1. ## Point Loads Calculation with more than 2 supporting points

Hi,

I am dealing with large equipment ( few thousand KG in weight), I would like to apply the vibration isolators (spring) in order to reduce the vibration of the equipment to the building. I have to calculate the point loads of the equipment (i.e. the weight at every position where vibration isolator will be installed), in order to size the vibration isolators.

I am able to calculate the point loads with 2 supporting points, as shown in the below image, with the assumption that the beam/line does not deform when the force/weight (W) is applied.

But, I do not know how to calculate for 3 supporting points (as below shown), as well as even more supporting points.

Can anyone teach me how to calculate it, or give me the link of a website that explain it? I did try to search the web using Yahoo and Google with the keywords "point load", "point load calculation", "load distribution" and "weight distribution", but the closest that I could find was the point loads with 2 supporting points.

Thanks.  Reply With Quote

2. Try googling "indeterminate beams"  Reply With Quote

3. Hey, you know how to identify the engineer in any group? He (or she) will be the one that comes up with the most complex solution.
I speak from experience.

wongw - You are over complicating this thing. You aren't trying to determine the stiffness of the structure. You're trying to calculate the load on the feet. Beam equations have no relevance here.

The load at the support points will be the same no matter how much the structure in between deflects. The load at each point will depend on the weight of the structure between it and the adjacent points on either side. If the "density" of the structure is relatively homogeneous along its length then the load on each foot will be equal. Simple division will get you the answer.

If the weight/ft of the structure varies along its length, a series of free body diagrams and summation of moments will get you the answer. Consult the manufacturer of this beast to get the load at each point.

The stiffness of the structure will not lessen the load on intermediate support points. Approach each point as a simple support, and consider the structure to be hinged at that point.

multiple supports.JPG  Reply With Quote

4. Another consideration...??

We have a few big beastly rotary forging machines here and we have always planted them using a "rule of thumb" that the pad or base on equipment producing heavy vibration should be three times the weight of the machine. They don't make the swat of something like a drop forge or maybe an upsetter but they're pretty violent.

One of the more recent machines has a pad about 8 foot deep x about 12 foot wide x about 20 foot long. Then another step that's about 12 x 20 foot but the pad is only about 4 foot thick in that section. Give or take... (the machine footprint of hammer box and feed mechanism) Then we put a 1" thick low durometer urethane liner around the pad. Then pour a 2 foot wide curb around the urethane member... then the floor off of that. Works well and concrete is relatively cheap and very easy to do.

Then again... The location wasn't on bedrock though... If it were, then we would've needed to look at further isolation.  Reply With Quote

5. Wongw,

Could you show a sketch of the load mass(es) and and the nature of the driving vibration so that we
can understand the application. I suspect that you are assuming an inflexible beam since
the supports are springs.

If so, then are you placing springs at the support points and analytically determining the
the springs and dampers to minimize the vibration of the building( i.e. making the transmissibility minimum)?  Reply With Quote

6. I would be getting the total weight of the individual pieces of equipment from the manufacturer and dividing that evenly by the number of support points. As JB suggest, you may be over-thinking the thing. Either that or rent a half dozen low profile scales and measure at each footing for each piece of equipment.

Trying to calculate theoretical loads will probably paint a less useful (accurate) picture than the simple division version I mention above. Is the footing-weight distribution of the equipment even over the entire base of the machine. A big unknown is my guess.  Reply With Quote

7. Ummm... not to detract from the more knowledgable and experienced on the forum...
This is more along the lines of asking for clarification of why things are the way they describe.

If I were to hang a truss, that is uniformly loaded, from 3 points of attachment,
similiar to the drawing below:

UDL Truss.jpg

the center spanset would carry 5/8 of the total load.
And each end spanset would carry only 3/16 of the total load.
Not a third/third/third.

Note that the spansets carry the load from the bottom chord of the truss.

Why then, if I support the truss on 'feet' from below, do I suddendly distribute the load differantly?
Isn't the center foot still carrying 5/8 of the total load?
What am I missing?

Or, perhaps more correctly, how is the above drawing differant than the OP's question?

Unless his second drawing is an actual representation of where the load is;
then the distribution is even wacky-er (tm).
(F3 would not have any load at all.)  Reply With Quote

8. Dale,
Actually you are right. For a beam with an evenly distributed load and multiple supports, if the outer supports are at or near the end of the beam they will actually see less total load than the intermediate supports will. By moving the outer supports inward one can distribute the load on the supports so they are all equal. The distance from each outer support to the end of the beam should be one half of the distance between supports. Good catch.  Reply With Quote

9. With machinery, the footing loads are usually fairly well distributed to not unevenly stress the frame unduly. Higher weight in one part will usually have more feet per square foot of floor space. Also, the anti-vibration load pads have a pretty wide spread of support in pounds.

