Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations JAE on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
three point reaction loads 1
- Thread starter Bert2
- Start date
- Thread starter
- #22
If you are only interested in that degree of accuracy, I would just consider that all the load goes directly onto the middle axle. Then when those tires deflect, the load sheds to the outside. I'm not a truck designer, but the appropriate bending moment might be something like P/3 x axle spacing.
isn't the linkage between the cab and trailer a hinge ? ie you're not looking at the entire vehicle, so what are you looking at, i ask myself ?
it looks to me as though there's an extension on the back of the cab, which sort of looks like your original question, except of course the extension is simply supported so that isn't it. maybe it's the cab itself, and you're considering the load from the trailer being reacted by all the wheels.
if this is it, you need to consider the flexibility of the tires if you want to get the right answer.
again, another thread unravels as the OP eventually describes the problem as nothing like the original question. maybe if you'd posted "this is the problem i have" and "this is how i'm modelling it" and "this is the bit i need help with".
sigh
it looks to me as though there's an extension on the back of the cab, which sort of looks like your original question, except of course the extension is simply supported so that isn't it. maybe it's the cab itself, and you're considering the load from the trailer being reacted by all the wheels.
if this is it, you need to consider the flexibility of the tires if you want to get the right answer.
again, another thread unravels as the OP eventually describes the problem as nothing like the original question. maybe if you'd posted "this is the problem i have" and "this is how i'm modelling it" and "this is the bit i need help with".
sigh
- Thread starter
- #25
we're answered, several times, the original question.
it's "just" that the pic doesn't match the original question.
you've posted that you are in fact interested in the reactions of the cab and trailer ... but isn't the linkage between them a hinge ?
the trailer solves reasonably as a simply supported beam. there seems to be something between the cab and the trailer, again simply supported. the cab, again looks simply supported.
no ??
it's "just" that the pic doesn't match the original question.
you've posted that you are in fact interested in the reactions of the cab and trailer ... but isn't the linkage between them a hinge ?
the trailer solves reasonably as a simply supported beam. there seems to be something between the cab and the trailer, again simply supported. the cab, again looks simply supported.
no ??
- Thread starter
- #27
"neglect the hinge and the rear axles on the trailer" = neglect the trailer (no?, 'cause you're interested in the tractor unit)
the tractor has a point load from the dolly.
you can model the three rear tires as rigid supports, and we've posted how to solve this. IMHO, neglecting the tire flexibility introduces significant error into the results. neglecting the weight of the tractor and the neglecting trailer weight reacted by the fwd tractor tires is probably reasonable (unless you're particularly interested in these tires).
part of your problem is going to be that solving the rear tractor 3 tires as a beam will produce an up load on one of the outer tires, which is obviously unreasonable, hence you need to consider tire flexibility.
all this being said, there probably is a shortcut that has been developed for distributing a load on a rigid beam over flexible supports, with the same stiffness.
the tractor has a point load from the dolly.
you can model the three rear tires as rigid supports, and we've posted how to solve this. IMHO, neglecting the tire flexibility introduces significant error into the results. neglecting the weight of the tractor and the neglecting trailer weight reacted by the fwd tractor tires is probably reasonable (unless you're particularly interested in these tires).
part of your problem is going to be that solving the rear tractor 3 tires as a beam will produce an up load on one of the outer tires, which is obviously unreasonable, hence you need to consider tire flexibility.
all this being said, there probably is a shortcut that has been developed for distributing a load on a rigid beam over flexible supports, with the same stiffness.
Bert2:
Your problem is a classic beam on elastic springs problem. The front axle, and the three tandem axles, and their tires and suspension are spring reactions for your beam, with some spring constants; there may be a walking beam someplace in the rear three tandem arrangement; your beam (the tractor chassis frame) has a stiffness “EI”; and when you achieve compatibility between the beam deflection and the spring deflections, you will have solved your problem. You must consider these relative stiffness and flexibility issues to even start to get near a meaningful answer to your problem. And, that’s true however you want to simplify or screw up your presentation of the real question. Although, as in our previous meeting, your impressive picture may bear no relation to your actual question.
