Chapter 16, Problem 16.19P

Figure 16.21 shows a continuous foundation with a width of 1.8 m constructed at a depth of 1.2 m in a granular soil. The footing is subjected to an eccentrically inclined loading with e = 0.3 m, and α = 10°. Determine the gross ultimate load, Q_u(ei), that the footing can support using:

1. a. Meyerhof (1963) method [Eq. (16.52)]
2. b. Saran and Agarwal (1991) method [Eq. (16.53)]
3. c. Patra et al. (2012) reduction factor method [Eq. (16.54)]

To determine

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Meyerhof method.

Answer to Problem 16.19P

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Meyerhof method is $\underset{\_}{984\text{kN/m}}$.

Explanation of Solution

Given information:

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

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

The soil friction angle ${\varphi }^{\prime }$ is $34°$.

The location of depth of footing base $D_f$ is 1.2 m.

The width of the footing B is 1.8 m.

The value of eccentricity e is 0.3 m.

The inclined angle $\alpha$ is $10°$.

Calculation:

Determine the effective width of the footing using the relation.

$B^{\prime }=B-2e$

Substitute 1.8 m for B and 0.3 for e.

$\begin{array}{c}{B}^{\prime }=1.8-2\times 0.3\\ =1.2\text{m}\end{array}$

For the continuous foundation, all shape factors are equal to one $\left({\lambda }_{cs}={\lambda }_{qs}={\lambda }_{\gamma s}=1.0\right)$.

Determine the depth factor ${\lambda }_{cd}$ using the relation.

${\lambda }_{cd}=1+0.4{\tan }^{-1}\left(\frac{D_f}{B^{\prime }}\right)$

Substitute 1.2 m for $D_f$ and 1.2 m for $B^{\prime }$.

$\begin{array}{c}{\lambda }_{cd}=1+0.4\left(\frac{1.2}{1.2}\right)\\ =1.4\end{array}$

Determine the depth factor ${\lambda }_{qd}$ using the relation.

${\lambda }_{qd}=1+2\tan {\varphi }^{\prime }{\left(1-\sin {\varphi }^{\prime }\right)}^2\left(\frac{D_f}{B^{\prime }}\right)$

Substitute $34°$ for ${\varphi }^{\prime }$, 1.2 m for $D_f$, and 1.2 m for $B^{\prime }$.

$\begin{array}{c}{\lambda }_{qd}=1+2\tan 34°{\left(1-\sin 34°\right)}^2\left(\frac{1.2}{1.2}\right)\\ =1.262\end{array}$

Determine the inclination factor ${\lambda }_{qi}$ using the relation.

${\lambda }_{qi}={\left(1-\frac{\alpha }{90°}\right)}^2$

Substitute $10°$ for $\alpha$.

$\begin{array}{c}{\lambda }_{qi}={\left(1-\frac{10°}{90°}\right)}^2\\ =0.79\end{array}$

Determine the inclination factor ${\lambda }_{\gamma i}$ using the relation.

${\lambda }_{\gamma i}={\left(1-\frac{\alpha }{{\varphi }^{\prime }}\right)}^2$

Substitute $10°$ for $\alpha$ and $34°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{\lambda }_{\gamma i}={\left(1-\frac{10°}{34°}\right)}^2\\ =0.498\end{array}$

Determine the ultimate bearing capacity of the soil $\left({q^{\prime }}_u\right)$ using the relation.

$\begin{array}{c}{q^{\prime }}_u=c^{\prime }{N}_c{\lambda }_{cs}{\lambda }_{cd}+q{N}_q{\lambda }_{qs}{\lambda }_{qd}+\frac{1}{2}\gamma B^{\prime }{N}_{\gamma }{\lambda }_{\gamma s}{\lambda }_{\gamma d}\\ =c^{\prime }{N}_c{\lambda }_{cs}{\lambda }_{cd}+\gamma D_f{N}_q{\lambda }_{qs}{\lambda }_{qd}+\frac{1}{2}\gamma B^{\prime }{N}_{\gamma }{\lambda }_{\gamma s}{\lambda }_{\gamma d}\end{array}$ (1)

Here, ${\lambda }_{\gamma d}$ is the depth factor and $N_c$, $N_q$, and $N_{\gamma }$ are bearing-capacity factors.

Refer Table 16.2, “Bearing-capacity factors $N_c$, $N_q$, and $N_{\gamma }$” in the textbook.

For ${\varphi }^{\prime }=34°$;

The values of $N_c$ is 42.16, $N_q$ is 29.44, and $N_{\gamma }$ is 41.06.

