Chapter 8, Problem 8.27CTP

To determine

Find the vertical stress increase at 4 m below A, B, and C.

Answer to Problem 8.27CTP

The increase in the vertical stress $\left(\Delta \sigma \right)$ at point A located at depth of 4 m is $\underset{\_}{38.964{\text{kN/m}}^2}$.

The increase in the vertical stress $\left(\Delta \sigma \right)$ at point B located at depth of 4 m is $\underset{\_}{14.586{\text{kN/m}}^2}$.

The increase in the vertical stress $\left(\Delta \sigma \right)$ at point C located at depth of 4 m is $\underset{\_}{14.322{\text{kN/m}}^2}$.

Explanation of Solution

Given information:

The depth of the point from the ground level, $z=4\text{m}$.

The intensity of the uniform pressure, $q=60{\text{kN/m}}^2$.

Calculation:

Vertical stress increase at point A:

Show the free-body diagram of the L-shaped raft as in Figure 1.

Rectangle 1:

Find the value of ${m^{\prime }}_1$ using the relation:

${m^{\prime }}_1=\frac{B_1}{z}$

Substitute 4 m for $B_1$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_1=\frac{4}{4}\\ =1\end{array}$

Find the value of ${n^{\prime }}_1$ using the relation:

${n^{\prime }}_1=\frac{L_1}{z}$

Substitute 6 m for $L_1$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_1=\frac{6}{4}\\ =1.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_1=1$ and ${n^{\prime }}_1=1.5$;

The value of variation ${\left(I_3\right)}_1$ is 0.1935.

Find the increase in vertical stress $\left(\Delta {\sigma }_1\right)$ using the relation:

$\Delta {\sigma }_1=q{\left(I_3\right)}_1$

Substitute $60{\text{kN/m}}^2$ for q and 0.1935 for ${\left(I_3\right)}_1$.

$\begin{array}{c}\Delta {\sigma }_1=60\times 0.1935\\ =11.61\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 2:

Find the value of ${m^{\prime }}_2$ using the relation:

${m^{\prime }}_2=\frac{B_2}{z}$

Substitute 8 m for $B_2$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_2=\frac{8}{4}\\ =2\end{array}$

Find the value of ${n^{\prime }}_2$ using the relation:

${n^{\prime }}_2=\frac{L_2}{z}$

Substitute 6 m for $L_2$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_2=\frac{6}{4}\\ =1.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_2=2$ and ${n^{\prime }}_2=1.5$;

The value of variation ${\left(I_3\right)}_2$ is 0.2234.

Find the increase in vertical stress $\left(\Delta {\sigma }_2\right)$ using the relation:

$\Delta {\sigma }_2=q{\left(I_3\right)}_2$

Substitute $60{\text{kN/m}}^2$ for q and 0.2234 for ${\left(I_3\right)}_2$.

$\begin{array}{c}\Delta {\sigma }_2=60\times 0.2234\\ =13.404\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 3:

Find the value of ${m^{\prime }}_3$ using the relation:

${m^{\prime }}_3=\frac{B_3}{z}$

Substitute 8 m for $B_3$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_3=\frac{8}{4}\\ =2\end{array}$

Find the value of ${n^{\prime }}_3$ using the relation:

${n^{\prime }}_3=\frac{L_3}{z}$

Substitute 8 m for $L_3$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_3=\frac{8}{4}\\ =2\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_3=2$ and ${n^{\prime }}_3=2$;

The value of variation ${\left(I_3\right)}_3$ is 0.2325.

Find the increase in vertical stress $\left(\Delta {\sigma }_3\right)$ using the relation:

$\Delta {\sigma }_3=q{\left(I_3\right)}_3$

Substitute $60{\text{kN/m}}^2$ for q and 0.2325 for ${\left(I_3\right)}_3$.

$\begin{array}{c}\Delta {\sigma }_3=60\times 0.2325\\ =13.95\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the increase in vertical stress $\left(\Delta \sigma \right)$ at point A using the relation:

$\Delta \sigma =\Delta {\sigma }_1+\Delta {\sigma }_2+\Delta {\sigma }_3$

Substitute $11.61\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_1$, $13.404\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_2$, and $13.95\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_3$.

