Finding the volume of an oblique prism

CivilSigma

Problem Statement
The current expected rain water uniform depth on a flat roof is 97 mm for a total volume of 69.84 m^3
Find the new equivalent water depth if the roof is expected to deflect downwards with a slope of 10m to 250mm according to the following image.
https://imgur.com/a/eXHnyw6
Relevant Equations
V= Base * Width * Height
The expected shape of the deflected roof will look like a pyramid with a water height of 'h'.
I am having a hard time to find the a formula for the new shape.
Can some one provide me some guidance please, will I have to use calculus to find the volume for this particular shape?

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LCKurtz

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I'm trying to visualize the shape of the roof from your diagram. Assuming the rectangle is a top view, are the blue shapes supposed to represent the side views? I'm going to assume so. Here's a 3D picture of what I think the roof looks like, with dimensions exaggerated:

This should make it possible to calculate your volume, I think. Maybe I will crunch some numbers later, if necessary.
Edit, added: I get $70\text{m}^3$ for the volume the roof will hold.

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CivilSigma

I got a total volume of 60 m^3 .
My professor posted a tutorial. He simplified the shape into 4 corners and a middle portion.

For the corners, (the 10m by 12 m plan segments that slope downwards), the volume of this oblique triangular prism is:

$$V= 12 \cdot 10 \cdot 0.250/2 \cdot 1/2 = 7.5 m^3 \cdot 4 = 30 m^3$$

I have to say that I don't understand why he divides the height by 2...

Then for the middle segment which is a triangular prism:

$$V= (12+12) \cdot 0.250 \cdot 10 \cdot 1/2 = 30 m^3$$

LCKurtz

Homework Helper
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Well, one of us has a mistake. Here's another figure with my analysis below it:

I'm going to focus in the volume of the block minus the volume under the roof. Look at the right end. There is a rectangular pyramid with green base and vertex O. Its volume is $V_1=\frac 1 3 2bha$. On the front is a rectangular pyramid with blue base and vertex O. Its volume is $V_2=\frac 1 3 ahb$. The pyramid with the tan base has the same volume as $V_2$. So the volume under the right third of the roof is $V_1+2V_2 = \frac 4 3 abh$. The left end has the same volume under the roof. So far we have $\frac 8 3 abh$. The yellow rectangle is one of the faces of a triangular prism with volume $V_3 = \frac 1 2 hba$ and there is another on the other side. Add these to what we already have for a total volume under the roof of $\frac {11} 3 hba$. The volume of the solid block is $V=3a2bh =6abh$. So the volume above the roof is $V=6abh -\frac{11} 3 abh = \frac 7 3 abh$. In your picture $a=10,~b=12,~h = \frac 1 4$. So $V = \frac 7 3\cdot 10 \cdot 12\cdot \frac 1 4 = 70$.

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CivilSigma

Wow, thank you. That was really well explained, I really appreciate your help

LCKurtz

Homework Helper
Gold Member
Wow, thank you. That was really well explained, I really appreciate your help
So did you figure out what your Professor had wrong and point it out to him?

CivilSigma

So did you figure out what your Professor had wrong and point it out to him?
To be honest, I did not figure out his error. I' am taking a condensed summer class and I kind of let it go as we have already moved on to other topics and assignments/projects.

"Finding the volume of an oblique prism"

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