HYDRAULICS Engr. Michelle D. Enriquez Lesson 4 Buoyancy and Stability of floating bodies TOPICS Buoyancy Immersed and Floating Bodeis Stability of Floating Bodies LEARNING OUTCOMES At the end of the lesson, you should be able to: Understand why bodies immersed and float in liquid Analyse and design of dams Understand the principle of buoyancy This lesson describes how to determine whether an object will tip over or remain in an upright position when placed in a liquid. This topic is important for the design of objects such as ships and buoys. Although civil engineers are not boat designers, they do have to deal with cases of buoyancy from time to time. Some typical examples are: 1. Buried gas pipelines in waterlogged ground 2. Exploration rigs used by oil or gas corporations 3. Towing large steel dock/lock gates by sea or river (assuming that the structure can float, of course) Buoyancy A fundamental principle affecting objects submerged in fluids was discovered by Greek mathematician and natural philosopher Archimedes. Archimedes’ principle can be stated as follows: “Any object completely or partially submerged in a fluid is buoyed up by a force With magnitude equal to the weight of the fluid displaced by the object” Everyone has experienced Archimedes’ principle. It’s relatively easy, for example, to lift someone if you’re both standing in a swimming pool, whereas lifting that same individual on dry land may be a difficult task. Water provides partial support to any object placed in it. We often say that an object placed in a fluid is buoyed up by the fluid, so we call this upward force the buoyant force. The buoyant force is not a mysterious new force that arises in fluids. In fact, the physical cause of the buoyant force is the pressure difference between the upper and lower sides of the object, which can be shown to be equal to the weight of the displaced fluid. The fluid inside the indicated sphere, colored darker blue, is pressed on all sides by the surrounding fluid. Arrows indicate the forces arising from the pressure. Because pressure increases with depth, the arrows on the underside are larger than those on top. Adding them all up, the horizontal components cancel, but there is a net force upward. This force, due to differences ⃗ . The sphere of water neither rises nor falls, so the vector sum of the in pressure, is the buoyant force 𝐵 buoyant force and the force of gravity on the sphere of fluid must be zero, and it follows that 𝐵 = 𝑀𝑔, where 𝑀 is the mass of the fluid HYDRAULICS Engr. Michelle D. Enriquez Replacing the shaded fluid with a cannon ball of the same volume, as in Figure b, changes only the mass on which the pressure acts, so the buoyant force is the same: 𝐵 = 𝑀𝑔,, where M is the mass of the displaced fluid, not the mass of the cannon ball. The force of gravity on the heavier ball is greater than it was on the fluid, so the cannon ball sinks. Archimedes’ principle can also be obtained, relating pressure and depth, using Horizontal forces from the pressure cancel, but in the vertical direction 𝑃2 𝐴 acts upward on the bottom of the block of fluid, and 𝑃1 𝐴 and the gravity force on the fluid, Mg, act downward, giving: 𝑩 = 𝑷𝟐 𝑨 − 𝑷𝟏 𝑨 = 𝑴𝒈 where the buoyancy force has been identified as a difference in pressure equal in magnitude to the weight of the displaced fluid. This buoyancy force remains the same regardless of the material occupying the volume in question because it’s due to the surrounding fluid. Using the definition of density: 𝑩 = 𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅 𝒈 where𝝆𝒇𝒍𝒖𝒊𝒅 is the density of the fluid and 𝑽𝒇𝒍𝒖𝒊𝒅 is the volume of the displaced fluid. This result applies equally to all shapes