ENGR-36_Lec-25_FluidStatics_HydroStatics
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Transcript ENGR-36_Lec-25_FluidStatics_HydroStatics
Engineering 36
Chp09:
FluidStatics
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Fluid Statics
The Fish Feels More
“Compressed” The Deeper
it goes – An example of
HydroStatic Pressure
Definition of Pressure
• Pressure is defined as the amount of force exerted
on a unit area of a surface:
P
force
area
N
m
2
Pa
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Direction of fluid pressure on
boundaries
Furnace duct
Pipe or tube
Heat exchanger
Pressure is a Normal
Force
• i.e., it acts perpendicular
to surfaces
Dam
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• It is also called a
Surface Force
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Absolute and Gage Pressure
Absolute Pressure, p: The pressure of a
fluid is expressed relative to that of a
vacuum (absence of any substance)
Gage Pressure, pg: Pressure expressed
as the difference between the pressure
of the fluid and that of the surrounding
atmosphere (the BaseLine Pressure)
which has an Absolute pressure, p0.
• p0 ≡ BaseLine
Pressure:
Engineering-36: Engineering Mechanics - Statics
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p p0 p g
Bruce Mayer, PE
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Pressure Units
U n it
1 p a s c a l (P a )
1 k g m -1 s -2
1 bar
1 x 105 Pa
1 a tm o s p h e re (a tm )
1 0 1 ,3 2 5 P a
1 to rr
1 / 7 6 0 a tm
760 m m Hg
1 a tm
1 4 .6 9 6 p o u n d s p e r
s q . in . (p si)
1 a tm
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D e fin itio n o r
R e la tio n s h ip
Bruce Mayer, PE
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Static Fluid Pressure Distribution
Consider a Vertical
Column of NonMoving
Fluid with density, ρ, and
Geometry at Right
Taking a slice of fluid at
vertical position-z note
that by Equilibrium:
The sum of the z-directed
forces acting on the fluidslice must equal zero
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Bruce Mayer, PE
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The z-Direction Fluid Forces
Consider the Fluid Slice at z
under the influence of gravity
Let pz and pz+Δz denote the
pressures at the base and x
top of the slice, where the
elevations are z and z+Δz
respectively
7
z z
z
• S is the Top & Bot
Slice-Surface Area
Engineering-36: Engineering Mechanics - Statics
Sp
y
mg S z g
Sp
z
Bruce Mayer, PE
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The z-Direction Fluid Forces
A Force Balance in the z-Dir
F
z
0 Sp z Sp
S p
or
z z
z z
p
z
S zg
Solving for Δp/Δz and
Taking the limit Δz→0
Δp
Δp dp
g and lim
z 0 z
z
dz
8
z z
S zg
S Δp S zg
Engineering-36: Engineering Mechanics - Statics
Sp
z
y
x
Sp
so
dp
S zg
z
g
dz
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Constant Density Case
For Constant ρ
Δp z g
or
p 2 p1 z 2 z1 g
Next Consider the
typical Case of a
“Free Surface” at
Atmospheric
Pressure, p0, and
total Fluid Depth, H
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z
H
If h is the DEPTH,
then the z↔h reln:
z H h
so
z 2 z1 H h 2 H h1
Bruce Mayer, PE
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Constant Density Case
Then z 2 z1 h1 h 2
or
z h
z
H
Sub into Δp Eqn
Δp z g h g
or
Δp g h
Δp g h g h 0
Now if at
h = 0, p = p0,
and Δh = h2 – 0 = h
or
p h p 0 gh
In this Case
or
p h p 0 gh
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Constant Density Case
The Gage Pressure,
pg, is that pressure
which is Above the
p0 Baseline
z
H
p p 0 gh p 0 p g
or
p g gh
p p 0 liq gh
Thus the typical
or p g liq gh
Formulation for
Liquids at
• The Gage Pressure is used
atmospheric pressure for Liquid-Tanks or Dams
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Specific Weight
Since the
acceleration of
gravity, g, is almost
always regarded as
constant, the ρliqg
product is often
called the Specific
Weight
liq g liq
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z
H
Then the pressure eqns
p p 0 liq h
or
p g liq h
Bruce Mayer, PE
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Bouyancy
A body immersed in
a fluid experiences a
vertical buoyant
force equal to the
weight of the fluid
it displaces
A floating body
displaces its own
weight in the fluid in
which it floats
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h1
Free liq surf
F1
Δh
h2
F2
Bruce Mayer, PE
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Bouyancy
h1
The upper surface of
the body is subjected
to a smaller force
than the lower surface
F1
Free liq surf
y
x
Δh
h2
Thus the net bouyant
force acts UPwards
F2
The net force due to pressure in the vertical
direction
