I need help with a lifting ring calc/research

[longude17]

Hi everyone, I have just started a new job as an entry level mech eng working with steam turbines and steam compressors. Today I was asked to verify a proposed design for a two-pieced lifting collar / ring.

I have no clue on where to begin to analyze this and see if it will with stand the 7000lb that it will be lifting...so any help on how you would do the calcs would be greatly appreciated

ok heres the jist of it....it is a two piece collared lifting ring that will be placed around a machined notch on the rotor shaft. (picture a circular donut 1 in thick, 6 inches o.d.) the two pieces when placed around the shaft will be bolted together using a drilled and tapped 1/2 inch hex head bolt....90 degrees to the bolts is two bolted on swivel lifting trunions where the lifting rigging will attach.

I know the fail points will be either at the lifting trunions, or the bolts, prob is its been so long I dont know how to do the calcs for this.

how would you do the calcs to see if this thing is safe to use?

I will attach pics and a shop drawing if needed.

Thanks again

 

[desertfox]

Hi longude

Will the bolt axis that fasten the collar be perpendicular to the shaft lift? if so the bolts will be in shear, if the axis is parallel then bolts are in tension, maybe a sketch would help.

desertfox

 

[dooron]

Interesting problem, especially since the Rotor shaft offers support in bending/buckling to the Ring.
I would break down the problem to a number of different failure modes and tasks.
1. Draw your lifting force on the trunnion pins-assume 60%-40% i.e. 60% on one side.
2. Assume the load is applied at the pin outer end-therefore max bending moment on the pin. then look at the ring and pin subject to this bending moment and shear. Combined Vom Mises stress, to be less then yield.
By the way-give a safety factor of 3 on yield, or multiply the load by 3, and design to Limit State.
Now that you looked at the beam and pin in bending and shear,
2. do a similar thing with the bolts and flange at the bolt, find the forces and analysed bending shear and combined. At the bolt allow some additional forces for prying action. becuase all your elements, pins/flanges are short beams Use Roark Formulas for stress and strain, especially the section on short beams.Hope this helps

 

[dhengr]

Like desertfox I would like to see some good pics. and sketches so we all know what you’re talking about, because your word-picture leaves way too much to the imagination on something I can’t see from your vantage point.
Dooron is probably on the right tract, but he’s got long distance x-ray vision and the rest of us don’t.

 

[Twoballcane]

If you have or can get a Shigley's Machine Design book, all you need is there...

Tobalcane
"If you avoid failure, you also avoid success."

 

[chicopee]

If these lifting collar/ring are commercially available get the manufacturer's or distributor catalog which should show you the information needed.
Another important piece of info is the sling configuration which may impose more tension on the collar/rings if the leg spread between end loops is too far apart.

 

[chicopee]

show us a sketch of the lifting collar/ring and how it fits on the swivel trunion

 

[longude17]

sorry for the delay , here is a pic of it, and here is the sketch given to me from the machinist.

I ran the calc on the bolts with a sf of 3 and it failed with a 5,000lb load (15,000 lb load after s.f) the material is 1018 steel with grade 8 bolts

 

[longude17]

here is another pic of the sketch.

How would each of you do the calc? I need help on this one as i dont have any software at work to help.

discriptions, even sketchs and equations would be helpful at this point. Im VERY resistant to let the machinst use the lift ring bc i ran a bending calc on the bolt with the 7500lb distributed load, and it exceeded the yeild stress of the 5/8 grade 8 bolt.

any help?

 

[rb1957]

distributed laod vs point load ... doesn't matter as far as the bolt loads are concerned.

2 bolts are carrying the load in tension, 7,000 lbs lift, 3,500 lbs per. if you want SF = 3, then you want bolts with a tensile strength of 10,500 lbs. [this is easy to check for the bolts you use]

note, it's a another question whether the ring is strong enough for this load; a section 1.88" x 1.25" sounds pretty reaonable.

 

[rb1957]

looking at your picture again, i think i was a bit hasty ... the parting plane isn't at right angles to the lift. thus the closer bolt will reaction more load, or at least a different combination of shear and tension.

my book on rings says for this loading (2 diametrically opposite loads ...
Tension = P/2*sin(x)
Shear = P/2*cos(x)
Moment = PR*(0.3183-sin(x)/2)
the datum for angle x is the diameter the loads are applied along.

Now unfortunately you'll need both shear and tension allowables for your bolt and the interaction of shear and tension on the bolt (Rs^2+Rt^2 = 1, or simpler and more conservative Rs+Rt = 1) R = applied/allow.

your machinist hasn't helped you out any ! it'd be much easier if the parting plane had be at 90deg to the lift axis. btw, what are the lift lugs rated for ? maybe the weakest part is the connection for the lift lugs to the ring (a thread in tension). what material is the ring ? (Al ??) [sorry if you've already posted this detail]

 

[longude17]

the machinist made the ring this way to move the bare rotor and now wants to know if the coupling is safe to hang the 5000lb completed rotor...the material is 1018 steel.(ys
= 53700psi, us=63800psi) attached with gr8 5/a allen bolts (ys=120ksi us=150ksi)

im not too worried about the lifting lugs as they are off the shelf lifting lugs that are a screw in application and not lifting there load limits (i think they are 5000lb each)

I talked to the machinist and he made it in this way bc he feels it would be stronger....he felt that putting the bolted connections at 90 degrees to the lifting lugs would create a pivot point, thus he moved them closer to the lifting lugs.

ive assumed this in pure tension and calculated the bolt hole tear out forces and bolt strip forces, but now im stuck.......

any ideas on how to calculate this properly?