$\begin{array}{c}\Delta \sigma =11.61+13.404+13.95\\ =38.964{\text{kN/m}}^2\end{array}$

Therefore, the increase in the vertical stress $\left(\Delta \sigma \right)$ at point A located at depth of 4 m is $\underset{\_}{38.964{\text{kN/m}}^2}$.

The vertical stress increase at point B:

Show the free-body diagram of the L-shaped raft as in Figure 2.

Rectangle 1:

Find the value of ${m^{\prime }}_1$ using the relation:

${m^{\prime }}_1=\frac{B_1}{z}$

Substitute 12 m for $B_1$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_1=\frac{12}{4}\\ =3\end{array}$

Find the value of ${n^{\prime }}_1$ using the relation:

${n^{\prime }}_1=\frac{L_1}{z}$

Substitute 6 m for $L_1$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_1=\frac{6}{4}\\ =1.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_1=3$ and ${n^{\prime }}_1=1.5$;

The value of variation ${\left(I_3\right)}_1$ is 0.2280.

Find the increase in vertical stress $\left(\Delta {\sigma }_1\right)$ using the relation:

$\Delta {\sigma }_1=q{\left(I_3\right)}_1$

Substitute $60{\text{kN/m}}^2$ for q and 0.2280 for ${\left(I_3\right)}_1$.

$\begin{array}{c}\Delta {\sigma }_1=60\times 0.2280\\ =13.68\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 2:

Find the value of ${m^{\prime }}_2$ using the relation:

${m^{\prime }}_2=\frac{B_2}{z}$

Substitute 8 m for $B_2$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_2=\frac{8}{4}\\ =2\end{array}$

Find the value of ${n^{\prime }}_2$ using the relation:

${n^{\prime }}_2=\frac{L_2}{z}$

Substitute 14 m for $L_2$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_2=\frac{14}{4}\\ =3.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_2=2$ and ${n^{\prime }}_2=3.5$;

The value of variation ${\left(I_3\right)}_2$ is 0.2385.

Find the increase in vertical stress $\left(\Delta {\sigma }_2\right)$ using the relation:

$\Delta {\sigma }_2=q{\left(I_3\right)}_2$

Substitute $60{\text{kN/m}}^2$ for q and 0.2385 for ${\left(I_3\right)}_2$.

$\begin{array}{c}\Delta {\sigma }_2=60\times 0.2385\\ =14.31\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 3:

Find the value of ${m^{\prime }}_3$ using the relation:

${m^{\prime }}_3=\frac{B_3}{z}$

Substitute 8 m for $B_3$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_3=\frac{8}{4}\\ =2\end{array}$

Find the value of ${n^{\prime }}_3$ using the relation:

${n^{\prime }}_3=\frac{L_3}{z}$

Substitute 6 m for $L_3$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_3=\frac{6}{4}\\ =1.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_3=2$ and ${n^{\prime }}_3=1.5$;

The value of variation ${\left(I_3\right)}_3$ is 0.2234.

Find the increase in vertical stress $\left(\Delta {\sigma }_3\right)$ using the relation:

$\Delta {\sigma }_3=q{\left(I_3\right)}_3$

Substitute $60{\text{kN/m}}^2$ for q and 0.2234 for ${\left(I_3\right)}_3$.

$\begin{array}{c}\Delta {\sigma }_3=60\times 0.2234\\ =13.404\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the increase in vertical stress $\left(\Delta \sigma \right)$ at point B using the relation:

$\Delta \sigma =\Delta {\sigma }_1+\Delta {\sigma }_2-\Delta {\sigma }_3$

Substitute $13.68\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_1$, $14.31\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_2$, and $13.404\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_3$.

$\begin{array}{c}\Delta \sigma =13.68+14.31-13.404\\ =14.586{\text{kN/m}}^2\end{array}$

Therefore, the increase in the vertical stress $\left(\Delta \sigma \right)$ at point B located at depth of 4 m is $\underset{\_}{14.586{\text{kN/m}}^2}$.

