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Resulting forces in the Z axis.

Resulting forces in the Z axis.

Resulting forces in the Z axis.

(OP)
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.

RE: Resulting forces in the Z axis.

These are easy to figure out but I am wondering why would someone in the Mechanical forum would ask this type of high school problem? I would be willing to give you answers only if I am convince that these are not school problems.  My answers  would be in general terms.  So, in what direction is the Z axis.

RE: Resulting forces in the Z axis.

i think the OP means that the slings are inclined in both directions, ie not in the X-Y or Y-Z planes.

I agree the question is PDB (PrettyDamnBasic) ...
let the load in the sling = P
[do all the slings have the same load ?]
calc components of P in the 3 axes (yes?)
sum the (presumably)Z components to equal the applied load
solve of P,
and it's components.

RE: Resulting forces in the Z axis.

I can't believe it's not a student enquiry

RE: Resulting forces in the Z axis.

I'm pretty sure he's not a student, based on his other posts and replies.  More probably, this wasn't his specialty area and he's forgotten.

Patricia Lougheed

******

Please see FAQ731-376: Eng-Tips.com Forum Policies: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

RE: Resulting forces in the Z axis.

loking at it a 2nd time, it's more complicated than i originally thought, 'cause each dim'n is different (ie each sling will have a different load).

you've got 4 unknowns (the load in each sling) and 6 equations of equilibrium, so it does solve.  it is possible that the solution may be impractical, if one of the slings is required to carry a -ve load.

an important practical point, the CG of the load needs to be on the line of action of the lift.

RE: Resulting forces in the Z axis.

(OP)
@ rb1957 yes inclined in both planes.

i understand the basics but am i making this problem harded than it is? i guess i am ....  

RE: Resulting forces in the Z axis.

(OP)
@ rb1957 correct each sling has a different length and the CoG is not centered about the rigging... only to the lift itself.

i dont understand how one sling may carry a -ve load?

RE: Resulting forces in the Z axis.

In practice, with the load c/g not centered under the lifting attachments, you will not be able to rig in this fashion to obtain a completely level load; it would require 8 slings of precisely measured and different lengths.

Because riggers carry sets of matched length slings, the load will go off level to get the c/g under the line of action.  The riggers will often insert some shackles in the slings on the high side to roughly even out the load.

If the load has to be maintained level, the riggers will likely use a different, and more adjustable rigging.

RE: Resulting forces in the Z axis.

I can not open files due to company protocal, but have you done a FBD?  

Tobalcane
"If you avoid failure, you also avoid success."
"Luck is where preparation meets opportunity"  

RE: Resulting forces in the Z axis.

You can even use the cut and sum method to figure out the loads in each strap...

Tobalcane
"If you avoid failure, you also avoid success."
"Luck is where preparation meets opportunity"  

RE: Resulting forces in the Z axis.

i think there's a problem with the angles, 74deg and 72deg don't balance in Fx.  74deg requires the 72deg to be 72.5deg.

it was a bigger difference when i started, got smaller as i worked on it some more ;)

i (now) think the problem is redundant (the lower portion has 4 unknowns and only three equations).  You can still solve this with other methods (i like the unit force method, but there are probably other methods better suited to this problem).

the slings can't carry -ve load, that's what i meant by "impractical solution" ... the math will quite happily put a -ve load on a sling, but you can't in real life.
 

 

RE: Resulting forces in the Z axis.

yeah, this is a redundant problem.  thinking (a little) about it, maybe an easy way to solve it is to remove each of the slings in turn, solving the remaining three (easy to do, don't worry if any sling loads are -ve), then combining the 4 different solutions (sum/4); but now -ve sling loads is a problem.

RE: Resulting forces in the Z axis.

yeah, that approach gave me reasonable numbers pretty quick.  now you know the Z component at each corner, the upper sling loads can by found easily-ish.

btw, your other angles are a little off ... add a decimal place or two.

RE: Resulting forces in the Z axis.

I don't understand


















Unless I am missing something this problem looks trivially simple. So please correct me if my simple analysis doesn't stack up.

First, look at the bottom set of straps. You can easily draw a vector diagram from the left side view forcing the  the vertical resultant to 1/2 the load. The length of the vector would be in true length.

