196.NA August 2022 Q.10

 A box-shaped vessel is 20 m long and 10 m wide. The weight of the vessel is uniformly distributed throughout the length and the draught is 2.5 m. The vessel contains ten evenly spaced double bottom tanks, each having a depth of 1m. Draw the shear force diagrams: 

With No.1 and No.10 tanks filled; 

With No.3 and No.8 tanks filled; 

With No.5 and No.6 tanks filled. 

Which ballast condition is to be preferred from the point of view of strength?

GIVEN 

L = 20m

B = 10m

d = 2.5m

D = 1m

Solution

So we have 10 tanks and total length = 20 m

So length of one tank l = 20/10 = 2m

mass added/tank = l x B x D x 1.025

                            = 2 x 10 x 1 x 1.025

                            = 20.5 t

total mass added = 20.5 x 2 = 41 t (2 tank filled condition)

mass/m = 20.5/2 = 10.25 t

weight = 10.25 g kN

buoyancy required = 41g/20 = 2.05 g

Excess load = 10.25 g - 2.05 g = 8.2 g kN

a)With No 1 and 10 tank filled

SF @ the end = 0

SF @ the end of No 10 = -8.2 g x 2 = -16.40 g kN

SF @ midship = -16.4 g + (2.05 g x 8) = 0

SF @ end of No 1 = 2.5 g x 8 = 16.4 g

SF @ fore end = 16.4 g - (8.2 g x 2) = 0

b)With No 3 and 8 tanks filled

SF @ the aft end = 0

SF @ the aft end of No 8 = 2.05 g x 4 = 8.2 g kN

SF @ the fore end of No 8 = 8.2 g - (8.2 g x 2) = -8.2 g kN

SF @ midship = -8.2 g + (2.05 g x 4) = 0

SF @ aft end of No 3 = 2.05 g x 4 = 8.2 g

SF @ fore end of No 3 = 8.2 g - (8.2 g x 2) = -8.2 g

SF @ fore end = 0

c)With No 5 and 6 tanks filled

SF @ the aft end  = 0

SF @ aft end of no 6 = 2.05 g x 8 = 16.4 g

SF @ fore end of no 5 = 16.4 g - (8.2 g x 4) = -16.4 g

SF @ the fore end = 0

The max shear force in b is half of the max values in case of b and c. So b is the best loaded condition