# Physics meets art history

It is not often that artists need to deal with physics questions, but here's a good one (more than one actually). I hope someone out there will be in the mood for addressing this. It has to do with stone-carving techniques.

If anyone wants to know more context, I'll be glad to fill it in, but here's the problem.

A sculptor's punch is a square bar sharpened to a point in such a way that the tapering sides are four triangles. The degree of taper varies from tool to tool, and the angle matters for my purposes. The tool is driven into the stone at an oblique angle with one of the flat faces toward the block. It is hit one or more times to drive it far enough under the surface (a few cm) that a big chip bursts out in whatever direction the stone is thinner. You do this repeatedly to plow a wide furrow across the stone, ideally, one blow, one chip. (Actually, the sides often aren't quite flat, presumably in order get more expansive force deeper in the stone before the rupture starts.)

The first thing I want to understand is the equation tells:

* How much pressure is applied at the tip.
* How much expansive pressure is applied to the sides (and how much of that is outward.)
* How these vary with the depth of the penetration and the angle of the taper.

Obviously, the pressure on the sides starts at zero when the chisel is on the surface, and increases as it gets deeper until the stone yields.

Incidentally, measures of the compressive strength of stone come in two different flavors: units of force(e.g., daN) or units of pressure (kg/cm^2). I'm not clear on how one translates one to the other, because one is vector quantity and the other is sclar, no? (Hey, I already said I'm no physicist.)

The reason the answer would be interesting, and the harder question below, have to do with the historical origin of the practice of using the tool in this way, which does not occur when you would expect it to, given the date of the advent of the rest of the standard tool set. (Formerly, punches used to be used at 90 degrees, not obliquely. This produced a cone shaped hole facing the direction the hammer is coming from, and is less efficient (and not good for other reasons), but stresses the tool differently (or so I think.))

My intuition is that the sideways stresses are much less at 90 degrees as follows below, but if anyone can explain this in mathematical terms, I'd be grateful:

The chip is up to half-dollar size, and is a shallow truncated cone shape. The tool tip is below the center of the cone, but the truncated part of the cone usually jumps over the tool tip, which is of course the part that was most deeply embedded in the stone. In other words, the area covered by the chip surrounds the tool tip, and the missing chip exposes most of the tool, but the last little bit of the tip remains embedded---maybe a cm deep. (You can see this in the marks left behind.)

That much can be observed, but is the following true:

Just before the chip blows,there is tremendous pressure on the tool from all sides, but at the exact instant the chip blows out on one side, the pressure on that side drops to zero (correct?) so presumably the tool is bing banged sideways, even as the last cm of the tip is still stuck in solid rock.

My ultimate goal is to understand the forces (that I think) are trying to break the tip at this point. It seems like both sheering and bending are occurring at the same time. bending: the embedded tip is preventing the tip of the chisel from moving sideway as pressure hits further up the shaft. Sheering: at least three things are resisting its ability to pivot despite the uneven pressure (the inertia of the chisel, the mass of users hand at gripping the other end, and the friction from hammer's momentum still pressing on the end of the tool) but I don't know how much they count against the abrupt pressure change.

So would anyone care to illuminate what forces are in play in this situation, and how they could be calculated?

Discussion of forces and geometries associated with cutting tools is really a mechanical or production engineering issue.
Many books and sections of books have been written about the subject in these disciplines.
I will try to look some up to post some references.

Andy Resnick
It is not often that artists need to deal with physics questions, but here's a good one (more than one actually). I hope someone out there will be in the mood for addressing this. It has to do with stone-carving techniques.

You are asking perfectly reasonable questions, but the answers are not so simple. First, please forget what you know- for example, the quantities of interest are not vectors *or* scalars, but something else entirely.

One important conceptual simplification is that the stone is *uniform*- that is, the crystalline structure of the rock is much smaller than the chip and the tool- and the tool point. Also, the material the tool is made of is uniform, and the facets are flat and edges straight, etc. The tip can be a hemisphere or some other simple shape.

