Mass hanging - equilibrium equations using rod length

[Pawel27]

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.

 

[GregLocock]

That's a nice enough drawing but either the answer is so simple you should go back to school, or you haven't described the problem properly.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[ALREMAN]

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.

 

[Pawel27]

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

 

[rb1957]

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

 

[handleman]

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)

 

[Pawel27]

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

 

[Pawel27]

I must change picture when we consider code above.

 

[Pawel27]

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

 

[desertfox]

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

 

[IDS]

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/

 

[Pawel27]

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.

 

[handleman]

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)

 

[rb1957]

"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

 

[GregLocock]

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


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[IDS]

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/

 

[Pawel27]

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.

 

[Pawel27]

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.

 

[desertfox]

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/9789...011eb8/example-problems-on-static-equilibrium

 

[IDS]

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/