Stress analysis of ore particle flow behaviour under the influence of a flexible barrier

Abstract

Additional stresses caused by a flexible barrier, drawpoints and the self-weight of ore particles were investigated based on a physical experiment of the shrinkage stoping method of synchronous filling, and the flow behaviour of ore particles is described by the component forces. This study shows that when ore particles are drawn from multiple drawpoints under the influence of a flexible barrier, the stress distribution of the ore particles is sinusoidal, with an amplitude that gradually decreases with increasing height. The values of σ_x and σ_y gradually decrease with increasing height; the maximum σ_y values occur at x = 0, the maximum σ_x values occur at x = ± 10 and the maximum shear stresses occur near the lateral walls. The shape of the draw column is determined by the additional vertical stress, and its contour determines the essential morphology of the drawbody. The surface of the ore particles is smoothed by the additional stress caused by the barrier, and the self-weight of the ore particles is the inherent cause of the additional stress.

Appendices

Appendix 1. General solution to the stress of a semi-planar body subjected to a concentrated force at the boundary

The problems described by Eqs. (3), (5) and (8) belong to the problem of a semi-planar body subjected to a concentrated force at the boundary. Therefore, we first introduce the general solution of the stress to the problem of a semi-planar body subjected to a concentrated force at the boundary.

Suppose a semi-planar body is subjected to a concentrated force F (dimension MT⁻²) on its straight boundary, and the coordinates of any point P in the plane are (ρ, φ), as shown in Fig. 15.

Fig. 15

Semi-planar body subjected to a concentrated force at the boundary

The semi-inverse method is used to solve the problem, and the stress function of the model is assumed to be ϕ. From dimensional analysis (Xu 1982), we know that ϕ is a function of φ times ρ, namely:

$$ \phi =\rho f\left(\varphi \right) $$

(13)

Substituting Eq. (13) into the compatibility equation of the stress function in polar coordinates provides:

$$ \frac{1}{\rho^3}\left[\frac{{\mathrm{d}}^4f\left(\varphi \right)}{d{\varphi}^4}+2\frac{{\mathrm{d}}^2f\left(\varphi \right)}{d{\varphi}^2}+f\left(\varphi \right)\right]=0 $$

(14)

Normalising by the factor $\frac{1}{\rho^3}$ and solving this ordinary differential equation provides:

$$ \phi = A\rho \cos \varphi + B\rho \sin \varphi +\rho \varphi \left(C\cos \varphi +D\sin \varphi \right) $$

(15)

The stress is the second partial derivative of the stress function ϕ, while the first two terms Aρ cos φ + Bρ sin φ = Ax + By in Eq. (15) are the first terms of x and y, so these two terms do not affect the stress. Therefore, we need to consider only the following:

$$ \phi =\rho \varphi \left(C\cos \varphi +D\sin \varphi \right) $$

(16)

According to the calculation formula of the stress component (Xu 1982), the following can be obtained:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma}_{\rho }=\frac{1}{\rho}\frac{\partial \phi }{\partial \rho }+\frac{1}{\rho^2}\frac{\partial^2\phi }{\partial {\varphi}^2}=\frac{2}{\rho}\left( Dcos\varphi -C\sin \varphi \right)\\ {}{\sigma}_{\varphi }=\frac{\partial^2\phi }{\partial {\rho}^2}=0\\ {}{\tau}_{\rho \varphi}={\tau}_{\varphi \rho}=-\frac{\partial }{\partial \rho}\left(\frac{1}{\rho}\frac{\partial \phi }{\partial \varphi}\right)=0\end{array}} $$

(17)

Next, the stress boundary conditions are investigated, and the unknown coefficients in Eq. (17) are determined. There are no normal or tangential surface forces along φ = 0and φ = π; thus, the stress components σ_φ and τ_ρφ need to satisfy (σ_φ)_{φ = 0, π, ρ ≠ 0} = 0, (σ_φ)_{φ = 0, π, ρ ≠ 0} = 0. According to Eq. (17), stress components σ_φ and τ_ρφ satisfy these two boundary conditions.

