hydrostatics on surface_fluid mechanics.

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1. Hydraulics CEHYDR40 ( Principles of Hydrostatics) Lesson Content:

2. Hydraulics CEHYDR40 ( Total hydrostatic force on surfaces ) Lesson Content: Total hydrostatic force on plane surface

3. Hydraulics CEHYDR40 ( Principles of Hydrostatics) Lesson Content: Total hydrostatic force on plane surface Total hydrostatic force on curved surface

4. Hydraulics CEHYDR40 ( Principles of Hydrostatics) Lesson Content: Total hydrostatic force on plane surface Total hydrostatic force on curved surface Dams

5. Hydraulics CEHYDR40 ( Principles of Hydrostatics) Lesson Content: Total hydrostatic force on plane surface Total hydrostatic force on curved surface Dams Bouyancy

6. Total Hydrostatic Force on Plane Surfaces For horizontal plane surface submerged in liquid, or plane surface inside a gas chamber, or any plane surface under the action of uniform hydrostatic pressure, the total hydrostatic force is given by F= A   where  p  is the uniform pressure and  A  is the area.   In general, the total hydrostatic pressure on any plane surface is equal to the product of the area of the surface and the unit pressure at its center of gravity. F= A   where  p cg  is the pressure at the center of gravity. For homogeneous free liquid at rest, the equation can be expressed in terms of unit weight  γ  of the liquid. F= γ hA where   ℎ   is the depth of liquid above the centroid of the submerged area.

7. From the figure above,    = +e, thus, the distance between cg and cp is   Eccentricity,   e =   Where, - is centroidal moment of inertia of an object   - the distance from cg to the liquid surface  

8. Total Hydrostatic Force on Curved Surfaces In the case of curved surface submerged in liquid at rest, it is more convenient to deal with the horizontal and vertical components of the total force acting on the surface. Note: the discussion here is also applicable to plane surfaces. Horizontal Component The horizontal component of the total hydrostatic force on any surface is equal to the pressure on the vertical projection of that surface. = A   Vertical Component The vertical component of the total hydrostatic force on any surface is equal to the weight of either real or imaginary liquid above it. = V   Total Hydrostatic Force F =   Direction of F Tan =  

9. Case 1: Liquid is above the curve surface The vertical component of the hydrostatic force is downward and equal to the volume of the real liquid above the submerged surface. Case 2: Liquid is below the curve surface The vertical component of the hydrostatic force is going upward and equal to the volume of the imaginary liquid above the surface.

10. Analysis of Gravity Dam Dams are structures whose purpose is to raise the water level on the upstream side of river, stream, or other waterway. The rising water will cause hydrostatic force which will tend the dam to slide horizontally and overturn about its downstream edge or toe. The raised water level on the upstream edge or heel will also cause the water to seep under the dam. The pressure due to this seepage is commonly called hydrostatic uplift and will reduce the stability of the dam against sliding and against overturning. Gravity Dam Analysis The weight of gravity dam will cause a moment opposite to the overturning moment and the friction on the base will prevent the dam from sliding. The dam may also be prevented from sliding by keying its base into the bedrock.  

11. Step 1 Consider 1 unit length (1 m length) of dam perpendicular to the cross section. Step 2 Determine all the forces acting:  Vertical forces 𝑊 = Weight of dam 𝐹𝑉  = Weight of water in the upstream side (if any) 𝑈 = Hydrostatic uplift Weight of permanent structures on the dam Horizontal forces 𝐹𝐻 = Horizontal component of total hydrostatic force Wind pressure, wave action, floating bodies, earthquake load, etc. Step 3 Solve for the reaction Horizontal component of the reaction 𝑅𝑥=Σ𝐹𝐻 Vertical component of the reaction R y =ΣF V Step 4 Moment about the toe Righting moment, 𝑅𝑀 𝑅𝑀 = Sum of all rotation towards the upstream side Overturning moment, 𝑂𝑀 𝑂𝑀 = Sum of all rotation towards the downstream side

12. Step 5 Location of 𝑅𝑦 as measured from the toe Factors of Safety Factor of safety against sliding, FS s Factor of safety against overturning, 𝐹𝑆𝑜 Where μ𝜇 = coefficient of friction between the base of the dam and the foundation. Foundation Pressure Eccentricity, 𝑒 If 𝑒≤𝐵/6,  R y  is within the middle third and the foundation pressure is trapezoidal acting from heel to toe. If 𝑒 is exactly 𝐵/6, the shape of foundation pressure is triangular also acting from heel to toe.

