# How to calculate the force of a cantilever beam at a given point of deflection?

I have put together a simple cantilever beam using finite difference modeling according to the following standard wave equation

$$ρAy_{tt} = -EIy_{xxxx}$$

where $$ρ$$ is density, $$A$$ is cross sectional area, $$E$$ is Young's modulus, and $$I$$ is the moment of inertia.

I have run this simulating a simple deflection where this beam (clamped to horizontal at points 1 & 2) is moving upwards in the positive y-axis until it strikes an immovable object at point 12.

The deflection looks reasonable.

I am now wondering how I would calculate the force exerted by the beam at point 12 based on my finite difference model.

In the finite difference wave equation, local force at any given node would be given by:

$$(ΔxρA)y_{tt} = -ΔxEIy_{xxxx}$$

$$F = -ΔxEIy_{xxxx}$$

Because $$F = ma$$ and $$m = ΔxρA$$ and $$a = y_{tt}$$.

The problem is I think this is just the local force for that node relative to the neighbors. It's not giving a reasonable/useful total force output just by calculating the force at that one point.

This makes sense as the force it calculates is based on the fourth spatial derivative, and at the collision point it's almost perfectly straight so this will be near zero. Most of the restoring beam force is coming from the strong bend closer to nodes 1 and 2.

If I want the total force exerted on the "immovable object" striking the beam at point 12, how would I get that?

I know that in a general approximation, this is given by:

$$F = \frac{6EI(y)}{x^2(3x-a)}$$

Where $$y$$ is y deflection of a given point, $$a$$ is the distance to point of collision contact, and $$x$$ is the position of the point you're assessing. But I actually want to figure this out from my finite difference model so it can account for funnier shapes/motions.

I wonder if this is given by bending moment and stress force:

$$M = -EIy_{xx}$$

$$F = \frac{M (y)}{IA}$$

Where $$y$$ is the deflection from neutral and $$A$$ is cross sectional area.

However, I'm getting nothing useful from that. For a simple small plastic lever which should only generate 12 N or so of force from the general estimate equation I'm getting 10s of thousands of newtons using this approach.

Any help? Thanks.

This is not really an answer but a hint.

I agree with you that the impact is not limited to the small collision zone and propagates back and forth throughout the length of the beam and the vibration speeds can easily become supersonic, with the air cracking sound.

What you are asking is called whiplash vibrations. Your beam is considered a whip and the rigid point should have some nominal flexibility otherwise it will break or tear the whip.

Before the era of computers fifty-some years ago we used the energy methods and calculus of variations to find and equate the lagrangian of the system to zero. It would require several times integration by parts and be very hairy and error-prone.

I guess modern simulator programs can easily model that.

The problem, as stated, has no solution, due to the combination of two factors. First, your support is unyielding (i.e., it has infinite stiffness) and second, your beam was reduced to a 1-D object.

Recall that force is the rate of change of momentum, $$F=\frac{\partial p}{\partial t}$$. In this setting, the point that comes into contact with the support (point 12 in your notation) has a non-zero momentum until it hits the support, then it's momentum must become exactly zero. This is a discontinuous function; therefore, its derivative is undefined. In your simulation, it should blow up to infinity inversely proportional to your timestep.

You can fix this by either making the object at point 12 a (very stiff) contact spring or modelling your beam as a 2D (or 3D) object.

The second approach would allow you to (in principle) model the elastic waves travelling on the body, but it would require a change in constitutive model and in solution scheme (finite differences can be made to work for 2D or 3D elasticity, but the treatment of boundary conditions becomes very cumbersome).