This document discusses calculating moments of inertia for planar areas using integration. It describes: 1) Choosing a differential element for integration that has size in only one direction to simplify the calculation. 2) The procedure involves specifying a rectangular differential element and orienting it parallel or perpendicular to the axis of rotation. 3) Moments of inertia are calculated through single or double integration, depending on whether the element has thickness in one or two directions.

1. 10.4 Moments of Inertia
for an Area by Integration
When the boundaries for a planar area are
expressed by mathematical functions,
moments of inertia for the area can be
determined by the previous method
If the element chosen for integration has a
differential size in two directions, a double
integration must be performed to evaluate
the moment of inertia
Try to choose an element having a
differential size or thickness in only one
direction for easy integration

2. 10.4 Moments of Inertia
for an Area by Integration
Procedure for Analysis
If a single integration is performed to determine
the moment of inertia of an area bout an axis, it
is necessary to specify differential element dA
This element will be rectangular with a finite
length and differential width
Element is located so that it intersects the
boundary of the area at arbitrary point (x, y)
2 ways to orientate the element with respect to
the axis about which the axis of moment of
inertia is determined

3. 10.4 Moments of Inertia
for an Area by Integration
Procedure for Analysis
Case 1
Length of element orientated parallel to the axis
Occurs when the rectangular element is used to
determine Iy for the area
Direct application made since the element has
infinitesimal thickness dx and
therefore all parts of element
lie at the same moment
arm distance x from the y axis

4. 10.4 Moments of Inertia
for an Area by Integration
Procedure for Analysis
Case 2
Length of element orientated perpendicular to
the axis
All parts of the element will not lie at the same
moment arm distance from the axis
For Ix of area, first calculate moment of inertia of
element about a horizontal
axis passing through the
element’s centroid and x axis
using the parallel axis theorem

5. 10.4 Moments of Inertia
for an Area by Integration
Example 10.1
Determine the moment of
inertia for the rectangular area
with respect to (a) the
centroidal
x’ axis, (b) the axis xb passing
through the base of the
rectangular, and (c) the pole or
z’ axis perpendicular to the x’-y’
plane and passing through the
centroid C.

6. 10.4 Moments of Inertia
for an Area by Integration
Solution
Part (a)
Differential element chosen, distance y’ from
x’ axis
Since dA = b dy’
h/2
I x = ∫ y ' dA = ∫
2 2
y ' dy
A −h / 2
1 3
= bh
12

7. 10.4 Moments of Inertia
for an Area by Integration
Solution
Part (b)
Moment of inertia about an axis passing
through the base of the rectangle obtained
by applying parallel axis theorem
I xb = I x + Ad 2
2
1 3 h 1 3
= bh + bh  = bh
12 2 3

8. 10.4 Moments of Inertia
for an Area by Integration
Solution
Part (c)
For polar moment of inertia about point C
1 3
I y ' = hb
12
JC = I x + I y'
1
= bh(h 2 + b 2 )
12

9. 10.4 Moments of Inertia
for an Area by Integration
Example 10.2
Determine the moment of
inertia of the shaded area
about the x axis

10. 10.4 Moments of Inertia
for an Area by Integration
Solution
A differential element of area that is parallel
to the x axis is chosen for integration
Since element has thickness dy and
intersects the curve at arbitrary point (x, y),
the area
dA = (100 – x)dy
All parts of the element lie at the same
distance y from the x axis

11. 10.4 Moments of Inertia
for an Area by Integration
Solution
I x = ∫ y 2 dA
A
= ∫ y 2 (100 − x)dy
A
200  y2 
= ∫ y 100 −

2
dy
0
 400 

200 1 200 4
= 100∫ y dy −
2
∫0 y dy
0 400
= 107(10 6 )mm 4

12. 10.4 Moments of Inertia
for an Area by Integration
Solution
A differential element parallel
to the y axis is chosen for
integration
Intersects the curve at
arbitrary point (x, y)
All parts of the element do not
lie at the same distance from
the x axis

13. 10.4 Moments of Inertia
for an Area by
Integration
Solution
Parallel axis theorem used to determine moment
of inertia of the element
For moment of inertia about its centroidal axis,
1 3
Ix = bh
12
For the differential element shown
b = bx h = y
Thus,
1
dI x = dxy3
12

14. 10.4 Moments of Inertia
for an Area by Integration
Solution
For centroid of the element from the x axis
~ = y/2
y
Moment of inertia of the element
2
~ 2 = 1 dxy 3 + ydx y  = 1 y 3dx
dI x = dI x + dAy  
12 2 3
Integrating
1 3 100 1
I x = ∫ dI x = ∫ y dx = ∫
A3 0 3
(400 x )3 / 2 dx
( )
= 107 106 mm 4

15. 10.4 Moments of Inertia
for an Area by Integration
Example 10.3
Determine the moment of inertia with respect
to the x axis of the circular area.

16. 10.4 Moments of Inertia
for an Area by Integration
Solution
Case 1
Since dA = 2x dy
I x = ∫ y 2 dA
A
= ∫ y 2 (2 x)dy
A
a
(
= ∫ y 2 2 a 2 − y 2 dy
−a
)
πa 4
=
4

17. 10.4 Moments of Inertia
for an Area by Integration
Solution
Case 2
Centroid for the element lies
on the x axis
Noting
dy = 0
For a rectangle,
1 3
I x ' = bh
12

18. 10.4 Moments of Inertia
for an Area by Integration
Solution
1
dI x = dx(2 y )
3
12
2 3
= y dx
3
Integrating with respect to x
Ix = ∫
a2 2
−a 3
(
a − x2 )
3/ 2
dx
πa 4
=
4

19. 10.4 Moments of Inertia
for an Area by Integration
Example 10.4
Determine the moment of inertia of the
shaded area about the x
axis.

20. 10.4 Moments of Inertia
for an Area by Integration
Solution
Case 1
Differential element parallel to x axis chosen
Intersects the curve at (x2,y) and (x1, y)
Area, dA = (x1 – x2)dy
All elements lie at the same distance y from
the x axis
I x = ∫ y dA = ∫ y (x1 − x2 )dy = ∫ y 2
A
2
1
0
2
1
0
( )
y − y dy
2 7/2 1 4 1
I x = y − y = 0.0357m 4
7 4 0

21. 10.4 Moments of Inertia
for an Area by Integration
Solution
Case 2
Differential element parallel
to y axis chosen
Intersects the curve at (x, y2)
and (x, y1)
All elements do not lie at the
same distance from the x axis
Use parallel axis theorem to
find moment of inertia about
the x axis

22. 10.4 Moments of Inertia for
an Area by Integration
Solution
1 3
I x' = bh
12
Integrating
2
1  y −y 
dI x = dI x + dA~ 2 = dx( y2 − y1 ) + ( y2 − y1 )dx y1 + 2 1 
3
y
12  2 
(
1 3
) 1
(
= y2 − y13 dx = x 3 − x 6 dx
3 3
)
1 1 3
(
I x = ∫ x − x 6 dx
3 0
)
1 4 1 71
= x − x = 0.0357 m 4
12 21 0