13. For the sign of 6𝑒/𝐵, use (+) at point where Ry is nearest. From the diagram above, use (+) for 𝑞𝑇 and (-) for 𝑞𝐻. A negative 𝑞 indicates compressive stress and a positive 𝑞 indicates tensile stress. A positive 𝑞 will occur when 𝑒>𝐵/6. In foundation design, soil is not allowed to carry tensile stress, thus, any +𝑞 will be neglected in the analysis. If 𝑒>𝐵/6, 𝑅𝑦 is outside the middle third and the foundation pressure is triangular.

14. Buoyancy Archimedes Principle Any body immersed in a fluid is subjected to an upward force called buoyant force equal to the weight of the displaced fluid.        Archimedes (287-212 B.C.) Where 𝐵𝐹 = buoyant force 𝛾 = unit weight of fluid V D  = volume of fluid displaced by the body Buoyant force acting on a body submerged in fluid is merely the resultant of two vertical hydrostatic forces. Consider the cylindrical body shown below to have some length perpendicular to the drawing. The horizontal components of hydrostatic force acting on the body are in equilibrium because the vertical projection of the body in opposite sides is the same.  

15. The upward force F V1  is the total force exerted by the fluid on the under surface of the body; the downward force F V2  is the total force exerted by the fluid on the upper surface of the body. Since liquid pressure increases by depth, F V1  is greater than F V2 . The difference F V1 −F V2   is therefore upward, and this difference is the buoyant force. For homogeneous body of volume  𝑉 "floating" in a homogeneous liquid at rest, the volume displaced is

16. For a floating body whose cross-sectional area   𝐴   is perpendicular to the liquid surface, the area submerged is given by For a floating body of height   𝐻   and constant cross-sectional area parallel to the liquid surface, the submerged length   𝐷   is given by

17. Problem 1: A vertical rectangular gate 1.5m wide and 3m high is submerged in water with its top edge 2m below the water surface. Find the total pressure acting on one side of the gate and its location from the bottom. Problem 2: A vertical triangular gate with top base horizontal and 1.5m wide is 3m high. It is submerged in oil having sp. Gr. Of 0.82 with its top base submerged to a depth of 2m. Determine the magnitude and location of the total hydrostatic pressure acting on one side of the gate.

18. Problem 3: A vertical rectangular plate is submerged half in oil (sp.gr. = 0.8) and half in water such that its top edge is flushed with the oil surface. What is the ratio of the force exerted by water acting on the lower half to that by oil acting on the upper half? Problem 4: A vertical circular gate in a tunner 8m in diameter has oil (sp.gr. 0.8) on one side and air on the other side. If oil is 12m above the invert and the air pressure is 40 kPa, where will a single support be located (above the invert of the tunnel) to hold the gate in position?

19. Problem 5: The tank shown in figure is 3m wide into the paper. Neglect atmospheric pressure. Unit weight of water is 9.79 kN / . Determine the nearest value to the: Vertical component of the total hydrostatic force on the quarter-circle panel AB, in kN. Resultant hydrostatic force acting on the quarter-circle panel AB, in kN. Angle that the resultant hydrostatic force make with the horizontal, in degrees.   Problem 6: Gate AB in the figure is semicircular, hinged at B, and held by horizontal force P at A: Determine the hydrostatic force acting on the gate. Determine the location of the hydrostatic force from point A. What force P is required for equilibrium?

20. Problem 7 A piece of wood 305 mm (1 ft) square and 3 m (10 ft) long, weighing 6288.46 N/m 3  (40 lb /ft 3 ), is submerged vertically in a body of water, its upper end being flush with the water surface. What vertical force is required to hold it in position? Problem 8 A concrete dam has the profile shown in Figure P-911. If the density of concrete is 2400 kg/m 3  and that of water is 1000 kg/m 3 , determine the maximum compressive stress at section  m-n  if the depth of the water behind the dam is  h  = 15 m.