Download presentation

Presentation is loading. Please wait.

1
Static Surface Forces hinge water ? 8 m 4 m

2
Static Surface Forces ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability submerged bodies ä Forces on plane areas ä Forces on curved surfaces ä Buoyant force ä Stability submerged bodies

3
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

4
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 = volume P = 500 kPa What is p?

5
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

6
Forces on Plane Areas: Inclined Surfaces A’ B’ O O O O x x y y Free surface centroid center of pressure The origin of the y axis is on the free surface Where could I counteract pressure by supporting potato at a single point?

7
Magnitude of Force on Inclined Plane Area p c is the pressure at the __________________ y y centroid of the area h c is the vertical distance between free surface and centroid

8
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

9
Second Moments Also called _______________ of the area I xc is the 2 nd moment with respect to an axis passing through its centroid and parallel to the x axis. The 2 nd moment originates whenever one computes the moment of a distributed load that varies linearly from the moment axis. moment of inertia

10
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

11
Properties of Areas ycyc b a I xc ycyc b a R ycyc d

12
Properties of Areas a ycyc b I xc ycyc R R ycyc

13
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 = sum of moment of distributed forces

14
Center of Pressure: x R Product of inertia Parallel axis theorem y x

15
Center of Pressure: y R Sum of the moments Parallel axis theorem p = 0 when y = 0 You choose the pressure datum to make the problem easy

16
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

17
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 Example using Moments ä ä ä ä ä ä ä ä Location of the resultant force Find F using moments about hinge teams

18
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

19
Location of Resultant Force hinge water F F 8 m 4 m FrFr FrFr 12.5 m Slant distance to surface 0.125 m b = 2 m a = 2.5 m cp

20
Force Required to Open Gate How do we find the required force? F = ______ 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 809 kN cp

21
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?) ___________ ___________________ The gate was symmetrical about at least one of the centroidal axes. Pressure increases with depth.

22
Forces on Plane Areas: Pressure Prism ä A simpler approach that works well for areas of constant width (_________) 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 (_________) 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

23
Forces on Plane Areas: Pressure Prism O AA Force = Volume of pressure prism Center of pressure is at centroid of pressure prism Free surface h2h2 h1h1

24
Example 1: Pressure Prism Dam h=10 m Dam is 50 m wide FRFR 24º F R = 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 hh

25
Example 2: Pressure Prism hinge water 8 m 4 m x 4 m (square conduit) 5 m 12 m) 8 m) O O y y hh

26
Solution 2: Pressure Prism 5 m (h 2 ) (h 1 ) Magnitude of force Location of resultant force a ypyp w F R = ________ y R = _______ 1.96 MN measured from hinge 2.667 m 4 m FRFR FRFR

27
Forces on Curved Surfaces ä Horizontal component ä Vertical component ä Tensile Stress in pipes and spheres ä Horizontal component ä Vertical component ä Tensile Stress in pipes and spheres

28
Forces on Curved Surfaces: Horizontal Component ä What is the horizontal component of pressure force on a curved surface equal to? (Prove it!) 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? (Prove it!) 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 teams

29
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?

30
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

31
water = (3 m)(2 m)(1 m) + (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

32
= 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

33
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 a y x (1 m)(2 m) 3 /12 = 0.667 m 4 4 m

34
Example: Forces on Curved Surfaces 78.5 kN 89.7 kN 4.083 m 0.948 m 119.2 kN horizontal vertical resultant

35
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.

36
Curved Surface Trick water 2 m 3 m A W1W1 W2W2 F F O O W 1 + W 2 ä Find force F required to open the gate. 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 The pressure forces and force F pass through O. Thus the hinge force must pass through O! ä Hinge carries only horizontal forces! (F = ________)

37
Tensile Stress in Pipes: High Pressure ä pressure center is approximately at the center of the pipe T1T1 T2T2 FHFH b r F H = ___ T = ___ = ____ (p c is pressure at center of pipe) 2rp c (e is wall thickness) rp c p c r/e is tensile stress in pipe wall per unit length

38
Tensile Stress in Pipes: Low pressure ä pressure center can be calculated using moments ä T 2 __ T 1 ä pressure center can be calculated using moments ä T 2 __ T 1 T1T1 T2T2 FHFH b r h > h=2r b Projected area y c is distance to centroid from virtual free surface F H = ___ 2p c r

39
Solution Scheme ä Determine pressure datum and location in fluid where pressure is zero (y origin) ä Determine total acceleration vector (a) including acceleration of gravity ä Define h tangent to acceleration vector (call this vertical!) ä Determine if surface is normal to a, inclined, or curved ä Determine pressure datum and location in fluid where pressure is zero (y origin) ä Determine total acceleration vector (a) including acceleration of gravity ä Define h tangent to acceleration vector (call this vertical!) ä Determine if surface is normal to a, inclined, or curved

40
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

41
Buoyant Force 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 _________ zero different

42
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? _____________ _________ 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? _____________ _________

43
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

44
Buoyant Force: Applications F1F1 F1F1 W W 11 11 F2F2 F2F2 W W 22 22 Weight Volume Specific gravity 1 > 2 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

45
Buoyant Force: Applications Suppose the specific weight of the first fluid is zero (force balance) Equate weights Equate volumes

46
----------- ________ 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. A sailboat is sailing on Cayuga Lake. 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 Cayuga Lake increase or decrease? Why?_______________________________ ____________________________________ ____________________ A sailboat is sailing on Cayuga Lake. 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 Cayuga Lake increase or decrease? Why?_______________________________ ____________________________________ ____________________

47
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

48
End of Lecture 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

49
End of Lecture 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

50
Team Work ä How will you define a coordinate system? ä What are the 3 major steps required to solve this problem? ä What equations will you use for each step? ä How will you define a coordinate system? ä What are the 3 major steps required to solve this problem? ä What equations will you use for each step? hinge water F 8 m 4 m

51
Gates

53
Radial Gates

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google