General Cantilever Equations

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Main Question or Discussion Point

I am trying to determine some general cantilever equations.

I have an aluminum beam extending out 235 mm (L) from an aluminum
block. The beam is 25 mm wide (W) and 3 mm thick (H). A force is
applied at a point approx. 200 mm from the block and I am interested
at a point 25 mm away from the block, which is the active grid of a strain gauge. I know the exact measurement of vertical deflection at a point 100 mm from the block.

I am assuming the modulus of elasticity E is 10*10^6 psi.

In general, I have the following diagram:
http://www.brentless.com/Images/station2.jpg

In the diagram

A: a driving rod that moves up and down from a loudspeaker setup not
shown, this guides the cantilever on this end
B: a measuring caliper to measure the vertical displacement at point
alpha measured from the aluminum block
C: a mounted strain gauge, the point of interest, centered at delta
from the aluminum block
D: an aluminum block mounted the cantilever on one end

Z1: a known, measurable displacement at alpha distance
Z2: a displacement not known, at the end of the beam

alpha: the measurement from the block to the measuring caliper
beta: the measurement from the caliper to the end of the beam
gamma: the measurement from the caliper to the driving rod
delta: the measurement from the block to the strain gauge


My problems to this point, most formulas I have found assume that the
measurement of deflection is actually taken at the end of the beam.
So how can I use the measurement at the 100 mm point above. I don't
specifically know the value of the force being applied (my cantilever
is being driven up and down by a loudspeaker, so if possible I would
like to leave out the force value and determine an equation based on
the measurement of deflection. Thoughts?

I am trying to determine a general equation of which I can relate to
stress and strain the values and measurements I have stated above for
a testing model. I am not a mechanical engineer, so I don't really
understand this stuff, so I appreciate any advice or help anyone can
give me.
 
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Answers and Replies

  • #2
FredGarvin
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The basic cantillever equation you are looking for is one of the two following:

[tex]\delta=\frac{Pa^2}{6EI}(3x-a)[/tex] for x>a

[tex]\delta=\frac{Px^2}{6EI}(-x+3a)[/tex] for x<=a

where:
[tex]\delta[/tex]= deflection
[tex]P[/tex] = applied load
[tex]E[/tex] = Young's Modulus (modulus of elasticity)
[tex]I[/tex] = area moment of inertia
[tex]x[/tex] = measured distance to load from the cantillever (restrained) end
[tex]a[/tex] = distance from the load to the free end of the beam

You can see that up to the point of load application, the deflection is linear with x. Once you reach the point of application, the deflection then becomes a function of [tex]x^3[/tex]. This will fall out nicely in a spreadsheet.

So to calculate the deflection at a point prior to the load application, use the second equation. For the deflection from the load application to the end, use the first. Don't forget to keep your units correct.

EDIT:
I just went back and noticed you added some more information. I would caveat the above equations by saying that the differential equations used to solve the general beam bending for this case assume that there are no constraints on the free end. Looking at your diagram, you have constrained the free end of the beam so that [tex]\theta[/tex]=0. I'm not quite sure how much that will come into play for your situation. I'd say to try the above route and see how close to measured values you are. There may be some tweaking needed.
 
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  • #3
Mech_Engineer
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It looks to me like your case might be best approximated as a "left end guided, right end fixed" for a concentrated intermediate load scenario as shown in Roark's formulas for stress and strain.

I suck at Latex, let me try to figure it out for a few minutes here and if I can I'll post up the equations. :uhh:
 
  • #4
Mech_Engineer
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Ok, here goes, I'm assuming the "end of the beam" is where the speaker is coupled to the beam, since we don't really care what is going on to the left of the attachment, only to the right. The definition of the "left end guided, right end fixed" beam is simply that it's right hand point (B) is fixed in both up and down translation, as well as angular twist, while the left hand point (A) is allowed to translate up and down, but not allowed to twist; sort of a semi-cantilever joint.

