What is the largest diameter a vertical mine shaft may have in bedrock?

Question

What are the design constraints to the diameter of a mine shaft in bedrock, considering the pressure of the ambient rock, its compression strength, the width of the liner, and its compression strength? The ambient rock of the Kola borehole was under such great pressure and strained so easily because it was so hot that it just spontaneously sealed itself. How could one calculate whether that will happen?

Answer 1

You could try to estimate stresses at the edge of the borehole. Assuming bedrock material density $\rho$ and gravitational acceleration $g$, the pressure at a depth $h$ would be:

$$P_d = \rho\cdot g\cdot h$$

Now consider vertical cylindrical borehole with diameter $r_i$ and a horizontal section at depth $h$. At this section at infinite radius from the hole center, the internal stresses should be in equilibrium with the pressure $P_d$. We can modify Lamé solution of stresses in thick cylinder under pressure assuming that outer diameter is infinity. For zero pressure at internal radius $r_i$ and a cylinder with external radius $r_e$:

$$\sigma_r(r) = -P_d\cdot \frac{1-\frac{r_i^2}{r^2}}{r_e^2-r_i^2}\cdot r_e^2$$

Radial stress limit for external radius going to infinity : $$\sigma_{r,\infty}(r) = \lim\limits_{r_e\rightarrow\infty}\sigma_r(r) = -P_d\cdot \left(1-\frac{r_i^2}{r^2}\right)\cdot \lim\limits_{r_e\rightarrow\infty}\frac{r_e^2}{r_e^2-r_i^2} = -P_d\cdot \left(1-\frac{r_i^2}{r^2}\right)$$

Tangential stress can be derived from equilibrium:

$$\sigma_{t,\infty}(r) = \sigma_{r,\infty}(r)+\frac{\partial \sigma_{r,\infty}(r)}{\partial r}\cdot r = -P_d\cdot \left(1+\frac{r_i^2}{r^2}\right)$$

Lastly, the vertical stress component should be constant: $$\sigma_z(r) = -P_d$$

Finally, the stresses will be most extreme at the edge:

$$\sigma_{r,\infty}(r=r_i) = 0$$ $$\sigma_z(r=r_i) = -P_d$$ $$\sigma_{t,\infty}(r=r_i) = -2P_d$$

As you can see, the stresses are independent of the hole radius, so the limiting factor would be just the depth. However, this is purely from the stress perspective, with many simplifications and omitting factors like erosion.

Answer 2

It depends on the rock and bore size. If I remember right chalk is a pain for flowing into the bore ,also salt. Any serious hole will use steel casing , likely a few concentric strings. Steel casing strength to 125 ksi is standard and 150 ksi is optional. Welded line pipe is an option for diameters over about 3ft. Then strength will be up to about 80 Ksi . Wall thickness will depend on the manufactures capability. I presume cost is no concern. Collapse strength of steel casing is calculated using API bul. 5C3. There are also other techniques like two concentric casing strings with cement filling the annulus.