calculate force distribution
calculate force distribution
(OP)
Hello Everybody,
I have a question about force distribution calculation.
Suppose there is a rectangular with a force perpendicular to its surface at some point, but not center.
how to calculate the force distributations at the four corners of the rectangular?
My problem is actually a weight above a base located not at the center of the base. there are four supports at the four corners of that rectangular. I need to know how much force for each support?
thank you very much for your attention!
I have a question about force distribution calculation.
Suppose there is a rectangular with a force perpendicular to its surface at some point, but not center.
how to calculate the force distributations at the four corners of the rectangular?
My problem is actually a weight above a base located not at the center of the base. there are four supports at the four corners of that rectangular. I need to know how much force for each support?
thank you very much for your attention!





RE: calculate force distribution
You can set the sum of vertical forces equal to zero. You can sum moments about two different axes. That gives you three equations. If you just have three unknown forces, you can solve for them. With four, you can't.
If you have an axis of symmetry, you could eliminate an unknown from the mix.
RE: calculate force distribution
thank you very much for your reply!
I do not need to consider the deflection and there is no internal forces in the structure.
I do have four unknown here. two of them are one alxe symmetry. I am thinking if there is no symmetry, how to find four equations to solve four unknown?
thank you very much !! :)
RE: calculate force distribution
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
in the first site below there is a sheet for calculating those reactions under Plates -> Simple bending -> Rectangular -> 4 corners -> Conc.load. Keep in mind however that the solution is approximate, the approximation being not very good if the load is on one side.
JStephen,
my answer to the question would have been exactly the same as yours. However, by thinking a bit more to this problem, it appears evident that the reactions do not really depend on the elastic properties of the plate. This is simple to experiment with the sheet above, but is also evident: why should the distribution of reactions change as the plate becomes more and more rigid, and towards what values would the reactions tend when the plate becomes substantially rigid? The intuitive answer is that they would remain the same.
By playing a bit around, I found a simple rule to which the reactions calculated by the sheet above are in quite a good approximation, and that gives the exact answer when there is a symmetry that allows for calculating the reactions by equilibrium.
It is very simple: divide the rectamgle into 4 rectangles by tracing two lines parallel to the sides and intersecting at the load. Now each reaction is proportional to the area ratio of the partial rectangle containing the opposite vertex to that of the full rectangle.
So, hello2006, this would be a simple answer to your question, as this is a supplemental and independent equation with respect to the three equilibrium equations.
Is there someone that can give a proof that this equation is incorrect, or that indeed it is correct?
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
With a thin plate, shear and beam deformation must be considered. With a thick plate, deformations remain small enough to neglect.
The reactions change according to the rigidity of the plate, since shear deformation and excessive bending takes place in a thin plate. Beam theory is based on relativly small deflections and plane sections remaining relatively plane. Given a 4 point reaction with both an Mx & My acting on the plate, the reactions would not vary directly with distance, but vary in accordance with the bending resistance and differential rotation of an effective beam between the loading point and the reaction point under consideration. A thick rigid plate's reaction at a corner would be quite high in relation to that of a thin plate, where the effective beam "strip" running between load and reaction point is long and thin in depth and can deflect and rotate along its length. A far corner could much more easily deflect and rotate than a close corner, thus reducing the far corner's effective contributing resistance to the applied load in relation to a corner having a shorter effective length and subsequent high stiffness of its effective beam.
Some possible reaction distributions depending on plate stiffness,
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
However I can't follow your way of reasoning.
I think we can assume the plate is thick enough to make its behavior linear, and also that shear deflections may be neglected, just as in the ordinary theory of beams and flat plates subject to simple bending.
What I found by very simple numerical tests is that the values of the 4 reactions do not change by changing the stiffness of the plate (varying either the thickness or the elastic modulus or both).
Said in other words: if you calculate two plates with the same geometry, except that one has a maximum deflection of, say, half the thickness (hence a relatively large, though still small deflection) and the other one has a totally negligible deflection (for example because it is done with a very stiff material), then you'll find exactly the same corner reactions (for whatever load position, but of course the same in the two plates).
This is a fact.
Unfortunately I cannot explain why that simple rule of area ratios gives results that are exact in some simple cases, and closely match those obtained by the calc sheet in xcalcs, at least in the few cases that I tested.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Its not my way of thinking. I confess I learned it from, Blodgett's "Design of Welded Structures" about 30 years ago.
http
here's another good one,
http://
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
Roark doesn't cover rectangular plates supported at four corners, don't know about Blodgett.