Substitute 0 for $c^{\prime }$, 42.16 for $N_c$, 1.0 for ${\lambda }_{cs}$, 1.4 for ${\lambda }_{cd}$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1.2 m for $D_f$, 29.44 for $N_q$, 1.0 for ${\lambda }_{qs}$, 1.262 for ${\lambda }_{qd}$, 0.79 for ${\lambda }_{qi}$, 1.2 m for $B^{\prime }$, 41.06 for $N_{\gamma }$, 1.0 for ${\lambda }_{\gamma s}$, 1.0 for ${\lambda }_{\gamma d}$, and 0.498 for ${\lambda }_{\gamma i}$ in Equation (1).

$\begin{array}{c}{q^{\prime }}_u=\left\{\begin{array}{l}\left(0\times 42.16\times 1.0\times 1.4\right)+\left(17\times 1.2\times 29.44\times 1.0\times 1.262\times 0.79\right)+\\ \left(\frac{1}{2}\times 17\times 1.2\times 41.06\times 1\times 1\times 0.498\right)\end{array}\right\}\\ =807.33{\text{kN/m}}^{\text{2}}\end{array}$

Determine the gross ultimate load $Q_{u\left(ei\right)}$ using the relation.

$Q_{u\left(ei\right)}=\frac{{q^{\prime }}_u{B}^{\prime }}{\cos \alpha }$

Substitute $807.33{\text{kN/m}}^{\text{2}}$ for ${q^{\prime }}_u$, 1.2 m for $B^{\prime }$, and $10°$ for $\alpha$.

$\begin{array}{c}{Q}_{u\left(ei\right)}=\frac{807.33\times 1.2}{\cos 10°}\\ =984\text{kN/m}\end{array}$

Therefore, the gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Meyerhof method is $\underset{\_}{984\text{kN/m}}$.

To determine

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Saran and Agarwal method.

Answer to Problem 16.19P

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Saran and Agarwal method is $\underset{\_}{1,237\text{kN/m}}$.

Explanation of Solution

Given information:

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

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

The soil friction angle ${\varphi }^{\prime }$ is $34°$.

The location of depth of footing base $D_f$ is 1.2 m.

The width of the footing B is 1.8 m.

The value of eccentricity e is 0.3 m.

The inclined angle $\alpha$ is $10°$.

Calculation:

Determine the ratio of $\left(\frac{e}{B}\right)$.

Substitute 0.3 for e and 1.8 m for B.

$\begin{array}{c}\left(\frac{e}{B}\right)=\frac{0.3}{1.8}\\ =0.167\end{array}$

Determine the gross ultimate load $Q_{u\left(ei\right)}$ using the relation.

$\begin{array}{c}{Q}_{u\left(ei\right)}=B\left[c^{\prime }{N}_{c\left(ei\right)}+q{N}_{q\left(ei\right)}+\frac{1}{2}\gamma B{N}_{\gamma \left(ei\right)}\right]\\ =B\left[c^{\prime }{N}_{c\left(ei\right)}+\gamma D_f{N}_{q\left(ei\right)}+\frac{1}{2}\gamma B{N}_{\gamma \left(ei\right)}\right]\end{array}$ (2)

Here, $N_{c\left(ei\right)}$, $N_{q\left(ei\right)}$, and $N_{\gamma \left(ei\right)}$ are the bearing capacity factors.

Refer Figure 16.14, “Variation of $N_{c\left(ei\right)}$-(b)” in the textbook.

Take the $N_{c\left(ei\right)}$ value as 24.0 for ${\varphi }^{\prime }$ value of $34°$ and the $\left(\frac{e}{B}\right)$ value of 0.167.

Refer Figure 16.15, “Variation of $N_{q\left(ei\right)}$-(b)” in the textbook.

Take the $N_{q\left(ei\right)}$ value as 22.8 for ${\varphi }^{\prime }$ value of $34°$ and the $\left(\frac{e}{B}\right)$ value of 0.167.

Refer Figure 16.16, “Variation of $N_{\gamma \left(ei\right)}$-(b)” in the textbook.

Take the $N_{\gamma \left(ei\right)}$ value as 14.5 for ${\varphi }^{\prime }$ value of $34°$ and the $\left(\frac{e}{B}\right)$ value of 0.167.

Substitute 0 for $c^{\prime }$, 24.0 for $N_{c\left(ei\right)}$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1.2 m for $D_f$, 22.8 for $N_{q\left(ei\right)}$, 1.8 m for B, and 14.5 for $N_{\gamma \left(ei\right)}$, in Equation (2).

$\begin{array}{c}{Q}_{u\left(ei\right)}=1.8\left(0\times 24.0+17\times 1.2\times 22.8+\frac{1}{2}\times 17\times 1.8\times 14.5\right)\\ =1,237\text{kN/m}\end{array}$

Therefore, the gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Saran and Agarwal method is $\underset{\_}{1,237\text{kN/m}}$.

