Assume a 48′ long simply supported beam.
One end of the beam (point A) is supported at a point 2′-6″ in from that end of the beam, the other end of the beam (Point B) is supported at a point 7’=6″ in from that end of the beam. The span is 38′-0″.
Assume a uniformly dense load of 48,000 pounds imposed upon the beam. The load is 30 feet long. So we have 8 feet difference between the supported span and the length of the load.
Assume that the centerline (G/G) of the load is equidistant from the support points.
Question #1) is the force exerted on the supporting points equal? (ignore the unequal weight of the beam cantilevers). I would think so if the load is centered.
Question #2) What is the effect of moving the load one foot toward Point A? Two feet toward Point A?
Question #3) Is the answer to Question #2 a linear function?
Question #4) What is the effect of placing some part of the load beyond Point B?
For those saavy and astute observers, yes, this is a semi-trailer and I’m trying to calculate axle loads.
Thanks.
Replies
Don't take my word for it this is just my gut feeling.
Question #1) If the load is centered I would think 24,000 pounds per point.
Question #2) Every foot you move adjusts the load by 1600 one foot 25,600& 22,400
two feet 27200 & 20800 etc
Question #3) Yes I think it is linear
Question #4) Once you go beyond I would think for every foot beyond it is counter balanced by that amount. So when the edge of the load is above point B you have 17600 at A 30400 at B. once you go beyond B you are counterbalancing the load but for every foot beyond B you still added 1600 pounds to point B. Now you have a mechanical lever. This is where I LOOSE IT This late at night.
Wallyo
On #4 since it is a simple lever, every pound of downward force has an equal upward force at the other point at the same distince from the fulcrum & multiplied for every foot farther away right???????
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1. yes
2. A>
3. yes
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(1) As long as the load is centered, the loads will be equal. 24,000 # each support.
(2) The effect of shifting the load one foot does not quite change the load at each point by the amount of the linear load per foot (1600 #) because of the shift in the moment arms, or lever arms as Wallyo called them. The actual load shift is 1264 # change at each support point for each foot the load is shifted, provided the total length of the load remains constant.
(3) yep.
(4) Any shift you make in the load toward or past either support will increase/decrease the load at the two support points by the same amount listed in (2) above (1264 #) times the number of feet moved.
Catskinner:
The general formula for the force at point B is 48,000 divided by (D/38), where D is the distance from the CENTER of the load to point A. From that formula you can get the load whereever the you place it.
for question 1, D=19 feet, so D/38 is exactly one half, so force on B is same as force on A
for question 2, each foot moving towards point A DECREASES the distance D, so the force at point B decreases by 48,000/38 pounds, about 1263 pounds. similarly, 2 feet would be 2526 pounds
for question 3, you can see that yes, it is a linear function
for question 4, you can see from the equation that the force on B keeps increasing as the distance from A increases. (The most it can be the total load, where the center of the load is centered over point B.) The edge of the load is over point B when D=23 feet, so the at that position the force on B is 29,050 pounds, one foot past B is 1263 more, or 30,313 pounds.
As you know, the sum of the axle loads is equal to the total load, so as force at B goes up, the force at A goes down the same amount.
Rich, that is exactly what I was looking for.And especially thanks for showing the calculations, I'm going to print out your reply and probably make up a spreadsheet to mess around with.Loads vary dimensionally and by density, the laws for axle weights do not. The resulting difficulties which can arise . . . you get it. <G>Thank you.
Rich, I just followed along with your work, and I have a question;<<The general formula for the force at point B is 48,000 divided by (D/38), where D is the distance from the CENTER of the load to point A. From that formula you can get the load whereever the you place it. >>So I understand that given a load (L), the force at point B can be expressed as ((L/(D/38))<<for question 2, each foot moving towards point A DECREASES the distance D, so the force at point B decreases by 48,000/38 pounds, about 1263 pounds. similarly, 2 feet would be 2526 pounds>>This is where I got lost. Why is it now 48,000/38 rather than ((48,000/(18/38)) ?Or, why is the weight shift 1,263 rather than 2,526?
Ok, I didn't refresh before I saw all the answers. I'll give you my formulas and you can stick it in a spreadsheet. Let's say you have a uniformly distributed load of "weight"
Distance from the rear axle to the center of the load is L1
Distance from the rear axle to the tongue is L2
F1 is the weight on the rear axle
F2 is the weight on the tongueF1 = weight * (L2 - L1)/L2
F2 = (weight * L1)/L2
if the center of the load is behind the rear axle, L1 is negative.