FB = F2- F1 = (pbottom - ptop)(ΔxΔy) = Δp(ΔxΔy)
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Bouyancy
h1
The pressure
difference
y
x
pbottom – ptop = ρg(h2-h1) = ρgΔh
Δp = ρgΔh
Subbing for Δp in the
FB Eqn
FB = (ρgΔh)(ΔxΔy)
But ΔhΔxΔy is the
Volume, V, of the
fluid element
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F1
Free liq surf
Δh
h2
F2
So Finally the Bouyant
Force Eqn
FB gV V
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Equivalent Point Load
Consider Fluid
Pressure acting on a
submerged FLAT
Surface with any
azimuthal angle θ
In this situation the Net
Force acting on the Flat
Surface is the pressure
liq
liq cent
at the Flat Surface’s
GEOMETRIC Centroid i.e., Fliq = the AVERAGE
Pressure times
times the Surface area
F gh
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the Area
Bruce Mayer, PE
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A
Equivalent Point Load
Fliq is NOT
Positioned at the
Geometric Centroid;
Instead it is located at
the CENTER of
PRESSURE
Calculation of the
Center of Pressure
Requires Moment
Analysis as was
described last lecture
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Fliq liqhcent A
Fliq P avg A
Bruce Mayer, PE
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Force on a Submerged Surface
Pressure acts as a
function of depth
F y p y A y A
F y y b y
1
2
base height
2
19
1
R d d b Pavg
2
Engineering-36: Engineering Mechanics - Statics
y
Δy
Then the Magnitude of
the Resultant Force is
Equal to the Area under
the curve R 1 p d d b
Atriangle
0
Resultant, R
d
d
Face Width of
Structure INTO
the Screen = b
A
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Force on a Submerged Surface
The location of the Resultant
of the pressure dist. is found
the by same technique as
used on the horizontal beam
OS R
by
before
ydF
0
y p dA y y bdy
OS R
y
d
y b y dy
2
0
d
d
b y dy
2
OS R
yR y
y
R
0
bd
2
2
b y dy
2
0
A Pdist
Thus the Location of the
Resultant Hydrostatic
force Passes Thru the
Pressure-Area Centroid
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S
Resultant, R
d
d
The Distance OS is
Also Called the
Center of Pressure
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Angled Submerged Surface
How does one find the
forces on a SUBMERGED
surface (red) at an angle?
Construct the FBD
Detach The Triangular
Fluid Volume from the
bulk fluid to Reveal
Forces:
• Weight of the Volume
• Pressure at NORMAL
Surfaces
Can also ATTACH the
Fluid to the Body if desired
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Bruce Mayer, PE
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Angled SubMerged Surface
The resulting force distribution without the
weight of the water would show
Equal to the
Trapezoidal Area
Under the
Pressure Curve
Times the Width b
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Bruce Mayer, PE
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NonLinear Submerged Surface
Consider The
P-distribution on a
non-linear surface.
Liquid Generates
Resultant, R on The
Surface
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Use Detached Fluid
Volume as F.B.D.
R R1 R 2 W 0
The Force Exerted
by the Surface is
Simply −R
Bruce Mayer, PE
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HydroStatic Free Body Diagram
Examine Support Structure for
Non-Rectilinear Surfaces
Detach From Surrounding Liquid, and
And from the Structure, a LIQUID
VOLUME that Exposes Flat X&Y
Surfaces exposed to Liquid Pressure
• Pressure is
– Constant at X-Surfaces (Horizontal)
– Triangular or Trapezoidal at
Y-Surfaces (Vertical)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Centroids of Parabolas
Sector
Engineering-36: Engineering Mechanics - Statics
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Spandrel
Bruce Mayer, PE
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WhtBd Example: P9-122
Determine the
resultant horizontal
and vertical force
components that the
water exerts on the
side of the dam.
Find for R the PoA
on the Dam-Face
• The dam is 25 ft long
• SW for Water = 62.4
lb/cu-ft
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Resultant Location
P9-122 Dam-Face Force Location
25
20
y
15
10
5
0
0
1
2
3
Engineering-36: Engineering Mechanics - Statics
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4
5
x
6
7
8
9
10
Bruce Mayer, PE
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Solve By MATLAB
% Bruce Mayer, PE
% ENGR36 * 03Dec12
% ENGR36_Dam_Face_P9_122_1212.m
%
W = 260; % kip
P = 487;.5 % kip
%
Deqn = [P/4 W -(25*P/3+30*W/8)]
x = roots(Deqn)
y = (x(2))^2/4
xp = linspace(0,10, 500);
yp1 = polyval(Deqn, xp);
yp = xp.^2/4;
xD = x(2)
yD = y
plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid,
xlabel('x'),...
