ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material... · ME 201...
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ME 201 Engineering Mechanics: Statics Unit 9.3 Center of Gravity, Center of Mass, and Centroid for a Body Moments of Inertia by Integration
Transcript of ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material... · ME 201...
ME 201Engineering Mechanics: Statics
Unit 9.3
Center of Gravity, Center of Mass, and
Centroid for a Body
Moments of Inertia by Integration
Centroids
Given:
If we
increase the number of elements into which the area A is divided
decrease the size of each element
Then
A
AxX
~
A
AyY
~
A
dAA
A
dAxAx ~~ A
dAyAy ~~
Centroids
And finally
A
A
dA
dAx
A
AxX
~~
A
A
dA
dAy
A
AyY
~~
SolutionGiven:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve x
y
3 m
6 m
y
x
)2
,()~,~(y
xyx
dx
3
0
3
0
ydx
xydx
|
|3
0
2
3
0
32
226
32
26
xx
xx
(x,y)
A
A
dA
dAxX
~
3
0
3
0
2
26
26
xdx
dxxx
3
0
3
0
)3(2
)3(2
dxx
dxxx
m19
9
Solution
3
0
3
0
2
26
21218
xdx
dxxx
|
|3
0
2
3
0
32
226
32
21218
xx
xxx
A
A
dA
dAyY
~
3
0
3
0
22
)3(2
2
)3(2
dxx
dxx
3
0
3
0 2
ydx
ydxy
m29
18
Given:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve x
y
3 m
6 m
y
x
)2
,()~,~(y
xyx
dx
(x,y)
Alternate SolutionGiven:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its thickness?what’s its length?what’s its location from the y axis?what’s its centroid?
Define integral & solvex=3-y/2
x
y
3 m
6 m
y
x
),2
()~,~( yx
yx
(x,y)
6
0
6
0 2
xdy
xdyx
|
|6
0
2
6
0
32
43
)122
39(2
1
yy
yyy
dy
A
A
dA
dAxX
~
6
0
6
0
2
)2
3(
)2
3(2
1
dyy
dyy
6
0
6
0
2
)2
3(
)4
39(2
1
dyy
dyy
y
m19
9
Alternate Solution
A
A
dA
dAyY
~
|
|6
0
2
6
0
32
43
)62
3
yy
yy
6
0
6
0
)2
3(
)2
3(
dyy
dyy
y
6
0
6
0
xdy
yxdy
6
0
6
0
2
)2
3(
)2
3(
dyy
dyy
y
m29
18
Given:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its thickness?what’s its length?what’s its location from the y axis?what’s its centroid?
Define integral & solvex=3-y/2
x
y
3 m
6 m
y
x
),2
()~,~( yx
yx
(x,y)dy
Moments of Inertia
Given
If we
increase the number of elements into which the area A is
divided
decrease the size of each element
Then
About X and Y axes, respectively
A
x dAyI 2
2AyI x 2AxI y
A
y dAxI 2
Example ProblemI of a General Area
Given:
y2=400x
Find:
Ix, Iy about
axes
x
y
100 mm
20
0 m
m
Example ProblemGiven:
y2=400x
Find:
Ix, Iy about axes
x
y
100 mm
200 m
m
Example Problem Solution
400
2yx
A
dyxy )100(2
A
dyy
y )400
100(2
2
x
y
y2=400x
(100,200)
dyx(100-x)
y
Ax dAyI 2
200
0
42
400100 dy
yy
200
0
53
)54003
100(
yy
4610107 mm
Given:
y2=400x
Find:
Ix, Iy about axes
Example Problem Solution
Adxyx )(2
Adxxx 5.2 )400(
x
y
y2=400x(100,200)
dx
x
y
Ay dAxI 2
100
0
5.220 dxx
100
0
5.3
5.320
x
461057 mm
Composite Centroid Review
Find the centroid, y of the
beam’s cross section.
