whiz kid mathematical types please

[hazmat]

<font color=blue>Thanks for the great explanation and formulas for calculating the forces involved. I wish had known these for a few past projects. Now the next question, is there a formula or table somewhere to calculate the strength of steel tubing/flat/pipe/etc? So that you know it will be able to withstand 3500lbs without bending.</font color=blue>

Actually anything will bend when you apply a force, the question is how much will it bend? It is a bit detailed to get into here, I recommend you pick up a "mechanics" textbook at your local library. You'll have better luck finding one at a college library. The sub- subjects are statics (stationary) and dynamics (moving).

<A target="_blank" HREF=http://www.amazon.com/exec/obidos/tg/detail/-/0534371337/qid=1034085130/br=1-12/ref=br_lf_b_12//002-7437373-6458468?v=glance&n=491348>Mechanics Book at Amazon</A>

The "Bible" for mechanical engineering is <A target="_blank" HREF=http://www.amazon.com/exec/obidos/tg/detail/-/0070049971/qid=1034085430/sr=8-1/ref=sr_8_1/002-7437373-6458468?v=glance&n=507846> Mark's Handbook </A> However, this is a reference book, it assumes you have already learned the stuff in it. The textbook is a better way to educate yourself.

here is a web site with some info <A target="_blank" HREF=http://www.efunda.com/home.cfm>Efunda</A>

 

[hazmat]

<font color=blue>This is off the subject but what converter did you use. A DWG to jpg? I could use somthing like this quite a bit.</font color=blue>

Actually I used a tiff to jpeg converter. ProE will let me save a drawing as a tiff. Do a google search, I'm sure what you want is out there somewhere. Autocad may even have a plug in.

 

[tpaulson]

Hazmat
Thanks for the suggested books and the website. Too bad Efunda charges for access to their website. It looks like it contains a lot of useful information. I may have became a paid member.

I was hoping there was a relatively easy formula/calculator were you could enter a constant for a particular steel type & shape from a table, the force that would be applied, and length of steel. That would give the amount of deflection for a span and usage type(ex. cantilever). After looking at efunda I see that I was probably over simplifying the math and physics involved.

Tim

 

[hazmat]

Here is the formula for a cantilevered beam with a point load on the end

deflection = (Load*length^3)/(3*E*I)

E = Modulus (material specific) typical value for steel = 30,000,000 lb/in^2

I = Moment of inertia (geometry specific ie I beam vs. rect. tubing etc.) for a rectangle = (BH^3 -bh^3)/12

B = base of outside
H = Height of outside
b= base of inside rectangle
h= height of inside rectangle

 

[tpaulson]

Thanks Hazmat!!! I guess that doesn't look too difficult. But there must be a large number of formulas depending on material shape and usage (cantilevered, simple span with 2 end points, etc.)

So is following correct for Harv's example?

3500 lbs of force
4' back from the center of gravity
If he used a single 3" x 3" x 1/4" steel tube the end of the tube would deflect .106 inches.
If he used 2 equal tubes and the force was equal on both, each tube would deflect .053 inches.

I = (BH^3 -bh^3)/12
I = ((3" x 3")^3 - (2.5" x 2.5")^3) / 12
I = (729 - 244) / 12
I = 40.4

Note: I presume the length needs to be in inches to keep the units consistant, so I multiplied 4' x 12
deflection = (Load*length^3)/(3*E*I)
deflection = (3500 x (4' x 12)^3 ) / (3 x 30,000,000 x 40.4)
deflection = .106 in

Would a .106" deflection on a 4' piece of steel tubing cause a perminent deformation (bend) after the force was removed?

 

[hazmat]

<font color=blue>But there must be a large number of formulas depending on material shape and usage (cantilevered, simple span with 2 end points, etc.)</font color=blue>

Yup, different formula for each application. The equation changes for cantilevered vs. simple span & point loading vs. uniform loading etc.

The moment of Inertia (I) changes based on the geometry of the beam (rectangle vs. I beam vs. pipe).

<font color=blue>Would a .106" deflection on a 4' piece of steel tubing cause a perminent deformation (bend) after the force was removed?</font color=blue>

My first guess is no. But we need to caculate the stress. According to my Marks handbook, The shear = W = 3500 lbs. Then we need to divide by the cross sectional area to get the shear stress = 3500/2.75in^2 = 1272 PSI. A36 steel has a yield point of 36,000 psi. So WHarv can use even smaller steel if he wants to.

If we were really good we'd use Mohr's circle to calculate the combined stresses (tension and torsion), but since 1272 psi is so low compared to 36,000 there is no need to.

So Harv, if this works, & I need I new fence some day, you may need to plan a vacation w/ IRIS to New England/w3tcompact/icons/grin.gif

 

[wroughtn_harv]

Thanks Stephen,

I figured the thirty five hundred pound load was where I wanted to be, nice round number.

I was looking at the tractor yesterday and decided how I was going to build the assembly I needed.

The two primary pieces are going to be attached and pivot two hundred and seventy degrees along the rear roof line of the canopy. Right next to where the lift hooks are located. You have to figure if that place is stout enough to lift the tractor then it's also stout enough to attach a jack for lifting the tractor/w3tcompact/icons/smile.gif.

They will collapse, one tube into another, and lay along the roof when not needed. When needed they will fold over and down and then extend out to the desired length for attaching the pad.

Right now I'm thinking two rods attached at the bottom back of the tractor are what I need to locate the pads front to back. After all when the pads are functioning to lift the back of the tractor the rods will be in tension which is probably they're strongest trait. The two arms coming off the back top will be in compression, the second strongest trait of the tubing if I'm thinking straight.

This allows me to use smaller lighter than if I was wanting to rely on twisting or bending of the tube to do work.

The pads will either be bowl shaped or if I can find some, short squatty body wheels. The tractor lifts in an arc. I can move it back and forth on the pads if they're bowl shaped or even better wheels with just the actions of the boom to keep the hole plumb.

 

[hazmat]

wharv,

Sounds like you've got a solid plan. Let us know how it works out.