[1:14,5] How much force?

Guten tag :-)

First of all, sorry for my non-German. I hope that I one day will be able to meet you all in your native language, but until then I have to stay with English.

I am in the design phase of the crane boom cylinders, on my 1:14 LTM 1500-8.1... but I have a huge problem. While I can calculate how much I will be able to lift at the end of the boom, I can not find out how to include the weight of the boom it self in the calculation.
My question therefore is: How much force will the orange vector need to push with, in order to lift the boom?



Data:
The boom pivots around the red spot to the right.
Boom length: 4 meter
Boom weight: 12 kg (mass)
Distance pivot-point to point where orange vector lifts: 0,6 meter

Can anyone help me please? My head is spinning out of control over this, and I cannot agree with my self how to calculate.

Bis dan,
Robert Holsting **** www.robse.dk

Hi there,

you need the center of gravity! If the boom is straight and equal from the beginning to the end then you find center of gravity right in the middle. Now you have the length (i.e. 2m) and a force (12kg x 9.81) round about 120N --> 240Nm. To get the force at the cylinder pivot point you need to calculate this way: 240Nm : 0,6m = 400N (40Kg).

this calculation is for a horizontally boom and a vertical direction of Force!

> how to include the weight of the boom it self in the calculation

There are detailed ways to do this, but a rough approximation is to assume the boom is weightless and half its actual weight is placed at the one end and the other half is at the other end (the pivot point and hence does not contribute to the required lifting force).

Thank you all very much. I was afraid that the force needed was as high as 400N, but also happy that I did not go ahead and build without this taken into consideration. :-) I need to re-think my design.

Amazing how far and wide our model building takes us..

The force you need for the cylinders also depends from what you want to lift up at the hook. The 400N are additional to weight at the hook x radius. simply expressed...

This evening I will have a look at my notes, I´ve calculated sth like this already for my project. Maybe I can explain it better then.

Hi,
Oh, great. :-)


I am aiming at the full picture, but was missing the boom as an isolated load.
I am actually working on a formula that includes the following information:

Mass, boom,
Mass, hook,
Center of (boom)gravety,
Angle between cylinder and boom,
Angle of boom vs. horizontal,
Cylinder power,
Distance, pivot point of boom -> point of lift (cylinder)
Distance, pivot point of boom -> tip of boom,

-all this to find out how strong my cylinders must be, and later make a load-chart. If I make it, I'll make an online version like the other (crane)engineering tools on my web.

I look forward to your input, and thank you very much for taking the time. :-)

Hi Robert,
I found some draft with angles and distances, but not the calculation... tomorrow I will have more time to look after it. In the worst case I have to calculate again, but that would take a bit more time.
By the way, the time I spend on it will be a benefit for me as well

Hi,
Thank you very much !
When I get close enough to my own pc again, I will try to show what I got so far.
Brgds, Robert

Hi Robert,

at first I calculated a "simple" draft with the boom in transport position. You will be surprisedwhich force the cylinders will need just for erecting the boom
The reason of this high force is just this "worse" angle between the cylinders and the boom.

I used the dimensions of my draft, but the weight of the boom you told us.

I hope you can identify something
the force of the cylinder is called Fc...

At this weekend, I will try to complete the calculation with an erected boom and some weight at the hook.

Hi Dave,
Thank you VERY much!
I will sit down with my calculator and make a calculation for my setup (Need to redo a lot, I can see), and see what kind of numbers I reach., but I can already agree with you: I AM supprised: That IS a lot of force, just to get the boom up. :-)


I'll be back.. and once Again: thank you for taking the time.
Brgds,
Robert

Hi Dave,
Two small questions, please:

The "80" on the left, distance between center of boom and lower point of cylinder is not used, right?

Where are the "7,13 degree" measured?
(Cylinder angle in ref to boom?)

Brgds,
Robert

The '80' is a '40'
On the bottom of the paper is a triangle of the forces/dimensions, that's what I needed to get the 7,13°. Yes, it's the angle between boom and cylinder.

Excellent, thanks :-)

Hi Again,
It looks as if I might get away with it a little easier than the example, unless I messed up that is.. :-)
This example is for when the boom is not extended, and horizonal.

The strange PI() and /180 in the SIN and COS calculation is caused by Excel. Excel reads radians as standard, everyone else uses degrees... Thus the conversion.

I'll Work on it more, and add more variables... load capacity, boom angle, extended boom etc.
Brgds,
Robert

Hi again,
Well... I have got a little longer, but not quite there yet.
First I can calculate the power needed to raise the boom if it is fully horizontal, and fully extended (3,4m max length) (Worst case scenario):

Pic 1

I then use that number to calculate the available power used for hooks and load:

Pic 2

What would be nice is to merge the two... so you can input cylinder power, boom angle and level of extension, and then get "max load". To get to that, the first formula must be extended to include boom angle.. and then the two formulas must merge.

Best regards,
Robert

Good morning Robert,

the same problem as I had, merging tese calculations

Sorry for not-answering in the last few days, I have to get my workshop ready until the electrcian comes on saturday to connect my new fuse box