HOLE VOLUME

Correct concrete weights are important when loading a trailer or determining lifting required for the product.  You can not calculate the correct hole volume using the standard pi*r^2*wall thickness because the hole formed by intersecting two cylinders is not a round shape.  We will reference previous calculations from the intersecting cylinders page to start our calculations.

For all calculations below a= Hole Radius,  b = Manhole Radius and all dimensions are in inches

From the front view we can make the following determinations for \[0<=t<=2\pi\]

Front View
Front View

\[cos(t)= \frac{x}{a}\]

\[x = a * cos(t)\]

and

\[sin(t) = \frac{z}{a}\]

\[z = a * sin(t)\]

 

 

 

 

Manhole OD for hole volume
Top View

From the top view we can solve for the y value.  Since we are trying to find the volume we will also need to solve for the y for the outside diameter of the manhole.

\[b^2 = y^2 + x^2\]

\[y = \pm \sqrt{b^2-x^2}\]

\[y_{ID}(t) = \pm \sqrt{b_{ID}^2-x^2}\]

\[y_{OD}(t) = \pm \sqrt{b_{OD}^2-x^2}\]

 

 

To find the hole volume we will integrate in terms of x from 0 to a – the hole radius.  We will need to get x(t) in terms of t and substitute t into the equation for z(t) to have everything in terms of x.  We will be calculating the area throughout the bounds of the integral of a rectangle formed by the y and z axis.

\[x = a * \cos(t)\]

\[\frac{x}{a} = \cos(t)\]

\[\arccos(\frac{x}{a}) = t\]

Now we need to substitute t into the equation for z.

\[z = a * \sin(t)\]

\[z = a * \sin(\arccos(\frac{x}{a}))\]

For simplicity we will integrate 1/4 of the hole and multiply the result x 4.

\[4*\int_0^a (y_{OD}-y_{ID})*z \, dx\]

\[4*\int_0^a (\sqrt{b_{OD}^2-x^2}-\sqrt{b_{ID}^2-x^2})*a * \sin(\arccos(\frac{x}{a})) \, dx\]

The hole volume for a 30″ Diameter hole in a 4′ diameter structure the integral would look as follows:

\[V=4*\int_0^{15} (\sqrt{29^2-x^2}-\sqrt{24^2-x^2})*15 * \sin(\arccos(\frac{x}{15})) \, dx\]

\[V=3968.46 in^3 \]

Using the standard formula for a circle and wall thickness you would get the following:

\[V= \pi*r^2*Wall Thickness\]

\[V= \pi*15^2*5\]

\[V= 3532.5 in^3\]

The difference in the example holes is small but it can really make a difference when dealing with larger structures and holes.

Click Here to Try the Hole Volume Calculator