Calculator Help

[wsdave]

Hey folks,

New guy here, and I'm trying to do some figuring using these calculators:

http://www.engineersedge.com/calcula...are_case_4.htm

http://www.engineersedge.com/beam_be...m_bending9.htm

I'm designing an engine crane similar to this:

http://www.harborfreight.com/2-ton-f...ane-35915.html


The problem is that using the section calculator, I get a Moment of Inertia of .8 and Mid point of 1.125 based on a 2.25" square tube with .125" walls.

When I plug this into the beam bend calculator with 1,000# at 48", my stress at a specific point of 12" (where the ram connects) is over 50,000#, and the fulcrum is over 67,000#.

That is already higher than the yeild point of A500B square tube (about 46,000#), and that's at HALF the rated lifting capacity of a 1-ton crane, much less the 2-ton unit.

So either the engine cranes are made of magic, or I'm doing my math wrong.

Suggestions?

Thanks,

Dave

[PinkertonD]

Hi Dave,

Well, a few things to get you started. The calculator you are using is for a cantilever and this is not one.

The part you are calling the "fulcrum" is actually a counter-balance to the load. The ram is the point load on your beam.

Try studying a sketch of what you are trying to do and visualize the forces. What you need to think about is what load do you want the ram to lift in that configuration before the beam fails at that ram connection point. The load at the ram will be 3:1 of the desired load-lift. Also, don't forget to factor in a Safe Working Load (SWL) for the beam sizing.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

[Kelly_Bramble]

One of these calculators are closer for determining the reactive forces - not exact


http://www.engineersedge.com/simple_levers_menu.shtml

[Kelly_Bramble]

Ok, when I put the numbers in for the small extension square tube with the wall thickness you suggest:

Moment on Inertia = 0.8

Extension of the small tube only, because the moment of I is different for the larger tube - the wall thickness is likely double because there is a tube within a tube and it has a larger outside crossection. Boom extends from 42" to 63-1/2", therefore small tube extends (63.5 - 42) = 21.5 inches max. = Distance (l)

Load = 1000#

Distance (x) = whatever....

Distance to nuetral axis = 1.125

I get stress at opposite end of small single tube lift loading = 30,234 psi (This is where the small tube enters the big tube).

So, if material yield is 48ksi there is plenty of fos...

[wsdave]

One of these calculators are closer for determining the reactive forces - not exact


http://www.engineersedge.com/simple_levers_menu.shtml

When I said "engine crane", that's not like a normal crane: It has the fulcrum at the "rear" and a lifting ram about 12"-18" down the arm. I believe the term is a class 3 lever.

[wsdave]

Hi Dave,

Well, a few things to get you started. The calculator you are using is for a cantilever and this is not one.

The part you are calling the "fulcrum" is actually a counter-balance to the load. The ram is the point load on your beam.

Try studying a sketch of what you are trying to do and visualize the forces. What you need to think about is what load do you want the ram to lift in that configuration before the beam fails at that ram connection point. The load at the ram will be 3:1 of the desired load-lift. Also, don't forget to factor in a Safe Working Load (SWL) for the beam sizing.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

You mention the counter-balance to the load instead of fulcrum. That's brings up the vision of a see-saw, where more weight is applied to one side to offset the extra length of arm on the other side (for example). Basically like a crane that builds a skyscraper.

The engine "crane" isn't that. It's hinged at one end of the lifting arm, with a lifting ram connected to the arm about 12"-18" from the hinged end. I think it's referred to as a 3rd class (3rd order?) lever.

I figured that using the beam calculator would tell me how much force would be pressing down at a point, as well as at the end, so I could tell if the metal would be strong enough to withstand a given weight at the end.

[wsdave]

Ok, when I put the numbers in for the small extension square tube with the wall thickness you suggest:

Moment on Inertia = 0.8

Extension of the small tube only, because the moment of I is different for the larger tube - the wall thickness is likely double because there is a tube within a tube and it has a larger outside crossection. Boom extends from 42" to 63-1/2", therefore small tube extends (63.5 - 42) = 21.5 inches max. = Distance (l)

Load = 1000#

Distance (x) = whatever....