So, to start, you hit the tool, imparting a force. This creates pressure at the region of contact between the tool and stone (pressure = force/area of tip) which must be quite large in order to create a crack into which the tool is guided. So a tool with a sharp tip will be easier to 'drive' than a dull tool.

The resistance of the stone (or any solid) can be characterized using several different parameters, such as the Youngs modulus, the tensile strength, the ultimate strength, the hardness, the toughness, etc. etc. For your application, it's probably the tensile strength- which can vary from about 150 MPa (granite) to 20 MPa (sandstone). YMMV for your stone.

to get 50 MPa pressure using a tool tip of 1 mm^2 (a sharp tip) means you need to hit the tool with 50 N (11 lb) of force- not much at all.

Now, fracture dynamics are not completely understood (and I don't know anything other than the absolute basics), so I can't really explain why driving the tool at an angle is more efficient than head-on, but in any case, as you are driving the tool in at an angle, it is generating more and more tensile stress in the stone- it's trying to pry apart the stone. The stone resists this- stone may be an elastic solid, but the modulus of elasticity is *very* high and so it does not yield much before the crack propagates radially out from the tip (that's why the ship is conical), ejecting the chip. Why the tip of the cutting tool remains ahead of the large fracture, I can't say; most likely, the stress at the tip not sufficient to exceed the tensile strength. Clearly, the angle of the taper will control the depth of the tool penetrates before the rock fractures.

The same process happens during machining: a metal lathe, drill, some kinds of saw blades for woodworking (the carbide ones), etc. etc. The cutting tool is forced into the material, creating chips as the material fractures.

Metalworking can easily explore the whole range of brittle and ductile fracture and the size and shape of the chip, and so there is quite a bit of literature out there- anything involving "tool engineering" will have tons of material.

does this help?

Drakkith
Staff Emeritus
Seems to me that hitting the chisel in at an angle would allow you to control the direction that your force is going. If you hit the rock at a 90 degree angle, then the force goes straight down into the rock, which spreads out from the impact point and causes the immediate area around the tip of the chisel to burst or chip because there is very little pressure on the surface to absorb the force.

Now, if you hit it in at an angle, then the force of your blow is directed more towards the surface of your rock. Instead of a circle of chipped rock, you get a gouge in the direction that your chisel went in at.

When you initially hit the chisel, the tip breaks the surface of the rock and starts to move through it as the rock absorbs the force of the blow. Since your chisel is pointed at the tip and the diameter increases as it moves into the stone, it also exerts a force perpendicular to the movement, which is out on all sides with the greatest force at the surface of the rock where the diameter of the chisel is greatest. As the tip continues to move through the stone, the force that the chisel is exerting to its sides builds up as well. The chisel acts like a lever; the rock that is furthest up the chip is being forced outwards by the chisel, and much like a lever or wedge, it is exerting an increasing amount of force down through the rock towards the tip of the chisel. At a certain point, the leverage action of the chisel finally builds up enough sideways force that the rock breaks off at the tip end. (As that is where the force is the greatest thanks to the lever action.)

By this point, your chisel has expended most of its energy moving into the rock and only barely moves any more, leaving it slightly embedded in the remaining rock. If you were to hit your chisel in at a deeper point in the rock, or with less force, you would have to hit it multiple times before it could take a chip out of the rock. If you hit it harder, you have multiple chips and your tip ends up deeper.

Hopefully all that is correct. It makes sense to me. Someone correct me if im wrong.

Thanks for these very illuminating replies---extremely clear with respect to how the tool acts on the marble. This is indeed the first part of what I need to know. The second part is actually the critical thing: how the marble acts on the tool.

For context, the use of the punch came in early in the Iron Age. Bronze and iron both work at 90 degrees but both are too soft for an oblique attack--they just skid off. For quite a few centuries, the punch was all there was, and was only used at the right angle.

