CEB Catenary Vaults

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This is a discussion of the safety factors involved in using CEB vaults.

Visualize a catenary arch formed between two 4 foot high 2.5' square columns on 16 foot centers

A catenary using equation y=-5*cosh(x/5)+17 spans 13.5; feet at y=4, is 11.5 feet wide at y=6, and peaks at y=12. (cosh is Hyberbolic Cosine)

The thrust from this force is tangential to the 4 foot mark, about 18-19 degrees off vertical, transferring the force more or less to the center of the 2.5 foot column four feet down.

134 bricks form the arch, with a max angle of about 5 degrees across the top, and decreasing from there.

134 bricks @ 30 pounds a piece, divided by two for each direction is just over a ton of downward force from the arch alone. The surface of the brick has 0.5 square feet, or 72 square inches. 2010 pounds / 72 inches = 27.9 PSI

The infill (the wall that will be above the arch) has volume of 30 cubic feet approx, which at a density of 180 pounds per cubic foot (6 bricks) would translate onto the arch up to an additional 180*30, or 5400 pounds, divided by two = 2700 raising the maximum possible pressure on the bottom brick to 4800 pounds, at 72 square inches makes 67 psi.

The nominal minimum CEB strength per wiki page is something like 1200 PSI, giving us a safety factor of 18. Or, to say, we could add an additional 4800*2*18=172800 pounds, or 86.4 tons or 960 cubic feet compressed earth or 5760 bricks two thick above that 16 foot span, which would be one hundred and twenty FEET straight up. (visualize a 120 foot tower of bricks layed flat, two thick (4" x12" facing out) over this 16 foot wide arch, before we were in serious danger of it failing.

Suffice it to say, our bricks are plenty strong enough to build structures even several stories tall using this kind of brick and method using direct vault construction.

Were we to use a groin vault design, on another 16 center offset, 134 * 16= 2144 bricks. There are about 40 cubic feet taken out for the cross vault which at 6 per foot gives 1904 bricks, which weigh 57120 pounds. Divided by four (for each of the four columns) puts the load on the corner at 14280, divided 72 square inches gives a psi of 198.33. 1200/120 = 10 - a reasonable safety factor.

Put compressed earth density fill atop that...

8.5 high by 14.5 by 16 long is volume 1972 cf, volume of the face is 85 square feet time 16 is 1360 cf. one way vault fill is 612 cubic feet for the span.the vault takes about 30% out (36 face feet fill vs. 85 face feet of vault) so I'll estimate 184 cf * 180 pounds per foot = 33048 divided by four corners 8262 pounds. That raises the corner load to 22542 pounds, over 72 inches makes 313 psi, or a 3.8 compression strength safety factor if the bricks only have 1200 psi of strength - and thats the worst case - that load isn't transferred within the fill *beyond* the arch itself!

The corners would risk failure upon the addition of 3.8 times 33048 = 125582 pounds *to the roof* - that would be 4186 more bricks worth of weight - enough to create a CEB floor using the *narrow* face of the bricks *on top* of the room almost 5 layers deep.

If the PSI of the bricks was only 1200.

This is about centering from an engineering book: [civil engineers' handbook Page 41]

Keep the thrust line in the center third of the arch - but doesn't say how to calculate that line. [[1]]

[theory of arches]

[On PSI and loading]

[Calculating Pressures]