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8/13/2019 Stability Problems 5
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1. A box-shaped vessel 200 m in leng th ,32 m breadth , f loats in SW at an even
keel draft o f 9.0 m. The KG is 10.0 m. Thevessel has a con t inuous center line
bu lkhead which is watert ight . What is
the bod i ly s inkage i f an emptycompartment 20.0 m in leng th and
symmetr ical about am idsh ips is bi lged
on one s ide?
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1. Solu t ion:
INC. IN DR = VOL. OF LOST BOUYANCY
AREA OF INTACT W.P.
= 20 x 16 x 9
200 x 32 - 20 x 16
= 28806400 - 320
INC. IN DR = 0.474 m
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2. You r vessel arr ives in po rt w ith
su ff ic ient fuel to steam 550 m iles at 13 kts .
If you are unable to load bunkers, at whatspeed mus t you p roceed to reach you r
next po rt wh ich is 683 m iles away?
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2. Solu t ion:
N Cons = N Spd x N Dist
O Cons O Spd x O Dist
N Spd = N Cons x O Spd x O Dist
O Cons x N Dist
N Spd = x (x) 13 x 550 nm
x (x) 683 nm
= 136.09077 N Spd = 11.66 Kno ts
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3. You have steamed 1,124 miles at 21knots . and consumed 326 T of fuel . If you
have 210 T o f usable fuel remaining, howfar can you steam at 17 kno ts.?
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3. Solu t ion :
N Cons = N Spd x N Dist
O Cons O Spd x O Dist.
N Dist = N Cons x O Spd x O Dist
O Cons x N Spd
N Dist = 210 tons x 21 x 1,124 nm
326 tons x 17
N Dis t = 1,104.86 nm
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4. A box -shaped vessel 200 m length , 32 mbreadth , f loats in SW at an even keel d raft
o f 9.0 m. The KG is 10.0 m. It has acont inuous centre l ine bulkhead which is
watert ight . An empty compartment 20.0 m
in leng th and symmetr ica l about
am idsh ips is b i lged on one side. What is
the KM if the BM is 8.989 m?
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INC. IN DR = VOL. OF LOST BOUYANCY
AREA OF INTACT W.P.
= 20 x 16 x 9
200 x 32 - 20 x 16
= 2880
6400 - 320
INC. IN DR = 0.474 mOLD DR = 9.000 m
NEW DR = 9.474 m
NEW KB = x DR= (9.474)
NEW KB = 4.737 m
NEW BM = 8.989 m
NEW KB = 4.737 m (+)
NEW KM = 13.726 m
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5. A box -shaped vessel 200m in leng th,
30m breadth , is f loating in SW at even
keel d raft o f 9.0m . What w il l be the changeof tr im i f the forward end compartment
10.0m long is b i lged w ith 2767.5 tons SW
assum ing an MCTC = 878.79 tons meters?
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5. Solu t ion
Ch. of trim = W X D
MCTC = 2767.5 x 10
878.79
Ch. Of Trim = 31.492 cm
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6. A doub le-bo ttom tank , when ful l , has
i ts center of g ravi ty at a height o f 60 cm
above the keel and can hold 380 tons o fwater. The KG o f the sh ip is 9.4 meters
and her disp lacement is 3700 tons when
the tank is empty . What w i l l be her KG
when the tank is f i l led?
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7. A sh ip of 45,000 d isp lacement has a
KG of 9.49m , KM=12.53m, GGo = 0.18m .
Find the value of GZ fo r a 20 degreesheel. (KN = 4.45m)
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7. Solu t ion :
GZ = KN KG X SIN
= KN( KG + GG0 X SIN ) = 4.45(9.49 + 0.18 X SIN 20)
GZ = 1.1427 m
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8. What is the bodi ly s inkage of a box -shaped vessel 80m x 14m float ing at an
even keel draf t of 4m if an emptym idsh ips DB tank is bi lged 16m x 14m x
4.2m?
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8. Solu t ion : TPC = 1.025 x A
100 = 1.025 (80) (14) 100
TPC = 11.48 BODILY SINKAGE = WT. TPC
= (16)(14)(4.2)(1.025) 11.48BODILY SINKAGE = 84 cm s
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9. A box -shaped l igh ter is 25 meters long,
6 meters w ide and floats at a draft of 1.10
meters fo re and aft . What w il l be her newdraft after 30 tons o f pig -iron have been
sp read evenly over the bot tom?
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9. Solu t ion :
= L x B x Dr x Dens.
= 25 x 6 x 1.10 x 1.025 = 169.125
+ 30.000 (Pig Iron)
= 199.125 = L x B x Dr x Dens.Dr =
L x B x Dens.
= 199.12525 x 6 x 1.025
Dr = 1.295 m
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10. A ship o f 6,000 tonnes displacement is
f loat ing in fresh water and has a deep
tank (10m x 15m x 6m) which is undiv idedand is part ly f i l led w ith nut oi l of relat ive
density 0.92. Find the vir tual loss o f GM
due to the free surface.
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10. Solu t ion :
VIRTUAL LOSS
OF GM = LB x d1 12V d2
= 10(15) x .92
(12)(6,000) 1.000
Virtual Loss o f GM = 0.431 m
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11. You r sh ip o f 12,000 tonsdisplacement has a center o f gravi ty o f
21.5 ft . above the keel. You run agroundand est im ate the weigh t aground is 2,500
tons. The vir tu al r ise in the center of
gravi ty is:
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11. Solu t ion :
GG = W X D
= 2,500 x 21.5
9,500
GG = 5.66 UPWARD
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12. When a weigh t o f 800 lbs. issuspended, what is the stress on the
haul ing part when using a gun tacklerove to least advan tage?
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12. Solu t ion :
Force = Weight x ( 1 + 10% N.O.S )
Mechanical AdvantageForce = Weight
Mechanical Advantage
Force = 800 lbs
2
Force = 400 lbs
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13. On arr ival at the d ischarg ing po rt, thed isp lacement was 7,800 t. A fter
Discharging 3,200 t of cargo w ith anaverage KG of 5.8 m the new KG was
found to be 6.14 m . What was the
vessels KG prior to discharge?
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13. Solu t ion :
WT DIST MOMENT
78003,200 = 4600 DISCH = 3,200 x 5.8 = 18,560
FINAL DISPL= 4,600 x 6.14 = 28,244 (+)
INITIAL DISPL = 7,800 46,804
KG = MOMENT / WEIGHT
= 46,804
7,800
OLD KG = 6.00 m
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14. You r vessel tank measu re 30 ft. long ,20 feet w ide and 15 ft. deep and the
speci f ic gravi ty of l iqu id in the tank is0.63. Find the free surface constan t if
yo ur vessel is f loat ing to a dens ity 1.024?
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14. Solu t ion:
r = .63
1.024 r = .615
FSK = r x l x b
420
FSK = .615 x 30 x 20
420
FSK = 351.42
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15. Compu te for the free su rfaceco rrect ion fo r vessel having a dimension
o f 45 ft. long , 36 ft. w ide and 25 ft. deep ,the free surface constant is 4,272 and the
vessel has d isp lacement o f 12,500 T.
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15. Solu t ion :
FSC = FSK
FSC = 4,272
12,500 t
FSC = 0.342 ft .