To determine

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Patra et al. reduction factor method.

Answer to Problem 16.19P

The gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Patra et al. reduction factor method is $\underset{\_}{1,006\text{kN/m}}$.

Explanation of Solution

Given information:

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

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

The soil friction angle ${\varphi }^{\prime }$ is $34°$.

The location of depth of footing base $D_f$ is 1.2 m.

The width of the footing B is 1.8 m.

The value of eccentricity e is 0.3 m.

The inclined angle $\alpha$ is $10°$.

Calculation:

For the continuous foundation, all shape factors are equal to one $\left({\lambda }_{cs}={\lambda }_{qs}={\lambda }_{\gamma s}=1.0\right)$.

Determine the depth factor ${\lambda }_{cd}$ using the relation.

${\lambda }_{cd}=1+0.4{\tan }^{-1}\left(\frac{D_f}{B}\right)$

Substitute 1.2 m for $D_f$ and 1.8 m for B.

$\begin{array}{c}{\lambda }_{cd}=1+0.4\left(\frac{1.2}{1.8}\right)\\ =1.26\end{array}$

Determine the depth factor ${\lambda }_{qd}$ using the relation.

${\lambda }_{qd}=1+2\tan {\varphi }^{\prime }{\left(1-\sin {\varphi }^{\prime }\right)}^2\left(\frac{D_f}{B}\right)$

Substitute $34°$ for ${\varphi }^{\prime }$, 1.2 m for $D_f$, and 1.8 m for B.

$\begin{array}{c}{\lambda }_{qd}=1+2\tan 34°{\left(1-\sin 34°\right)}^2\left(\frac{1.2}{1.8}\right)\\ =1.175\end{array}$

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

$\begin{array}{c}{q}_u=c^{\prime }{N}_c{\lambda }_{cs}{\lambda }_{cd}+q{N}_q{\lambda }_{qs}{\lambda }_{qd}+\frac{1}{2}\gamma B{N}_{\gamma }{\lambda }_{\gamma s}{\lambda }_{\gamma d}\\ =c^{\prime }{N}_c{\lambda }_{cs}{\lambda }_{cd}+\gamma D_f{N}_q{\lambda }_{qs}{\lambda }_{qd}+\frac{1}{2}\gamma B{N}_{\gamma }{\lambda }_{\gamma s}{\lambda }_{\gamma d}\end{array}$ (3)

Refer Table 16.2, “Bearing-capacity factors $N_c$, $N_q$, and $N_{\gamma }$” in the textbook.

Take the $N_c$ as 42.16, $N_q$ as 29.44, and $N_{\gamma }$ as 41.06 for the ${\varphi }^{\prime }$ value of $34°$.

Substitute 0 for $c^{\prime }$, 42.16 for $N_c$, 1.0 for ${\lambda }_{cs}$, 1.26 for ${\lambda }_{cd}$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1.2 m for $D_f$, 29.44 for $N_q$, 1.0 for ${\lambda }_{qs}$, 1.175 for ${\lambda }_{qd}$, 1.8 m for B, 41.06 for $N_{\gamma }$, 1.0 for ${\lambda }_{\gamma s}$, and 1.0 for ${\lambda }_{\gamma d}$, in Equation (3).

$\begin{array}{c}{q}_u=\left\{\begin{array}{l}0\times 42.16\times 1.0\times 1.26+17\times 1.2\times 29.44\times 1.0\times 1.175+\\ \frac{1}{2}\times 17\times 1.8\times 41.06\times 1\times 1\end{array}\right\}\\ =1,333.89{\text{kN/m}}^{\text{2}}\end{array}$

Determine the gross ultimate load $Q_{u\left(ei\right)}$ using the relation.

$Q_{u\left(ei\right)}=B{q}_u\left[1-2\left(\frac{e}{B}\right)\right]{\left(1-\frac{\alpha }{{\varphi }^{\prime }}\right)}^{2-\left(D_f/B\right)}$

Substitute 1.8 m for B, $1,333.89{\text{kN/m}}^{\text{2}}$ for ${q^{\prime }}_u$, 0.3 m for e, $10°$ for $\alpha$, $34°$ for ${\varphi }^{\prime }$, and 1.2 m for $D_f$.

$\begin{array}{c}{Q}_{u\left(ei\right)}=1.8\times 1,333.89\left[1-2\left(\frac{0.3}{1.8}\right)\right]{\left(1-\frac{10°}{34°}\right)}^{2-\left(1.2/1.8\right)}\\ =1,006\text{kN/m}\end{array}$

Therefore, the gross ultimate load $Q_{u\left(ei\right)}$ that the footing can support by using Patra et al. reduction factor method is $\underset{\_}{1,006\text{kN/m}}$.