Distance to nuetral axis = 1.125

I get stress at opposite end of small single tube lift loading = 30,234 psi (This is where the small tube enters the big tube).

So, if material yield is 48ksi there is plenty of fos...

My tube info was actually based on the larger tube, and while this specific crane I linked to happens to have a 21.5" max, I was looking for a more generic solution as to why the numbers don't work.

The 2-ton version can support 1000# at the end of the 63.5" arm, with the two sections overlapping a mere 12". While that only puts 30,000# at the end of the small tube, it's bound to be more at the end of the big tube, which has the MAXIMUM 46,000# yield. And that's assuming they are using A500B high strength steel, which (based on the $200 price) I'm not certain they are.

[Kelly_Bramble]

My tube info was actually based on the larger tube, and while this specific crane I linked to happens to have a 21.5" max, I was looking for a more generic solution as to why the numbers don't work.

The 2-ton version can support 1000# at the end of the 63.5" arm, with the two sections overlapping a mere 12". While that only puts 30,000# at the end of the small tube, it's bound to be more at the end of the big tube, which has the MAXIMUM 46,000# yield. And that's assuming they are using A500B high strength steel, which (based on the $200 price) I'm not certain they are.

Well, the top truss strap support on your example engine lift changes the math and loading considerably. The strap will be tension during lifting - absorbing a considerable amount of the loading that would other wise is applied to the tubing section at the fulcrum. Without the two top straps the lift would likely have some strength issues. Have you studied or are familiar with "statics"?

The calculators here on Engineers Edge are typical; however do not represent this case exactly. You really need a free-body diagram associated with the different structural cross sections to calculate the applied loading at each of the different sections.

To calculate the moment of I and stress at the lifting hydraulic actuator fulcrum, you would need to include the vertical strap and the resulting re-distributed loading.

[PinkertonD]

The engine "crane" isn't that.

Oh, OK, my mistake.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

[wsdave]

Well, the top truss strap support on your example engine lift changes the math and loading considerably. The strap will be tension during lifting - absorbing a considerable amount of the loading that would other wise is applied to the tubing section at the fulcrum. Without the two top straps the lift would likely have some strength issues. Have you studied or are familiar with "statics"?

The calculators here on Engineers Edge are typical; however do not represent this case exactly. You really need a free-body diagram associated with the different structural cross sections to calculate the applied loading at each of the different sections.

To calculate the moment of I and stress at the lifting hydraulic actuator fulcrum, you would need to include the vertical strap and the resulting re-distributed loading.

I didn't include the top strap, because I'm hoping to not use one. My intention was to use thicker metal, of a larger diameter, and keep the top clear. I'll need that space to run a winch cable that will be mounted at the rear and run along the top to the front on rollers.

http://www.northerntool.com/shop/too...3726866-_-Hein

http://www.harborfreight.com/1-2-hal...nch-37555.html

As for the differing structural cross sections, I realize that the telescoping nature makes things a little more complicated, but that that's why I'm trying to build for the weakest link.

So I guess my question comes down to: Am I doing the calculations wrong, or does the top strap really make 10's of thousands of pounds worth of strength difference?

[wsdave]

Oh, OK, my mistake.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

Honest mistake, though. As an engineer, "crane" probably has a specific meaning to you. I'm NOT an engineer. Heck, I've never even ridden a train, much less driven one...

Seriously though: As an auto mechanic, we call that a "crane", because it lifts stuff up, and because we don't know any better.

I want to build one that has an extra long lifting arm, but don't know which math/calculator to use to figure out how long the arm can be, based on metal stresses.

[wsdave]

Well, I managed to get some time of a Physics instructor at my local community college, and he mentioned a number of components I hadn't thought of. I also heard back from a metal supplier about some rectangular tubing. Turns out that rectangular is stronger and lighter than square for this application.

The Physics instructor offered to put me on the right path, but the time he can devote to my course-correction is limited, so I'll be making the most of the calculators here.