ylabel('y'), title('P9-122 Dam-Face Force Location')
Deqn =
1.0e+03 *
0.1218
0.2600
x =
-7.5856
5.4500
y =
7.4257
xD =
5.4500
yD =
7.4257
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
-5.0333
Example Small Dam
Given Fresh Water
dam with Geometry
as shown
If the Dam is 24ft
wide (into screen)
Find the Resultant
of the pressure
forces on the
Dam face BC
• Assume
water 62.4 lbs/ft
Engineering-36: Engineering Mechanics - Statics
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3
Detach the
Parabolic Sector
of Water
Bruce Mayer, PE
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Example: Small Dam
Create FBD for
Detached
Water-Chunk
The gage Pressure
at the Bottom
p18 h
lb
62.4 3 18 ft
ft
lb
1123.2 2 psf
ft
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Example: Small Dam
The Force dF18 at
the base of the Dam
with Face Width, b
(into paper),
consider a vertical
Distance dh located
at 18ft Deep
dF18 p18dA p18 dh b
lb
1123.2 2 dh b
ft
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Example: Small Dam
Now to Covert dF to
w (lb/vertical-Foot)
Simply Divide by the
vertical distance that
generated dF
dF 18
dh
1123 . 2
2
dh b
ft
dh
w18
1123 . 2
lb
lb
ft
2
b
In this case b = 1ft
Engineering-36: Engineering Mechanics - Statics
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lb
w18 hmaxb 1123.2
ft
Bruce Mayer, PE
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Example: Small Dam Digression
For a Submerged
Flat surface, with
Face Width, b, in a
constant density
Fluid the Load per
Unit-Length profile
value, w(h) can
found as
wh b h b p
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• In Units of lb/ft or
lb/in or N/m
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Example: Small Dam
The Load Profile is a
TriAngle → A = ½BH
• B = 1123 lb/ft
• H = 18 ft
Then the Total Water
Push, P, is the Area
under the Load Profile
1 1123 lb
P
18 ft
2 ft
10108 lb
Engineering-36: Engineering Mechanics - Statics
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Previous Slide
2
Aparabolic ab
3
sector
Bruce Mayer, PE
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Example: Small Dam
Then the Volume of
the Detached Water
V Aparab Thickness
2
18 ft 10 ft 1 ft
3
3
120 ft
And the Weight, W4,
of the Attached
Water
Engineering-36: Engineering Mechanics - Statics
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W4 V 62.4120
7488 lb
Bruce Mayer, PE
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Example: Small Dam
And the Location of
the of P is the Center
of Pressure which is
located at the
Centroid of the
LOAD PROFILE
• In this case the CP is
6ft above the bottom
Also W4 is applied
at the CG of the
parabolic sector
Engineering-36: Engineering Mechanics - Statics
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From “Parabola” Slide
X C 2b 5 210 ft 5
4 ft to left of vertical edge
Bruce Mayer, PE
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Example: Small Dam
Now the Dam also
pushes on the
Detached Water
For Equilibrium the
Push by the Dam on
the Water-Chunk
must be the negative
of Resultant of the
Load ON the Dam
The Applied Loads
are P and W4
Engineering-36: Engineering Mechanics - Statics
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R
Then the FBD for
the Water-Chunk
R P W 0
Bruce Mayer, PE
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Example: Small Dam
The Water Chunk FBD
Notice that This is a
3-Force Body
• Thus the Forces are
CONCURRENT and
no Moments are
Generated
The Force Triangle
Must CLOSE
• Use to Find
Mag & Dir for R
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Example: Small Dam
The Force Triangle:
The the Load per
Unit Width of the
Dam
• 12580 lb/ft-Width
• Directed 36.5° below
the Horizontal and to
the Left relative to
the drawing
Engineering-36: Engineering Mechanics - Statics
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12580 lbs
Bruce Mayer, PE
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Resultant Location on Dam
Determine the
CoOrdinates,
Horizontally &
Vertically, for the
Point of Application
of the 12 580 lb
M
C
x, y
12580 lbs
Take the ΣMC=0
about the upper-left
Corner where the
water-surface
touches the dam
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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10 '
18 '
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
W =
7488
MATLAB Results
P =
10109
k =
0.0309
Deqn =
1.0e+05 *
0.0023
x =
-56.4769
12.7360
y =
5.0064
0.1011
-1.6624
x canNOT be
Negative in this
Physical
Circumstance
xD =
12.7360
yD =
5.0064
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
10
9
8
7
6
5
4
3
2
1
0
PoA for R on Dam Face
y
0
2
4
8
Dam Face Force Location
6
x
10
12
14
16
12580 lbs
18
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
MATLAB Code
% Bruce Mayer, PE
% ENGR36 * 23Jul12
% ENGR36_Dam_Face_1207.m
%
W = 7488
P = 10109
k = 10/18^2
Deqn = [k*W P -(6*W+12*P)]
x = roots(Deqn)
y = k*(x(2))^2
xp = linspace(0,18, 500);
yp = polyval(Deqn, xp);
yp = k*xp.^2;
xD = x(2)
yD = y
plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid,
xlabel('x'),...
ylabel('y'), title('Dam Face Force Location‘)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
WhiteBoard Work
Barrel Dam
Problem
ATTACH some Water
Segments to the Drum
Engineering-36: Engineering Mechanics - Statics
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A 55-gallon, 23-inch diameter
DRUM is placed on its side to
act as a DAM in a 30-in wide
freshwater channel. The drum is
anchored to the sides of the
channel. Determine the resultant
of the pressure forces acting on
the drum and anchors.
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Engineering 36
Appendix
dy
dx
sinh
µx
T0
µs
T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
Dam Face Force Location
10
9
8
7
y
6
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
18
x
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx