Question 5 of 7: Damage Radius from a Charge Attenuation Relationship

Question 5: Damage Radius from a Charge Attenuation Relationship (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Charge diameter D=102 mm, density ρ=1.2 g/cm³, length L=4 m; attenuation law $$u=700\,W^{0.7}/R^{1.4}$$.

Find. The expected damage radius R.

Approach. Compute the total charge weight W from the cylindrical charge's volume and density, adopt a critical particle-velocity threshold marking the onset of rock damage, and solve the given attenuation law for R at that threshold.

1. Charge weight. $$W=\frac{\pi}{4}D^2L\,\rho=\frac{\pi}{4}(0.102)^2(4)(1200)=\boxed{39.2\ \text{kg}}$$
2. Adopt the damage threshold. No damage-criterion PPV is stated in the question, so the widely used Bauer-Calder threshold for the onset of new/extended rock fracturing (u_crit≈700 mm/s, the boundary between "minor cracking" and "extensive cracking/damage") is adopted – which happens to equal the formula's own leading coefficient.
3. Solve for R at u=700 mm/s. Because the threshold equals the coefficient, the equation collapses to a clean closed form independent of the charge's actual size: $$700=700\frac{W^{0.7}}{R^{1.4}}\ \Rightarrow\ R^{1.4}=W^{0.7}\ \Rightarrow\ R=W^{0.5}=\sqrt{W}$$ $$R=\sqrt{39.2}=\boxed{6.26\ \text{m}}$$

Discussion of approximations. (1) The charge is a 4 m long cylindrical column, not a true point/spherical source, yet the formula given is explicitly a spherical charge attenuation law – applying it with the full column weight W treats the whole charge as a single point source, which overstates the near-field particle velocity (and therefore understates R only mildly, since R≈6.3 m is already several charge lengths away, where the point-source approximation becomes reasonable) but would be poor practice for evaluating points close to the charge (R<L). (2) The damage threshold itself (u_crit=700 mm/s) is an assumption, not given by the source; a stricter crushing threshold (≈2500 mm/s) would give a smaller damage radius ($$R=\sqrt{W}\times(700/2500)^{1/1.4}\approx3.4\ \text{m}$$ for reference), while a looser minor-cracking threshold (≈500 mm/s) would give a larger one – the reported 6.26 m radius should therefore be read as the boundary of the "extensive cracking/damage" zone, not the boundary of total destruction nor of zero effect. (3) Ground/rock-mass properties (jointing, anisotropy) are assumed uniform and isotropic around the charge, which is rarely exactly true underground or in a jointed rock mass – actual damage extent will vary by direction, generally further along discontinuities and less across them.