General Cantilever Equations

There are equations for the deflected shape of a beam depending on how it is loaded.

There are "Beam Tables"....that is, compilations of beam types with various loading schemes.

If I understand your description........

and if you're not interested in the deflection of the beam between the loading point & the free end of the beam then you can consider it a

200mm long cantilever loaded at the end

you just need the eqaution for the deflected shape for an end loaded cantilever.

use the measured delfection at 100mm (mid way between the load & the support?) to solve for the "force" needed to cause this delfection.

Then use this force in the eqaution for the deflected shape to get the deflection at your other points of interest.

btw Moment of inertia, is I = (1/12)*W*H^3 H (thickness) is cubed not squared, the units of I are Length^4

I think this link to a lab exercise should do the trick for you

clte.asu.edu/active/Beam%20Bending_Lab%20Procedure.doc

your "beam" is a little wide compared to it's depth (thickness) ...I forget the width to depth ratio where a correction might be needed but the beam model is probably close enough

at least for a first try analysis

I couldn't find any beam tables online

cheers Bob

This might make things a little easier to understand what I have, I would like to show a diagram of what I have: Link:

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

Depending on the required accuracy for your measurements, your situation is a little more complicated than a simple cantilever. The use of opposing wingnuts means that a resisting moment can be generated where "A" meets the beam. This is even reflected in your drawing as evidenced by the horizontal piece of the beam (especially outboard) at the attachment point. The value of the moment will be dependent on many factors (e.g. clamping force, hole tolerance for "A", stiffness characteristics of "A", etc). Even if you replaced the wingnuts with a ball/socket arrangement so as to emulate a point contact, you'd still have some analysis issues based on amount of deflection. For example: if the deflection is relatively large then does the contact point on the beam remain fixed (i.e. gamma doesn't change) OR is "A" constrained to move vertically (i.e. gamma is allowed to increase as the beam moves away from under the contact point). At the very least, see if you can get your hands on a copy of "Formulas for Stress and Strain" by Roark and Young. It's pretty much a standard when trying to define stress/strain relationships when transitioning from analysis to measurement. Hope this helps a little.

Bob's advice was to the point. Let me amplify it a little and dump the engineering gobledegook which I suspect is unhelpful to you.

Supposing that you would like to know the displacement at the point at which a point force is applied to your cantilever beam, knowing the displacement at a measuring point.

I see that your point force is at 200 mm and your measuring point is at 100 mm from the fixed support, so first expressing the displacement in the usual engineering form:

Displacement at a point L distant from the fixed end, at which a force F is applied. D1 = F L^3 / 3 E I and for a point X distant from the fixed support along that same loaded beam: D2 = F X^2 ( 3 L - X) / 6 E I

So the ratio in which you are most interested is D1 / D2

This is the same as writing F L^3 / 3 E I times 6 E I / F X^2 ( 3 L - X ) which boils down to

2 L^3 / X^2 ( 3 L - X) This is just geometry. ***1***

You mentioned that X = 100 mm and L is 200 mm In other words, L = 2 X And putting this into the equation ***1*** above, it turns out that D1 / D2 = 16 / 5 or 3.2 ***2*** so if your voice coil loads the end of a cantilever with one pound and it moves say 120 thousands inch you would be able to measure a deflection of 37.5 thou at your measuring piont (half way)

Unfortunately, your diagram does not show a cantilever, but instead, something that is rather stiffer, because your force attachment seems to be keeping the end horizontal. That would force the lever to a kind of "S" shape, what is called a logistics curve in another context. So instead of a 3.2 multiplier on deflection from the measuring point to the load point, the factor could be as low as 2. The exact factor would depend on the details of your attachment. But I hope that meets your current question.

Before finishing, I should say that a flexure was an early way to support a voice-coil for a loud speaker - but more recently a similar design has been used in lab scales (i.e. a weighing machine) The strain gage is used to drive the voice coil with enough force to balance the weight set in a pan attached to the voice coil. The drive current is a measure of the weight in the pan.

The current required to null the cantilever with the pan empty is used to set the indicator to zero.

Hoping this will serve your purpose

Brian Whatcott Altus OK p.s. anonymous names tend to suggest student writing.

On 2007-02-14, 13:47 Z, in news: snipped-for-privacy@a75g2000cwd.googlegroups.com, Whatever_I_Fear wrote (paraphrased):

Whatever_I_Fear: I assumed your fully threaded driving rod has thread size M2.5 x 0.45, length 100 mm, modulus of elasticity 200

000 MPa, and has a pinned slider connection at its upper end where the vertical driving force, P, is applied. I assumed your aluminum beam has modulus of elasticity E = 68 950 MPa. Please let us know if these assumptions are incorrect.

Using the above assumptions, the bending moment on your aluminum beam at point A is Ma = 22.062*P, where P = vertical load applied at threaded rod upper end, positive downward (N), and Ma = moment, positive clockwise (N*mm). Stress at the point C strain gauge is sigma = 20.694*z1, where z1 = beam vertical displacement at point B, positive downward (mm), and sigma = stress at strain gauge (MPa). Strain at strain gauge is sigma/E. Applied load P = 5.364*z1, where P and z1 are defined above.

The above solution (small deflection theory) applies only if z1 doesn't exceed 3.25 mm, which corresponds to P = 17.4 N.

On 2007-02-14, 13:47 Z, in news: snipped-for-privacy@a75g2000cwd.googlegroups.com, Whatever_I_Fear wrote (paraphrased):

Whatever_I_Fear: I assumed your fully threaded driving rod has thread size M2.5 x 0.45, length 100 mm, modulus of elasticity

200 000 MPa, and has a pinned slider connection at its upper end where the vertical driving force, P, is applied. I assumed your aluminum beam has modulus of elasticity E = 68 950 MPa. Please let us know if these assumptions are incorrect.

Using the above assumptions, the bending moment on your aluminum beam at point A is Ma = 22.062*P, where P = vertical load applied at threaded rod upper end, positive downward (N), and Ma = moment, positive clockwise (N*mm). Stress at the point C strain gauge is sigma = 21.86*z1, where z1 = beam vertical displacement at point B, positive downward (mm), and sigma = stress at strain gauge (MPa). Strain at strain gauge is sigma/E. Applied load P = 5.364*z1, where P and z1 are defined above. Beam vertical displacement at point A is z2 = 3.078*z1.

The above solution (small deflection theory) applies only if z1 doesn't exceed 3.25 mm, which corresponds to P = 17.4 N.