Simple cantilever beam question

In summary: I'm sorry that this is confusing for you. It is a difficult topic and it definitely requires some knowledge in order to fully understand it.
  • #1
I've been looking at the equations describing different aspects of cantilever beam dynamics, but I can't seem to find what I'm looking for. (And I'm not smart enough to derive it.) If I pluck the tip of the beam and measure the location (x) of the tip over time (t), it should look something like this picture I found on google image search: beam L=0.6.jpg

What is the ODE that describes dx/dt? Is this so simple that nobody actually talks about it? Thanks for any help~
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  • #2
Well what is dx/dt? It is the change in position over time. So you will have to use positions at incrementally increasing times (i.e. v2-v1, v3-2, etc)

You aren't "not smart enough" to derive the equation, something like this isn't that tough. "deriving" this equation is just thinking it through and understanding what output you want.
  • #3
OK, I appreciate the encouragement but this is not a homework question and physics/engineering is not my field. Surely this is in a textbook somewhere? I'm not sure why I can't find it. Any ideas?
  • #4
I have seen a lot of questions on cantilever beams, but I am not familiar with this type. generally we're asked only to find out the maximum deflection at the tip, but you need the change in deflection with respect to time. so you have a case of SHM.

suppose you bend a cantilever by loading it at one end. the cantilever will now have an angle θ, (say) with respect to the original position. so if we consider really small angles (i.e. really small loads) we can approximate the deflection(x) as:

x= length of the cantilever(L)*θ

now if we take the moments about the pivot point (the point where the cantilever is attached to the wall), we have from Newton's second law ( Ʃtorque= Iα)
mass moment of inertia of the system about the point (J)* angular acceleration(double differential of θ, let's give it a symbol DDθ) = algebraic sum of moments about the pivot point.

at this stage I'm assuming that the mass of the rod is uniformly distributed, and therefore the centre of mass lies at the centre of the rod i.e. at a distance of L/2 from the pivot point. ( this assumption maybe wrong, but this is the only way i know how to do this)

(m*L*L*cosθ*cosθ)/(4) *DDθ= m*g*l*θ

this shd be your ODE. correct me if I'am wrong.

sorry for the stupid symbols and stuff like L*L. don't knw how to work this.
  • #5
waiittt.. we need to eliminate theta... i'll get back on tht after i take a nap
  • #7
Thanks for the replies~

So at any rate this does seem to be somewhat beyond "Cantilever Beams 101" and the type of thing that would be listed in any textbook. OK, well let me take a look at what's been posted here and see if it makes sense.
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  • #8
alpha zero;
I know that its damping. it would be damping anywhere on Earth that's without a vacuum. but don't you think we should ignore the damping caused by air?... it will be pretty small.
if there was a damper attached somewhere then you should have considered damping.
this case is very similar to a simple pendulum undergoing oscillations. now the differential equation describing its motion is l*DDθ+gθ=0

but how does the pendulum stop if it isn't damped?.. it is damped but only a little by the air friction.. maybe the time period for such a case is huge...!

if we do take in the damping into consideration i'am getting this equation:
F=-mgcosθ/2-mlDDθ/4... where F is the damping force...

i can show you how i got this if you guys are interested... now that we have this though how do we find the damping force of air?
  • #9
the more i post on this the more confused i get
  • #10
I'm confused too, and I do appreciate the time that you're spending on this.

I've taken basic diff eqs, but this seems to be much harder than I would have expected, very different from forced springs and similar things. Now I'm starting to see why it's not just right there in any textbooks. Do you people who claim that it's a straightforward derivation actually know for a fact that it is, or are you just assuming?

What is a simple cantilever beam?

A simple cantilever beam is a type of structural element that is supported at only one end. It is commonly used in engineering and construction to support loads and resist bending.

What are the main components of a simple cantilever beam?

The main components of a simple cantilever beam include the beam itself, the support at one end, and any loads applied to the other end. The beam is typically made of a rigid material such as steel or concrete, and the support can be a wall, column, or other structure.

How do you calculate the bending moment of a simple cantilever beam?

The bending moment of a simple cantilever beam can be calculated using the equation M = WL/2, where M is the bending moment, W is the applied load, and L is the length of the beam.

What factors affect the strength of a simple cantilever beam?

The strength of a simple cantilever beam is affected by several factors, including the material properties of the beam, the length of the beam, the magnitude and location of the applied load, and any other external forces acting on the beam.

What are the common applications of simple cantilever beams?

Simple cantilever beams are commonly used in construction for supports, such as in bridges, balconies, and roofs. They are also used in mechanical and aerospace engineering for structural components and load-bearing elements.

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