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Mass hanging - equilibrium equations using rod length

Mass hanging - equilibrium equations using rod length

Mass hanging - equilibrium equations using rod length

(OP)
I am looking for equation describing situation like on the picture. It should be equilibrium equation using Rod Length and Rod Young modulus and concerns gravity center.
The problem is that I have ready quite long Visual Basic code made by other author but I cannot understand it. Let's start my question without this code.

RE: Mass hanging - equilibrium equations using rod length

1) Free body diagram of system
2) Equilibrium equations (sum forces in each direction and then sum moments about a point)
3) Now that you have found the tensile force in each rod as a function of W and X1 and X2, you can use the length, modulus, and area to find displacement (stretch) of each rod.

As Greg has pointed out, this will not be too difficult to solve if you have had training to get you through steps 1, 2, and 3 above. I hope this helps.

RE: Mass hanging - equilibrium equations using rod length

(OP)
No, it is not this solution. This is much more complicated. When I have more time I will try to prepare something more.

RE: Mass hanging - equilibrium equations using rod length

how many rods support the weight ?

is it a 2D problem as shown ?

is it a static equilibrium problem or something different (dynamic, vibration, ...) ?

if the answers are 2, yes, yes ... then it is a simple problem, independent of E.

Quando Omni Flunkus Moritati

RE: Mass hanging - equilibrium equations using rod length

I'm guessing that since E is involved, it's a static equilibrium problem, but it's shown in an initial condition of zero weight. Of course, it's unsolvable as shown with no deformation because the tension of the horizontal wires are infinite for any nonzero weight. Just looking at it, I think it may require an iterative approach (hence the VBA) because as each rod deflects, it will change the angle of the wire attached to it. As the wire's angle changes, it changes how the rod is bent, therefore changing the angle. As the angle of each wire changes, the amount of tension shared between wires due to the free-hanging weight also changes.

You will certainly need a few FBDs to solve this one.

-handleman, CSWP (The new, easy test)

RE: Mass hanging - equilibrium equations using rod length

(OP)
Hangleman - it can be something in this direction You described. Rb1957 - it is 3-dimensional subject. There is no vibration or other dynamic interactions. There are many rods about 20, 30, 40 rods - it depends from the element. These are boiler Power Station suspended elements. It is very interesting that this macro include rod length effect. For example when one rod is shorter than others, it carry much more load than others rod. Below is a fraction of VBA code. There is "Inverse" used. Matrix is created from
F11 to F33 values and inversed. I do not know this method of weight distribution calculation. If You could describe it in some lines generally (ideolically) or indicate books,
publications where such solutions are described I will appreciate it. I am particulary interested this Matrix Inversion and alfa, beta, delta quantity.

"Area column" is cross section area of the rod
"Length column" is length of the rod
Cells(i, xColumn)); Cells(i, yColumn)) - there are rods coordinates
(Cells(TotalRow + 1, j); (Cells(TotalRow + 2, j) - these are correspondingly x and y coordinates of the structure center
--------------------------------------------------------------------------------------------------------------------------

MassSum = 0
MomentSum_x = 0
MomentSum_y = 0
MomentSum_z = 0


For j = LoadCaseColumnStart To LoadCaseColumnEnd
For i = RowsMassStart To RowsMassEnd
If Cells(i, j) <> 0 Then
MassSum = MassSum + Cells(i, MassColumn) * gravity * Cells(i, j)
MomentSum_x = MomentSum_x + Cells(i, MassColumn) * gravity * Cells(i, j) * Cells(i, xColumn)
MomentSum_y = MomentSum_y + Cells(i, MassColumn) * gravity * Cells(i, j) * Cells(i, yColumn)
MomentSum_z = MomentSum_z + Cells(i, MassColumn) * gravity * Cells(i, j) * Cells(i, zColumn)
End If
Next
Cells(TotalRow, j) = MassSum
If MassSum = 0 Then
Cells(TotalRow + 1, j) = 0
Cells(TotalRow + 2, j) = 0
Cells(TotalRow + 3, j) = 0
Else
Cells(TotalRow + 1, j) = MomentSum_x / MassSum
Cells(TotalRow + 2, j) = MomentSum_y / MassSum
Cells(TotalRow + 3, j) = MomentSum_z / MassSum
End If

