Calculation of Crane Boom Loads 2

[Fencer01]

we are doing a crane boom analysis and are confused about the application of the loadings at the tip of a the boom.

Our scenario is –

6m SWH, we have computed the dynamic factor to be 2.81 for an SWL of 12 tonnes. Thus the actual load at the crane boom tip is about 34 tonnes (12 x 2.81). However we are unsure if this load is to be applied as individual to the crane boom tip or is to be combined with the tension in the wire rope on the other side of the sheave (see below)

As u can see the confusion.. the 12 tonnes becomes 60 tonnes for structural analysis.. seems a bit over the top. Would highly appreciate if this can be clarified.

 

[SwinnyGG]

The cable tension is the only thing applying any load to the boom- so it must be accounted for in both directions.

The simplified way to look at this is each leg of the cable applying the same force, centered at the tip of the boom, with some angle between them.

Those two vectors are resisted by a single force vector along the boom axis.

 

[Compositepro]

You need to do a free body diagram showing all the force vectors (a vector is a quantity that has magnitude and direction). When you add all the force vectors they will sum up to zero (everything is in balance and nothing is accelerating, F=ma, a=0. So this is a statics problem, not dynamics). You have two forces of 12 tons due to rope tension. The resultant load vector bisects the angle between the ropes. This is resisted by a compression load on the boom and a bending load on the boom. You do not show how the boom is supported so this bending load cannot be calculated with the information given.

 

[chicopee]

Is the dynamic load due to the boom being raised while the load is not being hoisted or due to the load being hoisted while the boom is stationary or both condition with boom being raised and load hoisted? the question is being asked since if the load is being hoisted, there will be rotational resistance at the boom tip sheave that needs to be included to the load applied at the portion of the wire rope between boom tip and the hoisting mechanism. It all depends how accurate you want to be and if accuracy is your goal than the 34 tonnes will need to be increased as 34/(1-Uk) where Uk is the dynamic coefficient of friction of the sheave bearing. Uk may vary from 4.4% to 0.9%. Arbitrarily you can use 2%. An excellent reference is Crane and Derricks by Shapiro.

 

[Jboggs]

Think of it this way. Sounds simple but it works for me. Ask yourself "what does the crane know?" The crane doesn't know there is a 34 ton load down there somewhere. It also doesn't know some cable is anchored anywhere. All it knows is what it "sees", or feels, from what is touching it. What is touching it? A cable wrapped to a known degree around a pulley, and a support fixture holding it fixed. Both ends of that cable are exerting a force. The resultant vector of those two forces is what the crane feels. The fixture exerts an equal and opposite force. That's all it knows.

 

[IFRs]

Your geometry determines the load on the crane boom. I think 60T may not be correct. What is the angle between the vertical load and the boom? The boom load would be 34T x Sin(49) plus 34T x Sin(angle to the vertical load). As mentioned above, a free body diagram really makes things obvious.

 

[rb1957]

draw a FBD and I think you'll see that the boom is in bending.

your sketch doesn't have enough data to draw a FBD ... it looks like the angle between the boom and the inclined cable is 49deg (it doesn't look like it, but maybe that's artistic license). what is the angle between the boom and the vertical tension ? looks less than 41deg (ie the inclined tension isn't horizontal) but then it looks more than 49deg ?? whatever, the boom doesn't look to bisect the angle made by the cables.

another day in paradise, or is paradise one day closer ?

 

[IFRs]

I would not expect the boom to bisect the angle, I see them as 2-force members with nearly idealized pin connections at both ends. Where do you see bending?

 

[saplanti]

So you are calculation the axial load on the boom structure. However you do not mention of the boom angle with vertical/horizontal.
Assuming 49 deg is the angle with horizontal the boom axial load should be :

1. from the horizontal wire rope: = 34 * Sin (90-49)
2. from vertical load: = 34 * Sin 49

Total Boom axial load = 34 * ( sin(90-49) + Sin49 )

Note that this is not resultant vertical load, but axial load on the boom structure. I guess you are after this since you are going to check the boom against buckling first.

If you are after crane overall stability the resultant vertical load is 34 tonnes only.

 

[saplanti]

I made a mistake in the first load from the horizontal wire rope. It should be = 34/Sin(90-49).

This makes the total boom axial load = 34 [ 1/sin(90-49) + Sin49 ].

 

[rb1957]

if the boom doesn't bisect the angle made by the cable (which obviously it'll only do at one angle) then the result isn't aligned along the boom, so there's a transverse component, and so there's bending.

but I guess I'm assuming that the cable tension is the same on both sides of the boom. I can see that it doesn't have to be, so a force polygon (three lines of action, one known length) is the way to approach the solution.

another day in paradise, or is paradise one day closer ?

 

[IFRs]

An idealized two force member does not support bending.

 

[rb1957]

aye, and therein the rub ... is the boom a two force member or not ?

I initially thought not (that it was in bending) but can see (if the cable tension changes around the end of the boom) that it could be.

another day in paradise, or is paradise one day closer ?

 

[IFRs]

With pins at both ends, it has no choice.

 

[SwinnyGG]

There's nothing in the original statement that indicates that the boom of this mysterious crane is being idealized in any way, or is pinned at both ends.

 

[Fencer01]

Hey everyone..

thanks for all your replies. I have also got the confirmation from lloyds for the same. Appreciate your help.

thanks a lot.

 

[rb1957]

ok, what did they say to do ?

another day in paradise, or is paradise one day closer ?