91Ó°ÊÓ

1. A tunnel of length L=150 m and height H=7.2 m and width 5.8 m which is at distance, d=60 m
2. Area of tunnel roof is 960 cm²
3. Mass of 1.0 cm³ of the ground material is 2.8 g

Consider the formula for the density as:

$\mathrm{ÒÏ}=\frac{\mathrm{m}}{\mathrm{v}}$ ….. (i)

Here, $\mathrm{ÒÏ}$is density of the material, m is mass, and v is volume.

Consider the formula for the weight as:

w=mg ……. (ii)

Here, w is weight, m is mass, and g is gravitational acceleration.

Determine the area of the roof as:

$\begin{array}{rcl}\mathrm{A}& =& 150\mathrm{m}×5.8\mathrm{m}\\ & =& 870{\mathrm{m}}^2\end{array}$

Consider the volume of the material is obtained as:

$\begin{array}{rcl}\mathrm{V}& =& \mathrm{A}×\mathrm{d}\\ & =& 870{\mathrm{m}}^2×60\mathrm{m}\\ & =& 52200{\mathrm{m}}^3\end{array}$

Solve for the density of the material as:

$\begin{array}{rcl}\mathrm{ÒÏ}& =& 2.8\frac{\mathrm{g}}{{\mathrm{cm}}^3}\\ & =& 2800\frac{\mathrm{kg}}{{\mathrm{m}}^3}\end{array}$

Determine the mass using equation (i) as,

$\begin{array}{rcl}\mathrm{ÒÏ}& =& \frac{\mathrm{m}}{\mathrm{v}}\\ \mathrm{m}& =& \mathrm{ÒÏ}×\mathrm{v}\\ & =& 2800\mathrm{kg}/{\mathrm{m}}^3×52200{\mathrm{m}}^3\\ & =& 1.46×{10}^8\mathrm{kg}\end{array}$

Use the equation (ii) to find the weight.

$\begin{array}{rcl}\mathrm{w}& =& \mathrm{mg}\\ & =& 1.46×{10}^3\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2\\ & =& 1.4×{10}^9\mathrm{N}\end{array}$

Therefore, the weight of the ground material the column must support is $1.4×{10}^9\mathrm{N}$.

Consider the ultimate strength of the steel is$400×{10}^6\frac{\mathrm{N}}{{\mathrm{m}}^2}$. Multiplying it by area of the column would give us the strength of the columns. All the columns should support the weight of the ground and compressive stress on each column should be less than half of the ultimate strength, so we have:

$\mathrm{n}=\frac{\mathrm{w}}{{\displaystyle \frac{1}{2}}×\mathrm{P}×\mathrm{A}}$

Substitute the values and solve as:

$\begin{array}{rcl}\mathrm{n}& =& \frac{1.4×{10}^9\mathrm{N}}{{\displaystyle \frac{1}{2}}×400×{10}^6\mathrm{N}/{\mathrm{m}}^2×960×{10}^{-4}{\mathrm{m}}^2}\\ & =& 75\end{array}$

Therefore, the number of columns that are needed to keep compressive stress on each column should be less than half of the ultimate strength is 75.