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Lifting Device Problem 1

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purplemonkey

Automotive
May 22, 2007
31
Hello everyone,

I am currently in the process of designing a lifting device for a rear axle. I have attached a file of the proposed design. The load carried is 13,000 lbs. The lifting device picks up the rear axle from the C.G.


Forgive the fundemental nature of this problem however I am a little confused about how to calculate the forces at the weld joint. If I take into consideration "View A", the forces acting on the weld are due to vertical shear and bending which gives me a resultant that does not hit my FOS (4) for the weld. If I take into consideration "View B", the forces acting on the same weld, are due to twisting and vertical shear, which now meet my required factor of safety.

This device is in static equilibrium so shouldn't the forces balance regardless of what direction I am taking my calculation from? What am I doing wrong here and am I complicating this or missing a key fundemental concept all together? Any and all help is greatly appreciated!
 
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you need to balance all the forces. looking on different planes shows you different things, but they're acting at the same time.

if you have to do something silly, like picking up the beam off-set from the load (don't over react, we all do silly things sometimes for the best of reasons; in your case it looks as though the red lug behind the beam is preventing you from being sensible) you have to consider how the beam will react (to the off-set).

where's the CG of the lift ?

just looking at the beam in isolation, there's an off-set between the load and the reactions. maybe you're taking this out at the lift point, between the two lug fittings at the top of the brkt.

i think the only way for the beam to be in equilibirium is for the welds to the beam are only reacting the shear loads applied to the beam, and the plate is having to deal with the torque due to the off-set.

but in the real world, you'll have to worry how the lift is going to move, it'll want to align the lift to the weight (lift over the CG). i don't think it works as shown (with a single lift off-set from the loads). if you have a single lift force, then you'll need to take the torque out at the attachment to the load.
 
Thanks for the quick response. To clarify the attached picture:

1) Disregard the red lug behind the beam. The pick up point text refers to a chain attached to the various holes in the yellow plate (reference View B)

2) The C.G. of the rear axle is directly in line with the middle hole of the yellow plate, of the first hole pattern (if that makes any sense) as shown in View B.

As per your reply:

"if you have a single lift force, then you'll need to take the torque out at the attachment to the load." Could you please clarify?

My calculations for reference are as follows:

As per View A

Forces acting on weld: shear + bending

Load = 13000 lbs
Reaction due to Load = 13000/2 = 6500 lbs
Length to weld from reaction = 43"

Area of Weld = Aw = 7+4+7+4 = 22 in^2
Section modulus = Sw = 44.33

Vertical Shear = P/Aw = 6500/22 = 295.45
Bending = M/Sw = 6500(43)/44.33 = 6304.98 <--- limiting force

-----------

As per View B

Load = 13000 lbs
Reaction due to Load = 13000/2 = 6500 lbs
Length to weld from reaction = 9.12

Area of Weld = Aw = 7+4+7+4 = 22 in^2
Section modulus for Twisting = Jw = 221.83

Vertical Shear = P/Aw = 6500/22 = 295.45
Twisting Horizontal = 534
Twisting Vertical = 935

All calcs performed treating weld as lines (formulas omitted in View B). as you can see the forces are far less considering twisting over bending.
 
the lift force needs to be aligned to the weight load. maybe you have a helper line to take out a small imbalance. if not the load will rotate to align them

the liftng beam pick-ups should be aligned to the same line of action. if they aren't they need to react the offset moment (ie you can't use a single point pick-up (from the beam to the load as you are); Unless there is another lifting point, then you can balance the load between the two lifts, yes?

draw your free body, lift and load, first. then add the lifting beam. i think it'll work out that the beam will balance the lift force with an off-set force and a moment at the interface to the load. then i think it is clear that view B is giving you a better picture of the loads on the welds of the brkt. if you have a point force lift, then the reactions to the beam will be a shear force and a moment. this moment needs to be reacted by the load, consider a free body of the beam loaded by a force and a moment, the reactions must be a force and a moment (if the beam is straight and the load is aligned with the reactions); clear as mud.

another way to react the off-set is to have two lifting forces picking up the brkt, reacting the offset. one force will be adding to the load, and the other will be something like double the weight of the load. i don't see this as a preactical arrangement.

 
looking at your pic, i'd suggest picking up the bore (the way you are), and picking up the two lugs, fwd and aft of the bore, to provide a stable lift.