I'm obviously not going to ask you folks to design this project for me, nor am I going to ask you to teach me the engineering involved. I will ask, however, if anyone would be willing to check my concepts, and tell me if I'm using the right calculators. I won't ask you to do the math, just to make sure my brain is working properly.

Thanks for your time and consideration.

Dave

[jboggs]

You will have to analyze each telescoping section independently, using worst case scenarios (maximum extension, maximum load, minimum telescoping overlap). The reason is that each piece has to withstand its own load. Failure of one does not imply failure of the other. The inner tube is loaded at the end and supported at each end of the "overlap" length. The outer tube has no idea what the actual load is. It just sees the load imposed on it by the inner tube - at the overlap section. Kelly asked if you are familiar with the study of "statics" (forces, reactions, moment arms, etc). That is the key to determining what your actual loads are BEFORE you even start to analyze tube bending.

And yes, the top strap can make a huge difference in strength. But there are other ways to accomplish the same stiffening effect and still leave room for a winch cable to pass through.

[RWOLFEJR]

There are Cherry Pickers out there already being made with longer reach and higher tonnage capabilities. Some made from rectangular tube etc... You can almost certainly buy one for far less than you will spend building one... unless this one you need has to be to some exact height and lift etc.

One on "the" online auction site for lifting tractor trailer type diesel engines with 9 foot boom... very husky looking unit. It's in New Jersey and they're taking offers. Another option would be to buy a heavy duty one and beef it up. Get you closer to what you need and probably for less $$...

Another quick thought... Remember that there's always more than one way to skin a cat. An overhead beam and a nice electric hoist is a dandy thing for pulling motors! A gantry type crane can work O.K. too. You might want to consider some of the other options that a few hundred more bucks might get you? If you start getting in pretty deep on a picker you might want to upgrade right off and skip the picker and go electric etc.?

[wsdave]

There are Cherry Pickers out there already being made with longer reach and higher tonnage capabilities. Some made from rectangular tube etc... You can almost certainly buy one for far less than you will spend building one... unless this one you need has to be to some exact height and lift etc.

One on "the" online auction site for lifting tractor trailer type diesel engines with 9 foot boom... very husky looking unit. It's in New Jersey and they're taking offers. Another option would be to buy a heavy duty one and beef it up. Get you closer to what you need and probably for less $$...

Another quick thought... Remember that there's always more than one way to skin a cat. An overhead beam and a nice electric hoist is a dandy thing for pulling motors! A gantry type crane can work O.K. too. You might want to consider some of the other options that a few hundred more bucks might get you? If you start getting in pretty deep on a picker you might want to upgrade right off and skip the picker and go electric etc.?

The closest thing I've found to date runs about $6,000 and isn't as long as I need. the advantage of my design is that it doesn't require any electricity or overhead structure to be able to lift material quite high.

[wsdave]

You will have to analyze each telescoping section independently, using worst case scenarios (maximum extension, maximum load, minimum telescoping overlap). The reason is that each piece has to withstand its own load. Failure of one does not imply failure of the other. The inner tube is loaded at the end and supported at each end of the "overlap" length. The outer tube has no idea what the actual load is. It just sees the load imposed on it by the inner tube - at the overlap section. Kelly asked if you are familiar with the study of "statics" (forces, reactions, moment arms, etc). That is the key to determining what your actual loads are BEFORE you even start to analyze tube bending.

And yes, the top strap can make a huge difference in strength. But there are other ways to accomplish the same stiffening effect and still leave room for a winch cable to pass through.

My system will be 3 tubes long. The third tube will be 72" long, 12" of which is overlap into the second tube. I choose this 12" length arbitrarily based on existing designs, so it is NOT critical.

That piece weighs 31.62 pounds. I plugged the info into the "Cantilevered Beam with Uniform Load" calculator, giving it's entire 6' as the length, and choosing 12" as the stress at a specific point (I figured that the greatest stress would be at the closest unsupported edge).

