Question

A construction company estimates the volume of concrete to pour in a house foundation with rectangular slab containment sized as: length = 18.30 +/- 0.02 m; width = 12.20 +/- 0.02 m; height = 0.50 +/- 0.05 m. calculate the uncertainty of the necessary volume of concrete, and the amount that would not fall short for the job.

Answer 1

IF the estimates are 100% accurate, then the maximum volume of concrete is the product of the maximum dimensions, namely
Vmax=18.32*12.22*0.55=123.13 m^3
and the minimum volume of concrete is the product of the minimum dimensions
Vmin=18.28*12.18*0.45=100.19 m^3
the uncertainty is therefore 123.13-100.19=22.94 m^3

On the practical side, to this uncertainty must be added to 
- squareness of the formwork
- evenness of the surface of the crushed stones on which the foundation sits on
- excess of concrete delivered by the truck
- volume of air bubbles trapped in the the concrete, ...

Answer 2

Answer:

•Uncertainty = 0.10

• Amount that will not fall short = 122.79 m³

Explanation:

Given:

L= 18.30 +/- 0.02 m

W = 12.20 +/- 0.02 m

H = 0.50 +/- 0.05 m

We know Volume = L*W*H

For log V, we have:

ln V = ln L + In W + ln H

Lets diferentiate, we have:

$\frac{dV}{V} = \frac{dL}{L} + \frac{dW}{W} + \frac{dH}{H}$

$= \frac{0.02}{18.30} + \frac{0.02}{12.20} + \frac{0.05}{0.50}$

= 0.1027

=> 0.10

For V = LWH +/- dv:

V = (18.30 * 12.50 * 0.50) +/- (0.10 * 18.30 * 12.20 * 0.50)

= 111.63 +/- 11.16

Amount that wil not fall short=

111.63 + 11.16

= 122.79 m³