Imagine a perfectly symmetrical shaft clamped rigidly and horizontal on one end and an imaginary vertical plane aligned to the axis of the shaft. If you look down the shaft you are looking at the vertical plane edge on. If we apply an axial load (a load in the direction of the axis of the shaft) to the end of the shaft it will create a stress in the shaft material that is the same all the way around the shaft, because the shaft is perfectly symmetrical. This stress will be tensile or compressive depending on the direction we load the shaft, and the magnitude of this uniform stress times the cross sectional area of the shaft wall (we'll assume the shaft is hollow) will equal the load applied (the force) because the shaft is not moving. A static situation requires that the forces all be in balance.
Now let's remove the axial force and hang a weight on the tip of the shaft, so that it bows downward, but still in the vertical plane. Pick a point on the shaft and imagine an axis that is perpendicular to the vertical plane and passes through the center of the shaft. Now because the shaft is being bent down, the material on the top of the shaft is in tension and the material on the bottom of the shaft is in compression, with the greatest stresses in the material at the very top of the shaft and the very bottom of the shaft (tension and compressive stresses, respectively). The stress levels transition from maximum tension at the very top of the shaft to maximum compression at the very bottom of the shaft, passing through zero at the location of the axis we have imagined perpendicular to the plane of bending and passing through the center of our symmetrical shaft. This is the neutral bending axis and is characterized by being the location of zero stress in the walls of the shaft and the axis about which the moments created by the tensile forces in the upper half of the shaft and the moments created by the compressive forces in the lower half of the shaft equal each other (in bending the moments have to sum to zero for static equilibrium). Remember that a moment is a force times the distance away from the axis you are calculating it about. In the wall of the shaft each little element of the shaft wall has a stress on it and that stress times the area of the element (which is a force) times the distance that element is from the neutral axis defines that elements contribution to the total moment about the neutral axis. (this is really tough without being able to draw figures) If instead of pulling down on the tip of the shaft we pull up the geometry of the shaft cross section has not changed so the neutral axis remains in the middle of the shaft, the moments from the stresses in the top half of the shaft still equal the moments in the bottom half (with the signs reversed) and the shaft shows the same stiffness when being bent up as it did when it was being bent down.
Now let's suppose that the material in the top of the shaft is thicker than in the bottom half of the shaft, the shaft is now asymmetric (a 'strong' side and a 'weak' side if you apply an axial load, but not, as we will see, with a bending load). But the shaft is still in static equilibrium so the moments created by the stresses above the neutral axis still have to equal those below the neutral axis. What happens is that the stress distribution changes and the neutral axis moves closer to the upper (the thicker) wall of the shaft. The shaft stiffness remains the same whether the shaft is being bent up or being bent down (within the assumed bending plane) because the geometry of the shaft does not change (at least not appreciably with the amount of bending in a golf club) and the neutral axis remains in the same location regardless of the direction of bending.
The stiffness of a shaft depends on it's geometry and the plane of bending you are talking about, but within that plane of bending the shaft is equally stiff in both directions. When engineers talk about bending stiffness we aren't talking about direction we are talking about the plane in which the bending occurs. Dave T. understands this very well and when he talks about NBP being 180* apart he is trying to put this concept into words someone without the mechanics background who thinks of shafts as having a strong side and a weak side in bending (as they can when loaded with an axial force) can relate to. NBP's (or directions of minimum bending stiffness) are 180* apart because that puts them in the same plane and the bending stiffness of a shaft in any given plane of bending is the same in both directions. The same is true of 'spines'.
Consider a shaft with an elliptical cross section. The plane of maximum stiffness will be that through the major (larger) diameter of the ellipse and the plane of minimum stiffness through the minor (smaller) diameter. These planes will be 90* apart (exactly if it's a true ellipse). Consider a yardstick as an extreme example of this. Most shafts of any reasonable symmetry exhibit this property: the maximum and minimum stiffness planes are about 90* apart.
I noticed that Dave T. responded to your questions about material strength being different in compression and tension. As Dave pointed out strength refers to the failure of material, with bending (without failure) we are concerned with the 'stiffness' of the material, or its modulus in engineering terms. I believe there are some aligned-long-chain polymers that exhibit some differences, but for the materials we are talking about in golf clubs the modulus is the same in both directions.
I just realized how long this had gotten, but I hope it helps. I'll go quietly now.
Alan Brooks
At 04:41 PM 10/9/2003 -0700, you wrote:
On Thu, October 9, 2003 at 2:23 pm, Dave Tutelman sent the following > > SPINE is the stiffest plane of a shaft in bending. NBP is the most > flexible plane. The concept of plane is the first area of > disagreement among the protagonists. Some say that it is possible for > the stiff side to be 180* opposite the flexy side. But the engineers > in the group say that is not so; it is an artifact of an imperfect > instrument used to find the spine. In fact, spines will be 180* apart > every time, hence I talk about "plane". Similarly NBPs will be 180* > apart from one another every time. BTW, that is what John Kaufman > proved. He proved nothing about how to align the spine or NBP when > you make a club, just that the order of things as you go around the > shaft is spine-NBP-spine-NBP at about 90* intervals. This also agrees > with: * Theory that every mechanical and structural engineer learns > in school. * Tests that others have done, including one I've > witnessed involving a FlexMaster.
Not being a mechanical or structural engine who is familiar with the tensile properties of the materials used in golf shafts, I may be way off base here, but here goes:
Isn't it possible for a material to have different compression and tensile strengths?
For example, we'll use some poor ascii art below depicting a side view of a shaft where one side of the shaft is significantly thicker than the other:
--------------------------------
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The top side drawn with the - is the thin side, the bottom side drawn with the = is the thick side.
Assume (which is probably a bad assumption but will get the point across) that the material does not compress at all, but does stretch linearly depending on the thickness of the material.
If you pull down on the shaft the club will deflect some amount. But if you push up on the shaft the club will only deflect 1/2 the amount as it deflected down. Measured in a frequency meter you will only get one reading as a frequency meter can only measure stiffness in planes, but in a deflection meter you will find the shaft bends different amounts in each direction obviously affecting the position you might want to orient the shaft. It also seems that if the shaft started off bent in one direction or the other, it would affect the stiffness of the shaft in each direction. How much? I don't know, you guys who know materials better than can will probably tell me what I've described isn't possible. ;-) Please do if I'm totally off base.
So from what I can tell, the main drawback of a frequency meter is that you can only measure stiffness in planes, when a shaft may actually have a different stiffness in each direction.
The main drawback of your typical spine finder (don't know if this applies to the NF2 or not) is that you don't take into account any bend of the shaft. It seems to me that a method which could be used to measure the stiffness of a shaft would be done as follows:
Clamp one end of the shaft in a bearing device with no load on the other end. Take 2-d deflection measurements at multiple points around the shaft, making sure that each measurement is made using a constant force. An easy way would be to use a weight free hanging to prevent side-forces from affecting measurements. 2-d measurements may not be necessary if shafts always bend in the vertical plane, but I think if they did, you would always get FLO no matter which way you aligned a shaft.
Once you've got this data, you will be able to predict what direction the shaft will flex and how much based on the weight used. The tricky part will then to use that data to align the shaft in the position which most consistently returns the clubhead to the spot the golfer expects it to.
I think Dan's NF2 can do all these types of measurements but the 2-d measurements, maybe it can be easily adapted to make 2-d deflection measurements as well.
-Dave