Your post of 13Aug10, 7:58, is really out of line; your original question or OP bears very little resemblance to the truck picture you have posted. You got good answers to your original question. Now, you show a very impressive picture of a truck which does not relate to your original question, and blame us for not being mind readers. The fact is, you asked the question wrong, but these guys, for the most part, answered your question correctly and with good advice. This is the very same thing we went around about on your post #404-265967 in FEB10. If you don’t really understand your problem, and then post a question which doesn’t relate properly to the real problem, and then post a picture, to impress, which may or may not have prompted your question; you can not be mad at us for not guessing what you want.
Again, you need a local, friendly mentor of your very own, who can help you through these problems, with sketches and text books and the like. This kind of instruction, guidance and advice is very difficult to give in words alone, when you don’t understand the problem well enough to ask the question correctly. Many of the people here are smart enough and experienced enough to be led immediately to give the correct answer for the question you asked. However, they can’t be blamed if you asked the wrong question, in the wrong way, or you are ignoring important fact of the matter and don’t even know it.
Your problem is a classic beam on elastic springs problem. The front axle, and the three tandem axles, and their tires and suspension are spring reactions for your beam, with some spring constants; there may be a walking beam someplace in the rear three tandem arrangement; your beam (the tractor chassis frame) has a stiffness “EI”; and when you achieve compatibility between the beam deflection and the spring deflections, you will have solved your problem. You must consider these relative stiffness and flexibility issues to even start to get near a meaningful answer to your problem. And, that’s true however you want to simplify or screw up your presentation of the real question. Although, as in our previous meeting, your impressive picture may bear no relation to your actual question.
Your post of 13Aug10, 7:58, is really out of line; your original question or OP bears very little resemblance to the truck picture you have posted. You got good answers to your original question. Now, you show a very impressive picture of a truck which does not relate to your original question, and blame us for not being mind readers. The fact is, you asked the question wrong, but these guys, for the most part, answered your question correctly and with good advice. This is the very same thing we went around about on your post #404-265967 in FEB10. If you don’t really understand your problem, and then post a question which doesn’t relate properly to the real problem, and then post a picture, to impress, which may or may not have prompted your question; you can not be mad at us for not guessing what you want.
Again, you need a local, friendly mentor of your very own, who can help you through these problems, with sketches and text books and the like. This kind of instruction, guidance and advice is very difficult to give in words alone, when you don’t understand the problem well enough to ask the question correctly. Many of the people here are smart enough and experienced enough to be led immediately to give the correct answer for the question you asked. However, they can’t be blamed if you asked the wrong question, in the wrong way, or you are ignoring important fact of the matter and don’t even know it.
msquared48
Structural
I agree fully with dhengr here. Classic beam on elastic foundation analysis is needed. And the the model you choose DOES affect the results. ![[hammer]](/data/assets/smilies/hammer.gif "[hammer] [hammer]")
Garbage in, garbage out. [flush]
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
Garbage in, garbage out. [flush]
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
depending on the accuracy required, you could assume ...
1) that the wheels are equally spaced,L,
2) that each wheel unit has the same stiffness, # of tires, tire pressure, etc
3) that the "beam"/chasis/bogie stiffness is much greater than the tire stiffness
then you could apply the direct load equally to the three wheel units ... P/2/3 (/2 for both sides of the chasis)
and account for the offset between the load and the central tire, d, as a couple on the extreme tires ... P/2*d/(2L)
so the maximum load on a wheel unit is P/2*(1/3+d/2L)
but that is pretty rough ...
