Three-dimensional numerical and analytical study of horizontal group of square anchor plates in sand

Abstract

In this paper, numerical and analytical methods are used to evaluate the ultimate pullout capacity of a group of square anchor plates in row or square configurations, installed horizontally in dense sand. The elasto-plastic numerical study of square anchor plates is carried out using three-dimensional finite element analysis. The soil is modeled by an elasto-plastic model with a Mohr–Coulomb yield criterion. An analytical method based on a simplified three-dimensional failure mechanism is developed in this study. The interference effect is evaluated by group efficiency η, defined as the ratio of the ultimate pullout capacity of group of N anchor plates to that of a single isolated plate multiplied by number of plates. The variation of the group efficiency η was computed with respect to change in the spacing between plates. Results of the analyses show that the spacing between the plates, the internal friction angle of soil and the installation depth are the most important parameters influencing the group efficiency. New equations are developed in this study to evaluate the group efficiency of square anchor plates embedded horizontally in sand at shallow depth (H = 4B). The results obtained by numerical and analytical solutions are in excellent agreement.

Appendices

Appendix 1: Analytical solution of break-out factor $\varvec{N}_{{\boldsymbol{\gamma} \,{\textbf{isolated}}}}$ for an isolated square anchor plate

Volumes of the portions 1, 2 and 3 shown in Fig. 25:

Fig. 25

Failure mechanism for isolated square plate

$$ V_{1} = B^{2} H $$

(21)

$$ V_{2} = 0.5BH^{2} \tan \theta $$

(22)

$$ V_{3} = \frac{1}{6}H^{3} \tan^{2} \theta $$

(23)

The ultimate pullout load $Q_{u}$ is equal to the weight of the soil located within the failure mechanism:

$$ Q_{u} = \gamma \left( {V_{1} + 4V_{2} + 4V_{3} } \right) $$

(24)

Using Eqs. 21, 22 and 23, we finally get

$$ Q_{u} = B^{2} \gamma H + 2\gamma H^{2} B\tan \theta + \frac{2}{3}\gamma H^{3} \tan^{2} \theta $$

(25)

By definition, the ultimate pullout capacity $q_{u}$:

$$ q_{u} = \frac{{Q_{u} }}{A} = \frac{{Q_{u} }}{{B^{2} }} $$

(26)

$$ q_{u} = \gamma H\left( {1 + 2\frac{H}{B}\tan \theta + \frac{2}{3}\left( {\frac{H}{B}} \right)^{2} \tan^{2} \theta } \right) $$

(27)

By convention, $q_{u}$ is also given by:

$$ q_{u} = \gamma HN_{\gamma } $$

(28)

So that the break-out factor $N_{\gamma }$ for an isolated square plate anchor is:

$$ N_{{\gamma \,{\text{isolated}}}} = 1 + 2\frac{H}{B}\tan \theta + \frac{2}{3}\left( {\frac{H}{B}} \right)^{2} \tan^{2} \theta $$

(29)

Appendix 2: Analytical solution of break-out factor $\varvec{N}_{{\boldsymbol{\gamma} \,{\textbf{end}}}}$ for a square anchor plate located at an end

For two or n square anchor plates with $S < 2H\tan \theta$, the ultimate pullout load of one anchor plate at the end is equal to the weight of the soil located within its failure mechanism defined by points a, b, c, d, e shown in Fig. 21.

Volumes of the portions 4 and 5 shown in Fig. 26:

Fig. 26

Interference detail of two failure mechanisms

$$ V_{4} = \frac{1}{2}\left( {2H\tan \theta - S} \right)\left( {H - \frac{S}{2}\cot \theta } \right)B $$

(30)

$$ V_{4} = BH^{2} \tan \theta + B\frac{{S^{2} }}{4}\cot \theta - BSH $$

(31)

$$ V_{5} = \frac{1}{3} \times \frac{1}{2}\left( {\sqrt 2 H\tan \theta - S\frac{\sqrt 2 }{2}} \right)^{2} \left( {H - \frac{S}{2}\cot \theta } \right) $$

(32)

$$ V_{5} = \frac{1}{3}H^{3} \tan^{2} \theta - \frac{1}{2}SH^{2} \tan \theta - \frac{{S^{3} }}{24}\cot \theta + \frac{1}{4}S^{2} H $$

(33)

The volume V corresponding to points a, b, c, d, e shown in Fig. 16 is equal to:

$$ V = V_{\text{T}} - \frac{1}{2}V_{4} - 2\left( {\frac{1}{2}V_{5} } \right) = V_{\text{T}} - \frac{{V_{1} }}{2} - V_{5} $$

(34)

where V _T is the total volume of soil located within the failure mechanism for an isolated square anchor plate.