A. 125 mm
B. 132 mm
C. 162 mm
D. 200 mm
E. None of the above
Centroids
Given:
If we
increase the number of elements into which the area A is divided
decrease the size of each element
Then
A
AxX
~
A
AyY
~
A
dAA
A
dAxAx ~~ A
dAyAy ~~
Centroids
And finally
A
A
dA
dAx
A
AxX
~~
A
A
dA
dAy
A
AyY
~~
Example Problem
Given:
y=2(3-x)
Find:
X, Y of area under curve
x
y
3 m
6 m
SolutionGiven:
y=2(3-x)
Find:
X, Y of area under curve
x
y
3 m
6 m
SolutionGiven:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve x
y
3 m
6 m
y
x
)2
,()~,~(y
xyx
dx
3
0
3
0
ydx
xydx
|
|3
0
2
3
0
32
226
32
26
xx
xx
(x,y)
A
A
dA
dAxX
~
3
0
3
0
2
26
26
xdx
dxxx
3
0
3
0
)3(2
)3(2
dxx
dxxx
m19
9
Solution
3
0
3
0
2
26
21218
xdx
dxxx
|
|3
0
2
3
0
32
226
32
21218
xx
xxx
A
A
dA
dAyY
~
3
0
3
0
22
)3(2
2
)3(2
dxx
dxx
3
0
3
0 2
ydx
ydxy
m29
18
Given:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve x
y
3 m
6 m
y
x
)2
,()~,~(y
xyx
dx
(x,y)
Alternate SolutionGiven:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its thickness?what’s its length?what’s its location from the y axis?what’s its centroid?
Define integral & solvex=3-y/2
x
y
3 m
6 m
y
x
),2
()~,~( yx
yx
(x,y)
6
0
6
0 2
xdy
xdyx
|
|6
0
2
6
0
32
43
)122
39(2
1
yy
yyy
dy
A
A
dA
dAxX
~
6
0
6
0
2
)2
3(
)2
3(2
1
dyy
dyy
6
0
6
0
2
)2
3(
)4
39(2
1
dyy
dyy
y
m19
9
Alternate Solution
A
A
dA
dAyY
~
|
|6
0
2
6
0
32
43
)62
3
yy
yy
6
0
6
0
)2
3(
)2
3(
dyy
dyy
y
6
0
6
0
xdy
yxdy
6
0
6
0
2
)2
3(
)2
3(
dyy
dyy
y
m29
18
Given:y=2(3-x)
Find:Centroid of area under curve
Solution:Define element
what’s its thickness?what’s its length?what’s its location from the y axis?what’s its centroid?
Define integral & solvex=3-y/2
x
y
3 m
6 m
y
x
),2
()~,~( yx
yx
(x,y)dy
In Class Exercises
Given:
y1=x2
y2=x
Find:
X, Y of area between curves
x
y
1 ft1 ft
y1
y2
ftX 5.0
ftY 4.0
SolutionGiven:y1=x2
y2=xFind:
Centroid of area between curvesSolution:
Define elementwhat’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve
dx
A
A
dA
dAxX
~
|
|1
0
32
1
0
43
32
43
xx
xx
x
y
(1,1)
y1
y2
y2 -y1 )~,~( yx
),( 1yx
),( 2yx
1
0
2
1
0
32
)(
)(
dxxx
dxxx
1
012
1
012
)(
)(
dxyy
dxyyx
1
0
2
1
0
2
)(
)(
dxxx
dxxxx
ft5.0
6
112
1
x
)2
,( 21 yyx
Solution
A
A
dA
dAyY
~
txx
xx
|
|1
0
32
1
0
53
32
)53
(2
1
1
012
1
012
12
)(
)(2
dxyy
dxyyyy
1
0
2
1
0
42
)(
2
dxxx
dxxx
1
0
2
1
0
22
)(
)(2
dxxx
dxxxxx
ft4.0
6
115
1
Given:y1=x2
y2=xFind:
Centroid of area between curvesSolution:
Define elementwhat’s its width?what’s its height?what’s its location from the x axis?what’s its centroid?
