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ajk1 (Structural) (OP)
15 May 13 21:14
To use the Westergaard equations to design slab on grade, I need to know for a fire engine weighing 75,000 pounds (maximum axle load = 27,000 pounds), what is:
- the number of wheels on the axle most heavily loaded
- the tire pressure
- the tire width

Can anyone help?

Ron (Structural)
15 May 13 21:24
ajk1...not sure of the manufacturer, but generally, I would conclude 3 axles with the loading you described (tandem rear axles, each having load of 27k, front axle having load of 21k...though if given 75k truck, I would probably break it down as 30k, 30k, 15k) the wheel loads would be about 7000 lbs for each wheel (dual wheels, tandem axle would be the same). If you use 30k per axle, the load would be 7.5k per wheel.

Tires will have a center-to-center distance of about 14 to 16 inches, so you will have overlapping wheel stresses.

Width of tire will be about 8 to 10 inches, depending on tire size.

Tire pressure would be 100 to 110.
Ron (Structural)
15 May 13 21:28 might also model this using elastic layer analysis. That will do the overlapping stresses for you since you'll have overlaps from tire to tire and from tandem to tandem.
Splitrings (Structural)
16 May 13 8:35
I have a truck in my station that has a GVW of 42,000#. It has a single 30,000# rear axle(dual tires) and a 12,000# front axle(single tire). The rear tires are inflated to 130 psi and the front are 120 psi. Now this truck doesn't weight 42,000#. It probably weighs 40,0000# or so.

If your truck truely weighs 75,000# it will have four axles. It will have a set of rear dual axles with a GAWR of 50,000, a pup axle and then a 20,000# front axle, I would guess. Tire pressure will bepend on the vehicle weight.

msquared48 (Structural)
16 May 13 16:31
If this is a ladder truck, the deployed outrigger loads will control over the wheel loads, by far. Check with the local fire marshall to see the configuration of what he is using.

Mike McCann
MMC Engineering

rb1957 (Aerospace)
16 May 13 16:49
you're 1/2 way there already !? you'll saying a 75000 lbs truck with a maximum axel load of 27000 lbs ... maybe two at 27000 and one at 21000 ?

so "the number of wheels on the axle most heavily loaded" ? ... well, pick a tire ... talk to a tire manufacturer.
and "the tire pressure, the tire width" ? ... see above (ie the tire manufacturer)

bear in mind you might not want to assume all the tire on the axle are equally effective ... 4 tires doesn't mean (IMHO) a tire load of 27000/4 = 6500 lbs; 'cause of non-uniformities ... grade, tire pressure, ...
so you might use 4 8000 lbs tires to support a 27000 lbs load.

Quando Omni Flunkus Moritati

slickdeals (Structural)
16 May 13 16:53
I have designed a couple for a job in India with the following distributions
slickdeals (Structural)
16 May 13 16:53
Is there a way to put multiple attachments on a single post?
ajk1 (Structural) (OP)
16 May 13 18:20
Thanks msquared, r1957 and slickdeals for the help and information.
I will follow up with the fire department of the municipality where the building is the meantime does anyone have a feel for the magnitude of the outrigger load?
ajk1 (Structural) (OP)
17 May 13 7:10
Ron - can you suggest an elastic layer analysis program to use?
ajk1 (Structural) (OP)
17 May 13 7:13
Splitrings - I am curious how the inner tire of dual tires, ia pumped up? Does the valve stem extend out to the outside of the truck, or does someone have to get under the truck?
Splitrings (Structural)
17 May 13 9:19
The inner dual tire should have a long valve stem that is accessible from the outside tire. No need to crawl under a truck.
Splitrings (Structural)
17 May 13 9:40
Here is some info from a manufacturer on one of there vehicles. The link to the complete spec is below. It looks like NFPA requires the stabilizers not to exceed 75 psi ground pressure, assuming the operator is using the pads. This specific truck comes with 1/2" thick aluminum pads that are 26" square. If this pad is stiff enough to evenly spread the load, you could back calculate what the pressure would be on the 10"x14" pads if the operator choose not to use them in the station or on the apron.

"The stabilizer ground contact area for each foot pad shall be 10” x 14” without auxiliary pads
and 26” x 26” with auxiliary pads deployed. The ground pressure shall not exceed 75 psi with
auxiliary pads deployed when the apparatus is fully loaded and the aerial device is carrying its
rated capacity in every position. This shall be accomplished with the stabilizer pads deployed, as
outlined in NFPA"

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