DETERMINATION OF IMMEDIATE
SETTLEMENT
Suez Halder
Email: suez_301@yahoo.com
By
SIMPLIFIED DEFINITION
When the settlement of the footing takes place
immediately after the load is applied over the footing, it
is termed as immediate settlement.
When Occurs?
Immediate settlement occurs in the case of shallow
footings, having sand as base of the footing.
IMMEDIATE SETTLEMENT CALCULATION
According to the theory of elasticity,
Si=
𝑞𝐵
𝐸
1 − µ2 If
Where,
q= Net foundation pressure
B= Width of the foundation
𝜇= Poisson’s ration
E= Young’s modulus of soil
If= Influence factor
SIGNIFICANCE OF INFLUENCE FACTOR
(If)
We have to calculate Influence Factor (If) for different
shapes of footing, i.e. square, circular, rectangular and
for different types of foundation. Such as-
1. Flexible type foundation. Ex: Isolated Footing
2. Rigid Type Foundation. Ex: Raft or Matt Foundation
ELASTIC SETTLEMENT OF FLEXIBLE AND
RIGID FOUNDATIONS
EXPLANATION
For Flexible Foundation-
Settlement value at the corner of the footing and the center of the
footing are not the same. Here, settlement value of the center is
more than the corner of the foundation. So we have to calculate
influence factor individually for corner and center. Also we have to
calculate the average If .
For Rigid Foundation-
The settlement value is uniform throughout corner and center. So
the influence factor (If ) will be 1.0
It is observed that-
At center,
Flexible Foundation Settlement > Rigid Foundation
Settlement
INFLUENCE FACTORS FOR
FOUNDATIONS
Influence Factor, 𝑰 𝒑
Flexible
Shape 𝐿
𝐵
Center Corner Rigid
Circle ---- 1.00 0.64 0.79
Square 1 1.12 0.56 0.88
Rectangle 1.5 1.36 0.68 1.07
2 1.53 0.77 1.21
3 1.78 0.89 1.42
5 2.10 1.05 1.70
10 2.54 1.27 2.10
20 2.99 1.49 2.46
50 3.57 1.80 3.00
100 4.01 2.00 3.43
TWO TYPES OF CORRECTIONS
1. Rigidity Correction (for rigid foundation)
If for Rigid Foundation= 0.8 × If for the Flexible Foundation at
Center
We have to first calculate the settlement considering the
influence factor of the flexible foundation at the center, then
we will multiply that settlement calculation by 0.8 and will
get settlement of rigid foundation.
2. Depth Correction
PROBLEM
+0m
-2.5m
-7m
-19m
50 kN/m2
G.L
G.W.T
10m
4.5m
12m
16.5
2.5m
Cu= 35 kN/m2
Cu= 20 kN/m2
Layer- 1
Layer- 2
Footing Base, Df= 2.5m
Raft Dia= (10 × 15)m
Immediate Settlement=??
Hard Strata or Rock
SOLUTION
Given,
qn= 50 kN/m2, B= 10m, 𝜇= 0.5 (for clay)
and E= 700 Cu
We know,
Si=
qn 𝐵
𝐸
1 − µ2 Ip
From figure, 𝐿
𝐵 =
15
10
= 1.5 So, If = 1.6 (center for flexible
foundation)
Again given,
For Layer-1: Cu= 35 kN/m2
For Layer-2: Cu= 20 kN/m2
∴ E1= 700 × 35= 24500 kN/m2
And, E2= 700 × 20= 14000 kN/m2
∴ Weighted Eavg=
24500×4.5 +14000×12
16.5
= 16864 kN/m2
From Equation,
Si=
50×10
16864
(1 − 0.52) × 1.36 = 30.24 (without
correction)
We Know,
Rigidity Correction Factor= 0.8
Depth Correction Factor= ??
From Problem,
𝐿
𝐵 = 1.5
∴
𝐷
√𝐿𝐵
=
2.5
√10×15
= 0.2
As footing base, Df= 2.5m
From graph,
Depth Factor=
97
100
= 0.97
∴ Si(corrected) = 30.24× 0.97× 0.8
= 23.47 mm

Determination of Immediate Settlement

  • 1.
