 # Static Surface Forces hinge water ? 8 m 4 m . Static Surface Forces ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability of.

## Presentation on theme: "Static Surface Forces hinge water ? 8 m 4 m . Static Surface Forces ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability of."— Presentation transcript:

Static Surface Forces hinge water ? 8 m 4 m 

Static Surface Forces ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability of floating and submerged bodies ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability of floating and submerged bodies  

Forces on Plane Areas ä Two types of problems ä Horizontal surfaces (pressure is _______) ä Inclined surfaces ä Two unknowns ä ____________ ä Two techniques to find the line of action of the resultant force ä Moments ä Pressure prism ä Two types of problems ä Horizontal surfaces (pressure is _______) ä Inclined surfaces ä Two unknowns ä ____________ ä Two techniques to find the line of action of the resultant force ä Moments ä Pressure prism constant Total force Line of action

Side view Forces on Plane Areas: Horizontal surfaces Top view A p =  h F is normal to the surface and towards the surface if p is positive. F passes through the ________ of the area. h What is the force on the bottom of this tank of water? weight of overlying fluid! F R = centroid h = _____________ _____________ Vertical distance to free surface

Forces on Plane Areas: Inclined Surfaces ä Direction of force ä Magnitude of force ä integrate the pressure over the area ä pressure is no longer constant! ä Line of action ä Moment of the resultant force must equal the moment of the distributed pressure force ä Direction of force ä Magnitude of force ä integrate the pressure over the area ä pressure is no longer constant! ä Line of action ä Moment of the resultant force must equal the moment of the distributed pressure force Normal to the plane

center of pressure Forces on Plane Areas: Inclined Surfaces   A’A’ B’B’ B’B’ O O Free surface centroid O O x x y y The origin of the y axis is on the free surface

Magnitude of Force on Inclined Plane Area   y y p c is the pressure at the __________________ centroid of the area h c is the vertical distance between free surface and centroid

First Moments For a plate of uniform thickness the intersection of the centroidal axes is also the center of gravity Moment of an area A about the y axis Location of centroidal axis

Second Moments Also called _______________ of the area moment of inertia I xc is the 2 nd moment with respect to an axis passing through its centroid and parallel to the x axis. Parallel axis theorem

Product of Inertia ä A measure of the asymmetry of the area If x = x c or y = y c is an axis of symmetry then the product of inertia I xyc is zero. y x y x Product of inertia I xyc = 0

Properties of Areas ycyc b a I xc ycyc b a R ycyc d

Properties of Areas a ycyc b I xc ycyc R R ycyc See Figure 2.18 in text

Forces on Plane Areas: Center of Pressure: x R ä The center of pressure is not at the centroid (because pressure is increasing with depth) ä x coordinate of center of pressure: x R ä The center of pressure is not at the centroid (because pressure is increasing with depth) ä x coordinate of center of pressure: x R Moment of resultant force = sum of moment of distributed forces

Center of Pressure: x R Product of inertia Parallel axis theorem y x

Center of Pressure: y R Sum of the moments Parallel axis theorem

Inclined Surface Findings ä The horizontal center of pressure and the horizontal centroid ________ when the surface has either a horizontal or vertical axis of symmetry ä The center of pressure is always _______ the centroid ä The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid ä What do you do if there isn’t a free surface? ä The horizontal center of pressure and the horizontal centroid ________ when the surface has either a horizontal or vertical axis of symmetry ä The center of pressure is always _______ the centroid ä The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid ä What do you do if there isn’t a free surface? coincide below decreases 0 >0 (y c increases)

ä ä ä ä ä ä ä ä 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. hinge water F 8 m 4 m Solution Scheme Magnitude of the force applied by the water (F R ) Example using Moments Location of the resultant force (x R, y R ) Find F using moments about hinge

Depth to the centroid Magnitude of the Force b = 2 m a = 2.5 m p c = ___ F R = ________ h c = _____ hinge water F F 8 m 4 m FRFR FRFR 10 m 1.54 MN (3.46  10 5 lb)

Location of Resultant Force hinge water F F 8 m 4 m FRFR FRFR 12.5 m Slant distance to surface 12.625 m b = 2 m a = 2.5 m cp (or 0.125 m below y c )

Force Required to Open Gate How do we find the required force? b = 2 m 2.5 m l cp =2.625 m l tot hinge water F F 8 m 4 m FRFR FRFR Moments about the hinge =Fl tot - F R l cp F = ______ 809 kN (1.82  10 5 lb) cp

Forces on Plane Surfaces Review ä The average magnitude of the pressure force is the pressure at the centroid ä The horizontal location of the pressure force was at x c (WHY?) ____________________ ___________________________________ ä The vertical location of the pressure force is below the centroid. (WHY?) ___________ ___________________ ä The average magnitude of the pressure force is the pressure at the centroid ä The horizontal location of the pressure force was at x c (WHY?) ____________________ ___________________________________ ä The vertical location of the pressure force is below the centroid. (WHY?) ___________ ___________________ Pressure increases with depth. The gate was symmetrical about at least one of the centroidal axes.