The vertical stress increase at point C:

Show the free-body diagram of the L-shaped raft as in Figure 3.

Rectangle 1:

Find the value of ${m^{\prime }}_1$ using the relation:

${m^{\prime }}_1=\frac{B_1}{z}$

Substitute 12 m for $B_1$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_1=\frac{12}{4}\\ =3\end{array}$

Find the value of ${n^{\prime }}_1$ using the relation:

${n^{\prime }}_1=\frac{L_1}{z}$

Substitute 14 m for $L_1$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_1=\frac{14}{4}\\ =3.5\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_1=3$ and ${n^{\prime }}_1=3.5$;

The value of variation ${\left(I_3\right)}_1$ is 0.2447.

Find the increase in vertical stress $\left(\Delta {\sigma }_1\right)$ using the relation:

$\Delta {\sigma }_1=q{\left(I_3\right)}_1$

Substitute $60{\text{kN/m}}^2$ for q and 0.2447 for ${\left(I_3\right)}_1$.

$\begin{array}{c}\Delta {\sigma }_1=60\times 0.2447\\ =14.682\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 2:

Find the value of ${m^{\prime }}_2$ using the relation:

${m^{\prime }}_2=\frac{B_2}{z}$

Substitute 12 m for $B_2$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_2=\frac{12}{4}\\ =3\end{array}$

Find the value of ${n^{\prime }}_2$ using the relation:

${n^{\prime }}_2=\frac{L_2}{z}$

Substitute 8 m for $L_2$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_2=\frac{8}{4}\\ =2\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_2=3$ and ${n^{\prime }}_2=2$;

The value of variation ${\left(I_3\right)}_2$ is 0.2385.

Find the increase in vertical stress $\left(\Delta {\sigma }_2\right)$ using the relation:

$\Delta {\sigma }_2=q{\left(I_3\right)}_2$

Substitute $60{\text{kN/m}}^2$ for q and 0.2385 for ${\left(I_3\right)}_2$.

$\begin{array}{c}\Delta {\sigma }_2=60\times 0.2385\\ =14.31\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Rectangle 3:

Find the value of ${m^{\prime }}_3$ using the relation:

${m^{\prime }}_3=\frac{B_3}{z}$

Substitute 8 m for $B_3$ and 4 m for z.

$\begin{array}{c}{m^{\prime }}_3=\frac{8}{4}\\ =2\end{array}$

Find the value of ${n^{\prime }}_3$ using the relation:

${n^{\prime }}_3=\frac{L_3}{z}$

Substitute 8 m for $L_3$ and 4 m for z.

$\begin{array}{c}{n^{\prime }}_3=\frac{8}{4}\\ =2\end{array}$

Refer Figure 8.16 “Variation of $I_3$ with $m^{\prime }$ and $n^{\prime }$” in the textbook.

For ${m^{\prime }}_3=2$ and ${n^{\prime }}_3=2$;

The value of variation ${\left(I_3\right)}_3$ is 0.2325.

Find the increase in vertical stress $\left(\Delta {\sigma }_3\right)$ using the relation:

$\Delta {\sigma }_3=q{\left(I_3\right)}_3$

Substitute $60{\text{kN/m}}^2$ for q and 0.2325 for ${\left(I_3\right)}_3$.

$\begin{array}{c}\Delta {\sigma }_3=60\times 0.2325\\ =13.95\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the increase in vertical stress $\left(\Delta \sigma \right)$ at point B using the relation:

$\Delta \sigma =\Delta {\sigma }_1-\Delta {\sigma }_2+\Delta {\sigma }_3$

Substitute $14.682\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_1$, $14.31\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_2$, and $13.95\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_3$.

$\begin{array}{c}\Delta \sigma =14.682-14.31+13.95\\ =14.322{\text{kN/m}}^2\end{array}$

Therefore, the increase in the vertical stress $\left(\Delta \sigma \right)$ at point C located at depth of 4 m is $\underset{\_}{14.322{\text{kN/m}}^2}$.