because any irregular shape can be approximated by a large number of infinitesimal cubes. A totally submerged object that is less dense than the fluid in which it is submerged is acted upon by a net upward force. (b) A totally submerged object that is denser than the fluid sinks. HYDRAULICS Engr. Michelle D. Enriquez Case I: A Totally Submerged Object. When an object is totally submerged in a fluid of density 𝝆𝒇𝒍𝒖𝒊𝒅, the upward buoyant force acting on the object has a magnitude of 𝑩 = 𝝆𝒇𝒍𝒖𝒊𝒅𝑽𝒐𝒃𝒋 𝒈, where 𝑽𝒐𝒃𝒋 is the volume of the object. If the object has density 𝝆𝒐𝒃𝒋 , the downward gravitational force acting on the object has a magnitude equal to 𝑤 = 𝑚𝑔 = 𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈, and the net force on it is 𝐵 − 𝑤 = (𝝆𝒇𝒍𝒖𝒊𝒅 − 𝝆𝒐𝒃𝒋 )𝑽𝒐𝒃𝒋 𝒈. Therefore, if the density of the object is less than the density of the fluid, the net force exerted on the object is positive (upward) and the object accelerates upward, as in figure a; whereas if the density of the object is greater than the density of the fluid, as Figure b, the net force is negative and the object accelerates downward. Case II: A Floating Object. Now consider a partially submerged object in static equilibrium floating in a fluid, as in the figure above. In this case, the upward buoyant force is balanced by the downward force of gravity acting on the object. If 𝑽𝒇𝒍𝒖𝒊𝒅 is the volume of the fluid displaced by the object (which corresponds to the volume of the part of the object beneath the fluid level), then the magnitude of the buoyant force is given by 𝑩= 𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅 𝒈. Because the weight of the object is 𝑤 = 𝑚𝑔 = 𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈, and because 𝑤 = 𝐵, it follows that𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅𝒈 = 𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈. 𝝆𝒐𝒃𝒋 𝑽𝒇𝒍𝒖𝒊𝒅 = 𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒐𝒃𝒋 Immersed and Floating Bodies Immersed Bodies When a body is completely immersed in a liquid, its stability depends on the relative positions of the center of gravity of the body and the centroid of the displaced volume of fluid, which is called the center of buoyancy. If the center of buoyancy is above the center of gravity, such as in the figure shown below, any tipping of the body produces a righting couple, and consequently, the body is stable. However, if the center of gravity is above the center of buoyancy, any tipping produces an increasing overturning moment, thus causing the body to turn through 180°. This is the condition shown in c. Finally, if the center of buoyancy and center of gravity are coincident, the body is neutrally stable—that is, it lacks a tendency for righting or for overturning, as shown in b. Conditions of stability for immersed bodies: (a) Stable. (b) Neutral. (c) Unstable HYDRAULICS Engr. Michelle D. Enriquez Floating Bodies The question of stability is more involved for floating bodies than for immersed bodies because the center of buoyancy may take different positions with respect to the center of gravity, depending on the shape of the body and the position in which it is floating. For example, consider the cross section of a ship shown in a. Here the center of gravity G is above the center of buoyancy C. Therefore, at first glance it would appear that the ship is unstable and could flip over. However, notice the position of C and G after the ship has taken a small angle of heel. Ship stability relations As shown in b, the center of gravity is in the same position, but the center of buoyancy has moved outward of the center of gravity, thus producing a righting moment. A ship having such characteristics