Back to Basics!!!!

[CSCPE24]

Hi All,

I'm even embarassed to have to ask this. But can someone remind me of how I can calculate the reactions at the 4 corners of the rectangle if the CG of the part is offset as shown? The weight of the part is 10 lbs. If the CG is at the center, then we know the reactions should be 10/4 = 2.5 lbs at each corner. But since it's offset, I'm kinda stumped.

I know I have to sum up values in x and y direction from the original CG but I just can't seem to remember everything.

Thanks for all the help in advance!

 

[reidh]

Sum of forces, sum of moments = 0, assuming everything is static.

 

[GregLocock]

The reason you can't solve it is that it is statically indeterminate.

If you make the correct additional assumptions then you will be able solve it.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

well you can always do what we do ... assume (and don't tell me the joke about "assume", i've heard it).

reidh is right (so too is greg in an absolute way).

assuming you don't know what sum of forces and moments means ('cause it is basic) you've got four unknowns and only three equations (which is why it is redundant). assume your part is rigid (and so remains a plane). a simple way is to assume that each reaction is inversely proportional to the distance for the load CG (the closest one is going to see the highest load).

 

[CSCPE24]

reidh, Greglocock, rb1957,

Thank you so much guys! A light bulb just turned on!

Cheers!
CSCPE24

 

[zekeman]

quote"rb1957 (Aerospace)
9 Feb 08 7:41
well you can always do what we do ... assume (and don't tell me the joke about "assume", i've heard it).

reidh is right (so too is greg in an absolute way).

assuming you don't know what sum of forces and moments means ('cause it is basic) you've got four unknowns and only three equations (which is why it is redundant). assume your part is rigid (and so remains a plane). a simple way is to assume that each reaction is inversely proportional to the distance for the load CG (the closest one is going to see the highest load)."

You make the assumptions that each support is equally rigid and the plate does not distort under the load.

In general there are problems with these assumptions, particularly the rigidity of the plate.

 

[40818]

Careful Zekeman, your in danger of invalidating 95% of the aircraft flying about today!!

 

[zekeman]

40818,
Show me where. I would be happy to learn.

 

[JStephen]

"assume your part is rigid (and so remains a plane). a simple way is to assume that each reaction is inversely proportional to the distance for the load CG (the closest one is going to see the highest load)."

Suppose you have a square plate supported at each corner. You apply equal loads at two diagonal corners. The CG is then at the center of the plate. All four supports are equally distant from the center of the plate. Thus the reaction is the same at each corner per this method.

 

[GregLocock]

Or, indeed there could be no force at two of the corners.

Hence demonstrating the problem.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

JStephen,
if the CG of th eload is over the cg of the plate, yeah, i'd approximate the reactions (based on the assumptions of equal stiffness and rigid plate) as 4 equal loads in the corners.

the +ve i'd take from greg's comment, is that yes, if two of the supports were ineffective, then two (diagonal ones) would see 1/2 the applid load ... so to be safe you could work to this instead; myself, i'd consider this overly conservative.