Shanghai Maritime University Chapter 1 What is

Shanghai Maritime University

Chapter 1 What is stability? Stability is the ability of a vessel to return to its original condition or position after it has been disturbed by an outside force. Think: The differences between list and heel

Types of initial stability Positive stability Neutral stability Negative stability

Six motions of a vessel z Stability in motion ahead or astern Directional stability Lateral motion stability x y Longitudinal stability Transverse stability Positional motion stability

Archimedes’ principle (law of floating bodies) Floating body displaces a weight of water equal to its own weight; that is, weight equals buoyancy.

Centers of gravity and buoyancy Weight Water line G B △ displacement K

The couple (righting moment) Exterior force M WL WL 1 . . G B K . Z B 1

Metacentre M Metacentre height G Moment of inertia Volume of displacement Vertical centre Of gravity (VCG) B Vertical centre of buoyancy (VCB) Keel K = I V

Ship stability The Concept GM BM KB and BM depend on Geometrical form of ship KG KB KG depends on loading of ship. KM=KG+GM KM=KB+BM

Righting arm=GZ Righting tendency WL WL 1 B 1 Righting moment= GZ x ∆ Foot-tons or meter-metric tons ∆=displacement of vessel B K . . G Z B’

Initial stability The resistance of a boat to a small amount of lateral tilting from its equilibrium WL position. It is determined by the WL 1 difference in the volume of water displaced on each side of the boat as it tilts and is relatively independent of hull shape for small angles of tilt. M θ . . G B K Z B’

Stable equilibrium M WL M is above G WL 1 . . G B K Z B’

Neutral equilibrium G coincides with M G WL . M WL 1 . . B K B’

Unstable equiblibrium WL G is above M Z . . . M WL 1 B K . . B’ G

Metacentric height (GM) KM is the height of Metacenter. GZ=GM sinθ M GM = KM - KG KM is available in hydrostatic data. KG is calculated by officer θ G Center line GM shows the initial stability of ship Z . B 1

Thinking Why GM is not available as measure of stability for All angles of inclination? WL M WL 1 . . . G B K Z B’

Metacentric radius(BM) BM is the radius of the circle KB depends upon the shape of the immersed portion of the hull. WL WL 3 WL 2 M WL 4 . . . WL 1 When KM is not available, KB and BM can be found to use. G Z B B 4 B 3 B 2 B 1 K

Summary Stability is the tendency of a vessel to return to an erect position It can be determined for all angles of inclination by the value of the righting moment ∆GZ or righting arm GZ For small angle of inclination or initial stability, stability could be decided by metacentric height GM. While KM is not available, it maybe necessary to find KB and BM.

Chapter 2 Calculation of the Ship’s vertical center of gravity, KG This chapter we will learn ü how to caculate the vertical center of the ship when ship is loaded ü the main method of calculating KG

Where is the center of Gravity of the Ship? Position of G is in three dimensional space Vertical center of Gravity KG Transverse center of Gravity TCG WL A P TCG G + G KG Longitudinal center of gravity, LCG Transverse stability

Light ship KG and weight Light ship weight and KG are calculated by naval architects After loading , every weight (w) will affecting the original light ship KG and Weight. The theory of moment must be employed by officers Thinking How to get the original data of KG of the ship?

Definition of moment A moment is mathematical term that indicates something, (such as a weight, force, area, or volume), multiplied by its distance from a particular point or axis. A moment of a force (weight) can be expressed in forcedistance units, like tons-meters

A F A B F B 8 m 5 t A 8 m 5 t F B 8 m 5 t

A 5 m F 8 t A B 8 m 5 t 3 m 2 m F 6 t 11 t B 8 m 5 t

A F d 2 d 1 w 1 A D C W 3 w 2 F C w 1 W d 3 w 2 W 3 W replacing w 1, w 2, w 3, the position at which it is placed will be the center of gravity of the weights w 1, w 2, w 3. W x D= (w 1 x d 1) + (w 2 x d 2) + (w 3 x d 3) W

Taking moments about the keel W is the weight of the ship W 1, w 2 are weights loaded/discharged K g 1 w 1 G G’ g 2 w 2 W W’ Loading weights W’= W + w 1 + w 2 KG’=Sum of the moments above K Sum of the weights =W x KG+(w 1 x Kg 1)+(w 2 x Kg 2) W’ Discharging weights W’= W-w 1 -w 2 KG’=’=Sum of the moments about K Sum of the weights =W x KG-(w 1 x Kg 1)-(w 2 x Kg 2) W’

Example 1 A vessel floating at her light draft displaces (weights) 5000 tons. The light ship KG is 20 feet above the keel. Two hundred tons are loaded 10 feet above the keel and 300 tons 5 feet above the light ship KG. What will be the position of the new KG? Weights in tons KG in feet 5000 20 100, 000 200 10 2, 000 300 25 7, 500 5500 total weight Dividing total moments by total weights Moments in foot-tons 109500 total moment KG=19. 9 feet

Example 2 The vessel in example 1 now has the following weights removed from the location listed below. Find the new KG after discharging. Weights in tons KG in feet 700 5 3500 300 2 600 15 2250 1150 total weight Moments in foot-tons 6350 total moment Original 5500 tons 109500 foot tons discharge -1150 tons -6350 Final 4350 tons 103150 total moment New KG= 103150 foot tons/4350 tons =23. 7 feet

