Transcript ENGR-36_Lec-23_Center_of_Gravity

Engineering 36
Chp09: Center
of Gravity
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-232_Center_of_Gravity.pptx
Introduction: Center of Gravity
 The earth exerts a gravitational force on
each of the particles forming a body.
• These forces can be replaced by a
SINGLE equivalent force equal to the
weight of the body and applied at the
CENTER OF GRAVITY (CG) for the body
 The CENTROID of an AREA is
analogous to the CG of a body.
• The concept of the FIRST MOMENT of
an AREA is used to locate the centroid
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Total Mass – General Case
dVn at xn , yn , zn 
 Given a Massive
Body in 3D Space
 Divide the Body in to
Very Small Volumes
→ dV
 Each dVn is located at
position (xn,yn,zn)
r  xniˆ  yn ˆj  zn kˆ
 The DENSITY, ρ, can
be a function of
POSITION → n   xn , yn , zn 
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Total Mass – General Case
 Now since m = ρ•V, then the
incremental mass, dm
dVn at xn , yn , zn 
dmn   xn , yn , zn   dVn
r  xniˆ  yn ˆj  zn kˆ
 Integrate dm over the entire
body to obtain the total
Mass, M
M
 dm
body
n
  x , y , z  dV

n
n
n
volume
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n
Bruce Mayer, PE
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Total WEIGHT – General Case
 Recall that W = Mg
dVn at xn , yn , zn 
 Using the the previous
expression for M
W  Mg 
 gdm

n
body

 g x , y , z  dV
n
n
n
r  xniˆ  yn ˆj  zn kˆ
n
volume
  x , y , z g  dV
n
n
n
n
volume

  x , y , z g  dV
n
n
n
n
volume
 Now Define SPECIFIC
WEIGHT, γ, as ρ•g →
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W
  x , y , z g  dV
n
n
n
n
volume

  x , y , z  dV
n
n
n
volume
Bruce Mayer, PE
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n
Uniform Density Case
 Consider a body with
UNIFORM DENSITY; i.e.
dVn at xn , yn , zn 
 xn , yn , zn    xm , ym , zn   
 xn , yn , zn    xm , ym , zn   
r  xniˆ  yn ˆj  zn kˆ
 Then M & W→
M
   dV
   dV
n
volume
W
n


volume
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 dV
 dV
n
 V
n
 V
volume
volume
Bruce Mayer, PE
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Center of Mass Location
 Use MOMENTS to Locate
Center of Mass/Gravity
dVn at xn , yn , zn 
 Recall Defintion of a
MOMENT
r  xniˆ  yn ˆj  zn kˆ
Moment leverArm  Intensity
 In the General Center-of-Mass Case
• LeverArm ≡ Position Vector, rn, or its components
• Intensity ≡ Incremental Mass, dmn
RM  LeverArms  Intensities
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Intensities
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Center of Mass Location
 RM in Component form
dVn at xn , yn , zn 

RM  X M iˆ  YM ˆj  ZM kˆ
r  xniˆ  yn ˆj  zn kˆ
 Now Define the Incremental
Moment, dΩn


ˆ
ˆ
ˆ
dΩn  xni  yn j  zn k gdVn 
LeverArm
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Intensity
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Center of Mass Location
 Integrating dΩ to Find Ω for
the entire body
Ω
 dΩ
n
dVn at xn , yn , zn 

r  xniˆ  yn ˆj  zn kˆ
body
 x iˆ  y ˆj  z kˆgdV 
n
n
n
n
volume
Ω   xiˆ   y ˆj   x kˆ 
giˆ
ˆj  y dV  gkˆ z dV


x

dV

g
 n n
 n n
 n n
all x
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all y
all z
Bruce Mayer, PE
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Center of Mass Location
 Now Equate Ω to RM•Mg