Based on the above information, the 12" point has 697.1# of stress on it. My "logic" (I use the term loosely) tells me that the stresses on that tube would be calculated as weight on the end ("Cantilevered Beam with One Load Applied at End"), with the resulting stress at the 12" point, PLUS the 697.1# of stress caused by it's own weight.

This combined stress would be compared to the yield point of the metal.

This is obviously not including any modifications based on the initial 12" of sheathed tube.

The alternative would be to only call the beam 60" (the unsheathed portion) and place the stress at point value at 0.0". That gives a higher stress of 836.5#.

Am I headed in the right direction?

Thanks

[Kelly_Bramble]

[QUOTE=wsdave;392]My system will be 3 tubes long. The third tube will be 72" long, 12" of which is overlap into the second tube. I choose this 12" length arbitrarily based on existing designs, so it is NOT critical.

Yes, you are going in the right direction. I like the telescoping tube design for storage and handling reasons. You will need to make sure there is plenty of contact bearing area between the tubes as that the fits are not so loose to cause point loading resulting in an over stressed location.

BTW, the stress at "x" or "Stress at a specific point" is interesting; however the number that you should pay attention to is the "stress at the support". That load (stress) will be the maximum loading realized by the loading configuration.

Also, don't forget that you will need to determine the loading carrying capacity of the connection points. I'm assuming that you are going to use a few welds as well as a couple of pivot shear pins to support the loads. These will need to be understood as well.

See:
http://www.engineersedge.com/weld_design_menu.shtml

[wsdave]

[QUOTE=Kelly Bramble;393]

My system will be 3 tubes long. The third tube will be 72" long, 12" of which is overlap into the second tube. I choose this 12" length arbitrarily based on existing designs, so it is NOT critical.

Yes, you are going in the right direction. I like the telescoping tube design for storage and handling reasons. You will need to make sure there is plenty of contact bearing area between the tubes as that the fits are not so loose to cause point loading resulting in an over stressed location.

BTW, the stress at "x" or "Stress at a specific point" is interesting; however the number that you should pay attention to is the "stress at the support". That load (stress) will be the maximum loading realized by the loading configuration.

Also, don't forget that you will need to determine the loading carrying capacity of the connection points. I'm assuming that you are going to use a few welds as well as a couple of pivot shear pins to support the loads. These will need to be understood as well.

See:
http://www.engineersedge.com/weld_design_menu.shtml

You are correct: With 3 tubes weighing anywhere from 32# to 125# each, being able to separate them for storage, handling, and transport is vital.

My tube "slop" on the 12" overlap is .125" top to bottom and side to side. In engineering terms that may be a chasm, but it seems pretty tight from a fitting standpoint.

As for the stress-at-point numbers: I find that, curiously, the stress at the support is a fraction of the stress at the "connection" point, that's why I used the higher number (I assumed that the point of the highest stress would be the weakest link, and thus what I must design for).

As for welding, there will be as little as possible. Unfortunately, there WILL be some holes for locating/shear pins. My physicist tells me that the moment you add a hole, you've compromised the structure and need to adjust the math accordingly, but says that we can work that out AFTER we've worked out the baseline numbers.

[jboggs]

You said you were using the "Cantilevered Beam with Uniform Load" calculator. I think that is the wrong calculator, for two reasons. Your beam is not a cantilever. It is supported by a point support at each end of the overlap. Also "Uniform Load" means the load is spread out over the whole beam. I think your load is concentrated at the end of the beam. Isn't that correct? I think your loading diagram would show one force downward at the end (the load), one force upward in the middle (at the end of the overlap), and one force downward at the other end of the overlap to balance the load. You will have to use statics to calculate what those forces are. The calculator I would use is "Supported at Both Ends Loaded at any Location". That name sounds wrong but if you look at the diagram and turn it upside down you will see what I mean. If this is unclear let me know and I'll post a diagram for you.

[PinkertonD]

I think that is the wrong calculator, for two reasons. Your beam is not a cantilever.

JB, you are not alone, see my post, #2 in this thread. I elected to drop out of this race when I learned that my knowledge gleaned over 40 years of Engineering did not apply. You may want to conserve your energy too.

Dave