Calculating Lightweight of a Barge: A Newbie's Guide

[lyranaval18]

hi I am a newbie here,
from Philippines,

I need you help guys i want to know how to calculate the lightweight of the barge since i only have the length =28 m, breadth=10m, depth=3m, light draft=0.8m,debsity of sea water=1.025t/m3, cb=0.9 and grt=237 tons. I am so confused how to compute since i don't know also the deadweight and displacement of the barge.

Thank you for immediate response.

[Mentor's note: Thread moved from engineering forum to homework forum]

 

[lyranaval18]

help me also to get the deadweight and the displacement especially light weight of this barge,the barge is box type only. please help me guys.

thank you.

 

[Baluncore]

Displacement of unladen vessel = lightweight ?
A rectangular box displaces volume = length * beam * draft = 28 * 10 * 0.8 = 224 cubic metres.
Then convert volume to weight. volume * density = 224 * 1.025 = 229.6 tonne.

What is cb=0.9 ?

 

[lyranaval18]

Cb=block coefficient of the barge,
yes I am looking for a lightweight, of the barge.
but, my problem is i already used that formula but still my answer is still big for the light weight of that barge, so I am confused for the correct computation of lightweight of the barge.

 

[Baluncore]

but, my problem is i already used that formula but still my answer is still big for the light weight of that barge,

What is the correct lightweight of the barge ?

 

[lyranaval18]

What is the correct lightweight of the barge ?

I don't know. only I know my Boss told me the lightweight i give to him is to big for that barge, and he didn't give me any estimated value of lightweight, that's why I am so confused. the exact value of lightweight for that barge. ; (

 

[gneill]

Take a look at this website:

https://www.tc.gc.ca/eng/marinesafety/tp-tp14609-3-stability-180.htm

It covers the subject of ship displacements and tonnage with examples. It also includes the definition of the block coefficient which, if I understood what I read there, you'll need to use in your calculation.

 

[Baluncore]

i want to know how to calculate the lightweight of the barge since i only have the length =28 m, breadth=10m, depth=3m, light draft=0.8m,debsity of sea water=1.025t/m3, cb=0.9 and grt=237 tons. I am so confused how to compute since i don't know also the deadweight and displacement of the barge.

The empty hull has a lightweight draft of 0.8 m, the total displacement will then be ;
28 * 10 * 0.8 * 0.9 * 1.025 = 206.64 tonne lightweight.
If your Boss thinks that is wrong then ask him to explain why.

The GRT is the internal volume measured in multiples of 100 cubic feet = 2.832 m³.
GRT = 237 means 237 * 2.832 = 671.184 m³ internal volume.
The external dimensions give an external volume of 28 * 10 * 3 * 0.9 = 756 m³
The 85 m³ difference is the volume of the hull structure and the unusable space.
Those figures are consistent, which suggests the 3 m depth is the height of the barge wall, not the fully laden draft.

Remember that GRT is a volume, not a weight.
But if the GRT volume available was loaded with a deadweight of 237 tonne,
then the total mass would be 206.64 + 237 = 443.64 tonne displacement.
The draft would then be 443.64 / (28 * 10 * 0.9 * 1.025) = 1.717 m. There would be 1.28 m freeboard.

Without better defined information we cannot tell anything more about this problem.

 

[lyranaval18]

The empty hull has a lightweight draft of 0.8 m, the total displacement will then be ;
28 * 10 * 0.8 * 0.9 * 1.025 = 206.64 tonne lightweight.
If your Boss thinks that is wrong then ask him to explain why.

The GRT is the internal volume measured in multiples of 100 cubic feet = 2.832 m³.
GRT = 237 means 237 * 2.832 = 671.184 m³ internal volume.
The external dimensions give an external volume of 28 * 10 * 3 * 0.9 = 756 m³
The 85 m³ difference is the volume of the hull structure and the unusable space.
Those figures are consistent, which suggests the 3 m depth is the height of the barge wall, not the fully laden draft.

Remember that GRT is a volume, not a weight.
But if the GRT volume available was loaded with a deadweight of 237 tonne,
then the total mass would be 206.64 + 237 = 443.64 tonne displacement.
The draft would then be 443.64 / (28 * 10 * 0.9 * 1.025) = 1.717 m. There would be 1.28 m freeboard.

Without better defined information we cannot tell anything more about this problem.

Good Morning,

Thank you so much Baluncore, I really appreciate your help for me to solve this problem.

Have a Merry Christmas :-)
God Bless you

 

[lyranaval18]

Take a look at this website:

https://www.tc.gc.ca/eng/marinesafety/tp-tp14609-3-stability-180.htm

It covers the subject of ship displacements and tonnage with examples. It also includes the definition of the block coefficient which, if I understood what I read there, you'll need to use in your calculation.

Good morning,

Thank you also Gneill. Have a Merry Christmas ☺