But you still need to lift above the CG (and probably you are), just stating the obvious.
 
I am constrained to my lift points since the purpose of the lifting device is to lift the rear axle (red) out of it's fit up fixture. The lift is a 1 point lift with a supporting chain that will be connected to the nose of the rear axle via a swivel hoist ring in order to raise or lower the nose if required. I omitted this in calculation however since that chain will essentially not support any of the load if not in use.
 
then draw the free body ... you've got the load reacted by the beam, off-set from the CG, and the beam reacted at the lift point (above the load).

because the beam reactions are off-set from the load, they have to carry moment, torque on the beam. you need to pick up the load at two points at each end to carry this moment.

if you pick it up as shown, when you lift the load will rotate to put the CG the lift (chain) and the beam pick-ups along the same line of action.
 
How many of these rear axles are you going to lift, that you would make such a complicated little lifting beam? Is it a good idea to be hooking and lifting from a machined end and inside of that housing? Why not use a three legged sling; one hook and clevis to the hole in the red left torque arm; and two longer chain slings which wrap around the outside of the housing near its round ends and back onto themselves; all up to a pear shaped link for lifting. Cable slings could replace the chains. The C.G. of the housing is apparently 9.12" left of the center of the housing in your view “B”. Why not roll the housing clockwise 15-20° to get the C.G. and centerline to line up. Then take an appropriately sized steel beam with a wood block on the top flange to mate with the inside of the housing; this beam goes through the housing, and is longer than the housing, and you lift from each end of the beam, near the end of the housing. Why not make a Tee shaped lifting frame, with hooks at the ends as you sketch shows. But, then provide a Tee stem which goes left to be over the hole in the red torque arm. Provide a short link (turnbuckle) and clevis down to the hole, and a lifting eye off the top of the Tee stem.

That little lifting beam you show would not be that difficult to design and build, but do you need that complicated lifting beam? You need to know some Statics and Strength of Materials, Structural Engineering and some welding design, and you seem to be having troubles with the fundamentals of the first of these. So what happens with the actual stresses and welds? You place a torque of (13k)(9.12") on the lifting beam, and that will have to be taken out as a reaction at your two end housing hook points. You place a bending moment of (13k +)(43" x 2)/4 on the lifting beam, with 6.5k reactions at the end hooks. Is the axle housing (43" x 2) wide? Until you have resolved these design considerations, the exact weld size in not that important. How have you designed the main beam, the sliding pinned end members and the hooks, pins and lifting eye? CAD drawings are wonderful, but they don’t do your design and stress analysis, and member sizing for you. OSHA will want you to put some locking feature on all of your lifting hooks and lift points, for safety reasons.
 
Purplemonkey, this is not a critcism of your analysis on the weld, however,I think that the stress analysis on the welds should be based on the torsional stress formula and not the bending stress formula; then the resulant stress will be based on the shear stress and maximum torsional stress. You'll need to identify the highest torsional stress point to calculate the resulant stress. You also have to figure out the polar moment of inertia, the center of gravity and the external moment or torque on the weld layout which I imagine will consist of top and bottom horizontal welds and vertical welds on the box beam.
 
Thank you everyone for your responses. Everything was noted and I have been banging my head against a wall ever since rb1957 responded. I, for the life of me, can't understand why I failed to omit the moment on the FBD when I initially started solving this problem and further more, why I assumed a reaction moment to counter act the moment being created during the lift in the first place.

Regardless of weld strength the design doesn't work as I intended. I'd like to say thank you to everyone for their help, but I think I wasted more of everyone's time than anything! [thumbsdown]
 
purplemonkey

time was not wasted. Brain storming is of great importance.
I like deng idea. I would suggest following his suggestion, & as constructive advice.
a more positive lock down & a beefer design of the lifting device is prudent.

Take Care
Mfgenggear
 
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