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### Transcript of Pressure Prism

Chapter 2. FLUID STATICS

Pressure Prism Method Draw the pressure loading diagram, or pressure prism, directly on the plane surface. The total force equals the volume of the pressure prism. The line of action is at the centroid of the pressure prism.

h1 h2

1 FC.G. of prism

p1 = h1F

p 2 = h2

C.G. of prism

p1 = h 1l = length w = width

2

p2 = h 2

w

l

2 51

Chapter 2. FLUID STATICS

From the figure:

h1 + h2 ) (l)(w) 2 The location of the line of action is at the centroid of the pressure prism. The total force equals:

F = p = (

h1

Fy = F1 y1 + F2 y 2 ( p )y = (h 1lw )y1 + ( y = F1 y1 + F2 y 2 p (h 2 h 1 ) lw )y 2 2

FF2y1

(h2 h1 )

yh2y2 l

2 52

l l 2 2 y2 = l 3 y1 =

Chapter 2. FLUID STATICS

Class examples: What is the force required to hold the gate closed? Neglect the gate weight. Solve by using the cent of pressure formula and check by the pressure prism method.44 cg prism cg gate

pivot

60

4 / sin60= 4.61y = 8.6'

8sin60 =6.93

F8

yp = ?

10. 93

R= ?2 53

Chapter 2. FLUID STATICS

Method No. 1: Center of Pressure

A = 8(5) = 40 ft 2 y = 8 / 2 + 4 / sin 60 = 8.61 ft I 213 = 8.61+ = yp = y + 8.61(40) yA = 8.61+ .617 yp = 9.227 ft

2 54

Chapter 2. FLUID STATICS

Pressure:Force on gate = F = p avg A = p A 4 + 10 .93 lb . p = = 7 .47 = 7.47 (62 . 4 ) = 465 2 ft . 2

F = p A = 465 (8) (5 ) = 18 , 620 lb. Take moments about the pivot+

MH = 0: 18 , 620 ( 4.617 ) R(8 ) = 0 R = 10 , 750

2 55

Chapter 2. FLUID STATICS

Method No. 2: Pressure Prism MethodF=?resultant force F F2

ypF1

4

Force F1 F2 R

Magnitude 4 (8)(5) = 160 6.93 + 0 ( )(5)(8) 2 = 138.6

Arm 4.0' 16' /3 8

Moment

6.93 4

2/ 3x8 8

+640 +740 8R

R

= ?

2 56

Chapter 2. FLUID STATICS

+ MH = 0: 8R + 640 + 740 = 0

R=

1380 (62 .4 ) = 10 , 750 lb. 8

Find where the resultant force F acts by moments about the pivot

+640 + 740 = Fy p

+1380 = (160 + 138 .6 ) y p 1380 = = 4.61 7 298 .6

yp2 57

Chapter 2. FLUID STATICS

Class examples:

(a)

(b)

2 58

Chapter 2. FLUID STATICS

Class examples:

(c)

(d)

2 59

Chapter 2. FLUID STATICS

Class examples:

oil

XH 2O

What would happen if a hole were here?

(e)

(f)

2 60

Chapter 2. FLUID STATICS

Example C: Determine the magnitude and location of the hydrostatic pressure force acting on one side of the vertically placed triangle shown. Use center of pressure method.x

+

yp 6m xp 10m cp

y

S = 0.82

2 61

Chapter 2. FLUID STATICS

Force Components on Curved Surfaces The pressure force at any point on the curved surface AB is a function (p = h) of the depth below the free surface.y

AdF1 dF2 dF3 dF4 + dF5X

dA

h5 B

Consider the force at the level, h 5 , below the free surface: dF5 = h5dAdF5 can be resolved into components dFx ,dFy2 62

fluid dFn

Chapter 2. FLUID STATICS

The total force on the surface is: Therefore,

F dFii

F dFi = lim 0 dF = dFx + dFy . dAi

The simplest way to analyze forces on curved surfaces is to consider the horizontal and vertical components separately.

