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


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


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 = ?

4γ

10. 93γ

pivot 60°

8sin60° =6.93

8’

4’

F

R=?

4 / sin60°= 4.61

y = 8.6'

cg prismcg gate

2 54


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


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


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


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


(a) (b)

Class examples:

2 59


(c) (d)

Class examples:

2 60


(e) (f)

H 2O

oil

X

What would happen if a hole were here?

Class examples:

2 61


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


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


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


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


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


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


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


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.

6γ

3γ

2 69


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


Resultant Force: Vector sum of FH and FV

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

Fθ

5050

6000

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

Tan θ = 5 .056 .0

= .842

θ = 40 °

2 71


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


B

A

2γ

FAB

verticalprojectio

Horizontal forces

+ + 2γ

4γC

D

FCD

B

FBC

2γ

4γ C

2 73


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


Vertical forces

A

B

FAB

+ +

C

B D

A

∀ ABCD

Net Virtual Volume

=

D

C

FCD2

FCD1

C

B

FBC1

FBC 2

2 75


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


Example: Sketch the pressure prism on the gate

(a) (b)

B

A B

A

2 77


(c) (d)

A

B

A

B

2 78


(e)

(f)

(g)

A B

water

mercury

oil

Pressure Prism

Documents

Transcript of Pressure Prism