Square Pyramid Stack

What about the brass monkey ? What weight does he feel ?

In naval terms there's no such thing as a Brass Monkey.

Term: brass monkey - Naval History and Heritage Command:

It has often been claimed that the "brass monkey" was a holder or storage rack in which cannon balls (or shot) were stacked on a ship. Supposedly when the "monkey" with its stack of cannon ball became cold, the contraction of iron cannon balls led to the balls falling through or off of the "monkey." This explanation appears to be a legend of the sea without historical justification. In actuality, ready service shot was kept on the gun or spar decks in shot racks (also known as shot garlands in the Royal Navy) which consisted of longitudinal wooden planks with holes bored into them, into which round shot (cannon balls) were inserted for ready use by the gun crew.

.............and you knew it was cold, when it was cold enough to "freeze the balls off a brass monkey"

Although not the correct "brass monkey" it suitably illustrates the problem...........it's bloody cold...........or freezing.

You are, of course, talking about the vertical component of the weight of the cannon balls. Since the weight of the balls in the next layer up is not centered over the balls in the current layer, I suspect the distribution to the layer below will not be equal, but I don't have time to study it. You may get into messy vector diagrams; maybe things will neatly cancel out, but my intuition says "no comment!"

This is a nice problem that seems trivial until you think about it. From your diagram, I see that each ball is supported in four places. You cannot assume perfect inelastic spheres and use force vectors to get a solution because you have more unknowns than parameters. A tetrahedral stack would have a unique solution.

I agree with Lehman 57 and Osborne83 GAs.

In Usbport's picture the top ball is applying an outward force to the four balls beneath it: that means that the force between the "corner balls" in the second and third rows is greater than the other forces (between the second and third rows).

The answer is going to depend a lot on the coefficient of friction: between the balls, and, between the balls and the monkey. How deep do you want the analysis to be?

The OP just asked for the weight. It doesn't matter what the lateral forces are because they ultimately cancel; the weight of the cannonballs will yield a downward vector force equal to their weight.

If you were to stack the cannonballs in the normal manner or with a thin ('massless') sheet between each layer, the net weight on the layer below is the same, either way they are stacked.

The OP also said

What pressure is felt by the brass monkey?

In the simplest case take five rusty cannonballs: arrange 4 touching each other in a square, and, place one on top; they will probably stay where you put them. Now take five large oily ball bearings and try the same thing. The monkeys holding the two sets of balls will "feel" two different types of load.

You may well have answered the OPs question adequately: it depends on what he is really trying to calculate.

An experiment most everyone can try:

Stack billiard balls similarly and see how high they'll stack...

The old problem of supporting a weight on more than three supports, it's indeterminate as a slight dimensional difference changes everything. Or we assume some flexibilty as in a steel bridge, suppose the balls were rubber then the weight would spread equally, but wobbly iron cannon balls could go any way. But for this problem we assume the weight is equally spread four ways.

Next, the bottom layer would need a solid barrier to hold the balls in the formation, is this the monkey? It isn't possible to put one pool ball on top of four because the horizontal component forces them apart.

If this is a square base then what is the reference to a "spherical arrangement of cannonballs"

The effective weight on cannonballs at the base depends on the row they are in from the outside and if on a corner. Consider a 2D pyramid, and look at the weight distribution.

If all that is required is the total weight of balls then it seems to go 1 squared +2 squared + 3 squared + 4 squared + 5 squared and so on, until x = the number of layers, multiplied by the weight of a ball.