2 63

Chapter 2. FLUID STATICS

Horizontal Forces The horizontal force on a curved surface A-B equals the force produced on a vertical projection ( A B ) of the curved surface subjected to the hydrostatic pressure. Calculation proceeds as in the plane surface analysis.A A

vertical projection cg F x B hB B 2 64

Chapter 2. FLUID STATICS

Vertical Forces The vertical force on a curved surface is equal to the weight of the volume of liquid that stands (or would stand) between the curved surface and the free surface.

The location of this force is through the centroid of the virtual volume. The total force is found by vector addition F = Fx + FyAFx F Fy2 65

Acg

FyB

B

Chapter 2. FLUID STATICS

Example B: Determine the magnitude and location of the hydrostatic pressure force acting on the ellipsoidal gate shown below.b. min.

F 2.5m H2 O 4m Hingexa. maj.

2m

A=2 66

ab, Ix = a3b 4 4

Chapter 2. FLUID STATICS

Example D: The width of the surface is six feet into the figure. Find: The resultant force on the surface A-B Solution: Analyze by resolving into horizontal and vertical components.

3 ft C3 ft

WATER

A 6 ft gate width

B2 67

Chapter 2. FLUID STATICS

Horizontal Force: Force on vertical projection of curve.

3

A3 ft.

FH6

BPr ismFH

FH = Volume of Pr essure

3 +6 ) (3)(6 ) = ( 4 .5 )(18 ) = 81(62 .4 ) = 5050 lb = FH = ( 22 68

Chapter 2. FLUID STATICS

Vertical Force: Weight of water between the shape and free surface. Break volume into known shapes:Fv1 Fv 2 A

Vol 1 = (3 )(3)(6 ) = 54 lbs = Fv 12

B

Vol 2 = 1 / 4[ (3 ) ]6 = .785 (9)(6 )( ) = 42 .4 lbs = Fv 2Fv = Fv1 + Fv2 = p = (54 + 42.4)(62.4)2 69

Fv = 6000 lb

Chapter 2. FLUID STATICS

Resultant Force: Vector sum of FH and FV5050

6000

F

F=

6 + 5 .05 (1000 ) = 7842 .35 lb. 5 .05 = .842 Tan = 6 .0 = 40

2

2

Acts through the center of gate curve, i.e. C2 70

Chapter 2. FLUID STATICS

Example E (2.12 pg. 52): Find For a 1 meter length of cylinder, determine the weight of the cylinder and the force against the wall. Solution: A Analysis Concept:r=D

B

C

Lets break the curved surface up into its simplest non-compound surfaces and then calculate the components of Fx and Fy on each surface

2 71

2

m

Chapter 2. FLUID STATICS

Horizontal forcesvertical projectio

A

FAB 2 B

+FBC

2

B

+C C

D

2 FCD

4

4

2 72

Chapter 2. FLUID STATICS

FAB FBC FCD Fx

= = = =

AB BC CD

= = =

2 (2)(1) = 19 ,600 2 2 + 4 ( )(2)(1) = 2 2 + 4 ( )(2)(1) = 2 19 ,600 N

N 58 ,836 N 58 ,836 N

FAB + FBC + FCD =

2 73

Chapter 2. FLUID STATICS

Vertical forces FAB AFBC1

FCD

1

+B

B

+FBC C2

D FCD C2

=B

A

D C

Net Virtual Volume ABCD

2 74

Chapter 2. FLUID STATICS

Vertical forces continued2 2 2 r Fy = ABCD = (r ) + (r ) (1) = 131,562 N 4

The location of the line of action is through the centroid of ABCD. It is easier to locate the centroids and forces of the elementary shapes forming ABCD than it is to determine the single value location.

2 75

Chapter 2. FLUID STATICS

Example: Sketch the pressure prism on the gate

B

A

B A

(a)2 76

(b)

Chapter 2. FLUID STATICS

AA B

B

(c)2 77

(d)

Chapter 2. FLUID STATICS

A

B

oil(e)

water mercury(f)

(g)2 78