Load distribution to three supports due to concentrated load 3

[token]

I am an architect and it has been a long time since my structures courses at school. So some help with the following problem would be greatly appreciated:

Given a triangular plate pinned at each vertex, how do you calculate the reactions at each support to a point load applied at an arbitrary location on the plate. My intuitive response to the problem suggests that the load at any given vertex would vary inversely with the area defined by the following four points: the location of the point load, the vertex and two points along the sides adjacent to the vertex defined by extending lines from the other two vertices through the position of the point load.

This obviously works if the point load falls on the centroid - but I am not convinced that I am on the right track given the problems that arise at the limit conditions on the perimeter of the triangle. Is there a simpler method? If not, where might I find the equations for the solution? This problem also represents the simple case, what if the supports are not at the vertices or the plane is not a triangle (but is still supported at three points)?

Many thanks,

Token

 

[chicopee]

Is the plate to remain horizontal while being loaded?

 

[Mixtli]

Hi Token,

There are no direct equations or formulas that would solve the problem. What you have to do is carry on with the free body diagram in two different directions. I.e. define the XY plane coplanar with your plate, being, say X, colinear with one side of the triangle and the Z perpendicular to the plane (obviously), then on the plane YZ obtain the reactions, one would be for one vertex, end the other reaction would be for the other two vertexes (which are aligned, right?), then on the plane XZ obtain the other two reactions by taking into account only the previously obtained reaction corresponding to the aligned reactions as acting force.
To obtain the reactions at each step you have to obtain the free body diagram, then determine the equilibrium equations and solve them (by pure math).

This is considering the plate to be horizontal and with no deflections that can cause horizontal displacements on the supports.

Hope this helps

Tony

 

[GregLocock]

Assuming all the forces are parallel (they aren't, there are membrane forces as well, as described).

Take moments about each pair of vertices, to give you 3 equations of the form

applied load*distance from hinge line =reaction at the other support*distance from hinge line

and a 4th equation is that the sum of the reaction forces is equal to the applied load.

I'd be astonished if any more was needed because 3 legged stools are stable.


Cheers

Greg Locock

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

 

[chicopee]

Yes stools are stable but joints are not hinged; if they were then each leg would be tied with chains or members to the other legs.

 

[GregLocock]

By hinge line I meant the line connecting the two vertices.

Cheers

Greg Locock

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

 

[rb1957]

a simple approach would be to distribute the load according to the ratios of the distance to the three vertices ... if the load point is equidistant to all vertices, equal reactions; if one vertex is closer to the load, it'd react more.

 

[prex]

Well, there is a geometric relationship, and:
Greg, you only need three equations, as there are three unknowns...,
rb, it can't be a relationship based onto the distances of the load to the vertices, as the case of the load being on a side of the triangle sees obviously the third vertex with no reaction, but the distance is not infinite...
The relationship is easily derived by making use of moment equations. Say we want the reaction on vertex 3: we write a moment equation about the line joining the vertices 1-2. It is:
R₃=Pd₁₂/h₃
where
P is the load
d₁₂ is the distance of the load to the side 1-2
h₃ is the height of the triangle with respect to side 1-2.
Now
A=s₁₂h₃/2
A_12P=s₁₂d₁₂/2
where
A is the area of the triangle
A_12P is the area of the triangle formed by side 1-2 and the load
s₁₂ is the length of side 1-2
Now the conclusion is already clear:
the reactions at the vertices of a triangle that equilibrate a load are proportional to the ratio: area of the triangle formed by the side opposite to the vertex under consideration and the load / area of the triangle.
It could of course be also expressed as a function of the ratio of distances point to line.
This relationship should still hold if the load is outside the triangle, but the triangle formed by the load and the closest side (entirely outside the base triangle) should be counted as having a negative area.

prex

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[rb1957]

true enough ... i was thinking of the general case where the load is somewhere inside the triangle

 

[BAretired]

If the point load is not normal to the plate, the solution is a bit more complicated. The force may be resolved into a normal and parallel component, say N and P respectively.

For the normal component N, the reactions may be calculated as outlined above.

For the parallel component P, the applied force has a moment about the centroid of the plate, P*e where e is the eccentricity of P from the centroid. The moment about a normal axis through the centroid is M = P*e.

If P1, P2 and P3 are the in-plane forces at each corner (normal to a line from centroid to corner) and a, b and c are the distances from each corner to the centroid, then:

P1 = P/3 ++ M*a/(a^2 + b^2 +c^2)
P1 = P/3 ++ M*b/(a^2 + b^2 +c^2)
P1 = P/3 ++ M*c/(a^2 + b^2 +c^2)

In the above, ++ indicates vector addition

And finally, assuming N1, N2 and N3 are the three reactions normal to the plate found by the method outlined above by prex, the three reactions are:

R1 = P1 ++ N1
R2 = P2 ++ N2
R3 = P3 ++ N3


BA

 

[msquared48]

When all else fails, cut a piece of 1/2" plywood to the shape of the triangle and place it over three scales, one at each corner. Place a 10 pound weight in various places in the surface and read the scales. Make a table to record the results.


Mike McCann
MMC Engineering

 

[GregLocock]

prex - I know my 3 moment calculations aren't truly independent, I only need to solve two of them. Normally I'd just take moments around any two non parallel axes, and I know what the total of the reaction forces is.



Cheers

Greg Locock

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

 

[token]

Thanks for all the replies. This is extremely useful.