MassSum = 0
MomentSum_x = 0
MomentSum_y = 0
MomentSum_z = 0

F11 = 0
F12 = 0
F13 = 0
F21 = 0
F22 = 0
F23 = 0
F31 = 0
F32 = 0
F33 = 0

For i = RowsRodsStart To RowsRodsEnd
If Cells(i, PointingColumn - 1) = "x" Then
F11 = F11 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn))
F12 = F12 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn))
F13 = F13 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn))
F21 = F21 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn))
F22 = F22 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn))
F23 = F23 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn))
F31 = F31 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 1, j) - Cells(i, xColumn))
F32 = F32 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * (Cells(TotalRow + 2, j) - Cells(i, yColumn))
F33 = F33 + (E * Cells(i, AreaColumn) / Cells(i, LengthColumn))
End If
Next

massForce = Cells(TotalRow, j)

Sheets("Inverse").Cells(3, 1) = F11
Sheets("Inverse").Cells(3, 2) = F12
Sheets("Inverse").Cells(3, 3) = F13
Sheets("Inverse").Cells(4, 1) = F21
Sheets("Inverse").Cells(4, 2) = F22
Sheets("Inverse").Cells(4, 3) = F23
Sheets("Inverse").Cells(5, 1) = F31
Sheets("Inverse").Cells(5, 2) = F32
Sheets("Inverse").Cells(5, 3) = F33

alfa = Sheets("Inverse").Cells(7, 3) * massForce
beta = Sheets("Inverse").Cells(8, 3) * massForce
telta = Sheets("Inverse").Cells(9, 3) * massForce

For i = RowsRodsStart To RowsRodsEnd
If Cells(i, PointingColumn - 1) = "x" Then
Cells(i, j) = (E * Cells(i, AreaColumn) / Cells(i, LengthColumn)) * ((Cells(TotalRow + 1, j) - Cells(i, xColumn)) * alfa + (Cells(TotalRow + 2, j) - Cells(i, yColumn)) * beta + telta)
End If
Next

RE: Mass hanging - equilibrium equations using rod length

(OP)
Some description more:
Trying describe F11, F12 and F13 generally, it looks like below:

F11:
M*L/EA = x1^2+x2^2+x3^2... ; M is instead of F11. F11 seems to be Moment.

F12:
M*L/EA = x1y1+x2y2+x3y3... ; M is instead of F12. F12 seems to be Moment.

F13:
F*L/EA = x ; F is instead of F13. F13 seems to be Force.

Assumption E1=E2=E3..=E
A1=A2=A3..=A
L1=L2=L3..=L

RE: Mass hanging - equilibrium equations using rod length

Hi Pawel27

I am not familiar with power stations so can you explain what these rods actually suspend off and how they get loaded by whatever force, your first posted diagram is very misleading as it depicts a 2D simple problem which now turns into a much more complicated problem four days later.

desertfox

RE: Mass hanging - equilibrium equations using rod length

Quote:

Of course, it's unsolvable as shown with no deformation because the tension of the horizontal wires are infinite for any nonzero weight.

I see no horizontal wires.

If we could have a reasonably complete description of the problem, it would help to get some meaningful answers.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Mass hanging - equilibrium equations using rod length

(OP)
Wires (rods) are vertically oriented (red colour). Suspended element is a rectangle (black colour). The pictures above are 2-D simplification. In reality it is 3-d problem with many rods. Below is 3-d sketch - maybe now it will be more clear. If You have any questions don't hesitate to ask.