1) that the wheels are equally spaced,L,
2) that each wheel unit has the same stiffness, # of tires, tire pressure, etc
3) that the "beam"/chasis/bogie stiffness is much greater than the tire stiffness
then you could apply the direct load equally to the three wheel units ... P/2/3 (/2 for both sides of the chasis)
and account for the offset between the load and the central tire, d, as a couple on the extreme tires ... P/2*d/(2L)
so the maximum load on a wheel unit is P/2*(1/3+d/2L)
but that is pretty rough ...
-
1
- #32
Bert2...still good for $US1000.00? LOL! I see nobody posted the answer.
Here's the tip, five equations in five unknowns; a guy needs to make an assumption about the stress of the beam.
This is an indeterminate, nonsymmetrical loading case so you cannot approach it in the classical sense.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Here's the tip, five equations in five unknowns; a guy needs to make an assumption about the stress of the beam.
This is an indeterminate, nonsymmetrical loading case so you cannot approach it in the classical sense.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
I've chosen to solve the problem using Singularity Functions, noting the Boundary Conditions as the supports are immovable, fixed.
Integration of the loaded beam starting from the Intensity Function introduces two system constants, one of which becomes zero noting the extreme left hand reaction does not deflect, consistent with the above boundary conditions.
This therefore reduces the five equations in five unknowns to four equations in four unknowns. I note three boundary conditions at the support have no deflection, plus sum of the forces in the vertical and sum of moments of the beam each equal zero. I now have the necessary system of similtaneous equations required to draw upon the closed form solution set.
I have used a matrix to manipulate and solve for the reaction supports. I find the extreme LHS is -1.201, the middle is 26.659, and the RHS as 1.772 units each.
Looking at the shear and bending moment diagram, I note the extreme points of bending provides a zero rotation just to the right of the middle support. This is the point of maximum rotation. Further analysis would reveal the maximum point of deflection, which is of particular interest to the problem in your case.
I would design the beam to resist maximum deflection at the point in question and ensure shear above the supports is acceptable.
This problem is a typical static indeterminate form since you have more supports than required. A guy could of also tried this by Virtual Work, I seem to have remembered the Singularity Functions better from the college days, and thanks for the memories!
Fun little question.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Integration of the loaded beam starting from the Intensity Function introduces two system constants, one of which becomes zero noting the extreme left hand reaction does not deflect, consistent with the above boundary conditions.
This therefore reduces the five equations in five unknowns to four equations in four unknowns. I note three boundary conditions at the support have no deflection, plus sum of the forces in the vertical and sum of moments of the beam each equal zero. I now have the necessary system of similtaneous equations required to draw upon the closed form solution set.
I have used a matrix to manipulate and solve for the reaction supports. I find the extreme LHS is -1.201, the middle is 26.659, and the RHS as 1.772 units each.
Looking at the shear and bending moment diagram, I note the extreme points of bending provides a zero rotation just to the right of the middle support. This is the point of maximum rotation. Further analysis would reveal the maximum point of deflection, which is of particular interest to the problem in your case.
I would design the beam to resist maximum deflection at the point in question and ensure shear above the supports is acceptable.
This problem is a typical static indeterminate form since you have more supports than required. A guy could of also tried this by Virtual Work, I seem to have remembered the Singularity Functions better from the college days, and thanks for the memories!
Fun little question.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
- Thread starter
- #34
Doesn't matter, are you suggesting that the tires significantly alter the reaction loads? Hardly. Besides, this would be the "steady state solution", transient effects would have dampened out to give you the static load conditions I have noted. You need to remember that adding elastic supports greatly increases the complexity of the model, we're not looking for six decimal significant digit solutions here.
In keeping it in the general case the reactions show that the middle one takes most of the load and actually acts as a fulcrum point for the bed. The front tires actually lift up slightly and the rear tires take a little bit better load downwards.
So the shape of the beam is like an "S" as drawn in the diagram, top left hand side.
What I would suggest is now do a second solution in the transverse direction right at the middle support. You know that the tires on the left and right support half the computed middle load, you can now solve for deflection directly and get the overall solution for the plate at the point underneath the load.
