Settlement of Shallow Foundations

The settlement of shallow foundations can be divided into two categories based on the time frame of occurrence. The settlement of a structure cannot be avoided even we build the structure on a rock.

In the pile design also a certain amount of settlement could occur due to the failures of construction with the increase of the loading.

Buildings constructed on a soil definitely settle and it should be within the limit considered in the recommendation to foundation design.

The designer’s responsibility is to maintain a uniform settlement over the building as the different settlement creates issues.

Soil act as a spring which referred to as by subgrade reaction in calculations and does the vertical movement of the structure with the application of loads. As explained above, it could occur in two-stage as below.

1. Elastic settlement or immediate settlement
2. Consolidation settlement

In this article, we discussed the elastic settlement or the immediate settlement of shallow foundations with a worked example.

Elastic Settlement of Shallow Foundations

Elastic settlement occurs during the construction of the structure and immediately after the construction of the structure.

However, consolidation settlement occurs over a period of time. It occurs due to the reduction of the pour water pressure in the saturated clay. Consolidation settlement occurs in two phases namely they are primary and secondary consolidation.

The calculation of the elastic settlement is discussed in this article. The method of calculating elastic settlement based on the theory of elasticity is discussed with the worked example for ease of understanding.

Method of Calculation of Elastic Settlement

Elastic Settlement, Se, as per the book Principles of Foundation Engineering

S_e =q₀(αB’)[(1-μ²_s)/E_s]I_sI_f

Where,

q₀ – Net applied pressure on the foundation

μ_s – Poisson’s ratio of soil

E_s – Average modulus of elasticity of the soil under the foundation, measured form Z=0 to about Z =4B

B’ – B/2 for the center of the foundation and ‘B’ for conner of foundation

I_s – Shape factor (Steinbrenner, 1934)

I_s = F₁ + [(1-2μ_s )/(1- μ_s )]F₂

F₁ = (1/π)[A₀ + A₁]

F₂ = (n’/2π) tan^-1A₂

A₀ = m’ ln { [ 1+(m’²+1)^0.5](m’² + n’²)^0.5 } / { m'[1+(m’² +n’² + 1)^0.5]}

A₁ = ln { [ m’ + (m’²+1)^0.5 ] (1 + n’²)^0.5 } / [m’+(m’² +n’² + 1)^0.5]

A₂ = m’ / [ n’ (m’² +n’² + 1)^0.5 ]

I_f = depth factor (Fox, 1948) = f (D_f/B, μ_s , and L/B

α = a factor that depends on the location on the foundation where settlement is being calculated

Following values are used to calculated the settlement at the center of the foundation.

α = 4, m’ = L/B, and n’ = H / (B/2)

Settlement at the corner of the foundation can be calculated using follwining values.

α = 1, m’ = L/B, and n’ = H / B

Due to the existance of deffenet layers of soil under the foundation, E_s will vary layer to layer. Weighted averation of the E_s is considered for the calculations as recommended by Bowels (1987). E_s can be calculated following equation.

E_s = [ ∑ E_s(i) ∆z ] / Z₀

Where,

E_s(i) = soil modulus of elasticity within a depth ∆z

Z₀ = H or 5B, whichever is smaller

The above equation and theories are extracted from the book Principles of Foundation Engineering.

The Wikipedia article on Foundation (Engineering) discusses the type of foundations to be used in construction.

Worked Example Calculation of Settlement of Shallow Foundations at the Center

Data

• Dimensions of foundation 1.5m x 2m
• Net applied pressure on the foundation, q₀ = 175 kN/m²
• Poisson’s ratio of soil μ_s = 0.3

B = 1.5 m

L = 2m

Average, E_s

E_s = ( 8000 x 2 + 12000 x 2 + 10000 x 2) / 6 = 10000 kN/ m²

α = 4

m’ = L/B = 2 / 1.5 = 1.333

n’ = H / (B/2) = 6 / (1.5/2) = 8

F₁ and F₂ can be calculated from the above equations after calculating A₀ , A_1, and A2. Or the tables given in the book Principles of foundation engineering could be referred.

A₀ = m’ ln { [ 1+(m’²+1)^0.5](m’² + n’²)^0.5 } / { m'[1+(m’² +n’² + 1)^0.5]}

A₀ = 1.333 x ln { [ 1+(1.333²+1)^0.5](1.333² + 8²)^0.5 } / { 1.333[1+(1.333² +8² + 1)^0.5]}

A₀ = 0.760

A₁ = ln { [ m’ + (m’²+1)^0.5 ] (1 + n’²)^0.5 } / [m’+(m’² +n’² + 1)^0.5]

A₁ = ln { [ 1.333 + (1.333²+1)^0.5 ] (1 + 8²)^0.5 } / [1.333+(1.333² +8² + 1)^0.5]

A₁ = 0.934

A₂ = m’ / [ n’ (m’² +n’² + 1)^0.5 ]

A₂ = 1.333 / [ 8 (1.333² +8² + 1)^0.5 ]

A₂ = 0.020

F₁ = (1/π)[A₀ + A₁] = (1/π)[0.760 + 0.934] = 0.539

F₂ = (n’/2π) tan^-1A₂

Firstly, calcuate the tan^-1A₂ in degrees and it conver to radins

tan^-1A₂ = tan^-1(0.02) = 1.146⁰

Radian value = 1.146⁰ x ( π / 180 ) = 0.02

F₂ = (n’/2π) tan^-1A₂ = (8 / 2π) x 0.02 = 0.025

I_s = F₁ + [(1-2μ_s )/(1- μ_s )]F₂ = 0.539 + [(1-2×0.3 )/(1- 0.3 )] x 0.025 = 0.553

I_f can be evaluated from the Table given in the principles of foundation engieering.

For μ_s = 0.3, Df / B = 1/1.5 = 0.667 and B/L = 1.5/2 = 0.75

I_f = 0.755

Foundation Settlement

S_e =q₀(αB’) [(1-μ²_s) / E_s ]I_sI_f

at center of the foundaiton

S_e =q₀(αB/2)[(1-μ²_s)/E_s]I_sI_f = 175(4×1.5/2)[(1-0.3²)/10000] x 0.553 x 0.755 = 0.02m

Settlement of the foundation is 20mm.

Reference: Principles of Foundation Engineering