Sum of the moments formula

[tspco]

I have looked all over to find a formula I have seen in the past, one phrase I have heard is to calculate the sum of the moments, using that phrase I haven't found what I was looking for or at least it wasn't the same as I recall it. Heres what I want to do, I am working on a project similar to a front mounted zero turn lawn mower Grasshopper(r)is one of this type, ok I need to know the calculated weight on the front (drive)axle, I don't have the exact measurements available yet, If I have 200 pounds mounted forward of the drive axle, and 100 pounds half way between the front axle and the rear axle, how much weight is actually on the front wheels centerline?, for example the 200 pounds attached 18" ahead of of the front wheels, and the 100 pounds is 18" behind the front axle, what is the weight on the front wheels (2)?

 

[SnTMan]

Assuming no other loads and a distance of "l" from the 100 lb load to the rear wheel CL the load on the front wheel CL is: R = (100(l)+200(36+l))/(18+l)

if l = 18, R = 350, Load at rear wheel CL = -50

Regards,

Mike

 

[tspco]

I think I understand, the load on the front wheel(s) is 350 and the load on the rear wheels (s) is-50, so based on that the thing would probably nosedive am I reading it right? I will try some variations in load placement, I can move the CL of the 100 pound weight to the rear, a foot or so, and then add some other weight in that rear area as well, not dead weight but useful weight like a hydraulic tank with say a 10 gallon capacity, which would be roughly 110 pounds, I can reduce the nose dive tendency, but clearly more needs to be done. I will play with this and see what transpires. Thanks Mike!
Ryan

 

[SnTMan]

tspco yes that is correct based ONLY on the two loads given. Any other loads applied including the weights of any other components or structure (and operator) will alter that result. For best accuracy they must ALL be taken into account. Good job for a spreadsheet.

Good luck,

Mike

 

[tspco]

I do spreadsheets, mainly hydraulic stuff. I will build one to day for this particular problem. Thanks again, Ryan

 

[berkshire]

tspco (Industrial)
Turn all of your loads into moments measured from a common datum point, load times distance, (inch pounds or metre kilograms). In your case your datum would appear to be the front wheels. You would like to have a greater moment to the rear so the thing does not nosedive.
Google calculating center of gravity.

 

[rb1957]

it sounds like you want the weight to be carried by one axle, with the other axle providing a balancing load (like the horizontal stabilizer on a plane).


pose the question arround what's fixed ... suppose the distance between the axles is varible, but the forces are fixed (at 200lbs ahead of the fwd axle and 100 lbs 18" behind). it might be easier to understand the problem to consider each load separately and superimpose (add them) later ...
the 100 lbs (down) between the axles produces down loads (up reactions) at each axle, both < 100 lbs);
the 200 lbs ahead produces a large down load on the fwd axle (> 200 lbs) and an up load on the aft axle (it's trying to tip the vehicle over).

maybe the distance between the axles is fixed, but the load positions can be changed ?

 

[tspco]

Not really, I need to know the total weight, on the drive axles, because the hydraulic wheel motors I am planning to use have a maximum radial load rating, and I need to stay well below that rating, most of the weight will be in the form of implements (attachments) that will be positioned just ahead of the front axle, no more than 12-15" ahead of the front axle CL
is the attachment point the cg of the implements will vary by implement design, I need to factor that in also.


Better lives through fluid power, or Necessity, no laziness in the mother of invention, What don't you want to do today?

 

[rb1957]

that sounds like the weight is fixed (well, some different versions), and the cg is known for each version, and the fwd axle position is known (and it's load limit is known), so the rear axle position is variable.

so you should be able to set up a spreadsheet describing each configuration (weight and cg) and play with the distance between the axles to solve the problem ...

maybe look at the problem this way ...
suppose you had only one reaction point, it'd have to be under the cg (and would react the weight). but you have two supports. assume the weight is reacted at the front axle. but this causes a couple (moment) between the load and the reaction (=Wt*d, where d is the distance between the cg and the fwd axle). this couple is +ve if the load is behind the fwd axle. this couple is reacted by a couple at the two reactions points, L apart. if Wt*d is positive then the reaction at the rear axle is +ve (up) and = (Wt*d)/L; and the reaction at the fwd axle is = Wt-(Wt*d/L), also +ve up. with +ve (up) reactions at both axles, there is down load at both axles (which is what you want).

if the cg is ahead of the fwd axle, then the reaction at the rear axle is -ve (down) so that the load applied here is upwards ... tipping the vehicle over ... not what you want.