is stable. The reason for the change in the center of buoyancy for the ship is that part of the original buoyant volume, as shown by the wedge shape AOB, is transferred to a new buoyant volume EOD. Because the buoyant center is at the centroid of the displaced volume, it follows that for this case the buoyant center must move laterally to the right. The point of intersection of the lines of action of the buoyant force before and after heel is called the metacenter M, and the distance GM is called the metacentric height. If GM is positive—that is, if M is above G—the ship is stable; however, if GM is negative, the ship is unstable. Quantitative relations involving these basic principles of stability are presented in the next paragraph (a) Plan view of ship at waterline. (b) Section A-A of ship Consider the ship shown above, which has taken a small angle of heel 𝛼. First evaluate the lateral displacement of the center of buoyancy, then it will be easy by simple trigonometry to solve for the metacentric height GM or to evaluate the righting moment. Recall that the center of buoyancy is at the centroid of the displaced volume. Therefore, resort to the fundamentals of centroids to evaluate the displacement. From the definition of the centroid of a volume, 𝑥̅ 𝑉 = ∑ 𝑥𝑖 ∆𝑉𝑖 Where: 𝑥̅ = 𝐶𝐶 ′ (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑏𝑜𝑢𝑡 𝑤ℎ𝑖𝑐ℎ 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑡𝑜 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑉) w 𝑉 = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 ∆𝑉𝑖 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 𝑥𝑖 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 HYDRAULICS Engr. Michelle D. Enriquez Take moments about the plane of symmetry of the ship. Recall from mechanics that volumes to the left produce negative moments and volumes to the right produce positive moments. A convenient way to do this is to consider the moment of the volume before heel, subtract the moment of the volume represented by the wedge AOB, and add the moment represented by the wedge EOD. In a general way this is given by the following equation: 𝑥̅ 𝑉 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑉 𝑏𝑒𝑓𝑜𝑟𝑒 ℎ𝑒𝑒𝑙 − 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑉𝐴𝑂𝐵 + 𝑚𝑜𝑚𝑒𝑛𝑡 𝑉𝐸𝑂𝐷 Because the original buoyant volume is symmetrical with 𝑦 − 𝑦, the moment for the first term on the right is zero. Also, the sign of the moment of 𝑉𝐴𝑂𝐵 is negative; therefore, when this negative moment is subtracted from the right-hand side, the result is: 𝑥̅ 𝑉 = ∑ 𝑥𝑖 ∆𝑉𝑖𝐴𝑂𝐵 + ∆𝑉𝑖𝐸𝑂𝐷 In addition to the force of gravity or weight, all objects submerged in a fluid are acted on a force BF. The buoyant force acts upward and is equal to the weight of the fluid displaced by the object. Buoyancy force can be simply defined as: 𝐵𝐹 = 𝛾𝑤 𝑉𝐷 Where: 𝐵𝐹 = 𝑏𝑢𝑜𝑦𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝛾𝑤 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑉𝐷 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒 For a freely floating object (with no external forces) the weight of the object is equal to the buoyant force on the object (acting upward) 𝑊=𝐵𝐹 𝑊 = 𝛾𝑤 𝑉𝐷 Example 1: A driver and his suit weight 890N. It requires 130N of lead to sink him in fresh water. If the volume of the lead is 0.00117𝑚3 . What is the volume of the driver and his suit? Solution: ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹1 + 𝐵𝐹1 = 𝑊1 + 𝑊2 𝛾𝑤1 𝑉𝐷1 + 𝛾𝑤2 𝑉𝐷2 = 890𝑁 + 130𝑁 𝛾𝑤 (𝑉𝐷1 + 11.3𝑉𝐷2 ) = 1020𝑁 𝑁 9810 3 (𝑉𝐷1 + 0.00117) = 1020𝑁 𝑚 𝑉𝐷1 = 0.1028𝑚3 . HYDRAULICS Engr. Michelle D. Enriquez Example 2: Cube A is 30cm long along edge and weighs 1445N. It is attached to the square prism B, which is 15cm x 15cm x 2.4m 𝑁 and weighs4700 𝑚3. What length of B will project above the liquid surface is the relative density of the liquids is 1.50. ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹1 + 𝐵𝐹1 = 𝑊1 + 𝑊2 1.5 (9810 𝑁 𝑁 𝑁 ) (0.30)2 (1.0𝑚) + 1.5 (9810 3 ) (0.15)2 (2.4 − ℎ) = 1445𝑁 + 4700 3 (0.15m x 0.15m x 2.4m) 𝑚3 𝑚 𝑚 1324.5𝑁 + 1.5 (9810 𝑁 ) (0.15)2 (2.4 − ℎ) = 1698.8𝑁 𝑚3 𝑁 ) (0.15)2 (2.4 − ℎ) = 374.3𝑁 𝑚3 ℎ = 1.2694𝑚 1.5 (9810 Example 3: A block of wood is 20mm thick floating in seawater. The specific gravity of wood is 0.65 while that of the seawater is 1.03. Find the area of a block of wood, which will just support a woman weighing 80kg when the top of the surface is just at the water surface. Solution: ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹1 = 𝑊𝑤𝑜𝑜𝑑 + 𝑊𝑤𝑜𝑚𝑎𝑛 𝑆𝐺𝑠𝑤 𝛾𝑤 𝑉𝐷 = 𝑆𝐺𝑤 𝛾𝑤 𝑉 + 𝑊𝑤𝑜𝑚𝑎𝑛 (1.03) (9810 ℎ = 0.095𝑚3 𝑁 𝑚 3 ) 𝐴(0.02𝑚) = 0.65 (9810 𝑁 𝑚 3 ) (𝐴)(0.02𝑚) + 80𝑘𝑔 𝑚 (9.81 2 ) 𝑠 Example 4: The weight of certain crown in air is 14N and its weight in water is 12.7N. Assuming that the crown is an alloy of gold (SG= 19.3) ad silver (SG = 10.5). Assume unit weight of 𝑘𝑁 water is 9.79 𝑚 3 a. Compute the volume of the crown b. Compute the specific gravity of the brown c. Compute the fraction of silver in the crown. HYDRAULICS Engr. Michelle D. Enriquez Solution: ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊𝑎𝑖𝑟 − 𝑊𝑤𝑎𝑡𝑒𝑟 9,790 𝑘𝑁 𝑉 𝑚 3 𝑐𝑟𝑜𝑤𝑛 = 14𝑁 − 12.7𝑁 𝑽𝒄𝒓𝒐𝒘𝒏 = 𝟎. 𝟎𝟎𝟎𝟏𝟑𝟐𝟕𝒎𝟑 Specific gravity of the crown: 𝑊 𝛾𝑐𝑟𝑜𝑤𝑛 = 𝑉 14𝑁 𝛾𝑐𝑟𝑜𝑤𝑛 = 0.0001327𝑚3 𝑘𝑁 𝛾𝑐𝑟𝑜𝑤𝑛 = 105.422 3 𝑚 𝑘𝑁 105.422 3 𝛾𝑐𝑟𝑜𝑤𝑛 𝑚 = 10.768 𝑆𝐺𝑐𝑟𝑜𝑤𝑛= = = 𝑘𝑁 𝛾𝑤 9.79 3 𝑚 Fraction of silver in the crown Let 𝑥 = 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑖𝑙𝑣𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑟𝑜𝑤𝑛 10.5 𝑥 + (1 − 𝑥 )19.3 = 10.768 𝒙 = 𝟎. 𝟗𝟔𝟗𝟓 Example 5: A wooden buoy of specific gravity 0.75 floats in a liquid with specific gravity of 0.85. 1. What is the percentage of the volume above the liquid surface to the total volume of the buoy 2. If the volume above the liquid is 0.0145𝑚3, what is the weight of wooden buoy? 3. What load will cause the buoy to be fully submerged? Solution: a. % of volume above the liquid surface ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 (0.75) (9.810 𝑘𝑁 𝑚3 ) 𝑉 = (0.85) (9.810 𝑉 = 1.13333𝑉2 𝑉2 = 0.8824𝑉 𝑉1 = 𝑉 − 𝑉2 𝑉1 = 𝑉 − 0.8824𝑉 𝑉1 = 0.1177𝑉 𝑉1 = 0.1178 𝑉 𝑉1 = 11.78% 𝑉 𝑘𝑁 𝑚3 ) 𝑉2 HYDRAULICS Engr. Michelle D. Enriquez b. Weight of the wooden buoy 𝑊 𝛾𝑤𝑜𝑜𝑑 = 𝑉 𝑊 = 𝛾𝑤𝑜𝑜𝑑 𝑉 Since 𝑉1 = 0.1177𝑉 = 0.0145𝑚3 = 0.1177𝑉; 𝑉 = 0.1232𝑚3 𝑊 = (0.75) (9.810 𝑘𝑁 𝑚3 ) (0.1232𝑚3 ) 𝑊 = 0.9064𝑘𝑁 c. Load that will cause the buoy to be fully submerged 𝑃 = 𝛾𝑏𝑢𝑜𝑦 𝑉1 𝑃 = (0.85) (9.810 𝑘𝑁 𝑚3 ) (0.0145𝑚3 ) 𝑃 = 0.1209 𝑘𝑁 Example 6: A block of wood 0.60m x 0.60m x h meters in dimension was thrown into the water, it floats with 0.18m projecting above the water surface. The same block was thrown into a container of a liquid having a specific gravity of 0.90 and it floats with 0.14cm projecting above the liquid surface. 1. Determine the value of “h” 2. Determine the specific gravity of the block 3. Determine the weight of the block Solution: a. Value of h ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝛾𝑏𝑙𝑜𝑐𝑘𝑠𝑢𝑏 𝑉 𝑘𝑁 𝑘𝑁 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 (9.810 3 ) (0.60𝑚)(0.60𝑚)ℎ = (9.810 3 ) (0.60𝑚)(0.60𝑚)(ℎ − 0.18) 𝑚 𝑚 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = (ℎ − 0.18) − 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 Thrown in other liquid: ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝛾𝑏𝑙𝑜𝑐𝑘𝑠𝑢𝑏 𝑉 𝑘𝑁 𝑘𝑁 (9.810 3 ) (0.60𝑚)(0.60𝑚)ℎ = (0.90) (9.810 3 ) (0.60𝑚)(0.60𝑚)(ℎ − 0.14) 𝑚 𝑚 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = 0.90(ℎ − 0.14) − 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 