When steel began to be more available for humble uses, the other basic tools quickly came in: claw chisels, straight chisel, etc. But despite all the innovations, the use of the punch at 90 degrees persisted for a few more centuries. This has always been explained as "tradition" which seems pretty weak, given that all the other tools were immediately accepted. So why would sculptors get all traditional about one technique when they had just quintupled the size of their tool kit? Something is fishy....

My hypothesis is that everything can be explained more simply by the technology alone as follows: The forces on the 90 degree punch are symmetrical, but the more efficient oblique approach produces very uneven pressure at the moment of the stone ruptures. Suddenly, this big uneven sideways pressure acting on only one side of the tool, with the tip still stuck in the stone. My guess is that this tends to snap the tips off if the steel isn't great. That is what I want to confirm, hopefully by knowing what the forces might be, and whether it is easier to make hard steel than tough steel.

BTW, only the punch goes deep--the other tools only go very shallowly into the rock. Steel was good enough for the less demanding claw, straight chisel, etc., and for the 90 degree punch, before it got good enough to handle the sheer forces that the punch gets hit with on the oblique stroke.

BTW, this is a more important point than you'd think, because the vertical stroke also injures the stone, "bruising" it, which leaves a deep permanent milky cloud under the impact point. While the Greeks usually painted the stone anyway, they often did not paint representations of female flesh (the men were generally painted brown.) Bruised stone might as well be limestone. Not very pretty.

I don't know how sculptors work, but another thought from mining practice.

Miners and construction workers use a tool called a rock drill. Yes this is for drilling holes, rather than sculpting, but its method of use is interesting.

It was discovered long ago that such a tool is more effective if it is turned 1/4 turn between each blow. This applies to both the older manual drill and more modern powered ones.

Last edited:
The use of the rock drill is interesting. We're talking about the Fourth to Sixth C. BCE with these tools. They seem to have mostly used drills that twist---in the early days, spun between the palms, but usually spun by a bow pulled back and forth. You can tell by the marks.

But they apparently also had a drill that works exactly as the one you describe, that just pulverizes it's way into the stone, tap, tap, tap, and yes, it also gets randomly turned when you use it. I had not idea this was still done. I don't know how early it appears, but I imagine not until steel came along. The twist drills could be made to work with flint and corundum, apparently.

Andy Resnick
BTW, this is a more important point than you'd think, because the vertical stroke also injures the stone, "bruising" it, which leaves a deep permanent milky cloud under the impact point. While the Greeks usually painted the stone anyway, they often did not paint representations of female flesh (the men were generally painted brown.) Bruised stone might as well be limestone. Not very pretty.

This was useful for me.

Ok, so we have basic notions of the tool entering the stone, creating a crack, and as the tool goes deeper, the crack is forced to propagate, eventually fracturing the stone. But that's not the only thing that is going on.

Pressure and shear are components of *stress*. Stress is not a vector or a scalar- it's something called a 'tensor', and allows us to discuss 3D geometry very easily. Personally, I visualize a tensor as the surface of a cube. Each face has three directions: 1 perpendicular to the face and 2 that lie in the face. The perpendicular components relate to the pressure, while the others relate to the shear.

Solids, unlike fluids, can support a shear stress. That means that if I apply a shear stress to a solid, that stress energy can be absorbed by the solid without producing a deformation- this is unlike a fluid that flows when a shear is applied.

Now, when you whack the stone with your sharp tool, that energy is dissipated within the stone- and the specifics of how that energy is dissipated depend on the details of the structure of the stone and the way the stress is applied to the stone. One way for the energy to dissipate is to form a crack. Other ways are to create dislocations (which could be thought of as microcracks which can move-that's annealing), plastic deformation, heating,.... I suspect the 'bruising' is due to the creation of many dislocations locally, which will scatter light differently than the uncracked stone, creating a different appearance. I suspect the bruised stone is more fragile- there's a lot of stress energy locked up in the dislocations.

