Today's problem involves loss of ship stability caused by lifting a weight on a crane which has an effect on the ship's centre of gravity height.
I think where I am struggling may be a misuse of Mathcad but I'm not sure.
Below is the problem:
Here is some of the explanatory equations to use:
And this is where I have got to.
I've obviously got the correct change in centre of gravity from the keel, however, I am .1 of a degree out with the angle of heel caused by swinging the crane out by 30 degrees.
I have assumed that the lateral distance from the crane to the mass is the opposite of a right triangle of the radius of the crane's jib as the radius forms the hypotenuse of this triangle.
Not sure I have got the trig functions correct with regards finding the angle of heel?
Also I've not used the lift height i.e. the weight is lifted 10m but I don't think this is relevant? I say this as the virtual centre of gravity of the weight is at the suspension point which is 20m?
Help as always welcomes.
Andy Wibroe wrote:
How? there is no insert attachment option?
Unfortunately its more complicated to attach files ion this forum than it should be.
First you have to chose "Use advanced Editor" (upper right corner of the edit window).
Then you can chose "Attach" at the lower right corner to attach your file.
If you edit a post (you can do this as long as nobody has replied to your post by choosing "Actions" at the lower left corner) you are automatically in "advanced" editor,
It is a long standing wish of community members that the advanced editor should be the standard one (and that they add the ability to cite there!).
Although I am starting to think the calculation of H is obviously wrong as the swing will not give a right triangle it self. I believe there are 2 right triangles which make up the space beneath the distance H I am after but I am now still not at the right answer.
Hello again everyone,
I have now convinced myself that the movement of the crane is a pivoting movement of a crane arm which was originally in line with the ship centre line and now moves outward by way of pivoting 30 degrees toward the side of the ship (looking from a bird's eye view).
I also believe that the H value referred to in the calculations is essentially the lateral distance of movement of the suspension point therefore giving the linear dimension component of the moment in the numerator in the ultimate formula for calculating heel of the ship. However, this still leaves me with an erroneous answer in comparison to the 1.36 degree heel given as the model answer.
im given the instructions you see in my original post.
r is the radius of the crane arm
Dlets (the symbol) is the displacement weight of the ship
w is the weight of the object picked up by the crane
turn is the angle the crane arm is swung
Volker - it's a puzzle with the specialized symbols of ship design. As presented it is more difficult for anyone who hasn't done this before than for the OP to figure this out. I tried to help on a previous problem and spent about an hour scouring Google results to see how the formulas were derived.
While a picture is worth 1000 words, this is more like 20 questions.
yeah I appreciate that so apologise. I think I've had it easy on here so far.
the formula tan theta= weight of cargo * transverse distance / weight of ship * distance from centre of gravity to meta centre is correct and can be seen in many text books.
i think where I'm going wrong is my derivation of H I.e. transverse distance the crane's jib moves i.e. The suspension point where the weight of the cargo acts.
There is always the possibility the textbook answer is incorrect.
It's what make this a task to first understand how the equations came about so that it's possible to understand where errors in their application could come from. For example: by reversing the problem to see what turn angle would give the textbook result.
Any text that includes the word 'clearly' is not a friendly one. If it was clear they would not feel compelled to point out how clear it is.
The irritation on my part is from rusty neurons exposed to a salt-water problem, not the request for help.
I have contacted the originator of the problem to confirm that as I too am considering possibility of rounding errors causing the difference.
i will upload some further info on how the equations have been derived.
Hello again to all,
So firstly my sincerest apologies for being quite presumptuous and not taking the time to outline my understanding of the background. I will now try to do this.
Firstly, the topic is initial ship stability. It is called initial stability because for small angles of heel the righting moment for a ship is generated by the transverse height of the metacentre which is the pivoting point the ship pivots about for small angles of heel whereby the metacentre is assumed to be static and does not move. For large angles the metacentre moves up and down, but this is not relevant.
The basic theory is that:
The short story is that as a ship heels the centre of buoyancy will move as the under water shape changes and equilibrium occurs when the centre of buoyancy is beneath the centre of gravity. In this case the metacentre (subscript T for transverse metacentre), Centre of Gravity (G) and centre of Buoyancy (B) will be aligned.
There are two moments of interest which are the righting moment and the heeling moment.
The righting moment being the mass of the ship (usually denoted by greek symbol Delta) multiplied by GZ which is a theoretical righting lever given by the distance from the centre of gravity to the line between the metacentre and the centre of buoyancy (as the ship heels) which is seen in figure 8.1 above (except that the centre of buoyancy is not shown on the diagram but B would be would be below Z in line with Mt.
The heeling moment in this context being by a shit of mass transversely is the weight of the object shifted (w) multiplied by the distance of transverse shift (d) multiplied by cos (theta) where theta is the angle of heel of the ship.
Combining these together to give the equilibrium state of final heel it can be seen that the final heel angle can be obtained as tan (theta) = wd/Delta GMt where GMt is the distance from the centre of gravity to the transverse metacentre.
Regarding this specific problem, I am of the understanding that the above equation for movement of objects has been adapted noting that the centre of gravity of the weight to be shifted transversely i.e. by the action of the crane swinging a weight outboard acts through the crane's jib irrespective of the length of the crane wire suspending the weight.
Therefore the adapted formula is:
tan (theta) = w x H / Delta GMt'
Where theta is the final angle of heel, w is the weight of the cargo, H in this context is the transverse movement of the crane's jib as this is where the weight of the suspended cargo acts, Delta is the total weight of the ship and GMt' is the 'new' distance between the transverse metacentre and the new centre of gravity. the new centre of gravity having risen due to the lifting of the weight.
I have drawn some crude hand sketches to further explain my understanding.
Looking at my sketches one by one:
The first is a schematic of my understanding of the problem where the shift in the cargo's centre of gravity instantaneously shifts from its actual position at G1 to G2 when it is lifted i.e. it acts from its position to the suspension point instantly on lift.
Then an idea of how the ship's overall position of centre of gravity shift upward when the cargo is lifted and this effectively reduces GMt to GMt'
Then an overhead view of the ship and how the crane swings, and finally my understanding of how to calculate the H distance i.e. the transverse shift of the crane's jib i.e. the distance the weight of the cargo moves transversely.
Hopefully I have done my bit in further explaining the background and happy to go into more detail as needed. I will say that now I have gone through this the more I think I might be right.
No unfortunately not. I do now have an answer for this one which I am in the process of writing up to upload for interest.
You probably can help me with graphs again in my other thread on stability and graphing though...
I have made myself calculations and sketchings to the problem you posted.
I think, that there is no exact solution and your approximately solution seems to be the best way to be practically applicable.
Neverthless i attached my calculations and sketchings.
The sketching is a exe file (no worry about viruses) and you can see it if you doubleclick on it. You don't need the program e-Drawings to open and see it.
best regards, Volker