If you had a non-uniform load, you could split it up into uniform pieces and sum the forces after you do each calculation separately.I usually just use TLAR myself, except that I usually visualize the whole operation to see if there are unbalanced moments or forces involved. TLAR without considering things like that have probably contributed significantly to the number of crushing accidents that happen every year -- it's one of the top reasons for industrial deaths. I get to move some heavy stuff at work, because nobody stops me. I wish I got to drive a train, I'm the kind of engineer that wears a pocket protector and has glasses that are held together with tape.Eric
Thanks, Eric. That's going to work right now.Just so you have a context for this, the Federal Motor Carrier Safety Administration allows 34,000# on a tandem set (like tractor drive axles, about a 4'2" spread) and 40,000# on a 10'-2" spread axle (like what you see on flatbeds, that is, the axles are 10'-2" apart).My tare on the drive axles is 11,520 pounds.Tare of the 10-2 spread (trailer axles) is 9,080 pounds.Total tare for drivers and trailer is 20,600 pounds.A load of 48,000 pounds is common, what is even more common is brokers lying about the weight and it's really 49,000 to 49,500. They make more money that way, and the more the shipper can stuff onto the truck, the happier he is, too.So while the FMCSA axle allowance considered by itself suggests that up to 74,000 between these axle sets is permissible, that is not actually the case. I have a steer axle, too, and a legal limitation to 80,0000 pounds gross (CGVW).So back to the trailer;As you can see from the formula you provided me with, the location of the load on the trailer is crucial. If the load is incorrectly placed,and an axle is overweight, and the DOT discovers this at the scale house, if it's lumber and I need to move a few sticks to balance the load, that's a bummer, but it isn't tragic.If it's steel or something else a human can't pick up, then I have a fairly serious and expensive problem.So now with the help of everyone here I can determine the balance point on this trailer and where the centerline of the load needs to go to keep the axle weights legal and safe.BTW, I'm thinking of going back to school to be the pocket-protector-and taped-eyeglasses kind of engineer. I think it's a pretty cool job. I'd probably miss trucks and tractors, but the money has GOT to be better. <G>
Edited 2/2/2008 7:36 pm by Catskinner
One more engineering question;Given that the trailer axle set is a 10-2 spread, are my calculations to be done using the distance to the C/L of the rearmost axle, or to the point halfway between the set of rear axles (that is an imaginary point 5'-1" forward of the rear-most axle)?Thanks.
Edited 2/2/2008 9:33 pm by Catskinner
Catskinner:
Sorry I haven't checked messages in a few weeks. It seems like Eric has answered your questions. If you still need some help, please post again and I will keep checking in on this thread.
Rich
Thanks, Rich.I've been gone (in more ways than one <G>) for a few weeks, myself.Yes, I think I've got it. And this is really useful -- I took the tare weight of the drive axles and the trailer axles and calculated the center of balance for the trailer. Now I have a mark on the trailer for loader operators to work with.Hopefully that means no over-axle fines. I probably owe you guys some beer now, right? <G>
Glad to be of help. Drink one for me.
Rich
Thanks to everyone else who answered, also.
Next, we gotta look at this "lever arm" thing. <G>
to solve a problem like this in general, you have to calculate the moments around one of the support points. In order for your trailer not to flip over or fly through the air, sum of the forces and moments is zero. When the load is centered over the rear axle, the load will be zero on the tongue. You can treat it as a point load unless you are interested in the deflection of the trailer beam. If you solve this problem, you'll find the loads are linearly related to the position relative to the axle. But since you're asking for a loads analysis on the internet, please don't drive past my house :)
This is a physics problem, but I never saw it until I got to statics in engineering school.
<<But since you're asking for a loads analysis on the internet, please don't drive past my house :)>>As an engineer, you may have some appreciation for this; so far I have been using the same method as the early railroad engineers, that is, the revered and time-honored TLAR algorithm."That Looks About Right."Those guys didn't have the math, either. Have you ever read J.E. Gordon's work? It's wonderful.Like the more successful railroad engineers, I'm surprisingly close, usually. Within a few hundred pounds.I just had no idea why, and now I do. This stuff is really interesting.