dVn at xn , yn , zn 

R M  Mg  Mg  X M iˆ  YM ˆj  Z M kˆ 
giˆ  xn dVn  g ˆj   yn dVn  gkˆ  zn dVn
all x
all y
r  xniˆ  yn ˆj  zn kˆ
all z
 Canceling g, and equating
Components yields, for example, in the X-Dir
XMM 
 x dV 
n
n
all x
LeverArm
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Intensity
Bruce Mayer, PE
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Center of Mass Location
 Divide out the Total Intensity,
M, to Isolate the Overall
X-directed Lever Arm, XM
XM 
dVn at xn , yn , zn 
 xn dVn 
r  xniˆ  yn ˆj  zn kˆ
all x
M
 And the Similar expressions
for the other CoOrd Directions
  yn dVn 
 zn dVn 
YM 
all y
M
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ZM 
all x
M
Bruce Mayer, PE
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Center of Gravity of a 2D Body
 Centroid of an Area
 Taking Incremental Plate
Areas, Forming the Ωx &
Ωy, Along With the Fz=W
Yields the Expression for the
Equivalent POINT of W
application
• Note Ω Units = In-lb or N-m
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 Centroid of a Line


x

x W   x W
  x dW
y

y W   y W
  y dW
Bruce Mayer, PE
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Centroids of Areas & Lines
 Centroid of an Area
x W   x dW

x At    x t dA
x A   x dA   x
For Plate of Uniform
Thickness
•
•
•
  Specific Weight
t  Plate Thickness
dW =  tdA
 1st moment w.r.t. y  axis
yA   y dA   y
 1st moment w.r.t. to x  axis
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 Centroid of a Line
x W   x dW

x  La    x  a dL
Wire of
Uniform
Thickness
x L   x dL
yL   y dL
Bruce Mayer, PE
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•
  Specific
Weight
•
a  X-Sec
Area
•
dW = a(dL)
First Moments of Areas & Lines
• An area is symmetric with respect to an axis
BB’ if for every point P there exists a point
P’ such that PP’ is perpendicular to BB’ and
the Area is divided into equal parts by BB’.
• The first moment of an area with respect to
a line of symmetry is ZERO.
• If an area possesses a line of SYMMETRY,
its centroid LIES on THAT AXIS
• If an area possesses two lines of symmetry,
its centroid lies at their INTERSECTION.
• An area is symmetric with respect to a
center O if for every element dA at (x,y)
there exists an area dA’ of equal area
at (−x, −y).
• The centroid of the area coincides
with the center of symmetry, O.
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Centroids of Common Area Shapes
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Centroids of Common Line Shapes
 Recall that for a SMALL Angle, α
sin   
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 xr
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Composite Plates and Areas
• Composite plates
X W   xk Wk
Y  W   yk W k
X W  x1W1  x 2W2  x3W3
• Composite area
X  A   xk Ak
Y  A   yk Ak
Y  A  y1 A1  y 2 A2  y 3 A3
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Bruce Mayer, PE
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Example: Composite Plate
 Solution Plan
 For the plane area
shown, determine the
first moments with
respect to the x and y
axes, and the location
of the centroid.
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• Divide the area into a triangle,
rectangle, semicircle, and a
circular cutout
• Calculate the first moments of
each area w/ respect to the axes
• Find the total area and first
moments of the triangle,
rectangle, and semicircle.
Subtract the area and first
moment of the circular cutout
• Calc the coordinates of the area
centroid by dividing the
NET first moment by
the total area
Bruce Mayer, PE
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Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Example: Composite Plate

Find the total area and first moments of the
x
triangle, rectangle, and semicircle. Subtract
the area and first moment of the circular cutout  y
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 757.7  103 mm3
 506.2  103 mm3
Bruce Mayer, PE
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Example: Composite Plate
 Solution
3
3
x
A

757
.
7

10
mm
X 

 A 13.828103 mm2
X  54.8 mm
 Find the coordinates
of the area centroid
by dividing the
first moment totals
by the total area
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3
3
y
A

