Simpson rules for

3 co-ordinate= 1-4-1

4 co-ordinate= 1-3-3-1

5 co-ordinate= 1-4-2-4-1

6 co-ordinate= 1-3-3-2-4-1

8 co-ordinate=1-3-3-2-4-2-4-1

How came for 6 co-ordinate

By spilt that co-ordinate in firat 4 and last 3

Ex 1-2-3-4-5-6 co ordinates are there

Simpson rule come by splitting is

1-2-3-4-5-6 section

1-3-3-1 first 4

1-2-1 last three

1-3-3-2-2-1 by adding it (Simpson coordinates)

For 8

1-2-3-4-5-6-7-8 section

1-3-3-1

1-4-2-4-1

1-3-3-2-4-2-4-1 Simpson coordinates

Find the TPC when offset of water plane is as follows:

Station AP 2 3 4 5 FP

?? breadth (m) 0.9 1.5 2.1 1.6 0.9 0.1 Length of water plane is 15 meters.  

A. 0.2 t/cm

B. 0.3 t/cm

C. 0.4 t/cm

D. 0.5 t/cm

Soln:

0.9 1 = 0.9

1.5 3 = 4.5

2.1 3 = 6.3

1.6 2 = 3.2

0.9 4 = 3.6

0.1 1 = 0.1

---------------------------------

18.6

H= length of waterplane/(no of coordinates -1)

H=15/(6-1)

H= 3

Aw=H/3*18.6*2 ---------------(*2 is for its half section)

Aw=37.2

Tpc=(Aw*ro)/100= (37.2*1.025)/100=0.38=0.4

Find CWP and TPC of a vessel whose ?? breadths of the water plane as follows:

Station: AP 1 2 3 4 FP

?? Breadths (M) 1.0 1.44 1.55 1.33 0.89 0.2

Length between perpendiculars is 14 meters.Consider the nearest value. 

A. 0.56 tonnes/cm

B. 0.34 tonnes/cm

C. 0.43 tonnes/cm

D. 0.65 tonnes/cm

Soln:

H= 14/(6-1)= 2.8

1.0 1 = 1.0

1.44 3 = 4.32

1.55 3 = 4.65

1.33 2 = 2.66

0.89 4 = 3.56

0.2 1 = 0.2

-------------------------------

16.39

Aw=(2.8/3)*16.39*2

=30.53

Tpc= (30.53*1.025)/100

=0.31 = 0.34

Find the TPC of the water plane. Length between perpendiculars is 15 m.

Station: AP 2 3 4 FP?? breadth (m): 0.6 1.1 1.6 0.9 0.1

 

A. 0.5 tonnes/cm

B. 0.4 tonnes/cm

C. 0.3 tonnes/cm

D. 0.2 tonnes/cm.

Soln:

H=15/4=3.75

0.6 1 =0.6

1.1 4 =4.4

1.6 2 =3.2

0.9 4 =3.6

0.1 1 =0.1

----------------------------

11.9

Aw=(3.75/3)*11.9*2

Aw= 29.75

Tpc=(29.75*1.025)/100

=0.3

Find the Cw of the following

AP-2-3-4-Fp

0.6-1.1-1.6-0.9-0.1

Breadth is 15

a) 0.82 b)0.72 c)0.62 d)0.52

Soln:

0.6 1 =0.6

1.1 4 =4.4

1.6 2 =3.2

0.9 4 =3.6

0.1 1 =0.1

----------------------------

11.9

Here ½ section is not given

AW=H/3*11.9 = 14.87

Cw=Aw/(l*b)=14.87(15*1.6) --------------------------(biggest coordinate)= 0.62

A vessel loaded to summer load line. Summer load line Displacement= 15,000 t, TPC=26.2, ship’s mean draft= 6.2 m. Ship is going in Winter Region…what is the New Displacement?

a) 14112.3

b) 14878.8

c) 14655.1

Soln :

Tpc=26.2

Draft= 6.2

So, tpc=Aw*ro/100

Change in draft =mass added /tpc

So,

Mass added =(6.2/48)*100*26.2

=338.4

New disp=totoal disp –chamge in disp

= 1500-338.4

=14655.1

(Note: summer load line is 1/48 of winter load line so we divided to 48)

A box barge length =35, B=7m, d=3m, in brackish water of density=1.012t/m3 , the length of mid ship compartment is 10m and full depth. the compartment is bilged, find new draft ?

a) 4.2 b)4.4 c) 4.3 d)4.1

Soln :

Increase in draft=(volume of lost bouncy/area of intact water plane)

Volume of lost bouncy=L*B*D

=10*7*3

= 210

Now area of intact waterplane= (35-10)*7

= 175

Increase in draft=(volume of lost bouncy/area of intact water plane)

=(210/175)

=1.2

New draft = 3+1.2 = 4.2

The main dimension of a vessel is 150 m long, 30 m beam and is floating at draft 10 m. The block coefficient and mid-ship section coefficients are 0.86 and 0.95 respectively. A constant section, length 30 m, is added at mid-ship. Find the new block coefficient. 

A. 0.925

B. 0.900

C. 0.875

0.85

Cb=dV/l*b*h0.86=dV/150*30*10

dV= 38700

Cm=Am/b*dAm=0.95*30*10Am=285

Am=285

*30= 8550New Displacement =38700+8550=47250

New Cb= d

V/l*b*d=47250/(150+30)*30*10

=0.875

A barge of length 6.5 m beam 10.2 m and draft 4m in even keel condition the ship is in water of density 1.006t/m^3 find the bodily sincage of the mean draft if the displacement is constant

a) 0.086 b)0.076 c)0.095 d)0.075

Soln:

L=6.5M

B=10.2M

D=4M

Ro=1.006

Disp in sea water= Disp in brackish water

6.5*10.2*4*1.025=6.5*10.2*d*1.006

D=4.0755

D=4.0755-4=0.0755=0.76

The RD solid block of wood is 0.75. All sides are equal to 1.0m. The block is floating in mercury. The RD of mercury is 13.4. What will be weight of a steel block just to submerge the block?

A. 12.6 tonnes;

B. 13.5 tonnes;

C. 14.2 tonnes;

Soln

Density of wood = 0.75*1000------- (multiply water density)

= 750

Density = mass/volume

Mass =750*1*1*1

mass =750 kg

Density of mercury= 13.4*1000

=13400

Weight= 13400-750

=12650

=12.65tonn

Transverse areas at different stations are given as follows:

AP 1 3 4 FP

3.4 m2 4.1 m2 5.8 m2 2.8 m2 1.4 m2

Length of the vessel is 24m. Find the Displacement of vessel

A. 360.1 tonnes

B. 345.2 tonnes

C. 382.8 tonnes

D. 350.0 tonnes

Don’t hav soln but answer is right

D. Filling them with water up to maximum head, which will come on them, i.e. to the top of the air pipe or 2.45 above the crown of the tank which ever is lower.

Table 1

Station

AP 1 2 3 4 5

FP

?? Breadths

2.5 4.73 5.88 6.0 5.09 3.11 0.0

Table 2

Station

AP -1 -2 -3

?? Breadths

2.5 1.88 1.25 0.63

The tables 1 and 2 of a vessel are ??breadths of water plane between perpendiculars and appendage respectively. The lengths between perpendiculars and appendages are 60m and 7.5m respectively. Find the LCF of the water plane from mid ship.

A. 4.87 m towards ford

B. 4.23 m towards ford

C. Midship

D. 4.04 m towards aft

(don’t know the ans also)