Stability of a barge- Fluid Mechanics

[ttb90]

Homework Statement

A large barge of overall width 4m to be used at sea has a uniform empty weight per meter length of M=6327.19kg/m (length is into the page). This empty weight is uniformly distributed along the length of the barge. The center of gravity of the empty barge is at position G, 2m above the flat bottom of the vessel. The barge has a rectangular water-line area and the density of seawater is 1025 kg/m^3. Neglect the thickness of the hull.

Ballast of mass Mb kg per meter length is added to the barge in order to effect the stability. The weight of the ballast acts at point A (on the barge centerline at the bottom of the hull) and is uniformly distributed along the length of the barge. The addition of the ballast will shift the center of gravity from G to G (prime). Write out an expression for the height of the new center of gravity above the bottom of the hull in terms of Mb (distance AG(prime))

Homework Equations

I know that the key word in the question is 'acting on point A'. Adding Mb will shift G to G(prime) however G(prime) is lower than G. Because the weight was added below and not above.

I am aware that i need to use moments to find G (prime) but am not sure how to start it off..Really stuck

The Attempt at a Solution

I have found the draft value of h=1.5432m.
m of barge= 6327.19 kg
length (l) of barge =1m
width= 4m
ρ= 1025kg/m^3

For equilibrium, buoyancy force= 6327.19g
volume of displaced seawater V=wxhxl= 4xhx1=4h
If we equate the two equations above we get h= 1.5432m..
This value is calculated for G not sure if it stays the same when we have G (prime)??

 

[SteamKing]

The location of G is given as 2 m above the bottom of the barge, and this is for the empty barge. If Mb of ballast is added with a center of gravity of zero meters above the bottom of the barge, what is the height of the c.g. of the barge and ballast together?

 

[ttb90]

If you look at the attached diagram it shows the height from point A to the new center of gravity to be h.. That's the value I found in the above calculations which is, h=1.5432m

 

[SteamKing]

I think you are misreading the figure attached with the problem. The height of G for the empty barge is 2 m above the bottom of the barge. You are asked to find out how much ballast to add to the barge so that G', the center of gravity of the barge with ballast, will produce a stable condition. The value of G' may not necessarily be equal to the draft h.

 

[db725]

Hi SteamKing, just realized that you are completely right. I assumed the height stayed the same but the G' has moved so h is probably not equal to the draft h.. Because the question specifically asks 'write the expression for the height of the new center of gravity above the bottom of the hull in terms of Mb(distance AG')) i have no idea how to obtain this. Any starting points you would recommend?

 

[SteamKing]

The height of the center of gravity of the empty barge plus ballast can be found by writing the vertical moment equation, using point A as the reference.

 

[db725]

Definitions:
Mb=mass of ballast
M= mass of barge (empty)=6327.19kg

0=ƩMa=-(AG' X (M+Mb)+(AG X M)
since we know that AG=2m;
0=ƩMa=-(AG' X (M+Mb)+(2 x M)

The negative sign is due to the anticlockwise moment

0=ƩMa=-(AG' X (6327.19+Mb)+(2 x 6327.19)
AG'(6327.19+Mb)=12654.38
therefore Ag'=12654.38/(6327.19+Mb)

is this right?

 

[SteamKing]

Looks good.

 

[db725]

Also i have started doing the next question and it asks: 'Write out an expression for the height of the center of buoyancy above the bottom of the hull (the distance AB), again in terms of Mb.

For this question I though that the draft h value must be different from part (i) as we have added Mb on point A. However this makes things complicated, how do we know if B changed or not.. It is not marked on the diagram. Does this mean it stays the same and we just recalculate the new h value and assume its half way through the new h value?

 

[ttb90]

That's how i did mine as well. Took a while to get the answer. I am stuck with this new buoyancy B though, do we assume it stays at the same point after the MB is added to point A. What exactly changes here?

 

[SteamKing]

When ballast is added, the barge will sink. In order to find the new position of B, the draft of the ballasted barge is required. In general, when the displacement of a vessel changes, the location of the center of buoyancy changes as well.