RE: Mass hanging - equilibrium equations using rod length

Looks like I read the sketch way wrong, especially now that we have the 3D one. I interpreted the big black box as some sort of stand, with the red rods being like posts sticking up from it and the CG symbol being a weight that is hung from a wire suspended between the tops of the posts on two unequal length wires.

-handleman, CSWP (The new, easy test)

RE: Mass hanging - equilibrium equations using rod length

"M is instead of F11. F11 seems to be Moment." ? ... looking at "M*L/EA" i'd suggest that "M" is a force.

the problem, while complex to solve, is simple to explain. you have many axial members supporting a plane face. each axial member strains according to the load it carries, and deflects PL/EA. all the deflections are co-ordinated so that the plane face remains plane; all the loads sum to the applied load. out of this you should get an undefined set of equations ... more variables than equations; the "best" way to solve this is using matrix methods ... goggle search should find the solution.

why do you think the horizontal wires carry infinite force for an applied vertical load? initially they'd carry nothing (no effective component) but maybe there's a preload, then they'll be deflected out of their initial horizontal position so they'll have a small vertical component.

considering your initial pic, and what we're talking about now ... another badly described problem. it would have helped if you had said initially "a highly redundant (or hyperstatic) problem". Note, if all the wires have the same E, then the solution is independent of E.

Quando Omni Flunkus Moritati

RE: Mass hanging - equilibrium equations using rod length

Hmm, so my first post was on the money.

In practice how are these things tensioned up? How accurate are their lengths? How flat is the platform that is being supported? Are you sure it doesn't deflect under load?

Does the structure you are hanging this from, deflect?


Cheers

Greg Locock


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RE: Mass hanging - equilibrium equations using rod length

Quote:

all the deflections are co-ordinated so that the plane face remains plane; all the loads sum to the applied load. out of this you should get an undefined set of equations ... more variables than equations; the "best" way to solve this is using matrix methods ... goggle search should find the solution.

If the plane face remains plane then it can be analysed in the same way as a beam under combined axial load and bending:

- Find the total area of the rods, A (using transformed section properties if the E value varies)
- Find the centroid
- Find the second moment of area about the centroid, I (again using transformed section properties if the E value varies)

Stress due to axial load = P/A
Stress due to moment at y from the centroid = My/I

Of course in reality the plane face will not remain plane, but it may be an acceptable approximation. It's probably the approach used in your code anyway (although I haven't worked through it to check).

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Mass hanging - equilibrium equations using rod length

(OP)
Handleman, Greg, IDS we can assume that this is a stiff box suspended into rods. When I showed VBA code maybe it will be better to look at it as a 3-D problem.

Rb1957 - I also tried to relate it with known equation sigma=FL/EA, but looking at units at both sides of equation F11 seems to be a moment [with Nmm unit]. Please look at my F11 description. If F11 would be force then units disagree at both sides of equation. What is probably important and can be helpful in solving the problem:

F11 is related with x^2
F12 is related with x*y
F13 is related with x
F21 is related with y*x
F22 is related with y^2
etc...

Below screenshot from more real CAD model. You will get maybe better look at the problem. Of course there are rods with unequal length. We don't have to worry about it in our considerations. We can assume that rods have equal lengths. We can focus at general picture "Suspended_element_3D.png".

What I am interested in is getting general equations describing the problem basing on VBA code.

RE: Mass hanging - equilibrium equations using rod length

(OP)
Rb1957 - looking at matrix methods it seems to be Direct Stiffness method used, but I am not sure yet. I must look at it more precisely. There is tension rod stiffness EA/L used like in VBA.

RE: Mass hanging - equilibrium equations using rod length

Hi Pawel27

I believe it might follow the direct stiffness method checkout example 4 on the link by using the scroll bar on the right hand side.
I think the case you have is more complicated as the example is only 2D

http://www.learningace.com/doc/978932/83fe4795baa7...

RE: Mass hanging - equilibrium equations using rod length

Quote:


What I am interested in is getting general equations describing the problem basing on VBA code.

Did you have some problems with the equations I gave you?

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Mass hanging - equilibrium equations using rod length

Hi Pawel27

If all the rods are of the same material which I assume they are and the rods were symmetrically spaced around the centre of gravity then each rod would carry a load relative to its stiffness however you indicted in your first post that the centre of gravity wasn't central to the supporting rods.

RE: Mass hanging - equilibrium equations using rod length

(OP)
Yes, gravity center is usually misaligned (not central to the supporting rods). I just try to write down relations (equations) to describe such kind of structures looking at VBA code.

RE: Mass hanging - equilibrium equations using rod length

then i have no idea what ML/EA represents.

PL/EA is the extension due to a force P.

i think Doug(IDS) posts a reasonable solution strategy. if the load CG aligns with the rods' centroid then the rods will be extending an equal amount (the weight will simply translate). if not the weight will rotate (slightly) and the off-set moment will be reacted by varying loads in the rods. As the rods see tension from the weight, and some will see compression (-ve tension if you will) from the moment, you need to check that the compression from the moment doesn't exceed the tension from the weight ... if it does then the rod becomes ineffective in both calcs (it can't carry a resultant compression load). I can't see this check in your posts. if the rods have some I, now i see only A, then you could react some compression (as an euler column).

Quando Omni Flunkus Moritati

RE: Mass hanging - equilibrium equations using rod length

Pawel27:
I can’t see your sketches unless they are pdf files, so I’m guessing a bit here, but based on the posts above.... I think you have to find the C.G. of the mass, and write a set of equations which compare its location to the locations of all of the supporting rods. The rods are support springs with a stiffness or elongation of PL/AE and they will carry the load in proportion to their distance from the C.G. and their section properties, mainly ‘A & L’, and they should sum-up to the total weight. But, then a set of additional equations will relate these spring support forces to the stiffness of the mass structure. As if it were an inverted spread footing/a plate (some structure) supported by randomly located springs. It’s a beam or plate supported by springs. The rods are probably not going to change their vertical location very much unless the stiffness of the supported structure allows it to change dimensions in plan by a significant amount. The plate curvature, under load, should probably not cause much dimensional change in plan. You are looking for compatibility of these two systems, the rod’s elongation at each node point, vs. the deflection of the supported structure due to these support loads. Are there temperature affects too?

RE: Mass hanging - equilibrium equations using rod length

Using the coordinate system in your second-posted diagram, the one with Z vertical, and assuming that the deflections are small, the deflection of the top surface of your "box" is entirely vertical and is given as a linear equation
Dz = a + b.X + c.Y
where (X,Y) is the location of the point of interest.

This equation has three unknowns.

You also have three equations of equilibrium:  vertical forces to sum to zero;  moments about the X axis to sum to zero;  and moments about the Y axis to sum to zero.  The only forces acting on the system are the weight of the box and the Z-direction forces in every hanger.  The force in any hanger is proportional to the product of its axial stiffness and the Dz value at its lower end.

So, regardless of how many hangers in your problem you have a linear problem in only three unknowns.  It is a fairly easy task to set up a matrix-based representation of the equilibrium equations, then solve it by inverting the appropriate 3x3 stiffness matrix.  All easily done in VBA, possibly also able to be done without resort to VBA.

This is exactly the approach IDS was recommending, just worded a bit differently.

RE: Mass hanging - equilibrium equations using rod length

i agree with denial ... the three variable are
1) the amount of translation (equal extension of all rods),
2) the amount of tilt due to Mx, and
3) the amount of tilt due to My.

to get the moment terms, you should strictly use the principal axes of the rods (ie a set of areas, calc Ixx, Iyy, and Ixy, calc the principal axes of inertia I11, I22). the moment due to the load is easy to determine, the load X the distance from the load to the centroid of the rods. then this moment will have components in the 11 and 22 directions.

Quando Omni Flunkus Moritati

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