Stability Problems 5

8/13/2019 Stability Problems 5

1/31


2/31

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?


3/31

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


4/31

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?


5/31

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


6/31

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.?


7/31

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


8/31

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?


9/31

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


10/31

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?


11/31

5. Solu t ion

Ch. of trim = W X D

MCTC = 2767.5 x 10

878.79

Ch. Of Trim = 31.492 cm


12/31

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?


13/31


14/31

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)


15/31

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


16/31

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?


17/31

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


18/31

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?


19/31

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


20/31

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.


21/31

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


22/31

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:


23/31

11. Solu t ion :

GG = W X D

= 2,500 x 21.5

9,500

GG = 5.66 UPWARD


24/31

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?


25/31

12. Solu t ion :

Force = Weight x ( 1 + 10% N.O.S )

Mechanical AdvantageForce = Weight

Mechanical Advantage

Force = 800 lbs

2

Force = 400 lbs


26/31

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?


27/31

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


28/31

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?


29/31

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


30/31

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.


31/31

15. Solu t ion :

FSC = FSK

FSC = 4,272

12,500 t

FSC = 0.342 ft .

Date post:	04-Jun-2018
Category:	Documents
Upload:	staicu-anghel-elena
View:	219 times
Download:	0 times

Stability Problems 5

Documents