Q9 (10 Marks) Hull Construction 🔥 Repeated 2x in exams
SC&S • Written Exam

The 1⁄2 ordinates of a waterplane at 15m intervals, commencing from aft are 1, 7, 10.5, 11, 11, 10.5, 8.4 and Om. Calculate:

(a) TPC (6)

(b) Distance of the centre of flotation from midships (5)

(c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.(5)

Appeared In: Jul 2022Apr 2018

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready

1/2 Ord

SM

Product

Lever

Product

Lever

Product

1

1

1

_4

4

+4

16

7

4

28

+3

84

+3

252

10.5

2

21

+2

42

+2

84

11

4

44

+1

44

+1

44

11

2

22

0

ΣMA = 174

0

0

10.5

4

42

-1

-42

-1

42

8

2

16

-2

-32

-2

64

4

4

16

-3

-48

-3

144

0

1

0

-4

0

-4

0

ΣA = 190

ΣMF = -122

ΣI = +646

$$Common\:interval\:\left(h\right)=15$$

$$Waterplane\:area\:\left(A_{w}\right)=2\times\frac{h}{3}\sum A$$

$$=2\times\frac{15}{3}\times190$$

$$=1900m^2$$

$$Longitudinal\:centre\:of\:flotation\:LCF=h\times\frac{\sum M_{A}+\sum M_{F}}{\sum A}$$

$$=15\times\frac{174-122}{190}$$

$$LCF=4.105m$$

$$A_{w}=\frac{100\:\times\:TPC}{\rho}$$

$$TPC=\frac{1.025\times1900}{100}$$

$$TPC=19.475m^2$$

$$Second\:moment\:about\:midships\:\left(I_{m}\right)=2\times\frac{h^3}{3}\sum I$$

$$=2\times\frac{15^3}{3}\times646$$

$$I_{m}=1453500m^4$$

$$Second\:moment\:of\:area\:about\:centroid\:\left(I_{F}\right)=I_{m}-A_{w}\times TPC^2$$

$$1453500-\left(1900\times41.05^2\right)$$

$$I_{F}=1421483m^4$$

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