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Loads in the slings and calculate the reactions at bottom 1
- Thread starter StrP88
- Start date
- Thread starter
- #2
MikeHalloran
Mechanical
You have to know where the CG of the load is, relative to the lift points on the load.
Then you can figure individual sling loads.
Search on 'BTH' calculations. [below the hook]
Mike Halloran
Pembroke Pines, FL, USA
Then you can figure individual sling loads.
Search on 'BTH' calculations. [below the hook]
Mike Halloran
Pembroke Pines, FL, USA
you have a weight supported by three equal slings ?
Can you do a force resultant ? The vertical component of each sling will be 1/3 of the applied load. From this calculating the load in the sling is simple ...
load in sling = weight/3*(length of sling)/(height of sling)
another day in paradise, or is paradise one day closer ?
Can you do a force resultant ? The vertical component of each sling will be 1/3 of the applied load. From this calculating the load in the sling is simple ...
load in sling = weight/3*(length of sling)/(height of sling)
another day in paradise, or is paradise one day closer ?
- Thread starter
- #6
rb1957 (Aerospace)
Thank you for your response
Yes I agree the vertical compontent of sling force is 1/3 of the 12.07k (4.03 kips) but how to find the two other components in the horizontal plane
My bad that my geometry is rusty - I appreciate your guidance and help
Thank you for your response
Yes I agree the vertical compontent of sling force is 1/3 of the 12.07k (4.03 kips) but how to find the two other components in the horizontal plane
My bad that my geometry is rusty - I appreciate your guidance and help
This isn't even a college level problem, should be able to do this with highschool level mathematics.
Your sling lengths are some fixed length, start there. You will know the horizontal distance (plan view) from each sling to the CG which is where the hook on the crane lines up. You can solve for the tension using trigonometry or similar triangles at that point. If you don't know how to do it, google it.
Your sling lengths are some fixed length, start there. You will know the horizontal distance (plan view) from each sling to the CG which is where the hook on the crane lines up. You can solve for the tension using trigonometry or similar triangles at that point. If you don't know how to do it, google it.
MikeHalloran
Mechanical
The 1/3 assumption is only supportable if the CG of the load is equidistant from the three lifting points in plan view.
Real loads are rarely symmetrical.
Mike Halloran
Pembroke Pines, FL, USA
Real loads are rarely symmetrical.
Mike Halloran
Pembroke Pines, FL, USA
"Can you do a force resultant ?" ... I guess the answer to that is "no"
read up on force components, forces as vectors.
each sling is reacting (approximately) 1/3rd of the weight. This is the vertical component of the sling load (inclined to the vertical). the length of the sling is ... L = sqrt(X^2+Y^2+Z^2), or if the sling points are on a PCD, then L = sqrt(R^2+Z^2).
then the sling reaction is W/3*L/Z.
another day in paradise, or is paradise one day closer ?
read up on force components, forces as vectors.
each sling is reacting (approximately) 1/3rd of the weight. This is the vertical component of the sling load (inclined to the vertical). the length of the sling is ... L = sqrt(X^2+Y^2+Z^2), or if the sling points are on a PCD, then L = sqrt(R^2+Z^2).
then the sling reaction is W/3*L/Z.
another day in paradise, or is paradise one day closer ?
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1
- #13
StrP88:
The way these questions are being asked suggests that this would be a very risky lifting operation. You don’t need more software, but you do need a few pounds of basic engineering knowledge and experience, or you shouldn’t be messin around with this project.
You can do all the.... “I am wondering ifs” in the world, but if you can’t do a little basic engineering thinking you shouldn’t be messing around with this engineering assignment. Don’t you have a boss or senior engineer who you can talk this over with. That local and personal interaction on such learning exercises is much better than trying to get the answers from someone unknown, over the internet. You can actually be looking at the same plans, draw a few sketches explaining how things work, etc. What you are doing is really scarey engineering.
The way these questions are being asked suggests that this would be a very risky lifting operation. You don’t need more software, but you do need a few pounds of basic engineering knowledge and experience, or you shouldn’t be messin around with this project.
You can do all the.... “I am wondering ifs” in the world, but if you can’t do a little basic engineering thinking you shouldn’t be messing around with this engineering assignment. Don’t you have a boss or senior engineer who you can talk this over with. That local and personal interaction on such learning exercises is much better than trying to get the answers from someone unknown, over the internet. You can actually be looking at the same plans, draw a few sketches explaining how things work, etc. What you are doing is really scarey engineering.
- Thread starter
- #14
BAretired (Structural)
Thank you very much ffor your kind help rather than criticizing
May I kindly ask since you had the good intention to help and to really make me understand how u got the numbers for the horizontal force dram the dimentions and angles that you used when derived 0.118W
If is not too much asking if possibly can draw a stekch and dimensionsand angles I would tremendously appreciate it
Sincerely,
StrP88
Thank you very much ffor your kind help rather than criticizing
May I kindly ask since you had the good intention to help and to really make me understand how u got the numbers for the horizontal force dram the dimentions and angles that you used when derived 0.118W
If is not too much asking if possibly can draw a stekch and dimensionsand angles I would tremendously appreciate it
Sincerely,
StrP88
- Thread starter
- #15
Here you go
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1464876011/tips/201606020851_bw0rjo.pdf[/url]
vertical base reactions: 1/3(12.07) = 4.03 k
horizontal base reactions: (4.27^2-4.03^2)^0.5 = 1.42 k
NOTE: these forces are assuming this object is static and is symmetrically loaded at the base. It doesn't account for the forces seen if it sways, as it's hoisted, or impact (stopped suddenly while being lowered).
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1464876011/tips/201606020851_bw0rjo.pdf[/url]
vertical base reactions: 1/3(12.07) = 4.03 k
horizontal base reactions: (4.27^2-4.03^2)^0.5 = 1.42 k
NOTE: these forces are assuming this object is static and is symmetrically loaded at the base. It doesn't account for the forces seen if it sways, as it's hoisted, or impact (stopped suddenly while being lowered).
- Thread starter
- #17
AnimusVox (Structural),
Thank you very much, following your calc I figured out where I am making error.
Again thank you very much for taking your time and scaning the cals.
I now believe there are good people willing to help around.
Thank you very much
Thank you very much, following your calc I figured out where I am making error.
Again thank you very much for taking your time and scaning the cals.
I now believe there are good people willing to help around.
Thank you very much
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