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9.6 Fluid PressureAccording to Pascal’s law, a fluid at rest creates apressure ρ at a point that is the same in alldirectionsMagnitude of ρ measured as a force per unitarea, depends on the specific weight γ or massdensity ρ of the fluid and the depth z of the pointfrom the fluid surface ρ = γz = ρgzValid for incompressible fluidsGas are compressible fluids and thus the aboveequation cannot be used
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9.6 Fluid PressureConsider the submerged plate3 points have been specified
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9.6 Fluid PressureSince point B is at depth z1 from the liquidsurface, the pressure at this point has amagnitude of ρ1= γz1Likewise, points C and D are both at depth z2and hence ρ2 = γz2In all cases, pressure acts normal to thesurface area dA located at specified pointPossible to determine the resultant forcecaused by a fluid distribution and specify itslocation on the surface of a submerged plate
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9.6 Fluid PressureFlat 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
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9.6 Fluid PressureFlat 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 ρ1= γz1 at depth z1 and ρ2 = γz2 at depth z2
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9.6 Fluid PressureMagnitude of the resultant force FR = volume ofthis loading diagram and FR has a line of actionthat passes through the volume’s centroid, CFR does not act at the centroid of the plate but atpoint P called the center ofpressureSince plate has a constantwidth, the loading diagramcan be viewed in 2D
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9.6 Fluid PressureFlat Plate of Constant Width Loading intensity is measured as force/length and varies linearly from w1 = bρ1= bγz1 to w 2 = bρ2 = bγz2 Magnitude of FR = trapezoidal area FR has a line of action that passes through the area’s centroid C
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9.6 Fluid PressureCurved 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 FR and location of center of centroid C or pressure P
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9.6 Fluid PressureCurved Plate of Constant WidthExample Consider distributed loading acting on the curved plate DB
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9.6 Fluid PressureCurved Plate of Constant WidthExample For equivalent loading
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9.6 Fluid PressureCurved Plate of Constant Width The plate supports the weight of the liquid Wf contained within the block BDA This force has a magnitude of Wf = (γb)(areaBDA) 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 FAD’s magnitude = area under trapezoid and acts through centroid CAD of this area
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9.6 Fluid PressureCurved 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 FAB is simply the area of the rectangle This force acts through the area centroid CAB or the midpoint of AB Summing three forces, FR = ∑F = FAB + FAD + Wf
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9.6 Fluid PressureCurved Plate of Constant Width Location of the center of pressure on the plate is determined by applying MRo = ∑MO 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
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9.6 Fluid PressureFlat Plate of Variable Width Consider the pressure distribution acting on the surface of a submerged plate having a variable width
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9.6 Fluid PressureFlat 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’)
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9.6 Fluid PressureFlat Plate of Variable Width FR = ∫ ρdA = ∫ dV = V A V 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 x= ∫ ~dV y = ∫ ~ dV V x V y ∫ dV V ∫ dVV This point should not be mistaken for centroid of the plate’s area
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9.6 Fluid PressureExample 9.13Determine the magnitude and location of theresultant hydrostatic force acting on the submergedrectangular plate AB. Theplate has a width of 1.5m;ρw = 1000kg/m3.
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9.6 Fluid PressureSolution The water pressures at depth A and B are ρ A = ρ w gz A = (1000kg / m 3 )(9.81m / s 2 )(2m) = 19.62kPa ρ B = ρ w gz B = (1000kg / m 3 )(9.81m / s 2 )(5m) = 49.05kPa Since the plate has constant width, distributed loading can be viewed in 2D For intensities of the load at A and B, wA = bρ A = (1.5m)(19.62kPa) = 29.43kN / m wB = bρ B = (1.5m)(49.05kPa) = 73.58kN / m
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9.6 Fluid PressureSolution For magnitude of the resultant force FR created by the distributed load FR = area of trapezoid 1 = (3)(29.4 + 73.6) = 154.5 N 2 This force acts through the centroid of the area 1 2(29.43) + 73.58 h= (3) = 1.29m 3 29.43 + 73.58 measured upwards from B
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9.6 Fluid PressureSolution Same results can be obtained by considering two components of FR defined by the triangle and rectangle Each force acts through its associated centroid and has a magnitude of FRe = (29.43kN / m)(3m) = 88.3kN Ft = (44.15kN / m)(3m) = 66.2kN Hence FR = FRe + FR = 88.3kN + 66.2kN = 154.5kN
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9.6 Fluid PressureSolution Location of FR is determined by summing moments about B ∑(M R )B = ∑ M B ; (154.5)h = 88.3(1.5) + 66.2(1) h = 1.29m
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9.6 Fluid PressureExample 9.14Determine the magnitude of the resultanthydrostatic force acting on the surface of a seawallshaped in the form of a parabola. The wall is 5mlong and ρw = 1020kg/m2.
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9.6 Fluid PressureSolution The horizontal and vertical components of the resultant force will be calculated since ρ B = ρ w gz B = (1020kg / m 2 )(9.81m / s 2 )(3m) = 30.02kPa Then wB = bρ B = 5m(30.02kPa) = 150.1kN / m Thus 1 Fx = (3m)(150.1kN / m) = 225.1kN 3
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9.6 Fluid PressureSolution Area of the parabolic sector ABC can be determined For weight of the wafer within this region Fy = ( ρ w gb)(area ABC ) 1 = (1020kg / m )(9.81m / s )(5m)[ (1m)(3m)] = 50.0kN 2 2 3 For resultant force FR = Fx2 + Fy2 = (225.1) 2 + (50.0) 2 = 231kN
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9.6 Fluid PressureExample 9.15Determine the magnitude and location ofthe resultant force acting on the triangularend plates of the wafer of the water trough.ρw = 1000 kg/m3
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9.6 Fluid PressureSolution Magnitude of the resultant force F = volume of the loading distribution Choosing the differential volume element, dF = dV = ρdA = ρ w gz (2 xdz ) = 19620 zxdz For equation of line AB x = 0.5(1 − z ) Integrating 1 F = V = ∫ dV = ∫ (19620) z[0.5(1 − z )]dz V 0 1 = 9810 ∫ ( z − z 2 )dz = 1635 N = 1.64kN 0
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9.6 Fluid PressureSolution Resultant passes through the centroid of the volume Because of symmetry x =0 For volume element 1 z= ∫z ~dV V = ∫0 z (19620) z[0.5(1 − z )]dz ∫VdV 1635 1 9810∫ ( z 2 − z 3 ) dz = 0 = 0.5m 1635
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