As you seem to know the answer, can you state your solution to the original question: formula to calculate the four corner reactions in a rectangular plate supported at four corners only, with a concentrated load at a defined off center location.
Note that your formula above P/4 + Mx/2dx + My/2dy is only valid for loads at center; also discard the moments, they are zero.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
where Ex, Ey = eccentricity in x direction and/or y direction
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Here's four more equations for you, + n more
R1/d1 = R2/d2 = R3/d3 = R4/d4 = Rn/dn
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
And I'm still without answers to my questions.
Will you please take this very simple example:
-square plate a x a=1000x1000 mm
-a single load of say P=10 kN at Ex=300 and Ey=200.
Counting the vertices ccw from the first quadrant, my formula gives:
R1=5.6 kN
R2=1.4 kN
R3=0.6 kN
R4=2.4 kN
It's easy to check that these values satisfy the resultant equation (ΣR=10 kN) and the two moment equations:
P·Ex=(R1+R4)·a/2-(R2+R3)·a/2
P·Ey=(R1+R2)·a/2-(R3+R4)·a/2
Will you please make a similar calculation with your method and give the results?
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Deal or no deal?
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
The point of application of the load is at E=√(2002+3002)=360 mm from centerline.
Let's assume the bars are equispaced and placed on a circle D=240 mm dia., and also (as a minor detail) that 4 bars cross the axes of symmetry.
Calling R1, R2, R3, R4, R5 the unknown reactions (because of symmetry R6=R4, R7=R3, R8=R2), with R1 farthest from the load, we simply have:
R1=P/8-PE/(2D)=-6.25 kN
R2=P/8-PE/(2√2D)=-4.05 kN
R3=P/8=1.25 kN
R4=P/8+PE/(2√2D)= 6.55 kN
R5=P/8+PE/(2D)= 8.75 kN
It's your turn now!
Note however that I assumed above, as you probably expected, the behavior of a section in a bent beam (coming from the assumption that the bent sections remain plane) with linear distribution of stresses. This is a rough approximation for example for the base ring of a column (though it is customarily used), but is not valid for a plate, even if infinitely rigid.
To be convinced of that think to another example: a continuous beam with uniform section over for example 3 supports. With your way of reasoning you would expect there too, I suppose, a linear distribution of the support reactions, but this is far from being true, and for whatever stiffness: try a sample case to be convinced of that.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Now reduce one bar by 1mm^2. Or add 1mm^2 of concrete to the south side of the column. Does that invalidate your answer, or is it still (approximately) a good solution? At what point does the solution become invalid?
Here's my simple statics solution. Not the exact problem you gave me, and its solved backwards, so turn the solution around... OK, but then I gave you one way too easy. I have no doubt that it can be worked in reverse... just no time to do it... and I don't like structural calculations. I'd rather make stress diagrams and take the integral of the stress over any given area to arrive at an equivalent force when such is necessary.
BigInch
-born in the trenches.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
You found the following corner reactions:
R1=600, R2=0, R3=500, R4=1200
You might check that also the following combinations satisfy the moment and resultant equations:
R1=470, R2=130, R3=370, R4=1330
R1=340, R2=260, R3=240, R4=1460
R1=1100, R2=-500, R3=1000, R4=700
and infinite others may be obtained.
Of course the last one is quite physically unrealistic, but look at the first one: is perfectly realistic and is safer than yours, as the maximum reaction is higher.
My point remains the same: you assume a linear distribution of the reactions around a so called neutral axis, but there is nothing that supports such an assumption in a general situation. This assumption is more or less valid only when we have something that can be assimilated to the section of a beam: there the constraint of the parts of the beam adjacent to our section will force the section to remain plane after deformation, and this is the basis for assuming a linear distribution of reactions.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
I may be wrong, but I also think minimum internal energy is going to be expended by the first group of (my) reactions, as it appears that the other solutions seem to create a greater sum of moments on various similar elements. I think that is the basis of the theory I'm using, but I can't say that I'm sure about that. It would be interesting to confirm, but I'm not going to volunteer. Would you?
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
R1=489 R2=126 R3=374 R4=1310
As you see this is not very different from my first set above, that was hand calculated by the simple rule of areae. Of course we are speaking of an elastic plate of whatever stiffness here, as in the original question, not of a pipe flange, that sure behaves much more like you calculate.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Wonder how the OP is doing? Appologies for the thread hijack.
http://virtualpipeline.spaces.msn.com
RE: calculate force distribution
For a square plate with load at center or certain cases of symmetry, you can deduce that one load equals another and then solve. But for odd-shaped plate, or oddly-located loads, or structure with varying stiffness, you can't find the loads without consideration of the stiffness.
