Static Surface Forces hinge 8m water 4m Static Surface Forces Forces on plane areas Forces on curved surfaces Buoyant force Stability submerged bodies Forces on Plane Areas Two types of problems Horizontal surfaces pressure is constant Two unknowns Inclined surfaces dp dz Total force Line of action Two techniques to find the line of action of the resultant force Moments Pressure prism Forces on Plane Areas Horizontal surfaces P 500 kPa What is the force on the bottom of this tank of water FR pdA p dA pA FR g hA volume h What is p Side view p h FR weight of overlying fluid h Vertical distance to free surface F is normal to the surface and towards the surface if p is positive F passes through the centroid of the area A Top view Forces on Plane Areas Inclined Surfaces Direction of force Normal to the plane Magnitude of force integrate the pressure over the area pressure is no longer constant Line of action Moment of the resultant force must equal the moment of the distributed pressure force Forces on Plane Areas Inclined Surfaces Free surface FR hc A hc A O x xc xR Where could I counteract pressure by supporting potato at a single point centroid B center of pressure O y yR yc The origin of the y axis is on the free surface Magnitude of Force on Inclined Plane Area FR pdA p h y sin FR sin ydA 1 yc ydA A A y FR Ay c sin FR hc A FR pc A hc is the vertical distance between free surface and centroid centroid of the area pc is the pressure at the First Moments xdA A xc 1 xdA A A Moment of an area A about the y axis Location of centroidal axis 1 yc ydA A A 1 3 h For a plate of uniform thickness the intersection of the centroidal axes is also the center of gravity Second Moments moment of inertia of the area Also called I x y 2 dA A I x I xc Ay 2 Ixc is the 2nd moment with respect to an axis passing through its centroid and parallel to the x axis The 2nd moment originates whenever one computes the moment of a distributed load that varies linearly from the moment axis Product of Inertia A measure of the asymmetry of the area I xy xydA A Product of inertia I xy xc yc A I xyc Ixyc 0 y Ixyc 0 y x x If x xc or y yc is an axis of symmetry then the product of inertia Ixyc is zero Properties of Areas Ixc Ixc a b yc ab yc A 2 a b d Ixc A ab R yc a yc 2 a yc 3 b d xc 3 A p R 2 yc R ba 3 I xc 12 3 I xyc 0 ba 2 b 2d 72 ba I xc 36 I xyc p R4 I xc 4 I xyc 0 Properties of Areas Ixc yc 4R yc 3p p R4 I xc 8 I xyc 0 A p ab yc a p ba 3 I xc 4 I xyc 0 p R2 A 4 4R yc 3p p R4 I xc 16 p R2 A 2 R b a yc Ixc R yc Forces on Plane Areas Center of Pressure xR The center of pressure is not at the centroid because pressure is increasing with depth x coordinate of center of pressure xR xR FR xpdA A Moment of resultant sum of moment of distributed forces p y sin FR yc A sin 1 xR xpdA FR A 1 xR xy sin dA A yc A sin 1 xR xydA A yc A Center of Pressure xR xR 1 xydA yc A A I xy xR yc A xc yc A I xyc xR yc A I xyc xR xc yc A I xy xydA A I xy xc yc A I xyc Product of inertia Parallel axis theorem y x Center of Pressure yR y R FR ypdA A Sum of the moments 1 y R ypdA FR yc A sin FR A 1 2 yR y sin dA A yc A sin 1 2 yR y dA A yc A I x y 2 dA A Ix yR yc A I x I xc yc2 A I xc yc2 A yR yc A I xc yR yc yc A p y sin p 0 when y 0 You choose the pressure datum to make the problem easy Parallel axis theorem Inclined Surface Findings 0 I xyc The horizontal center of pressure and the xR xc yc A coincide when the surface horizontal centroid has either a horizontal or vertical axis of symmetry 0 below the y I xc y The center of pressure is always R c y A c centroid The vertical distance between the centroid and the center of pressure decreases as the surface is lowered deeper into the liquid What do you do if there isn t a free surface Example using Moments An elliptical gate covers the end of a pipe 4 m in diameter If the gate is hinged at the top what normal force F applied at the bottom of the gate is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side Neglect the weight of the gate teams Solution Scheme Magnitude of the force applied by the water 8m Location of the resultant force Find F using moments about hinge water F hinge 4m Magnitude of the Force FR pc A 8m A ab hc 10 m Depth to the centroid water FR F hc pc FR hc ab N FR 9800 3 10 m 2 5 m 2 m m a 2 5 m FR 1 54 MN b 2m hinge 4m Location of Resultant Force I xc yR yc yc A yc hc yc 12 5 m ba 3 y R yc 4 yc ab 2 a y R yc 4 yc m y R yc 0 125 8m Slant distance to surface p ba 3 I xc 4 Fr F 4m A ab 2 5 m y R yc 4 12 5 m 2 xc xR water hinge a 2 5 m cp b 2m Force Required to Open Gate How do we find the required force 8m Fr F Moments about the hinge M hinge 0 Fltot FRlcp FR lcp F ltot water lcp 2 625 m 4m 2 5 m cp 1 54 x 10 N 2 625 m F 6 5 m F 809 kN hinge b 2m ltot Forces on Plane Surfaces Review The average magnitude of the pressure force is the pressure at the centroid The horizontal location of the pressure force gate was symmetrical was at xc WHY The about at least one of the centroidal axes The vertical location of the pressure force is Pressure below the centroid WHY increases with depth Forces on Plane Areas Pressure Prism A simpler approach that works well …