2015/12/20 林再興教授編 1 §2.9 pressure prism area of the plane is rectangular since p 0 =0...
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112/04/22 林再興教授編 1
§2.9 Pressure Prism
Area of the plane
is rectangular
Since p0=0 at surface p = p0+ρgh at bottom
FR= pCGA = (p0+ρghCG)A =ρg (h/2) A where p0=0
=ρg (h/2) b·h = 1/2 (ρgh)·b·h = (Volume of pressure prism)
The resultant force must pass through the centroid of the pressure prism (1/3 h above the base)
CGCGCG
xycR
CGCG
XCR
xxAy
Ix
hhhh
bhh
bh
yAy
Iy
3
2
262)(2
12
3
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FR= (Volume of trapezoidal
pressure prism )
= (volume of pressure prism ABDE)
+(volume of pressure prism BCD)
= F1+F2= (ρgh1A) + [(1/2)ρg(h2-h1)A]
FRyR = F1y1+F2y2 (F1+F2) yR = F1[(h2-h1)/2]+F2 [(h2-h1)/3]
Only if the area of plane
is rectangular
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Example 2.8
Given ： as figure
SG=0.9(oil)
square 0.6m * 0.6m
Pgage = 50Kpa = PS
Find: the magnitude and location of the
resultant force on the attached plate.
Solution：Trapezoidal pressure prism, ABCDO
FR = (vol. of trapezoidal pressure
prism)
= (vol. of ABDO)+(Vol. of BCD)
= F1+F2
A B
D C
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F1=(PS+ρgh1)A = {50*103+[0.9*(9.81*103N/m3)*2m]}*
0.6 * 0.6 = 24.4 * 103 N
F2=(Vol. of BCD)=[ ] A
= *(0.6m*0.6m)
=0.954*103N
FR=F1+F2=25.4*103N
By summing moments around an axis through point O,so that FRy0=F1*0.3m+F2*0.2m
(25.4*103N)y0=24.4*103*0.3+0.954*103N*0.2m y0 = 0.296m
2
)hh(g 12
2
)26.2(/1081.99.0 33 mmmN
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§2.10 Hydrostatic Force on a Curved Surface
FH = F2 = ρghCACAAC = PCG(AV)proj
FV = F1 + w = ρg (aboveAB) +ρg ( ABC)
FR = 22
VH FF
V
H
F
F1tan
VVcp XdFFX
HHcp YdFFY
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Example2.6
Given :
Length(b)
=1m
Determine : The magnitude and line of FR
Solution:
FH = F2 = ρghCACAAC = 62.4
FV=W=ρg(ABC)=
lbf8.280)ft1ft3)(2
ft3(
ftlbf
3
lbfftftft
lbfRft
lbf 4411)4
3(4.621)
4(4.62 3
2
3
2
3
011
2222
5.32441
281tantan
523)441()281(
V
H
VHR
F
F
lbflbflbfFFF
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§2.11 Buoyancy, Flotation, and Stability
§2.11.1 Archimedes’ Principle
─Buoyant force : when a body is completely submerged in a
fluid, or floating so that it is only partially submerged, the
resultant fluid force acting on the body is called the
buoyant force.
─Buoyant force direction: upward vertical force
because pressure increases with depth and pressure forces
acting from below are larger than the pressure force acting
from above.
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─The first law (immersed body)
In equilibrium
dFB= dFup –dFdown
=ρgy2dA–ρgy1dA
=ρg(y2-y1)dA = ρghdA = γhdA
= weight of fluid equilibrium to body volume
For unform density of body immersed point B = Center of mass
For variable density of body immersed point B≠ Center of mass
gVVhdAhdAdFF BB
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─The second law (floating body)
dFB=dFup-dFdown
=[γ1h+ γ2(y2-h)]dA- γ 1y1dA
= γ1(h-y1)dA+ γ2(y2-h)dA
= γ1dV1+ γ 2dV2
22112211 )( vvdvdvdFF BB
For a layered fluid i
iiii
iiLFB VrvolumedisplacedgF )()(
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Example 2.10
Given:
As the figures on right
(ρg)seawater =10.1*103N/m3
Note:(ρg)fresh water
= 9.80*103N/m3
Weight of buoy = 8.50*103N
Dbuoy = 1.5m
Determine:
What is the tension of the cable?