Forces on Plane Areas: Pressure Prism ä A simpler approach that works well for areas of constant width (_________). It works especially well if the surface intersects with the free surface. ä If the location of the resultant force is required and the area doesn’t intersect the free surface, then the moment of inertia method is about as easy. ä A simpler approach that works well for areas of constant width (_________). It works especially well if the surface intersects with the free surface. ä If the location of the resultant force is required and the area doesn’t intersect the free surface, then the moment of inertia method is about as easy. rectangles

Forces on Plane Areas: Pressure Prism  O AA Force = Volume of pressure prism Center of pressure is at centroid of pressure prism Free surface h2h2 h1h1

Example 1: Pressure Prism Dam h=10 m Dam is 50 m wide FRFR  24º F R = Volume F R = (h/cos  )(  h)(w)/2 F R = (10 m/0.9135)(9800 N/m 3 *10 m)(50 m)/2 F R = 26 MN h/cos  w hh FRFR width = 50 m

Example 2: Pressure Prism hinge water 8 m 4 m x 5 m (rectangular conduit) 5 m   12 m 8 m8 m O O y y hh 4 m

Solution 2: Pressure Prism 5 m h2h2 h1h1 Magnitude of force Location of resultant force a w F R = ________ y R = _______ 1.96 MN 2.667 m 4 m FRFR FRFR measured from hinge

Forces on Plane Areas: Pressure Prism ä A simpler approach that works well for areas of constant width (rectangles). ä Use pressure prism method if the rectangles intersect the free surface. ä If the location of the resultant force is required and the area doesn’t intersect the free surface, then the moment of inertia method is about as easy. ä A simpler approach that works well for areas of constant width (rectangles). ä Use pressure prism method if the rectangles intersect the free surface. ä If the location of the resultant force is required and the area doesn’t intersect the free surface, then the moment of inertia method is about as easy.

Forces on Curved Surfaces ä Horizontal component ä Vertical component ä Horizontal component ä Vertical component

Forces on Curved Surfaces: Horizontal Component ä What is the horizontal component of pressure force on a curved surface equal to? ä The center of pressure is located using the moment of inertia or pressure prism technique. ä The horizontal component of pressure force on a closed body is _____. ä What is the horizontal component of pressure force on a curved surface equal to? ä The center of pressure is located using the moment of inertia or pressure prism technique. ä The horizontal component of pressure force on a closed body is _____. zero

Forces on Curved Surfaces: Vertical Component ä What is the magnitude of the vertical component of force on the cup? r h p =  h F =  h  r 2 = W! F = pA What if the cup had sloping sides?

Forces on Curved Surfaces: Vertical Component The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the (virtual or real) free surface. Streeter et al. The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the (virtual or real) free surface. Streeter et al.

water = (3 m)(2 m)(1 m)  +1/4  (2 m) 2 (1 m)  Example: Forces on Curved Surfaces Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc. F V = F H = 2 m 3 m W1W1 W2W2 W 1 + W 2 = 58.9 kN + 30.8 kN = 89.7 kN =  (4 m)(2 m)(1 m) = 78.5 kN y x

= 0.948 m (measured from A) with magnitude of 89.7 kN Take moments about a vertical axis through A. Example: Forces on Curved Surfaces The vertical component line of action goes through the centroid of the volume of water above the surface. water 2 m 3 m A W1W1 W2W2

Example: Forces on Curved Surfaces water 2 m 3 m A W1W1 W2W2 The location of the line of action of the horizontal component is given by b=1 a=2 y x (1 m)(2 m) 3 /12 = 0.667 m 4 4 m

Example: Forces on Curved Surfaces 78.5 kN 89.7 kN 4.083 m 0.948 m 119.2 kN horizontal vertical resultant

C (78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0 0 0.948 m 1.083 m 89.7kN 78.5kN Cylindrical Surface Force Check ä All pressure forces pass through point C. ä The pressure force applies no moment about point C. ä The resultant must pass through point C. ä All pressure forces pass through point C. ä The pressure force applies no moment about point C. ä The resultant must pass through point C.

