Title: Ch7 - Distributed forces
1Distributed Forces
C.G. / C.M. / Centroid
2Center of Gravity, Center of mass , Centroid
C.G. Location of equivalent (single)
resultant of gravity.
C.M. Location of equivalent mass (point
mass).
Centroid Location of equivalent
geometric shape.
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4Gravity Force
Equivalent Gravity force (resultant weight) on
body
direction, magnitude, point of application
W
Center of Gravity (C.G.)
Experiment Finding C.G.
For each body, there exist a unique C.G.
(Assume direction of gravity is the same for all
point)
How can we find C.G. by math?
2-force member
5C.G. Location
Equivalent System
Sys I
Sys II
6Finding a center of gravity
Sys 1
Sys 2
y
y
W
by equivalent system technique (finding
average value of moment)
x
x
Assumption gravity field is parallel
Moment about y-axis
z
C.G.
Moment about x-axis
Moment about z-axis
y
Center of Gravity (Force) C.G.
x
(Object-attached Axis Rotating)
7dW gdm and W mg
The point calculated from
is called the Center of Mass
Center of mass is useful for studying problems
about the motion of the body under the influence
of force (Dynamics)
Assumption gravity (g) is constant over the
object,
the center of mass the center of gravity.
8Centroids
dm
the density ( ? ) of the body
The point calculated from
is called the Centroid
Geometric Center Centroid of a body depends on
its geometric shape only.
Assumption denity (r) is uniform,
the center of mass the centroid.
9dV
Center of Gravity
Gravity is parallel over object
Resultant Weight
Equal if gravity is uniform over object
Center of Mass
Motion of object
Equal if density is uniform over object
Centroid
Depends on geometric shape
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115/3 Centroid
Volume Centroid
x-z plane
How far the whole object is far away from the
x-z plane.
How far the whole object (evaluated by its
volumn) is far away from the axiss origin.
The centroid location is only up to the shape
of the body (not depends on axis selection)
Geometric Center
y
centroid by xy -coord
y
centroid by xy-coord
x
x
12Volume Centroid
Area Centroid
Uniform thickness (t)
Line Centroid
Uniform section area (A)
13Centroid and Symmetry
0 (symmetry or z0 for all point)
0 (symmetry)
you have to find only this
14Parabola
Note hh(x)
dA
dA
15- Find centroid C of area A
dA
dA
How much dA is far away from x-axis?
how much the centroid of dA is far away from
x-axis
Where is the centroid of DA ?
16 Find Area Centroid
dA
dA
centroid
If p0, C.M Centroid ?
17- Find centroid C of line L, given a b 100 mm
Parabola
147.9 mm
8483.6 mm
x,y has some relation
dx,dy has some relation
18- Find centroid C of line L, given a b 100 mm
147.9 mm
x,y has some relation
dx,dy has some relation
19L length, it depends on y
?
L
Area Centroid
0
20Sample5/1 Locate the centroid of a circular arc.
y
Good axis-selection make it easy.
dL
0
Line Centroid
(Symmetric)
dq
q
x
r cosq
Better Using Coordinate
21Locate the centroid of the area of a circular
sector
y
dA
dq
x
Area Centroid
22Locate the centroid of the area of a circular
sector
23Locate the centroid of the area of a hemisphere
(radius-r)
?
Volume Centroid
24Example Centroids 4 1
Center
Built around 2560 BC, the Great Pyramid of Khufu
(Cheops) is one of the Seven Wonders of the
Ancient World. It was 481 ft high the horizontal
cross section of the pyramid is square at any
level, with each side measuring 751 ft at the
base. By discounting any irregularities, find the
position of Pharaohs burial chamber, which is
located at the heart centroid of the pyramid.
(http//ce.eng.usf.edu/pharos/wonders/pyramid.html
)
25Example Centroids 4 3
Center
26Example Centroids 4 2
Center
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28Basic Shape (by intergration)
Composite Shape
Rotated Shape
29Composite Bodies
C.G.
C.G.
(Principle of moment , Equivalent Moment at
origin)
30Composite Bodies 2
C.G.
C.G.
Think that Weight is minus . (easy to
remember)
31C.G.
Centroid
C.M.
32Example Centroids 5 1
Center
- Find C.G. C of the following homogenous body.
C.G. C.M. Centroid
33Example Centroids 5 2
Center
34(3032010,2020)
(60,50)
(303030201060/3, 100/3)
(60,12.73)
35r constant
C.G. centroid
Unit kg/m2
5/69 A cylindrical container with an extended
rectangular back and semi-circular ends are all
fabricated from the same sheet-metal stock.
Calculate the angle a made by the back with the
vertical when the container rests in an
equilibrium position on a horizontal surface.
36FBD
r constant
C.G. centroid
c2
a
c1
a
W2
W1
2 unknown (N, a) 2 Eqs (sum-force-on-y-axis
automatically 0)
Moment at N
Ans
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38Fluid Statics
Fluid Statics
- Required in studies and designs of pressure
vessels, piping, ships, dams and off-shore
structures, etc. - Topics
- Definitions
- Fluid pressure
- Hydrostatic pressure on submerged surfaces
- Buoyancy
39Fluid Statics Definitions
Fluid Statics
- Fluid is any continuous substance which, when at
rest, is unable to support shear force. - Fluid Statics studies pressure of fluid at rest.
- Hydrostatics stationary liquid
- Aerostatics stationary gas
- Pascals Law the pressure at any given point in
a fluid is the same in all directions. - Pressure p in fluid at rest is a function of
vertical dimension and its density r. - Resultant force on a body from pressure acts at
the center of pressure P.