506
.
2

10
mm
Y 

 A 13.828103 mm2
Y  36.6 mm
Bruce Mayer, PE
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Centroids by Strip Integration
xA   xdA   x dxdy   xel dA
yA   ydA   y dxdy   yel dA
x A   xel dA
  x  ydx
yA   yel dA
y
   ydx
2
x A   xel dA
x A   xel dA
ax
 a  x dy

2
yA   yel dA
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• Double integration to find the first
moment may be avoided by defining
dA as a thin rectangle or strip.
  y a  x dy
2r
1 2 
  cos  r d 
3
2

yA   yel dA

2r
1

sin   r 2 d 
3
2

Bruce Mayer, PE
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Example: Centroid by Integration
 Solution Plan
 Determine by direct
integration the
location of the
centroid of a
parabolic spandrel.
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• Determine Constant k
• Calculate the Total Area
• Using either vertical or
horizontal STRIPS,
perform a single
integration to find the
first moments
• Evaluate the centroid
coordinates by dividing
the Total 1st Moment
by Total Area.
Bruce Mayer, PE
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Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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Example: Centroid by Integration
 Solution
• Determine Constant k
y  k x 2 ; and y  b when x  a
b
2
bka k  2
a
b
a
y  2 x 2 or x  1 2 y1 2
a
b
• Calculate the Total Area
A   dA  Use VerticalStrips
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a
 b x3 
b 2
  y dx   2 x dx   2 
a
 a 3 0
0
ab

3
Bruce Mayer, PE
a
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Example: Centroid by Integration
 Solution
• Calc the 1st Moments, Ωi
 b 
 x   xel dA   x  y dx   x 2 x 2 dx
a

0 
a
a
b x 
b 3
a 2b
  2 x dx   2  
a
4
 a 4 0
0
a
4
 Using vertical strips,
2
a
y
1
b


perform a single
 y   yel dA    y dx    2 x 2  dx
2
2a

integration to find
0
a
2
5
the first moments.
b x 
ab2
 4  
 2a 5 0 10
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Bruce Mayer, PE
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Example: Centroid by Integration
 Finally the Answers
xA  x
2
ab a b
x 
3
4
 Evaluate the
centroid coordinates
• Divide Ωx and Ωy
by the total Area
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yA   y
2
ab ab
y 
3 10
3
x a
4
3
y b
10
Bruce Mayer, PE
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Theorems of Pappus-Guldinus
 Surface of
revolution is
generated by
rotating a plane
curve about a fixed
axis.
 Area of a surface of revolution
is equal to the length of the
generating curve, L, times the
distance traveled by the
centroid through the rotation.
A   2ydL  2  ydL
A  2 yL as
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yL   y   ydL
Bruce Mayer, PE
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Theorems of Pappus-Guldinus
 Body of revolution is
generated by
rotating a plane
area about a fixed
axis.
 Volume of a body of revolution
is equal to the generating area,
A, times the distance traveled
by the centroid through the
rotation.
V   dV  2ydA
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V  2  ydA  2 yA
Bruce Mayer, PE
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Example: Pappus-Guldinus
 Solution Plan
 The outside diameter of a
pulley is 0.8 m, and the
cross section of its rim is as
shown. Knowing that the
pulley is made of steel and
that the density of steel,  =
7850 kg/m3, determine the
mass and weight of the rim.
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• Apply the theorem of PappusGuldinus to evaluate the
volumes of revolution for the
rectangular rim section and the
inner cutout section.
• Multiply by density and
acceleration of gravity to get
the mass and weight.
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-232_Center_of_Gravity.pptx
Example: P-G
 Apply Pappus-Guldinus
to Sections I & II
 Subtract: I-II



3
 9 3
m  V  7.85 10 kg m 7.65 10 mm 10 m mm 


W  589 N
W  mg  60.0 kg  9.81 m s 2
3

3
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6

3
m  60.0 kg
Bruce Mayer, PE
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WhiteBoard Work
Find the
Areal &
Lineal
Centroids
Find for Area
XA & YA
Find for OutSide Line
XL & YL
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Bruce Mayer, PE
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Engineering 36
Appendix
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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