Chapter 11, Problem 11.1P

To determine

Calculate the elastic settlement due to the net applied pressure on the foundation.

Answer to Problem 11.1P

The elastic settlement due to the net applied pressure $\left(S_{e,\text{rigid}}\right)$ is $\underset{\_}{12.1\text{mm}}$.

Explanation of Solution

Given information:

The vertical column load $\left(P\right)$ is $600\text{kN}$.

The length of the foundation $\left(L\right)$ is $2\text{m}$.

The breadth of the foundation $\left(B\right)$ is $1\text{m}$.

The depth of foundation $\left(D_f\right)$ is $0.75\text{m}$.

The average modulus of elasticity $\left(E_s\right)$ is $20,600\text{kN}/{\text{m}}^2$.

The Poisson’s ratio $\left({\mu }_s\right)$ is $0.3$.

The depth below the foundation $\left(H\right)$ is $5\text{m}$.

Calculation:

Calculate the elastic settlement for the perfectly flexible foundation $\left(S_{e,\text{flexible}}\right)$ using the relation.

$S_{e,\text{flexible}}=\Delta \sigma \left(\alpha B^{\prime }\right)\frac{1-{\mu }_s^2}{E_s}{I}_s{I}_f$ (1)

Here, $\Delta \sigma$ is the net applied pressure on the foundation, $\alpha$ is the factor depends on the foundation where settlement is being calculated, $B^{\prime }$ is $\frac{B}{2}$ for center of foundation or B for corner of foundation, $I_s$ is the shape factor, and $I_f$ is the depth factor.

Calculate the net applied pressure $\left(\Delta \sigma \right)$ as shown below.

$\Delta \sigma =\frac{P}{L\times B}$

Substitute $600\text{kN}$ for P, $2\text{m}$ for L, and $1\text{m}$ for B.

$\begin{array}{c}\Delta \sigma =\frac{600}{2\times 1}\\ =300\text{kN}/{\text{m}}^{\text{2}}\end{array}$

For the center of the foundation:

The factor $\left(\alpha \right)$ is $4$ at the center of the foundation.

Calculate $B^{\prime }$ as shown below.

$B^{\prime }=\frac{B}{2}$

Substitute $1\text{m}$ for B.

$\begin{array}{c}{B}^{\prime }=\frac{1}{2}\\ =0.5\text{m}\end{array}$

Calculate the ratio $\left(m^{\prime }\right)$ as shown below.

$m^{\prime }=\frac{L}{B}$

Substitute $2\text{m}$ for L and $1\text{m}$ for B.

$\begin{array}{c}{m}^{\prime }=\frac{2}{1}\\ =2\end{array}$

Calculate the ratio $\left(n^{\prime }\right)$ as shown below.

$n^{\prime }=\frac{H}{\frac{B}{2}}$

Substitute $5\text{m}$ for H and $1\text{m}$ for B.

$\begin{array}{c}{n}^{\prime }=\frac{5}{\frac{1}{2}}\\ =10\end{array}$

Calculate the shape factor $\left(I_s\right)$ using the relation.

$I_s=F_1+\frac{1-2{\mu }_s}{1-{\mu }_s}{F}_2$ (2)

Here, $F_1$ and $F_2$ are the factors and a function of $m^{\prime }$ and $n^{\prime }$.

Calculate the factor $\left(F_1\right)$ as follows.

Refer Table 11.1 “Variation of $F_1$ with $m^{\prime }$ and $n^{\prime }$” in the Text Book.

Take the value of $F_1$ as $0.641$, for the values $m^{\prime }$ of $2$ and $n^{\prime }$ of $10$.

Calculate the factor $\left(F_2\right)$ as follows.

Refer Table 11.2 “Variation of $F_2$ with $m^{\prime }$ and $n^{\prime }$” in the Text Book.

Take the value of $F_2$ as $0.031$, for the values $m^{\prime }$ of $2$ and $n^{\prime }$ of $10$.

Calculate the shape factor $\left(I_s\right)$ as shown below.

Substitute $0.641$ for $F_1$, $0.031$ for $F_2$, and $0.3$ for ${\mu }_s$ in Equation (2).

$\begin{array}{c}{I}_s=0.641+\left(\frac{1-2\times 0.3}{1-0.3}\right)\times 0.031\\ =0.641+0.018\\ =0.659\end{array}$

Calculate the ratio $\frac{D_f}{B}$ as shown below.

Substitute $0.75\text{m}$ for $D_f$ and $1\text{m}$ for B.

$\begin{array}{c}\frac{D_f}{B}=\frac{0.75}{1}\\ =0.75\end{array}$

Calculate the ratio $\frac{L}{B}$ as shown below.

Substitute $2\text{m}$ for L and $1\text{m}$ for B.

$\begin{array}{c}\frac{L}{B}=\frac{2}{1}\\ =2\end{array}$

Calculate the depth factor $\left(I_f\right)$ as follows.

Refer Table 11.3 “Variation of $I_f$ with $\frac{L}{B}$ and $\frac{D_f}{B}$” in the Text Book.

Take the value of $I_f$ as $0.75$, for the values $\frac{L}{B}$ of $2$, $\frac{D_f}{B}$ of $0.75$, and ${\mu }_s$ of $0.3$.

Calculate the elastic settlement for the perfectly flexible foundation $\left(S_{e,\text{flexible}}\right)$ as shown below.

Substitute $300\text{kN}/{\text{m}}^2$ for $\Delta \sigma$, $4$ for $\alpha$, $0.5\text{m}$ for $B^{\prime }$, $0.3$ for ${\mu }_s$, $20,600\text{kN}/{\text{m}}^2$ for $E_s$, $0.659$ for $I_s$, and $0.75$ for $I_f$.

$\begin{array}{c}{S}_{e,\text{flexible}}=300\times 4\times 0.5\left(\frac{1-{0.3}^2}{20,600}\right)\times 0.659\times 0.75\\ =296.55\times 4.417\times {10}^{-5}\\ =0.013\text{m}\end{array}$

Calculate the elastic settlement for the rigid foundation $\left(S_{e,\text{rigid}}\right)$ as shown below.

$S_{e,\text{rigid}}=0.93S_{e,\text{flexible}}$

Substitute $0.013\text{m}$ for $S_{e,\text{flexible}}$.

$\begin{array}{c}{S}_{e,\text{rigid}}=0.93\times 0.013\\ =0.01209\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =12.1\text{mm}\end{array}$

Therefore, the elastic settlement due to the net applied pressure $\left(S_{e,\text{rigid}}\right)$ is $\underset{\_}{12.1\text{mm}}$.