A tunnel of length L = 141 m, height H = 6.9 m high, and width 6.0 m (with a flat roof) is to be constructed at distance d = 60 m beneath the ground. 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.
What is the total weight of the ground material the columns must support?
totalweight= densityEarth*volume
= density*141*60*6.0 m^3
and the density=2.8g/cm^3 which is 2800kg/m^3
How many columns are needed to provide a safety factor of two against rupture.
To find the total weight of the ground material that the columns must support, we need to calculate the volume of the ground material in the tunnel and then multiply it by the mass of 1.0 cm^3 of the ground material.
First, let's calculate the volume of the tunnel. The volume of a rectangular prism (which represents the tunnel) is given by the formula:
Volume = Length x Width x Height
Substituting the given values, we have:
Volume = 141 m x 6.0 m x 6.9 m
Next, we need to convert the volume to cm^3 because the mass of the ground material is given in grams per cm^3. Since 1 m^3 is equal to 1,000,000 cm^3, we have:
Volume = (141 m x 6.0 m x 6.9 m) x (1,000,000 cm^3 / 1 m^3)
Now we have the volume of the tunnel in cm^3. Let's calculate the total weight of the ground material that the columns must support by multiplying the volume by the mass of 1.0 cm^3 of the ground material:
Total weight = Volume x Mass per cm^3
Substituting the given value for the mass per cm^3 (2.8 g/cm^3), we have:
Total weight = (141 m x 6.0 m x 6.9 m) x (1,000,000 cm^3 / 1 m^3) x 2.8 g/cm^3
Calculating this expression will give us the answer to the question.