Transcript Static Surface Forces

Static Surface Forces
hinge
8m
water
?
4m
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 _______)
constant
dp
 Inclined surfaces
 
 Two
unknowns
dz
Total force
 ____________
of action
 Line
____________
 Two
techniques to find the line of action of
the resultant force
 Moments
 Pressure
prism
Forces on Plane Areas:
Horizontal surfaces P = 500 kPa
What is the force on the bottom of this
tank of water?
FR   pdA  p  dA  pA
FR = g hA
= volume
h
What
is p?Side view
p = h
FR = weight of overlying fluid!
h = _____________
Vertical distance
to free surface
_____________
F is normal to the surface and towards
the surface if p is positive.
F passes through the ________
centroid of the area.
A
Top view
Forces on Plane Areas: Inclined
Surfaces
 Direction
of force Normal to the plane
 Magnitude of force
 integrate
the pressure over the area
 pressure is no longer constant!
 Line
of action
 Moment
of the resultant force must equal the
moment of the distributed pressure force
Forces on Plane Areas: Inclined
Surfaces
Free surface
FR  hc A
q
hc
A’
O
x
xc
xR
Where could I
counteract
pressure by
supporting
potato at a
single point?
centroid
B’
O
center of pressure
yR
y
yc
The origin of the y
axis is on the free
surface
Magnitude of Force on Inclined
Plane Area
FR   pdA
p  h  y sin q
FR   sin q  ydA
1
yc   ydA
A A
q
y
FR  Ayc sin q
FR  hc A
hc is the vertical distance between free
surface and centroid
FR  pc A
centroid of the area
pc is the pressure at the __________________
First Moments

A
xdA
1
xc   xdA
A A
Moment of an area A about the y axis
Location of centroidal axis
1
yc = ò ydA
A A
1
h
3
For a plate of uniform thickness the intersection of the centroidal
axes is also the center of gravity
Second Moments
moment of inertia of the area
Also called _______________
I x   y 2 dA
A
I x  I xc  Ay 2
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.
Product of Inertia
 A measure
I xy   xydA
A
of the asymmetry of the area
Product of inertia
I xy  xc yc A  I xyc
Ixyc = 0
Ixyc = 0
y
y
x
x
If x = xc or y = yc is an axis of symmetry then the product of
inertia Ixyc is zero.
Properties of Areas
b
Ixc
Ixc
a
yc
ab
A
=
yc
2
a
b
d
Ixc
A = ab
R
yc
a
yc =
2
a
yc =
3
b+d
xc =
3
A = p R 2 yc = R
ba 3
I xc =
12
I xyc = 0
3
ba 2
=
(b - 2d )
72
ba
I xc =
36
I xyc
p R4
I xc =
4
I xyc = 0
Properties of Areas
Ixc
yc
4R
yc =
3p
p R4
I xc =
8
I xyc = 0
A = p ab
yc = a
p ba 3
I xc =
4
I xyc = 0
p R2
A=
4
4R
yc =
3p
p R4
I xc =
16
p R2
A=
2
R
b
a
yc
Ixc
R
yc
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
xR FR   xpdA
A
Moment of resultant = sum of moment of
distributed forces
p  y sin q
FR  yc A sin q
1
xR 
xpdA

A
FR
1
xR 
xy sin qdA

A
yc A sin q
xR 
1
xydA

yc A A
Center of Pressure: xR
xR 
1
xydA

yc A A
xR 
I xy
xR 
xR 
I xy   xydA
A
I xy  xc yc A  I xyc
yc A
Product of inertia
Parallel axis theorem
xc yc A  I xyc
yc A
I xyc
yc A
y
 xc
x
Center of Pressure: yR
y R FR   ypdA
A
Sum of the moments
1
yR 
ypdA
FR  yc A sin q

A
FR
1
2
yR 
y
 sin qdA

yc A sin q A
1
2
yR 
y
dA

A
yc A
I x   y 2 dA
A
Ix
yR 
yc A
I x  I xc  yc2 A
I xc + yc2 A
yR =
yc A
I xc
yR =
+ yc
yc A
p  y sin q
p = 0 when y = 0
You choose the
pressure datum to
make the problem easy
Parallel axis theorem
Inclined Surface Findings
0
I xyc
The horizontal center of pressure and the
xR 
 xc
yc A
coincide when the surface
horizontal centroid ________
has either a horizontal or vertical axis of
symmetry
>0
 The center of pressure is always _______
below the y  I xc  y
R
c
y
A
centroid
c
 The vertical distance between the centroid and
the center of pressure _________
decreases as the surface
is lowered deeper into the liquid
 What do you do if there isn’t a free surface?

