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.

 

[desertfox]

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

 

[rb1957]

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.

 

[Bert2]

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

 

[desertfox]

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

 

[zekeman]

"........ 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.

 

[rb1957]

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 ...

 

[zekeman]

"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.

 

[desertfox]

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

 

[Bert2]

Thanks dessert fox.

 

[rb1957]

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

 

[Bert2]

@rb1957

did you use the same method as desertfox?

thanks for the effort looking into this.

 

[rb1957]

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 ?

 

[zekeman]

For the bottom set,I get:

182,167,207,226

 

[rb1957]

that's good for 85 Te ??

 

[rb1957]

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

 

[zekeman]

"182,167,207,226 "

Correction by 10 factor.
Should read:

18.2,16.7,20.7,22.6

 

[IDS]

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/

 

[IDS]

And for the top cables:

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


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[zekeman]

"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?

 

[IDS]

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)

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/