In addition, point O should be considered subject to the concentrated force F. According to Saint-Venant’s principle, a small detached body Oabc is cut out near point O (Fig. 15). Then, considering the equilibrium condition of the detached body, three equilibrium equations are listed:

$$ \Big\{{\displaystyle \begin{array}{l}\sum {F}_x=0,\kern0.75em {\int}_0^{\uppi}\left[{\left({\sigma}_{\rho}\right)}_{\rho =\rho}\cos \varphi\ \rho \mathrm{d}\varphi +{\left({\tau}_{\rho \varphi}\right)}_{\rho =\rho}\sin \varphi\ \rho \mathrm{d}\varphi \right]=0\\ {}\sum {F}_y=0,\kern0.75em {\int}_0^{\uppi}\left[{\left({\sigma}_{\rho}\right)}_{\rho =\rho}\sin \varphi\ \rho \mathrm{d}\varphi -{\left({\tau}_{\rho \varphi}\right)}_{\rho =\rho}\cos \varphi\ \rho \mathrm{d}\varphi \right]+F=0\\ {}\sum {M}_o=0,{\int}_0^{\uppi}{\left({\tau}_{\rho \varphi}\right)}_{\rho =\rho}\rho \mathrm{d}\varphi \cdot \rho =0\end{array}} $$

(18)

Substituting the stress component in Eq. (17) into Eq. (18), it can be obtained that:

$$ \Big\{{\displaystyle \begin{array}{l}D=0\\ {}C=\frac{F}{\uppi}\end{array}} $$

(19)

Substituting Eq. (19) into Eq. (17) to obtain the final solution of the stress component (i.e. the Flamant solution) provides:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma}_{\rho }=-\frac{2F}{\uppi}\frac{\sin \varphi }{\rho}\\ {}{\sigma}_{\varphi }=0\\ {}{\tau}_{\rho \varphi}={\tau}_{\varphi \rho}=0\end{array}} $$

(20)

Using the coordinate transformation formula, the stress component in rectangular coordinates can be obtained from the above formula:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma}_x={\sigma}_{\rho }{\cos}^2\varphi =-\frac{2F}{\uppi}\frac{\sin \varphi co{s}^2\varphi }{\rho}\\ {}{\sigma}_y={\sigma}_{\rho }{\sin}^2\varphi =-\frac{2F}{\uppi}\frac{\sin^3\varphi }{\rho}\\ {}{\tau}_{xy}={\sigma}_{\rho}\sin \varphi co s\varphi =-\frac{2F}{\uppi}\frac{\sin^2\varphi co s\varphi}{\rho}\end{array}} $$

(21)

This is the stress component in rectangular coordinates expressed in polar coordinates. The polar coordinates in Eq. (21) can also be transformed into rectangular coordinates to obtain:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma}_x=-\frac{2F}{\uppi}\frac{x^2y}{{\left({x}^2+{y}^2\right)}^2}\\ {}{\sigma}_y=-\frac{2F}{\uppi}\frac{y^3}{{\left({x}^2+{y}^2\right)}^2}\\ {}{\tau}_{xy}=-\frac{2F}{\uppi}\frac{x{y}^2}{\rho}\end{array}} $$

(22)

Appendix 2. Eq. (3)

In Fig. 7, the projected component in the vertical direction at point N is Q(ξ), and the concentrated force acting on the hypotenuse of the differential element is $d{F}_1=\sqrt{1+{\left(D\hbox{'}\right)}^2}Q\left(\xi \right) d\xi$. Its stress on any point M of the ore particles belongs to the problem of the semi-planar body subjected to a concentrated force at the boundary, which can be solved by Eq. (22). Equation (3) can be obtained by substituting the relevant parameters of the model into Eq. (22):

$$ \Big\{{\displaystyle \begin{array}{l}d{\sigma}_x\hbox{'}=-\frac{2d{F}_1}{\uppi}\frac{{\left(x-\xi \right)}^2\left(y-D\left(\xi \right)\right)}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\\ {}d{\sigma}_y\hbox{'}=-\frac{2d{F}_1}{\uppi}\frac{{\left(y-D\left(\xi \right)\right)}^3}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\\ {}d{\tau}_{xy}\hbox{'}=-\frac{2d{F}_1}{\uppi}\frac{\left(x-\xi \right){\left(y-D\left(\xi \right)\right)}^2}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\end{array}} $$

