Chapter 16, Problem 16.1CTP

The following table shows the boring log at a site where a multi-story shopping center would be constructed. Soil classification and the standard penetration number, N₆₀, are provided in the boring log. All columns of the building are supported by square footings which must be placed at a depth of 1.5 m. Additionally, the settlement (elastic) of each footing must be restricted to 20 mm. Since the column loads at different location can vary, a design chart is helpful for quick estimation of footing size required to support a given load.

1. a. Prepare a chart by plotting the variation of maximum allowable column loads with footing sizes, B = 1 m, 1.5 m, 2 m, and 3 m. Use a factor of safety of 3.
2. b. If the gross column load from the structure is 250 kN, how would you use this chart to select a footing size?
3. c. What would be the design footing size for the column in Part (b) if you use Terzaghi’s bearing capacity equation? For the well graded sand, assume that ϕ′ = 33°. Use F_s = 3.
4. d. Compare and discuss the differences in footing sizes obtained in Parts b and c.

To determine

Plot the variation of maximum allowable column loads with size of footings to prepare a chart.

Explanation of Solution

Given information:

The location of depth of footing $D_f$ is 1.5 m.

The given size of the footing B is 1 m, 1.5 m, 2 m, and 3 m.

The settlement of each footing $S_e$ is 20 mm.

The given factor of safety $F_s$ is 3.

Calculation:

For B value is 1 m:

Determine the depth factor using the relation.

$F_d=1+0.33\frac{D_f}{B}$

Substitute 1.5 m for $D_f$ and 1.0 m for B.

$\begin{array}{c}{F}_d=1+0.33\times \frac{1.5}{1.0}\\ =1.495\end{array}$

The $F_d$ value should be less than 1.33. The calculated $F_d\left(1.495\right)>1.33$. Take $F_d$ as 1.33.

The field standard penetration number $N_{60}$ should be averaged up to a distance of 2 m $\left(2\times 1\right)$ below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

$\text{depth}=D_f+2B$

Substitute 1.5 m for $D_f$ and 1 m for B.

$\begin{array}{c}\text{depth}=1.5+2\left(1\right)\\ =3.5\text{m}\end{array}$

Determine the averaged $N_{60}$ value using the relation.

$N_{60}=\frac{{\left(N_{60}\right)}_{\text{SW}}+{\left(N_{60}\right)}_{\text{ML}}}{2}$

Here, ${\left(N_{60}\right)}_{\text{SW}}$ is the field standard penetration number of well graded sand and ${\left(N_{60}\right)}_{\text{ML}}$ is the field standard penetration number of sandy silts.

Substitute 12 for ${\left(N_{60}\right)}_{\text{SW}}$ and 7 for ${\left(N_{60}\right)}_{\text{ML}}$.

$\begin{array}{c}{N}_{60}=\frac{12+7}{2}\\ =9.5\\ \approx 10\end{array}$

Determine the net allowable bearing capacity of the soil $\left(q_{\text{net}}\right)$ using the relation.

$q_{\text{net}}=\frac{N_{60}}{0.05}{F}_d\left[\frac{S_e}{25}\right]$

Substitute 10 for $N_{60}$, 1.33 for $F_d$, and 20 mm for $S_e$.

$\begin{array}{c}{q}_{\text{net}}=\frac{10}{0.05}\times 1.33\left[\frac{20}{25}\right]\\ =212.8{\text{kN/m}}^{\text{2}}\end{array}$

Determine the maximum allowable column load $Q_{\text{all-net}}$ using the relation.

$Q_{\text{all-net}}=\frac{q_{\text{net}}\times B^2}{F_s}$

Substitute $212.8{\text{kN/m}}^{\text{2}}$ for $q_{\text{net}}$, 1 m for B, and 3 for $F_s$.

$\begin{array}{c}{Q}_{\text{all-net}}=\frac{212.8\times 1^2}{3}\\ =71\text{kN}\end{array}$

For B value is 1.5 m:

Determine the depth factor using the relation.

$F_d=1+0.33\frac{D_f}{B}$

Substitute 1.5 m for $D_f$ and 1.5 m for B.