 

[ttb90]

Does that mean that when I do find the draft of the ballasted barge we assume that the new position of B is still halfway of the new draft value? Is this assumption correct?

 

[ttb90]

SteamKing: This is following I did to find the new position of B.

For the height of the center of buoyancy above the bottom of the hull i did the following.

Volume of displaced water= V= 4 x h x 1= 4h
The weight of the displaced seawater = (6327.19+Mb)g =ρ x g x4h

Therefore h= (6327.19+Mb)g/(ρxgx4)= (6327.19+Mb)/4100

However the distance AB= (1/2)x((6327.19+Mb)/4100)=(6327.19+Mb)/8200

The reason I divided it by 1/2 is because the new buoyancy point must be located half way through the new draft value.

Are my calculations correct?

 

[SteamKing]

The calculation of B is correct, although you are actually multiplying by 1/2 instead of dividing.

 

[ttb90]

Thanks very much for that.

There is now a change of scenario and I attached it as figure 2.

Question states that 'If the total ballast acting at point A is now increased to 3000 kg/m, what is the maximum mass of cargo per unit length Mc that could be safely carried at position C with the barge still stable?'

In this question do we use a similar approach before in using the stability formula? What's the best way to solve this problem?

 

[SteamKing]

The procedure is similar. Write a vertical moment equation and find the new height of the c.g. of the loaded barge. Determine the new draft for the loaded barge and calculate B.
Io depends only on the shape of the waterplane. With the new locations for B and G', evaluate GM.

 

[ttb90]

So I'm finding this last section really hard to start off. This is what I have done so far:

Mb=3000kg
M= mass of barge (empty)=6327.19kg
g=9.81
Mc= mass of cargo
ρ=1025kgm^(-3)

I struggled to find the moment equation so i started off by doing the following:

1. New draft for the load barge
Volume of displaced water= V= 4xhx1= 4h
The weight of the displaced water= (M+Mb+Mc)g=ρg(4h)
(9327.19+Mc)g=ρg(4h)
h=(9327.19+Mc)/4100

New location of B, has to be located half way between the new draft h

So the new distance of AB= (1/2)x(9327.19+Mc)/4100=(9327.19+Mc)/(8200)

CG'=AC-AG'
AG' was found in the previous sections to be = 12654.38/(6327.19+Mb)
Substituting our new Mb value gives , AG' =1.3567

CG'=4-1.3567=2.6433 m

Vsubmerged= 4 x 1 x h= (9327.19+Mc)/1025

BG'(bar)= CG'+AB = 2.6433+(9327.19+Mc)/(8200)

Therefore using G'M(bar)= Io/(V submerged) - BG'(bar)
= 5.333/((9327.19+Mc)/1025) -(2.6433+(9327.19+Mc)/(8200))

For the maximum I was thinking of using calculus but not sure if that is possible to use on this?? or if there is a better way ?

Not sure if what I have done is right for this question,very confused... didn't use moments as you said earlier because i was getting confused with all the different variables.. Am I on the wrong track?

 

[SteamKing]

The vertical moment of a mass is this:
moment of mass = mass times (distance of c.g. of mass from point A)

The total vertical moment = sum of the vertical moments of the individual masses

To find the maximum amount of cargo, calculus is not required. The minimum stability is G'M = 0. Solve for Mc.

 

[ttb90]

Definitions:
Mb=mass of ballast
M= mass of barge (empty)=6327.19kg

0=ƩMa=-(AG' X (M+Mb)+(AG X M)
since we know that AG=2m;
0=ƩMa=-(AG' X (M+Mb)+(2 x M)

The negative sign is due to the anticlockwise moment
M=mass of barge = 6327.19 kg
Mb=mass of ballast = 3000 kg
Mc=mass of cargo
Distance of C from A: 4 m

0=ƩMa: -(AG' x g x(M+Mb+Mc))+(AG x M x g)+(4 x Mc x g)
AG'(6327.19+3000+Mc)=(2 x 6327.19) + (4 x Mc)
AG'(9327.19+Mc)=12654.38 + 4Mc
AG' =(12654.38 + 4Mc)/(9327.19+Mc)