    DETERMINATION OF IMMEDIATE SETTLEMENT SuezHalder Email: suez_301@yahoo.com By
  • 2.
    SIMPLIFIED DEFINITION When thesettlement of the footing takes place immediately after the load is applied over the footing, it is termed as immediate settlement. When Occurs? Immediate settlement occurs in the case of shallow footings, having sand as base of the footing.
  • 3.
    IMMEDIATE SETTLEMENT CALCULATION Accordingto the theory of elasticity, Si= 𝑞𝐵 𝐸 1 − µ2 If Where, q= Net foundation pressure B= Width of the foundation 𝜇= Poisson’s ration E= Young’s modulus of soil If= Influence factor
  • 4.
    SIGNIFICANCE OF INFLUENCEFACTOR (If) We have to calculate Influence Factor (If) for different shapes of footing, i.e. square, circular, rectangular and for different types of foundation. Such as- 1. Flexible type foundation. Ex: Isolated Footing 2. Rigid Type Foundation. Ex: Raft or Matt Foundation
  • 5.
    ELASTIC SETTLEMENT OFFLEXIBLE AND RIGID FOUNDATIONS
  • 6.
    EXPLANATION For Flexible Foundation- Settlementvalue at the corner of the footing and the center of the footing are not the same. Here, settlement value of the center is more than the corner of the foundation. So we have to calculate influence factor individually for corner and center. Also we have to calculate the average If . For Rigid Foundation- The settlement value is uniform throughout corner and center. So the influence factor (If ) will be 1.0 It is observed that- At center, Flexible Foundation Settlement > Rigid Foundation Settlement
  • 7.
    INFLUENCE FACTORS FOR FOUNDATIONS InfluenceFactor, 𝑰 𝒑 Flexible Shape 𝐿 𝐵 Center Corner Rigid Circle ---- 1.00 0.64 0.79 Square 1 1.12 0.56 0.88 Rectangle 1.5 1.36 0.68 1.07 2 1.53 0.77 1.21 3 1.78 0.89 1.42 5 2.10 1.05 1.70 10 2.54 1.27 2.10 20 2.99 1.49 2.46 50 3.57 1.80 3.00 100 4.01 2.00 3.43
  • 8.
    TWO TYPES OFCORRECTIONS 1. Rigidity Correction (for rigid foundation) If for Rigid Foundation= 0.8 × If for the Flexible Foundation at Center We have to first calculate the settlement considering the influence factor of the flexible foundation at the center, then we will multiply that settlement calculation by 0.8 and will get settlement of rigid foundation.
  • 9.
  • 10.
    PROBLEM +0m -2.5m -7m -19m 50 kN/m2 G.L G.W.T 10m 4.5m 12m 16.5 2.5m Cu= 35kN/m2 Cu= 20 kN/m2 Layer- 1 Layer- 2 Footing Base, Df= 2.5m Raft Dia= (10 × 15)m Immediate Settlement=?? Hard Strata or Rock
  • 11.
    SOLUTION Given, qn= 50 kN/m2,B= 10m, 𝜇= 0.5 (for clay) and E= 700 Cu We know, Si= qn 𝐵 𝐸 1 − µ2 Ip From figure, 𝐿 𝐵 = 15 10 = 1.5 So, If = 1.6 (center for flexible foundation) Again given, For Layer-1: Cu= 35 kN/m2 For Layer-2: Cu= 20 kN/m2
  • 12.
    ∴ E1= 700× 35= 24500 kN/m2 And, E2= 700 × 20= 14000 kN/m2 ∴ Weighted Eavg= 24500×4.5 +14000×12 16.5 = 16864 kN/m2 From Equation, Si= 50×10 16864 (1 − 0.52) × 1.36 = 30.24 (without correction) We Know, Rigidity Correction Factor= 0.8 Depth Correction Factor= ??
  • 13.
    From Problem, 𝐿 𝐵 =1.5 ∴ 𝐷 √𝐿𝐵 = 2.5 √10×15 = 0.2 As footing base, Df= 2.5m From graph, Depth Factor= 97 100 = 0.97 ∴ Si(corrected) = 30.24× 0.97× 0.8 = 23.47 mm