equating 1 and 2: (ℎ − 0.18) = 0.90(ℎ − 0.14) (ℎ − 0.18) = 0.90ℎ − 0.126 ℎ = 0.54𝑚 : HYDRAULICS Engr. Michelle D. Enriquez b. Specific gravity of block From equation 1: 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = (ℎ − 0.18) 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 (0.54) = (0.54 − 0.18) 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 = 0.6667 c. Weight of the block 𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝑊 𝑆𝐺𝑏𝑙𝑜𝑐𝑘 𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝑊 𝑊 = (0.667) (9.810 𝑊 = 1.272𝑘𝑁 𝑘𝑁 ) (0.60𝑚)(0.60𝑚)(0.54) 𝑚3 Stability of Floating Bodies The stability of floating bodies is defined as its ability to return to its neutral position after the external force has been applied and removed. The location of the metacentre is important in determining the stability. The metacentre is a point on the vertical neutral axis through which the buoyant force always acts for small angles of tilt. For a large angle of tilt, the metacentre may move along the neutral axis. 1. For stability to exist, the objects center of gravity must below its metacentre. 2. For objects submerged in fluids, the metacentre is located at its center of buoyancy. 3. Where: For partially objects the location of the metacentre is found from the following relation: 𝐴𝑀 = 𝐴𝐵𝑜 + 𝑀𝐵𝑜 𝐴𝑀 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑒𝑡𝑎𝑐𝑒𝑛𝑡𝑒𝑟 𝐴𝐵𝑜 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡𝑒ℎ 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡𝑦ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑦𝑒𝑟 𝑜𝑓 𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦 The location of the buoyancy 𝐵𝑜 is located at the geometric center of the displacement volume. The magnitude and geometry of the displaced volume maybe found from the buoyancy equation. The distance from the center of the buoyancy to the center of the metacentre 𝑀𝐵𝑜 = 𝐼𝑠 𝑉 Where: 𝐼𝑠 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 HYDRAULICS Engr. Michelle D. Enriquez When the center of gravity is below the metacentre, there is a couple acting to return the vessel to its neutral position but if the center of gravity is above the metacentre, the couple will continue to tip the vessel. Vertical Stability of Floating Bodies: 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑒𝑑 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝐵𝐹 2 𝐵𝐹𝑥 = 𝐹 ( 𝐵) 3 𝐵 𝐻 (𝑦) 2 𝐵𝐹𝑥 = ( 𝐵) 𝛾𝑤 2 2 3 𝑦 tan 𝜃 = 𝐵 2 𝐵𝑡𝑎𝑛 𝜃 𝑦= 2 𝐵𝐹 = 𝛾𝑤 𝑉 𝑥 = 𝑀𝐵0 𝑠𝑖𝑛𝜃 𝐵𝑡𝑎𝑛 𝜃 𝐵𝐻( 2 ) 2 ( 𝐵) 𝛾𝑤 2 2 3 𝐵 𝐻 (𝐵𝑡𝑎𝑛 𝜃) 𝐵 𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 = ( ) 𝛾𝑤 2 2 3 3 𝐵 𝐻 𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 = 12 𝑡𝑎𝑛 𝜃𝛾𝑤 𝐵3 𝐻 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 = 𝑡𝑎𝑛 𝜃 12 𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 = But 𝑠𝑖𝑛𝜃 = tan 𝜃 ( 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒𝑠0 𝑉𝑀𝐵0 = 𝐼 𝐼 𝑀𝐵0 = 𝑉 Where: 𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑙𝑜𝑛𝑔 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑉 = 𝑚𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 Metacentric Height: 𝑀𝐵0 = 𝐵2 𝑡𝑎𝑛2 𝜃 (1+ ) 12𝐷 2 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 HYDRAULICS Engr. Michelle D. Enriquez Example 1: A rectangular barge has dimensions of 30m long, 10m wide and 3m deth. It weighs 4500kN when loaded, with its center of gravity along the vertical axis is 4m from the botoom of the barge. The barge is floating in seawater havinga a secific gravity of 1.01 Determine the draft of the barge in seawater Determine the metacentric height of the barge dur to rolling Determine the metacentruic height of the barge due to pitching Solution: a. Draft of the barge in seawater ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 