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Grain pile

How can I derive the formula of the volume of a grain pile in terms of w,l,h,t?

5xum
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orangevsbanana
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    Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. – 5xum Mar 02 '20 at 10:11
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum Mar 02 '20 at 10:11
  • Try turning it into 3 shapes to solve - a triangular based prism in the centre, and two pyramids, one at either end – lioness99a Mar 02 '20 at 10:27
  • Work out the area of a cross-section horizontal rectangle at height x, then integrate from 0 to h – Cato Mar 02 '20 at 10:32

2 Answers2

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We can divide this shape into $3$ parts in order to find the total volume - one triangular based prism in the centre and two rectangular based pyramids, one at either end. We do this by dropping verticals from either end of the length $t$.

We can then find the volumes of these shapes separately - the volume of a prism is:

\begin{align}V_{\text{prism}}&=\text{area of cross section}\times\text{length of side}\\ &=\frac 12 wh \times t\\ &=\frac{wht}2\end{align}

The volume of one pyramid is:

\begin{align}V_{\text{pyramid}}&=\frac 13 \times \text{base area}\times\text{height}\\ &=\frac13 \times \left(w\times \frac{l-t}2\right)\times h\\ &=\frac{hw(l-t)}6\end{align}

Now we can calculate the total area:

\begin{align}T_{\text{total}}&=V_{\text{prism}}+2\times V_{\text{pyramid}}\\ &=\frac{wht}2+2\times\frac{hw(l-t)}6\\ &=\frac{hw(2l+t)}6\\ &=\frac{hlw}3+\frac{htw}6 \end{align}

lioness99a
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At height $x$ from the base, the horizontal cross section is a rectangle whose dimensions range linearly from $w \times l$ to $t \times 0$

so, area of horizontal cross-section at height x is

$w(h-x) / h \times (l \times (h -x) + t \times x) / h$

$=(wh - wx) (lh - lx + tx) / h^2$

$=(lwh^2 - lwxh + txwh - lwhx + lwx^2 - wtx^2) / h^2$

now integrate x from zero to h, meaning that we are summing the areas of infinitesimal rectangles whose areas we know at a height x from the base

$\int_0^h (lwh^2 - lwxh + txwh - lwhx + lwx^2 - wtx^2) / h^2 dx$

$=[lwx - \frac{1}{2}lwx^2/h + \frac{1}{2}twx^2/h - \frac{1}{2}lwx^2/h + \frac{1}{3}lwx^3/h^2 - \frac{1}{3}wtx^3/h^2]_0^h$

$= lwh - \frac{1}{2}lwh + \frac{1}{2}twh - \frac{1}{2}lwh + \frac{1}{3}lwh - \frac{1}{3}wth$

$= \frac{1}{2}twh + \frac{1}{3}lwh - \frac{1}{3}twh$

$ = wh(\frac{1}{3}l + \frac{1}{6}t)$

Cato
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