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worked examples

Elastic Foundation Settlement Example

Worked SI-unit example estimating immediate elastic settlement and evaluating the assumptions that control the result.

Problem Statement

Estimate the immediate elastic settlement of a shallow footing using:

  • Net service pressure qnet=150 kPaq_{\mathrm{net}}=150\ \mathrm{kPa}.
  • Footing width B=2.5 mB=2.5\ \mathrm{m}.
  • Representative soil modulus Es=30,000 kPaE_s=30{,}000\ \mathrm{kPa}.
  • Poisson ratio ν=0.30\nu=0.30.
  • Combined influence factor Is=1.12I_s=1.12.

The influence factor is assumed to represent the selected method's shape, rigidity, embedment, and evaluation-point corrections. The example does not calculate consolidation or secondary settlement.

Step 1: State The Model

Use the simplified elastic expression:

si=qnetBEs(1ν2)Iss_i=\frac{q_{\mathrm{net}}B}{E_s}\left(1-\nu^2\right)I_s

Pressure and modulus use the same units, so their ratio is dimensionless. Multiplication by footing width gives settlement in metres.

Step 2: Poisson-Ratio Term

1ν2=1(0.30)2=0.911-\nu^2=1-(0.30)^2=0.91

Step 3: Pressure-Modulus Ratio

qnetEs=15030,000=0.005\frac{q_{\mathrm{net}}}{E_s}=\frac{150}{30{,}000}=0.005

The calculation assumes the selected modulus represents the footing's stress and strain range and the relevant depth of influence.

Step 4: Calculate Settlement

si=(0.005)(2.5)(0.91)(1.12)=0.01274 ms_i=(0.005)(2.5)(0.91)(1.12)=0.01274\ \mathrm{m}

Convert to millimetres:

si=0.01274(1000)=12.74 mms_i=0.01274(1000)=12.74\ \mathrm{mm}

The estimated immediate settlement is approximately 12.7 mm12.7\ \mathrm{mm}.

Step 5: Sensitivity To Modulus

If the representative modulus were 20,000 kPa20{,}000\ \mathrm{kPa} rather than 30,000 kPa30{,}000\ \mathrm{kPa}:

si=150(2.5)20,000(0.91)(1.12)(1000)=19.11 mms_i=\frac{150(2.5)}{20{,}000}(0.91)(1.12)(1000)=19.11\ \mathrm{mm}

If it were 45,000 kPa45{,}000\ \mathrm{kPa}:

si=8.49 mms_i=8.49\ \mathrm{mm}

This range shows why modulus selection deserves more attention than extra decimal places in the final calculation.

Engineering Interpretation

The computed 12.7 mm12.7\ \mathrm{mm} is one settlement component at one representative point. Acceptance requires comparison with project-specific total and differential movement criteria.

Before using the result, verify:

  • The net pressure and excavation-stress convention.
  • Whether the modulus is drained or undrained as appropriate.
  • Whether it represents the full influence depth or only a near-surface test zone.
  • Whether layered-soil integration is needed.
  • Whether consolidation, creep, collapse, or heave adds movement.
  • Whether adjacent footing interaction changes the stress field.
  • Whether the structural system can tolerate the predicted settlement pattern.

Limits Of The Example

The calculation idealizes the soil as homogeneous, isotropic, and elastic. Real soil stiffness is nonlinear and often increases with confining stress. Construction disturbance, groundwater, anisotropy, and stress history can change the response.

For a mat or strongly layered deposit, an integrated or numerical analysis may be more appropriate. A sophisticated model still requires defensible stiffness parameters and construction stages.

References And Further Reading

  • USACE EM 1110-1-1904, Settlement Analysis.
  • FHWA GEC 6, Shallow Foundations.
  • Project-specific field and laboratory stiffness data.

FAQ

Why use net pressure?

Settlement is driven by stress change relative to the preconstruction condition. Net pressure can represent that change when excavation, replacement, groundwater, and load history are modeled consistently.

Is 12.7 millimetres acceptable?

The number cannot be judged without structure-specific total and differential movement criteria and comparison with other supports.

Can the default influence factor be used for every footing?

No. Select the factor from the calculation method being used and confirm which shape, rigidity, embedment, and location corrections it represents.