Center of Gravity and Centroid 9 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010...
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Transcript of Center of Gravity and Centroid 9 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010...
Copyright © 2010 Pearson Education South Asia Pte Ltd
Center of Gravity and Centroid99
Engineering Mechanics: Statics in SI Units, 12e
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Objectives
• Concept of the center of gravity, center of mass, and the centroid
• Determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape
• Theorems of Pappus and Guldinus• Method for finding the resultant of a general distributed
loading
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Center of Gravity and Center of Mass for a System of Particles
2. Composite Bodies
3. Theorems of Pappus and Guldinus
4. Resultants of a General Distributed Loading
5. Fluid Pressure
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Center of Gravity • Locates the resultant weight of a system of particles• Consider system of n particles fixed within a region of
space• The weights of the particles can be replaced by a
single (equivalent) resultant weight having defined point G of application
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Center of Gravity • Resultant weight = total weight of n particles
• Sum of moments of weights of all the particles about x, y, z axes = moment of resultant weight about these axes
• Summing moments about the x axis,
• Summing moments about y axis,
nnR
nnR
R
WyWyWyWy
WxWxWxWx
WW
~...~~
~...~~
2211
2211
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Center of Gravity • Although the weights do not produce a moment about
z axis, by rotating the coordinate system 90° about x or y axis with the particles fixed in it and summing moments about the x axis,
• Generally,
mmz
zmmy
ymmx
x
WzWzWzWz nnR
~,
~;
~
~...~~2211
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Center Mass • Provided acceleration due to gravity g for every
particle is constant, then W = mg
• By comparison, the location of the center of gravity coincides with that of center of mass
• Particles have weight only when under the influence of gravitational attraction, whereas center of mass is independent of gravity
mmz
zmmy
ymmx
x
~,
~;
~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Center Mass • A rigid body is composed of an infinite number of
particles • Consider arbitrary particle having a weight of dW
dW
dWzz
dW
dWyy
dW
dWxx
~;
~;
~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Centroid of a Volume• Consider an object subdivided into volume elements
dV, for location of the centroid,
V
V
V
V
V
V
dV
dVz
zdV
dVy
ydV
dVx
x
~
;
~
;
~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Centroid of an Area• For centroid for surface area of an object, such as
plate and shell, subdivide the area into differential elements dA
A
A
A
A
A
A
dA
dAz
zdA
dAy
ydA
dAx
x
~
;
~
;
~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.1 Center of Gravity and Center of Mass for a System of Particles
Centroid of a Line• If the geometry of the object takes the form of a line,
the balance of moments of differential elements dL about each of the coordinate system yields
L
L
L
L
L
L
dL
dLz
zdL
dLy
ydL
dLx
x
~
;
~
;
~
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Example 9.1
Locate the centroid of the rod bent into the shape of a parabolic arc.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.1
Differential element
Located on the curve at the arbitrary point (x, y)
Area and Moment Arms
For differential length of the element dL
Since x = y2 and then dx/dy = 2y
The centroid is located at
yyxx
dyydL
dydydx
dydxdL
~,~
12
1
2
222
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.1
Integrations
m
dyy
dyyy
dL
dLy
y
m
dyy
dyyy
dyy
dyyx
dL
dLx
x
L
L
L
L
574.0479.18484.0
14
14~
410.0479.16063.0
14
14
14
14~
1
02
1
02
1
02
1
022
1
02
1
02
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.2 Composite Bodies
• Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular
• A body can be sectioned or divided into its composite parts
• Accounting for finite number of weights
W
Wzz
W
Wyy
W
Wxx
~~~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.2 Composite Bodies
Procedure for Analysis
Composite Parts• Divide the body or object into a finite number of
composite parts that have simpler shapes• Treat the hole in composite as an additional
composite part having negative weight or size
Moment Arms• Establish the coordinate axes and determine the
coordinates of the center of gravity or centroid of each part
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.2 Composite Bodies
Procedure for Analysis
Summations• Determine the coordinates of the center of gravity by
applying the center of gravity equations• If an object is symmetrical about an axis, the centroid
of the objects lies on the axis
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Example 9.10
Locate the centroid of the plate area.
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Solution
Composite Parts
Plate divided into 3 segments.
Area of small rectangle considered “negative”.
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Solution
Moment Arm
Location of the centroid for each piece is determined and indicated in the diagram.
Summations
mmA
Ayy
mmA
Axx
22.15.11
14~
348.05.11
4~
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.3 Theorems of Pappus and Guldinus
• A surface area of revolution is generated by revolving a plane curve about a non-intersecting fixed axis in the plane of the curve
• A volume of revolution is generated by revolving a plane area bout a nonintersecting fixed axis in the plane of area
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.3 Theorems of Pappus and Guldinus
• The theorems of Pappus and Guldinus are used to find the surfaces area and volume of any object of revolution provided the generating curves and areas do not cross the axis they are rotated
Surface Area• Area of a surface of revolution = product of length of
the curve and distance traveled by the centroid in generating the surface area
LrA
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.3 Theorems of Pappus and Guldinus
Volume• Volume of a body of revolution = product of generating
area and distance traveled by the centroid in generating the volume
ArV
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Example 9.12
Show that the surface area of a sphere is A = 4πR2 and its volume V = 4/3 πR3.
