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

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Discussion Overview

The discussion revolves around calculating the capacity and displacement of a rake barge, specifically focusing on the effects of loading a 350-ton cargo onto a barge with given dimensions. Participants explore the relevant equations and concepts related to buoyancy and stability, while clarifying their understanding of the calculations involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant provides the dimensions and capacity of the barge, along with initial calculations for draft after loading.
  • Another participant questions the source of the number 50.868, which appears multiple times in the calculations, indicating confusion about its origin.
  • Some participants suggest keeping equations in symbolic form until the end to avoid confusion.
  • Concerns are raised about the precision of the barge dimensions and whether they account for factors like wall thickness and wave pressure.
  • One participant notes that the effects of trim were not included in the calculations, emphasizing the importance of having zero trim and heel for accurate loading assessments.
  • Another participant mentions that the stability calculation assumes the internal compartments of the barge are dry.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the calculations, with no consensus on the interpretation of specific numbers or the approach to the problem. Multiple competing views on how to handle the calculations and the importance of symbolic representation are present.

Contextual Notes

Some calculations may depend on assumptions not explicitly stated, such as the uniformity of barge dimensions and the conditions of loading. The discussion reflects a mix of theoretical exploration and practical considerations without resolving the underlying uncertainties.

Who May Find This Useful

Individuals interested in marine engineering, naval architecture, or those studying buoyancy and stability principles in practical applications may find this discussion relevant.

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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