How much will it hold w/o bending?

[Artisan]

Guys does anyone have any engineering savvy and can you suggest to me
how much weight this " cantilevered " (?) 2" solid bar will hold w/o bending?

FWIW, the solid bar is actually 61"long, it slips into another tube on the left.

Bonus points if you can link me to an online calculator so I can try to figure out other
issues meself!

TIA!

 

[hosspuller]

Need the material specs. And what is your max deflection? Or do you just want to know the yield point? (where it will stay bent) All steels will deflect with just the weight of material. Some a little and some a lot.

 

[Gary Fowler]

You need the material specs for yield strength and also what kind of heat treatment the material has undergone to get a good figure. FWIW, some times hollow tube is harder to bend than solid bar stock.

 

[Artisan]

Need the material specs. And what is your max deflection? Or do you just want to know the yield point? (where it will stay bent) All steels will deflect with just the weight of material. Some a little and some a lot.

Well...

"What is my max deflection?" you asked....
Ummm, I think I would pefer no bending, if that is what deflection means.
Then you say it will all "deflect". Gosh I am silly stupefied here... if it
bends just a little bitty bit and straightens back out I am ok w/ that. For alls
I know this will hold 15,000 pounds, in the real world if I get 4000 I think
I would be happyhappyhappy.

The "Yield" point...well, I will be off trying to figure out what that means when I finish this reply
but it is spec'd in the spec's below. I do not want it to bend and stay bent, THAT would be real bad!

What I do not know about steel will fill volumns and volumns but
I kinda think I would go with the A108 material available fairly easily
to me as spec'd below.



2"x2" Square Solid Bar


◦Specifications: ASTM A108
◦AKA: Cold Roll Square, CRS Square, KeyStock
◦Applications: shafts, machine mounts, fixtures, etc.
◦Workability: Easy to Weld, Cut, Form and Machine.
◦Mechanical Properties: Brinell = 126, Tensile = 64,000 +/-, Yield = 54,000 +/-

or

◦Specifications: ASTM A36
◦AKA: HR square, square bar
◦Applications: frame work, braces, supports, shafts, axels, etc.
◦Workability: Easy to Weld, Cut, Form, and Machine
◦Mechanical Properties: Brinell = 112, Tensile = 58-80,000 +/-, Yield = 36,000 +/-

 

[Artisan]

Gary if 2" x 2" tube is tougher than solid bar in this situation I would
certainly go w/ it to save on weight. As for the heat treating I have no
info, just off the shelf stuff from Metals Depot? - Buy Metal Online! Steel, Aluminum, Stainless, Brass

 

[Tractor Seabee]

I have an engineer friend I use for these problems but he charges $150/HR. You can find the formulas in engineer handbooks if you know enough math including calculus to enter all the values required. There are even online calculators but you need to understand all the terminology to find the known values and enter in the right place. From practical experience; your 2 x 2 bar will not take even a fraction of the loads you are discussing. Yes 2 X 2 tube will hold more than solid bar but not that much. The vertical dimension of the member is where the strength is not the width or the weight. Going wider does not increase strength in a linear manner. Your weak link may be the fastening point at the fixed end. A lot of stress develops there. We all my be able to apply some of our practical experience if you told us more about what you are trying to accomplish; pictures may help. Naturally we DIY guys tend to over design just to be safe; or conversely, under design due to lack of knowledge or experience. A practical observation; look at a HF gantry crane, its 6" I beam at 8' long is rated 2,000# at the 4' center with no deflection. There shop crane is a 2" square tube inside a 2.5" tube and is rated at 1,000# at the short length of about 2', at 4' it is only 250# if my memory still works. Food for thought.

Ron

 

[Artisan]

Yes Ron,

I can find the pro engineer as well and I have done tried to figure all the Youngs Modulous and Intertia and moments and
everything else out and it ain't happening.

Where I say it is Fixed on the left is not real world.
I just need to know what a 2" square anything will hold "IF"

It is actually the exposed part of a telescoping widget, 45" is exposed the rest of the widget to the left
is designed rock solid, the stinger is the weak link and I am trying to figure out the max weight it will
hold w/o bending.

If I were to show the whole shootin' match it would make everyone brains explode! ;-)

I just need to worry about the weakest part. When will it fail?

As I wrote in post #1, it actually extends 16" to the left inside another tube,
there will be no failure there, it is way overbuilt. I can not overbuild this
stinger, that is why I seek the failure number.

I deleted all the bracing and a LOT of other stuff in an attempt to not cloud
my original question, cause it will... ha! Everyone will see it eventually... ;-)

 

[Artisan]

...to be exact, here is a little more;

 

[Sodo]

From Stress and Deflections in Beams
you have the lower condition, except only half of it. So I put 8000 lbs in the middle, and then what you're looking at is 4,000 lbs at the end.

See Area Moment of Inertia to get area moment of inertia. You can fool around with larger tubing to reduce the weight. A 2" solid rod is much stronger than a 2" pipe. Think of it this way, rod will be a little stronger than a 2" pipe with a solid rod inside the pipe.

