Who is REALLY smart here, and can work this out for me pretty please... beg, grovel, beg!

[Ray C]

I didn't see anyone mention it but there is another consideration when figuring the total load capacity of the system and that is the "dynamic" load. All the calculations I saw referenced were for the static load on the arm (RayC at least mad mention that the mass of the arm was significant as well). When the load starts moving, or "bouncing" as it may be, the mass, as seen by the pivot point, can increase a great deal. The mass accelerates at a rate of 9.8 m/s² due to gravity.

Say that the load/winch got hung up or slipped even 10cm. Using the simple F=ma (force = mass x acceleration) and figuring the acceleration of the mass in a 10cm drop - F = 20kg x 9.8/1.01² (I calculated the time as just over a second) the Force = 20kg x 9.6 = 192n (so the equivalent of a 192kg load for this discussion's purpose). The you have to figure out the torque force at the pivot point and the added force of the arm moving, etc. etc. I'll spare you that much as you see where I'm going with this.:biggrin:

I'm not a mech. engineer (electronics engineer, minor in physics a looong time ago) so I am not sure what standards are used to rate lifting devices. I am sure they are available somewhere and hopefully someone can provide them.

-Ron

Don't worry... I was going to get to that part when we add-in the safety margins.

... FYI: I was going to use impulse reactive forces assuming worst case of dropping the load from full height which I believe was stated as 30cm. BTW, the only reason I'm entertaining this is because A) 30cm is no-where near what classifies as an "overhead lifting device" which my insurance company forbids me from providing consult and B) we're talking about a 20kg (feather-weight) load. If the load was significant, I wouldn't respond. This could all be a wash depending on the arm beam being used. Hate to say this but, I won't give details about beam design or selection.

 

[Dave Smith]

Ian--building a jib crane may be handy but if your shop could handle a portable adjustible shop gantry crane it could handle much heavier loads and be a lot less expensive--it would not be just confined to one area also---I know HF has them very reasonable and some other member went that route---just another thought----Dave

 

[Ian Bee]

There is an app for that!!!


Chris

What is it called?

Cheers

- - - Updated - - -

Okay guys, a wealth of info out there for sure!

I have now got enough info so I can call the specs on the two 60 mm. bearings, as well as the 70 mm. thrust bearing.

As far as the fab is concerned, not a problem, this is my main line of work, as can be seen on my website appliedtectonics.com

I simply want a boom to hold my hand held spot welder, nothing more.

Again, thanks guys... much appreciated!

Cheers

Ian

 

[Marco Bernardini]

I simply want a boom to hold my hand held spot welder, nothing more.

Ian, then instead of a bulky rigid boom what about a balanced arm like the Anglepoise lamp?
Hats off for you incredible signs, they are masterpieces!

 

[Shadowdog500]

There is an app for that!!!


Chris

What is it called?

Cheers

- - - Updated - - -

Okay guys, a wealth of info out there for sure!

I have now got enough info so I can call the specs on the two 60 mm. bearings, as well as the 70 mm. thrust bearing.

As far as the fab is concerned, not a problem, this is my main line of work, as can be seen on my website appliedtectonics.com

I simply want a boom to hold my hand held spot welder, nothing more.

Again, thanks guys... much appreciated!

Cheers

Ian

It is called force effect.

I used to do these by hand, but force effect is great for quick stuff like this. It also shows you all of the calculations.


chris

 

[Vavet]

A few things here...

1) General observation: 20 kg is almost trivial compared to the weight of the beam arm. For the US-System folks, 20kg is about 45 lbs (1/3 the weight of a typical person). Question: What is the beam material you intend to use? The weight of the beam factors heavily in this design. Question: How will you attach the arm beam to the rotating collar that holds the bearings? This is where a piece of the problem lies. Bolts can shear and welds can crack.

2) You're not really interested in the moment about the bearing. Yes, Torque = Force x Moment but, what you really want are the lateral forces applied to the top and bottom bearing. This problem is by definition a "leverage" problem not a torque problem. In this problem, the primary force will be generated by the sum of the mass of the arm beam and the maximum intended load -both considered as "point-loads" at the end of the arm beam. The leverage ratios are between the distance of the two bearings vs the length of the arm beam.

3) What is the vertical supporting column made of and what is it's diameter? This is where we to calculate torque. We need to make sure it doesn't bend -just the same we must make sure the bolts or welds don't snap.

Once we calculate the "theoretical" forces involved, we can then add some safety margins and then move on to determining what your weld or bolt strengths must be.

Ray

1 - what he said - don't forget to factor in the weight of the beam and any associated lifting devices - chains, hooks, load levelers, etc. not already included in the mass of the 20 kg load you're trying to lift. If there is none, you can probably assume the mass distribution of the beam to be fairly uniform and count the weight of it as a single vector acting on the halfway point of the beam itself.

There is an app for that!!!


Chris

Almost, but 20 kg does not equal 20 N. It's more like 200N. Force = m*a as others have mentioned so we need 10kg * 9.81 m/s^2. The actual math works out to be 196.2 N acting on a lever arm of 3 meters, so the torque then becomes 196.2 N * 3 m = 588.6 N-m for the 20 kg mass suspended at the end.

As I mentioned above, you still haven't accounted for the weight of the beam. Let's say that beam has a mass of another 20 kg. If you assume it is acting on a moment arm of 1.5m, then that is (20kg * 9.81m/s^2) * 1.5m = 294.3 N-m. That needs to be added to the 588.6 for a total of 882.9 N-m. Add in a safety factor as mentioned.

Finally, it's a really good idea to wear a hard hat and safety shoes whenever working with overhead lifting apparatus.

 

[Marco Bernardini]

Since the load is applied just at the end of the boom, I'd consider some lightening holes to reduce the overall weight of the structure in the less solicited parts.
And the calculation of these is, of course, once again mechanical engineers bread

 

[Shadowdog500]

Almost, but 20 kg does not equal 20 N. It's more like 200N. Force = m*a as others have mentioned so we need 10kg * 9.81 m/s^2. The actual math works out to be 196.2 N acting on a lever arm of 3 meters, so the torque then becomes 196.2 N * 3 m = 588.6 N-m for the 20 kg mass suspended at the end.

As I mentioned above, you still haven't accounted for the weight of the beam. Let's say that beam has a mass of another 20 kg. If you assume it is acting on a moment arm of 1.5m, then that is (20kg * 9.81m/s^2) * 1.5m = 294.3 N-m. That needs to be added to the 588.6 for a total of 882.9 N-m. Add in a safety factor as mentioned.

Finally, it's a really good idea to wear a hard hat and safety shoes whenever working with overhead lifting apparatus.

Good catch! I thought I typed 200N!!

Chris

 

[Ian Bee]

Ian, then instead of a bulky rigid boom what about a balanced arm like the Anglepoise lamp?
Hats off for you incredible signs, they are masterpieces!

Marco.. Thank you!

Cheers

Ian

 

[TOOLMASTER]

i used 2" 1/4 wall dom tube for my hinge. w 2x4" 1/8" wall for the 10 ft arm.. made it about 15 years ago...


most of the stress will be on how you attach it to your wall...i put hole through the double brick with a steel plate on the other side.

i use a chain hoist on the end....i built it to hold 250 but it can hold way more