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Finding the volume of an oblique prism

  • Thread starter CivilSigma
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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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fresh_42

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Please upload images. If the link will get broken, the thread might become unreadable.
 

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:
roof.jpg

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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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

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Well, one of us has a mistake. Here's another figure with my analysis below it:
roof2.jpg

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

LCKurtz

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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?
 
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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.
 

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