hinge
Statics
?
Monroe L. Weber-Shirk
Surface Forces
School of Civil and
Environmental Engineering
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
p a
Inclined surfaces 
  g
Two unknowns
dz
Total force
____________
Line of action
____________
Two techniques to find the line of action of
the resultant force
Moments
Pressure prism
Forces on Plane Areas:
net Horizontal surfaces P = 500 kPa
What is the force on the bottom of this
tank of water?
gage
FR   pdA  p  dA  pA
FR   g hA
p = gh
= volume
FR = weight of overlying fluid!
h
What
is p?Side view
FR
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.
p

p a
   ax  0
x
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
y
FR  pc A
g
pc q
x
yR
center of pressure
Where could I
counteract
pressure by
supporting
potato at a
single point?
y
centroid
xR
The coordinate
system origin is at
the centroid (yc=0)
Magnitude of Force on Inclined
Plane Area
FR   pdA
Change in pressure due
to change in elevation
g
y
q
p  pc   gy cosq
FR   pc dA    gy cosq dA
A
A
FR  pc A   g cosq  ydA
A
FR  pc A
òydA = 0
for y origin at centroid
A
centroid of the area
pc is the pressure at the __________________
First Moments

A
Moment of an area A about the y axis
xdA
1
xc   xdA
A A
1
yc   ydA
A A
Location of centroidal axis
Plate thickness
yc  gAt   y  gtdA
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
Could define i as I/A…
A
I x  I xc  Ayc2
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 of the asymmetry of the area
I xy   xydA
A
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
(the resulting force will pass through xc)
inertia Ixyc is zero.______________________________________
Properties of Areas
b
Ixc
Ixc
a
yc
ab
yc A  2
a
b
d
Ixc
A  ab
R
yc
a
yc 
2
a
yc 
3
bd
xc 
3
A   R 2 yc  R
I xc 
3
ba
12
I xyc  0
3
ba
I xc 
36
I xc 
 R4
4
I xyc
I xyc
I xc a 2

A 12
ba 2

 b  2d 
72
I xc a 2

A 18
2
I
R
 0 xc 
A
4
Properties of Areas
Ixc
yc
A
R
 R2
2
4R
yc 
3
 R 4 I  0 I xc R 2
I xc 

xyc
8
A
4
b
a
yc
Ixc
R
yc
A   ab
A
 R2
4
yc  a
4R
yc 
3
I xc 
 ba
I xc 
3
4
 R4
16
I xyc
I xc a 2

0
A
4
I xc R 2

A
4
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  pc   gy cosq
FR  pc A
1
xR 
xpdA

A
FR
1
xR 
x  pc   gy cosq  dA

A
pc A
1
1
xR 
xpc dA 
x  gy cosq dA


pc A A
pc A A
Center of Pressure: xR
1
 g cosq
xR   xdA 
xydA

AA
pc A A
I xyc   xydA
For x,y origin at centroid
A
1
xdA  0

A A
xR  
 g cosq I xyc
pc
A
xR is zero if the x axis or y axis is a line of symmetry
Center of Pressure: yR
y R FR   ypdA
Sum of the moments
A
1
yR 
ypdA
FR  pc A

A
FR
1
yR 
y  pc   gy cosq  dA

A
pc A
yR 
1
1
2
yp
dA


gy
cosq dA
c


pc A A
pc A A
1
 g cosq
yR   ydA 
A A
pc A

A
y 2 dA
p  pc   gy cosq
g
Center of Pressure: yR
FR
1
 g cosq 1
2
yR   ydA 
y
dA

A A
pc
A A
I xc   y 2 dA
1
ydA  0

A A
 g cosq I xc
pc
g cosq  g y
For y origin at centroid
A
yR  
q
yR
A
yR FR   I xc  g cosq
Location of line of action is below
centroid along slanted surface.
│yR │ is distance between
centroid and line of action
The moment about the centroid is
independent of pressure!
Location of average pressure vs.
line of action
0 1 2 3 4 5 6 7 8 9 10
What is the average depth of blocks? 3 blocks
Where does that average occur?
5
Where is the resultant? Use moments
yR FR  1m  4blocks  3m  8blocks  5m 12blocks  7m 16blocks  9m  20blocks
yR FR  380m  blocks
380m  blocks
yR 
 6.333m
60blocks
Inclined Surface Findings
 The horizontal center of pressure and the
horizontal centroid ________
coincide when the x or
y axis is a line of symmetry for the surface
 The center of pressure is always _______
below
the centroid
 The vertical distance between the centroid
and the center of pressure _________
decreases as the
surface is lowered deeper into the liquid
 The center of pressure is at the centroid for
horizontal surfaces
0
xR  
 g cosq I xyc
pc
A
>0
yR  
 g I xc
pc A
cos (90) = 0
cosq
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
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?
hinge
8m
water
F
4m
Magnitude of the Force
g
y
Pressure datum? _____
atm Y axis?
FR  pc A
8m
A  ab
water
q
FR
F
hc = _____
10 m Depth to the centroid
g hc
pc = r___
a = 2.5 m
FR   ghc ab
kg 
m

