Chapter 10, Problem 10.1P

Refer to the rectangular combined footing in Figure 10.1, with Q₁ = 100 kip and Q₂ = 150 kip. The distance between the two column loads L₃ = 13.5 ft. The proximity of the property line at the left edge requires that L₂ = 3.0 ft. The net allowable soil pressure is 2500 lb/ft². Determine the breadth and length of a rectangular combined footing.

To determine

Find the length and breadth of the combined footing.

Answer to Problem 10.1P

The length of the combined footing is $\underset{\_}{22.2\text{ft}}$.

The width of the combined footing is $\underset{\_}{4.50\text{ft}}$.

Explanation of Solution

Given information:

The distance between two column loads $\left(L_3\right)$ is $13.5\text{ft}$.

The column loads $\left(Q_1\right)$ and $\left(Q_2\right)$ are 100 kip and 150 kip.

The distance $\left(L_2\right)$ is $3.0\text{ft}$.

The net allowable soil pressure $\left(q_{\text{all}\left(\text{net}\right)}\right)$ is $2,500{\text{lb/ft}}^{\text{2}}$.

Calculation:

Find the location of the resultant of the column loads $\left(X\right)$ using the relation:

$X=\frac{Q_2{L}_3}{Q_1+Q_2}$

Substitute 150 kip for $Q_2$, 13.5 ft for $L_3$, and 100 kip for $Q_1$.

$\begin{array}{c}X=\frac{150\text{kip}\times 13.5\text{ft}}{150\text{kip}+100\text{kip}}\\ =8.1\text{ft}\end{array}$

Find the length of the foundation $\left(L\right)$ using the relation:

$L=2\left(L_2+X\right)$

Substitute 3 ft for $L_2$ and 8.1 ft for X.

$\begin{array}{c}L=2\left(3\text{ft}+8.1\text{ft}\right)\\ =22.2\text{ft}\end{array}$

Therefore, the length of the combined footing is $\underset{\_}{22.2\text{ft}}$.

Find the length $\left(L_1\right)$ using the relation:

$L_1=L-L_2-L_3$

Substitute 22.2 ft for $L$, 3 ft for $L_2$, and 13.5 ft for $L_3$.

$\begin{array}{c}{L}_1=22.2\text{ft}-3\text{ft}-13.5\text{ft}\\ =5.7\text{ft}\end{array}$

Find the area of the combined footing $\left(A\right)$ using the relation:

$A=\frac{Q_1+Q_2}{q_{\text{all}\left(\text{net}\right)}}$

Substitute 100 kip for $Q_1$, 150 kip for $Q_2$, and $2,500{\text{lb/ft}}^{\text{2}}$ for $q_{\text{all}\left(\text{net}\right)}$.

$\begin{array}{c}A=\frac{100\text{kip}+150\text{kip}}{2,500{\text{lb/ft}}^{\text{2}}}\\ =\frac{250\text{kip}\times \frac{1,000\text{lb}}{1\text{kip}}}{2,500{\text{lb/ft}}^{\text{2}}}\\ =100{\text{ft}}^{\text{2}}\end{array}$

Find the width of the rectangular footing $\left(B\right)$ using the relation:

$B=\frac{A}{L}$

Substitute $100{\text{ft}}^{\text{2}}$ for A and 22.2 ft for $L$.

$\begin{array}{c}B=\frac{100{\text{ft}}^{\text{2}}}{22.2\text{ft}}\\ =4.50\text{ft}\end{array}$

Therefore, the width of the combined footing is $\underset{\_}{4.50\text{ft}}$.