10.1 Moments of Inertia by Integration
10.1 Moments of Inertia by Integration Example 1, page 1 of 2
1. Determine the moment of inertia of the rectangle
about the x axis, which passes through the centroid C.
y
b
h/2
C
x
h/2
10.1 Moments of Inertia by Integration Example 1, page 2 of 2
y
1 We want to evaluate
Ix = y2 dA
b
where the differential element dA is located a
distance y from the x axis (y must have the same
value throughout dA).
dy
h/2
y
C
x
2 dA = area of rectangle
=b
dy
h/2
h 2
3 Ix =
-h 2
y2dA
h 2
=
y2(b
-h 2
dy)
4 Limits of integration
=
by3
3
h 2
-h 2
3
=
bh
12
Ans.
10.1 Moments of Inertia by Integration Example 2, page 1 of 2
2. Determine the moment of inertia of the rectangle
about its base, which coincides with the x axis.
y
b
h
x
10.1 Moments of Inertia by Integration Example 2, page 2 of 2
y
1 We want to evaluate
Ix = y2 dA
b
where the differential element of area dA is located a distance y from the
x-axis (y must have the same value throughout the element dA).
dy
2 dA = area of rectangle
h
y
=b
x
3 Ix = y2dA
h
=
4 Limits of integration
=
=
0
y2(b
by3
3
bh3
3
dy)
h
0
Ans.
dy
10.1 Moments of Inertia by Integration Example 3, page 1 of 3
3. Determine the moment of inertia of the
right triangle about the x axis.
y
h
x
b
10.1 Moments of Inertia by Integration Example 3, page 2 of 3
y
1 We want to evaluate
x
Ix = y2 dA
(x,y)
where y has the same value throughout
differential element dA.
dy
4 Equation of line
h
y = slope
y
x + intercept
h
= ( )x+h
b
x
5
Solve for x to get
b
2 dA = area of rectangle
= x dy
3 Since we will integrate with respect to y,
we must replace x by a function of y.
x= ( b )y+b
h
so,
dA = x
dy
= ( b y + b) dy
h
(1)
10.1 Moments of Inertia by Integration Example 3, page 3 of 3
6 Ix = y2 dA
=
7 Limits of integration are from
bottom of triangle to top.
h
0
=b
= b[
y2 ( b y + b) dy
h
h
0
(
y3 2
+ y ) dy
h
y4 y3 h
+ ]
4h 3 0
1
= bh3[ 1 + ]
4 3
bh3
= 12
Ans.
10.1 Moments of Inertia by Integration Example 4, page 1 of 3
4. Determine the moments of inertia of the area
bounded by an ellipse about the x and y axes.
x2 y2
+ =1
a2 b2
y
b
x
b
a
a
10.1 Moments of Inertia by Integration Example 4, page 2 of 3
1 We want to evaluate
Ix = y2 dA
where y has the same value throughout the
differential element dA.
2 x = half the length of the
differential element,
so 2x = entire length
y
x
3 dA = area of rectangle
= 2x dy
(x, y)
dy
y
x
4 Since we will integrate with
respect to y, we must replace x
by a function of y
5
x2 y2
+ =1
a2 b2
Solve for x to get
x=
a 1
(y/b)2
6 Since x locates a point to the right of the y axis, choose
the plus sign:
x = +a 1
(y/b)2
(1)
10.1 Moments of Inertia by Integration Example 4, page 3 of 3
7 Using Eq. 1 in the expression for dA gives:
dA = 2x
dy
= 2a 1
(y/b)2
dy
8 Ix = y2 dA
9 Limits of integration from
bottom to top of ellipse
=
b
-b
y2(2a 1
ab3
=
4
(y/b)2 ) dy
Ans.
11 To calculate Iy, use symmetry in the following way: in all the
above equations, replace x's by y's, original y's by x's, a's by b's
and original b's by a's. Then the result would be
Iy =
a3b
4
Ans.
10 Evaluate the integral either with a
calculator that does symbolic
integration or use a table of integrals.
10.1 Moments of Inertia by Integration Example 5, page 1 of 4
5. Determine the moments of inertia of the
crosshatched area about the x and y axes.
y
4 ft
y=4
x
2 ft
x2
10.1 Moments of Inertia by Integration Example 5, page 2 of 4
y
1
We want to evaluate
I x = y2 dA
where y has the same value throughout the differential element dA.
