How to Determine increase in Vertical Effective Stress

Reading Time: 3 minutes

There are standard methods to determine the vertical effective stress σ_oˡ (always calculated from the ground surface to the required depth) and increase in the vertical effective stress, Δσˡ (usually calculated from the lower face of the foundation). These can be found in standard textbooks.

For the quick determination of Δσˡ, the method known as 2:1 method would be explained below. This method is based on the assumption that the stress from the foundation spreads out with the vertical-to-horizontal slope of 2:1.

2:1 method explained

For rectangular borings, the diagram in Figure below applies;

From the Figure above, Δσ_Dˡ = Load (P)/Area of depth at D (A) = P/ [(B + D) (L + D)]

Where, P = the load (in ‘kN’) applied on the foundation and A = the area (‘m²’) of the stress distribution at the depth (D).

For circular borings, the diagram in Figure below applies;

From the Figure above, Δσ_Dˡ = Load (P)/Area of depth at D (A) = P/ [π/4 (B + D)²]

Where, P = the load (in ‘kN’) applied on the foundation and B = diameter (in ‘m’) of the foundation.

Example (adapted from Example 4 of Foundation Engineering text: subsoil exploration by Ahmed S. Al-Agha)

Site investigation is to be made for 2500 kN load carried on (3.0 m x 2.0 m)footing. The foundation will be built on layered soil as shown in the figure below, estimate the depth of bore hole. (Assume ?_?= 10 KN/m³).

Given: P = 2500 kN, foundation dimensions = (3 x 2) m

q = P/A=2500/ (3 x 2) =416.67 kN/m²; D_f=1.5m

D₃ = 100−1.5=98.5m

Step 1. Calculating the depth (D₁) at which Δ?_?1 = (?/??) x ?:

(1/ 10) x q = (1/10) x 416.67 = 41.67 kN/m².

The following figure showing the distribution of stress under the foundation at depth (D₁):

The increase in vertical stress (Δσˡ) at depth (D₁) is calculated as follows:

Δσ_D1ˡ = P/A = 2500/ (3 + D₁) x (2 + D₁)

At D₁ ↠ Δσˡ= (1/10) x q ↠ 2500/ (3 + D₁) x (2 + D₁) = 41.67 ↠ D₁ = 5.26 m.

Step 2. Calculating the depth (D₂) at which (Δ?ˡ/?_?ˡ) = ?.??∶

Let D₂ = 3.5 + T (T: distance from the clay layer to reach (D₂).

The effective stress(σ_oˡ) at depth D₂ is calculated as following:

σ_o,D2ˡ = (17 x 1.5) + (17 x 2) + (18.5 − 10) x 1.5 + (16.9 − 10) x T

↠σ_o,D2ˡ = 72.25 + 6.9T ↠ σ_o,D2ˡ = 72.25 + 6.9 x (D₂ − 3.5)

↠ σ_o,D2ˡ = 48.1 + 6.9 D₂

The increase in vertical stress (Δσˡ) at depth (D₂) is calculated as follows: Δ_σD2′=P/A = 2500 (3 + D₂) x (2 + D₂)

At D₂ ↠ (Δσ′/σ_o′) = 0.05 ↠ 2500 (3 + D₂) x (2+D₂) = 0.05 x (48.1 + 6.9 D₂)

↠ D₂ = 15.47 m

So, the value of (D) is the smallest value of D₁, D₂, and D₃ ↠ D = D₁=5.26 m.

↠ D_boring = D_f + D ↠ D_boring = 1.5 + 15.26 = 6.76 m.