$\begin{array}{c}{F}_d=1+0.33\times \frac{1.5}{1.5}\\ =1.33\end{array}$

The field standard penetration number $N_{60}$ should be averaged up to a distance of 3 m $\left(2\times 1.5\right)$ below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

$\text{depth}=D_f+2B$

Substitute 1.5 m for $D_f$ and 1.5 m for B.

$\begin{array}{c}\text{depth}=1.5+2\left(1.5\right)\\ =4.5\text{m}\end{array}$

Determine the averaged $N_{60}$ value for 4.5 m depth using the relation.

$N_{60}=\frac{{\left(N_{60}\right)}_{\text{4}}+{\left(N_{60}\right)}_{\text{6}}}{2}$

Here, ${\left(N_{60}\right)}_{\text{4}}$ is the field standard penetration number for 4 m depth and ${\left(N_{60}\right)}_{\text{6}}$ is the field standard penetration number for 6 m depth.

Substitute 7 for ${\left(N_{60}\right)}_{\text{4}}$ and 8 for ${\left(N_{60}\right)}_{\text{6}}$.

$\begin{array}{c}{N}_{60}=\frac{7+8}{2}\\ =7.5\\ \approx 8\end{array}$

Determine the net allowable bearing capacity of the soil $\left(q_{\text{net}}\right)$ using the relation.

$q_{\text{net}}=\frac{N_{60}}{0.08}{\left(\frac{B+0.3}{B}\right)}^2{F}_d\left[\frac{S_e}{25}\right]$

Substitute 8 for $N_{60}$, 1.33 for $F_d$, and 20 mm for $S_e$.

$\begin{array}{c}{q}_{\text{net}}=\frac{8}{0.08}\times {\left(\frac{1.5+0.3}{1.5}\right)}^2\times 1.33\left[\frac{20}{25}\right]\\ =153.2{\text{kN/m}}^{\text{2}}\end{array}$

Determine the maximum allowable column load $Q_{\text{all-net}}$ using the relation.

$Q_{\text{all-net}}=\frac{q_{\text{net}}\times B^2}{F_s}$

Substitute $153.2{\text{kN/m}}^{\text{2}}$ for $q_{\text{net}}$, 1.5 m for B, and 3 for $F_s$.

$\begin{array}{c}{Q}_{\text{all-net}}=\frac{153.2\times {1.5}^2}{3}\\ =115\text{kN}\end{array}$

For B value is 2 m:

Determine the depth factor using the relation.

$F_d=1+0.33\frac{D_f}{B}$

Substitute 1.5 m for $D_f$ and 2 m for B.

$\begin{array}{c}{F}_d=1+0.33\times \frac{1.5}{2.0}\\ =1.248\end{array}$

The field standard penetration number $N_{60}$ should be averaged up to a distance of 4 m $\left(2\times 2\right)$ below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

$\text{depth}=D_f+2B$

Substitute 1.5 m for $D_f$ and 2.0 m for B.

$\begin{array}{c}\text{depth}=1.5+2\left(2\right)\\ =5.5\text{m}\end{array}$

Determine the averaged $N_{60}$ value for 5.5 m depth using the relation.

$N_{60}=\frac{{\left(N_{60}\right)}_{\text{1}}+{\left(N_{60}\right)}_{\text{4}}+{\left(N_{60}\right)}_{\text{6}}}{3}$

Substitute 12 for ${\left(N_{60}\right)}_{\text{1}}$, 7 for ${\left(N_{60}\right)}_{\text{4}}$, and 8 for ${\left(N_{60}\right)}_{\text{6}}$.

$\begin{array}{c}{N}_{60}=\frac{12+7+8}{3}\\ =9\end{array}$

Determine the net allowable bearing capacity of the soil $\left(q_{\text{net}}\right)$ using the relation.

$q_{\text{net}}=\frac{N_{60}}{0.08}{\left(\frac{B+0.3}{B}\right)}^2{F}_d\left[\frac{S_e}{25}\right]$

Substitute 9 for $N_{60}$, 1.248 for $F_d$, and 20 mm for $S_e$.