So:

$$ V = B^{2} H + \frac{3}{2}H^{2} B\tan \theta + \frac{1}{3}H^{3} \tan^{2} \theta - \frac{1}{8}BS^{2} \cot \theta + \frac{1}{2}BSH + \frac{1}{2}SH^{2} \tan \theta + \frac{1}{24}S^{3} \cot \theta - \frac{1}{4}S^{2} H $$

(35)

The ultimate pullout capacity of anchor plate located at the end is then:

$$ q_{{u\,{\text{end}}}} = \frac{{Q_{{u\,{\text{end}}}} }}{A} = \frac{{Q_{{u\,{\text{end}}}} }}{{B^{2} }} = \frac{\gamma \times V}{{B^{2} }} $$

(36)

Using Eq. 35, we get

$$ q_{{u\,{\text{end}}}} = \gamma H\left( {1 + \frac{3}{2}\left( {\frac{H}{B}} \right)\tan \theta + \frac{1}{3}\left( {\frac{H}{B}} \right)^{2} \left( {\tan \theta } \right)^{2} - \frac{1}{8}\left( {\frac{S}{B}} \right)\left( {\frac{S}{H}} \right)\cot \theta + \frac{1}{2}\left( {\frac{S}{B}} \right) + \frac{1}{2}\left( {\frac{S}{B}} \right)\left( {\frac{H}{B}} \right)\tan \theta + \frac{1}{24}\left( {\frac{S}{B}} \right)^{2} \left( {\frac{S}{H}} \right)\cot \theta - \frac{1}{4}\left( {\frac{S}{B}} \right)^{2} } \right) $$

(37)

Since

$$ q_{{u\,{\text{end}}}} = \gamma HN_{{\gamma \,{\text{end}}}} $$

(38)

We finally obtained $N_{{\gamma \,{\text{end}}}}$

$$ N_{{\gamma \,{\text{end}}}} = 1 + \frac{1}{3}\left( {\frac{H}{B}} \right)^{2} \left( {\tan \theta } \right)^{2} + \frac{1}{2}\left( {3 + \frac{S}{B}} \right)\frac{H}{B}\tan \theta + \frac{1}{8}\frac{S}{H}\frac{S}{B}\left( {\frac{1}{3}\frac{S}{B} - 1} \right)\cot \theta - \frac{1}{4}\left( {\frac{S}{B}} \right)^{2} + \frac{1}{2}\frac{S}{B} $$

(39)

Appendix 3: Analytical solution of break-out factor $N_{{\gamma \,\text{inter}}}$ for an intermediate square anchor plate

For n square anchor plates in row configuration with $S < 2H\tan \theta$, the ultimate pullout load of an intermediate anchor plate is equal to the weight of the soil located within its failure mechanism defined by points a, b, c, d, e, f as shown in Fig. 23. The break-out factor corresponding to this volume is noted as $N_{{\gamma \,{\text{inter}}}}$

The volume V corresponding to points a, b, c, d, e, f shown in Fig. 18 is equal to:

$$ V = V_{T} - 2\left( {\frac{1}{2}V_{4} } \right) - 4\left( {\frac{1}{2}V_{5} } \right) = V_{T} - V_{4} - 2V_{5} $$

(40)

where V _T is the total volume of soil located within the failure mechanism for an isolated square anchor plate.

Using Eqs. (31 and 33), we obtain:

$$ V = B^{2} H + H^{2} B\tan \theta - \frac{1}{4}BS^{2} \cot \theta + BSH - SH^{2} \tan \theta + \frac{1}{12}S^{3} \cot \theta - \frac{1}{2}S^{2} H $$

(41)

The ultimate pullout capacity of one anchor plate located between two anchor plates is given by:

$$ q_{{u\,{\text{inter}}}} = \frac{{Q_{{u\,{\text{inter}}}} }}{A} = \frac{{Q_{{u\,{\text{inter}}}} }}{{B^{2} }} = \frac{\gamma \times V}{{B^{2} }} $$

(42)

$$ q_{{u\,{\text{inter}}}} = \gamma H\left( {1 + \left( {\frac{H}{B}} \right)\tan \theta - \frac{1}{4}\left( {\frac{S}{B}} \right)\left( {\frac{S}{H}} \right)\cot \theta + \frac{S}{B} + \left( {\frac{S}{B}} \right)\left( {\frac{H}{B}} \right)\tan \theta + \frac{1}{12}\left( {\frac{S}{B}} \right)^{2} \left( {\frac{S}{H}} \right)\cot \theta - \frac{1}{2}\left( {\frac{S}{B}} \right)^{2} } \right) $$

(43)

Since

$$ q_{{u\,{\text{inter}}}} = \gamma HN_{{\gamma \,{\text{inter}}}} $$

(44)

We get

$$ N_{{\gamma \,{\text{inter}}}} = 1 + \left( {1 + \frac{S}{B}} \right)\frac{H}{B}\tan \theta - \frac{1}{3}\frac{S}{B}\frac{S}{H}\left( {\frac{1}{4}\frac{S}{B} - 1} \right)\cot \theta - \frac{1}{2}\left( {\frac{S}{B}} \right)^{2} + \frac{S}{B} $$

(45)