10.1 Moments of Inertia by Integration - Civil · PDF file10.1 Moments of Inertia by Integration Example 1, page 1 of 2 1. Determine the moment of inertia of the rectangle about the

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.
C
b
h/2
x
y
h/2

x
yWe want to evaluate
Ix = y2 dA
where the differential element dA is located a
distance y from the x axis (y must have the same
value throughout dA).
yh/2
h/2
Ix = y2 dA
= y2 (b dy)
by3
3
bh3 Ans.
12
=
=
Limits of integration
dA = area of rectangle
= b dy
1
2
3
4h 2
-h 2
dy
h 2
-h 2
h 2
-h 2
C
b

h
b
y
x
2. Determine the moment of inertia of the rectangle
about its base, which coincides with the x axis.

b
y
x
We want to evaluate
Ix = y2 dA
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).
hy
2 dA = area of rectangle
= b dy
3
=
0
h
Ix = y2 dA
= y2 (b dy)
by3
3
= Ans.
Limits of integration4
1
h
0
dy
bh3
3

b
3. Determine the moment of inertia of the
right triangle about the x axis.
x
y
h

x
y
b
x
y
(x,y)
We want to evaluate
Ix = y2 dA
where y has the same value throughout
differential element dA.
Equation of line
y = slope x + intercept
( ) x + h (1) =
1
= x dy
h
Since we will integrate with respect to y,
we must replace x by a function of y.
Solve for x to get
x = ( ) y + b
so,
dA = x dy
= ( y + b) dy
4
5
3
2
dy
b
h
bh
hb

Ix = y2 dA
= y2 ( y + b) dy
= b ( + y2) dy
= b[ + ]
= bh3[ + ]
= Ans.
6
Limits of integration are from
bottom of triangle to top.7
h
0
bh
y3
h
y4
4h
y3
3
h
0
311
4
bh3
12
0
h

x
yx2 y2
a2 b2 = 1+
4. Determine the moments of inertia of the area
bounded by an ellipse about the x and y axes.
b
b
a a

x
y
dy
x
y
We want to evaluate
Ix = y2 dA
where y has the same value throughout the
differential element dA.
x = half the length of the
differential element,
so 2x = entire length
1
(x, y) dA = area of rectangle
= 2x dy
Since we will integrate with
respect to y, we must replace x
by a function of y
Since x locates a point to the right of the y axis, choose
the plus sign:
x = +a 1 (y/b)2 (1)
+ 1=a2 b2 x2 y2
Solve for x to get
x = a 1 (y/b)2
2
4
3
6
5

Evaluate the integral either with a
calculator that does symbolic
integration or use a table of integrals.
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 = Ans.
Limits of integration from
bottom to top of ellipse
8
9
11
10
b
-b
4ab3
a3 b4
7 Using Eq. 1 in the expression for dA gives:
dA = 2x dy
= 2a 1 (y/b)2 dy
Ix = y2 dA
= y2(2a 1 (y/b)2 ) dy
= Ans.

2 ft
5. Determine the moments of inertia of the
crosshatched area about the x and y axes.
y = 4 x2
x
y
4 ft

2 ft
We want to evaluate
I x = y2 dA
where y has the same value throughout the differential element dA.
= x dy
Since we will integrate with respect to y,
we must replace x by a function of y.
Solving for x to get
x = 4 y
Since x locates a point to the right of the y axis, choose the plus sign:
x = + 4 y (1)
1
2
3
4
5
y = 4 x2
(x,y)
y
dy
x
y
4 ft

Using Eq. 1 in the expression for dA gives
dA = x dy
= 4 y dy
6
Ix = y2 dA
= y2 4 y dy
= 19.50 ft4 Ans.
Use the integral function
on your calculator
7
8
9
y = 4 at the top of
the crosshatched area
4
0

2 ft
Next we want to evaluate
Iy = x2 dA
where x has the same value throughout the differential element dA. Thus, we
choose a vertical rectangular strip.
= y dx
= (4 x2) dx
Iy = x2 dA = x2(4 x2 ) dx = 4.27 ft4 Ans.
dx
x
10
11
120
2
yy = 4 x2
x
y
4 ft

crosshatched area about the y axis.
x = 2y6 50y5 y3 + 100
x
y
1.156 m
100 m
Scales on
the x and y axes are
not the same.

100 m
1 We want to evaluate
Iy = x2 dA
where x has the same value throughout the differential
element dA. Thus, we choose a vertical rectangular strip.
= y dx
But this approach won't work because we
can't express y as a function of x.
2
3
y
x = 2y6 50y5 y3 + 100
x
y

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) :
(1)
B
Bh
b
100 m
1.156 m
yApplying Eq.1 to the differential element
gives the differential moment of inertia.
dIy
(dy)x3=
3
5
Replacing x by the function of y gives
dIy = x3dy
= (2y6 50y5 y3 + 100)3 dy
Iy = dIy = (2y6 50y5 y3 + 100)3 dy
= 2.72 105 m4 Ans.
Use the integral function on a calculator to evaluate
this integral.
6
7
8
x = 2y6 50y5 y3 + 100
x
x
dy
bh3
3I BB =
=3bh3
31
31
31
0
1.156

y = 3x x2
y = 9 x2
crosshatched area about the x axis.
9 m
3 m
x
y

x
y
We want to evaluate
Ix = y2 dA
where y has the same value throughout the differential
element dA. Thus it appears that we should use a
horizontal rectangular strip.
But using a horizontal strip is awkward we would
have to use three different expressions for dA,
depending on the position of the strip.
2
1
A better approach is to use a vertical strip and then
apply the parallel-axis theorem to the strips.
Parallel-axis theorem
for a general region
Ix = Ic + Ad2
d
Centroid
Area A
x
y
For a rectangle in particular,
Ix = bh3 + (b h)d2
= ( h3 + hd2 )b
y
x
y
x
C
d
C
h
b
h
b dx
d
The moment of inertia of a strip of width
b dx is then
dIx = ( h3 + hd2 ) dx (1)
3
121
112
121
C

x
y
(xel,yel)
(x,y1)
(x,y2)
The y-coordinate of the element
centroid is the average of y1

Documents

10.1 Moments of Inertia by Integration - Civil · PDF file10.1 Moments of Inertia by Integration Example 1, page 1 of 2 1. Determine the moment of inertia of the rectangle about the