How would I go about calculating deflection of the long beams in the model? I am assuming a 400 lb weight spread uniformly across the top of the platform (grid of 1x2 either steel or al. rect. tube). Long tubes (beams) are 60" long. Their will be 4" caster at the four corners bringing the load to the ground.
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Assessing deflection in a pair of beams with cross beams. 1
- Thread starter chopnhack
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1. Yes, at the four corners the 4" casters will be meeting the ground.
2. That is were I am getting stuck at - if the load is spread across the entire frame how will I calculate load on the cross members, what equation should I use to evaluate the model?
3. Yes
2. That is were I am getting stuck at - if the load is spread across the entire frame how will I calculate load on the cross members, what equation should I use to evaluate the model?
3. Yes
Whatever the flooring is I would say that it says onto the short members so the load is applied to the short members. For the total deflection, I would look at the total deflection relative to the cross panel deflection and limit it to span/250 for total working load and span/500 for live load alone.
msquared48
Structural
I doubt that this diagram will look correct when i post it, but it looks good in the message box.
P 2P 2P 2P P
____________________________________________
^ Rail ^
200# / 8P (each rail takes 1/2 of the 400# load)
P = 25#
2P = 50#
You will get deflection from the 2P Loads. Solve for the centerline deflection by superposition. Use the first equation for the 2P load at the centerline, then the next equation for the other two 2P loads.
Simple.
Mike McCann
MMC Engineering
a simple solution would be to assign a uniform distributed load along all the members, so the long members are loaded by a UDL and a pattern of discrete loads. max deflection is either at the middle of the long rail or at the middle of the short rail closest to the middle of the long rail ... if you put a short cross rail at the middle of the long rail, then this'll have the largest deflection.
paddingtongreen
Structural
You don't say why you need this. If you are looking for an upper bound number, put half the weight in the middle of each long beam and figure the actual deflection will be less.
If you are putting a cabinet on the platform, it will have stiffness of it's own, if you are putting small items on the platform, msquared48's solution works.
Michael.
Timing has a lot to do with the outcome of a rain dance.
If you are putting a cabinet on the platform, it will have stiffness of it's own, if you are putting small items on the platform, msquared48's solution works.
Michael.
Timing has a lot to do with the outcome of a rain dance.
- Thread starter
- #8
Thanks guys, the loading will be UDL because their will be 3/4" plywood fastened to the frame with cabinetry on top. The major concern was the deflection at mid span of the long rails.
Based on 2x1 Al. rect. tube with a section mod. of 0.3317, elasticity of 10mil psi, and 3.3lb/in (200lb per rail) load I am coming up with just under 3/16" deflection at center - this is assuming udl of 200lb over 60" of a single rail.
What I would like help to figure out is how to determine deflection at center after adjusting for the cross rails and secondly with the 3/4" plywood top. This is purely academic, its for a shop table, nothing critical, but I would like to learn the math behind it so that I can evaluate other systems, for instance switching to steel, using c channel, subtracting cross pieces, etc.
Based on 2x1 Al. rect. tube with a section mod. of 0.3317, elasticity of 10mil psi, and 3.3lb/in (200lb per rail) load I am coming up with just under 3/16" deflection at center - this is assuming udl of 200lb over 60" of a single rail.
What I would like help to figure out is how to determine deflection at center after adjusting for the cross rails and secondly with the 3/4" plywood top. This is purely academic, its for a shop table, nothing critical, but I would like to learn the math behind it so that I can evaluate other systems, for instance switching to steel, using c channel, subtracting cross pieces, etc.
you've assumed a 200 lbs UDL load on each long rail to conservatively (and easily) analyze the distributed load applied to the panel. less conservative (but not by much, i think) would be to allocate some load to the short rails; simply done by assuming a UDL along the entire run of the rails (long and short).
the short rails will then be applying load to the long rails. so the loading of the long rail is now a UDL + a number of discrete loads. you critical deflection is still near the mid-span of the long rails. deflection of the long rails is now the sum of deflections due to the UDL and each of the point loads (a bunch more work than previously, particularly as you have to solve for the general deflection of the long rail to get the deflection at the critical area for each; generally this won't be the maximum deflection for the individual point loads). now you know how the long rails have deflected, you need to add the deflection of the short rail (under it's UDL).
if this is an exercise, you could look at how to place the short rails so the critical deflection occurs on the long rail. you might also consider the effect of fixing the ends of these beams.
of course, we ignoring the deflection of the panel itself !? (which doesn't affect the deflection of the rails but could be critical !?)
the short rails will then be applying load to the long rails. so the loading of the long rail is now a UDL + a number of discrete loads. you critical deflection is still near the mid-span of the long rails. deflection of the long rails is now the sum of deflections due to the UDL and each of the point loads (a bunch more work than previously, particularly as you have to solve for the general deflection of the long rail to get the deflection at the critical area for each; generally this won't be the maximum deflection for the individual point loads). now you know how the long rails have deflected, you need to add the deflection of the short rail (under it's UDL).
if this is an exercise, you could look at how to place the short rails so the critical deflection occurs on the long rail. you might also consider the effect of fixing the ends of these beams.
of course, we ignoring the deflection of the panel itself !? (which doesn't affect the deflection of the rails but could be critical !?)
chopnhack,
Sounds to me like you know how to calculated deflection on a simple beam with a udl so why not stick to that. If you apply half the total weight to each long beam as a u.d.l. then what you calculate will be within about 15% of the theoretical exact solution.
