Resulting forces in the Z axis. 1

[Bert2]

Please can someone help me in the solution for calculating the resulting force in the Z axis. attached is the situation (rigging diagram) specifiaclly the top slings are in question. the end elevation shows them to have an angle 65-69deg.interior angle and in the side elevation 72-74deg. interior angle. Basically what is the resulting force acting on the top slings given the two angles their at?

Using a force off 75Te for the object itself.

thanks.

 

[IDS]

Having reviewed the thread, I still don't think that the upper cables are indeterminate.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[zekeman]

"Having reviewed the thread, I still don't think that the upper cables are indeterminate."

4 vector forces passing thru a single point adding to only the z component means 4 unknowns and 3 equations, one for each direction; this makes it indeterminate.

 

[IDS]

zekeman - remove any one of the cables (upper or lower) and you have a mechanism.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

IDS, nice catch with the Te

my answer is the same, near enough 'cause you used the angles provided.

zeke, the upper cables are determinate 'cause the only load to be determined is the lateral load since the lower cable loads define everything else.

 

[zekeman]

You guys are still not getting my point.
Forget the lower half.
If I take this contraption and look a the external loads, I find 750T weight at the mass and 750 T at the upper hook in a straight line.
Now the 4 cable force vectors, whose directions are known, emanating from the upper tie, added vectorially, equal that force vector.

Only 3 equations can get you there, but you have 4 unknown vector forces.

That is the problem.

BTW, I did the problem, minimizing the energy and get answers very close to both of you, but I disagree with any method that does not use minimization of energy as a 4th criterion.

Rb, as a matter of fact you have already shown that by cutting one of the upper ties (there are 2 3tie solutions) you can get a valid solution.
Does that not offer evidence of the indeterminate nature of the problem.

 

[rb1957]

zeke, the lower slings are determinate.
then the upper slings Y- and Z- components are defined for the upper slings (consider the Fz balance at the transition, as an example) and the only thing that changes in the upper slings is that there is an X-component. This is carried by the strut between the two slings at an end, therefore equal and opposite forces are applied. therefore this component can be determined by static equilibrium.

 

[zekeman]

OK,
You finally got me. The error I made is in saying that the upper cable vectors meet only 3 conditions for equilibrium.
In fact the struts preclude that assumption and 2 pairs of vectors are held at angles determined by the strut and cable geometry thus guaranteeing a unique solution.

And BTW, the upper set of tensions are greater than the lower ones by a factor equal to the cosine of the angles between the conmtact pairs.

My solutions are:

Lower:18.2,16.7,20.7,22.6
Upper:20.0,18.4,22.1,24.1
units in Te

 

[IDS]

Not a bad discussion for a "homework" question


On the question of why the upper four cables are determinate when there is a stable condition with only three cables; the three cable stable condition has a different geometry to the four cable stable condition, so if we cut one of the four cables it is not stable until it has undergone a large movement.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

And by the way:

"Te, te
non-standard symbols for the tonne (metric ton), used by some engineers in the UK. The only proper symbol for the tonne is t. If it is necessary to distinguish the tonne from the British Imperial ton, use tn for the British unit. "

http://www.unc.edu/~rowlett/units/dictT.html

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

from wiki (yeah i know) ...
"has a maximum take-off weight of 107.5 metric tons"

so just as kg is used for weight by the general public, maybe the OP meant tonne.f (= 9810N?, but a "valid" force) rather than 1000kg (mass)

 

[Bert2]

IDS; That is the same way i calculate the the lower forces with moments etc just to get an idea of the forces.

RB1957; Yes the angles are rounded. sorry should have made them to one d.p.

Te as i used for reference as one metric tonne (1000kg).

the solutions are interesting ill take my time to read the posts again. Great look from a different point of view with the solutions. thanks.

 

[desertfox]

hi Bert2

My solution as promised, I have only calculated the forces in the top slings using the theory of the link I posted last week.

Regards

desertfox

 

[zekeman]

Well, my take on this method is that itis an energy method that assumes a controlled small vertical displacement in the z direction which necessarily implies an induced moment at that point.

The method indeed conserves energy and vertical force equilibrium but does NOT satisfy horizontal force equilibrium.

As an example take the classical picture frame problem where the frame is supported by a nail at the top with two strings attached to the frame at different angles. You will get incorrect answers using this method, since you cannot satisfy horizontal equilibrium. As an extreme example, let one of the strings be vertical, then by classical methods it should take the ENTIRE weight of the frame and the other string at a an acute would take NO force. By the method above, the tensions would be quite different.

 

[Bert2]

@Zekeman;

Would you neglect the above method used even if just for an indication of the forces?

 

[desertfox]

zekeman

Just post your solution with full workings out, then we can see where I am incorrect.


desertfox

 

[zekeman]

"Would you neglect the above method used even if just for an indication of the forces?"

I would, if it couldn't pass muster like handling the simple picture frame problem I posed. How much confidence could you have in a solution that gives a different answer to a simple problem whose solution is well known.



"Just post your solution with full workings out, then we can see where I am incorrect."

Desertfox,

All I have to show is that it leads to incorrect results for a simple well-known problem. I gave a good example, but if that is not enough, I will show a more concrete simple example subsequently.

I would hope to see someone else weigh in with their view of this. Rb1957,IDS, where are you?

 

[desertfox]

zekeman

All you need to show is your solution to the original problem and how it balances out, no need to go to the trouble of calculating fresh examples, why not post the workings out for your answers you gave.
The solution is so well known that everyone as different answers, in addition my answers are as close as any others here, however I cannot tell from your post which tension is in which cable which if you post your workings and a diagram I can check the equilibruim for myself.

 

[IDS]

Regarding the picture frame problem - I was just about to point out that if one string was vertical then the other string would have zero force, then I re-read zekeman's post and realised that was exactly his point.

I'll have another look at the original problem and see if I have anything to add.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[desertfox]

Hi Bert2

Just comparing my results with that of IDS the % difference's are as follows:-

for force 1 = 3.7%

force 2 = 9.4%

force 3 = 2.1%

force 4 = 10.45%

Given that these sets of results have been calculated by two different methods I would say there very good bearing in mind I calculated the angles and geometry of the top slings and used the method given in the link.
I'll standby my answers for the forces in the slings as an estimate, as all the other answers are just that, having calculated any forces for lifting devices, one automatically puts a good safety factor onto them.

 

[IDS]

Desertfox's calc contains the assumption that the vertical deflections at each corner are equal, which is not correct. It doesn't make a huge difference in this case, but it might in other cases.

I have re-done my calculation assuming the given dimensions and the larger of the two angles at each level are exact (see attached spreadsheet for detailed calcs and diagram).

I now get:
Resultant Force, Strand7 Results, Convert to tonne-force
Fi = Vi/sin(NIP) 235.91 kN, 235.91 kN, 24.07 t-force
Fj = Vj/sin(NJP) 216.64 kN, 216.64 kN, 22.11 t-force
Fk = Vk/sin(NKP) 180.06 kN, 180.06 kN, 18.37 t-force
Fl = Vl/sin(NLP) 196.16 kN, 196.16 kN, 20.02 t-force

The forces are now in exact agreement with the FEA results (with corrected geometry), and very close to zekeman's.

As for the best way to do it in practice, I'd suggest doing the FEA or frame analysis (which apart from anything else is the easiest way to get the geometry), then do a "hand" (i.e. spreadsheet) calc to make sure it is in equilibrium.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/