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Course: MECHENG mecheng320, Fall 2009
School: Michigan
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for Pa-Pb=0.1 deltah=1ft Asin((Pa-Pb)/(deltah*(gamma-gammaB))) patm =gamma*h+pvapor I_xc=PI*R^4/4 y_R=I_xc/(y_c*A)+y_C distance below the shaft to the center of the pressure is y_R-y_C M_C=0=F_R*(y_R-Y_C) Example 2.8 A pressurized tank contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate bolted to its side. When the pressure gage on the top of the tank reads 50kPa, what is the magnitude and location...

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for Pa-Pb=0.1 deltah=1ft Asin((Pa-Pb)/(deltah*(gamma-gammaB))) patm =gamma*h+pvapor I_xc=PI*R^4/4 y_R=I_xc/(y_c*A)+y_C distance below the shaft to the center of the pressure is y_R-y_C M_C=0=F_R*(y_R-Y_C) Example 2.8 A pressurized tank contains oil (SG = 0.90) and has a square, 0.6-m by 0.6-m plate bolted to its side. When the pressure gage on the top of the tank reads 50kPa, what is the magnitude and location of the resultant force on the attached plate? The outside of the tank is at atmospheric pressure. F_1=(P_s+gamma*h1)*A F_2=1/2*gamma(h2-h1)/A F_R=F_1+F_2 M_0=-F_R*y_0+F_1*.3 Example 2.9 The 6-ft-diameter drainage conduit is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall. F1=gamma*hc*A Weight=gamma*volume Fh=F1 Fv=Weight FR=sqrt(Fh^2+Fv^2) Fbouyant=gamma*Volume Buoyant force passes through the centroid of the displaced volume. The point through which the buoyant force acts is called the center of buoyancy. Dz/dy=-a_Y/(g+a_Z) Slope of line of constant pressure dp=0 Dp/dz=-rho(g+a_z,) a_y=0, az is not=0 Volume of fluid in tank remains constant V_i=V PI*R^2*H= PI*2*R4/(4*g)+PI*R2*h_0 H-h_0= Relationship 2*R4/(4*g) between depth and speed is not linear A long vertical wall separates seawater from freshwater. If the seawater stands at a depth of 7m, what depth of freshwater is required to give a zero resultant force on the wall? When the resultant force is zero, will the moment due to the fluid forces be zero? Explain. yR I xc yc A yc Seawater: rho=1025kg/m^3, hsea=7m freshwater rho=998kb/m^3 hfresh=h Moment of Inertia Ixx about the center of gravity of a rectangular plate: 1/12 b L3; b = width, L = height of the plate Force = rho* g* hc* A For zero resultant force, FSW = FFW or Rho_SW*g*hSW*ASW = rhoFW*g*hFW*AFW For a unit length of the wall, (1025kg/m3) (9.81m/s2) (7/2 m) (7m x 1 m) = (998kg/m3) (9.81m/s2) (h/2 m) (h x 1 m) h = 7.11 m In order for the moment to be zero, the two forces must be collinear. However, they will act at different heights, based on the respective centers of pressure. Cable 8 ft Gate 6 ft 60o Hinge A homogeneous 4-ft wide 8 ft-long rectangular gate weighing 800 lb is held in place by a horizontal flexible as shown in the figure. Water acts against the gate which is hinged at point A. Friction in the hinge is Negligible. Specific weight of water = 62.4 lbf/ft3 Weight_gate=800lbf
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