𝛾𝑏𝑎𝑟𝑔𝑒 𝑉 = 𝑊 𝑆𝐺𝑏𝑎𝑟𝑔𝑒 𝛾𝑏𝑎𝑟𝑔𝑒 𝑉 = 𝑊 𝑘𝑁 (1.01) (9.810 3 ) (30𝑚)(10𝑚)(ℎ) = 4500 𝑚 𝐷 = 1.5139𝑚 b. Metacentric height if the barge is rolling 𝐼 𝑀𝐵0 = 𝑉 𝐵𝐻3 (30𝑚)(10𝑚)3 𝐼= = = 2500𝑚4 12 12 𝑉 = 𝐴ℎ = (30𝑚)(10𝑚)(1.5139𝑚) = 454.17𝑚3 𝑀𝐵0 = 𝐼 2500𝑚4 = = 5.5045𝑚 𝑉 454.17𝑚3 Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 5.5045𝑚 − 3.245𝑚 𝑀𝐺 = 2.2595𝑚 HYDRAULICS Engr. Michelle D. Enriquez c. Metacentric height if the barge is pitching 𝐼 𝑉 𝐵𝐻3 (10𝑚)(30𝑚)3 𝐼= = = 22500𝑚4 12 12 𝑉 = 𝐴ℎ = (30𝑚)(10𝑚)(1.5139𝑚) = 454.17𝑚3 𝑀𝐵0 = 𝑀𝐵0 = 𝐼 22500𝑚4 = = 49.5409𝑚 𝑉 454.17𝑚3 Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 49.5409 − 3.245𝑚 𝑀𝐺 = 46.2959𝑚 Example 2: A rectangular boat 8m long by 6m wide and 2.4m high is sibmerged in water by 1.8m a. Find the total draft after 𝑃 is placed at the edge of the boat without causing it to sink. Assume water is fresh with density equa to 1000 𝑘𝑔/𝑚3 b. What is the maximum P that can be placed at the edge of the boat without causing it to sink . assume water is fresh with density equal to 1000 𝑘𝑔/𝑚3 c. Solve the maximum P if the boat is submerged in sea water with a density at 1026 𝑘𝑔/𝑚3 Solution: a. Draft after P is placed at the edge of the boat ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑃 𝛾𝑤 𝑉 = 𝑃 𝑆𝐺𝜌𝑤 𝑉 = 𝑃 𝑘𝑔 𝑃 = (6.0𝑚)(8.0𝑚) (1000 3 ) (𝑥 ) 𝑚 𝑃 = 48,000𝑥 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑛′ 𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 (0.6 − 𝑥 )(3) 𝑘𝑔 2 𝑃 (3.0𝑚) = (8.0𝑚) (1000 3 ) ( ) (6.0𝑚) 2 𝑚 3 (0.6 − 𝑥 )(3) 𝑘𝑔 2 48,000𝑥 (3.0𝑚) = (8.0𝑚) (1000 3 ) ( ) (6.0𝑚) 2 𝑚 3 𝑘𝑔 48,000𝑥 (3.0𝑚) = (0.6 − 𝑥 )(12.0𝑚) (1000 3 ) (4.0𝑚) 𝑚 1 𝑥 = (0.6 − 𝑥 )( ) 3 𝑥 = 0.15𝑚 𝑇𝑜𝑡𝑎𝑙 𝑑𝑟𝑎𝑓𝑡 𝑎𝑓𝑡𝑒𝑟 𝑃 𝑖𝑠 𝑙𝑜𝑎𝑑𝑒𝑑 = 1.8 + 0.15 𝑻𝒐𝒕𝒂𝒍 𝒅𝒓𝒂𝒇𝒕 𝒂𝒇𝒕𝒆𝒓 𝑷 𝒊𝒔 𝒍𝒐𝒂𝒅𝒆𝒅 = 𝟏. 𝟗𝟓𝒎 HYDRAULICS Engr. Michelle D. Enriquez b. Maximum P at the edge of the boat 𝑃 = 48,000𝑥 𝑃 = 48,000(0.15) 𝑃 = 7200𝑘𝑔 c. Maixumum P if the boat is submerged ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 𝑆𝐺𝜌𝑤 𝑉 = 𝑊 𝑊 = (6.0𝑚)(8.0𝑚) (1000 𝑘𝑔 ) (1.8) 𝑚3 𝑊 = 86,400𝑘𝑔 ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑊 𝑆𝐺𝜌𝑤 𝑉 = 𝑊 86400𝑘𝑔 = (6.0𝑚)(8.0𝑚) (1026 𝐷 = 1.7𝟓𝟒𝟒𝒎 ∑ 𝐹𝑦 = 0 ↑ (+) 𝐵𝐹 = 𝑃 𝛾𝑤 𝑉 = 𝑃 𝑆𝐺𝜌𝑤 𝑉 = 𝑃 𝑃 = (6.0𝑚)(8.0𝑚) (1026 𝑃 = 49,248𝑥 𝑘𝑔 ) (𝑥 ) 𝑚3 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑛′ 𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 (0.65 − 𝑥 )(3) 𝑘𝑔 2 𝑃 (3.0𝑚) = (8.0𝑚) (1000 3 ) ( ) (6.0𝑚) 2 𝑚 3 (0.6 − 𝑥 )(3) 𝑘𝑔 2 49,248𝑥 (3.0𝑚) = (8.0𝑚) (1000 3 ) ( ) (6.0𝑚) 2 𝑚 3 𝑘𝑔 49,248𝑥 (3.0𝑚) = (0.65 − 𝑥 )(12.0𝑚) (1000 3 ) (4.0𝑚) 𝑚 1 ( ) 𝑥 = 0.65 − 𝑥 ( ) 3 𝑥 = 0.1625𝑚 𝑃 = 49,248𝑥 𝑃 = 49,248(0.1625) 𝑃 = 8002.8𝑘𝑔 𝑘𝑔 ) (𝐷 ) 𝑚3 HYDRAULICS Engr. Michelle D. Enriquez Example 3: A rectangualr scow 9.0m wide and 15.0m long and 3.6m high has a draft in seawater of 2.4m. its centyer of gravity is 2.70m above the bottom of the scow. a. Determine the initial metacentric height b. If the scow is tilted until one end is just submerged in water, find the sidewise shjifting of the center of buoyancy. c. Determine the final metacentric height Solution: a. Initial metacentric height MG 𝜃=0 (15𝑚)(9𝑚)3 𝑏ℎ3 𝐼 911.25𝑚4 12 12 𝑀𝐵0 = = = = 2.8125𝑚 𝑉 𝑙𝑤ℎ (15𝑚)(9.0𝑚)(2.4𝑚) 324𝑚3 𝐺𝐵0 = 2.70𝑚 − 1.20𝑚 = 1.50𝑚 Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 2.8125𝑚 − 1.50𝑚 𝑀𝐺 = 1.3125𝑚 b. Sidewise shifting of the center of buoyancy 1.2 ; 𝜃 = 14.93° 4.5 𝑥𝑏 𝑠𝑖𝑛𝜃 = 𝑀𝐵𝑜 𝐼 𝑡𝑎𝑛2 𝜃 ] 𝑀𝐵0 = [ 1 + 𝑉 2 𝑡𝑎𝑛2 14.93 ] 𝑀𝐵0 = 2.8125 [ 1 + 2 𝑀𝐵0 = 2.9125 𝑥𝑏 = 2.9125 sin 14.93 𝑥𝑏 = 0.7497𝑚 𝑡𝑎𝑛𝜃 = HYDRAULICS Engr. Michelle D. Enriquez c. Final metacentric height 𝑀𝐵0 = 𝐵2 𝑡𝑎𝑛2 𝜃 (1+ ) 12𝐷 2 𝑀𝐵0 = (9)2 𝑡𝑎𝑛2 14.93 (1+ ) 12(2.4) 2 𝑀𝐵0 = 2.9125 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 2.9125𝑚 − 1.50𝑚 𝑀𝐺 = 1.4125𝑚 Example 4: The center of