Example 3 A vessel is floating in ballast with a displacement of 32000 metric tons and a KG of 9. 25 meters. 22, 000 metric tons of ballast will be discharged with a KG of 8. 5 meters. 600 tons of bunkers will be taken with a KG of 1. 5 meters. 55, 000 tons of grain will be loaded at a height of 9. 7 meters. What will be the position of the new KG? Weights in metric tons KG in meters Initial condition 32, 000 9. 25 296, 000 -ballast -22000 8. 5 -187, 000 +bunkers 600 1. 5 900 55, 000 9. 7 533, 500 item + Grain 65, 600 total weights Moments in Meter-metric tons 643, 400 total moments Dividing total moments by total weights, KG= 9. 81 meters

Exercise 1 A vessel of 13000 tons, KM 10. 5 m (assumed constant), KG 9. 5 m. loads: 400 tons KG 2. 9; 900 tons KG 6. 0 m; 1500 tons KG 10. 6 m; 2000 tones KG 8. 3 m. she discharges: 700 tons KG 1. 5 m; 300 tones KG 12. 7 m. Calculate the moment of statical stability if she is now heeled 8⁰. Weights in tons KG in meters 13, 000 9. 5 123, 500 400 2. 9 1, 160 900 6. 0 5, 400 1, 500 10. 6 15, 900 2, 000 8. 3 16, 600 -700 1. 5 -1, 050 -300 12. 7 -3, 810 16, 800 total weights Moments in Meter- tons 157, 700 total moments New KG=157700/16800=9. 387 KM= 10. 500 m GM= KM – KG=1. 113 m Moment of statical stability= W x GM x sinθ = 16800 x 1. 113 x sin 8 ⁰ = 2602 meter-tons

Exercise 2 A vessel displacing 5800 tons KM 7. 0 m KG 6. 0 m has to load a quantity of deck cargo KG 11. 0 m. what is the maximum quantity that she can load so that her GM is not less than 0. 75 m. Weights in tons KG in feet 5800 6. 0 34800 w 11. 0 11 w (5800+ w tons) total weight KM=7. 00 m new GM=0. 75 Then, new KG=6. 25 Moments in meter-tons (34, 800 + 11 w) total moment New KG=34800+11 w = 6. 25 Then, w=305 tons 5800+ w GM now= KM-KG=7. 0 -6. 0 m= 1. 0 m GM required=0. 75 m GG’=0. 25 m upward GG’=(w x d)/(W+w) 0. 25=(w x 5)/(5800 + w) W= 305 tons

Taking moments about center of gravity Calculating GG’ is shift of center of gravity d A B G G’ w W W+w Weight distance Moment W 0 0 w d wxd W+w wxd GG’=w x d W+w W is the vessel’s displacement before loading the weight w is the weight loaded d is the distance of the loaded weight from the old center of gravity

Example 4 Suppose that on a 10, 000 -ton vessel with a KG of 25 feet, 200 tons are shifted vertically upwards a distance of 20 feet. Find the new KG for the vessel and the shift of G in terms of GG’=w x d W+w GG’= (200 tons X 20 feet )/(10, 000 tons) = 0. 4 feet up New KG= 25 feet + 0. 4 feet = 25. 4 feet

Example 5 The displacement of ship is 10, 000 tons, he head of the jumbo boom is 80 feet above keel. If you lift a container of 30 tons, which is in the lower hold, 1 feet clear of the deck. If GG’ increase when you lift the container more? What is the GG’? GG’= w x d ∆ = 30 tons x 80 feet 10, 000 =0. 24 foot up

Principle of suspended weights When a weight is suspended by a ship’s derrick, its center of gravity is to be considered as being at the derrick head, and it will remain at the derrick head as long as the weight is suspended.

Exercise 1 A vessel of 9920 tons displacement, KM 7. 8 m is to load two 40 tonne lifts on deck KG 13. 5 m and 5. 5 m each side of the center line, by means of her heavy lift derrick whose head is 21. 0 m above the keel and maximum swing out 15 m from the center line. What should be the vessel’s maximum KG before loading, if the list during loading is not to exceed 5⁰? (the inboard weight is to be loaded first) 15 m Derrick head G 9920 G G’ 40 10000 40 13. 5 m 21 m 5. 5 m G’ GMtan 5⁰=0. 082, so GM=0. 9373 KG=KM-GM=7. 800 -0. 9373=6. 8627 m

Basic rule about movement of G: When a weight is loaded on a vessel, G will directly toward the position of the loaded weight When a weight is discharge from a vessel, G will move directly away from the position of the discharged weight When a weight is shifted on a vessel, G will move parallel to the shifted weight. When a weight is suspended on a vessel, G will move toward the point of suspension.

Effect on stability when weight is loaded off the centre line Buoyancy G G’ w gravity

M M Θ=Angle of List θ G GG’= GMtanθ G’ B B ’ W W

Example 6 A vessel of 6500 tons displacement has a KM 7. 2 and KG 6. 8 m. A weight of 100 tons is shifted 2. 3 m to port and 3. 9 m upwards if the vessel is initially upright, calculate the resulting list. For vertical shift GG’= (100 x 3. 9)/6500 = 0. 06 m rise KG=6. 8 m Then, KG’=6. 86 m KM= 7. 20 m Then, G’M=+0. 34 For horizontal shift GG’= (100 x 2. 3)/6500 = 0. 0354 m to port Tan θ =GG’/GM =0. 0354/0. 34 = 0. 1041 Θ= 5. 95⁰ to port

Steps to find KG Find KG for every cargo, fuel, water, and stores compartment or tank on the vessel Multiply these distances by the weights in the respective locations Add total weights including weight of light ship Add total moments including the moment of light ship Divide total moments by total weights produce final KG