Vehicular Centifugal force on bridge

[4Boomers]

For a bridge structure on a horizontal curve does the centrifugal force result in an additional horizontal force and/or vertical force on the structure. I understand that the force is a horizontal force @ 6' above the deck, but am a little unsure if it results in an additional downward vertical load on the deck or if it is a horizontal force transmitted through the deck to the bearings, or both.


Any thoughts would be greatly appreciated.

Thank you.

 

[Leftwow]

I am not sure why people trying to assume G factors without some reference? If you want an acceleration it's velocity squared divided by the radius. At 70 mph convert into (102.67 ft/s)^2/(200 ft radius) you have 52.71 lbs/sec^2, which you would say 52.71 (lbs/sec^2)/32.2 (lbs/sec^2) would equal a g force and in that case would be 1.64Gs. I would definitely be checking the transportation engineers on that to make sure the radius doesn't create unsafe conditions. Also, not sure where calculating G forces would help you

 

[XR250]

At 70 mph convert into (102.67 ft/s)^2/(200 ft radius) you have 52.71 lbs/sec^2,

How did "lbs" get tied up in this?

 

[GregLocock]

Leftwow is in a units muddle i think. Dimensions of force are M L T-2.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[XR250]

Yea, 1.64 G's is not attainable on the average street car.

 

[IDS]

I see the original poster has not been back, and reading the question, he wasn't asking about the magnitude of the horizontal force anyway, but it would be defined in the applicable code. In the Australian Bridge Code it is defined as the force due to the radial acceleration, based on a full loading of design vehicles travelling at the design speed, with an upper limit of (0.35 + theta)Wc where theta is the superelevation slope and Wc the total weight of the design vehicles (in kN). So the Australian code uses a friction factor of 0.35. I don't know where that number came from, it seems quite low to me, but I expect other codes have a similar number.

Regarding the vertical effects, as stated by others, there will be no increase in the total vertical load but there will be an additional moment, so the distribution of loads will change. One difference between the Australian code and the requirements stated in the OP is that in Australia the centrifugal force is applied at deck level, so the effect of the vehicle centre of mass being some height above the road surface is ignored.

Another interesting difference in how friction forces are applied is that the factor for braking loads is 0.45 for a single vehicle, or 0.15 for multiple vehicles. This also seems quite low for an Ultimate load, especially the multi-lane load, since events that would cause all vehicles to brake simultaneously do happen.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[GregLocock]

When you look at loads on the roadway you don't really bother with cars, it's trucks that cause the issues. Hence the low mu values.

http://www.nhtsa.gov/DOT/NHTSA/NRD/Multimedia/PDFs/VRTC/ca/capubs/DOTHS809700.pdf

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[ACtrafficengr]

A significant number of drivers misjudge the radius, and vehicle path radii often exceed the centerline radius.