FR  1000 3  9.8 2  10 m  π  2.5 m  2 m 
m  s 

FR= ________
1.54 MN
b=2m
hinge
4m
Location of Resultant Force
g
yR  
 g I xc
cos q 
pc A
4
5
cos q
8m
g hc
pc = r___
water
FR
F
hinge
4m
I xc a 2

4
A
a = 2.5 m
 g a2  4 
yR  
 ghc 4  5 
a2
yR  
  0.125 m
5hc
cp
0
xR  __
b=2m
Force Required to Open Gate
g
How do we find the
required force?
8m
Moments about the hinge
 M hinge  0 =Fltot - FRlcp
F
water
FR
F
hinge
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 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 technique.
net
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 = gh
F = ghr2 =W!
r
What if the cup had sloping sides?
What if the cup bottom were a hemisphere?
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
surface where the pressure is equal to the
reference pressure.
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)g + /4(2 m)2(1 m)g
= 58.9 kN + 30.8 kN
= 89.7 kN
FH = p c A
= g(4 m)(2 m)(1 m)
= 78.5 kN
3m
water
2m
W1
2m
W2
Example: Forces on Curved
Surfaces
The vertical component line of action goes through Expectation???
A
the centroid of the volume of water above the surface.
Take moments about a vertical
axis through A.
4R
4(2 m)
3
x c FV  (1 m)W1 
W2
3
4(2 m)
(1 m)  58.9 kN  
 30.8 kN 
3
xc 
89.7 kN 
3m
water
W1
2m
2m
W2
= 0.948 m (measured from A) with magnitude of 89.7 kN
Example: Forces on Curved
Surfaces
pc   ghc
a2
yR  
12hc
hc  4 m
yR  0.083m
3m
water
b
I xc a 2

A 12
A
a
W1
2m
2m
W2
y
x
The location of the line of action of the horizontal
component is given by
 g I xc
cosq = 1
yR  
cosq
pc A
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
yR  
 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 = ___
rpc/e
s = ____
(e is wall thickness)
 g I xc
pc A
cosq
b
T1
r
FH
T2
s is tensile stress in pipe wall
How does pipe wall thickness change with diameter?
e
rpc
s tensile stress
Tensile Stress in Pipes:
Low pressure
pressure center can be
calculated using moments
T2 __
FH = 2p
___
> T1
cr
yR  
 g I xc
pc A
cosq
b
T1
2
I xc d

A 12
r
FH
d
T2
yR  
g d2
pc 12
Projected area
Use moments to calculate T1 and T2.
d
b
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
total acceleration on submerged surfaces
_______________)
horizontal surfaces (normal to total
acceleration) FR  pc A
inclined surfaces (y coordinate has origin at
 g I xc
centroid) FR  pc A
yR  
cosq
pc A
curved surfaces
Horizontal component FR  pc A A is projected area
Vertical component (________________________)
weight of fluid above surface
Questions
Why does FR = Weight?
h
What
is p?Side view
FR
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?
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)

p a
q
y
hc
FR  pc A
atotal
g
atotal hc
pc = _____
Alternate
method?
Where is y?
q = angle between atotal and y
 atotal I xc
yR  
ajet
Use total acceleration
pc
A
cosq
The jet is pressurized…
Circular Port
P=-2 kPa
0.5 m
0.5 m
Equivalent problem
air
 = 800 kg/m3
 = 1000 kg/m3
1m
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 ____.
(Two surfaces cancel, net horizontal force is zero.)
The vertical components of pressure on the top
and bottom surfaces are _________
different
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? ________
What is the buoyant force on
of water
the balloon? Weight
_____________
displaced
_________
FB
FB   g
Buoyant Force: Line of Action
The buoyant force acts through the centroid
of the displaced volume of fluid (center of
buoyancy)
g Vxc   xg dV
V
1
xc 
V
 xdV
Moment of resultant = sum of moments of
distributed forces
Definition of centroid of volume
V
 = volume
gd= distributed force
xc = centroid of volume
If g is constant!
Buoyant Force: Applications
Using buoyancy it is
possible to
determine:
_______
Weight of an object
Volume of an object
_______
Specific gravity of
_______________
an object
F1
g1 > g2
F2
g1
g2
W
W
Force balance
F1  V g 1  W
F2  V g 2  W
Buoyant Force: Applications
F1  V g 1  W
F2  V g 2  W
Equate weights
F1  V g 1  F2  V g 2
V  g 1  g 2   F2  F1
V 
F2  F1
g 1  g 2 
(force balance)
Equate volumes
V 
W  F1

W  F2
g1
g2
W g 2  F1g 2  W g 1  F2g 1
F1g 2  F2g 1
W
g 2  g1
Suppose the specific weight of the first fluid is zero
V 
F1  F2
g2
W  F1
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
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.
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
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.
(not including forces
from the side walls)
d1
Side view
L
d2
Gates
Gates
Radial Gates
Gates at Itaipu:
Why this shape?