gravity of the barge shown is on te mid sectios of 2.7m above the bottom of tyeh barge, the barge has a dimesnion of 12m long, 4.5m wide and 3.0m high. The draft is 2.40m . If an outside force (wind or wave action) hells the barge until point A is just the water surface. a. Comoute the moment at the wedge shift in kg.m b. What is the moment of the center of buoyancy shift? c. Compute the value of rigting or upsetting moment. Solution: a. Moment of wedge shift 𝐵𝐹 = 𝛾𝑤 𝑉 𝐵𝐹 = (1000 𝑘𝑔 ) (12𝑚)(4.5𝑚)(2.4𝑚) 𝑚3 𝐵𝐹 = 129,600 𝑘𝑔 𝑡𝑎𝑛𝜃 = 0.6 ; 𝜃 = 14.93° 2.25 𝑀𝐵0 = 𝐵2 𝑡𝑎𝑛2 𝜃 (1+ ) 12𝐷 2 𝑀𝐵0 = (4.5)2 𝑡𝑎𝑛2 14.93 (1+ ) 12(2.4) 2 𝑀𝐵0 = 0.7281m < 𝐺𝐵0 = 1.5 HYDRAULICS Engr. Michelle D. Enriquez 𝑀𝐺 𝑀𝐺 𝑀𝐺 𝑀𝐺 = 𝑀𝐵0 ± 𝐺𝐵0 = 2.9125𝑚 − 1.50𝑚 = −0.7719𝑚 = 0.7719𝑚 (𝑚𝑒𝑡𝑎𝑐𝑒𝑛𝑡𝑒𝑟 𝑖𝑠 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑔𝑟𝑎𝑣𝑖𝑡𝑦) 𝑦 = 𝑀𝐵0 𝑠𝑖𝑛𝜃 𝑦 = 0.7281sin(14.93) 𝑦 = 0.1876𝑚 𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃 𝑥 = 0.7719sin(14.93) 𝑥 = 0.1989𝑚 (0.6𝑚)(2.25𝑚) 𝑘𝑔 2 (12.0𝑚) (1000 3 ) (4.5𝑚) 2 𝑚 3 𝑴 = 𝟐𝟒, 𝟑𝟎𝟎 𝒌𝒈𝒎 𝑀= b. Moment of center of buoyancy shift 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑠ℎ𝑖𝑓𝑡 = 𝐵𝐹𝑦 𝑀 = 129,600 𝑘𝑔(0.1876𝑚) 𝑴 = 𝟐𝟒, 𝟑𝟏𝟐. 𝟗𝟔𝒌𝒈𝒎 c. Upsetting moment 𝑢𝑝𝑠𝑒𝑡𝑡𝑖𝑛𝑔 = 𝐵𝐹𝑥 𝑀 = 129,600 𝑘𝑔(0.1989𝑚) 𝑴 = 𝟐𝟓, 𝟕𝟕𝟕. 𝟒𝟒𝒌𝒈𝒎 HYDRAULICS Engr. Michelle D. Enriquez Example 5: The barge shown is 10.0m wide by 30.0 m long when loade, the barge diaplecs 5MN and its center of gravity is 0.5m abiove the water line. a. Find the value of 𝑀𝐵0 b. Determine the metacenter height above the center of gravity for a roll angle of 10° c. Compute the righting moment Solution: a. Value of 𝑀𝐵0 𝜃=0 𝐵𝐹 = 𝑊 𝛾𝑤 𝑉 = 5 𝑥 106 𝑘𝑔 𝑚 (1000 3 ) (9.81 2 )(10𝑚)(30𝑚)ℎ = 5 𝑥 106 𝑚 𝑠 ℎ = 1.699𝑚 (30𝑚)(10𝑚)3 𝑏ℎ3 𝐼 12 12 𝑀𝐵0 = = = 𝑉 𝑙𝑤ℎ (10𝑚)(30𝑚)(1.699𝑚) 2500𝑚4 𝑀𝐵0 = 509.7𝑚3 𝑴𝑩𝟎 = 𝟒. 𝟗𝟎𝟒𝟖𝒎 b. Metacenter height 𝐺𝐵0 = 0.5𝑚 + 0.85𝑚 𝐺𝐵0 = 1.35𝑚 Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 4.9048𝑚 − 1.35𝑚 𝑴𝑮 = 𝟑. 𝟓𝟓𝟒𝟖𝒎 HYDRAULICS Engr. Michelle D. Enriquez c. Righting moment 𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃 𝑥 = 3.5548𝑚(10) 𝑥 = 0.6173𝑚 𝑅𝑀 = 5 𝑥 106 (0.6173𝑚) 𝑹𝑴 = 𝟑. 𝟎𝟖𝟔𝟓 𝒙 𝟏𝟎𝟔 𝑵𝒎 Example 6: A cylindrical tank having a diamter of 40cm in diamter and 20cm in height floats in mercutry ina vertical positio, its deth of immersion being 8.0cm . a. Find the value of 𝑀𝐵0 b. Determine the metacenteric height c. If the water is pured uinto the vessel over the mercury until the cylinder is submerged; partly in mercury and partly in water, determine the depth of immersion of mercury Solution: a. Value of 𝑀𝐵0 𝜃=0 𝜋(40𝑐𝑚)4 𝜋𝑑 4 𝐼 64 64 𝑀𝐵0 = = = 𝜋(40𝑐𝑚)2 𝑉 𝜋𝑑 2 (8.0𝑐𝑚) 4 ℎ 4 40000𝜋𝑐𝑚 4 𝑀𝐵0 = 3200𝜋𝑐𝑚3 𝑴𝑩𝟎 = 𝟏𝟐. 𝟓𝟎𝒄𝒎 b. Metacentric height 𝐺𝐵0 = 10𝑐𝑚 − 4𝑐𝑚 𝐺𝐵0 = 6𝑐𝑚 Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 12.50𝑐𝑚 − 6.0𝑐𝑚 𝑀𝐺 = 6.50𝑐𝑚 c. Depth of immersion of mercury 𝐵𝐹 = 𝑊 𝛾𝑤 𝑉 + 𝛾𝐻𝑔 𝑉 = 𝑊 𝑔 𝜋(40𝑐𝑚)2 𝑔 𝜋(40𝑐𝑚)2 (20 − ℎ) + (1 3 ) (13.6) (ℎ ) (1 3 ) 𝑐𝑚 4 𝑐𝑚 4 𝑔 𝜋(40𝑐𝑚)2 (8.0𝑐𝑚) = (1 3 ) (13.6) 𝑐𝑚 4 (20 − ℎ) + 13.6ℎ = 108.8𝑐𝑚 𝒉 = 𝟕. 𝟎𝟒𝟕𝟔𝒄𝒎 HYDRAULICS Engr. Michelle D. Enriquez Example 7: For a ship with a waterline corss section as shown has adislacement of 5330kN a. Determine the metacentric height to remain in stable equilibrium b. Determine the maximum distance that the center of gravity may lie above the center of buoyancy if the ship is to remain stable c. Determine the horizontal displacement of the center of buoyancy if the ship tilts an angle of 10° without overturning Solution: a. metacentric height to ramin stable equilbirum 𝑀𝐺 = 0 b. Maximum distance that the center of gravity may lie above the centerof buoyancy 𝐼= 𝑏ℎ 3 𝑏ℎ 3 + 2 ( 36 ) (30𝑚0(7.5𝑚)3 7.5 ∗ 3.753 𝐼= + 2( ) 12 36 𝐼 = 1076.6602 𝑚4 12 𝛾𝑤 = 𝑉= 𝑊 𝑊 ;𝑉 = 𝑉 𝛾𝑤 5350𝑘𝑁 = 545.3619𝑚3 𝑘𝑁 