Appendix 3. Eq. (5)

In Fig. 7, the projected component in the horizontal direction at point N is S(ξ), and the concentrated force acting on the hypotenuse of the differential element is $d{F}_2=\sqrt{1+{\left(D\hbox{'}\right)}^2}S\left(\xi \right) d\xi$. Its stress on any point M of the ore particles belongs to the problem of the semi-planar body subjected to a concentrated force at the boundary, which can be solved by Eq. (22). Equation (5) can be obtained by substituting the relevant parameters of the model into Eq. (22):

$$ \Big\{{\displaystyle \begin{array}{l}d{\sigma}_x\hbox{'}\hbox{'}=-\frac{2d{F}_2}{\uppi}\frac{\left(x-\xi \right){\left(y-D\left(\xi \right)\right)}^2}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\\ {}d{\sigma}_y\hbox{'}\hbox{'}=-\frac{2d{F}_2}{\uppi}\frac{{\left(x-\xi \right)}^3}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\\ {}d{\tau}_{xy}\hbox{'}\hbox{'}=-\frac{2d{F}_2}{\uppi}\frac{{\left(x-\xi \right)}^2\left(y-D\left(\xi \right)\right)}{{\left[{\left(x-\xi \right)}^2+{\left(y-D\left(\xi \right)\right)}^2\right]}^2}\end{array}} $$

Appendix 4. Eq. (8)

In Fig. 8, there is an equivalent load F₃ at each drawpoint. The effect of any equivalent load F₃ on a point in the model can be regarded in terms of the problem of a semi-planar body subjected to a concentrated force at the boundary. Substituting the relevant parameters of the model into Eq. (22), the stress at a point of the model under any equivalent load F₃ can be obtained:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma_{x\hbox{'}}}^{\hbox{'}\hbox{'}\hbox{'}\hbox{'}}=-\frac{2{F}_3}{\uppi}\frac{{x_i}^2{y}_i}{{\left({x_i}^2+{y_i}^2\right)}^2}\\ {}{\sigma_{y\hbox{'}}}^{\hbox{'}\hbox{'}\hbox{'}\hbox{'}}=-\frac{2{F}_3}{\uppi}\frac{y^3}{{\left({x_i}^2+{y_i}^2\right)}^2}\\ {}{\tau_{xy\hbox{'}}}^{\hbox{'}\hbox{'}\hbox{'}\hbox{'}}=-\frac{2{F}_3}{\uppi}\frac{x_i{y_i}^2}{\rho}\end{array}} $$

(23)

By superimposing the stress caused by seven equivalent loads at the same point in the model, Eq. (8) can be obtained:

$$ \Big\{{\displaystyle \begin{array}{l}{\sigma}_x\hbox{'}\hbox{'}\hbox{'}\hbox{'}=-\frac{2F}{\uppi}\sum \limits_{i=1}^7\frac{x_i^2{y}_i}{{\left({x_i}^2+{y_i}^2\right)}^2}\\ {}{\sigma}_y\hbox{'}\hbox{'}\hbox{'}\hbox{'}=-\frac{2F}{\uppi}\sum \limits_{i=1}^7\frac{{y_i}^3}{{\left({x_i}^2+{y_i}^2\right)}^2}\\ {}{\tau}_{xy}\hbox{'}\hbox{'}\hbox{'}\hbox{'}=-\frac{2F}{\uppi}\sum \limits_{i=1}^7\frac{x_i{y_i}^2}{{\left({x_i}^2+{y_i}^2\right)}^2}\end{array}} $$