Example using Moments
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.
teams
Solution Scheme
 Magnitude of the force
applied by the water
hinge
8m
 Location of the resultant force
 Find F using moments about hinge
water
F
4m
Magnitude of the Force
FR  pc A
hinge
8m
A  ab
hc = _____
10 m Depth to the centroid
water
FR
F
 hc
pc = ___
FR  hcab
N

FR   9800 3 10 m π2.5 m 2 m 
m 

a = 2.5 m
FR= ________
1.54 MN
b=2m
4m
Location of Resultant Force
hinge
I xc
yR 
 yc
yc A
yc  hc
yc  ________
12.5 m
ba 3
y R  yc 
4 ycab
2
a
y R  yc 
4 yc
m
yR  yc 0.125
_______
8m
Slant distance
to surface
p ba 3
I xc =
4
Fr
F
4m
A  ab
2.5 m 
y R  yc 
412.5 m 
2
xc
xR  __
water
a = 2.5 m
cp
b=2m
Force Required to Open Gate
How do we find the
required force?
hinge
8m
Fr
F
Moments about the hinge
 M hinge  0 =Fltot - FRlcp
F
water
4m
FR lcp
ltot
lcp=2.625 m
2.5 m

1.54 x 10 N 2.625 m 
F
cp
6
5 m 
F = ______
809 kN
b=2m
ltot
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
gate was symmetrical
was at xc (WHY?) The
____________________
about at least one of the centroidal axes.
___________________________________
 The vertical location of the pressure force is
Pressure
below the centroid. (WHY?) ___________
increases with depth.
___________________
Forces on Plane Areas: Pressure
Prism
 A simpler
approach that works well for
areas of constant width (_________)
rectangles
 If the location of the resultant force is
required and the area doesn’t intersect the
free surface, then the moment of inertia
method is about as easy
Forces on Plane Areas: Pressure
Prism
Free surface
h2
h1
q
O
dF  hdA  d
dA
F 
Force = Volume
of pressure prism
g hdA = d "
1
xR 
xpdA

A
FR
1
xR   xd
 
1
y R   yd
 
Center of pressure
is at centroid of
pressure prism
Example 1: Pressure Prism
h=10 m
h/cosq
h
FR
q
24º
Dam
FR = 
FR = (h/cosq)(h)(w)/2
FR = (10 m/0.9135)(9800 N/m3*10 m)(50 m)/2
FR = 26 MN
Example 2: Pressure Prism
O
8m
water
4mx4m
(square conduit)
8 m)
12 m)
hinge
5m
y
Solution 2: Pressure Prism
Magnitude of force
h1  h2
(h1)
2
FR    (9800 N/m3 )(10 m)(5 m)(4 m) (h2)
MN
FR =1.96
________
yp
Location of resultant force measured from hinge
a  awh1  2a  awh2 
y R FR  
 

3 2  3  2 
5 m   8 m  12 m 
yR 


10 m   6
3 
5m
a
a 2 w
 h1  h2 
yR 
aw10 m    6 3 
FR
2.667 m
yR = _______
Forces on Curved Surfaces
 Horizontal
component
 Vertical component
 Tensile Stress in pipes and spheres
Forces on Curved Surfaces:
Horizontal Component
 What
is the horizontal component of
pressure force on a curved surface equal
teams
to?
(Prove it!)
 The center of pressure is located using
the moment of inertia or pressure prism
technique.
 The horizontal component of pressure
force on a closed body is _____.
zero
Forces on Curved Surfaces:
Vertical Component

What is the magnitude of the
vertical component of force on
the cup?
F = pA
h
p = h
F = hr2 =W!
What if the cup had sloping sides?
r
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 the (virtual or real) free
surface.
Streeter, et. al
Example: Forces on Curved
Surfaces
Find the resultant force (magnitude and location)
on a 1 m wide section of the circular arc.
F V = W1 + W 2
= (3 m)(2 m)(1 m) + /4(2 m)2(1 m)
= 58.9 kN + 30.8 kN
= 89.7 kN
W1
3m
water
2m
W2
2m
FH = p c A
x
= (4 m)(2 m)(1 m)
= 78.5 kN
y
Example: Forces on Curved
Surfaces
The vertical component line of action goes through
the centroid of the volume of water above the surface.
Take moments about a vertical
axis through A.
4R
4(2 m)
3p
x c FV = (1 m)W1 +
W2
3p
4(2 m)
(1 m) (58.9 kN ) +
(30.8 kN )
3p
xc =
(89.7 kN )
A
3m
water
W1
2m
2m
W2
= 0.948 m (measured from A) with magnitude of 89.7 kN
Example: Forces on Curved
Surfaces
The location of the line of action of the horizontal
component is given by
ba 3
I xc =
12
W1
3m
b
a
I xc = (1 m)(2 m)3/12 = 0.667 m4
yc = 4 m
0.667 m 4
yR =
+ (4 m ) = 4.083 m
(4 m)[(2 m)(1 m)]
water
2m
W2
2m
x
I xc
yR =
+ yc
yc A
A
y
4.083 m
0.948 m
Example: Forces on Curved
Surfaces
78.5 kN horizontal
89.7 kN vertical
119.2 kN resultant
Cylindrical Surface Force Check
0.948 m
C
1.083 m
78.5kN
89.7kN
All pressure forces pass
through point C.
 The pressure force
applies no moment about
point C.
 The resultant must pass
through point C.