(x,y)
dy
2 dA = area of rectangle
= x dy
4 ft
3 Since we will integrate with respect to y,
we must replace x by a function of y.
y
y=4
x2
x
4 Solving for x to get
x=
2 ft
4
y
5 Since x locates a point to the right of the y axis, choose the plus sign:
x=+ 4
y
(1)
10.1 Moments of Inertia by Integration Example 5, page 3 of 4
6 Using Eq. 1 in the expression for dA gives
dA = x
=
dy
4
y dy
7 Ix = y2 dA
8 y = 4 at the top of
the crosshatched
area
9 Use the integral function
on your calculator
4
=
0
y2 4
= 19.50 ft4
y dy
Ans.
10.1 Moments of Inertia by Integration Example 5, page 4 of 4
10 Next we want to evaluate
y
Iy = x2 dA
where x has the same value throughout the differential element dA. Thus, we
choose a vertical rectangular strip.
x
11 dA = area of rectangle
=y
= (4
4 ft
y=4
x2
y
x
dx
2 ft
12
Iy = x2 dA
=
2
0
x2(4
x2) dx = 4.27 ft4
Ans.
dx
x2) dx
10.1 Moments of Inertia by Integration Example 6, page 1 of 3
6. Determine the moment of inertia of the
crosshatched area about the y axis.
y
1.156 m
x = 2y6 50y5
Scales on
the x and y
axes are
not the
same.
x
100 m
y3 + 100
10.1 Moments of Inertia by Integration Example 6, page 2 of 3
1
We want to evaluate
Iy = x2 dA
y
where x has the same value throughout the differential
element dA. Thus, we choose a vertical rectangular strip.
2
dA = area of rectangle
=y
3
y
dx
But this approach won't work because we
can't express y as a function of x.
x = 2y6
x
100 m
50y5
y3 + 100
10.1 Moments of Inertia by Integration Example 6, page 3 of 3
4
An alternative approach is to use a horizontal
rectangular strip and employ the equation for the
moment of inertia of a rectangle about its base (BB)
:
3
IBB = bh
3
(1)
h
B
b
y
5
Applying Eq.1 to the differential element
gives the differential moment of inertia.
x
3
dIy = bh
3
=
dy
(dy)x3
3
x = 2y6
50y5
y3 + 100
B
1.156 m
6
Replacing x by the function of y gives
dIy = 1 x3dy
3
= 1 (2y6
3
50y5
y3 + 100)3 dy
x
1.156
7
Iy = dIy =
0
= 2.72
1
6
3 (2y
50y5
105 m4
100 m
y3 + 100)3 dy
Ans.
8
Use the integral function on a calculator to evaluate
this integral.
10.1 Moments of Inertia by Integration Example 7, page 1 of 3
7. Determine the moment of inertia of the
crosshatched area about the x axis.
y
y=9
x2
9m
y = 3x
x2
x
3m
10.1 Moments of Inertia by Integration Example 7, page 2 of 3
1
We want to evaluate
3
A better approach is to use a vertical strip and then
apply the parallel-axis theorem to the strips.
2
Ix = y dA
y
where y has the same value throughout the differential
element dA. Thus it appears that we should use a
horizontal rectangular strip.
y
2
Centroid
Parallel-axis theorem
for a general region
But using a horizontal strip is awkward we would
have to use three different expressions for dA,
depending on the position of the strip.
Area
A
C
Ix = Ic + Ad2
d
x
y
For a rectangle in particular,
Ix =
1 3
bh + (b
12
b
h)d2
C
h
d
=(
1 3
h + hd2 )b
12
x
y
b
The moment of inertia of a strip of width
b dx is then
dIx = ( 1 h3 + hd2 ) dx
12
x
(1)
h
dx
C
d
x
10.1 Moments of Inertia by Integration Example 7, page 3 of 3
y
5
The y-coordinate of the element
centroid is the average of y1 and y2
(x,y2)
d = yel = 1 ( y1 + y2 )
2
6 Eq. 1 becomes
y=9
y2 = 9
4
h = y2
y1
y1 = 3x
x
y = 3x
x2
Ix = dIx
3
= {1(9
0 12
3x )3 + (9
3x) ( 9 + 3 x
2 2
= 328 m
Ans.
