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hinge. Statics. ?. Surface Forces. Static Surface Forces. Forces on plane areas Forces on curved surfaces Buoyant force Stability submerged bodies. Forces on Plane Areas. Two types of problems Horizontal surfaces (pressure is _______) Inclined surfaces Two unknowns ____________ - PowerPoint PPT Presentation
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Monroe L. Weber-Shirk School of Civil and
Environmental Engineering
hingehinge
??
Statics
Surface Forces
Static Surface Forces
Forces on plane areas
Forces on curved surfaces
Buoyant force
Stability submerged bodies
Forces on Plane Areas
Two types of problemsHorizontal surfaces (pressure is _______)Inclined surfaces
Two unknowns________________________
Two techniques to find the line of action of the resultant forceMomentsPressure prism
constant
Total forceLine of action
dp gdz
r=- p a
Side view
Forces on Plane Areas: Horizontal surfaces
pAdAppdAFR
Top view
A
p = gh
F is normal to the surface and towards the surface if p is positive.
F passes through the ________ of the area.
h
What is the force on the bottom of this tank of water?
RF g hAr=weight of overlying fluid!FR =
centroid
h = _____________ _____________
Vertical distance to free surface
= volume
P = 500 kPa
What is p?
FR
p a 0xp ax
r¶- = =¶
gage
Forces on Plane Areas: Inclined Surfaces
Direction of forceMagnitude of force
integrate the pressure over the areapressure is no longer constant!
Line of actionMoment of the resultant force must equal the
moment of the distributed pressure force
Normal to the plane
Forces on Plane Areas: Inclined Surfaces
x y
RxRy
R cF p A=cp
centroid
center of pressureThe coordinate system origin is at the centroid (yc=0)
Where could I counteract pressure by supporting potato at a single point?
g
Magnitude of Force on Inclined Plane Area
pdAFR
ApF cR pc is the pressure at the __________________centroid of the area
g y
coscp p gyr q= -
cosR cA A
F p dA gy dAr q= -ò ò
cosR cA
F p A g ydAr q= - ò
Change in pressure due to change in elevation
0A
ydA =ò for y origin at centroid
First Moments
Ac xdA
Ax 1
AxdA
1c A
y ydAA
= ò
For a plate of uniform thickness the intersection of the centroidal axes is also the center of gravity
Moment of an area A about the y axis
Location of centroidal axis
h31
c Ay gAt y gtdA
Second Moments
Also called _______________ of the area
Ax dAyI 2
2x xc cI I Ay
Ixc is the 2nd moment with respect to an axis passing through its centroid and parallel to the x axis.
The 2nd moment originates whenever one computes the moment of a distributed load that varies linearly from the moment axis.
moment of inertia
Could define i as I/A…
Product of Inertia
A measure of the asymmetry of the area
xycccxy IAyxI
If x = xc or y = yc is an axis of symmetry then the product of inertia Ixyc is zero.______________________________________
Axy xydAI
y
x
y
x
Product of inertia
Ixyc = 0Ixyc = 0
(the resulting force will pass through xc)
Properties of Areas
yc
baIxc
yc
b
aIxc
A ab=2cay =
3
12xcbaI =
2abA =
3cb dx +=
3
36xcbaI =
2A Rp=4
4xcRI p=R
ycIxc
0xycI =
( )2
272xycbaI b d= -
0xycI =
3cay =
d
cy R=
2
12xcI aA=
2
18xcI aA=
2
4xcI RA=
Properties of Areas
3
4xcbaI p=A abp=
43c
Ryp
=
ayc
b
Ixc
2
2RA p= 4
3cRyp
=4
8xcRI p=
ycR
Ixc
0xycI =
0xycI =
4
16xcRI p=
2
4RA p=R
yc
cy a=
2
4xcI RA=
2
4xcI aA=
2
4xcI RA=
Forces on Plane Areas: Center of Pressure: xR
The center of pressure is not at the centroid (because pressure is increasing with depth)x coordinate of center of pressure: xR
ARR xpdAFx
A
RR xpdA
Fx 1
( )1 cosR cAc
x x p gy dAp A
r q= -òR cF p A=
Moment of resultant = sum of moment of distributed forces
coscp p gyr q= -
1 1 cosR cc cA A
x xp dA x gy dAp A p A
r q= -ò ò
Center of Pressure: xR
1 cosR
cA A
gx xdA xydAA p A
r q= -ò ò
1 0A
xdAA
=òFor x,y origin at centroidxyc A
I xydA=ò
cos xycR
c
Igxp A
r q=-
xr is zero if the x axis or y axis is a line of symmetry
Center of Pressure: yR
ARR ypdAFy
A
RR ypdA
Fy 1
R cF p A= coscp p gyr q= -
( )1 cosR cAc
y y p gy dAp A
r q= -ò
Sum of the moments
You choose the pressure datum to make the problem easy21 1 cosR cA A
c c
y yp dA gy dAp A p A
r q= -ò ò21 cos
R A Ac
gy ydA y dAA p A
r q= -ò ò
Center of Pressure: yR
2xc A
I y dA=ò1 0
AydA
A=ò
cos xcR
c
Igyp A
r q=-
21 cos 1R A A
c
gy ydA y dAA p A
r q= -ò ò
For y origin at centroid
Location of line of action is below centroid along slanted surface.│yR │ is distance between centroid and line of action
Ry
g
FR
cosR R xcy F I gr q=- The moment about the centroid is independent of pressure!
cos yg gq =
Location of average pressure vs. line of action
What is the average depth of blocks?Where does that average occur?Where is the resultant?
