Solving Elasticity Problem: Weight & Columns

[mikejones2000]

A tunnel of length L = 154 m, height H = 7.4 m high, and width 6.1 m (with a flat roof) is to be constructed at distance d = 60 m beneath the ground. (See Figure 12-50.) 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?
N
(b) How many columns are needed to keep the compressive stress on each column at one-half its ultimate strength?
columns

I know that for the first question the total weight must be equal to the volume x density x gravity but for some reason I keep screwing up finding the density and the number is much too low. Am I approaching this question badly?

 

[3trQN]

I got something near 157 Tonne of rock.

But my maths is weak.

Make sure your using the right units, wright the units out for each step and ensure that your density is correct ( g per square meter or centimeter ).

$$1m^{2} = 1m\times1m = 10cm\times10cm = 100cm^{2}$$
therefor the density is:
$$2.8g cm^{-2} = 2.8(100)g m^{-2}$$

 

[kyle8921]

Your approach to finding the density and the total weight is correct. The density of the ground material can be calculated by dividing the mass of 1.0 cm3 by its volume, which is equal to 2.8 g/1.0 cm3 = 2.8 g/cm3.

To find the total weight of the ground material, we need to first calculate the volume of the tunnel. The volume of a rectangular prism (the shape of the tunnel) is equal to its length x width x height. In this case, the volume of the tunnel is 154 m x 6.1 m x 7.4 m = 7,708.4 m3.

Now, we can use the formula for weight (W = m x g) to find the total weight of the ground material. The mass (m) is equal to the density x volume, so the total weight is equal to (2.8 g/cm3 x 7,708.4 m3) x 9.8 m/s2 = 211,222,848 N. So, the total weight that the columns must support is approximately 211,222,848 N.

For the second question, we need to find the ultimate strength of the columns. The compressive stress on each column is equal to the weight it is supporting divided by its cross-sectional area. In this case, the compressive stress should not exceed half of the ultimate strength. So, we can set up an equation as follows:

211,222,848 N / (960 cm2 x x) = 0.5 x ultimate strength

Where x represents the number of columns needed. Solving for x, we get x = 437. So, approximately 437 columns are needed to keep the compressive stress on each column at half of its ultimate strength.