$\begin{array}{c}{q}_{\text{net}}=\frac{9}{0.08}\times {\left(\frac{2+0.3}{2}\right)}^2\times 1.248\left[\frac{20}{25}\right]\\ =148.54{\text{kN/m}}^{\text{2}}\end{array}$

Determine the maximum allowable column load $Q_{\text{all-net}}$ using the relation.

$Q_{\text{all-net}}=\frac{q_{\text{net}}\times B^2}{F_s}$

Substitute $148.54{\text{kN/m}}^{\text{2}}$ for $q_{\text{net}}$, 2 m for B, and 3 for $F_s$.

$\begin{array}{c}{Q}_{\text{all-net}}=\frac{148.54\times 2^2}{3}\\ =198\text{kN}\end{array}$

For B value is 3 m:

Determine the depth factor using the relation.

$F_d=1+0.33\frac{D_f}{B}$

Substitute 1.5 m for $D_f$ and 3 m for B.

$\begin{array}{c}{F}_d=1+0.33\times \frac{1.5}{3}\\ =1.165\end{array}$

The field standard penetration number $N_{60}$ should be averaged up to a distance of 6 m $\left(2\times 3\right)$ below the foundation.

Determine the depth of foundation for the field standard penetration number gets averaged.

$\text{depth}=D_f+2B$

Substitute 1.5 m for $D_f$ and 3 m for B.

$\begin{array}{c}\text{depth}=1.5+2\left(3\right)\\ =7.5\text{m}\end{array}$

Determine the averaged $N_{60}$ value for 5.5 m depth using the relation.

$N_{60}=\frac{{\left(N_{60}\right)}_{\text{1}}+{\left(N_{60}\right)}_{\text{4}}+{\left(N_{60}\right)}_{\text{6}}+{\left(N_{60}\right)}_{\text{8}}}{4}$

Here, ${\left(N_{60}\right)}_{\text{8}}$ is the field standard penetration number at 8 m depth.

Substitute 12 for ${\left(N_{60}\right)}_{\text{1}}$, 7 for ${\left(N_{60}\right)}_{\text{4}}$, 8 for ${\left(N_{60}\right)}_{\text{6}}$, and 19 for ${\left(N_{60}\right)}_{\text{8}}$.

$\begin{array}{c}{N}_{60}=\frac{12+7+8+19}{4}\\ =11.5\\ \approx 12\end{array}$

Determine the net allowable bearing capacity of the soil $\left(q_{\text{net}}\right)$ using the relation.

$q_{\text{net}}=\frac{N_{60}}{0.08}{\left(\frac{B+0.3}{B}\right)}^2{F}_d\left[\frac{S_e}{25}\right]$

Substitute 12 for $N_{60}$, 1.165 for $F_d$, and 20 mm for $S_e$.

$\begin{array}{c}{q}_{\text{net}}=\frac{12}{0.08}\times {\left(\frac{3+0.3}{3}\right)}^2\times 1.165\left[\frac{20}{25}\right]\\ =169.15{\text{kN/m}}^{\text{2}}\end{array}$

Determine the maximum allowable column load $Q_{\text{all-net}}$ using the relation.

$Q_{\text{all-net}}=\frac{q_{\text{net}}\times B^2}{F_s}$

Substitute $169.15{\text{kN/m}}^{\text{2}}$ for $q_{\text{net}}$, 3 m for B, and 3 for $F_s$.

$\begin{array}{c}{Q}_{\text{all-net}}=\frac{169.15\times 3^2}{3}\\ =507.5\text{kN}\end{array}$

Summarize the calculated values as in Table (1).

Width B (m)	Column load (kN)
1	71
1.5	115
2	198
3	507.5

Plot the graph between the size of the footing and the column load as in Figure (1).

To determine

Find the footing size for the given gross column load of 250 kN.

Answer to Problem 16.1CTP

The footing size for the given gross column load of 250 kN is $\underset{\_}{2.25\text{m}}$.

Explanation of Solution

Given information:

The location of depth of footing $D_f$ is 1.5 m.