But what is interesting is the rotation maximizing just right of the middle support.
I note that the solution is consistent with everyday observations. If you have ever been on a loading dock, you will notice the immediate and first response of the truck once the load is placed on the bed is to have the front bumper slightly rise. Clearly the best place for the heavy load is over the axial, is it not? I would therefore move the load 155mm to the left of the indicated spot at present, noting the upward reaction of the LHS support is an influence on steering!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
In keeping it in the general case the reactions show that the middle one takes most of the load and actually acts as a fulcrum point for the bed. The front tires actually lift up slightly and the rear tires take a little bit better load downwards.
So the shape of the beam is like an "S" as drawn in the diagram, top left hand side.
What I would suggest is now do a second solution in the transverse direction right at the middle support. You know that the tires on the left and right support half the computed middle load, you can now solve for deflection directly and get the overall solution for the plate at the point underneath the load.
But what is interesting is the rotation maximizing just right of the middle support.
I note that the solution is consistent with everyday observations. If you have ever been on a loading dock, you will notice the immediate and first response of the truck once the load is placed on the bed is to have the front bumper slightly rise. Clearly the best place for the heavy load is over the axial, is it not? I would therefore move the load 155mm to the left of the indicated spot at present, noting the upward reaction of the LHS support is an influence on steering!
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
- Thread starter
- #37
"are you suggesting that the tires significantly alter the reaction loads" ... you better believe it ...
you solved the problem with rigid supports, right ?
how does a tire react -ve load ?? ... it can't
you Have to consider the flexibility of the supports. you can simplify the problem by assuming equal stiffness and that the beam stiffness is much greater than the tire (ie the beam is rigid).
talking about the front (steering) wheel confuses the problem ... i thought it was load dstribution about the three sets of tires at the back of the tractor. if it does involve the front tires, as the previous post suggests, then the attachment of the bogie (carrying the 3 sets of tires) to the tractor chasis comes into the problem ... can't the bogie pivot on the chasis (in the vertical plane) ??
you solved the problem with rigid supports, right ?
how does a tire react -ve load ?? ... it can't
you Have to consider the flexibility of the supports. you can simplify the problem by assuming equal stiffness and that the beam stiffness is much greater than the tire (ie the beam is rigid).
talking about the front (steering) wheel confuses the problem ... i thought it was load dstribution about the three sets of tires at the back of the tractor. if it does involve the front tires, as the previous post suggests, then the attachment of the bogie (carrying the 3 sets of tires) to the tractor chasis comes into the problem ... can't the bogie pivot on the chasis (in the vertical plane) ??
RB1957, I would suggest you are greatly over thinking the problem.
As posted, Bert2 admitted simplifying the problem to get the quick and dirty answer. Looks like most of the conversation centered about the wording of the problem, which correctly raised in later discussions, bought up the relevance of answering the question. Typically I like to cut to the chase and answer the question, let the politicians make the federal case as life moves on.
Of course Bert2, the fifth wheel. Didn't think of that at the time. Pretty hard to move! Interesting how that location induces a turning moment to the right about the axial. I'm sure there is a reason behind the original design, I'm oilfield and not automotives so hard to say.
Anyway, cute little problem. Hope it works out for you.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
As posted, Bert2 admitted simplifying the problem to get the quick and dirty answer. Looks like most of the conversation centered about the wording of the problem, which correctly raised in later discussions, bought up the relevance of answering the question. Typically I like to cut to the chase and answer the question, let the politicians make the federal case as life moves on.
Of course Bert2, the fifth wheel. Didn't think of that at the time. Pretty hard to move! Interesting how that location induces a turning moment to the right about the axial. I'm sure there is a reason behind the original design, I'm oilfield and not automotives so hard to say.
Anyway, cute little problem. Hope it works out for you.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
- Status
Similar threads
- Locked
- Question
- Locked
- Question
