Deflection of 30mm aluminium tube beam with 500N load

[EddieC147]

TL;DR

Calculating Beam Deflection with Moment of Inertia and Mass

Hi There,

I am wanting to calculate the amount of deflection (δ) from a simply supported Beam. My Beam is an Aluminium Tube ø30mm with a 3mm Wall Thickness.
Force (F) - 500N
Length (L) - 610mm
Youngs Modulus (E) - 68 Gpa
Moment of Inertia (I) - ?
δ = F L³ ∕ 48 E I

Q1: Is this the correct formula that I found online?
Q2: Put Simply why is the Moment of Inertia needed for this? ( I know this isn't relevant to solving the problem but I want to learn and understand)
Q3: What units should the Moment of inertia be measured into be entered into this formula?

The formula for the moment of inertia (I) of a tube I have found online is below.
Mass - M
Bore Radius - R1
Tube Radius - R2
I = 1/2 M (R1²+ R2²)

Q4: Is this the correct formula for finding the moment of inertia of a Hollow Tube?
Q5: What units should the Radius be measured into make the end unit match with Q3?
Q6: What units should the mass be measured into make the end unit match with Q3?

Finding the Mass needs the density of aluminium multiplied by the volume.

Q7: Do I need to multiply this by the volume of the full aluminium tube?

Any help is greatly appreciated.
Thanks
Ed

 

[Lnewqban]

Welcome, Ed!
A tube does not make a very strong beam.
In order to resist bending, the cross-section of any beam should have as much material as possible, as far as possible from its neutral axis, on the same plane the bending forces and moments are being applied.

That is the reason for the shapes of the cross-sections of steel I-beams, H-beams and C-beams, as well as structural elements with square and rectangular closed sections.
Such shapes have a greater moment of inertia (the term is confusing) than circular or oval cross-sections of similar dimensions and wall thickness.

Please, see:
https://en.m.wikipedia.org/wiki/I-beam#Design_for_bending

https://en.m.wikipedia.org/wiki/Section_modulus

https://en.m.wikipedia.org/wiki/Second_moment_of_area

https://en.m.wikipedia.org/wiki/First_moment_of_area
:)

 

[EddieC147]

Thanks for your input.

The Tube is not going to be used as a beam, just the beam bending calculation is the best way to represent the force that may act upon my Tube.

 

[FEAnalyst]

In this case, assuming that only the concentrated force in the middle is acting on the beam (ignoring self-weight), the maximum deflection will be: $$y_{max}=\frac{FL^{3}}{48EI}$$ where: $$I=\frac{\pi(D^{4}-d^{4})}{64}=\frac{\pi \cdot 30^{4} - 24^{4}}{64}=23474,765706 \ mm^{4}$$ Thus: $$y_{max}=\frac{500 \cdot 610^{3}}{48 \cdot 68000 \cdot 23474,765706}=1,48 \ mm$$

Here $I$ stands for area moment of inertia (also called second moment of area). It doesn't depend on the mass of the beam, only on its cross-sectional shape.
This video can help you understand the topic:



There's also one about the deflection but it discusses some advanced methods so try to get familiar with the concep of $I$ first.

 

[EddieC147]

Thanks for the reply, that is absolutely fantastic just what I wanted.

Thank you for your help