Sounds to me like you know how to calculated deflection on a simple beam with a udl so why not stick to that. If you apply half the total weight to each long beam as a u.d.l. then what you calculate will be within about 15% of the theoretical exact solution.
- Thread starter
- #11
Thanks Rb, factoring in the cross members reduces the psi in half from 3.333 to 1.63 dropping the deflection to a little over 1/16" at midspan assuming that the weight will transfer evenly over the entire area, but as you point out it will not and discrete loads will occur at points on the rail.
The ends of the beams will be fixed. There will be a cross beam at either end of the long rails fitted with a L shaped uneven angle 4x2 by 1/8" thick and bolted together. This detail will occur at each intersection of tubing unless a tig become available.
I'm not sure what equations to use together at this point. I have figured that the load at either end of the short rails is ~20lbs, which means there will be a point load on each long rail of 20 lbs starting at the beginning and then evenly spaced every 15 inches for a total of 5 rails in addition to the decreased udl of 98 lbs per long rail.
As for the panel, I believe it to be critical. I was thinking that the 3/4" plywood would provide some resistance to deflection since it will be fastened to the rail assembly and help to spread some of the load.
The ends of the beams will be fixed. There will be a cross beam at either end of the long rails fitted with a L shaped uneven angle 4x2 by 1/8" thick and bolted together. This detail will occur at each intersection of tubing unless a tig become available.
I'm not sure what equations to use together at this point. I have figured that the load at either end of the short rails is ~20lbs, which means there will be a point load on each long rail of 20 lbs starting at the beginning and then evenly spaced every 15 inches for a total of 5 rails in addition to the decreased udl of 98 lbs per long rail.
As for the panel, I believe it to be critical. I was thinking that the 3/4" plywood would provide some resistance to deflection since it will be fastened to the rail assembly and help to spread some of the load.
i'm very surprised it has that much effect ... all you're doing is replacing UDL load on the long rail with a set of discrete loads (that should add up to the same load (since it ain't got anywhere else to go).
sure the long rail UDL will drop (including the short rails), but 1/2 the load on each short rail is applied to both of the long rails.
you've got the solution for the long rail with the UDL (from a handbook i suspect). calculating the effect of the short rails means calculating the deflection of the long rail at the mid-span for each of the point loads (from the short rails). the easiest way is to derive the displacement curve for a SS beam with a single point load (what Roark, for example shows as a load P a distance a from one end of a beam; but you can't use the maximum deflection for these loads). then use superposition to sum the deflection at the mid-span for each load, then calculate the deflection of the short rail (SS beam with a UDL) and the total deflection is the sum of these two deflctions.
you'll do this once and then say (as others have above) that the simple loading gives a good enough answer ... but you'll know that for yourself 'cause you've worked it out (and not just "heard it on the 'web").
the plywood sheet is something to consider (ie don't forget that it has a part to play) but 3/4" thick 15" (the spacing of the short rails) x ?" sounds pretty stiff eough to transfer 100 lbs or so ... i expect you could stand on it and it won't deflect much (not more than the beams).
if your short rails are Ls, don't count on them fixing th long rails ... Ls are very weak in torsion which is the load you're applying to them if you want them to fix the end of the long rail.
sure the long rail UDL will drop (including the short rails), but 1/2 the load on each short rail is applied to both of the long rails.
you've got the solution for the long rail with the UDL (from a handbook i suspect). calculating the effect of the short rails means calculating the deflection of the long rail at the mid-span for each of the point loads (from the short rails). the easiest way is to derive the displacement curve for a SS beam with a single point load (what Roark, for example shows as a load P a distance a from one end of a beam; but you can't use the maximum deflection for these loads). then use superposition to sum the deflection at the mid-span for each load, then calculate the deflection of the short rail (SS beam with a UDL) and the total deflection is the sum of these two deflctions.
you'll do this once and then say (as others have above) that the simple loading gives a good enough answer ... but you'll know that for yourself 'cause you've worked it out (and not just "heard it on the 'web").
the plywood sheet is something to consider (ie don't forget that it has a part to play) but 3/4" thick 15" (the spacing of the short rails) x ?" sounds pretty stiff eough to transfer 100 lbs or so ... i expect you could stand on it and it won't deflect much (not more than the beams).
if your short rails are Ls, don't count on them fixing th long rails ... Ls are very weak in torsion which is the load you're applying to them if you want them to fix the end of the long rail.