0
1 4 3 8 5 12 7 16 9 20R Ry F m blocks m blocks m blocks m blocks m blocks= × + × + × + × + ×380R Ry F m blocks= ×
380 6.33360R
m blocksy mblocks×= =
1 2 3 4 5 6 7 8 9 103 blocks
5Use moments
Inclined Surface Findings
The horizontal center of pressure and the horizontal centroid ________ when the x or y axis is a line of symmetry for the surface
The center of pressure is always _______ the centroid
The vertical distance between the centroid and the center of pressure _________ as the surface is lowered deeper into the liquid
The center of pressure is at the centroid for horizontal surfaces
coincide
below
decreases
0
>0cosxc
Rc
Igyp Ar q=-
cos xycR
c
Igxp A
r q=-
( )cos 90 0=
An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force F applied at the bottom of the gate is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of the gate.
hingewater
F
8 m
4 m
Solution SchemeMagnitude of the force applied by the water
Example using Moments
Location of the resultant force
Find F using moments about hinge
teams
Depth to the centroid
Magnitude of the Force
ApF cR
abA
R cF gh abr p=( ) ( ) ( )3 2
kg m1000 9.8 10 m π 2.5 m 2 mm sRF æ öæ ö=è øè ø
b = 2 m
a = 2.5 mpc = ___
FR= ________
hc = _____
cg hr
10 m
1.54 MN
Pressure datum? _____ Y axis?atmhingewater
F
8 m
4 mFR
g
y
2
4a
Location of Resultant Force
__Rx b = 2 m
a = 2.5 mcp
cosxcR
c
Igyp Ar q=-
cosq =45
pc = ___ cg hr
2 44 5R
c
g ayghrr
æö=- è ø
02
5Rc
ayh
=- =- 0.125 m
xcIA=
hingewater
F
8 m
4 mFR
g
Force Required to Open Gate
How do we find the required force?
0hingeM
F = ______ b = 2 m
2.5 mlcp=2.625 m
m 5
m 2.625N 10 x 1.54 6
F
tot
cpR
llF
F ltot
Moments about the hinge=Fltot - FRlcp
809 kN
cp
hingewater
F
8 m
4 mFR
g
Forces on Plane Surfaces Review
The average magnitude of the pressure force is the pressure at the centroid
The horizontal location of the pressure force was at xc (WHY?) ____________________ ___________________________________
The vertical location of the pressure force is below the centroid. (WHY?) ___________ ___________________
The gate was symmetrical about at least one of the centroidal axes.
Pressure increases with depth.
Forces on Curved Surfaces
Horizontal componentVertical componentTensile Stress in pipes and spheres
Forces on Curved Surfaces: Horizontal Component
What is the horizontal component of pressure force on a curved surface equal to? (Prove it!)
The center of pressure is located using the moment of inertia technique.
The horizontal component of pressure force on a closed body is _____.zero
teams
Forces on Curved Surfaces: Vertical Component
What is the magnitude of the vertical component of force on the cup?
r
h
p = gh
F = ghr2 =W!
F = pA
What if the cup had sloping sides?
Forces on Curved Surfaces: Vertical Component
The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to thesurface where the pressure is equal to the reference pressure.
water= (3 m)(2 m)(1 m) + (2 m)2(1 m)
Example: Forces on Curved Surfaces
Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.
FV =
FH = cp A
2 m
2 m
3 m W1
W2
W1 + W2
= 58.9 kN + 30.8 kN= 89.7 kN
= (4 m)(2 m)(1 m)= 78.5 kN
= 0.948 m (measured from A) with magnitude of 89.7 kN
Take moments about a vertical axis through A.
Example: Forces on Curved Surfaces
The vertical component line of action goes through the centroid of the volume of water above the surface.
c V 1 24(2 m)x F (1 m)W W
3p= + water 2 m
2 m
3 m
A
W1
W2( ) ( )( )c
4(2 m)(1 m) 58.9 kN 30.8 kN3x
89.7 kNp
+=
43
Rp
Expectation???
Example: Forces on Curved Surfaces
water 2 m
2 m
3 m
A
W1
W2
The location of the line of action of the horizontal component is given by
b
a
ch = y
x
4 m
0.083Ry m=-
cosxcR
c
Igyp Ar q=- cosq = 1
cp = cghr2
12Rc
ayh
=-2
12xcI aA=
Example: Forces on Curved Surfaces
78.5 kN
89.7 kN4.083 m
0.94
8 m
119.2 kN
horizontal
vertical
resultant
C
(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0
0.948 m
1.083 m
89.7kN
78.5kN
Cylindrical Surface Force Check
All pressure forces pass through point C.