Question- you alluded to the function of taper angle in the OP, what's the effect?

So THAT'S what tensors are! You wouldn't believe the stuff they forget to teach in art school....

Actually, what happens in the ROCK isn't so much the issue, as what happens to the TOOL. When I look at the rock, it seems pretty clear that what you guys describe is exactly what is happening.

The historical issue is why they kept using the primitive stroke, when the oblique stroke works so much better despite having demonstrated plenty of willingness to adopt other new practices.

My hypothesis is very simple: even if steel is tough enough for the vertical stroke, it could still be too weak to stand up to the forces of the oblique stroke. I think to suppor this I need to be able to state that the sheer forces are increased as a function of the angle of attack. It seems like they would be, and that there are uneven sideways forces at the moment of bursting, but that's just an intuition.

The historical issue is why they kept using the primitive stroke, when the oblique stroke works so much better despite having demonstrated plenty of willingness to adopt other new practices.

How do we know this?

I suspect that more than one type of stroke was used and I offer the following thoughts.

If you strike a glancing blow to stone type material you are likely to chip off a longish sliver of varying length.

On the other hand if your tool impacts at right angles the effect (apart from the bruising) does not spread sideways very far.

More material is removed per stroke for low angle than for right angle impact, but the effect is much less controllable.

So if I were start with a block of stone I would chip away obliquely to remove the intial bulk, but keeping well back from the final boundary. Then I would change the angle of attack to cut back to the finish profile under much greater control.

I do not think that the cutting stroke has much to do with tool characteristics. Remember that in some parts of the world sculpting/carving was done with stone implements, not metal ones. Modern tools for cutting stone have a very bluff or blunt profile.

Good question #1: The historical usage of the tools is pretty well known: (a) There are a number of partially completed pieces from the Sixth through the Second C. BCE, that make it very clear. The marks of the tools are very easy to read---they look nothing alike, and they are exactly like those we make today. (b) plenty of completed pieces retain rough tool marks in places that weren't meant to be seen. BTW there were essentially no changes to carving technology between about 330 BCE and the early 20th century, when hard metals and power tools came in, and surprisingly few even then. None of the hand tools in my studio would be unfamiliar to a Hellenistic Period sculptor. They even had "die-grinders" (which historians called running-drills).

Good question #2 (first part): The relative efficiency difference of the strokes is huge. (a) the vertical blow removes much more stone per-hammer blow. (b) with the vertical stroke, you have to move the punch and start new each time, whereas with the oblique stroke, when the tool has moved forward, and the chip flies, it is already in place for the next chip. You plow a furrow that way. You don't take it off the stone till you get to the end of the furrow, making it very fast! I would say at least 5x as much stone removed per hour.

(second part) The bruising. Virtually all modern (post-Medaeval) marbles are carved taking great care not to bruise the stone, and have a deep translucent look. Bruised stone is opaque and permanent. Archaic sculpture and most classical marbles have a velvety opaque finish more like what we would think of as typical of limestone. The bruise is made of tiny cracks, which soften the stone and make it more absorbent of water, which is bad. They didn't care as much as we do because they painted most of it, but it would still have been an issue for say, women's faces and bodies.

Your observation about bluntness. Punches for hard stones are indeed blunt, as you point out, but they work in an entirely different way from marble punches. Punches for marble and limestone penetrate and blow out chips. Granite, diorite, basalt, etc are very hard, and the punches don't actually penetrate. They crush a pock into the surface and by occasionally splintering off shards. Reducing granite-like stones is more often done with a bush-hammer, which looks exactly like a meat tenderizer. It is basically a square matrix of punches. You simply hammer on the surface, and it crumbles the surface away by knocking many tiny pock marks into it over and over. Granite is so hard it throws sparks when hit with steel. Marble does not do this. Granite does not bruise either.