x2, we
x2
dIx = { 1 ( y2 y1 )3 + ( y2 y1 ) [ 1 ( y2 + y1 )]2} dx
12
2
1
= { [( 9 x2 ) ( 3x x2 )]3 + [( 9 x2 )
12
1
( 3x x2 )][ 1 ( 9 x2 ) + ( 3x x2 )]2} dx
2
2
9
3
= { 1 ( 9 3x )3 + ( 9 3x )[ + x x2 ]2} dx
2 2
12
x2 )2 } dx
8
4
x2
in Eq. 2. Thus Eq. 2 becomes
(x,y1)
7
(2)
x2, we
in Eq. 2. Similarly, since (x, y1) satisfies y = 3x
can substitute
yel
x ranges
from 0 to 3.
y1 ) [ 1 ( y1 + y2 )]2} dx
2
Since the point (x, y2) lies on the curve y = 9
can substitute
centroid
(xel,yel)
x2
dIx = ( 1 h3 + hd2 ) dx
12
= { 1 ( y2 y1 )3 + ( y2
12
Enter this expression directly into the
integral function of a calculator.
10.1 Moments of Inertia by Integration Example 8, page 1 of 4
8. Determine the moment of inertia of the
crosshatched area about the y axis.
y
1/4 m
4m
xy = 1
y = 2x
1/2 m
4m
1m
1m
x
10.1 Moments of Inertia by Integration Example 8, page 2 of 4
1
We want to evaluate
Iy = x2 dA
where x has the same value throughout the differential
element dA. Thus it appears that we should use
vertical differential strips.
y
2
x
But using a vertical strip is awkward
we would have to use three different
expressions for dA, depending on the
location of the strip.
10.1 Moments of Inertia by Integration Example 8, page 3 of 4
3
A better approach is to use a horizontal strip and then
apply the parallel-axis theorem to the strip.
centroid
y
area A
d
Parallel-axis theorem for a general region
Iy = Ic + Ad2
C
x
For a rectangle in particular,
Iy = 1 bh3 + (b
12
y
Area = b
d
2
h)d
C
1
= ( h3 + hd2 )b
12
b
h
x
y
The moment of inertia of a strip of width b
dIy = ( 1 h3 + hd2 ) dy
12
dy is then
d
(1)
C
b
h
x
dy
h
10.1 Moments of Inertia by Integration Example 8, page 4 of 4
y
h = x2
4
From the figure, we see that Eq. 1 can be written as
1
dIy = { h3 + hd2} dy
12
= { 1 ( x2 x1 )3 + ( x2
12
x1
Centroid of
strip
(Eq. 1 repeated)
x1 )[
1
( x + x )]2 } dy
2 2 1
(2)
(xel,yel)
(x1,y)
(x2,y)
C
5
Since the point (x1,y) lies on the line y = 2x,
we can solve for x and substitute
xy = 1
x1 =
y
2
in Eq. 2. Similarly, since (x2,y) lies on the
curve xy = 1, we can substitute
y = 2x
d = | xel | = | average of x1 and x2 | = |
1
( x + x )|
2 1 2
x2 = 1y
in Eq. 2. Thus Eq. 2 becomes
x
6
Iy = dIxy
=
7
Top of region
at y = 4
Bottom of
region at y = 1
4
1
y
1 y 1
[ 1 ( 1y + )3 + ( y + ) (
2
2 2y
12
8
y 2
) ] dy
4
Enter this expression directly into the
integral function of a calculator
= 2.81 m4
Ans.
1
dIy = { ( x2 x1 )3 + ( x2 x1 )[ 1 ( x2 + x1 )]2 } dy
12
2
y
y 1 1
y 2
={ 1 (1 (
))3 + ( 1y (
))[ ( y + (
) )] } dy
2
2
2
12 y
2
y
y
y 2
1 1
= { ( y + )3 + ( 1y + )[ 1
] } dy
2
2 2y
12
4
10.1 Moments of Inertia by Integration Example 9, page 1 of 3
9. Determine the moment of inertia of the
crosshatched area about the x axis.
y
2
y = 10e x
x
2
y = 10 e x
1m
1.5 m
10.1 Moments of Inertia by Integration Example 9, page 2 of 3
1
We want to evaluate
Ix =
2
3
Ix for region D
y2dA
3.6788 m
Because of the shape of the region, the integral has to
be evaluated over two sub-regions, D and E.
y
1m
3.6788 m
Region D
2
y(1) = 10 e (1) = 3.6788 m
4
x
1m
No integration needed.
