Question

In a bord and pillar panel, a square pillar of size 35 m \( \times \) 35 m (centre to centre) is extracted to form four equal square-shaped stooks as shown. The width of each gallery and crosscut is 5 m. The height of the working seam is 3 m. The reduction in safety factor after pillar splitting by using Bieniawski’s pillar strength formula, in %, is _______ (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{53.png}
Bieniawski's pillar strength formula is given by \( S_p = S_1 \left( 0.64 + 0.36 \frac{w}{h} \right) \), where \( S_1 \) is the strength of a 0.9 m\(^3\) coal block, \( w \) is the pillar width, and \( h \) is the mining height.

Hint:

Bieniawski’s pillar strength formula helps estimate the safety factor by considering the width and height of the pillar, and the strength of the coal. After pillar splitting, the safety factor typically reduces due to a decrease in pillar width.

Answer 1

Correct Answer: 0.6

Solution: 1. Initial Pillar Dimensions and Area: Centre to centre dimension of initial pillar = 35 m $\times$ 35 m Width of gallery/crosscut = 5 m Actual width of initial pillar ($w_1$) = $35 \, {m} - 5 \, {m} = 30 \, {m}$ Area of initial pillar ($A_1$) = $w_1^2 = (30 \, {m})^2 = 900 \, {m}^2$ 2. Dimensions and Area of the Stooks: Width of each stook ($w_2$) = $\frac{30 \, {m} - 5 \, {m}}{2} = \frac{25 \, {m}}{2} = 12.5 \, {m}$ Area of each stook ($A_{stook}$) = $w_2^2 = (12.5 \, {m})^2 = 156.25 \, {m}^2$ Total area of four stooks ($A_2$) = $4 \times A_{stook} = 4 \times 156.25 \, {m}^2 = 625 \, {m}^2$ 3. Mining Height: Height of the working seam ($h$) = 3 m 4. Bieniawski's Pillar Strength: Formula: $S_p = S_1 \left(0.64 + 0.36 \frac{w}{h}\right)$ 5. Initial Pillar Strength ($S_{p1}$): $S_{p1} = S_1 \left(0.64 + 0.36 \frac{30 \, {m}}{3 \, {m}}\right) = S_1 (0.64 + 0.36 \times 10) = S_1 (0.64 + 3.6) = 4.24 \, S_1$ 6. Strength of Each Stook ($S_{p2}$): $S_{p2} = S_1 \left(0.64 + 0.36 \frac{12.5 \, {m}}{3 \, {m}}\right) = S_1 (0.64 + 0.36 \times 4.1667) = S_1 (0.64 + 1.5000) = 2.14 \, S_1$ 7. Safety Factor: Initial Safety Factor ($SF_1$) $\propto \frac{S_{p1}}{A_1} = \frac{4.24 \, S_1}{900}$ Safety Factor after Splitting ($SF_2$) $\propto \frac{S_{p2}}{A_{stook}} = \frac{2.14 \, S_1}{156.25}$ (for one stook) 8. Reduction in Safety Factor: Ratio of safety factors: $\frac{SF_2}{SF_1} = \frac{\frac{2.14 \, S_1}{156.25}}{\frac{4.24 \, S_1}{900}} = \frac{2.14}{156.25} \times \frac{900}{4.24} = \frac{1926}{662.5} \approx 2.9072$ This ratio indicates an increase in the value of $\frac{S_p}{Area}$ for a single stook compared to the initial pillar. However, we need to consider the overall stability of the panel. Let's consider the safety factor in terms of average stress assuming constant load on the panel. Initial average stress ($\sigma_1$) $\propto \frac{1}{A_1} = \frac{1}{900}$ Average stress after splitting ($\sigma_2$) $\propto \frac{1}{A_2} = \frac{1}{625}$ (total area of stooks) $SF_1 \propto S_{p1} \times A_1 = 4.24 S_1 \times 900$ (considering capacity) $SF_2 \propto S_{p2} \times A_{stook} \times 4 = 2.14 S_1 \times 156.25 \times 4 = 1337.5 S_1$ (total capacity of stooks) $\frac{SF_2}{SF_1} = \frac{1337.5 S_1}{4.24 \times 900 \times S_1} = \frac{1337.5}{3816} \approx 0.3505$ Percentage reduction in safety factor = $\left(1 - \frac{SF_2}{SF_1}\right) \times 100 = (1 - 0.3505) \times 100 = 64.95 %$ Final Answer: The reduction in safety factor is $64.95 %$.