A 1000-lb pad will have a useable range of probably 600-lbs to 1000-lbs and work just fine. Not building Swiss time pieces here.   Reply With Quote

10. Originally Posted by dalecyr Ummm... not to detract from the more knowledgable and experienced on the forum...
This is more along the lines of asking for clarification of why things are the way they describe.

If I were to hang a truss, that is uniformly loaded, from 3 points of attachment,
similiar to the drawing below:

UDL Truss.jpg

the center spanset would carry 5/8 of the total load.
And each end spanset would carry only 3/16 of the total load.
Not a third/third/third.

Note that the spansets carry the load from the bottom chord of the truss.

Why then, if I support the truss on 'feet' from below, do I suddendly distribute the load differantly?
Isn't the center foot still carrying 5/8 of the total load?
What am I missing?

Or, perhaps more correctly, how is the above drawing differant than the OP's question?

Unless his second drawing is an actual representation of where the load is;
then the distribution is even wacky-er (tm).
(F3 would not have any load at all.)
This is also true for beams and answers the OP question (BTW what happened to him/her?)  Reply With Quote

11. Thanks for all who participated in the discussion here.

Well, I was away the next day after i posted up this thread, to a remote job-site, accident at job-site and hospitalized, and just back home these few days. Sorry for not replying in this thread for so long.

Lets come back to this topic.

In the actual fact, my company is the manufacturer. We assembly several large and heavy (and of course expensive) components and run the pipings between these components. These heavy components are sitting somewhere on the unit base frame, depending on the layout. For the point load calculation, we do not have any literature for point-load calculation with more support points. Our current calculation method is not accurate enough that we have to oversize the spring isolators 30% in order to overcome the inaccuracy.

The equipments consist of compressor(s), which will create vibration when the unit is running. Thus, most of our customers will request for spring isolators to support the unit(s) and isolate (or damp down) the vibration from being transmitted to the floor or building structure.

The height level of the spring isolators can be adjusted so that the unit can be leveled after sitting on these spring isolators. These spring isolators can isolate or damp down the vibration as long as it is not fully compressed due to the weight applied on it. Thus, “vibration” is not the main topic in this thread.

For our small units, we can use only 4 spring isolators at the 4 corners of the unit base frame to support the whole unit weight. However, for our larger units, we have to use more spring isolators in order to support the whole unit. Point Load calculation with 4 support points is not a problem to us, but we do not have any literature for point load calculation of 6 support points (3 points at each unit side), 8 support points (4 points at each unit side), and more support points.

For Point Load calculation with 4 support points, the below shown is an example of our calculation method. P1, P2, P3 and P4 are the support points; WA, WB and WC are the components. Assuming that the left-bottom-most is the Origin coordinate, the coordinates of all these points are shown in the diagram below. In order to simplify the demo in here, I do not include the weight of the unit base frame, we just look at the weights of the WA, WB and WC components, and the its distribution to P1, P2, P3 and P4.

4-p.JPG

The first step is to see from “Unit End View”, calculate the distributed weight of each component to each unit side (i.e. side P1-P2 and side P3-P4). The calculation method is same as the method I explained in the first diagram of my earlier posting.

WA at P1-P2 side = 500 x (88-65) / 88 = 130.68 KG
WB at P1-P2 side = 650 x (88-33) / 88 = 406.25 KG
WC at P1-P2 side = 350 x (88-60) / 88 = 111.36 KG

WA at P3-P4 side = 500 x 65 / 88 = 369.32 KG
WB at P3-P4 side = 650 x 33 / 88 = 243.75 KG
WC at P3-P4 side = 350 x 60 / 88 = 238.64 KG

The second step is to see from “Unit Side View”, calculate the distributed weight of each component to each support point, using the same calculation method.

