Help! Beam Caluclation

Please Help!
My boss gave me a Excel spreadsheet caluclator that calculates the increase in diameter of a cylindrical column needed to evenly distribute a load across that column. It doesnt work and I'm a moron when it comes to this stuff. Does anyone know what that calculation is? I'm desperate and would be greatful for any help. Thanks! ckozlik@newarchery.com

 

[DaveVikingPE]

>My boss gave me a Excel spreadsheet caluclator that calculates the increase in diameter of a cylindrical column needed to evenly distribute a load across that column.

Questions: Is it a beam or a column? A beam-column? What kind of loading - flexure (beam) or compression (column)?

>It doesnt work

What exactly doesn't work? Is the spreadsheet giving you an error or is the answer wrong? Post the spreadsheet cell formulae and we'll take a look.

>Does anyone know what that calculation is? I'm desperate and would be greatful for any help.

I'm not sure. From the e-mail address you've given, my guess is that you work for an archery equipment manufacturer, as such what is the calculation used for?

 

[ishvaaag]

If you simply want the vertical stress remain constant, simply state the equation for so to happen, accounting at some level under the (assumed uniformly distributed stress atop) load the weight of the trunk of cone of column above.

Even this -if usually more than acceptable- is but a simplification -in that we assume perfect uniformity of the stress at any level- since the column is not prismatic.

 

It's pretty clear that I'm way out of my area of expertise.

Specifically, I'm trying to calculate this:

I have a cylindrical beam that is tapered larger at the base and smaller at the end. It's supported at the base only. I'm trying to find the correct taper that will evenly distribute a bending load applied on the end of the beam when the base end is at a given dia. (lets say .165 dia.)

The spreadsheet that I have to do this MAC format from 1990 and it no longer opens correctly.

This calculation is to help me design an archery broadhead. The loads, both direct impact and side bending are high. I'm trying determine what diameter is needed in the main ferrule to evenly distriute the bending loads that it will be under during impact. (I'm NOT a structual engineer by trade, I'm a manufacturing engineer. Thats why I dont know a lot about these calculations OR terminology)

Thanks. Chris

 

[Imagineer]

What formula does his spreadsheet use?
Imagineer

 

[ishvaaag]

Mathcad solves the diameter of a column for constant stress at outermost fiber under Pv, Ph at tip both numerically and symbolically, giving by the way a quite daunting equation.

I may send to you both in doc format or mcd if you have access to. Put if so you want your e-mail address here.

 

[bengineer]

Just like the Lions and Tampa Bay playbooks, there are many unknowns here.

It sounds like you have a cantilever beam with a fixed base and linearly increasing diameter to the base. A good source for this type of problem is Roark and Young, 'Formulas for Stress and Strain'. If you are having problems with Excel, I would definitely steer clear of Mathcad.

 

[DaveVikingPE]

>The spreadsheet that I have to do this MAC format from 1990 and it no longer opens correctly.

I have a Macintosh... What spreadsheet was it originally created on? Wingz? I have a copy of Wingz that might open it. I have other spreadsheets on my Mac, too, and I've been able to open anything, just about, even M$ stuff.


>This calculation is to help me design an archery broadhead. The loads, both direct impact and side bending are high. I'm trying determine what diameter is needed in the main ferrule to evenly distriute the bending loads that it will be under during impact.

Wow! Can you post a picture so we can see what's going on? I'm not a bowhunter and I can only guess at the geometry, etc. But... I'll give it a try off the top of my head, you have a cantilevered plate, the plate is triangularlly-shaped. Roark & Young's Formulas for Strees and Strain should have the formula for this kind of plate. Of course, I don't have a copy of their book handy, wish I did. I'll see what I can come up with, though.

 

[DaveVikingPE]

I swear we posted nearly the same thing at the same time. That's twice... Great minds do think alike.

 

[LPPE]

Yeah, sure Dave. You just read his post and then repeated it. Just kiddin, but I also would like a picture, 'cause my mind is frazzled today (even havin trouble with a stiffened seat connection), and I cannot picture whats goin on here.

 

[Ron]

While you can approximate your conditions with static calculations, your problem is actually a dynamic one. I doubt there is any fixity at the base. Fundamentally, the arrow shaft and head are in a metastable condition, with stability given only by the relatively uniform vortex shedding and momentum (that's one reason arrows are round, not square).

When the arrow is first released, it is under significant compression (force from bowstring on butt, air resistance on tip). This causes a tendency to buckle (there's kl/r again!), but it doesn't, I suspect because the compression wave rebounds to tension very quickly. Then its off to the races with the compression wave alternating in the arrow like a sine wave.

Try getting someone to do an FEA model of this, throwing in the dynamics. Or you could design it like many other products....trial and error!