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Point Reactions

Point Reactions

Point Reactions

(OP)
Hi all!

Apologies if this is in the wrong spot as I am new here.

This also may seem pretty basic problem, but there is some argument about it going on in my office so I wanted to see what others thought.

Below are two scenarios.
Basically just looking for the reactions at A, B, C and D
All loads shown are acting into the page.
The reaction points are rollers that can roll towards the AB end or the CD end




Look forward to you help
any tips pointers about the problems or the forum please let me know

Thanks
Mat

RE: Point Reactions

(OP)
Dimensions are in mm

RE: Point Reactions

What is the structure here?

There is no information other than loads and dimensions. The structural layout is needed too.

Mike McCann, PE, SE (WA)


RE: Point Reactions

It depends on whatever actually constitutes the structure as Mike mentioned. Is it a slab, pilecap, something else?

I expect the answer you're looking for is what is the distribution of reactions under a point load load, P, at some x,y co-ordinate. This assumes a lot of things like:

1) Whatever provides the reactions are infinitely stiff, with no settlement
2) There is some kind of infinitely stiff structure connecting everything together like a slab.
3) The structure is statically determine i.e. one of the supports has a lateral and longitudinal restraint

Basically all you do is calculate the proportion of 'P' taken by rows AB and CD using simple statics/taking moments. This proportion of 'P' is then distributed to each reaction point A, B, C and D by taking moments in the opposite section. I'll do it for the 24.5kN point load:

Row AB = 24.5*312.5/586 = 13.0kN
Row CD = (24.5)(586-313)/586 = 11.4kN
Check AB + CD = 13.0 + 11.4 = 24.5kN, OK

For row CD say,

Push pull by inspection, RC = -11.4*152/369 = -4.7kN
RD = 11.4*(369+152)/369 = 16kN
Check RC + RD = 11.4kN, OK

RE: Point Reactions

if you're only interested in the reactions at the four points, and if loading 2 is equivalent to loading 1 (looks like it) then I'd say the four reactions should be exactly the same for the two loadings.

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

That's easy. Tranform the forces to a vertical-to-screen force at the centroid of the rectangle and to a couple of moments about both horizontal and vertical (on-screen) axes. Then calculate the reaction assuming force and moments equilibrium.

Jason McKee
proud R&D Manager of
Cross Section Analysis & Design
Software for the structural design of cross sections
Moment Curvature Analysis
Interaction Diagrams
Reifnorcement Design etc.

RE: Point Reactions

Aren't there four unknowns and three equations? What am I missing?

RE: Point Reactions

yes, 4 unknowns three equations. but you can
a) assume the supports are equally stiff so each support reaction 1/4 the load plus/minus couples from the two offset moments,
b) use a least squares solution to over defined equations (which should give the same result).

i'd get 12.25 +- 49*1.5/(369*2) +- 49*30.5/(2*586)

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

(OP)
Structure would look something like this

RE: Point Reactions

That's not very clear either. Do you mean there are beams running in the AC direction that are supported by beams running in the AB/CD direction? Simple spans, or is there continuity not shown?

RE: Point Reactions

(OP)
The single load of 49kN is at the centroid of the three other load points

So for the single load.
Taking moments around an axis BD
AC=(49*186)/369 = 24.699kN

Taking moments around an axis AC
BD=(49*183)/369 = 24.301kN

A=AC*(262.5/586) = 11.064kN
B=BD*(262.5/586) = 10.886kN
C=AC*(323.5/586) = 13.635kN
D=BD*(323.5/586) = 13.415kN
Sum of Reactions = 49kN


Looking at the multiple forces individually

Force1
F1AC=(12.25*(369+192.5))/369 = 18.641kN
F1BD=-(12.25*192.5)/369 = -6.391kN

F1A=F1AC*(157.5/586) = 5.010kN
F1B=F1BD*(157.5/586) = -1.718kN
F1C=F1AC*(428.5/586) = 13.631kN
F1D=F1BD*(428.5/586) = -4.673kN

Other Forces worked out in the same way

F2AC = 16.118kN
F2BD = -3.868kN

F2A = 7.358kN
F2B = -1.766kN
F2C = 8.760kN
F2D = -2.102kN

F3AC = -10.059kN
F3BD = 34.559kN

F3A = -5.364kN
F3B = 18.430kN
F3C = 4.695kN
F3D = 16.129kN

Sum of individuals

A = 5.010+7.358-5.364 = 7.004kN
B = -1.718-1.766+18.430 = 14.946kN
C = 13.631+8.760-4.695 = 17.696kN
D = -4.673-2.102+16.129 = 9.354kN
Sum of reactions = 49kN

RE: Point Reactions

having a "grillage" of beams, each statically determinate, greatly simplifies the problem.

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

(OP)
Do you mean like this?

RE: Point Reactions

Quoting reactions and forces to 3 S.F...student or secondary steelwork designer you pick ; )

RE: Point Reactions

(OP)
haha sorry

at this stage just using the extra figures to make sure they match exactly
will pull them back in final calcs

RE: Point Reactions

(OP)
So that's two way's of working it out.
with both giving the same answers.

But others in the office believe that using the centroid load should give the same results for the reactions.
i.e as I calculated earlier.

