I have been told that if I took two pieces of a flat flexable material the same length say 4 feet long and cut them into a pyramid shape that regardless of the width of the base that they would bend on the same radius if pulled down on the point end while locking in the wide base end. If this is rue what rule governs this? An example might be a 1" base as opposed to a 4" base 4 feet long.
Not sure what you are asking here. If you are asking would the same force bend them to the same radius, the answer is no; it would depend on the moment of the cross sectional area.
If you are asking about the neutral axis of the bend (tension above, compression below, or vice versa), it still depends on the moment of the cross section, and the centroid. I don’t have the formulas handy, but check any strength of materials textbook.
I don’t know about the engineering aspects of this, but isn’t there a geometric solution to this? Something about the relationship between the length of the arc and the radius? I’ve seen numerous algorithms for drawing circles and it sounds something like one of those solutions.
I wasn’t very clear in my question. I should have said will both limbs maintain an even bending radius throughout the limb as opposed to bending more at the inner limb or more in the outer limb.
IIRC, the radius of curvature at a given point along a cantlivered beam is governed by two parameters:
Bending moment. This is the distance between the point where you are applying your load (i.e. the free end of the cantilevered beam) and the point on the beam where you are assessing the radius of curvature. For any cantilevered beam with a single point load applied at the free end, the bending moment is zero at the point where the load is applied, and maximum at the fixed base.
Second moment of area. This is a mathematical description of how well the beam resists distortion when a bending moment is applied. I-beams are shaped the way they are (with a wide flange at top and bottom) because this shape (with a lot of material positioned far away from the neutral axis) results in a very high second moment of inertia with an economy of material, minimizing cost and weight.
If you look at the equation for the second moment of area of a beam with a rectangular cross section (scroll down on that page to “a filled rectangular area with a base width of b and height h”), you see that the SMoA is directly proportional to the width of the beam.
So if your bending moment is proportional to your distance X from the end of the beam, and your SMoA is also proportional to your distance X from the end of the beam, then the change in one parameter as you move along the beam exactly offsets the change in the other parameter, and you end up with a constant radius of curvature all the way from the tip of the beam to the base.
Have I correctly understood your question?
You answered very well! Thanks!