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Axle weights 1

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MikeMilburn

MikeMilburn

Mechanical
Mar 22, 2000
115
Does anyone know of any software that can be used to determine the axle weights of a truck if the load (in this case a dump body) is know. I would like to try different scenarios as far as center of body to center of tandems is concerned as well as other options, plows wings etc...Mike
 
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Don't know if there is any software out there (although I would guess that each of the manufacturers has some for their own chassis), but a simple configuration is fairly straightforward to calculate.

1. Determine the empty vehicle axle loads (typically a
trip to the scales)

2. Determine the differences between the empty load and the
max allowable axle loads.

3. Swing moments about each axle to determine the load
center:

(LCf)(Lf) - (LCr)(WB-Lf) = 0

where LCf = load capacity front
LCr = load capacity rear
Lf = distance to load center from
front axle
WB = wheelbase

4. Apply tests of reasonableness to the results
- typically in a set forward configuration the liming
factor is the rear axle load capacity because the
load center tends to move so far forward that you
can't reach the max front axle load and keep the
center reasonably located in the box.
- in a set back axle the load will tend to distribute
itself fairly reasonably in the load box.
- do I exceed state/county laws
- how do I handle things like heap ratio
- brakes, frame, etc...............

5. The calculations get a little more complex, but the
logic follows with pushers and tags.

All of this being said - your body builder can probably answer all of your questions in a few minutes if you are in process of purchase as can your dealer.
 
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Does anyone know as mentioned below.

How can I calculate axle weights if truck has four axles.
For example GVW=31t and load centre of mass is defined.

I know that I have two equations (first equations balance of vertical loads and second balance of moments) but there are four undefined axle weights.
 
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If you have 3 axles (tridem) at the rear , you should find the location of resultant (group of axles) reaction force which is the center line of axle 3. If you have four axles in the rear (quad) then the resultant reaction force is haf way betwee axles 3 and 4. This method assumes all rear axles carry the same load and front(steer) axle is #1.
 
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feadude,

That method only works accurately if the axles are closely coupled and on a bogie-type suspension (you would still need to account for the differences in the unsprung weight). A better approach is to consider compliances and solve the problem as a statically indeterminate system.

Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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mloew
The method that you are recommending assumes equal loads on the ground for each rear axle or not? How is suspension equalization (such as air ride) or lack of it for that matter incorporated into calculations? What do you mean by compliances?
Can you elaborate?
Thank You
 
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feadude,

No, the method actually does not assume equal load. That is the limitation of the conventional way of computing the axle loads by taking the reaction at an average position (bogie spring center) and then dividing the load in half for each tandem. A better way for non-coupled axles is to introduce compliances (a composite of the tire and suspension springs may work well) to account for the deflection of the vehicle. You can then solve for the system in terms of the deflection at one axle and the angle of the body. It should be clearer after you have the FBD. Another method is to use Finite Element Method (again using pseudo-springs).

None of these methods account for friction and hysteresis in the suspensions. Some differences in measured vs. calculated values should be expected.


Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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mloew
Do you know of a book, web site, or tech paper that I can read up on this subject.
Thank You
 
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If anyone out there is still looking for axle load software, Royal Nooteboom Trailers in the Netherlands ( have a useful product called Novab.

Aimed mainly at the (very) heavy haulage market.

A.
 
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mloew

Could you send me an example about tridem truck's calculation.
But that should be considered like that 2nd, 3rd and 4th axles' loads are different.


zeusfaber

I could not reach any software or demo that site. I saw only advertisement.

Regards.
 
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Truckker,

It is all statics. The derivation of the equations is dependent on the choice of datum elements. I will try to post a general derivation when I am in the office next.

A tridem might be best approximated by just using the clasical method with the reactions taken about the #3 axle and then split three ways if they are sufficiently close together, share spring loads, and have equal unsprung mass.

Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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Truckker,

You're right - I don't believe the software is downloadable.

Having said that, I think that if you get in touch with Nooteboom, they can provide a demo on CD ROM.

A.
 
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Update to an old thread:

The Dutch obviously like their axle weights. Nooteboom's northern competitor, Broshuis, has just introduced a free online axle load calculator at:


A.
 
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mloew,

Were you able to send info regarding calculations on Tridem drive trucks? What if tridem drive consists of Independent suspensions on each axle? (Here we are talking about Neway AD378 Barpin). I am actually in need to calculate more accurately the load on steering axle. I have used the average middle-point approach on the tridem drive side, but am not satisfied with results. Thank you.
 
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I can not send it out directly. Here is some code I developed in Matlab:

Code: 
% x = row vector of axle positions in x direction (mm)
% k = row vector of axle vertical spring rates (N/mm)
% m = vehicle mass
% xmc = x position of vehicle mass center (mm)

n = max(length(x),length(k));
g = 9.81; %m*s^-2 -- acceleration due to gravity 
    
K=zeros(2,2);

K(1,1)=sum(k);
K(1,2)=-sum(k*x');
K(2,1)=-K(1,2);
K(2,2)=-sum(k*(x.^2)');

F = (m*g).*[1;xmc];

U = inv(K)*F;

z(1) = U(1);
for i=2:n,
    z(i) = z(1)-x(i)*U(2);
end

load = k.*z;

The function is linear in the linear displacement of the spring at axle 1 and the cosine of the pitch angle.

Please let me know how my method compares to your result. I imagine the #1 axle load will be similar.


Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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If let say one the axles inthe tridem group carries less load than the other two, then the average point is not right at the mid axle center line but rather closer to the axle(s) that carry more load.

good57morning@netzero.com
 
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Hmmmm. No comments?
[ponder]
Must not have needed the code?

Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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The requestor hasn't been to the site since 8/24.

rmw
 
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My apologies for not replying... Got caught in many other issues with this project... And unfortunately I do not have Matlab, nor have I ever used it. I tried to translate the code to actual equations but could not. So I am still trying... I do gather that you are using spring loads at the axles, so there is a difference from what I was considering.

Sorry I could not give you comparison comments.
 
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