# Beam Bending 4 — The Right Answer

Last time, I asked if it was reasonable that we identify k as Well, the units check out.  Anyway, we’ll now do the problem a more “normal” way, then eventually relate it back to the method in the previous posts.

First we’ll treat the “hinge” as an actual three dimensional object that is bent. First, we note that the apex angle there is the same as the angle that we previously defined. (Proof)

The standard way of treating this problem is to assume that the hinge has bent into an arc of a circle.  Since this is a differential hinge, I don’t think it really matters if it’s a circle or a parabola or a catenary or even two lines with all the distorted area in a trapezoid between.  I think the differences would only matter to higher order.  Anyway, the math is pretty easy with a circle. We assume there is some line (blue) in the hinge that is unstressed.  The distance from the center of the circle to that line is r.  Everything closer to the center from that line is compressed (made shorter) while everything farther away is in tension (expanded).

The length of the blue line is r * theta. The length of any other part of the hinge is (r + y) * theta.  So the change in length is y * theta.

Hooke’s Law is stress is proportional to strain, E* delta L / L =  F / Area.  E is Young’s Modulus, F the force needed to stretch the material, and A the area (in this case, linear strain) perpendicular to the force.

We can manipulate that to F = (E * A / L) * delta L.  This is the other version of Hooke’s Law, F = k x, with k = (E * A / L), and x = delta L.  The energy stored is U = 1/2 k * x^2

Let’s look at the front view of the hinge:  So, let’s look at a differential area a distance y from the line that experiences no stress / strain.

Since it is a distance y, delta L = y * theta.  The energy is y changes, but everything else is a constant, so when we integrate to get the total energy, we get By identifying this with U = 1/2 K theta^2, we get k = EI / L.

(There was a factor of 2 error in the previous post, which we will resolve later).

L, by the way, is the length of this differential hinge.  It generally will thus be dx, not the entire length of the beam L.

Now we go back to the cantilevered beam.  We assume the weight on the end of the beam produces a torque on each differential hinge as shown: The torque is the force times lever arm.  We assume the deflection is small, so the cross product is simply the product of the two numbers.  Then it equals k theta by Hooke’s law.  We know k from above, but not theta, yet.

We used theta a lot in the previous posts.  On the other hand, the sum of all the angles equals the first derivative of the curve.  (Really, the tangent of this sum, but, again, we’re using small angle approximations) At x + dx, we add one more angle to get the first derivative there.  The difference of these first derivatives, divided by dx, gives the second derivative.  In other words, Plugging this into Hooke’s Law: So we have a simple equation for f.  All we have to do is integrate twice to get the beam’s shape (we don’t include integration constants, because we don’t care about the beam’s absolute height, and because the beam is held horizontal at the wall itself): But let’s try to connect this with the method of the previous posts.  What’s the energy in the beam?  Again, it’s 1/2 k theta^2. On the other hand, we can also find the potential energy lost by the weight.  It is simply W times the deflection at the end of the beam, giving W^2 L^3 / (3 EI).  These are not equal; in fact, twice as much energy has been lost in gravitational potential energy as has been gained in elastic potential energy.

Energy is not conserved.  The missing energy is lost as heat.  I think the energy must be lost through the contact to the wall; a wall that is not lowered at all and can hold a beam perfectly horizontal regardless of how much weight is applied to it must have potentially infinite energy.

This to some degree makes sense.  If energy was conserved, it would take no energy to raise the beam back to its original position (the lifting energy being supplied by the springs relaxing), whereas it must take some.  Or, to consider the guitar case, it should take no energy to straighten out a bent neck, because you can use the energy in the bent neck to straighten itself out again.

(On the other hand, for a situation where energy is conserved:  A superball that is converting potential energy to kinetic energy can be restored to (almost) its original height — by having it bounce off a hard surface)

So, how does this change the previous approach?  Well, obviously, energy is not conserved.  Also, it’s not one form of potential energy or another that’s at an extreme point, but the total potential energy that is at a minimum.

This is a little bit easier; we don’t need to use a Lagrange multiplier.  We just need to have the gradient of the total potential energy equal zero. This immediately solves for each theta (which is good, since we no longer have the conservation of energy equation to plug theta1 back into): Since each theta is exactly half of what it was when we assumed conservation of energy, the final answer will be half of what we got last time.

So everything works out.

We’re almost ready to move on to actually consider the bend in a neck, but we might work through a few more examples to be as confident as possible.