Calculation of Crane Boom Loads
Calculation of Crane Boom Loads
(OP)
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.
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.





RE: Calculation of Crane Boom Loads
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.
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
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 ?
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
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.
RE: Calculation of Crane Boom Loads
This makes the total boom axial load = 34 [ 1/sin(90-49) + Sin49 ].
RE: Calculation of Crane Boom Loads
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 ?
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
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 ?
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
RE: Calculation of Crane Boom Loads
thanks for all your replies. I have also got the confirmation from lloyds for the same. Appreciate your help.
thanks a lot.
RE: Calculation of Crane Boom Loads
another day in paradise, or is paradise one day closer ?