Define integral & solve
dx
x
y
(1,1)
y1
y2
y2 -y1 )~,~( yx
),( 1yx
),( 2yx
x
)2
,( 21 yyx
Alternate SolutionGiven:
y1=x2
y2=x
Find:
Centroid of area between curves
Solution:
yx 1
1
0
1
0
2
)(
2
dyyy
dyyy
|
|1
0
223
1
0
32
)223
(
)32
(2
1
yy
yy
x
y
(1,1)
)~,~( yx
),( 1 yx),( 2 yx
x2
x1 –x2
x1
dyyx 2
A
A
dA
dAxX
~
1
0
1
0
)(
))(2
(
dyyy
dyyyyy
1
021
1
021
21
)(
))(2
(
dyxx
dyxxxx
ft5.0
6
112
1
),2
()~,~( 21 yxx
yx
Alternate SolutionGiven:
y1=x2
y2=x
Find:
Centroid of area between curves
Solution:
yx 1
1
0
1
0
22
3
)( dyyy
dyyy
|
|
1
0
223
1
0
325
)223
(
)325
(
yy
yy
dy
x
y
x2
(1,1)
x1 –x2
)~,~( yx
),( 1 yx),( 2 yx
x1
yx 2
A
A
dA
dAyY
~
1
0
1
0
)(
)(
dyyy
dyyyy
1
021
1
021
)(
)(
dyxx
dyxxy
ft4.0
6
115
1
),2
()~,~( 21 yxx
yx
In Class Exercise
Solution
Moment of Inertia Review
Find the moment of
inertia of the beam
about the centroidal y
axis of the wood beam:
A. 171E6 mm4
B. 240E6 mm4
C. 463E6 mm4
D. 778E6 mm4
E. None of the above
Moments of Inertia
Given
If we
increase the number of elements into which the area A is
divided
decrease the size of each element
Then
About X and Y axes, respectively
A
x dAyI 2
2AyI x 2AxI y
A
y dAxI 2
Example ProblemI of a Rectangle
Given:
rectangle, b x h
Find:
Ix, Iy about
centroid and
base
b
h
Example ProblemGiven:
rectangle, b x h
Find:
Ix, Iy about centroid and base
b
h
Example Problem SolutionGiven:
rectangle, b x h
Find:
Ix, Iy about
centroid and
base
b
h
dy
y
x
y
dyby2
2
2
2h
hdyyb
Ax dAyI 2
)88
(3
33 hhb
2
2
3
3
h
h
yb
3
12
1bh
Example Problem SolutionGiven:
rectangle, b x h
Find:
Ix, Iy about
centroid and
base
b
h
dx
x
y
dxhx2
2
2
2b
bdxxh
Ay dAxI 2
)88
(3
33 bbh
2
2
3
3
b
b
xh
3
12
1hb
x
Example Problem SolutionGiven:
rectangle, b x h
Find:
Ix, Iy about
centroid and
base
b
h
dy
y
x
y
dyby2
h
dyyb0
2
Ax dAyI 2
)0(3
3 hb
h
yb
0
3
3
3
3
1bh
2AdII xcx 2
3
2)(
12
1
hbhbh
3
3
1bh
Example ProblemI of a General Area
Given:
y2=400x
Find:
Ix, Iy about
axes
x
y
100 mm
20
0 m
m
Example ProblemGiven:
y2=400x
Find:
Ix, Iy about axes
x
y
100 mm
200 m
m
Example Problem Solution
400
2yx
A
dyxy )100(2
A
dyy
y )400
100(2
2
x
y
y2=400x
(100,200)
dyx(100-x)
y
Ax dAyI 2
200
0
42
400100 dy
yy
200
0
53
)54003
100(
yy
4610107 mm
Given:
y2=400x
Find:
Ix, Iy about axes
Example Problem Solution
Adxyx )(2
Adxxx 5.2 )400(
x
y
y2=400x(100,200)
dx
x
y
Ay dAxI 2
100
0
5.220 dxx
100
0
5.3
5.320
x
461057 mm
Alternate Solution
In previous examples, the length of the element is oriented parallel to
the axis of interest
As an Alternate Solution, the length of the element may be oriented
perpendicular to the axis of interest
For this method, since all of the parts of the element do not lie the same
distance for the axis of interest, we must first use the parallel-axis
theorem to find the element’s moment of inertia about the axis, then
integrate this result to determine I.
Ax dAyI 2
Ax dAyI 2
x
y
x
y
Alternate Solution2AdII cx
How would you find I about centroidal
axis of the following problem?
x
y
y2=400x
(100,200)
Recall I = Ic + Ad2
In Class Exercise
Use parallel axis
theorem on 10-12
Solution
Solution
