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47P

Expert-verifiedFound in: Page 344

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A tunnel of length L=150 m, height H=7.2 m, and width 5.8 m (with a flat roof) is to be constructed at distance d=60 m beneath the ground. (See the Figure.) The tunnel roof is to be supported entirely by square steel columns, each with a cross-sectional area of 960 cm ^{2}. The mass of 1.0 cm^{3} of the ground material is 2.8 g. (a) What is the total weight of the ground material the columns must support? (b) How many columns are needed to keep the compressive stress on each column at one-half its ultimate strength?**

- The total weight of the ground material the columns must support is $1.4\times {10}^{9}\mathrm{N}$.
- Number of columns that are needed to keep are 75.

- 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
- Area of tunnel roof is 960 cm
^{2} - Mass of 1.0 cm
^{3}of the ground material is 2.8 g

** ****Consider the formula for the density as:**

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

Here, $\mathrm{\rho}$ 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:

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

Consider the volume of the material is obtained as:

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

Solve for the density of the material as:

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

Determine the mass using equation (i) as,

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

Use the equation (ii) to find the weight.

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

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

Consider the ultimate strength of the steel is $400\times {10}^{6}\frac{\mathrm{N}}{{\mathrm{m}}^{2}}$. 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:

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

Substitute the values and solve as:

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

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.

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