equilateral octagon help, you all awesome Mathematicians

[JTvede]

Hello, I have A Concrete Slab
thickness question. This will be a 10' tall equilateral Octagon ( 5" thick walls ) above ground Swimming Pool , with the Decking on Top.

Some Parameters are here.
1) Each side of Octagon is 10'
therefore the Width Across the entire Octagon is 20'

2) The Surface Area is 400sf
400 sf ( 8 triangular pies x .5B'x H' )

3) The Height of the Octagon is 10'

4) Weight of Walls is 75,000 lbs
The total amount ( 5" thick walls only ) of 5000 psi air entrained
concrete calculates to 12.34 cubic yards ( L'xW' divided by 12 x D" divided
by 27 = cubic yards ) ( then multiplied x 8 walls at 5" thick)

5) Weight of Pool Water is 263,840 lbs
400sf of Walls x 10' Height x 62.46lbs( 1cf water weight )

6) Concrete for Slab will be 5000psi with air

7) Rebar in Walls will be #4 (1/2") and weigh 1068.8 lbs
10 (20') 1/2" pcs in each of the 8 walls
Rebar is "named" in 1/8" increments.
#1 = 1/8"
#2 = 2/8"
#3 = 3/8" and so on

Question is : How Thick in inches does this 400sf slab made of Air Entrained 5000psi concrete
Need to be?

 

[Deleted member 4993]

Hello, I have A Concrete Slab
thickness question. This will be a 10' tall equilateral Octagon ( 5" thick walls ) above ground Swimming Pool , with the Decking on Top.

Some Parameters are here.
1) Each side of Octagon is 10'
therefore the Width Across the entire Octagon is 20'

2) The Surface Area is 400sf
400 sf ( 8 triangular pies x .5B'x H' )

3) The Height of the Octagon is 10'

4) Weight of Walls is 75,000 lbs
The total amount ( 5" thick walls only ) of 5000 psi air entrained
concrete calculates to 12.34 cubic yards ( L'xW' divided by 12 x D" divided
by 27 = cubic yards ) ( then multiplied x 8 walls at 5" thick)

5) Weight of Pool Water is 263,840 lbs
400sf of Walls x 10' Height x 62.46lbs( 1cf water weight )

6) Concrete for Slab will be 5000psi with air

7) Rebar in Walls will be #4 (1/2") and weigh 1068.8 lbs
10 (20') 1/2" pcs in each of the 8 walls
Rebar is "named" in 1/8" increments.
#1 = 1/8"
#2 = 2/8"
#3 = 3/8" and so on

Question is : How Thick in inches does this 400sf slab made of Air Entrained 5000psi concrete
Need to be?

I do not think it is a homework problem from a class.

If it is an actual construction problem, you need to consult a PE, who is conversant with local code and soil condition. The thickness of the slab (bottom of the pool, as I understand) will be dependent on to a large extent on the condition of the ground and quality of the concrete (beyond just compressive strength).

 

[mmm4444bot]

This will be a 10' tall equilateral Octagon ( 5" thick walls ) above ground Swimming Pool, with the Decking on Top.

An octagon is a 2-dimensional figure; your above-ground swimming pool is a 3-dimensional object.

Therefore, the pool walls do not comprise a 10-foot-tall octagon. Rather, the walls' outer footprint forms an octagon. Or, said another way, a horizontal cross-section of the pool is an octagon.

I'm assuming that the pool deck will extend significantly beyond the walls. Five inches seems somewhat thin, to support such decking.

Will you construct columns underneath the outside edge of the pool deck to support it? I'm curious as to whether any deck supports would rest on the ground; your description of the foundation slab seems to suggest so.

1) Each side of Octagon is 10'

therefore the Width Across the entire Octagon is 20'

20 feet is not the correct width, for an octagon with 10-foot sides.

The distance from the outside edge of a wall to the outside edge of the opposite wall is 24.1421 feet (rounded).

2) The Surface Area is 400sf

400 sf ( 8 triangular pies x .5B'x H' )

The surface area of a 10-foot regular octagon is 428.8427 square feet (rounded)

By the way, pie-shaped slices are not "triangular" because pies have a rounded edge.

Eight isosceles triangles comprise the area of the octagon.

3) The Height of the Octagon is 10'

4) Weight of Walls is 75,000 lbs
The total amount ( 5" thick walls only ) of 5000 psi air entrained
concrete calculates to 12.34 cubic yards ( L'xW' divided by 12 x D" divided
by 27 = cubic yards ) ( then multiplied x 8 walls at 5" thick)

Do you realize that the horizontal cross-section of each wall is a trapezoid, not a rectangle?

5) Weight of Pool Water is 263,840 lbs

With corrections above, the water volume will weigh over 300,000 pounds.

Foods for thought. Cheers :cool:

 

[mmm4444bot]

That's funny, JTvede. Even so, you'll clean that plate by Sunday, yes? :cool:

(The resolution is not sufficient for adequate differentiation, outside of the holiday.)

 

[wjm11]

I do not think it is a homework problem from a class.

If it is an actual construction problem, you need to consult a PE, who is conversant with local code and soil condition. The thickness of the slab (bottom of the pool, as I understand) will be dependent on to a large extent on the condition of the ground and quality of the concrete (beyond just compressive strength).

Please take this advice to consult a structural engineer seriously. Concrete ratings refer only to compressive strength. Tensile strength is only about 1/10 compressive strength, so all tensile loads must be carried by the steel reinforcement. Proper sizing and location of the steel is critical. On the walls, tension and compression surfaces (due to bending) at mid-wall will likely reverse at each wall joint. The bottom slab will have even load distribution across its surface except at the edges, which will have a line load over double the water load. There will also be bending at the edge of the slab going up into the walls. This is not a trivial problem. Failure to have proper rebar sizing and placement can result in cracking at all edges/corners.

 

[mmm4444bot]

JT has not "come back"

Not "true".

Check out the third paragraph, in this thread.

 

[JTvede]

No, No Spammer Here, just a do it all by yourselfer.

Please take this advice to consult a structural engineer seriously. Concrete ratings refer only to compressive strength. Tensile strength is only about 1/10 compressive strength, so all tensile loads must be carried by the steel reinforcement. Proper sizing and location of the steel is critical. On the walls, tension and compression surfaces (due to bending) at mid-wall will likely reverse at each wall joint. The bottom slab will have even load distribution across its surface except at the edges, which will have a line load over double the water load. There will also be bending at the edge of the slab going up into the walls. This is not a trivial problem. Failure to have proper rebar sizing and placement can result in cracking at all edges/corners.

Thanks guys, I have access to Concrete, Steel ,Forms, Labor, but only a little money for the Brains needed.
It's fun realizing through these comments, yes after reading your feedback, I will spend 1000 on an Archs simple structural plan
but not 20,000 on a pool. I have also decided on 15' deep, to keep the water cool. Thank you, until next time.