Spread Axle Weight Limit
 

I have a question and what better place to ask than here...........

After a discussion with another driver about axle weights, I looked in the front of the atlas and under the "National Weight & Size Provisions" it says that a 10' spread (in the U.S.) can have up to 43,500lbs.

My questions is this......am I reading that correctly? Assuming you are not over gross and you have a 10' spread axle, you can scale up to 43,500lbs on them.

In my opinion, the higher number may be the result of a calculation of "interior bridge" according to the formula, but it is limited by the National Single Axle limit of 20,000 lbs.



State roads may have higher single axle limits such as Connecticut (22,400 lbs), Florida (22,000 lbs), Hawaii, (22,500), and Georgia (20,340 lbs). My book is dated 2002.[/url]http://ops.fhwa.dot.gov/freight/sw/overview/index.htm

so would a spread axle be looked at as 2 single axles??

From my understanding it goes like this;

10 foot spread or split is 40k
9 " " " 39k
8 38k
7 37k
6 foot and under is 34k

Spread is measured center of axle to center of axle.

Be sure and double check this info.

based on what it states in the front of the atlas that is what I am thinking too.

Spread Axle
 

A ten foot spread is 20K per axle, or 40K total. Any more, the scale house monster will get you. Still can't go over 80K gross. Make sure the lady runnin' the CAT scale knows how to split the table. BOL

Federal law provides that any two or more consecutive axles may not exceed the weight computed by the Formula even though single axles, tandem axles, and gross weight are within legal limits. In other words, the axle group that includes the entire truck—sometimes call the "outer bridge" group—must comply with the Bridge Formula. But interior combinations of axles, such as the "tractor bridge" (axles 1, 2, and 3) and "trailer bridge" (axles 2, 3, 4, and 5), must also be in compliance with weights computed by the Formula.

In the formula, Weight equals 500 [(LN divided by N-1) + 12N+36]. L is the distance in feet between outer axles of any two cosecutive axle combinations, and N is the number of axles.

If you could get 48 feet between the center of the first drive axle and the center of the last trailer tandem axle, the formula would calculate the weight as follows:
500[(48x4 divided by 4-1) + (12 x 4) + 36]. The answer is 74,000 lbs.