Now for the upper region, you can do the same, but the vectors are not not shown in true form so they would  have to be rotated to get the answer.

If I can do it graphically, it can be done analytically, easily.














 

RE: Resulting forces in the Z axis.

yeah, i originally though the slings were inclined to the weight, and set up a s/sheet along those lines.  then i noticed the bttm slings have no Y component (inclined in the X-Z plane.  this relates Px1 and Px4, and Px2 and Px3 [figure out why ... hint, sum Mz)  

therefore Pz1 is relatd to Pz4, and Pz3 to Pz2 [geometry].  then sum moments in the elevation plane and the end elevation, and it should be statically determinate.

RE: Resulting forces in the Z axis.

"yeah, this is a redundant problem.  thinking (a little) about it, maybe an easy way to solve it is to remove each of the slings in turn, solving the remaining three (easy to do, don't worry if any sling loads are -ve), then combining the 4 different solutions (sum/4); but now -ve sling loads is a problem."

While this approach may be mathematically correct, it doesn't square with the physics. Of the 3 tri-vectors,I find that only one has 3 tensions and therefore it is the ONLY solution compatible with the physics and the mathematics.

When I look at the 3 equations with 4 unknowns,I call one unknown 0 and solve for the others. If you do this 4 times each with a different choice, you will get only one answer that squares with the   physics, namely three positive (tensions) values. I submit that that is the answer.

As for the lower support, the problem is mathematically tenable, since there are enough equations to match the unknowns.
 

RE: Resulting forces in the Z axis.

all i meant was it is possible for the math to give an impractical answer ... i didn't say it did.

having said that i noticed that one of trhe zero cases gave a negative load, but this is ok 'cause when you avaerage the different cases all the sling loads are positive.

Having said that, the lower portion solves easily, and this provided the Fz (& Fx) reactions for the upper portion, so all you need is the Fy component, again this solves easily.

 

RE: Resulting forces in the Z axis.

"having said that i noticed that one of trhe zero cases gave a negative load, but this is ok 'cause when you avaerage the different cases all the sling loads are positive."

That's my point. It is NOT ok. You have no license to average since it will give meaningless answers.

Any 4x3 linear set can have an arbitrary "solution set" by assigning an arbitrarily value to one unknown.
Does that make the answer correct?
 

RE: Resulting forces in the Z axis.

Hi Bert2

Does the part your lifting stay horizontal if you have the slings as shown on your diagram? if so I may have a solution.

regards

desertfox

RE: Resulting forces in the Z axis.

apart from being irrelevant ('cause the problem isn't redundant afteral), i think averaging the four cases with only three slings effective is a reasonable solution to the rudundant problem ... i thought it mirrored the unit force method.

RE: Resulting forces in the Z axis.

(OP)
@ Dersertfox yes the object will stay horizontal when lifted in the design of the slings used.

 

RE: Resulting forces in the Z axis.

hi Bert2

The only way to solve this I think is by using compatibility of deformation, so you need to know about the sling material elasticity and their cross sectional area's, have a look at this example, whilst I realise its only a 2D example I thinks its the way to solve your problem.
I have managed to calculate from your post the individual angles coming from the single hook down using trig but I am unsure about the two views you uploaded as to whether there drawn in first or 3rd angle projection, also what is the horizontal distance between the top slings in the side elevation? I have managed to calculate these using the 72 and 74 degree angles.

on the link go to page 49 example 2.9

http://books.google.co.uk/books?id=vwdApH7qneAC&pg=PA38&lpg=PA38&dq=statically+
indeterminate+force+systems&source=bl&ots=Gxv5jW8GmL&
sig=RHqYvYpXWUMfv6_pAzu2l27q47I&hl=en&ei=mZE8TYzMGdC
whQewytmNCg&sa=X&oi=book_result&ct=result&resnum=1&sqi
=2&ved=0CBYQ6AEwAA#v=onepage&q=statically%20
indeterminate%20
force%20systems&f=false

desertfox

RE: Resulting forces in the Z axis.

"........ i think averaging the four cases with only three slings effective is a reasonable solution to the rudundant problem ... i thought it mirrored the unit force method. "


Hate to belabor the point, but, in general, , maybe a more reasonable solution would be to minimize the energy stored in the slings.
So of the infinity of solutions to the 4X3 set, only one minimizes the energy stored.

