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An engineer designed an open box mold for concrete mix pouring.
By cutting the standard 4" x 8" G.I. sheet with the right pattern he
then bend and weld  to produce maximized weight concrete blocks,
how was it done?  What is the precast volume?
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1- cut sheet longitudinally , to form 2 pieces(2"x8").


2- bend these 2 pieces in the middle making a right angle.
3- welding them together to get an open box, with (4"x4"x2").
So the precast volume will be :32 cubic inches.
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1- cut sheet longitudinally , to form 2 pieces(2"x8").

2- bend these 2 pieces in the middle making a right angle.

3- welding them together to get an open box, with (4"x4"x2").

So the precast volume will be :32 cubic inches.

that is open bottom..the cement mix will leak or push the mold up as it flows down..

but it can be done on frying eggs.. +1

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@bonannova: 8.4125 cu.ft... is that the maximum using differential caculation?

that's okay but you wasted some surface material by cutting corners.

Ÿes.

And yes ... you did say weld, didn't you?

Back to the drawing board.

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cut into 3 pieces,bend and weld

By symmetry I'm thinking the pieces would be (1) bottom and (2-3) two side/end folded pieces all welded together. That relates the edge lengths and should either maximize the volume automatically or permit optimization by differentiation. I'm traveling but will give this some thought tomorrow. It's a nice puzzle, cuz it sounds at first like an old chestnut but isn't.

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cut into 3 pieces,bend and weld

 

Interesting.

 

Think it is as below..

Cut along the solid lines to get three parts, A, B and C. Bend to 90o along the dotted lines. B will be the base and one side of the box. B and C will each be 1.5 sides of the box (they will cover one side plus half an other side).

 

post-55015-0-84426100-1413190862_thumb.p

 

Since 4 * x = 4', x = 1', giving a volume of 7' x 2' x 1' = 14 cubic feet.

 

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cut into 3 pieces,bend and weld

 

Interesting.

 

Think it is as below..

Cut along the solid lines to get three parts, A, B and C. Bend to 90o along the dotted lines. B will be the base and one side of the box. B and C will each be 1.5 sides of the box (they will cover one side plus half an other side).

 

attachicon.gifprecast.png

 

Since 4 * x = 4', x = 1', giving a volume of 7' x 2' x 1' = 14 cubic feet.

 

 

Very good solution..

Few pieces,

it holds the cement mix and

used all the surface materials.

Yet lighter than precast produced 

by engineer's the pattern..

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Cut into 3 pieces of 1x4, 1x4, and 6x4
Bend the 6x4 so that it has a base of 4x4 and 2 sides of 1x4 each
Weld the 2 pieces of 1x4 on the remaining 2 sides of the base

You have a 4x4x1 = 16 cu ft volume box

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Cut along the solid lines, bend along the dotted lines. Weld to produce a 2'x4' mold. Volume is 16 cu.ft.

 

post-9659-0-94323200-1413208543_thumb.pn

 

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...theoretically, if we can ignore any lost material in cutting or welding (i.e. use all available area irregardless of how many pieces are required and ignoring how infinitesimally small some may be).

 

assuming a square bottom maximizes volume (which is pretty arbitrary but my calculus is too rusty to deal with two variables):

 

V = x * x * ((32-x^2)/4x)

V = x^2(8/x-x/4)

V = 8x-(x^3)/4

 

then differentiating and equating to zero to maximize V:

 

8-3/4x^2 = 0

x = (32/3)^.5

x = 3.266

V = 17.419

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Cut along the solid lines, bend along the dotted lines. Weld to produce a 2'x4' mold. Volume is 16 cu.ft.

 

attachicon.gifmold.png

 

 

 

Cut into 3 pieces of 1x4, 1x4, and 6x4

Bend the 6x4 so that it has a base of 4x4 and 2 sides of 1x4 each

Weld the 2 pieces of 1x4 on the remaining 2 sides of the base

You have a 4x4x1 = 16 cu ft volume box

 

Almost..but still lighter.

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...theoretically, if we can ignore any lost material in cutting or welding (i.e. use all available area irregardless of how many pieces are required and ignoring how infinitesimally small some may be).

 

assuming a square bottom maximizes volume (which is pretty arbitrary but my calculus is too rusty to deal with two variables):

 

V = x * x * ((32-x^2)/4x)

V = x^2(8/x-x/4)

V = 8x-(x^3)/4

 

then differentiating and equating to zero to maximize V:

 

8-3/4x^2 = 0

x = (32/3)^.5

x = 3.266

V = 17.419

good catch PG..Like Wolfgang proposal the 3 side mold below can work if the cement mix is a bit dry when poured ..

post-53237-0-59066100-1413305373_thumb.j

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...theoretically, if we can ignore any lost material in cutting or welding (i.e. use all available area irregardless of how many pieces are required and ignoring how infinitesimally small some may be).

 

assuming a square bottom maximizes volume (which is pretty arbitrary but my calculus is too rusty to deal with two variables):

 

V = x * x * ((32-x^2)/4x)

V = x^2(8/x-x/4)

V = 8x-(x^3)/4

 

then differentiating and equating to zero to maximize V:

 

8-3/4x^2 = 0

x = (32/3)^.5

x = 3.266

V = 17.419

the above spoiler was attempting to describe a open rectangular parallelepiped with a square base of length sqrt(32/3) and height of sqrt(8/3). such a mold would have surface area of 32 sq units and volume of 17.4186 cubic units.

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the above spoiler was attempting to describe a open rectangular parallelepiped with a square base of length sqrt(32/3) and height of sqrt(8/3). such a mold would have surface area of 32 sq units and volume of 17.4186 cubic units.

 

 

..turns out the assumption about square bottom is correct and and it maximizes the volume.

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