9.81 3 𝑚 𝐼 1076.6602 𝑚4 = 𝑉 545.3619𝑚3 𝑀𝐵0 = 1.9742𝑚 𝑀𝐵0 = Then the metacentric height: 𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0 𝑀𝐺 = 1.9742𝑚 − 0 𝑀𝐺 = 1.9742𝑚 c. Horizontal displacement of the center of buoyancy if the ship tilts an angle of 10° without overturning 𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃 𝑥 = 1.9742𝑚(10) 𝑥 = 0.3428𝑚 HYDRAULICS Engr. Michelle D. Enriquez Buoyancy and stability of floating bodies 1. A piece of wood floats in water with 50mm projecting above the water surface. When placed in glycerine of specific gravity of 1.35, the block projects 75mm above the liquid surface a. Find the height of the piece of wood b. Find the specific gravity of the wood c. Find the weight of wood if it has a cross sectional area of 200mm x 200mm 2. Two spheres each 1.20m in diameter are connected by means of a short rope. One weighs 4kN and the other weighs 12kN when placed in water a. Compute the tension in the rope b. Compute the depth of the 4kN sphere c. Compute the volume of sphere exposed above the water surface 3. A concrete block with a volume of 0.023cu,m is tied to one end of a wooden post 𝑘𝑁 having dimensions of 200mm x 200mm by 3.0m long. Weight of wood is 6.4 3 𝑘𝑁 𝑚 and that of concrete is 23.5 3. 𝑚 a. Determine the length of the wooden post above the water surface b. Determine the volume of additional concrete to be tied t the bottom of the post to make its top flush with the water surface c. Determine the total weight of concrete to make its top flush with the water surface 4. A rectangular barge weighing 200,000kg is 14.0m long, 8.0m wide and 4.5m deep. It will transport to 𝑘𝑔 Manila the 20mm diameter 6.0m long steel reinforcing bars having a density of 7850 3. Density of salt 𝑘𝑔 water is 1026 3. 𝑚 a. Determine the draft of the barge on sea water before the bars as loaded b. If a draft to be maintained at 3.0m how many pieces of steel bars it could carry c. What is the draft of the barge when one half of its cargo is unloaded 𝑘𝑔 in fresh water (1000 3) 𝑚 d. 5. A concrete cube 0.50m on each side is to be held in equilibrium under water by attaching a light foam buoy to its. Specific weight of concrete and foam are 𝑘𝑁 𝑘𝑁 𝑘𝑁 23.58 3. and 0.79 3. respectively. Assume weight of water 9.79 3. 𝑚 𝑚 𝑚 a. What is the minimum volume of foam required b. What is the weight of the foam c. What is the total weight of concrete and foam 𝑚 HYDRAULICS Engr. Michelle D. Enriquez 6. A loaded scow has a draft of 1.8m in fresh water. The scow is 6.0m wide and 12.0m long and 2.4m high. The center of gravity of the scow is 1.80m above the bottom along the vertical axis of symmetry a. Determine the initial metacentric height b. What is the maximum single weight that can be moved transversely from the center of the unloaded scow over the side without sinking the scow c. If the maximum weight obtained in (b) is doubled, at what distance from the center will the scow be on the verge of submergence. 7. A cylindrical caisson having an outside diameter of 9.0m floats in sea water with its axis vertical and its lower end submerged 9.0m below the water surface. If the center of gravity is on the vertical axis and is 3.6m above the bottom. a. Determine the value of 𝑀𝐵0 b. Find the true height c. Find the righting couple when the caisson is tipped through an angle of 8° 8. A scow having a length of 15m, 8.0 wide and 6.0m deep has a displacement of 164 metric tons. If the scow tilts at an angle of 14° without overturning a. find the metacentric height b. Determine the location of the center of gravity which lies on the vertical axis measured from the bottom of symmetry c. Determine the horizontal displacement of the center of buoyancy