- #1

ChemRocks13

- 6

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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

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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.