 

[rb1957]

i mentioned the lift ring only as a limit to the load that can be applied to the ring.

i think the bolts have shear and tension in them. i think your bolts are good for 46000 lbs (tension). based on that the maximum tension in them is P/2*SF = 7000/2*3 = 10500 lbs. Conservatively assume 10500 lbs shear; it's less than this, the equations i posted show tension and shear play off against one another.

Assume Fsu = .57*Ftu = 26300 lbs and assuming a conservative interaction Rs+Rt < 1 you have 10500/26300+10500/46000 < 1 ... gtg.

 

[zekeman]

"my book on rings says for this loading (2 diametrically opposite loads ..."

I don't think so.The loads are not diametrically opposite.

It is a ring with 2 side loads and a point load at the end. I'm sure the problem has been done. It involves taking 1/2 of the ring, cantilevering one end where there is a shear force P/2 and cantilevering the opposite end with no linear load and then adding a tangent off center P/2 load at the middle (trunnion location). The solution boundary at the trunnion is that the slopes of the upper 1/4 and bottom 1/4 be the same.

I would be very careful about using formulas that people think are accurate, especially when safety is the issue. This is not a problem for an entry level engineer or a machinist.

 

[rb1957]

i fugre the lift will be on one of the hooks, and the weight diametrically opposite. the parting plane is offset for this, so i gave the results in terms of angle.

"2 side laods" ... ??? is the weight going to be under the lift or beside it ? for the weight to be a side load it means the ring axis is horizontal and the weight is not in line with the lift. how does that work ??

 

[zekeman]

RB,

My take was the load was on the bottom of the ring and the 2 trunnions each in shear have 1/2 of the load; the top of the ring has no linear load. The upper hemisphere only has a constant moment.

Maybe a sketch would be nice.

 

[longude17]

Well ive done a worst case calc of converting the distributed load ( 1/2P = 7500lbs) to a point load directly in the bolt. But in this case the bending stress exceeds the allowable limits of the bolts. (w/o sf I calculate 117000 lb bending stress)..this is assuming both ends of the bolt will be fixed, since this is a bolted connection.

logically thinking, the biggest stresses will be in the bolts in a form of tension... ive calculated that it will take roughly 45000 lbs to strip the bolt hole treads (fus after sf)

while I know the ring itself and the shaft itself will counter act a portion of the bending forces via both axis, i have no clue how to do the study with out some sort of FEA modeling software which i dont have.

feel free to dazzle me with your own calcs.

 

[rb1957]

i doubt the bolts are bending, design the preload to keep the faces in contact so the bending is carried by the ring.

zeke has a different take on the loading ... i guess he's right, that the lift is shared by both of the lift links, yes?

in this case (a down load of P reacted by two side loads) ...
T = P/2*(0.3183cosx+sinx)
S = P/2*(cosx-0.3183sinx)
M = PR/2*(0.3183cosx+sinx-0.8183)
for 0<x<pi/2, x=0 at the applied load,P
and ...
T = P/2*0.3183cosx
S = P/2*-0.3183sinx
M = PR/2*(0.1817+.3183cosx) [moment is not constant in the upper 1/2]
for pi/2<x<pi

i'd apply a pretty significant preload, to keep the ring faces in contact. conservatively the tension is less than 10,500 lbs, so a preload of 1/2 ftu would be plenty, = 23,000 lbs, T = 0.2PD = 2900 in.lbs = 240 ft.lbs ... unreasonably high ??

 

[longude17]

yes the load is shared by both the lift rings, as the lifting collar fits around the shaft.

where are you getting the .3183 and .1817? and for what does T and S stand for?

 

[rb1957]

T is axial load
S is shear load
I'm using AFFDL-TR-69-42 "USAF Stress Analysis Manual", table 5-6, fig 5. i think you can pick up a copy on-line, try using the report number.

0.3183 and 0.1817 are just constants.

note, this doesn't include the off-set moments induced by the lift lugs, ie the lifts is not applied tangent to the ring but off-set slightly. calc the moment, Ma, = P/2*d (1/2 the weight (P) is reacted by a lift lug, off-set "d" from the ring CL). the effect on the internal loads in the ring is ...
T = 0.6366Mo/R*cosx
S = 0.6366Mo/R*sinx
M = Ma*(0.3183*(2*cosx+pi/2)-1)
Mo = M(0) = Ma*(0.3183*(2+pi/2)-1)
0<x<pi/2
and for pi/2<x<pi
T = 0.6366Ma/R*cosx
S = -0.6366Ma/R*sinx
M = 0.3183Ma(2cosx+pi/2)
this accounts for both moments; superimpose over the internal loads from the direct loads.