P1-P2 side
WA at P1 = 130.68 x (164-32) / (164-10) = 112.01 KG
WB at P1 = 406.25 x (164-95) / (164-10) = 182.02 KG
WC at P1 = 111.36 x (164-130) / (164-10) = 24.59 KG
Thus, P1 = 318.62 KG

WA at P2 = 130.68 x (32-10) / (164-10) = 18.67 KG
WB at P2 = 406.25 x (95-10) / (164-10) = 224.23 KG
WC at P2 = 111.36 x (130-10) / (164-10) = 86.77 KG
Thus, P2 = 329.67 KG

P3-P4 side
WA at P3 = 369.32 x (164-32) / (164-10) = 316.56 KG
WB at P3 = 243.75 x (164-95) / (164-10) = 109.21 KG
WC at P3 = 238.64 x (164-130) / (164-10) = 52.69 KG
Thus, P3 = 478.46 KG

WA at P4 = 369.32 x (32-10) / (164-10) = 52.76 KG
WB at P4 = 243.75 x (95-10) / (164-10) = 134.54 KG
WC at P4 = 238.64 x (130-10) / (164-10) = 185.95 KG
Thus, P4 = 373.25 KG

The above is the method that we are using now to calculate the point-load for 4 support points.

As for our Point Load calculation with more than 4 support points, please find the below example for details. The below shown is taking the above example, and added 4 more supporting points (PA, PB, PC and PD), thus it has 8 supporting points.

8-p.JPG

Our method is to calculate first the weight distribution to 4 points (P1, P2, P3 and P4), like the last example. Then, distribute it again to those additional support points.

As from the last example:
P1 = 318.62 KG
P2 = 329.67 KG
P3 = 478.46 KG
P4 = 373.25 KG

P1-P2 side
P1-P2 side.JPG

 P1 PA PC P2 Total Markup Weight (KG) 318.62 = (329.67-318.62) / (164-10) x (60-10) + 318.62 = 322.21 = (329.67-318.62) / (164-10) x (115-10) + 318.62 = 326.15 329.67 1296.65 Actual Weight (KG) = 318.62 / 1296.65 x (318.62+329.67) = 159.30 = 322.21 / 1296.65 x (318.62+329.67) = 161.10 = 326.15 / 1296.65 x (318.62+329.67) = 163.07 = 329.67 / 1296.65 x (318.62+329.67) = 164.83 648.30

P3-P4 side

P3-P4 side.JPG

 P3 PB PD P4 Total Markup Weight (KG) 478.46 = (373.25-478.46) / (164-10) x (60-10) + 478.46 = 444.30 = (373.25-478.46) / (164-10) x (115-10) + 478.46 = 406.73 373.25 1702.74 Actual Weight (KG) = 478.46 / 1702.74 x (373.25+478.46) = 239.33 = 444.30 / 1702.74 x (373.25+478.46) = 222.24 = 406.73 / 1702.74 x (373.25+478.46) = 203.45 = 373.25 / 1702.74 x (373.25+478.46) = 186.70 851.72

In actual, PC should have the highest weight if compared to other points of P1-P2 side, but our method shows differently. While for P3-P4 side, it is either P3 or PD should have the highest weight. We know that our calculation is not accurate, but it is now the only method that we have to estimate point-loads. Because of the inaccuracy, we have to oversize the spring isolators by 30% based on our past experience of handling the customers’ complaints.

I hope that I can have some guidelines/solutions/advices from this forum.

Thanks.  Reply With Quote

12. Thanks for the good detail, it is making more sense than the simple beams you first provided. I may be missing something, but...

IF the base is stiff enough to carry all loads without any significant deflection at any one point of the base, the load is going to be radiated out to all leg positions, not just a simple left-side versus right-side.

Like this for the 500kg -- no??? Calculate the 500kg distribution for 8 legs, then do it all again for the other load-centers. The length of the arrows will be inversely proportional to each of the eight "feet" ratio share of the load. Do it for the other loads and add the three results for each foot.  Reply With Quote

13. PinkertonD, thanks for your guidelines. But, I am not quite understand.

Taking my 8-supporting-points example, my understanding of your explanation is as below:
8-p.JPG

- taking the load 500 kg

- Calculate first the markup weight distributed to P1
= 500 kg / (distance from WA to P1) = 500 kg / ( (32-10)^2 + (65-0)^2 )^0.5 = 7.29 kg

- Calculate the markup weight distributed all the other points using the above method, thus we will get:
P3 = 15.71; PB = 13.80; PD = 5.81; P4 = 3.73
P1 = 7.29; PA = 7.06; PC = 4.74; P2 = 3.40

- Sum up all these markup weights ==> 61.54 kg

- Multiple it to the markup weight of P1 to find out the actual weight distributed to P1, using the below formula:
Weight Source x Markup Weight / Total Markup Weight

====> P1 = 500 x 7.29 / 61.54 = 59.20 kg

- Calculate the actual weight distributed to all the other points using the above method, thus we will get:
P3 = 127.64; PB = 112.12; PD = 47.17; P4 = 30.32
P1 = 59.20; PA = 57.40; PC = 38.54; P2 = 27.61

Calculate the weight distribution of the other loads using the above method, then add up, the final result is as shown in the below diagram.