RE: Resulting forces in the Z axis.

ok [waving white flag] ...

but we agree that the lower portion is statically determinate ('cause the slings are inclined in one plane only) ...

surprised that 72 and 74 deg worked, i needed 72.5 deg to balance the x components ...

RE: Resulting forces in the Z axis.

"but we agree that the lower portion is statically determinate ('cause the slings are inclined in one plane only) ..."

I'll give you that one.

RE: Resulting forces in the Z axis.

hi Bert2

Further to my last post I have calculated the tension in each leg using the theory above these are the magnitude's :-

      20.8559 Te
      22.15143 Te
      21.7367 Te
      20.4915 Te

I have checked these obtain equilibrium and they do to within 0.5 Te due to rounding off.
I will post a full solution on friday when I get to my main compiutor.

desertfox
       

RE: Resulting forces in the Z axis.

(OP)
Thanks dessert fox.  

RE: Resulting forces in the Z axis.

is tan74*3687 = tan72*4044 ?
tan74*3687 = tan72.54*4044

my lower reactions (based on 74 and 72.54) are ...
P1 = 20.6Te ... P1z = 19.86Te
P2 = 25.6Te ... P2z = 24.61Te
P3 = 23.5Te ... P3z = 22.43Te
P4 = 19.0Te ... P4z = 18.10Te

P1 and P2 are on the 74 deg slings
P3 and P4 are on the 72.54deg slings
P1 and P4 are on the 65 deg slings
P2 and P3 are on the 69 deg slings

RE: Resulting forces in the Z axis.

(OP)
@rb1957

did you use the same method as desertfox?

thanks for the effort looking into this.  

RE: Resulting forces in the Z axis.

static equilibrium equations

x- and z- compoents are related to one another

then sum Mz about pt3 ... relates P1x and P4x
similarly you can relate P3x and P2x
[i'm not going to give you the whole answer, just enough to help you ...]
then these allow you to relate P1z to P4z (and P3z to P2z)

then, using you EE view, sum Mx about pt2 relates P1z and P4z to the applied load, so you can now determine P1z (and P4z).  similarly sum Mx about pt1 gives you P2z and P3z.

btw, any comment on my posts about angles ?
 

RE: Resulting forces in the Z axis.

For the bottom set,I get:

182,167,207,226

RE: Resulting forces in the Z axis.

that's good for 85 Te ??

RE: Resulting forces in the Z axis.

oops, the lift is 75 Te, isn't it ? (so factor my posts accordingly)

RE: Resulting forces in the Z axis.

"182,167,207,226 "

Correction by 10 factor.
Should read:

18.2,16.7,20.7,22.6  

RE: Resulting forces in the Z axis.

Assuming Te is intended to indicate the mass of the load in tonnes, then kN would be the appropriate unit for the force in the cables.

I think the whole thing has been over complicated:

Take moments to get the distribution of vertcical forces along one axis.
Take moments in the perpendicular direction to get the distribution of these two forces, giving you a vertical force at each corner.
Resolve along the cable to get the axial force in the cable.

Load Mass 75t            
Total vertical force 735 kN        
Distribution along long axis 384, 351 kN    
Distribution along short axis 213, 172, 194, 157 kN
Resolve along cable 224, 179, 204, 163 kN


(My FEA program gives the same answer).

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

RE: Resulting forces in the Z axis.

And for the top cables:

Resolve along  top cable 240, 197, 218, 180 kN
 

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

RE: Resulting forces in the Z axis.

"And for the top cables:

Resolve along  top cable 240, 197, 218, 180 kN"

But the top cables are indeterminate. How does your FEA do it?

 

RE: Resulting forces in the Z axis.

Quote:

But the top cables are indeterminate.

I don't think that they are (but I suppose I'd better read the thread before getting definite about that)

 

Quote:

How does your FEA do it?


The same way it does any other problem, determinate or indeterminate, by solving for equilibrium and strain compatibility.  I just put the problem in as shown in the sketch and got the same answers as the the simple analysis (+- a kN or two, probably due to the angles being rounded to two figures).

