Question

A pad foundation 2 m square in plan bears on silty sand with the following properties: ∅'= 30°, y = 16.0 kN/m³ and Yₛₐₜ = 17.5 kN/m³. Foundation level is 1.5 m below ground level and the ground water level is 0.75 m below ground level. The engineer in charge has asked you to identify the safe vertical force that may be applied to the centre of this foundation design?

Answer 1

The safe vertical force that can be applied to the center of the 2 m square pad foundation on silty sand, considering a factor of safety of 3, is approximately 234.33 kN/m².

To determine the safe vertical force that may be applied to the center of the foundation design, we need to consider the bearing capacity of the soil. The bearing capacity is typically calculated using the Terzaghi's bearing capacity equation:

`\[ Q = cN_c + (1)/(2) \gamma B N_q + \gamma D N_\gamma \]`

where:

- Q is the bearing capacity,

- c is the cohesion of the soil (assumed to be zero for cohesionless soils like sand),

-
`\( \gamma \)` is the unit weight of the soil,

- B is the width of the foundation,

- D is the depth of the foundation below the ground level,

-
`\( N_c \), \( N_q \), and \( N_\gamma \)` are bearing capacity factors.

Let's go through the calculations step by step:

1. Determine Effective Unit Weight of Soil
`(\( \gamma' \))`:

`\[ \gamma' = y - \gamma_w \] \[ \gamma' = 16.0 \, \text{kN/m}^3 - 9.81 \, \text{kN/m}^3 \] \[ \gamma' = 6.19 \, \text{kN/m}^3 \]`

2. Determine Bearing Capacity Factors:

-
`\( N_c \), \( N_q \), and \( N_\gamma \)` can be obtained from bearing capacity charts. For a square footing with
`\( \phi' = 30^\circ \)`, they are approximately 9, 18, and 28, respectively.

3. Determine Depth Below Ground Level D':

`\[ D' = D + (B)/(2) \] \[ D' = 1.5 \, \text{m} + \frac{2 \, \text{m}}{2} = 2.5 \, \text{m} \]`

4. Calculate Bearing Capacity:

`\[ Q = cN_c + (1)/(2) \gamma' B N_q + \gamma' D' N_\gamma \] \[ Q = 0 + (1)/(2) * 6.19 \, \text{kN/m}^3 * 2 \, \text{m} * 18 + 6.19 \, \text{kN/m}^3 * 2.5 \, \text{m} * 28 \] \[ Q \approx 703 \, \text{kN/m}^2 \]`

5. Factor of Safety (FS):

Assume a factor of safety, let's say FS = 3.

6. Safe Vertical Force
`(\( Q_{\text{safe}} \))`:

`\[ Q_{\text{safe}} = (Q)/(FS) \] \[ Q_{\text{safe}} = \frac{703 \, \text{kN/m}^2}{3} \] \[ Q_{\text{safe}} \approx 234.33 \, \text{kN/m}^2 \]`

So, the safe vertical force that may be applied to the center of this foundation design is approximately
`\( 234.33 \, \text{kN/m}^2 \)`.