Oh yes--your last point. Yes, ancient peoples did use stone as we would use a bush hammer. The Egyptians for instance used diorite balls to reduce granite. But this was not really the practice in the Agean region because marble and limestone aren't that hard. They tended to use chisels to cut and emery to abrade. Emery is corundum, e.g., the stuff sapphires are made of. Second hardest mineral after diamond.

I don't think I made my point properly clear.

If the ancients used a two stage process to remove unwanted material you would only ever see the marks of the second stage which would remove the marks of the first.

This is a bit like a woodworker sanding out saw, chisel and plane marks to a finish line.

I do understand bush hammers. They have a vibration effect, not available to unpowered tools.

This is a most interesting discussion, keep going. Perhaps we could discuss the difference between cutting stone and metal?

Andy Resnick
Actually, what happens in the ROCK isn't so much the issue, as what happens to the TOOL. When I look at the rock, it seems pretty clear that what you guys describe is exactly what is happening.

Interesting. Well, cutting tools are often ground to a specific angle-for example, twist drill points have 4 different angles (point angle, chisel edge angle, lip relief angle, helix angle) and these are adjusted based on the size of the drill and the hardness/toughness of the material.

Single-point cutting tools have a more complex geometry (I count 6 angles: side and back rake angles, side and end cutting edge angles, side and edge relief angles), which is also controlled to optimize the cutting rate and wear.

AFAIK, these angles were all figured out by trial and error, not froma first-principles calculation.

Yes, you are right that normally the later stages of carving usually obliterate the evidence of the earlier, but
• Not all sculpture is completed all the way around, because the back is not meant to be seen.
• Lots of pieces get abandoned part way.
• Bruised stone leaves permanent evidence in the form of bruises and preferential erosion of formerly bruised areas. The effects run as much as several cm into the stone: deeper than you would ever grind off with the final smoothing.

Sculptor's bush hammers have an array of pyramidal points on the face. A given spot might be struck hundreds or thousands of times to remove even a few inches of stone (It's not vibration, although there are pneumatic versions that do it really fast---it's just tap, tap, tap...) The punches for hard stone mostly work like this too, so the tip angle is v. wide to withstand the impact.

The punches for marble are the issue, because they do NOT work this way. Tip angle is sharp. When a bush hammer fails, it fails because either the pyramids get worn flat, or they get broken off (by stone getting packed between them, usually). OTOH, when a marble punch fails, it is because the tip snaps off. anywhere from a few mm to a few cm.

It is the cause of the snapping off that is in question.

Some sketches / photos would be very handy.

High carbon iron or steel (greater than 4% carbon) can be made harder than marble, which is quite a soft rock.
However such steel is quite brittle.

As requested, here are three pix. One compares a marble punch to a granite punch. You can see that the granite punch cannot penetrate. (It is carbide tipped---even the hardest steel is rapidly destroyed by granite.) The other is a picture of a bush hammer---each point is quite similar to the granite punch; it just crushes superficial pocks in the stone.

The third shows the effect of punch on marble. This is fairly light punching.

Note the tracks of the sharp point, and the long stripes of broken stone between them where the chips have popped off. The question is, does the obviously non-symmetrical forces acting on the chisel mean imply that a sufficiently bad steel might be good enough for punching at 90, but not good enough to hold up at an oblique angle.

pix?

Dang---It looked like they uploaded! Must have been a button I didn't see. I'll have to do it this PM.

I have uploaded a couple of sketches about the way cuttin g tools work.

Since we have been talking tension/compression/shear I am assuming you know what these are, although you said you are not a physicist. If you have any doubts or queries, please ask for clarification.

Fig 1
Shows a cutting tool being forced along the surface of a workpiece.
The force from right to left causes a shear stress to develop across the neck of the top layer of the workpiece at AB.
This in turn causes a chip to shear off along BC, the upper face of the cutting tool.
This is also why it is easier to start with a shoulder or edge when cutting.