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Solution:
For equilibrium ΣF=0
T+W – FB = 0
T = FB – W
Where FB = (ρg)water Buoy
= (10.1*103N/m3)(4/3)π*(1.5/2)3(m3)
= 1.785*104N
W = 8.50*103N
T = 1.785*104 – 8.50*103
= 9.35*103N
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§2.11.2 Stability
“Distrub” the floating body slightly
(a)develops a restoring moment which will return it to its
origin positionstable
(b)otherwise unstable
The basic principle of the static-stability calculations
(1)The basic floating position
(2)The body is titled a small angle Δθ
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MG>0(M above CG)
=>Stable
MG<0(M below CG)
=>Unstable
M
M
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§2.12 pressure variation in a Fluid with Rigid-Body Motion
The general equation of motion,
)( agP
agP
avgP
or
avkgP
2
2
No shearing stress
z
x
ag
a
1tan
2/122 ])([ zx agaGwhereGds
dp
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§2.12.1 Linear Motion
Fluidat rest
Fluid withacceleration
P1 constantP3 P2 pressure lines
Free surface slope = dz/dy
az
ay
a
a
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)24.2........(
)24.2...(..........
)24.2........(
)2.2.......(
cz
by
ax
zyx
agz
P
ay
P
ax
P
kajaiakgkz
Pj
y
Pi
x
P
akgP
avkgP 2
No shearing stress
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dzagdyadzz
Pdy
y
Pdp
agz
P
ay
Px
Pa
zy
z
y
x
)(
)26.2)........((
)25.2........(
00
Along a line of constant pressure , dp=0
All lines of constant pressure will be parallel to the free surface
If from Eq(2.26)
)28.2........(z
y
ag
a
dy
dz
)( zagdz
dP 0&0 zy aa
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Example 2.11Given:Right figure
ay only
Questions:
(1) P = (ay) at point 1
(2) ay = ? if fuel level on point 1
Solution:
)(75.075.0
0
)28.2........()1(
11
g
az
g
az
g
a
dy
dz
afor
ag
a
dy
dz
yy
y
z
z
y
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since )25.2)........(( zagz
P
g
a
g
a
ftlbf
g
ag
zg
zgpp
zzgpp
azgp
y
y
y
sfc
sfcsfc
z
42.3028.20
]75.05.0)[(4.6265.0
]75.05.0[
]5.0[
)]()5.0[(
)(
)0(
3
1
11
25.212.3275.0
5.0
)2(
sftg
dy
dza
g
a
dy
dz
ag
a
dy
dz
y
y
z
y
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§2.12.2 Rigid-Body Rotation
since
agagP
)(
2ra
r
r
r
irkg
irkg
kz
Pi
P
ri
r
P
2
2 )()(
1
gz
P
P
rr
P
0
2
……(2.30 )
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at (r,z) = (0,0), P=P0
220
022
0
2
12
1
rgzPPor
PgzrP
CP
2rr
PFrom
cgzrP
cgzzf
gz
zf
gz
zf
z
P
zfrP
rrP
22
22
2
2
1
)(
)(
)(0
)(2
1
gz
P
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For a constant pressure
surface at air liquid surface,
P=P1
Paraboloidbraz
g
r
g
PPz
rgzPP
2
2210
2201
2
2
1
g
wRh
g
wRh
RrPPFor
42&
2
,2222
01
volume(ace)
Volume(abcde)
hR
g
rdrdzr
2)
2(
22222
hR2
e
b c d
a
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