Base Motion and Vertical Beam - Basic Reaction Forces?

[sabz333]

TL;DR

Trying to understand forces acting on vertical beam being accelerated at the base.

So I am trying to understand how to estimate the amount of deflection [D] the vertical beam shown above would experience if the base it is attached to is accelerating at a constant acceleration [a] of 9.81 m/s.

I assume the Force [F] would be equal to weight of the vertical beam (mass x gravity) as that would be the reaction force? or is this too much of an oversimplification?

I have the equations correct to calculate beam deflection I just need to understand the force it is seeing and can't wrap my head around it not simply being a reaction force to parts weight.

D = (F*Height^3) / (3*E*I)

E = Youngs Modulus
I = Second Moment of Area

Any help pointing me in the right direction would be appreciated!

 

[ergospherical]

The force $\mathbf{F}$ is an inertial force and satisfies $\mathbf{F} = -m\mathbf{a} = m|\mathbf{a}| \hat{\mathbf{x}}$; it acts at the centre of mass of the vertical beam. The point of application is important, because it determines the numerical pre-factor in the beam deflection formula. ($m$ is the mass of the vertical beam).

 

[sabz333]

Yes ok perfect, I just wanted to make sure I was thinking about this correctly!
So the Force being seen by the top of the vertical beam is equal to the acceleration of the base multiplied by the weight of the beam.

 

[ergospherical]

Corrected it for you :)

So the Force being seen by the top centre of the vertical beam is equal to the acceleration of the base multiplied by the weight mass of the beam.

 

[Baluncore]

In the longer term, you are modeling an inverted pendulum that will have stored energy and a resonant frequency.

 

[A.T.]

The force $\mathbf{F}$ is an inertial force and satisfies $\mathbf{F} = -m\mathbf{a} = m|\mathbf{a}| \hat{\mathbf{x}}$; it acts at the centre of mass of the vertical beam.

The inertial force acts unifomly on the entire beam. Just like uniform gravitation, so those two can be combined into a single uniform field. Uniform fields by themselves don't cause deformation, but the force acting on the base does.

 

[ergospherical]

Technically inertial forces don’t act anywhere, since they’re not actually forces but rather terms arising from the frame transformation. So just choose the most convenient description.

 

[A.T.]

Technically inertial forces don’t act anywhere, since they’re not actually forces but rather terms arising from the frame transformation. So just choose the most convenient description.

The most convenient description, that still gives you the correct answer to the question. For the question of deformation it makes a difference whether you assume the force acts at the CoM or is distributed over the entire beam.

 

[Lnewqban]

As your inertial load is not punctual at mid-span, your case is similar to the one shown in this picture:

 

[ergospherical]

The most convenient description, that still gives you the correct answer to the question. For the question of deformation it makes a difference whether you assume the force acts at the CoM or is distibuted over the entire beam.

That's fair, you're right!