Notice that in Fig 1 the width across the tool to shear is much greater than the AB so the workpiece will shear in preference to tool. Also the tool shear resistance should be greater than that of the work (but not always).

Now you were asking about tool failure so here goes in Fig2

The contact face BC between the tool and the work is subject to large forces and tends to heat up as a result of the energy transferred.

This has two effects.
Firstly as shown this face is subject to wear, what is known as crater wear, as shown dashed in the diagraram.
Secondly the local heating can locally soften the tool material ( and the work material but we are not considering this).

The first reduces the section to the tool available to resist the shear at EF
The second reduces its shear strength.

At some point a crack will initiate across EF.
Repeated blows will casue this to propagate in the weakened tool material.

#### Attachments

• tool1.jpg
10.6 KB · Views: 430
Ok, I think your picture illustrates a misunderstanding. Sorry, can't send you a picture at the moment, but perhaps your picture makes it unnecessary.

The picture is typical of a machine cutting tool, in that the back side of the tool is NOT under pressure from the material being cut, but the front is. Note, the back doesn't even touch the workpiece!

But the punch is not truly a cutting tool. It is a splitting tool, and works like a wedge for splitting logs. pressure from the two sides, not from the cutting tip, separate the stone.

The fact that the tip cuts into the stone is actually irrelevant because the punch would still work the same way even if you drove into a pre-drilled pilot hole. In such a case, the tip would never touch stone, but the result would be the same. (A wedge driven into a pre-drilled hole is in fact one of the main ways to divide stone.)

The problem with understanding the punch is that BOTH sides are under equal pressure until the instant of chip release. Until it breaks, there should be no shearing force across the tool other than from random irregularities in stone, etc.

The real question is, does a shearing force appear at the instant the chip is released, because the pressure from the chip side vanishes almost instantaneously, when the chip suddenly releases. This would be similar to the picture, but backwards, because the greater pressure would be from the back, not from the chip side!

The real question is, does a shearing force appear at the instant the chip is released, because the pressure from the chip side vanishes almost instantaneously, when the chip suddenly releases.

No I don't think it so.

I understand the splitting action you are talking about, but I think it is more complicated than you have described.

Yes I agree the diagram is a modified version of a metal working tool (could still be handraulic - doesn't have to be machine cutting). I did mention that cutting is easier against a shoulder - we call it scutching in brickwork.

When you apply a cold chisel to a surface that has no shoulder and try to drive it in, there is obviously initially no side pressure on either side, until it is partly embedded.

The actual embedding process must cause a slight local rise in the work material surface, there is nowhere else for it to go.

Once the crack has initiated (slitting action) it can propagate through the material with the grain. This is what happens in timber and stone with a definite grain such as slate, or it can be stopped by inhomgeneity (non isotropy) in the material. Wedges don't work well across the grain.

What I think happns is that the the tool encounters a harder bit and gets deflected one way or the orther, or worse still get twisted slightly.
Either of these combined with the shear, inherent in the axial force applied, are known to be the worst case loading for any component. Acceptable axial loading plus unintended torsion has broken many a component prematurely.

Incidentally I asked if you understood tension, compression and shear. Your comment about even pressure means no shear is not true. There is always shear associated with axial loading (tension or compression). This is easier to understand in diagram form I will draw one if you like.
In fact it is impossible for any material to fail in full (3 dimensional compression). The actual failure is always a shear failure.

Oh! I see. I seem to be picturing this in oversimplified terms. Yes, I do know what the definitions of tension, compression and shear are, but I can see that I wasn't picturing how one could give rise to the other.

So even though it looks like the hammer is only applying pressure in one direction, the steel is actually under forces from many directions due to it changing shape slightly, encountering unevenness and twisting, etc.

I'm getting the feeling that this might be more definitively answered by experiment. I could set up a hammer to drop from increasing distance until it hits hard enough to break the chisel held obliquely. Then see how much more (or less) it takes to break it on the vertical orientation. It might also be more convincing.

