Q6 (10 Marks) Ship Stability
SC&S • Written Exam

(a) Explain the purpose of non-watertight longitudinal subdivision of tanks. (6)

(b) A ship 90 m long displaces 5200 tonne and floats at draughts of 4.95 m forward and 5.35 m aft when in sea water of 1023 Kg/m3. The waterplane area is 1100m2, GM, 95m, LCB 0.6m forward of midships and LCF 2.2m aft of midships. Calculate the new draughts when the vessel moves into fresh water of 1002 Kg/m3 (10)

Appeared In: Oct 2024

Verified Model Answer (Text Solution)

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Exam Ready
Part (a)

(b) Given:

$$L\:=\:90m$$

$$\Delta\:=\:5200\:tonnes$$

$$d_{f}=\:4.95m$$

$$d_{a}=\:5.35m$$

$$\rho_{sw}=\:1023kg/m^3$$

$$A_{w}=1100m^2$$

$$G_{ML}=\:95cm$$

$$LCB=\:0.6m\:fwd\:of\:midship\:$$

$$LCF=\:2.2m\:aft\:of\:midship$$

$$when\:vessel\:moves\:into\:fresh\:water\:of\:density\:1002kg/m^3$$

$$new\:drafts\:=\:?$$

$$Change\:in\:mean\:draft\:due\:to\:change\:in\:density\:$$

$$=\:\frac{100\times\Delta}{A_{w}}\left\lbrack\frac{\rho_{s}\:-\:\rho_{r}}{\rho_{s}\times\rho_{r}}\right\rbrack$$

$$=\:\frac{100\:\times5200}{1100}\left\lbrack\frac{1.023\:-\:1.002}{1.023\:\times1.002}\right\rbrack=\:9.68\times10^{-3}m$$

$$Change\:in\:mean\:draft=\:9.7cm$$

$$MCT_{1cm}=\frac{\Delta\:\times GM_{}_{L}}{100\:\times L}$$

$$=\:\frac{5200\:\times95}{100\:\times90}$$

$$MCT_{1cm}=\:54.88\:ton.\:m$$

$$Change\:in\:trim\:when\:vessel\:moves\:from\:SW\:to\:FW$$

$$=\:\frac{\Delta\times FB}{MCT_{1cm}}\left\lbrack\frac{\rho_{s}-\rho_{r}}{\rho_{s}}\right\rbrack$$

$$FB\:=\:LCF\:+\:LCB$$

$$FB\:=\:2.2\:+\:0.6\:=\:2.8m$$

$$=\:\frac{5200\:\times2.8}{54.88}\left\lbrack\frac{1.023\:-\:1.002}{1.023}\right\rbrack\:=\:5.44\:\times10^{-3}$$

$$Change\:in\:trim\:=\:5.45cm\:by\:head$$

$$When\:trim\:by\:head,\:change\:in\:fwd\:draft$$

$$d_{f}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

$$=\:\frac{5.45}{90}\left\lbrack\frac{90}{2}+2.2\right\rbrack$$

$$d_{f}=\:2.858cm$$

$$When\:trim\:by\:head,\:change\:in\:aft\:draft$$

$$d_{a}=\:\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

$$=\:\frac{-5.45}{90}\left\lbrack\frac{90}{2}-2.2\right\rbrack$$

$$d_{a}=\:-2.59cm$$

New draught fwd = draft fwd + change in mean trim + change in fwd draft

$$=\:4.95\:+\:0.097+0.02858\:$$

$$D_{f}=\:5.076m$$

$$New\:aft\:draft\:=\:5.35+0.097-0.0259\:$$

$$D_{a}=5.421m$$

$$New\:fwd\:draft\:D_{f}=5.076m$$

$$New\:aft\:draft\:D_{a}=\:5.421m$$

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