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
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
RE: Point Reactions
There is no information other than loads and dimensions. The structural layout is needed too.
Mike McCann, PE, SE (WA)
RE: Point Reactions
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
another day in paradise, or is paradise one day closer ?
RE: Point Reactions
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
RE: Point Reactions
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
RE: Point Reactions
RE: Point Reactions
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
another day in paradise, or is paradise one day closer ?
RE: Point Reactions
RE: Point Reactions
RE: Point Reactions
at this stage just using the extra figures to make sure they match exactly
will pull them back in final calcs
RE: Point Reactions
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
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
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
Is this a plate or 5 beams or something else?
RE: Point Reactions
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
another day in paradise, or is paradise one day closer ?
RE: Point Reactions
RE: Point Reactions
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
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
The simple span beam structure is determinate and all four supports are required for stability.
BA
RE: Point 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
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
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
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
RE: Point Reactions
RE: Point Reactions
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