Overturning stability of L-shaped rigid barriers subjected to rockfall impacts

Abstract

Reinforced concrete barriers are commonly used as defence measures in hilly areas to contain falling boulders and landslide debris. These barriers are conventionally designed to satisfy the conditions of force and momentum equilibrium with a factor of safety. A major limitation of this approach is that the inertial resistance of the barrier is neglected such that the design could be over-conservative. This paper presents a novel displacement-based approach for the assessment of overturning stability of rigid L-shaped barriers subjected to rockfall impacts. Analytical solutions, which are derived based on conservation of momentum and energy, are used to take into account the contributions of the self-weight and, thus, the inertial resistance of the barrier in resisting an impact. The actual amount of energy transferred from the impacting boulder to the barrier is considered by including the coefficient of restitution between the two objects. The accuracy of the analytical solutions has been confirmed by laboratory impact experiments. Numerical assessments conducted using the new solutions indicate that a reasonably sized rigid barrier, due to its own inertial resistance, may adequately withstand the impact action of a heavy boulder rolling down a hillslope without relying on any anchorage to its support. A range of geometric design of the barriers with L-shaped cross sections also has been considered and analysed. The new approach presented in this paper is easy to apply in practice and will be useful for engineers designing concrete barriers as passive rockfall mitigation measures.

Appendices

Appendix A

Rectangular

For a rectangular block with height h, length l and mass M, the rotational inertia about its centroidal axis and its base corner can be respectively expressed as:

$$ {I}_{\theta \left(\mathrm{centroid}\right)}=M\left(\frac{h^2+{l}^2}{12}\right) $$

(A1)

$$ {I}_{\theta \left(\mathrm{base}\right)}=M\left(\frac{h^2+{l}^2}{3}\right) $$

(A2)

L-shaped

Figure 7 shows an isometric view of the L-shaped barrier previously shown in Fig. 2. The distances between the global axis z and the centroidal axis of the stem wall z_c(stem) and that of the base slab z_c(base) are r_stem and r_base respectively. The total rotational inertia is the sum of the moments of inertia of the wall I_z(stem) and of the base slab I_z(base). By the parallel axis theorem, the rotational inertia of the stem wall about axis z is:

Fig. 7

A schematic L-shaped barrier and reference axes

$$ {I}_{z\left(\mathrm{stem}\right)}={I}_{z_{\mathrm{c}}\left(\mathrm{stem}\right)}+{M}_{\mathrm{stem}}{r}_{\mathrm{stem}}^2 $$

(A3)

which can be expanded to:

$$ {I}_{z\left(\mathrm{stem}\right)}={M}_{\mathrm{stem}}\left(\frac{d^2+{w}_{\mathrm{stem}}^2}{12}\right)+{M}_{\mathrm{stem}}\left[\left(\frac{d^2}{4}+{dw}_{\mathrm{base}}+{w}_{\mathrm{base}}^2\right)+{\left(\frac{w_{\mathrm{stem}}}{2}\right)}^2\right] $$

(A4)

Since w_stem and w_base are typically much smaller than d, ignoring the contributions of their higher-order terms leads to:

$$ {I}_{z\left(\mathrm{stem}\right)}\approx {M}_{\mathrm{stem}}\left(\frac{d^2}{12}\right)+{M}_{\mathrm{stem}}\left(\frac{d^2}{4}+{dw}_{\mathrm{base}}\right) $$

(A5)

Substituting d = h − w_base (see Fig. 7) into Eq. A4 and rearranging thus gives:

$$ {I}_{z\left(\mathrm{stem}\right)}\approx \frac{M_{\mathrm{stem}}}{3}\left(h-{w}_{\mathrm{base}}\right)\;\left(h+2{w}_{\mathrm{base}}\right) $$

(A6)

According to Eq. A2, the rotational inertia of the rectangular base slab about the global axis z can be written as:

$$ {I}_{z\left(\mathrm{base}\right)}=\frac{M_{\mathrm{base}}}{3}\left({l}^2+{w}_{\mathrm{base}}^2\right) $$

(A7)

Therefore, the total rotational inertia of the barrier can be obtained by summing Eqs. A6 and A7:

$$ {I}_z\approx \frac{M_{\mathrm{stem}}}{3}\left(h-{w}_{\mathrm{base}}\right)\;\left(h+2{w}_{\mathrm{base}}\right)+\frac{M_{\mathrm{base}}}{3}\left({l}^2+{w}_{\mathrm{base}}^2\right) $$

(A8)

Expanding and ignoring the contribution of w_base2 leads to:

$$ {I}_z\approx \frac{M_{\mathrm{stem}}}{3}\left({h}^2+{hw}_{\mathrm{base}}\right)+\frac{M_{\mathrm{base}}{l}^2}{3} $$

(A9)

Equation A8 can be used to estimate the rotational inertia of an L-shaped barrier, but the result will be slightly less than the true value since the higher-order terms of w_stem and w_base have been neglected. For example, for a barrier with h and b equal to 5 m and w_stem and w_base equal to 0.5 m, the value of ${I}_{z_1{z}_1}$ is underestimated by 5% if estimated using Eq. A9. This is deemed acceptable for the purpose of routine design since the error is on the safe side. In the main text, I_z is written as I_θ to denote that it is the rotational inertia about the axis of rotation.

