Here is some information about a grain storage bin. Grain Storage Bin

Item	Value
Height (h)	6880 mm
Diameter (d)	4550 mm
Mass	1100 kg

Safety information: The bin can be filled to a maximum of 85% of its total volume. In this question, the grain storage bin can be modelled as a right cylinder with a right conical top. The height of the conical top is half the radius of the bin.

1. Find the base area using the radius in metres

Diameter $=4550\text{ mm}=4.55\text{ m}$

Radius $r=\dfrac{4.55}{2}=2.275\text{ m}$

Area of base
$A=\pi r^2=\pi(2.275)^2=16.258\ldots$

$A\approx 16.3\text{ m}^2$

1. Find the radius and the separate heights of the cone and cylinder in metres

$d=4550\text{ mm}=4.55\text{ m}$

$r=\dfrac{4.55}{2}=2.275\text{ m}$

Cone height $=\dfrac{1}{2}r=\dfrac{1}{2}(2.275)=1.1375\text{ m}$

Cylinder height $=6.880-1.1375=5.7425\text{ m}$

2. Calculate the volume of the cylindrical part

Volume of cylinder $=\pi r^2h$

$=\pi(2.275)^2(5.7425)$

$=93.369\ldots\text{ m}^3$

3. Calculate the volume of the conical top and add it to the cylinder volume

Volume of cone $=\dfrac{1}{3}\pi r^2h$

$=\dfrac{1}{3}\pi(2.275)^2(1.1375)$

$=6.165\ldots\text{ m}^3$

Total volume $=93.369\ldots+6.165\ldots=99.534\ldots\text{ m}^3$

$\approx 99.5\text{ m}^3$

1. Find the safe volume of grain using the 85% fill limit

From part (b), the modelled volume of the bin is $99.5\text{ m}^3$.

$\text{Safe volume}=85\%\times 99.5$

$=0.85\times 99.5=84.575\text{ m}^3$

2. Find the mass of grain at the safe volume

Density of grain $=410\text{ kg/m}^3$.

$\text{Mass of grain}=84.575\times 410$

$=34675.75\text{ kg}$

3. Add the mass of the empty bin to get the total mass

Mass of empty bin $=1100\text{ kg}$.

$\text{Total mass}=34675.75+1100$

$=35775.75\text{ kg}$

4. Convert the total mass to weight in kN

Given that $1000\text{ kg}$ is equivalent to $9.81\text{ kN}$,

$\text{Weight}=\frac{35775.75}{1000}\times 9.81$

$=350.56\text{ kN}\text{ approximately}$

5. Find the total weight per square metre on the ground

From part (a), the base area is $16.3\text{ m}^2$.

$\text{Weight per square metre}=\frac{350.56}{16.3}$

$=21.5\text{ kN/m}^2\text{ approximately}$

6. Compare with the limit and justify using the underestimation statement

$21.5\text{ kN/m}^2>20\text{ kN/m}^2$

The calculated value is already greater than the limit. Also, the model is an underestimation of the actual storage capacity, so the actual maximum grain weight would be greater than this calculated value.