Deflection at any point (x) along the beam, measured from the left:
[tex]Deflection=y=y_{a}+\theta_{a}x+\frac{M_{A}*x^2}{2EI}+\frac{R_{A}*x^3}{6EI}-\frac{W}{6EI}<x-a>^3[/tex]

Where since you are interested in [tex]x>a[/tex] (deflection of the beam to the right of where the force is applied), [tex]<x-a>^3=(x-a)^3<x-a>^0=(x-a)^3[/tex]. So the above equation goes to:

[tex]Deflection=y=y_{a}+\theta_{a}*x+\frac{M_{A}*x^2}{2EI}+\frac{R_{A}*x^3}{6EI}-\frac{W}{6EI}(x-a)^3[/tex]

Boundary Conditions:
[tex]R_{A}=0[/tex]
[tex]\theta_{A}=0[/tex]

[tex]R_{B}=W[/tex]
[tex]\theta_{B}=0[/tex]
[tex]y_{B}=0[/tex]

[tex]y_{A}=\frac{-W}{12EI}(l-a)^2(l+2a)[/tex]

Moment at the left (guided) end:
[tex]M_{A}=\frac{W(l-a)^2}{2l}[/tex]

Moment at the right (fixed) end:
[tex]M_{B}=\frac{-W(l^2-a^2)}{2l}[/tex]

EDIT: Added a semi-descriptive picture...
 

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  • #5
Mech_Engineer
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It's important to note that since [tex]\theta_{A}=0[/tex], the vertical deflection [tex]y[/tex] at any point left of [tex]A[/tex] (ignoring self-weight) will be the same as [tex]y_{A}[/tex].
 
  • #6
FredGarvin
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I was going to post that very same result. I noticed that as soon as I saw the picture of the problem. Good find.
 
  • #7
Mech_Engineer
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I was going to post that very same result. I noticed that as soon as I saw the picture of the problem. Good find.
Roark's is my friend :smile: That, and the Mechanical Engineering Package for MathCAD :cool:
 
  • #8
FredGarvin
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I have been meaning to find that. I use MathCad here as well. So I can take that as a good recommendation for it?
 
  • #9
Mech_Engineer
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I have been meaning to find that. I use MathCad here as well. So I can take that as a good recommendation for it?
The Roark's alone makes it worth it, but it also has a Machine Design and Analysis for analyzing geartrains and metalworking, and a Finite Element Analysis workbook with all of the basic equations laid out.

Definitely worth it :smile:

EDIT: What I would like to see is a package that quadruples the size of MathCAD's built-in reference tables, specifically in the area of material properties. Of course there's MatWeb, but it would be just so darn convenient to have it built in.
 
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  • #10
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So basically, let me make sure I get this, if I want to do anything at all with the problems I am gonna have to know either the value of the force applied or the amplitude of the displacement at the point at the "end of the beam"?

Ok, here goes, I'm assuming the "end of the beam" is where the speaker is coupled to the beam, since we don't really care what is going on to the left of the attachment, only to the right. The definition of the "left end guided, right end fixed" beam is simply that it's right hand point (B) is fixed in both up and down translation, as well as angular twist, while the left hand point (A) is allowed to translate up and down, but not allowed to twist; sort of a semi-cantilever joint.

Deflection at any point (x) along the beam, measured from the left:
[tex]Deflection=y=y_{a}+\theta_{a}x+\frac{M_{A}*x^2}{2EI}+\frac{R_{A}*x^3}{6EI}-\frac{W}{6EI}<x-a>^3[/tex]

Where since you are interested in [tex]x>a[/tex] (deflection of the beam to the right of where the force is applied), [tex]<x-a>^3=(x-a)^3<x-a>^0=(x-a)^3[/tex]. So the above equation goes to:

[tex]Deflection=y=y_{a}+\theta_{a}*x+\frac{M_{A}*x^2}{2EI}+\frac{R_{A}*x^3}{6EI}-\frac{W}{6EI}(x-a)^3[/tex]

Boundary Conditions:
[tex]R_{A}=0[/tex]
[tex]\theta_{A}=0[/tex]

[tex]R_{B}=W[/tex]
[tex]\theta_{B}=0[/tex]
[tex]y_{B}=0[/tex]

[tex]y_{A}=\frac{-W}{12EI}(l-a)^2(l+2a)[/tex]

Moment at the left (guided) end:
[tex]M_{A}=\frac{W(l-a)^2}{2l}[/tex]

Moment at the right (fixed) end:
[tex]M_{B}=\frac{-W(l^2-a^2)}{2l}[/tex]

EDIT: Added a semi-descriptive picture...
 