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___
0
Curved Surface Trick
 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
W1 + W2
forces! (F = ________)
A
water
3m
W1
O
F
2m
W2
Tensile Stress in Pipes:
High Pressure

pressure center is approximately at
the center of the pipe
per unit length
(pc is pressure at
FH = 2rp
___c
center of pipe)
rpc
T = ___
pcr/e
s = ____
(e is wall thickness)
s is tensile stress in pipe wall
b
T1
r
FH
T2
Tensile Stress in Pipes:
Low pressure
 pressure
center can be
calculated using moments
 T2 __
FH = 2p
___
> T1
cr
3
I xc
yR =
+ yc
yc A
ba
2br
I xc =
=
12
3
A  2br
b
T1
3
r
FH
h
T2
r2
Projected area
yR = yc +
3 yc
yc is distance to centroid from virtual free surface
h=2r
b
Solution Scheme
 Determine
pressure datum and location in
fluid where pressure is zero (y origin)
 Determine total acceleration vector (a)
including acceleration of gravity
 Define h tangent to acceleration vector (call
this vertical!)
 Determine if surface is normal to a,
inclined, or curved
Static Surface Forces Summary
 Forces
caused by gravity (or
_______________)
total acceleration on submerged surfaces
 horizontal
surfaces (normal to total
acceleration) FR = g hA
Location where p = pref
 inclined surfaces (y coordinate has origin at
free surface) FR  hc A y  I xc  y
R
c
y
A
 curved surfaces
c
component FR  hc A
 Vertical component (________________________)
weight of fluid above surface
 Horizontal
 Virtual
surfaces…
Buoyant Force
 The
resultant force exerted on a body by a
static fluid in which it is fully or partially
submerged
 The
projection of the body on a vertical plane is
zero
always ____.
 The vertical components of pressure on the top
different
and bottom surfaces are _________
Buoyant Force: Thought
Experiment
Place a thin wall balloon filled
with water in a tank of water.
What is the net force on the
zero
balloon? _______
Does the shape of the balloon
no
matter? ________
FB
What is the buoyant force on
of water
the balloon? Weight
_____________
displaced
_________
FB=V
Buoyant Force: Line of Action
 The
buoyant force acts through the centroid
of the displaced volume of fluid (center of
buoyancy)
1
g xdV = g Vx
ò
V
 = volume
d= distributed force
xc = centroid of volume
c
xc =
xdV
ò
V
V
Buoyant Force: Applications
F1
 Using
buoyancy it is
possible to
determine:
 _______
Weight
of an object
Volume of an object
 _______
Specific gravity of
 _______________
an object
1 > 2
F2
1
2
W
W
Force balance
F1 + V g1 = W
F2 + V g 2 = W
Buoyant Force: Applications
F1 + V g1 = W
F2 + V g 2 = W
Equate weights
F1 + V g1 = F2 + V g 2
V (g1 - g 2 ) = F2 - F1
F2 - F1
V =
(g1 - g 2 )
(force balance)
Equate volumes
W - F1 W - F2
V =
=
g1
g2
W g 2 - F1g 2 = W g1 - F2g1
F1g 2 - F2g1
W=
g 2 - g1
Suppose the specific weight of the first fluid is zero
F1 - F2
V =
g2
W  F1
Buoyant Force (Just for fun)
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?_______________________________
The anchor displaces less water when
____________________________________
it is lying on the bottom of the lake than it
____________________
did when in the boat.
Rotational Stability of
Submerged Bodies
 A completely
submerged body is
stable when its
center of gravity is
_____
below the center
of buoyancy
B
G
B
G
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.
d1
L
d2
End of Lecture
 What
didn’t you understand so far about
statics?
 Ask the person next to you
 Circle any questions that still need answers
Team Work
 How
will you define a coordinate system?
 What are the 3 major steps required to solve
this problem?
 What equations will you use for each step?
hinge
8m
water
F
4m
Gates
Gates
Radial Gates