- Thread starter
- #13
Sorry I misspoke! There is no free lunch. You are right the load has to transfer down to the four corners some how and the cross members will only help stiffen the unit some, therefore there can be a slight decrease over the theoretical ~3/16" deflection. The short rails were also to be 2x1 tube. I was thinking that the plywood would decrease deflection some. Will it be so negligible to be unaccounted for?
The equation I was using: ? = 5 q L4 / E I 384
I will work on this some more tomorrow, I appreciate the guiding hand. I will use ? = Pb/48EI*(3I^2-4b^2) to add up the point load deflections, then add that to the udl deflection of the long rail plus the udl of the short rails (which should be extremely tiny). Should be interesting to see what difference there will be. Thanks again.
The equation I was using: ? = 5 q L4 / E I 384
I will work on this some more tomorrow, I appreciate the guiding hand. I will use ? = Pb/48EI*(3I^2-4b^2) to add up the point load deflections, then add that to the udl deflection of the long rail plus the udl of the short rails (which should be extremely tiny). Should be interesting to see what difference there will be. Thanks again.
IMHO the main reason for the cross members is to support/stiffen the plywood sheet, by breaking it up into smaller panels, and by locally increasing the bending stiffness of the sheet. if they're the same section as the long rails they should be GTG; if they're smaller they should be looked at ... the moment in each will be much smaller than in the long rails, but if they're not "man" enough then there'll be a problem.
I think the main thin gyou want the sheet to do is to withstand some sort of punching load, a localised load rather than deflections due to a distributed load. i'd suggest sizing it by eye ... 1/8" 3-ply is obviously too small, 1" looks like overkill, 1/2" to 3/4" looks about right. when you biuld this bench try out some alternatives so you learn and don't just "hear it on the 'web".
d = 5/384... is the max deflection (at the mid span) due to a UDL. do you have a cross rail at the middle (an odd number of short rails) or not (an even number) ? what's the difference ?? if you don't have a cross rail at the middle then you're using a deflection that is (slightly) too large. you may be ok with this, but you need to understand the limitations of your equations.
d = 1/48... is the maximum deflection for a beam with a single point load. it is not the deflection at the mid-span, it is not at the same spanwise location for all of the point loads. using this equation for each of the point loads (and summing the results) will be very conservative ... again you may be ok with this, but the result will probably be quite a bit larger than the simple UDL on the long rail (and so misleading). if this is a learning opportunity for you, you might not be learning the right stuff.
I think the main thin gyou want the sheet to do is to withstand some sort of punching load, a localised load rather than deflections due to a distributed load. i'd suggest sizing it by eye ... 1/8" 3-ply is obviously too small, 1" looks like overkill, 1/2" to 3/4" looks about right. when you biuld this bench try out some alternatives so you learn and don't just "hear it on the 'web".
d = 5/384... is the max deflection (at the mid span) due to a UDL. do you have a cross rail at the middle (an odd number of short rails) or not (an even number) ? what's the difference ?? if you don't have a cross rail at the middle then you're using a deflection that is (slightly) too large. you may be ok with this, but you need to understand the limitations of your equations.
d = 1/48... is the maximum deflection for a beam with a single point load. it is not the deflection at the mid-span, it is not at the same spanwise location for all of the point loads. using this equation for each of the point loads (and summing the results) will be very conservative ... again you may be ok with this, but the result will probably be quite a bit larger than the simple UDL on the long rail (and so misleading). if this is a learning opportunity for you, you might not be learning the right stuff.
- Thread starter
- #15
Rb, the short rails will be the same section as the long rails.
I do have a beam falling dead center on the long rails (5 short rails).
Are you saying that using the max deflection formula while having a center cross rail is too conservative? ("if you don't have a cross rail at the middle then you're using a deflection that is (slightly) too large"
I may be swimming in the deep end, but I am trying to keep up ;-) Thanks for the explanation of the d = 1/48. From that I assume the equation that I would need to use is the more complicated y=PBx/6iEI(l^2-x^2-b^2) for 0<x<a (deflection at any section for a point load)
This is purely academic, I will build a workstation based off of this but it will be for personal use only, i.e. this is not howework, but its fascinating to me to know how and why things work. I tested out using mild steel of the same section and the deflection was tolerable. I would have preferred al. due to rust, but, c'est la vie. Who knows, perhaps welding the steel will be easier than bolting the al together.
Thanks again
I do have a beam falling dead center on the long rails (5 short rails).
Are you saying that using the max deflection formula while having a center cross rail is too conservative? ("if you don't have a cross rail at the middle then you're using a deflection that is (slightly) too large"
I may be swimming in the deep end, but I am trying to keep up ;-) Thanks for the explanation of the d = 1/48. From that I assume the equation that I would need to use is the more complicated y=PBx/6iEI(l^2-x^2-b^2) for 0<x<a (deflection at any section for a point load)
This is purely academic, I will build a workstation based off of this but it will be for personal use only, i.e. this is not howework, but its fascinating to me to know how and why things work. I tested out using mild steel of the same section and the deflection was tolerable. I would have preferred al. due to rust, but, c'est la vie. Who knows, perhaps welding the steel will be easier than bolting the al together.
Thanks again
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