Use the formula for the moment of inertia of a
rectangle about a centroidal axis.
Ic =
Region E
bh3
12
b
(1)
h/2
So
IxD =
C
(1 m)(2 3.6788 m)3
12
4
= 33.1915 m
h/2
(2)
10.1 Moments of Inertia by Integration Example 9, page 3 of 3
5
Ix for region E. Because the region is
symmetrical about the x axis, we can save
work by using vertical rectangular strips and
by applying Eq. 1 to these strips:
1
Ic = bh3
12
or,
y
1
dx
dIxE =
(dx)(2y)3
12
=
2
y = 10 e x
x
2
1
(dx)[2(10e x )]3
12
2
= 2000 e 3x dx
3
6
Thus, with the aid of the integral function on a
calculator, we have
IxE = dIx
1.5
=
1
2000 e 3x2 dx
3
= 4.7985 m4
(3)
Adding the results for regions D and E gives
Ix = IxD + IxE
by Eq. 2
by Eq. 3
= 33.1915 + 4.7985
1m
= 38.0 m4
x
1.5 m
Ans.
10.1 Moments of Inertia by Integration Example 10, page 1 of 3
10. Determine the moment of inertia of the
crosshatched area about the y axis.
y
2 in.
1 in.
4 in.
y = 3x
3 in.
y = x2
x
10.1 Moments of Inertia by Integration Example 10, page 2 of 3
1
We want to evaluate
3
Iy = x2dA
Integrate over region B.
(1)
4
2
where x has the same value throughout the differential
element dA.
dA = area of rectangle
y
Given the shape of the area, we have to evaluate the
integral in Eq.1 over two sub-regions, B and C.
= ( y2
y1 )
= ( 3x
x2 ) dx
x
y = 3x
(x,y2)
y
Region C
y2
y1
y = x2
(x,y1)
Region B
x
dx
1 in.
5
IyB = x2 dA
1
x
=
0
x2(3x
= 0.5500 in4
x2) dx
(2)
dx
10.1 Moments of Inertia by Integration Example 10, page 3 of 3
6
Integrate over region C.
7
y
x
y = x2
Write the equation of the line:
y 3
x 1
=
4 3
2 1
Solving gives
y=x+2
(x,y2)
8
y2
4 in.
3 in.
y1
= ( y2
(x,y1)
1 in.
dA = Area of rectangle
y1 )
= [(x + 2)
dx
x2 ] dx
dx
x
2 in.
9
IyC = x2 dA
=
2
1
x2( x + 2
x2 ) dx
= 2.2167 in4
(3)
10 Adding the results for regions B and C gives
Iy = IyB + IyC
by Eq. 3
by Eq. 2
= 0.5500 + 2.2167 = 2.77 in4
Ans.
10.1 Moments of Inertia by Integration Example 11, page 1 of 2
11. Given that the centroid C of the area bounded by a
quarter-circle lies a distance 4a/(3 above the base of the
quarter-circle, determine the moment of inertia of the cross
hatched area about an axis xc through the centroid.
y
yc
1
We want to evaluate
Ixc = yc2 dA
but because the equation of the circle is given in
terms of x and y instead of xc and yc, it is easier to
evaluate
x2 + y2 = a2
y2 dA
Ix
and then use the parallel-axis theorem:
C
xc
4a
3
Ix = Ixc + Ad2
= Ixc +
Area of quarter circle
a2 ( 4a )2
4 3
x
Solving gives
a
Ixc = Ix
4a4
9
Thus now we need to calculate Ix.
10.1 Moments of Inertia by Integration Example 11, page 2 of 2
y
2
dA = Area of rectangle
x
= x
x2 + y2 = a2
3
(x,y)
a
dy
Solve x2 + y2 = a2
to get
a2
x=
4
dy
y2
Substitute into the equation for dA
dA = x
dy
y
=
x
6
7
Evaluate the integral with a
calculator that does symbolic
integration or use a table.
Use the result given by Eq. 2 in Eq. 1:
4a4
9
4
a
4a4
16
9
IxC = Ix
(Eq. 1 repeated)
(
16
4 ) a4
9
Ans.
5
a2
y2 dy
Integrate
Ix = y2 dA
a
=
y2 a2
0
a4
=
16
y2 dy
(2)