Thanks for all the patient interaction. When I get the experiment done, I'll let you know how the cookie crumbles!

BTW, the particular stones in question are somewhat anisotropic, but not so much that the tools won't work in any direction. There's a "bedding" orientation that corresponds to the original horizontal sediment (even though the marble is metamorphic, the effect persists.) When you split a big piece, it definitely works much better either parallel to the bedding, or perpendicular. If you try to split at an angle, it will jump unpredictable and try to follow the bedding direction anyway, so you get a stair-step effect instead of a smooth planar break.

but I can see that I wasn't picturing how one could give rise to the other.

It's past the witching hour here, but I will post some more sketches tomorrow. I had intended to do these with the first lot, but thought it best to get something up quickly.

In the three attachments I am seeking to explore the following three questions.

1) How does shear arise from an applied tensile or compressive force?

2) How does a brittle material fail under a compressive load?

3) How does a brittle material fail under tensile load?

In attachment 1 I have shown specimens of material with subject to tensile and compressive loads.
If we take any plane at some random angle to the axis of the force, as shown dashed, this plane acts like half a wedge (=inclined plane).
If the two parts of the specimen, as separated by this plane, were physically separate the applied forces would either drage the upper section up the plane (tension) or push it down (compression0 as shown.
This implies a force acting parallel to the plane.
And indeed there is - it is the shear force.

The effect of the inclination of the plane is to tilt or rotate the effect of the applied force.

Now obviously unless the material actually breaks this remains a (shear) stress within the material, rather than a displacement.
The actual value of this shear stress obviously depends upon the actual value of the applied force. It also depends upon the angle the plane makes with the axis.
It is easier (requires less energy) to make the material slip sideways as shown than to pull the material physically apart, which is why the shear strength is (much) smaller than the direct stength of materials.

Since you want to concentrate on tool failure we should look at the compression loads next.

The response of an object to compression is considerably more influenced by its geometry that for tension.
So, as shown much exaggerated in attachemnt 2, a slender object, such as your chisel, bows or buckles slightly when struck.
Remember that impact forces are much greater than their static equivalents.
Tension appears along the stretched side of the now curved implement and, especially if the tool is not rotated (remember I mentioned this) microcracks will appear and start to grow (as in attachment 3), since chisels etc are made of brittle materials.

A more squat object under compression bulges out sideways, if allowed to, and eventually fails by what is known as central or diametral splitting. Again the action within the material translates the compression into internal tension.

If for some reason, for instance because there is more material to each side of our specimen confining it, shear stresses are also developed as shown in attachement 1. The maximum shear occurs at 45 degrees to the applied compressive force, and the material eventually fails in shear along these lines.
The last part of attachment 2 shows what happens to a cube of concrete when tested to destruction by crushing. The double pyramid result is quite characteristic.
This is what happens to the rock directly inder the punch or chisel point.

Both the stone and punch are brittle materials and attachment 3 shows what happens to a crack in such material when tension is applied across it.

In homogenous material, (metals or the matrix in concrete and marble) the end of the crack is at point T.

If the energy gained from the work done by the applied tension is greater than the energy required to tear the material between T and S asunder, the crack will propagate (open) from T towards S and eventually through the whole material.

This is quite a sudden process, that gives little or no prior warning before it happens, and is sometimes termed 'fast fracture'
It is also what happens to your tool tips.

On the other hand the last part of attachment 3 shows what can happen in a non homogeneous material.
Embedded obstacles, such as small stones, prevent the crack propagating directly. Directly means at right angles to the tension. At any other angle the tension has less effect, so less energy is availble to divert the crack off course. Obviously if the tension is great enough this is exactly what will happen.

#### Attachments

• tool2.jpg
8.4 KB · Views: 389
• tool3.jpg
13.5 KB · Views: 363
• tool4.jpg
10.3 KB · Views: 396