1. Pertemuan 05 - 07 Hydrostatics 2

2. Bina Nusantara Outline Pressure Forces on Plane Surface Pressure Forces on Curved Surface Pressure on Spillway Sections of Dam Stability of Dam Buoyancy

3. Bina Nusantara Pressure Forces on Plane Surface http://www.ce.utexas.edu/prof/kinnas/319LAB/Applets/PresPlane4.html

4. Bina Nusantara Hydrostatics Force on Immersed Surfaces  Immersed Surfaces are Subject to hydrostatics pressure  May generally be horizontal, vertical or inclined

5. Bina Nusantara Horizontal Surface The total weight of the fluid above the surface is equal to the volume of the liquid above the surface multiplied by the specific weight of the liquid F = g A h  g : specific weight of the liquid  A : the area of the surface  F : the force acting on the immersed surface h A horizontal surface immersed in a liquid

6. Bina Nusantara Hydrostatic Force on a Plane Surface The Center of Pressure Y R lies below the centroid - since pressure increases with depth F R =  A Y C sin  or F R =  A H c Y R = (I xc / Y c A) + Y c X R = (I xyc / Y c A) + X c but for a rectangle or circle: X R = X c For 90 degree walls: F R =  A H c

7. Bina Nusantara Hydrostatics Example Problem # 1 What is the Magnitude and Location of the Resultant force of water on the door?  W = 62.4 lbs/ft 3 Water Depth = 6 feet Door Height = 4 feet Door Width = 3 feet

8. Bina Nusantara Hydrostatics Example Problem #1 Magnitude of Resultant Force: F R =  W A H C F R = 62.4 x 12 x 4 = 2995.2 lbs Important variables: H C and Y c = 4’ X c = 1.5’ A = 4’ x 3’ = 12’ I xc = (1/12)bh 3 = (1/12)x3x4 3 = 16 ft 4 Location of Force: Y R = (I xc / Y c A) + Y c Y R = (16 / 4x12) + 4 = 4.333 ft down X R = X c (symmetry) = 1.5 ft from the corner of the door

9. Bina Nusantara Buoyancy Archimedes Principle: Will it Float? The upward vertical force felt by a submerged, or partially submerged, body is known as the buoyancy force. It is equal to the weight of the fluid displaced by the submerged portion of the body. The buoyancy force acts through the centroid of the displaced volume, known as the center of buoyancy. A body will sink until the buoyancy force is equal to the weight of the body. F B =  x V displaced = V disp FBFB FBFB W = F B F B =  W x V disp

10. Bina Nusantara Buoyancy Example Problem # 1 A 500 lb buoy, with a 2 ft radius is tethered to the bed of a lake. What is the tensile force T in the cable?  W = 62.4 lbs/ft 3 FBFB

11. Bina Nusantara Buoyancy Example Problem # 1 Displaced Volume of Water: V disp-W = 4/3 x  x R 3 V disp-W = 33.51 ft 3 Buoyancy Force: F B =  W x V disp-w F B = 62.4 x 33.51 F B = 2091.024 lbs up Sum of the Forces:  F y = 0 = 500 - 2091.024 + T T = 1591.024 lbs down

12. Bina Nusantara Will It Float? Ship Specifications: Weight = 300 million pounds Dimensions = 100’ wide by 150’ tall by 800’ long Given Information:  W = 62.4 lbs/ft 3

13. Bina Nusantara Assume Full Submersion: F B = Vol x  W F B = (100’ x 150’ x 800’) x 62.4 lbs/ft 3 F B = 748,800,000 lbs Weight of Boat = 300,000,000 lbs The Force of Buoyancy is greater than the Weight of the Boat meaning the Boat will float! How much of the boat will be submerged? Assume weight = Displaced Volume W B = F B 300,000,000 = (100’ x H’ x 800’) x 62.4 lbs/ft 3 H = Submersion depth = 60.1 feet