The pressure force applies no moment about point C.
The resultant must pass through point C.
Curved Surface Trick
water 2 m
3 m
A
W1
W2FO
W1 + W2
Find force F required to open the gate.
The pressure forces and force F pass through O. Thus the hinge force must pass through O!
Hinge carries only horizontal forces! (F = ________)
Tensile Stress in Pipes: High Pressure
pressure center is approximately at the center of the pipe
T1
T2
FH
b
r
FH = ___
T = ___
= ____
(pc is pressure at center of pipe)
2rpc
(e is wall thickness)
rpc
pcr/e
is tensile stress in pipe wall
per unit length
Tensile Stress in Pipes: Low pressure
pressure center can be calculated using moments
T2 __ T1
T1
T2
FH
b
r d
>
d
b
Projected area
FH = ___ 2pcr
cosxcR
c
Igyp Ar q=-
2
12Rc
g dypr=-
2
12xcI dA=
Use moments to calculate T1 and T2.
Solution Scheme
Determine total acceleration vector (a) including acceleration of gravity
Locate centroid of the surface Draw y axis with origin at the centroid (projection of
total acceleration vector on the surface) Set pressure datum equal to pressure on the other side of
the surface of interest Determine the pressure at the centroid of the surface Calculate total force (pcA) Calculate yr
Static Surface Forces Summary
Forces caused by gravity (or _______________) on submerged surfaceshorizontal surfaces (normal to total
acceleration)inclined surfaces (y coordinate has origin at
centroid)curved surfaces
Horizontal componentVertical component (________________________)
total acceleration
R cF p A=
weight of fluid above surfaceA is projected area
cosxcR
c
Igyp Ar q=-R cF p A=
R cF p A=
Review
How do the equations change if the surface is the bottom of an aquarium on a jet aircraft during takeoff? (accelerating at 4 m/s2)
g
ajet
atotal
p a Use total acceleration
The jet is pressurized…
atotal bottom = angle between and
ApF cR
cosxcR
c
Iyp A
r q=- atotal
_____cp =hctotal ca hr
y
Buoyant Force
The resultant force exerted on a body by a static fluid in which it is fully or partially submergedThe projection of the body on a vertical plane is
always ____.
The vertical components of pressure on the top and bottom surfaces are _________
zero
different
(Two surfaces cancel, net horizontal force is zero.)
Buoyant Force: Thought Experiment
FB
zero
no
Weight of water displaced
Place a thin wall balloon filled with water in a tank of water.What is the net force on the balloon? _______Does the shape of the balloon matter? ________What is the buoyant force on the balloon? _____________ _________
BF gr= "
Buoyant Force: Line of Action
The buoyant force acts through the centroid of the displaced volume of fluid (center of buoyancy)
cV
Vx x dVg g=ò1
cV
x xdVV
= ò = volumed= distributed force
xc = centroid of volume
Definition of centroid of volume
Moment of resultant = sum of moments of distributed forces
If is constant!
Buoyant Force: Applications
1 1F V Wg+ = 2 2F V Wg+ =
F1
W1
F2
W2
WeightVolumeSpecific gravity
1 2
Force balance
Using buoyancy it is possible to determine:_______ of an object_______ of an object_______________ of
an object
>
Buoyant Force: Applications
1 1F V Wg+ = 2 2F V Wg+ =
( )
( )
1 1 2 2
1 2 2 1
2 1
1 2
F V F VV F F
F FV
g gg g
g g
+ = +- = --= -
1 2
1 2
2 1 2 1 2 1
1 2 2 1
2 1
W F W FV
W F W FF FW
g gg g g g
g gg g
- -= =
- = --= -
1 2
2
F FVg-= 1FW
Suppose the specific weight of the first fluid is zero
(force balance)Equate weights Equate volumes
Rotational Stability of Submerged Bodies
B
G
BG
A completely submerged body is stable when its center of gravity is _____ the center of buoyancy
below
----------- ________
Buoyant Force (Just for fun)
The anchor displaces less water when it is lying on the bottom of the lake than it did when in the boat.
A sailboat is sailing on Cayuga Lake. The captain is in a hurry to get to shore and decides to cut the anchor off and toss it overboard to lighten the boat. Does the water level of Cayuga Lake increase or decrease?Why?_______________________________ ____________________________________ ____________________
End of Lecture Question
Write an equation for the pressure acting on the bottom of a conical tank of water.
Write an equation for the total force acting on the bottom of the tank.
L
d1
d2
Side view
End of Lecture
What didn’t you understand so far about statics?
Ask the person next to youCircle any questions that still need answers
Team Work
How will you define a coordinate system?
What is the pressure datum? What are the major steps
required to solve this problem?
What equations will you use for each step?
hingewater
F
8 m
4 m
Gates
Gates
Radial Gates
Gates at Itaipu:Why this shape?
Questions
Why does FR = Weight?
Why can we use projection to calculate the horizontal component?
How can we calculate FR based on pressure at the centroid, but then say the line of action is below the centroid?
Side viewhWhat is p?
FR