Example: Take a car. Put a jack under one point, start jacking. What happens? The farther you raise it, the more resistance there is to the jack- the load at that point is variable, and depends on the springiness of the suspension among other things.
RE: calculate force distribution
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RE: calculate force distribution
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RE: calculate force distribution
Can you give the size of your rectangle, load, mass, dimensions of bolt positions, thickness of plate etc
and is the "off center mass" only off centre in one plane?
I might be able to help with some more info on your problem.
regards
desertfox
RE: calculate force distribution
I am hello2006. I lost the password, therefore, now I am hello2007.
you draw a x- axle and y - axle. the cross of x and y alxes are the weight center, and the weight is perpendicular to the surface from up to down direction if you put that surface honrizontally. the surface of x and y alxes is to support the weight.
below that surface, there are four legs which are supporting the weight. But they are different distance from the x and y alxes.
it is like a table but not rectangular or square with four legs to support the weight.
Y-alxe is vertical alxe
x-alxe is horizontal alxe
Leg 1: the distance from X-alxe: 20"
from Y alxe: 15"
Leg 2: distance from x alxe: 20"
distance from y-alxe: 29"
Leg 3: distance from x alxe: 20"
distance from y alxe: 7"
Leg 4: distance from x alxe: 20"
distance from y alxe: 7"
I am looking for the force for each leg. I know leg 3 and leg 4 are symmetrical about y alxe. but I am thinking if they are not symmetrical, is there a way to calculate them too?
thank you very much for your help!
RE: calculate force distribution
based on the figures you gave me I calculated the load in each bolt as follows:-
assuming the rectangle was 1" thick and the load was 20 tons
I calc the leg 20" x and 29" y carries 18.6 tons
20" x and 7" y carries 0.3096 tons each
20" x and 15" y carries 0.746 tons
Now I will check my figures again as I don't feel that there right although if you add them up they come to 19.9652 tons
perhaps you can post your load and plate thickness and if someone else does another calc we can compare.
regards
desertfox
RE: calculate force distribution
as repeatedly stated above, there is simply no way to calculate those 4 reactions without calling for the intervention of some physical behavior, like the elastic deflection of the supported plate.
The only thing you can do is define a reasonable set of reactions.
In your example you didn't fully define the geometry, so I will assume that leg 1 is located in the first quadrant, leg2 in the second and so on.
Now P1+P2=P3+P4=P/2 because of the moment equilibrium.
We may also take P3=P4 as an arbitrary assumption: it is certainly incorrect but likely not too much.
So we get (assuming P=20 per desertfox):
P3=5
P4=5
P1=20x15/(15+29)=3.4
P2=6.6
I proposed above a simple rule to calculate such a repartition , but unfortunately it is only valid for a rectangle, and didn't find any obvious extension to the case of a quadrilateral.
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
My load is about 720 lbs. I am thinking to use a 7/8" plate. the deflection limitation is 0.002 in/in length of plate. if you drawing a x-alxe and y-alxe, four quadrants will be seperatedly by x and y alxes.
My first leg is at the left and upper area (quadrant), my second leg is at the right and upper quadrant, the third leg is at left and lower quadrant, and the fourth led is at right and lower quadrant. Their distances from the x and y alxes are shown in the above message.
I am thinking to use the 3" by 3" by 1/2" angle steel as legs. In my calculation, I use force on leg 3 equal to force on leg four. I thought that is correct. But Prex said that is not correct. why is that not correct?
Prex, I agree about your calculation for P3 and P4 if we are using 20 tons load as desertfox. But I will calculated P1+P2=10(tons) and P1*15=P2*29, therefore, P1=3.4 tons and P2=6.6 tons. What do you think?
hope to learn more!
RE: calculate force distribution
Your equation P1*15=P2*29 is correct, but the result is P1=6.6 and P2=3.4
prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: calculate force distribution
Big Inch
You've posted before how to put images into your posts. I've lost the thread. Would you please post it again (or possibly turn it into a FAQ.
Thanks
Patricia
RE: calculate force distribution
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RE: calculate force distribution
Patricia
RE: calculate force distribution
Greg Lamberson, BS, MBA
Consultant - Upstream Energy
Website: www.oil-gas-consulting.com
RE: calculate force distribution
I put downloadable instructions here,
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