1. 9.6 Fluid Pressure According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions Magnitude of ρ measured as a force per unit area, depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface ρ = γz = ρgz Valid for incompressible fluids Gas are compressible fluids and thus the above equation cannot be used

2. 9.6 Fluid Pressure Consider the submerged plate 3 points have been specified

3. 9.6 Fluid Pressure Since point B is at depth z 1 from the liquid surface, the pressure at this point has a magnitude of ρ 1 = γz 1 Likewise, points C and D are both at depth z 2 and hence ρ 2 = γz 2 In all cases, pressure acts normal to the surface area dA located at specified point Possible to determine the resultant force caused by a fluid distribution and specify its location on the surface of a submerged plate

4. 9.6 Fluid Pressure Flat Plate of Constant Width Consider flat rectangular plate of constant width submerged in a liquid having a specific weight γ Plane of the plate makes an angle with the horizontal as shown

5. 9.6 Fluid Pressure Flat Plate of Constant Width Since pressure varies linearly with depth, the distribution of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ 1 = γz 1 at depth z 1 and ρ 2 = γz 2 at depth z 2

6. 9.6 Fluid Pressure Magnitude of the resultant force F R = volume of this loading diagram and F R has a line of action that passes through the volume’s centroid, C F R does not act at the centroid of the plate but at point P called the center of pressure Since plate has a constant width, the loading diagram can be viewed in 2D

7. 9.6 Fluid Pressure Flat Plate of Constant Width Loading intensity is measured as force/length and varies linearly from w 1 = bρ 1 = bγz 1 to w 2 = bρ 2 = bγz 2 Magnitude of F R = trapezoidal area F R has a line of action that passes through the area’s centroid C

8. Curved Plate of Constant Width When the submerged plate is curved, the pressure acting normal to the plate continuously changes direction For 2D and 3D view of the loading distribution, Integration can be used to determine F R and location of center of centroid C or pressure P 9.6 Fluid Pressure

9. Curved Plate of Constant Width Example Consider distributed loading acting on the curved plate DB

10. 9.6 Fluid Pressure Curved Plate of Constant Width Example For equivalent loading

11. 9.6 Fluid Pressure Curved Plate of Constant Width The plate supports the weight of the liquid W f contained within the block BDA This force has a magnitude of W f = (γb)(area BDA ) and acts through the centroid of BDA Pressure distributions caused by the liquid acting along the vertical and horizontal sides of the block Along vertical side AD, force F AD ’s magnitude = area under trapezoid and acts through centroid C AD of this area

12. 9.6 Fluid Pressure Curved Plate of Constant Width The distributed loading along horizontal side AB is constant since all points lying on this plane are at the same depth from the surface of the liquid Magnitude of F AB is simply the area of the rectangle This force acts through the area centroid C AB or the midpoint of AB Summing three forces, F R = ∑F = F AB + F AD + W f

13. 9.6 Fluid Pressure Curved Plate of Constant Width Location of the center of pressure on the plate is determined by applying M Ro = ∑M O which states that the moment of the resultant force about a convenient reference point O, such as D or B = sum of the moments of the 3 forces about the same point

14. 9.6 Fluid Pressure Flat Plate of Variable Width Consider the pressure distribution acting on the surface of a submerged plate having a variable width

15. 9.6 Fluid Pressure Flat Plate of Variable Width Resultant force of this loading = volume described by the plate area as its base and linearly varying pressure distribution as its altitude The shaded element may be used if integration is chosen to determine the volume Element consists of a rectangular strip of area dA = x dy’ located at depth z below the liquid surface Since uniform pressure ρ = γz (force/area) acts on dA, the magnitude of the differential force dF dF = dV = ρ dA = γz(xdy’)

16. 9.6 Fluid Pressure Flat Plate of Variable Width Centroid V defines the point which FR acts The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations This point should not be mistaken for centroid of the plate’s area

17. 9.6 Fluid Pressure Example 9.13 Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; ρ w = 1000kg/m 3.

18. 9.6 Fluid Pressure Solution The water pressures at depth A and B are Since the plate has constant width, distributed loading can be viewed in 2D For intensities of the load at A and B,

19. 9.6 Fluid Pressure Solution For magnitude of the resultant force F R created by the distributed load This force acts through the centroid of the area measured upwards from B

20. 9.6 Fluid Pressure Solution Same results can be obtained by considering two components of F R defined by the triangle and rectangle Each force acts through its associated centroid and has a magnitude of Hence

21. 9.6 Fluid Pressure Solution Location of F R is determined by summing moments about B

22. 9.6 Fluid Pressure Example 9.14 Determine the magnitude of the resultant hydrostatic force acting on the surface of a seawall shaped in the form of a parabola. The wall is 5m long and ρ w = 1020kg/m 2.

23. 9.6 Fluid Pressure Solution The horizontal and vertical components of the resultant force will be calculated since Then Thus

24. 9.6 Fluid Pressure Solution Area of the parabolic sector ABC can be determined For weight of the wafer within this region For resultant force

25. 9.6 Fluid Pressure Example 9.15 Determine the magnitude and location of the resultant force acting on the triangular end plates of the wafer of the water trough. ρ w = 1000 kg/m 3

26. 9.6 Fluid Pressure Solution Magnitude of the resultant force F = volume of the loading distribution Choosing the differential volume element, For equation of line AB Integrating View Free Body Diagram

27. 9.6 Fluid Pressure Solution Resultant passes through the centroid of the volume Because of symmetry For volume element

28. Chapter Summary Center of Gravity and Centroid Center of gravity represents a point where the weight of the body can be considered concentrated The distance to this point can be determined by a balance of moments Moment of weight of all the particles of the body about some point = moment of the entire body about the point Centroid is the location of the geometric center of the body

29. Chapter Summary Center of Gravity and Centroid Centroid is determined by the moment balance of geometric elements such as line, area and volume segments For body having a continuous shape, moments are summed using differential elements For composite of several shapes, each having a known location for centroid, the location is determined from discrete summation using its composite parts

30. Chapter Summary Theorems of Pappus and Guldinus Used to determine surface area and volume of a body of revolution Surface area = product of length of the generating curve and distance traveled by the centroid of the curve to generate the area Volume = product of the generating area and the distance traveled by the centroid to generate the volume

31. Chapter Summary Fluid Pressure Pressure developed by a fluid at a point on a submerged surface depends on the depth of the point and the density of the liquid according to Pascal’s law Pressure will create a linear distribution of loading on a flat vertical or inclined surface For horizontal surface, loading is uniform Resultants determined by volume or area under the loading curve

32. Chapter Summary Fluid Pressure Line of action of the resultant force passes through the centroid of the loading diagram

33. Chapter Review