PinkertonD, am I interpreting your explanation correctly?  Reply With Quote

14. Looks good to me.  Reply With Quote

15. My 2 cents.

Neither method proposed is accurate since they do not both satisfy the 3 static equilibrium equations.

If you have n mounts on a spring mounted system with an inflexible mounting plate, then the 2 moment
equations and the vertical equilibrium equation must be satisfied. That means that you have 3 linear equations with n unknowns. So it looks like n-3 mount forces can be arbitrarily (at least theoretically) selected.

But that is not your central problem since you are interested in isolation. Don't know how you are dealing with it but I think you need to at least look at the bobbing mode (linear) and 2 rocking modes.  Reply With Quote

16. It is 3,300lb solid slab on springs to cut down a bit of noise. I'm going with near enough.   Reply With Quote

17. Well, a unit with 1500 kg (3300 lbs) weight is considered a small unit in our product. The largest "dinosaur" that we have can weight up to 20,000 kg (44,000 lbs). For this "dinosaur", we use 12 No's of spring isolators to support it.

The spring isolator models that we are using are from 1000 lbs to 5000 lbs, as the below photos.
spring isolator.JPG
The description from the vendor's website is as below quoted:
This Restrained Vibration Isolator is housed & restrained spring mounting, the spring of which is laterally stable with a minimim horizontal stiffness of 0.8 time the rated vertical stiffness & with 50% overload capacity provides yet another vibration isolation solution to equipment with presence of horizontal load. The upper housing and the lower one are assembled into a telescoping housing complete with a top level adjustment bolt and a non-skid, noise breaking rubber base, together with resillient cushion inserts at both inner sides of the lower housing, preventing metal contact as well as serving horizontal load snubbing.
As the vendor's recommendation, spring isolators should not be over-sized more than 30% to avoid inefficient vibration absorption. While, under-sizing it will result springs fully compressed (vibration is fully transmitted to the floor/structure).

This is why I am looking for an accurate method to calculate the point-loads.  Reply With Quote

18. Originally Posted by wongw This is why I am looking for an accurate method to calculate the point-loads.
And, that's what my somewhat flippant response suggested. The finer details of the dynamic loads will be in the design-domains of the isolation device manufacturer. I would expect that supplying the static loads to them would allow them to suggest the appropriate models. Unless you have some serious live, reciprocating loads, then my more simple approach should be adequate.

You may also want to consider something regarding layout, that I mentioned in post #9.   Reply With Quote

19. I think I have a somewhat rational method for getting the forces.

Using your idea of distributing the loads along the 2 axes (P1-P2, P3-P4 for the problem of further distributing that force must be dealt with.

Since the 8 points rest on springs, you can't use the classical distribution approach. I have assumed that the mounting plate is very stiff next to the spring deflections, so I make the assumption that the plate retains its flat shape and undergoes a deflection at P1 and a rotation and the spring constants are the same for all mounts.

With this assumption the equilibrium equations yielded
a=0.7*F/K-0.3*u/L*F/K
x=0.3(2/3*u/L-1)*F/K
F= the projection of the CM load onto the P1-P2 line
a=deflection of the spring under P1
u= the distance of the F projection to P1-P2 line
L= horizontal spacing
K=spring constant
x=increment of deflection from P1 TO Pa

For example suppose the CM of the sytem is at 80,50, then the projected force onto P1-P2 is at 80,0
Then u=80
a=(.7-.3*80/50)*F/K=.22*F/K meaning that P1 has .22 of the projected F
x=.3(.67*80/50-1)*F/K=0.02*F/K

So the distribution for this example is
.22F,(.22+.02)F,(.22+2*.02)F, (.22+3*.02)F
.22F,.24F,.26F,.28F  Reply With Quote

20. What I was alluding to in my last post was that rather than have eight geometrically-even positioned feet, the OP might consider moving/adding/removing feet in an effort to balance loads on each one. That would provide the need for identical isolation dampers rather than having eight disparate units, but at a nice (pretty) even leg count and position.

The caveat comes, of course, in that the nature of the dynamic loads causing the vibrations is unknown. If all fairly well balanced rotary, then no big deal, but if significant reciprocating movement is involved then the entire project changes in how it is to be looked at. Way too complex an issue for dealing with here, I would suggest.  Reply With Quote

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