Undergrad Help please -- Trying to figure out rake barge capacity and displacement

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SUMMARY

The discussion focuses on calculating the draft and stability of a barge with dimensions 54.864 m x 15.5448 m x 2.95 m and a capacity of 1500 tons, under a load of 350 tons. The initial draft is 470 mm, and after loading, the calculated draft is 1.33 m. Key equations used include volume calculations and stability checks, particularly focusing on the relationship between load, draft, and stability metrics such as GMT and KG. The conversation emphasizes the importance of using symbolic representations in calculations for clarity.

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ChemRocks13
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Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
Sorry to misunderstand. I am learning. Not to disturb others.
My load is 350 Ton size above shown.
What the draft after load the cargo on the barge?
The barge draft 470mm
Bow side after 5m has slope. I mean front side from the till 5 m has slope. This takes place in fresh water.

I was found these calculations and was trying to make sense of them but cannot figure out how the person had arrive at them. Initial equations will help. I understand the concepts but still have trouble fully understanding it. Physics is not my strongest subject.

Barge

L = 54.864 m

B = 15.5448 m

D = 2.95 mTLS = 470 mm

1500 ton capacity

5 m Bow RakeLoad:

350 T

L = 10 m

w = 3 m

====================================

Head log depth = 0.61 m

2.95 - 0.61 = 2.34 m

Rake slope = 2.34 / 5 = 0.468Barge light weight

V = (49.864 + 50.868)*0.5*15.5448*0.470

V = 367.98 m^3

W = 367.98 M.T.For 350 T Load:

V = (50.868 + dT / 0.468)*0.5*15.5448 * dT = 350

(50.868 + dT / 0.468) * dT = 45.03

50.868 dT + dt^2 / 0.468 - 45.03 = 0

2.1368 dT^2 + 50.868 dT - 45.03 = 0

dT = [-50.868 + SQRT (50.868^2 + 384.87)] / 4.2736

dT = 0.855 m

T = 0.47 + 0.86 = 1.33 m

================================

Check GMT:

IT –

LWL = 50.868 + 0.855 / 0.468 = 52.695 m

IT = 52.695^3 * 15.5448 / 12 = 189,544 m^4

BMT = 189,544 / (367.98 + 350) = 264 m

KB = 1.33 / 2 = 0.67KMT = 264.67 mKG = (367.98 * 0.6 * 2.95 + 350 *(2.95 + 3)) / (367.98 + 350)

KG = 3.81 mGMT = KMT - KG = 260.86 m

Barge is stable.
 
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Welcome to PF. :smile:

That's a big boat! Is this a real-life problem (and if so, don't you have help from others involved in the project)?

Or is it a schoolwork problem? I can move this to the schoolwork forums if it is. Or when I look at your Profile page, I get the impression that you may be involved in designing tests for students?

Any additional context that you can provide would be helpful. Thanks.
 
This was from an older post that I am using to learn off of. No it is not homework but I am not sure if the thread I got it from was a homework thread or if it was real life. However for me it is theoretical.
 
Okay, I'll leave your thread here for now. Other Mentors may decide to move it if appropriate.

Can you give a Cliffs Notes type summary of what is confusing you in the equations above? TBH, I'm not able to tell what you are asking about. Also, I'm not that familiar with barge load calculations, so can you post a link to the reading you've been doing about it so far? Thanks.
 
ChemRocks13 said:
I have no clue where the number 50.868 comes from
That's why you should leave things in symbolic form and only plug numbers in at the very end.
 
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ChemRocks13 said:
Barge size is 54.864 x 15.5448 x 2.95 m and Barge capacity 1500 ton
The dimensions of the barge are uniform within a mm or even fraction of mm? Is this including the thickness of the plate used to make the walls? :smile: Or the nonuniform pressure of the waves on the sides of the barge.
If you are just doing some theoretical analysis you should drop these bogus numbers of digits and take some realistically accurate dimensions. Or even better, use simbols for calculations, as suggested by Vanadium 50.
 
This calculation and problem was from a previous thread on here from 4 years ago. I was just trying to understand the equations and where the numbers were coming from. This is not a school work or real world problem for me. I was trying to learn from this and from the course notes I had found on a website (and posted in the thread above to aid in my understanding). The 7/18/15 thread I pulled this from on here is titled

Trying to calculate barge capacity and displacement​


https://www.physicsforums.com/threads/trying-to-calculate-barge-capacity-and-displacement.823814/
[Link added by a Mentor]

I completely agree with Vanadium 50 about keeping equations symbolic until the end or to write the symbolic equations with all the data that is given at the beginning before plugging and chugging them in.
 
The effects of trim in this calculation were not included in the calculations.

Obviously in an actual barge loading, you would want to have the loaded vessel with zero trim and zero heel.

Stability calculation assumes the internal compartments of the barge are dry.
 

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