ä Find force F required to open the gate. ä All the horizontal force is carried by the hinge ä The pressure forces and force F pass through O. Thus the hinge force must pass through O! ä Hinge carries only horizontal forces! (F = ________) ä Find force F required to open the gate. ä All the horizontal force is carried by the hinge ä The pressure forces and force F pass through O. Thus the hinge force must pass through O! ä Hinge carries only horizontal forces! (F = ________) Curved Surface Trick water 2 m 3 m A W1W1 W2W2 F F O O W 1 + W 2

Curved Surface Trick Traditional way: 0.948 m 1.083 m 89.7kN 78.5kN F Same as F v = W 1 +W 2 C

Static Surface Forces Summary ä Forces caused by gravity (or _______________) on submerged surfaces ä horizontal surfaces (normal to total acceleration) ä inclined surfaces (y coordinate has origin at free surface) ä curved surfaces ä Horizontal component ä Vertical component (________________________) ä Virtual surfaces… ä Forces caused by gravity (or _______________) on submerged surfaces ä horizontal surfaces (normal to total acceleration) ä inclined surfaces (y coordinate has origin at free surface) ä curved surfaces ä Horizontal component ä Vertical component (________________________) ä Virtual surfaces… total acceleration weight of fluid above surface Location where p = p ref

different ä The resultant force exerted on a body by a static fluid in which it is fully or partially submerged ä The projection of the body on a vertical plane is always ____. ä The vertical components of pressure on the top and bottom surfaces are _________ ä The resultant force exerted on a body by a static fluid in which it is fully or partially submerged ä The projection of the body on a vertical plane is always ____. ä The vertical components of pressure on the top and bottom surfaces are _________ Buoyant Force zero

Buoyant Force: Thought Experiment FBFB zero no Weight of water displaced FB=VFB=V Place a thin wall balloon filled with water in a tank of water. What is the net force on the balloon? _______ Does the shape of the balloon matter? ________ What is the buoyant force on the balloon? _____________ _________ Where is the line of action of the buoyant force? __________ Place a thin wall balloon filled with water in a tank of water. What is the net force on the balloon? _______ Does the shape of the balloon matter? ________ What is the buoyant force on the balloon? _____________ _________ Where is the line of action of the buoyant force? __________ Thru centroid of balloon

Buoyant Force: Line of Action ä The buoyant force acts through the centroid of the displaced volume of fluid (center of buoyancy)  = volume  d  = distributed force x c = centroid of volume

Buoyant Force: Applications F1F1 F1F1 W W 11 11 F2F2 F2F2 W W 22 22 Weight Volume Specific gravity Force balance ä Using buoyancy it is possible to determine: ä _______ of an object ä _______________ of an object ä Using buoyancy it is possible to determine: ä _______ of an object ä _______________ of an object  1 >  2 (With F 1, F 2,  1, and  2 given)

Buoyant Force: Applications Suppose the specific weight of the first fluid is zero (force balance) Equate weights Equate volumes

A sailboat is sailing on Lake Bryan. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Lake Bryan increase or decrease? Why?_______________________________ ____________________________________ ____________________ A sailboat is sailing on Lake Bryan. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Lake Bryan increase or decrease? Why?_______________________________ ____________________________________ ____________________ ----------- ________ Buoyant Force (Just for fun) The anchor displaces less water when it is lying on the bottom of the lake than it did when in the boat.

Rotational Stability of Submerged Bodies B G B G ä A completely submerged body is stable when its center of gravity is _____ the center of buoyancy below

End of Chapter Question ä Write an equation for the pressure acting on the bottom of a conical tank of water. ä Write an equation for the total force acting on the bottom of the tank. ä Write an equation for the pressure acting on the bottom of a conical tank of water. ä Write an equation for the total force acting on the bottom of the tank. L d1d1 d2d2

End of Chapter ä What didn’t you understand so far about statics? ä Ask the person next to you ä Circle any questions that still need answers ä What didn’t you understand so far about statics? ä Ask the person next to you ä Circle any questions that still need answers

Grand Coulee Dam http://users.owt.com/chubbard/gcdam/html/gallery.html

Gates

Download ppt "Static Surface Forces hinge water ? 8 m 4 m . Static Surface Forces ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability of."

Similar presentations