Solution
Surface Area
Generated by rotating semi-arc about the x axis
For centroid,
For surface area,
/2Rr
242
2
;~
RRR
A
LrA
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Solution
Volume
Generated by rotating semicircular area about the x axis
For centroid,
For volume,
32
34
21
34
2
;~
3/4
RRR
V
ArV
Rr
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.4 Resultant of a General Distributed Loading
Pressure Distribution over a Surface• Consider the flat plate subjected to the loading function
ρ = ρ(x, y) Pa• Determine the force dF acting on the differential area
dA m2 of the plate, located at the differential point (x, y)
dF = [ρ(x, y) N/m2](d A m2)
= [ρ(x, y) d A]N• Entire loading represented as
infinite parallel forces acting on separate differential area dA
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.4 Resultant of a General Distributed Loading
Pressure Distribution over a Surface• This system will be simplified to a single resultant force
FR acting through a unique point on the plate
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.4 Resultant of a General Distributed Loading
Magnitude of Resultant Force• To determine magnitude of FR, sum the differential
forces dF acting over the plate’s entire surface area dA• Magnitude of resultant force = total volume under the
distributed loading diagram
• Location of Resultant Force is
V
V
A
A
V
V
A
A
dV
ydV
dAyx
dAyxyy
dV
xdV
dAyx
dAyxxx
),(
),(
),(
),(
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure
• According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions
• Magnitude of ρ depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface
ρ = γz = ρgz• Valid for incompressible fluids • Gas are compressible fluids and the above equation
cannot be used
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure
Flat Plate of Constant Width• Consider flat rectangular plate of constant width
submerged in a liquid having a specific weight γ• Plane of the plate makes an angle with the horizontal
as shown
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9.5 Fluid Pressure
Flat Plate of Constant Width• As pressure varies linearly with depth, the distribution
of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ1= γz1 at depth z1 and ρ2 = γz2 at depth z2
• Magnitude of the resultant force FR = volume of this loading diagram
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure
Curved Plate of Constant Width• When the submerged plate is curved, the pressure
acting normal to the plate continuously changes direction
• Integration can be used to determine FR and location of center of centroid C or pressure P
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure
Flat Plate of Variable Width• Consider the pressure distribution acting on the surface
of a submerged plate having a variable width• Since uniform pressure ρ = γz (force/area) acts on dA,
the magnitude of the differential force dF
dF = dV = ρ dA = γz(xdy’)
Copyright © 2010 Pearson Education South Asia Pte Ltd
9.5 Fluid Pressure
Flat Plate of Variable Width• Centroid V defines the point which FR acts
• The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations
• This point should not be mistaken for centroid of the plate’s area
A VR VdVdAF
V
V
V
V
dV
dVyy
dV
dVxx
'~'
~
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 9.14
Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The water pressures at depth A and B are
For intensities of the load at A and B,
kPamsmmkggz
kPamsmmkggz
BwB
AwA
05.49)5)(/81.9)(/1000(
62.19)2)(/81.9)(/1000(23
23
mkNkPambw
mkNkPambw
BB
AA
/58.73)05.49)(5.1(
/43.29)62.19)(5.1(
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For magnitude of the resultant force FR created by the distributed load.
This force acts through the centroid of the area,
measured upwards from B
N
trapezoidofareaFR
5.154)6.734.29)(3(2
1
mh 29.1)3(58.7343.29
58.73)43.29(2
3
1
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Same results can be obtained by considering two components of FR defined by the triangle and rectangle.
Each force acts through its associated centroid and has a magnitude of
Hence
kNmmkNF
kNmmkNF
t 2.66)3)(/15.44(
3.88)3)(/43.29(Re
kNkNkNFFF RR 5.1542.663.88Re
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Location of FR is determined by summing moments about B
mh
h
MM BBR
29.1
)1(2.66)5.1(3.88)5.154(
;
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
1.The _________ is the point defining the geometric center of an object.
A)Center of gravity B) Center of mass
C)Centroid D) None of the above
2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________.
A) Center of gravity B) Center of mass
C) Centroid D) None of the above
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QUIZ
3. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ .
A) x B) y
C) z D) Any of the above.
4. If a vertical rectangular strip is chosen, then what are the values of x and y?
A) (x , y) B) (x / 2 , y / 2)
C) (x , 0) D) (x , y / 2)
~ ~
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QUIZ
5. A composite body in this section refers to a body made of ____.
A) Carbon fibers and an epoxy matrix
B) Steel and concrete
C) A collection of “simple” shaped parts or holes
D) A collection of “complex” shaped parts or holes
6. The composite method for determining the location of the center of gravity of a composite body requires _______.
A) Integration B) Differentiation
C) Simple arithmetic D) All of the above.
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QUIZ
7. Based on the typical centroid information, what are the minimum number of pieces you will have to consider for determining the centroid of the area shown at the right?
A)1 B) 2 C) 3 D) 4
8. A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over?
A) 30° B) 45 ° C) 60 ° D) 90 °
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
7. What are the min number of pieces you will have to consider for determining the centroid of the area?
A)1 B) 2 C) 3 D) 4
8. A storage box is tilted up by pulling C, with edge A remaining on the ground. What is the max angle of tilt possible before the box tips over?
A) 30° B) 45 ° C) 60 ° D) 90 °
3cm 1 cm
1 cm
3cm
30º
G
C
AB
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
9. For determining the centroid, what is the min number of pieces you can use?
A) Two B) Three
C) Four D) Five
10. For determining the centroid of the area,what are the coordinates (x, y ) of the centroid of square DEFG?
A) (1, 1) m B) (1.25, 1.25) m
C) (0.5, 0.5 ) m D) (1.5, 1.5) m
2cm 2cm
2cm
4cm
x
y
A1m
1m
y
E
F G
CB x
1m 1m
D