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calculate force distribution 1

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hello2006

Mechanical
Feb 2, 2007
26
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!
 
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In the general case, with no symmetry, you can support the plate at any three points. You can then make any assumption whatever about the load at the fourth point, and then proceed to solve for the loads at the other three points. And the set of 4 loads will then fit the sum of forces and sum of moments equations.

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.
 
Right! I thought it would be possible to evaluate using the stress analogy, if the plate could be considered totally rigid. Apparently ...NOT!



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Hi hello2006

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
 
Hi thanks everybody,

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!
 
Hi hello2007

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
 
hello2007,
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
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
 
thanks very much for everybody's reply!

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!
 
They are not equal simply because your quadrilateral is not symmetrical with respect to y axis. However, as I said above, it can be a fair guess.
Your equation P1*15=P2*29 is correct, but the result is P1=6.6 and P2=3.4

prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
 
I'm doing a short thread hijack-- sorry all.

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
 
Big Inch, sorry, your site is blocked; it's listed as "Social Networking;Dating/Social;Media Downloads"

Patricia
 
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