Building With Stone

Q: Where can I find information about building with stone? How thick a wall to support a given amount of weight? Etc.

A: Start by consulting your local building department and relevant local/state building code. The thickness of your wall will depend on the quality of materials and whether you choose to use steel reinforcing. The compressive strength of your stone will rarely be the controlling factor for walls built to human scale. Reinforced walls will require an engineer’s assistance. For an unreinforced wall, the required thickness is set so there is never a loss of compression on the stonework. Here’s an example:

Assuming solid sections…

Wall is 10 foot tall.

Wall weighs 120 pounds per square foot.

Wind load equals 20 pounds per square foot.

The “Section” of a 12″ length of wall is: (12)(thickness)(thickness)/6 = (2)(t^2).

“Moment” in wall = (20psf)(10ft)(10ft)/8 = 250 ft-lbs. = 3000 in-lbs.

The bending force (Fb) in the absence of vertical loads = Moment / Section = (1500)/(t^2) at each face (maximum at the faces, zero at the middle of the wall).

Weight of the wall at mid height – assuming no roof load – = (120)(5) = 600 lbs.

Unit compression per square foot = (600)/(12 * t).

Set up the equation such that the unit vertical weight at mid height is greater than or equal to the calculated bending stress…

(600)/(12t) >= (1500)/(t^2)

(600)/(12) >= (1500)/(t)

50 >= (1500)/(t)

t >= 30 inches.

Now go back and check the assumption about wall weight. Is a 30 inch thick wall going to weight just 120 pounds per square foot (elevation)? Revise and recalculate until you come close to the optimal thickness.

At 160 pounds per cubic foot for most stone, try a 16″ wall…213 psf.

(213)(5)/12 >= (1500)/(t)

t >= 16.9 inches.

Note we went a little under on that one. Looks like 18″ would work.

Check…

(240)(5)/12 >= (1500)/(t)

t>= 15.0 inches…ok

You can use this formula for various wall heights, wind loads and wall weights. You will note that axial load will add to the “weight” of the wall, thus further increasing the value to the left of the greater than or equal sign…which will permit you to use a thinner wall. Don’t go too far in this direction, as the roof weight (and forces tributary thereto will fluctuate (uplift, etc.)

For those of you in seismic / earthquake zones, the formulas are set up similarly, but the lateral force increases linearly with wall weight. Here’s how that formula looks:

Assuming solid sections…

Wall is 10 foot tall.

Wall weighs 240 pounds per square foot (starting with our 18″ wall from above). We’ll use a placeholder “w” to represent the weight of the wall in our formulas.

Lateral load equals 30% of wall weight.

The “Section” of a 12″ length of wall is: (12)(thickness)(thickness)/6 = (2)(t^2).

“Moment” in wall = (0.30)(w)(10ft)(10ft)/8 = 250 ft-lbs. = 45(w) in-lbs.

The bending force (Fb) in the absence of vertical loads = Moment / Section = (22.5(w))/(t^2) at each face (maximum at the faces, zero at the middle of the wall).

Weight of the wall at mid height – assuming no roof load – = (w)(5) = 5(w) lbs.

Unit compression per square foot = (5(w))/(12 * t).

Set up the equation such that the unit vertical weight at mid height is greater than or equal to the calculated bending stress…

(5(w))/(12t) >= (22.5(w))/(t^2)

“w” cancel; one “t” cancels

5/12 >= (22.5)/(t)

t >= 54 inches!

This is the reason we use steel reinforcing. By eliminating the “Tension = 0” limitation, we can design walls to utilize more of their allowable compressive strength…which allows us to make walls much thinner!

Scott McVicker, S.E.