Solving Elasticity Problem: Weight & Columns

  • Thread starter Thread starter mikejones2000
  • Start date Start date
  • Tags Tags
    Elasticity
AI Thread Summary
The discussion focuses on calculating the total weight of ground material that steel columns must support for a tunnel construction project. The tunnel dimensions are provided, and the density of the ground material is specified as 2.8 g/cm³. The participant struggles with the calculations, particularly in determining the correct density and applying the appropriate units. They initially estimated the weight to be around 157 tonnes but suspect their approach may be flawed. Emphasis is placed on ensuring correct unit conversions and calculations to accurately determine the total weight and the number of columns needed to maintain safe compressive stress levels.
mikejones2000
Messages
18
Reaction score
0
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?
 
Physics news on Phys.org
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 ).

<br /> 1m^{2} = 1m\times1m = 10cm\times10cm = 100cm^{2}<br />
therefor the density is:
<br /> 2.8g cm^{-2} = 2.8(100)g m^{-2}<br />
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top