 

[tspco]

Probably what I will do, is this, setup a counterweight system to place near the rear area of the machine, that way the axle distances will be fixed, then by moving the weight around can set up the weight at a nominal value that will work with all attachments, I have also decided to make the length of the mobile power unit longer, and move the prime mover as far rearward as possible, but still ahead of the rear axle. I may still have problems since the design is still preliminary, and the attachment weights are not calculated as yet. I am going to build the spreadsheet today, and see what comes out of it.
Thanks for your help so far!

Better lives through fluid power, or Necessity, no laziness in the mother of invention, What don't you want to do today?

 

[tspco]

Now that I have started on the spreadsheet, I got stuck, where the distance "1" has become variable, and the weights are also variable, I need to add new variables to the basic formula, is this where I say I flunked algebra?
What I have so far, is
L1=Wheelbase
L2=Overall length, base machine
L3=Length from front axle CL to attachment CL(CG) L4=Distance from rear axle CL to Counterweight CL(CG)
and
W1=Base weight
W2=Front attachment weight
W3=Rear weight
W4=Overall weight
Some of these may be unnecessary, I am not sure.
Help!


Better lives through fluid power, or Necessity, no laziness in the mother of invention, What don't you want to do today?

 

[desertfox]

Hi tspco

Use the rear wheel as your datum point ie take moments about the rear wheel to find the load on the front wheel.
so using your earlier post:-

200lb at 18" in front of front wheel therefore the front weight acts at 54" from the rear wheel:-

200*54"

then 100lb weight acts midway between wheels ie at 18" from
rear wheel:-

so 100lb * 18"

therefore 200 * 54" + 100lb * 18" = the reaction on front wheel * the distance from the rear wheel Ie:-

200*54" + 100lb * 18" = R1 * 36" where R1 is front
wheel load

This is how you build up the equation then transpose to find R1.

R1 = (200*54" + 100lb * 18)/36"= 350lb

load on rear wheel = 330lb -350lb = -50lb

The C of G for the overall weight of the machine cannot be determined untill you position all your other weights ie
counter balance weight etc...

You need to decide where everything is going to be and then
do your moment analysis if the results show the thing will tip then workout your counter balance weight bearing in mind that your counter balance weight will alter the position of the C of G of the overall machine together with its overall weight.

regards

desertfox

 

[tspco]

desert fox, I have a silly question in looking at your last post, I noticed that you had calculated the load on the rear wheel how did you get the 330lb (300lb)for the rear wheel value from? I can't see it.

Better lives through fluid power, or Necessity, no laziness in the mother of invention, What don't you want to do today?

 

[desertfox]

Hi tspco

My mistake a typo error should have read 300lb not 330lb

regards

desertfox

 

[tspco]

I figured that it was a typo, the actual question was actually, using the rear wheel as a datum, and running the numbers, I could not come up with the 300lb factor, I am probably just missing a step.
I will keep looking at it though.

Better lives through fluid power, or Necessity, no laziness in the mother of invention, What don't you want to do today?

 

[desertfox]

Hi tspco

The 300lb is the total sum of the 200lb and 100lb in front of and behind the front axle.
Once you have found one reaction assuming you only have two unknown reactions then the other reaction is found by subtracting it from the total downward forceie:-

100lb+200lb-350lb=-50lb

Which is the same anwser as SnTMan posted earlier.

regards

desertfox

 

[Cockroach]

This topic always makes me laugh! All engineers have been taught in first year to sum the moments, very religously we sum the moments ad infinitum.

Summing the moments comes from conservation of energy. We all know that energy is conserved (i.e. closed system), and work is energy just as energy is work. So a moment or torsion, expressed in units of work, is also conserved. We do this mathematically by summing the moments.

I would look in a playground, Tspco, and watch two (2) children on a teeder-taughter. Ask yourself where one child must sit to exactly balance off the other child sitting at the far end. Sounds a lot like your problem, aye? And energy is conserved about the axile.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[SnTMan]

Cockroach, I sometimes tell people that if you can solve triangles and teeter-totter with a child you can be a mechanical engineer

Regards,

Mike

 

[tspco]

"This topic always makes me laugh! All engineers have been taught in first year to sum the moments, very religously we sum the moments ad infinitum."

Thanks for the input cockroach, I have considered that after looking at a previous post, the one thing I will point out though, I realize that many if not all first year engineering already have had this drummed into them, but I am not an engineer, and I can't to go to school at 48, due to lack of funds etc. I am just a "idea man" someone comes to me with a problem and I do my best to help them solve their problem. This particular project is doable, not very practical however, it is a rich boys toy, but it will do as promised, as soon as I figure out the this particular issue.
Thanks again to everyone to replied!

Better lives through fluid power.

 

[Cockroach]

Absolutely buddy.....absolutely!

And thanks for the correct spelling with "teeter-totter"!

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada