I have a question on spread axle set up for doing flatbed work. Which is the best set up for running in the south east for spread distance.... Would I be better off with 10' 2" Spread or a 9ft spread? Im looking at two different trailers and am not sure which set up would work best.
Thanks In Advance For Any Input
Spread Axle Question For Flatbed
Discussion in 'Flatbed Trucking Forum' started by crackinwise, Sep 28, 2014.
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9' spread only allows you to run 39K on the trailer.
>10' spread buys you 40K.EverLuc, crackinwise and snowwy Thank this. -
It's all about the distance between the axels as far as bridge laws go. Not sure what you can gross out with a 9' spread, but with a 10-2 you can put 40,000 on that set up. It's all about the bridge laws as far as weight goes. Hope this helps you out.
crackinwise Thanks this. -
I liked the 10' 2" spread with 40,000 lbs on the axle spread with the ability to manually lift the axles on command. Of course this was in the 80's when I ran flat beds
crackinwise Thanks this. -
Florida doesnt recognize the spread, they allow 44k on your tandems, so its not needed for that state. Also, they've been cracking down on the king pin to center axle length for 53' spreads. The annual permit is only $25 and can be done online. The fine is $100. Soon as I saw him break out the tape measure, I knew what time it was.
As for the states that do recognize the spread,.. someone correct me if I'm wrong. But you need at least a 10' spread to to be allowed 40k lbs back there.
Your road atlas should show the axle weight limits per state for both straight tandem and spread axles.
Hurstcrackinwise Thanks this. -
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It may seem stupid that just 2" makes a different but it actually does. It's all about bridge laws.
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Thank you so much..This really helps.... I didnt realize the 9ft spread would reduce the weights.
MJ1657 Thanks this. -
Actually, the 2" don't matter unless you are only 9'10". The federal bridge formula works out to be 40,000 pounds allowed on 2 axles at a 10' spread.
W = 500(LN/N-1 + 12N + 36)
W = maximum weight of the axle group
L = feet between axles
N = number of axles
W = 500(10*2/2-1 + 12*2 + 36)
W = 500(20/1 + 24 + 36)
W = 500(80)
W = 40,000crackinwise and Boethel Thank this. -
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