Pressure Prism

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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 = ?
4γ
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 hydrostatic 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 AB 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 hydrostatic 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 ABSolution: 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.
6γ
3γ
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’
Fθ
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 noncompound surfaces and then calculate the components of
on each surface Fx and Fy
r =2 m
A
C
BD
2 72
Chapter 2. FLUID STATICS
B
A
2γ
FAB
verticalprojectio
Horizontal forces
+ + 2γ
4γC
D
FCD
B
FBC
2γ
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