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
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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
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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
Engineering-36: Engineering Mechanics - Statics
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• It is also called a
Surface Force
Bruce Mayer, PE
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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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
Sp
y
mg   S   z  g
Sp
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
Sp
z
y
x
Sp
so
dp
 S zg
z
  g
dz
Bruce Mayer, PE
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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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)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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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
Engineering-36: Engineering Mechanics - Statics
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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
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
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
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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
Engineering-36: Engineering Mechanics - Statics
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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
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
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 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
Engineering-36: Engineering Mechanics - Statics
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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
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
wh  b    h  b  p
Engineering-36: Engineering Mechanics - Statics
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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  62.4120
 7488 lb
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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  210 ft  5 
 4 ft to left of vertical edge
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
40
12580 lbs
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
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
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx
10 '
18 '
Engineering-36: Engineering Mechanics - Statics
42
Bruce Mayer, PE
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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
43
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
Engineering-36: Engineering Mechanics - Statics
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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
45
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
46
 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
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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
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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
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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
Engineering 36
Appendix
dy
dx
 sinh
µx
T0

µs
T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
53
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