * B * H^2 / 2
To find the base width "B" that will make the resultant of these forces pass through the outside third point of the bar, we need to find the location of the resultant. We can do this by equating the moments of the forces about the bottom of the wall.
Let X be the distance from the bottom of the wall to the location of the resultant. Then:
P * X + W_soil * (H/3 + X) = W_wall * (H/2 + 610mm)
Substituting the equations for P, W_soil, and W_wall and simplifying, we get:
0.216 * 18.9 * 5.8^3 / 6 * X + 18.9 * B * 5.8^3 / 18 * (1 + 3X/H) = 26.3 * B * 5.8^2 * (1 + 610mm/H)
Solving for "B", we get:
B = [26.3 * 5.8 * (1 + 610/5800) - 0.216 * 18.9 * 5.8^2 / 6] / [18.9 * 5.8 / 6 * (1 + 3X/5.8)]
To find the value of X, we can use the fact that the resultant passes through the outside third point of the bar, which is located at a distance of 2H/3 from the bottom of the wall. Therefore:
X = 2H/3 - H / [3 * (1 + tan(35°))]
Substituting the given values, we get:
X = 2.93m
Substituting X into the equation for "B", we get:
B = 3.98m
Therefore, the required base width for the wall to maintain stability is 3.98 meters.
To check the safety of the foundation against sliding, we need to calculate the resisting force and the driving force acting on the base of the wall. The resisting force is the weight of the soil and the frictional force acting on the base, given by:
R = γ * B * H^2 / 2 * (μ + tan(φ))
The driving force is the horizontal force acting on the wall, given by:
D = P
where P is the earth pressure calculated earlier.
If R > D, then the foundation is safe against sliding. Substituting the given values, we get:
R = 18.9 * 3.98 * 5.8^2 / 2 * (0.55 + tan(35°)) = 1089 kN
D = 347.7 kN
Since R > D, the foundation is safe against sliding.