Stability Problems 3

8/13/2019 Stability Problems 3

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2/31

1. A sh ip d isp laces 8,800 tons at a

summer d raught of 7.28m . Its Fresh

Water A l lowance is 315 mms. Find i tschange of d raught when proceeding to a

port o f UNIT DENSITY.

Answer = 31.5 cm s


3/31

1. Solu t ion :

A.I. = FWA x ( Change in Density )

25

= 315 mms. x ( 1,025 1,000 )

25

= 315 mms. x 25

25

A . I. = 315 mms. or 31.5 cms.


4/31

2. A box has a dimens ion of 2ft x 2.5f t x

3ft and weigh s 95 lbs . What is i ts

stowage factor?

Answer = 353.7


5/31

2. Solu t ion :

Volume = 2 ft. x 2.5 ft. x 3 ft.

Volume = 15 ft

Stowage Factor = Volume

Weight / 2240 lbs.

= 15 ft

95 lbs/ 2240 lbs

= 15 ft / 0.0424 lbs

Stowage Facto r = 353.7


6/31

3. You are go ing to load bales o f abaca

w ith SF 65 and lead w ith SF 18. The

Remaining space is 257,000 cu ft. and theto tal weigh t to be loaded is 5,400 ton s.

How much of each cargo shou ld be

loaded to make the vessel FULL AND

DOWN?

Answer = 2,000 t lead / 3,400 t abaca


7/31

3. Solut ion:

WLF = Weight of the cargo having the large SF

WLF = Volume

( Wt. of Cargo x Small SF ) Difference in SF

= 257,000 ft - ( 5,400t x 18 )

65

18 = 257,000 97,200

47

= 159,800 / 47

WLF = 3,400 t ( Abaca )

Wt. o f Lead = 5,400 t 2,400 t = 2,000 t


8/31

4. A vessel with a l ight d isplacement o f

6,500t loaded cargoes o f 1,300t, 1,400t

and 1,200t w ith a KG of 6.9m, 6.4m and8.4m respect ively. A fter loading, the new

KG was then found to be 7.07 m. What

was the KG pr ior to loading?

Answer = 7.00 m


9/31

WT DIST MOMENT LOADED = 1,300 x 6.9 = 8,970

LOADED = 1,400 x 6.4 = 8,960

LOADED = 1,200 x 8.4 = 10,080

INIT.DISPL = (+) 6,500

TOTAL DISPL= 10,400

(-) 8,970

(-) 8,960

(-)10,080

45,518

KG = MOMENT

WEIGHT

= 45,5186,500

OLD KG = 7.00 m

x 7.07 =73,528


10/31

5. A bu lk carr ier has a d isplacement o f

40,000 tons and the KG is 9 m .The KN at

10 degrees heel is 1.87 m. What is ther igh t ing arm at this heel?

Answer = 0.307 m


11/31

5. Solu t ion :

GZ = KN ( KG x Sin )

= 1.87 m ( 9 m x Sin 10 )

= 1.87 1.5628

GZ = 0.307 m


12/31

6. A bu lk carr ier has a displacement o f

25,000 tons . KG is 8.6 m. A t an ang le of

heel o f 15 degrees , KN is 2.98. Solve forthe r ight ing moment.

Answer = 18,854 mtr-ton s


13/31

6. Solu t ion :

GZ = KN ( KG x Sin )

2.98 ( 8.6 x Sin 15 )

2.98 2.2258

GZ = 0.754 m

MSS = W x GZ

= 25,000 x 0.754

MSS = 18,854 ton - meter


14/31

7. Wh i le inspect ing the log pond , youno t iced that the average leng th o f the logsto be loaded is 15 m and the averagediameter is 70 cm s. Wh i le f loat ing in pondo f dens ity 1,012 kgs ./m, 90 percent o ftheir vo lume is submerged. What is the

average weigh t of each log?

Answer = 5.26 ton s


15/31

7.Solut ion;

Vol. of Cylinder = x r x h

= 3.1416 x 0.35m x 15m

= 5.77m

x 0.9 (90%)

Volume = 5.2m

Weight = Volume x Density

= 5.2m x 1.012ton/m Weigh t = 5.26 tons


16/31


17/31

8. Solu t ion :

New Draft = x B x Sin List + ODr. x Cos

List

= x 12m x Sin 18 + 6.7m x Cos 18

= 1.854 + 6.372

New Draft = 8.226 m


18/31


19/31

9. Solu t ion :

Tan List = Wt x Dist

x GM

Weight = Tan List x x GM

Distance

Weight = Tan 3 x 8,800 tons x 0.45m

6m

Weigh t = 34.6 tons


20/31

10. The speed made good was 16 knots

and the eng ine speed 15.4 knots . What is

the percentage of sl ip?

Answer = - 3.9%


21/31

10. Solu t ion :

% Slip = Eng. Spd Obs. Spd

Eng. Spd

= 15.4 16.0

15.4

- 0.6

15.4

% Slip = - 3.9 %

X 100

X 100

X 100


22/31

11. A cyl ind r ical drum , 1.4m h igh has a

diameter o f 80 cms. It contains oi l o f

dens ity 0.78 ton /m Find the weigh t o f o ili f f i l led to capacity .

Answer = 549 kgs


23/31

11. Solu t ion :

Volume = x r x l

= 3.1416 x 0.40 x 1.4m

Volume = 0.70 m


= 0.70 m x 0.78 ton/m

Weight = 0.5489 ton x 1,000 kgs.

Weigh t = 549 kg s.


24/31

12. If the circum ference of a moo ring

rope is 250 mms, what is i ts diameter?

Answer = 80 mms


25/31

12. So lut ion:

Circumference = 3.1416 x Diameter

Diameter = C / 3.1416

= 250 mms / 3.1416

Diameter = 79.58 mms


26/31

13. A log is f loat ing in a pond of relat ive

dens ity 1.008. Its leng th is 12 m and its

diameter is 50 cm s. If 95% o f the log issubmerged, what is i ts weight?

Answer = 2.26 ton s


27/31

13. Solu t ion :

Vol. of Cylinder = x r x h

= 3.1416 x 0.25m x 12m Volume = 2.36m

x 0.95 (95%)

Vo lume = 2.24m


= 2.24m x 1.008m/ton

Weigh t = 2.26 tons


28/31

14. On arr ival at the discharg ing por t,

the d 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?

Answer = 6.00 m


29/31

WT DIST MOMENT 7800 3,200 = 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

= 46,8047,800

OLD KG = 6.00 m


30/31

15. A box-shaped vessel w ith 14 m

beam is f loat ing upr ight at a draught of

5.4 m even keel. Find the d raught i f thevessel is now lis ted 15.

Answer = 7.028 m


31/31

15. Solu t ion :

New Draft = x B x Sin List + ODr. x Cos List

= x 14m x Sin 15 + 5.4m x Cos 15

= 1.8133 + 5.22

New Draft = 7.028 m

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Stability Problems 3

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