3D reaction force calculation 2

[mdc1973]

Hello,
I wish to calculate the reaction force at Cz to keep the structure in equilibrium.

The linked diagram shows a platform that weighs 500kg.
The platform has a point load 1.9KN.
The platform size is 2.8m long x 0.9m wide and it is restrained by two shafts on opposing corners, free to rotate axially.

Link

 

[rb1957]

my numbers were a little off ...

ptC is off-set 0.857m from the axis 1-2,
but the load point is 1/6 of this (at point C the y-component of the offset is 0.9m, and the loading point it is 0.15m)

so the reaction at C is 1900/6 = 317N

Quando Omni Flunkus Moritati

 

[rb1957]

and B = 950N (+250kgf)
and A = 633N (+250kgf)

Quando Omni Flunkus Moritati

 

[BAretired]

rb1957, what you are missing is the fact that the structure is indeterminate so your method is not valid. It would be valid if supports A, B and C were hinge/rollers but they are not. Supports A and B are free to rotate about the X axis but not about the Y axis.

BA

 

[ThomasH]

If the boundary conditions that I assumed together with several others are correct, than that solution should also be correct. That is the solution that rb1957 has given.

I guess the only one who can clarify is the OP and he seems to be offline. We'll just have to wait and see I guess.

Thomas

 

[BAretired]

haynewp, I couldn't make any sense of your link, but it seems to me that if the platform is deemed rigid, there should be no deflections and no rotations, thus the reaction at point C should be zero.

BA

 

[haynewp]

Do you have RISA? You need it to open the file.

I agree. I am not sure how RISA treats "rigid" material. It gives a solution to the reactions with rotations and not deflections. I don't really get it. I am temporarily locked out of the program right now due to a VPN issue.

 

[rb1957]

@BA ... i guess we're free to read the sketch as we want. for all we know A and B could have spherical bearings, or a hoop of piano wire. the sketch could be a little off, and the dowels and A and B could be along the 1-2 axis

sure, to include the My moments at A and B the load causes a moment about the 1-2 axis, reacted by the vector sum of MyA, MyB and MC

Quando Omni Flunkus Moritati

 

[IDS]

The platform size is 2.8m long x 0.9m wide and it is restrained by two shafts on opposing corners, free to rotate axially.

Expressio Unius Est Exclusio Alterius : the expression of one thing is the exclusion of the other

In this case the statement that the shafts are free to rotate axially implies that they are not free to rotate about the Y or Z axes.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

See attached images from an analysis in Strand7.

Increments are:
1) Self weight
2) Point load
3) 1) + 2)

The plate was assumed to be steel 25 mm thick (with the density adjusted to give a mass of 500 kg).
Nodea A and B were restrained in position and for rotation about the Y axis.
Node C was restrained in the Z direction only.
Tabulated reactions are at point C.
Deflections are scaled by 100

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[msquared48]

No. Thhe shafts can still deflection t from gravity loads.

Mike McCann
MMC Engineering

 

[IDS]

No. Thhe shafts can still deflection t from gravity loads.

I don't know what that means.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

The platform has a mass of 500kg or a weight of about 4.88kN. The reaction at C was 0.1 kN, not very large compared to the total weight, so the bulk of the dead weight is carried by supports A and B.

The reaction of 0.3184kN at C due to the concentrated load is very close to the value found by others assuming completely hinged supports, namely P/6 or 0.3157kN. I would not have expected it to be larger than 0.3167 but smaller as it seems to me that any restraint at supports A and B would tend to reduce the reaction at C, so I am wondering why those restraints actually increased the reaction at C. That is not clear to me at the moment. Perhaps someone can shed some light on that.


BA

 

[BAretired]

*** 0.3157kN should read 0.3167kN

BA

 

[IDS]

BA - good question. It does seem anti-intuitive, but I presume it is due to the combination of the reaction due to the moment restraint plus the torsion between point A and point C.

I checked it with pinned supports at A and B, and it does give the correct reaction for that case (see attached image).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[haynewp]

And the real winner in all this is mdc1973!

 

[BAretired]

Thanks Doug. Nice presentation.

BA

 

[ThomasH]

Hi again

I would say that the OP needs to clarify the sketch. On the sketch there are five reactions implied and that's it. For some reason some posters have added some extra constraints. Maybe the rods at A and B is just something to attach a wire hoop.

If the five reactions are all there is then the solution came in one of the early posts. A simple hand calculation.

I'll wait for the OP to clarify.

Regards

Thomas

 

[BAretired]

ThomasH, Five reactions are two too many. Nobody has added extra constraints. The structure is indeterminate and cannot be analyzed by statics alone. Why do you persist in arguing?

BA

 

[ThomasH]

BAretired:

I thought if was a friendly discussion and not an argument. But never mind the semantics .

You say five reactions. In the figure there are five: Ay, Az, By, Bz and Cz. Do you have any others?
Some have mentioned constrained rotations but I cant see that in the figure.

If we agree on the five reactions, then i believe there is a solution.

Friendly Regards

Thomas

 

[BAretired]

ThomasH:

In that case, each joint has 6 potential constraints, namely Fx, Fy, Fz, φx, φy and φz. Joint A has 5 constraints as φx is zero (free to rotate about the x axis). φz does not enter the picture so it is assumed to be zero also, leaving only 4 constraints for Joint A. Joint B has 3 constraints, Fx, Fz and φy. Joint C has only Fz. So, looking at it that way, the number of constraints is 4 + 3 + 1 = 8.

BA