以下是关于一个谷物储藏仓的信息。 谷物储藏仓

项目	数值
高度 (h)	6880 mm
直径 (d)	4550 mm
质量	1100 kg

安全信息：该储藏仓最多可装载至其总容积的 85%。 在本题中，该谷物储藏仓可以建模为一个直圆柱体和一个直圆锥体顶部的组合。圆锥形顶部的高度是储藏仓半径的一半。

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.