Volume of displaced water= V=4 x h x 1= 4h
For equilibrium buoyancy force = (3000+6327.19+Mc)=9327.19+ Mc
Thus the weight of displaced seawater= (9327.19+Mc)xg=ρxgx(4h)
h=(9327.19+Mc)/(4100)

the distance AB is halfway between h (new draft), AB =((9327.19+Mc)/8200)

BG'(bar)= AG'-AB
= (12654.38 + 4Mc)/(9327.19+Mc) - ((9327.19+Mc)/8200)
Io=5.333
Vsubmerged= 4 x h x 1=4 x (9327.19+Mc)/(4100)=(9327.19+Mc)/1025

G'M= 5.33/((9327.19+Mc)/1025) - [(12654.38 + 4Mc)/(9327.19+Mc) - ((9327.19+Mc)/8200)]

When i tried to solve this, G'M=0 and I couldn't because I had a few root problems where it clashed with the Mc value.
Can you help please??

 

[SteamKing]

Your equation setup looks good. When G'M = 0, the equation will simplify to a quadratic in Mc. Check your signs and arithmetic when simplifying.

 

[ttb90]

Specifically speaking would you say that my moment equation is correct? I think that's the main part of the working out that has to be right as i use the result obtained in the following sections to calculate G'M=0

 

[SteamKing]

Your moment equation is correct. I suspect you earlier problem trying to solve for Mc was due to an incorrect sign or an arithmetical mistake. If you post your calculation, I can offer further help.

 

[ttb90]

G'M= 5.33/((9327.19+Mc)/1025) - [(12654.38 + 4Mc)/(9327.19+Mc) - ((9327.19+Mc)/8200)]

G'M=0

(12654.38+4Mc)/(9327.19+Mc)-(9327.19+Mc)/(8200)=5466.325/(9327.19+Mc)

(12654.38+4Mc -5466.325)/(9327.19+Mc)=(9327.19+Mc)/8200

(7188.055+4Mc)/(9327.19+Mc) = (9327.19+Mc)/8200

(9327.19+Mc)^2=58942051+ 32800Mc

(9327.19^2)+(2x9327.19xMc)+Mc^2=58942051+(32800xMc)

869964733+18654.38Mc+Mc^2=58942051+(32800xMc)

-50242403.67+Mc^2=14145.62Mc

Mc^2-14145.62Mc-50242403.67=0

I used my graphics calculator and found

Mc= 17086.153,-2940.533

So since we can't have a negative number the max the cargo can be is 17086.153kg..Would you be able to check this? Thank you soooooooooooooooo much!

 

[SteamKing]

There's an error in your equations when you collect terms to form the quadratic in Mc. Can you spot it. It changes your answer. BTW, always check to see if the barge has enough volume to support the empty weight plus ballast and cargo.

 

[ttb90]

I tried solving it again and don't think I succeeded. I have attached a scan of my working out so it's easier to read.. Please tell me what I am doing wrong. I think I've solved this like over 10 times and keep finding different answers each time :(

 

[SteamKing]

You have obtained the correct equation in Mc to solve. Although it is not obvious by inspection which of the two solutions for Mc is correct, you do have one condition which must be satisfied: the barge must remain floating when loaded with cargo and ballast.

 

[ttb90]

How do I go about showing that? I tried substituting the values into the previous expressions to see if all the conditions are met but couldn't get much out of it to come to a conclusion.. What shall I do?

 

[ttb90]

Read the question again and found that the question asks for the maximum mass of cargo, so since we found 2 masses that satisfy the G'M condition wouldn't that mean that the heavier mass is the answer to Mc. Thoughts?

So Mc=11760 kg/m ?

 

[SteamKing]

If you use the larger value of Mc as the amount of cargo which the barge carries, does the hull of the barge have enough buoyancy to float? (Hint: determine the draft required to support the weight of the barge, cargo, and ballast and see if the hull is sufficiently deep)

 

[ttb90]

I did all the calculations for it and found that Mc=2385.5699 kg/m. I really appreciate your help! Thank you soo much