  • #11
Mech_Engineer
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So basically, let me make sure I get this, if I want to do anything at all with the problems I am gonna have to know either the value of the force applied or the amplitude of the displacement at the point at the "end of the beam"?
I'm not 100% clear on where you're measuring vs. where you want to know the beam is deflecting, but it would seem to me with some algebraic manipulation and multiple equations with multiple unknowns, you should be able to achieve what you seek.

The first large equation can describe any point on the beam based on the boundary conditions provided, and a coordinate 'x'. You could relate one point on the beam with respect to the other by combining multiple equations.
 
  • #12
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So, let me clarify a few things and see if I am on the right track

I know the vertical displacement at a point 100 mm from the fixed end of the beam, i dont know the vertical displacement at the point where the load is applied or the force of the load, and I dont know the vertical displacement at the end of the beam

Do I have these values understood right?

a - the distance from the end of the beam to where the load is being applied?
x - the distance from the fixed end to a particular point?
l - the distance from the fixed end to where the load is being applied?

It looks like in the displacement equation you need your force, vertical displacement at A and then the location x to calculate the displacement at x. I know x and I know the vertical displacement y, but I dont know yA or W
 
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  • #13
radou
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If the only applied force is at the point A, then I can't say I understand Mech_Engineer's diagram and where the value W came from.
 
  • #14
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If the only applied force is at the point A, then I can't say I understand Mech_Engineer's diagram and where the value W came from.
Its the magnitude of the force at point A, is it not
 
  • #15
Mech_Engineer
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So, let me clarify a few things and see if I am on the right track

I know the vertical displacement at a point 100 mm from the fixed end of the beam, i dont know the vertical displacement at the point where the load is applied or the force of the load, and I dont know the vertical displacement at the end of the beam

Do I have these values understood right?

a - the distance from the end of the beam to where the load is being applied?
x - the distance from the fixed end to a particular point?
l - the distance from the fixed end to where the load is being applied?

It looks like in the displacement equation you need your force, vertical displacement at A and then the location x to calculate the displacement at x. I know x and I know the vertical displacement y, but I dont know yA or W
You know all of the beam's physical properties, so all you need to know is either the force that is applied, or a displacement somewhere on the beam; once you know either of these quantities, it is a single algebraic equation with one unknown. A little creative algebraic manipulation should get you where you want to go. My suggestion would be to use the known displacement to solve for W, and then use W to solve for the displacement of the point of interest. It would also be possible to solve directly for the displacement at the point of interest using the known displacement, and substituting W out of the equation. I have not yet tried to solve this problem but I'm pretty sure you now have everything you need to do it.

[tex]a[/tex]- Distance from the left hand side of the beam (point [tex]A[/tex]) to the point at which the force is being applied. Since your force is being applied at the end of the beam, [tex]a=0[/tex].

[tex]x[/tex]- X-coordinate of an intermediate point on the beam which you are interested in. Measured from point [tex]A[/tex], so the left hand side of the beam.

[tex]l[/tex]- The length of the beam, or the distance between [tex]A[/tex] and [tex]B[/tex].

[tex]y_{A}[/tex]- Deflection at the guided end of the beam, calculated through: [tex]y_{A}=\frac{-W}{12EI}(l-a)^2(l+2a)[/tex]

You should be able to solve for [tex]y_{A}[/tex] first using the substitution method, and then for [tex]W[/tex].

If the only applied force is at the point A, then I can't say I understand Mech_Engineer's diagram and where the value W came from.
[tex]W[/tex]- Magnitude of the force (assumed to be normal to the beam) pushing on the beam at point [tex]x=a[/tex].
 
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