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

RE: Resulting forces in the Z axis.

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

 

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

RE: Resulting forces in the Z axis.

"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.

RE: Resulting forces in the Z axis.

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

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

RE: Resulting forces in the Z axis.

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.

RE: Resulting forces in the Z axis.

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.

RE: Resulting forces in the Z axis.

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.

RE: Resulting forces in the Z axis.

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



   

RE: Resulting forces in the Z axis.

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/
 

RE: Resulting forces in the Z axis.

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/
 

RE: Resulting forces in the Z axis.

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)  

RE: Resulting forces in the Z axis.

(OP)
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.

 

RE: Resulting forces in the Z axis.

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.

RE: Resulting forces in the Z axis.

(OP)
@Zekeman;

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

RE: Resulting forces in the Z axis.

zekeman

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


desertfox

RE: Resulting forces in the Z axis.

"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?

 

RE: Resulting forces in the Z axis.

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.

RE: Resulting forces in the Z axis.

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/
 

RE: Resulting forces in the Z axis.

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.  

RE: Resulting forces in the Z axis.

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/
 

RE: Resulting forces in the Z axis.

Hi IDS

If the vertical deflections at each corner are not the same then the component being lifted cannot remain horizontal, which Bert2 stated it did remain horizontal during lift.
Good spreadsheet though.

desertfox

RE: Resulting forces in the Z axis.

Quote:

If the vertical deflections at each corner are not the same then the component being lifted cannot remain horizontal, which Bert2 stated it did remain horizontal during lift.

The deflections are small, so the load will stay very close to horizontal.

The only way the deflections would be the same is if the sling stiffness was proportional to the vertical load, but that wouldn't change the loads in each sling, because it is statically determinate.

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

RE: Resulting forces in the Z axis.

desertfox, how do you calc displacements when we don't know the sling areas ? (assume each sling the same ?)

RE: Resulting forces in the Z axis.

Hi rb1957

Yes the idea is that all the slings have the same area and modulus.

desertfox

RE: Resulting forces in the Z axis.

Here is a comparison of the 2 methods for the simple picture frame problem.
It is clear that for some problems, like this, the difference is huge.

RE: Resulting forces in the Z axis.

this here is one flogged dead horse !

the problem is statically determinate, so you don't need energy methods to solve it.

please, done and dusted, no?

RE: Resulting forces in the Z axis.

(OP)
@zekeman / rb1957

huge thanks for the time looking into this and the two different solutions.

If i was looking at it in a general case i would assume zero deflection in the slings.

to get a guide on the loads applied.

you could take it to the next level as zekeman has shown but in my case i dont need such an in depth approach-good even though it is.

RE: Resulting forces in the Z axis.

no, i wouldn't make any assumption about deflections ... the problem as posed is statically determinate.

notice though if you didn't have the parallel slings the problem would be indeterminate and you'd have to include deflections as a criteria for developing your loads.

the important thing to take from this is drawing a free body diagram will take you a long way towards solving the problem.

RE: Resulting forces in the Z axis.

My 2c worth. Posts 2 and 3 got it right. The problem is simple and can be solved from the bottom up summing forces only and nothing harder than solving 2 simultaneous equations in 2 unknowns.

The problem is definitely statically determinate even if the object being lifted is rigid and the slings are inelastic. If any of the sling lengths was to be changed slightly the geometry of the problem would change ie the sling lengths have been adjusted to produce the geometry shown (lower slings vertical in end view.)

Engineering is the art of creating things you need, from things you can get.

RE: Resulting forces in the Z axis.

"My 2c worth. Posts 2 and 3 got it right. The problem is simple and can be solved from the bottom up summing forces only and nothing harder than solving 2 simultaneous equations in 2 unknowns."

In hindsight you are probably right, but, save for the special geometry of the bottom slings, you would be looking at a messy problem. So what happened, on first blush the problem looks easy. On second blush, maybe indeterminate, and finally got it right on the 3rd look.

Personally, my performance was pretty shoddy, especially after everyone was screaming that the problem was determinate. I was hung up on the upper sling arrangement and disregarded the link connections to the lower half. After doing problems for over 50 years, I should know better.

In any event the thread was probably useful as an exercise in statics and determinate and indeterminate structures.  

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