If the goal is to have no significant deflection you won't have to concern yourself with yield. If you have "yield" it is bent and will stay bent. If you keep the deflections small don't worry about yield. However there are safety concerns, lots of them. You should be building to minimum double your design weight, maybe 4 or 5 times if there are people around this weight.

E for steel is always 29,000,000

===============

F - Load (lb)

L - Length of Beam (in)

I - Moment of Inertia (in4)

E - Modulus of Elasticity (psi)

y - Perpendicular distance from to neutral axis (in)



Total Load : 8000 (lb)
Length of Beam - L : 90 (in)
Moment of Inertia - I : .78 (in4) = (pi*D**4)/64 = 3.14159*2*2*2*2/64
Modulus of Elasticity - E : 29000000 (psi)
Perp. distance from neutral axis - y : 0 (in)
Support Force - R1 : 4000 (lb)
Support Force - R2 : 4000 (lb)
Maximum Stress - : 0 (psi)
Maximum Deflection - : 3.14 (in)

================

So 4,000 lbs at 45 inches out will deflect a 2" steel rod 5.34 inches.

Further engineering is necessary to check the stress to see if it will bend (or take a "set") or "yield" but my gut feel is that it would not spring back from 5.34"

This is all seat of the pants, maybe you can get a real engineer who remembers what I've forgotten.

 

[Artisan]

Initial thoughts are SWEET but it is not round it is Square...Oh man, see that is how much I know...

 

[Artisan]

Ugh... I just looked at Metals Depot? - Buy Metal Online! Steel, Aluminum, Stainless, Brass and they call round and square both BAR

 

[the old grind]

Do you have room above for a folding truss link hinged to the right end that could function when extended and be anchored/hinged/tucked above the junction of the telescoped sections when not? This would put a tension load on the truss and a compression load on what looks like an extendable boom or something. Either of these easier force distributions would seem to increase lift(?) capacity vs the bar alone with its typical deflection. (Not much to go on ... I imagined this..):

All the easier if the bar doesn't telescope to use/store, but a 1pc truss could be 'pinned on' at full extension, too.
(I claim insomnia ... sorry if I'm 'out there')

 

[Artisan]

It is a cantilevered beam not supported at both ends.

http://www.engineeringtoolbox.com/cantilever-beams-d_1848.html

I just can't figure out how to enter the data...

 

[Sodo]

ooops made a big mistake, (formula for intertia error) but I edited it in text above. They didn't have a cantilevered beam so I did a 90" rod - and look at the middle as your cantilever.

This does seem more feasible putting 4000 lbs in the center of a 90" rod.

deflection with 2" round rod is 5.3 inches
deflection with a 2" square solid bar is 3.14 inches

I suspect 3 or 5 inches would cause 'yield' or permanent bend.

i'm working pretty fast and loose here sorry. But this is how you can do it yourself with calculators on the internet.

 

[Artisan]

Do you have room above for a folding truss link hinged to the right end that could function when extended and be anchored/hinged/tucked above the junction of the telescoping sections when not? This would put a tension load on the truss and a compression load on what looks like an extendable boom or something. Either of these easier force distributions would seem to increase lift(?) capacity a good bit vs the bar with typical deflection. (Not much to go on ... I imagined this..):
View attachment 349408
All the easier if the bar doesn't telescope to use/store, but a 1pc truss could be 'pinned on' at full extension, too.
(I claim insomnia ...)

Now THAT is thinking outside the box! HA! I will be thinking about that ! Ha!

Gotta love TBN, there are some great minds here...

 

[Egon]

Might help??

[PDF]Bending Stresses for Simple Shapes - ATC Publications Home Page
www.atcpublications.com/Sample_pages_from_FDG.pdf
Cantilever Beam, Uniform Load ... The physical properties of the material used in the design of the ... A hollow pipe, square or round, is the easiest extrusion.

Once upon a time this type of calculation was well within my limited capabilities. However, as the hair disappeared so did the limited capabilities.

 

[Xfaxman]

...to be exact, here is a little more;

That didn't work, try again.

Can you use 2x4 rectangular tubing?

 

[Farmerford]

From memory:

Section modulus of a solid square = side cubed/6; 2x2x2/6 = 1.33
Stress in outer fiber = torque in inch pounds/section modulus
section modulus x stress = torque
1.33 x 36,000 = 48,000 in pounds torque to create 36,000 pounds of stress in outer fiber
48,000 inch pounds/45 inchs = approx 1,000 pounds force at 45 inches from fixed end of 2x2 square beam cantilever to stress A36 steel to tensile strength

 

[Artisan]

That didn't work, try again.

Can you use 2x4 rectangular tubing?

That was weird, that image is now visible. It shows I actually am dealing w/ about
38" of square tube or square solid bar w/ about 7" extended off the end of that
equaling about 45" of cantilever.

I could go 2x4 if there is a 2.5 x 4.5 .25wall counterpart to slip it into then redraw ALL. err.

 

[bcp]

A folding brace is an interesting idea but it would be stronger as a triangular design.

The rectangular brace will keep the end vertical as the bar bends but won't do much to stop the bending.



Bruce