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2 51 Chapter 2. FLUID STATICS Pressure Prism Method Pressure Prism Method Pressure Prism Method 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. 1 2 C.G. of prism F h 1 p 1 h 1 p 2 h 2 α h 2 F p 1 h 1 p 2 h 2 F C.G. of prism l = length w = width w l α

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### Transcript of Pressure Prism 2 51

Chapter 2. FLUID STATICS

Pressure Prism MethodPressure Prism MethodPressure Prism MethodPressure 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.

1

2

C.G. of prism

F

h1 p1 = γh1

p2 = γh2α

h2

F

p1 = γh1p2 = γh2

F

C.G. of prism

l = length w = width

w lα 2 52

Chapter 2. FLUID STATICS

Fy = F1y 1 +F2 y 2

(∀ p )y = (γh 1lw )y 1 + (γ(h 2 −h1 )

2lw )y 2

y =F1y 1 +F2 y 2

∀ p

y 1 = l2

l

y 2 = 23

l

From the figure:

• The total force equals:

• The location of the line of action is at the centroid of the pressure prism.

F = ∀ p = (

γh1 + γh2

2) (l)(w)

γh1

y

F

F2 y 1

y 2

l γh2

γ(h2 −h1) 2 53

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.

y p = ?

10. 93γ

pivot 60°

8sin60° =6.93

8’

4’

F

R=?

4 / sin60°= 4.61

y = 8.6'

cg prismcg gate 2 54

Chapter 2. FLUID STATICS

Method No. 1: Center of Pressure

A = 8(5) = 40 ft2

y = 8 /2 + 4 / sin 60° = 8.61 ft

yp = y + I

y A= 8.61+ 213

8.61(40)=

= 8.61+ .617∴ yp = 9.227 ft 2 55

Chapter 2. FLUID STATICS

Pressure:

Force on gate = F = p avg A = p A

p = 4 γ +10 .93 γ2

= 7 .47 γ = 7.47 (62 .4 ) = 465 lb .

ft .2

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

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

+ 2 56

Chapter 2. FLUID STATICS

Method No. 2: Pressure Prism Method

Force Magnitude Arm Moment

F1

F2

R

4.0'

16' /3

+640γ

+740γ

−8R 8

4γ(8)(5)=160γ6.93+0

2(γ)(5)(8)

= 138.6γ= ?

resultant

force ′ F F1F2

4 γ

6.93γ

yp

2/ 3x ′ 8

R

′ 8

′ 4

F=? 2 57

Chapter 2. FLUID STATICS

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

+640 γ +740 γ = Fy p

+1380 γ = (160 +138 .6 )γ yp

yp = 1380 γ298 .6 γ

= 4.61 ′ 7

MH∑ = 0: − 8R + 640 γ + 740 γ = 0

R = 13808

(62 .4 ) = 10 ,750 lb.

+ 2 58

Chapter 2. FLUID STATICS

(a) (b)

Class examples: 2 59

Chapter 2. FLUID STATICS

(c) (d)

Class examples: 2 60

Chapter 2. FLUID STATICS

(e) (f)

H 2O

oil

X

What would happen if a hole were here?

Class examples: 2 61

Chapter 2. FLUID STATICS

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

cp

xp

yp

6m

10m

+

y

x

S= 0.82 2 62

Chapter 2. FLUID STATICS

Force Components on Curved Surfaces

The pressure force at any point on the curved surface AB is a function

of the depth below the free surface.

Consider the force at the level,, below the free

surface:

can be resolved into components

(p = γh)

dF 5 = γh5dA

dFx,dFy

dF 5

dF1dF2dF3

dF4

dF5+

γ fluid

dFn

B

dA h5

A

X

y

h 5 2 63

Chapter 2. FLUID STATICS

The total force on the surface is:

Therefore,

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

F ≅ dF i

i∑

F ≅ dF i

i∑ = dA→0

lim dF∫ = dFx +∫ dFy.∫ 2 64

Chapter 2. FLUID STATICS

Horizontal Forces

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

′ A − ′ B

B

A

vertical projection

cg

Fx

hB

′ A

′ B 2 65

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.

F = Fx +Fy

The location of this force is through the centroid of the virtual volume. The total force is found by vector addition

B

A

cg

Fy

B

A

F

Fy

Fx 2 66

Chapter 2. FLUID STATICS

Example B: Determine the magnitude and location of the hydro-static pressure force acting on the ellipsoidal gate shown below.

A = π

4ab, Ix = π

4a3b

a. a. a. a. maj.maj.maj.maj.

bbbb. . . . min.min.min.min.

x2.5m

2m

4m

HingeH2O

F 2 67

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-BSolution: Analyze by resolving into horizontal and vertical components.

B

A

WATER3 ft

3ft

C

6 ft gate width 2 68

Chapter 2. FLUID STATICS

Horizontal Force: Force on vertical projection of curve.

FH = Volume of Pr essure Pr ism

FH = ( 3γ +6γ2

) (3)(6 ) = (4 .5γ)(18 ) = 81(62 .4 ) = 5050 lb = →FH

B

A

FH3 ft. 2 69

Chapter 2. FLUID STATICS

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

Vol1 = (3 )(3)(6 )γ = 54 γ lbs = Fv1

Vol2 = 1 / 4[π(3 )2 ]6 γ = .785 (9)(6 )(γ) = 42 .4 γ lbs = Fv2

Fv = Fv1+ Fv2

= ∀ pγ = (54 + 42.4)(62.4)

Fv = 6000 lb↓

Fv2

Fv1

A

B 2 70

Chapter 2. FLUID STATICS

Resultant Force: Vector sum of FH and FV

Acts through the center of gate curve, i.e. C’

5050

6000

F = 6 2 +5 .05 2 (1000 ) = 7842 .35 lb.

Tan θ = 5 .056 .0

= .842

θ = 40 ° 2 71

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:

Analysis Concept:

Let’s break the curved surface up into its simplest non-compound surfaces and then calcul-ate the components of

on each surface Fx and Fy

r =2 m

A

C

BD 2 72

Chapter 2. FLUID STATICS

B

A

FAB

verticalprojectio

Horizontal forces

+ + 2γ

4γC

D

FCD

B

FBC

4γ C 2 73

Chapter 2. FLUID STATICS

FAB = ∀ AB γ = 2γ2

(2)(1) = 19 ,600 N→

FBC = ∀ BC γ = (2γ +4γ2

)(2)(1) = 58 ,836 N→

FCD = ∀ CD γ = (2γ +4γ2

)(2)(1) = 58 ,836 N←

Fx = FAB + FBC + FCD = 19 ,600 N→ 2 74

Chapter 2. FLUID STATICS

Vertical forces

A

B

FAB

+ +

C

B D

A

∀ ABCD

Net Virtual Volume

=

D

C

FCD2

FCD1

C

B

FBC1

FBC 2 2 75

Chapter 2. FLUID STATICS

Vertical forces continued

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

∀ ABCD

∀ ABCD

Fy = ∀ ABCDγ = γ (πr2) + (r2 − πr2

4)

(1) = 131,562 N↑ 2 76

Chapter 2. FLUID STATICS

Example: Sketch the pressure prism on the gate

(a) (b)

B

A B

A 2 77

Chapter 2. FLUID STATICS

(c) (d)

A

B

A

B 2 78

Chapter 2. FLUID STATICS

(e)

(f)

(g)

A B

water

mercury

oil