Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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 cm². The mass of 1.0 cm³ 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 $1.4 \times 10^{9} N$ .
Number of columns that are needed to keep are 75.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the given quantities

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 cm²
Mass of 1.0 cm³ of the ground material is 2.8 g

Step 2: Determine the formulas:

Consider the formula for the density as:

$ρ = \frac{m}{v}$ ….. (i)

Here, $ρ$ 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.

Step 3: (a) Determine the total weight of the ground material the columns must support

Determine the area of the roof as:

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

Consider the volume of the material is obtained as:

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

Solve for the density of the material as:

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

Determine the mass using equation (i) as,

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

Use the equation (ii) to find the weight.

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

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

Step 4: (b) Determine the number of columns to keep the compressive stress on each column at one-half its ultimate strength

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

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

Substitute the values and solve as:

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

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.

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Determine the given quantities

Step 2: Determine the formulas:

Step 3: (a) Determine the total weight of the ground material the columns must support

Step 4: (b) Determine the number of columns to keep the compressive stress on each column at one-half its ultimate strength

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Determine the given quantities

Step 2: Determine the formulas:

Step 3: (a) Determine the total weight of the ground material the columns must support

Step 4: (b) Determine the number of columns to keep the compressive stress on each column at one-half its ultimate strength

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics