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.
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
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Mass hanging - equilibrium equations using rod length
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
RE: Mass hanging - equilibrium equations using rod length
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
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
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
RE: Mass hanging - equilibrium equations using rod length
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
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
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
RE: Mass hanging - equilibrium equations using rod length
-handleman, CSWP (The new, easy test)
RE: Mass hanging - equilibrium equations using rod length
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
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
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Mass hanging - equilibrium equations using rod length
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
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
RE: Mass hanging - equilibrium equations using rod length
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
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
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
RE: Mass hanging - equilibrium equations using rod length
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
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
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
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