AC=(49*186)/369 = 24.699kN
BD=(49*183)/369 = 24.301kN

A=AC*(262.5/586) = 11.064kN Compared to 7.00kN
B=BD*(262.5/586) = 10.886kN Compared to 14.94kN
C=AC*(323.5/586) = 13.635kN Compared to 17.70kN
D=BD*(323.5/586) = 13.415kN Compared to 9.36kN
Sum of Reactions = 49kN

RE: Point Reactions

it was alluded to earlier ... there is a difference if there is a rigid black box joining the loads and their reactions or if there are discrete loadpaths.

but i'm smelling a bit of a rat ... both solutions give the same sum AC and BD (24.7 and 24.3), comparing your centroid solution with the earlier solution. and the same sums AB and CD.

also the "shape" of your centroid solution ... 12.25+1.385, +1.185, -1.185, -1.364
compares with 12.25+5.45, +2.69, -2.89, -5.25 ...
this says to me that the centroid solution is reacting way less bending ...
check the moments about the CG of the reactions for both sets of loads

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

(OP)
So the load at the centroid position is attached to the individual load points via a steel wire rope.

Obviously this brings horizontal loads into the formula but for simplicity at this stage I was just working on the verticals as the horizontals should all cancel out.

So I guess my question is.. which is the best way to determine the correct reactions?

RE: Point Reactions

TTCorona132,
Is this a plate or 5 beams or something else?

RE: Point Reactions

(OP)
The structure is as per this diagram


The load at the centroid in the attached to the individual load points via a wire rope



As I said.. It has been simplified at this stage to remove an horizontal loads, and just concentrate on the verticals.

So 49kN at the centroid
will equal 24.5kN on each side

so the two loads on the left hand side will both equally see 12.25kN
and the load on the right will see the 24.5kN load.

apologies if I am not being too clear...

RE: Point Reactions

"So the load at the centroid position is attached to the individual load points via a steel wire rope." ... what the heck does that mean ? no-one has said anything about a rope before ? the way the problem was represented the load at the centroid is a theoretic equivalent for the set of applied loads. the solution should be the same for the set of loads and for the load at the centroid (of the loads). the external loads should eb the same, the internal loads will be very different (can't use a centroid load for internal loads)

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

Good luck!

RE: Point Reactions

(OP)
So if I treat it as 5 beams then I get this scenario


A = 7.66kN
B = 14.29kN
C = 17.04kN
D = 10.01kN

Much the same as my calcs above..

If I treat it as a solid box


A = 11.51kN
B = 11.08kN
C = 13.99kN
D = 13.00kN

which are much the same as using the centroid..

RE: Point Reactions

(OP)
Sorry.. no rat here... just a little clueless hahaha

So my actual structure is represented by the beam model shown above,
then these are the results of my reaction loads I should be using?

And not simplifying it to a "rigid black box"

RE: Point Reactions

A "rigid black box" is indeterminate when there are four supports. The solution requires a knowledge of stiffness of the box. The black box structure is determinate with only three supports. You can say, however that the centroid of the applied forces and the centroid of the reactions are identical.

The simple span beam structure is determinate and all four supports are required for stability.

BA

RE: Point Reactions

i have to admit that i'm surprised that the centroid solution is so different to the "grillage" solution ... why do the models have different couples in the reactions ??

no surprise that the grillage model matches the hand calc (it's determinate).
is there a moment reaction included in the black box model ?

as an exercise can you run those two models for each load separately

another day in paradise, or is paradise one day closer ?

RE: Point Reactions

(OP)
We are also a bit surprised by the answer which is where this discussion has come from.

I'm not sure what you mean by "why do the models have different couples in the reactions ??"

there are no applied moments added to either model..
only the loads and supports shown..
the supports are what Ansys calls "Simply Supported" in which they can pivot in any axis, but they cannot translate.

I have run the load results separately previously on the beam model in Ansys, and I get the same results as when I manually calculate each load separately.

RE: Point Reactions

(OP)
I think that treating the solution as a plate/box means the loads can propagate through the plate to each support evenly.

Where as within the actual structure of the product, when represented by a beam model,
the loads have discrete paths that they need to take and as such will load up particular supports more than others.

RE: Point Reactions

Quote (TTCorona)

I think that treating the solution as a plate/box means the loads can propagate through the plate to each support evenly.

What does that mean? The problem is indeterminate, so the calculation is invalid. It cannot be solved by statics alone.

Consider the 49kN load located as shown. Take moments about AD. The eccentricity of load from diagonal AD is about 16 mm toward C, practically on diagonal AD. If support B is removed, support A takes 262.5/586 * 49 = 21.95 kN and supports C and D together take 27.05 kN with C taking only 2.51 kN and D carrying 24.54 kN.

Alternatively, consider moments about diagonal BC and nearly all of the load is carried by supports B and C.

The correct analysis must take into account the stiffness of the box.

BA

RE: Point Reactions

(OP)
What if we assume that both the plate and the beams used in both methods are infinitely stiff?

RE: Point Reactions

(OP)
I am just looking at it as a free body diagram so no member strength/stiffness has been added at this stage

RE: Point Reactions

Quote (TTCorona)

What if we assume that both the plate and the beams used in both methods are infinitely stiff?

The beams are statically determinate. No matter how stiff they are, the answer is the same. If the supports are not precisely at the same elevation, the answer is the same.

The infinitely stiff plate is statically indeterminate to the first degree because there is one redundant reaction. If one of the supports is a hair's width higher or lower than the other three, the load is carried by three supports.

We have all sat at a wobbly table in a restaurant where one leg is shorter than the others. Lean on it one way and the short leg takes load. Lean on it another way and the diagonally opposite leg takes load. If leg A is shorter than the rest and the load is centered, the entire load is carried by B and C.

In the real world, we do not have infinitely stiff members and we do not have supports which are built precisely at the same elevation and even if they were, we cannot be certain there will not be differential settlement.

From statics, A+B, C+D, A+C, B+D, B+C and A+D can be determined but individual values for A, B, C and D cannot be determined without considering compatible deformations of the black box and the four supports.

BA

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