Chapter 8, Problem 8.1QPP

A 1.5 m square footing carries a column with a service load of 105 kN. It is founded at a depth of 2 m on a medium stiff clay with an undrained shear strength of 42 kPa, an overconsolidation ratio of 4, and a plasticity index of 35. The clay layer is 5 m thick and overlies a very stiff shale. Estimate the undrained settlement of the footing using the generalized elastic method with Christian and Carrier’s (1978) influence factors.

To determine

The undrained settlement of the footing using the generalized elastic method with Christian and Carriers (1978) influence factors.

Answer to Problem 8.1QPP

3.1 mm

Explanation of Solution

Given data:

Size of footing = 1.5m

Service load = 105kN

Depth = 2m

Undrained shear strength = 42kPa

Overconsolidation ratio = 4

Plasticity index = 35

Thickness of clay layer = 5m

Formula used:

  $\begin{array}{l}\delta =I_0{I}_1\frac{qB}{E}\\ where,\delta isundrainedsettlementoffooting\\ I_{0}is\inf luencefactoraccountingfordepthoffooting\\ I_{1}is\inf luencefactoraccountingforshapeoffootingandcompressiblelayerthickness\\ qisaveragebearingstress\\ Biswidthoffooting\\ Eisaveragesoil\mathrm{mod}ulus\end{array}$

The ratio of depth to width of footing is

  $\begin{array}{l}\frac{D_f}{B}=\frac{2}{1.5}\\ \frac{D_f}{B}=1.3\end{array}$

The influence factor accounting for depth of footing from the following figure is

  $I_0=0.94$

The aspect ratio of footing is

  $\begin{array}{l}\frac{L}{B}=\frac{1.5}{1.5}\\ \frac{L}{B}=1\end{array}$

The relative thickness of compressible layer is

  $\begin{array}{l}\frac{z_h}{B}=\frac{5}{1.5}\\ \frac{z_h}{B}=3.3\end{array}$

The influence factor accounting for shape of footing and compressible layer thickness from the above figure is

  $I_1=0.6$

The average soil modulus over depth of compressible layer is

  $\begin{array}{l}E=\beta s_u\\ where,Eisaveragesoil\mathrm{mod}ulus\\ \beta is\dim ensionlessparameter\\ s_{u}isundrainedshearstrength\end{array}$

Substituting the values in the above equation,

  $\begin{array}{l}E=300\times 42\\ E=12600kPa\end{array}$

The undrained settlement of the footing is

  $\begin{array}{l}\delta =I_0{I}_1\frac{qB}{E}\\ \delta =I_0{I}_1\left(\frac{P}{A}\right)\frac{B}{E}\\ \delta =I_0{I}_1\left(\frac{P}{B\times L}\right)\frac{B}{E}\\ \delta =I_0{I}_1\left(\frac{P}{LE}\right)\\ \delta =0.94\times 0.6\times \left(\frac{105}{1.5\times 12600}\right)\\ \delta =0.0031m\\ \delta =3.1mm\end{array}$

Conclusion:

The undrained settlement of the footing is 3.1 mm.