Rectangular side walls

Figure 8 shows an L-shaped barrier with two rectangular side walls. The side walls have height d and length c, and the mass of a single side wall is M_side. The rotational inertia of a single rectangular side wall about its centroidal axis z_c is:

Fig. 8

A schematic L-shaped barrier with rectangular side walls and reference axes

$$ {I}_{z_{\mathrm{c}}\left(\mathrm{side}\right)}={M}_{\mathrm{side}}\left(\frac{c^2+{d}^2}{12}\right) $$

(A10)

where d = h − w_base and c = l − w_stem. By the parallel axis theorem, the side wall’s rotational inertia about the axis z is:

$$ {I}_{z\left(\mathrm{side}\right)}={I}_{z_{\mathrm{c}}\left(\mathrm{side}\right)}+{M}_{\mathrm{side}}{r}_{\mathrm{side}}^2 $$

(A11)

where r_side is the distance between axes z_c and z, and can be expressed as:

$$ {r}_{\mathrm{side}}^2={\left(\frac{d}{2}+{w}_{\mathrm{base}}\right)}^2+{\left(\frac{c}{2}+{w}_{\mathrm{stem}}\right)}^2 $$

(A12)

Substituting Eqs. A10 and A12 into Eq. A11 and rearranging gives:

$$ {I}_{z\left(\mathrm{side}\right)}={M}_{\mathrm{side}}\left[\left(\frac{c^2+{d}^2}{12}\right)+{\left(\frac{d}{2}+{w}_{\mathrm{base}}\right)}^2+{\left(\frac{c}{2}+{w}_{\mathrm{stem}}\right)}^2\right] $$

(A13)

So far, only a single side wall has been considered. For barriers with two identical side walls or more, the combined rotational inertia is therefore:

$$ {I}_{z\left(\mathrm{side}\right)}={nM}_{\mathrm{side}}\left[\left(\frac{c^2+{d}^2}{12}\right)+{\left(\frac{d}{2}+{w}_{\mathrm{base}}\right)}^2+{\left(\frac{c}{2}+{w}_{\mathrm{stem}}\right)}^2\right] $$

(A14)

where n is the number of side walls. Summing Eqs. A9 and A14 gives the rotational inertia of the whole barrier:

$$ {I}_z\approx \frac{M_{\mathrm{stem}}}{3}\left({h}^2+{hw}_{\mathrm{base}}\right)+\frac{M_{\mathrm{base}}}{3}\times {b}^2+{nM}_{\mathrm{side}}\left[\left(\frac{c^2+{d}^2}{12}\right)+{\left(\frac{d}{2}+{w}_{\mathrm{base}}\right)}^2+{\left(\frac{c}{2}+{w}_{\mathrm{stem}}\right)}^2\right] $$

(A15)

Equation A15 can be used to calculate the rotational inertia of an L-shaped barrier with any number of rectangular side walls or counterforts about axis z, which is the assumed axis of rotation. Equation A15 is only approximate but the error is on the safe side as previously discussed. In the main text, I_z is shown as I_θ to denote that it is the rotational inertia about the axis of rotation.

Appendix B

Figure 9 shows an L-shaped barrier at the critical overturning condition, that is, when the barrier’s centre of gravity lies immediately above its point of rotation. The following derivation is applicable to any shape so long as the barrier’s centre of gravity is known. As shown in Fig. 9, the initial centre of gravity is located at the horizontal distance $\overline{x}$ measured from the outer edge of the stem wall and the vertical distance $\overline{y}$ measured from the bottom of the base slab. The distance between the point of rotation and the centre of gravity is thus $\sqrt{{\overline{x}}^2+{\overline{y}}^2}$. At the critical overturning condition, the barrier’s angle of rotation and the rise of the centre of gravity are denoted as θ_crit and Δ_{C. G. (crit)} respectively. From geometry, it can be shown that:

Fig. 9

Cross-section of a schematic L-shaped barrier at critical overturning condition

$$ {90}^{{}^{\circ}}={\theta}_{\mathrm{crit}}+\beta $$

(B1)

Substituting $\beta ={\tan}^{-1}\left(\frac{\overline{y}}{\overline{x}}\right)$ and rearranging thus gives:

$$ {\theta}_{\mathrm{crit}}={90}^{{}^{\circ}}-{\tan}^{-1}\left(\frac{\overline{y}}{\overline{x}}\right) $$

(B2)

The rise of the centre of gravity, Δ_{C.G. (crit)}, can be expressed as:

$$ {\varDelta}_{\mathrm{C}.\mathrm{G}.\left(\mathrm{crit}\right)}=\sqrt{{\overline{x}}^2+{\overline{y}}^2}-\overline{y} $$

(B3)