405/9 Fluid Statics
- Fluids at rest cannot support shear forces
- (i.e., a pressure force is always
perpendicular to a surface.)
Fluid at rest can exert only normal forces on a
bounding surface
Pressure
- Pressure at any given point in a fluid at
rest is the same in all directions (Pascals law)
p1 (dy)(dz) p3 (ds) (dz) sin ?
p2 (dx)( dz) p3 (ds) ( dz) cos ?
?
(ds) (sin ?) dy
(ds) (cos ?) dx
p1 p2 p3.
41Fluid Pressure
a fluid which is essentially incompressible.
Fluid gaseous , Liquid
Hydrostatic a study of static liquid
p dA ?g dA dh (pdp) dA 0
dp ? g dh
? constant
Incomplete FBD (only focus on vertical forces)
p po ?gh
For liquid at rest, the pressure is a function
of the vertical dimension
How to sum up Force due to These pressure?
Pressure
po pressure at the fluid surface ?gh gage
pressure
_ Area
Pressure Distribution
42Case I (a)
Case I
Flat surface (Plate)
constant width
Special case
Case III
Case II
Curve surface
submerged object
43Case I
Fluid pressure on plate (flat surfaces)
We want to know the resultant force (R) acting on
the plate (on one side)
Direction perpendicular to the surface
Magnitude the sum of the forces produces by
these pressure (pdA)
x
y
R volume of the pressure-plate object
x
y
R average pressure A (over plate)
Plate Centroid
R from integration
Using the special condition that pressure is
linearly increased due to the water depth only
Same R (same A)
const
R pressure at plate centroid A
44Case I
Location (point of application)
x
y
- In y direction, we apply the principle of
equivalent systems moment.
y-component of centroid of the volume
y
dR
( x , y ) coordinate of Point of the resultant R
(i.e. ) is the same as Volume centroid
R
45Fluid pressure on plate (flat surfaces)
Case I
Fluid pressure on rectangular plate
Case I(a)
Special case Of case I
Magnitude
Magnitude
P at plate centroid
Location
y-component of volume centroid of trapezoidal box
P at plate centroid
Location
x-component of Volume centroid
y-component of area centroid of trapezoidal area
y-component of Volume centroid
46Fluid pressure on rectangular plate
Recommend Method
Case I(a)
Think of 2 forces instead of 1 force.
Special case Of case I
If you really want to find only one force
y
Magnitude
Location
y-component of centroid of area
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48SP5/18 A rectangular plate, shown in vertical
section AB, is 4-m high and 6-m wide, and blocks
the end of a fresh-water channel 3-m deep. Find
the force B exerted on the plate by the ridge.
Air-pressure?
mg
R
1 m
49OK!, linear
WRONG, why?
P at plate centroid
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51Basic concept The normal force from floor is
dependent of the weigh of the dams. At the moment
overturning, the normal force will be at point C.
Employ this condition to find the required dam
weight.
52Triangular-section dam
Rectangular-section dam
A requires concrete of 22.2 Mg/m less than B.
53z
Find magnitude of R
plates centroid
?
Volume Analogy
( Finding volume integrate the following terms )
Direct Intergration
Find Location of R
0(sym)
Volume Analogy
Finding the centroid of volume
Equivalent Moment
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55Case II
Fluid pressure on cylindrical surface
virtual plate
You have to sum force (dR) at each axis.
fluid blocks equilibrium
Virtual plate
Direct Integration
Water
d
dR
dy
ds
dx
56SP5/20. Determine the resultant force R exerted
on the cylindrical dam surface by the
fresh-water. The dam has a length b, normal to
the paper, of 30 m
x
y
57x?
w
R
x
585/185 Determine the total force R exerted by the
water on the dam face (only the arch part).
R
Magnitude of R (R , D)
R
D
fluid-block method.
Water Weight
In the same line
Location of R
Location of D
R
90/3 m (from the buttom)
D
Vertical distance of Location of D vertical
distance of Location of R ?
Yes!
Moment about this axis
-gt R must pass this intersect point
2 line intersect with each other
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60Case III
Fluid Buoyancy of immersed object
same shape
Wood and Ceramics have the same buoyancy force
exerted by the surrounding water?
dF
Buoyancy
same h
Fluid in Equilibrium
Ceramics in equilibrium
Wood in equilibrium
two force member!
dFs direction up to surface shape only (same
surface same direction)
dFs magnitude up to height only (same
height, same magnitude)
The force of buoyancy (F) must pass
through the center of mass of the fluid portion
(in the opposite direction of gravity)
F is not depend on the material inside.
Similarly, for an object immersed in a fluid
The force of buoyancy is equal to the
weight of fluid displaced and passes through the
center of mass of the displaced fluid
Left Case I Right Case I Buttom Case II
61Sample5/22 A buoy of 8-m long and 0.2-m diameter
has mass of 200-kg. Determine the angle ?
?
G
B
Wmg
?
y
A
F
x
T
Ans
62Buoyancy Stability of a floating object
M meta-center h meta-centric height
If M is above G, stable
If M is below G, unstable
C.G. Line of normal position (fixed with boat)
Buoyancy line (always vertical)
H
H
unstable
stable
G center of mass B center of buoyancy
Buoyancy tries to recover the boat to normal
position
Buoyancy tries to sink the boat
63Recommended Problem
5/179 5/189 H9/116 5/209 5/210