The given size of the footing B is 1 m, 1.5 m, 2 m, and 3 m.

The settlement of each footing $S_e$ is 20 mm.

The given factor of safety $F_s$ is 3.

Calculation:

Refer Figure (1).

The size of the footing is 2.25 m for the gross column load of 250 kN.

Therefore, the footing size for the given gross column load of 250 kN is $\underset{\_}{2.25\text{m}}$.

To determine

Find the design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation.

Answer to Problem 16.1CTP

The design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation is $\underset{\_}{2,173\text{kN}}$.

Explanation of Solution

Given information:

The value of cohesion $c^{\prime }$ is 0.

The soil friction angle ${\varphi }^{\prime }$ is $33°$.

The location of depth of footing base $D_f$ is 1.5 m.

The width of the footing B is 2.25 m.

The given factor of safety $F_s$ is 3.

Calculation:

Determine the net ultimate bearing capacity of the soil $\left(q_u\right)$ using the relation.

$\begin{array}{c}{q}_{u-\text{net}}=1.3c^{\prime }{N}_c+q{N}_q+0.4\gamma B{N}_{\gamma }-q\\ =1.3c^{\prime }{N}_c+\gamma D_f{N}_q+0.4\gamma B{N}_{\gamma }-\gamma D_f\end{array}$ (1)

Here, $N_c$ is the contribution of cohesion, $N_q$ is the contribution of surcharge, $\gamma$ is the unit weight of soil, and $N_{\gamma }$ is the contribution of unit weight of soil.

Take the unit weight of the soil $\gamma$ is $17{\text{kN/m}}^{\text{3}}$.

Refer Table 16.1, “Terzaghi’s bearing-capacity factors–$N_c$, $N_q$, and $N_{\gamma }$” in the textbook.

Take the $N_c$ as 48.09, $N_q$ as 32.23, and $N_{\gamma }$ as 31.94 for the ${\varphi }^{\prime }$ value of $33°$.

Substitute $0$ for $c^{\prime }$, 48.09 for $N_c$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1.5 m for $D_f$, 32.23 for $N_q$, 2.25 m for B, and 31.94 for $N_{\gamma }$.

$\begin{array}{c}{q}_{u-\text{net}}=1.3\times 0\times 48.09+17\times 1.5\times 32.23+0.4\times 17\times 2.25\times 31.94-17\times 1.5\\ =1,287.59{\text{kN/m}}^{\text{2}}\end{array}$

Determine the net allowable bearing capacity $\left(q_{\text{all-net}}\right)$ that the footing can carry using the relation.

$\begin{array}{c}{Q}_{\text{all-net}}=q_{\text{all-net}}{B}^2\\ =\frac{q_{\text{u-net}}{B}^2}{F_s}\end{array}$

Substitute $1,287.59{\text{kN/m}}^{\text{2}}$ for $q_u$, 2.25 m for B, and 3.0 for $F_s$.

$\begin{array}{c}{Q}_{\text{all-net}}=\frac{1,122.15\times {2.25}^2}{3}\\ =2,173\text{kN}\end{array}$

Therefore, the design column load for the footing size of 2.25 m using the Terzaghi’s bearing capacity equation is $\underset{\_}{2,173\text{kN}}$.

To determine

Compare and discuss the differences in footing sizes obtained in parts (b) and (c).

Explanation of Solution

Given information:

The soil friction angle ${\varphi }^{\prime }$ is $33°$.

The location of depth of footing base $D_f$ is 1.5 m.

The width of the footing B is 2.25 m.

The given factor of safety $F_s$ is 3.

Calculation:

The net allowable column load obtained by using Terzaghi’s bearing capacity equation (2,173 kN) is significantly higher than the method based on the $N_{60}$ and limiting settlement value (250 kN). In actual foundation design, the bearing capacity based on shear strength and the settlement criteria need to be satisfied. It is possible that the allowable column load obtained in part c will decrease if the settlement limit is imposed. Considering the uncertainty in sampling and testing of granular materials for the shear strength determination, the $N_{60}$ method and settlement criteria may be preferable for the given condition.