1. 10.6 Inertia for an Area
Moment of inertia for an area is different for
every axis about which it is computed
In some applications of structural or
mechanical design, it is necessary to known
the orientation of those axes which give the
maximum and minimum moments of inertia
for the area
First, compute the product of the inertia for
the area as well as its moments of inertia for
given x, y axes

2. 10.6 Inertia for an Area
Product of inertia for an element of area dA
located at a point (x, y) is defined as
dIxy = xydA
Thus, for the entire area, for product of inertia,
I xy = ∫ xydA
A
If element of area chosen has
a differential size in 2 directions,
a double integration must be
performed to evaluate Ixy

3. 10.6 Inertia for an Area
Easier to choose an element having a differential
size or thickness in only one direction in which
case, the evaluation requires only one integration
Like the moment of inertia, product of inertia has
units of length raised to the forth power
However, since x or y may be a negative
quantity while the element of area is always
positive, the product of inertia may be positive,
negative or zero, depending on the location and
orientation of the coordinate axes

4. 10.6 Inertia for an Area
Example
Consider shaded area whereas for every element
dA located at (x, -y)
Since the products of inertia for these elements
are xy dA and –xy dA, the algebraic sum
or integration of all the
elements that are chosen
in this way will cancel
each other out

5. 10.6 Inertia for an Area
Consequently, the product of inertia for the
total area becomes zero
It also follows from the definition of Ixy that
the sign of this quantity depends on the
quadrant where the area is located
If the area is rotated from one quadrant to
another, the sign of Ixy will change

6. 10.6 Inertia for an Area
Parallel Axis Theorem
Consider shaded area where x’ and y’ represent
a set of axes passing through the centroid of the
area, and x’ and y’ represent
a set of axes passing through
the centroid of the area and x
and y represent a
corresponding set of parallel
axes

7. 10.6 Inertia for an Area
Parallel Axis Theorem
For the product of inertia of dA with respect
to the x and y axes
dI = ∫ ( x'+ d )( y '+ d )dA
xy x y
A
For the entire area,
dI = ∫ ( x'+ d )( y '+ d )dA
xy x y
A
= ∫ x' y ' dA + d ∫ y 'dA + d ∫ x'dA + d d ∫ dA
x y x y
A A A A
First term represent the product of inertia of
the area with respect to the centroidal axis

8. 10.6 Inertia for an Area
Parallel Axis Theorem
The integrals in the second and third terms are
zero since moments of the area are taken about
the centroidal axis
Forth integral represent the total area A
Thus,
I xy = I x ' y ' + Ad x d y
Important to maintain algebraic
signs for dx and dy while applying
equations

9. 10.6 Inertia for an Area
Example 10.7
Determine the product Ixy of the triangle.

10. 10.6 Inertia for an Area
Solution
Method 1
Differential element has thickness dx and area dA = y dx
Using parallel axis theorem for product of inertia of element
about the x, y axes
dI xy = dI xy + dA~~
xy
(~, ~ ) locates centroid of the element
x y
or origin of x’, y’ axes
Due to symmetry,
dI xy = 0 ~ = x, ~ = y / 2
x y

11. 10.6 Inertia for an Area
Solution
 y h   h 
dI xy = 0 + ( ydx) x  =  xdx  x x 
 2 b   2b 
h2 3
= 2 x dx
2b
Integrating
h2 b
I xy = 2
2b ∫
0
x 3dx
b2h 2
=
8

12. 10.6 Inertia for an Area
Solution
Differential element has thickness dy and
area dA = (b - x) dy
For centroid,
~ = x + (b − x) / 2 = (b + x) / 2, ~ = y
x y
For product of inertia of element
~ b+ x
dI xy = dI xy + dA~~ = 0 + (b − x)dy
xy y
 2 
 b   b + (b / h ) y  1  2 b2 2 
=  b − y dy   y = 2 y b − h 2 y dy
 
 h   2   

13. 10.6 Inertia for an Area
Solution
Integrating
1 h  2 b2 2 
I xy = ∫ y b − 2 y dy
2 0 
 h 

b2h2
=
8

14. 10.6 Inertia for an Area
Example 10.8
Compute the product of inertia of the beam’s
cross-sectional area about the x and y
centroidal axes.

15. 10.6 Inertia for an Area
Solution
Cross-sectional area considered as 3
rectangles area A, B and D
Due to symmetry, product
of inertia of each rectangle is
zero about each set of x’, y’
axes that passes through
the rectangle’s centroid

16. 10.6 Inertia for an Area
Solution
Parallel Axis Theorem
Rectangle A
I xy = I x ' y ' + Ad x d y
( )
= 0 + (300)(100)(− 250)(200) = −1.50 109 mm 4
Rectangle B
I xy = I x ' y ' + Ad x d y
= 0 + 0 = 0 mm 4

17. 10.6 Inertia for an Area
Solution
Parallel Axis Theorem
Rectangle D
I xy = I x ' y ' + Ad x d y
( )
= 0 + (300)(100 )(250 )(− 200 ) = −1.50 109 mm 4
For product of
inertia
( ) ( )
I xy = −1.50 109 + 0 + −1.50 109
= −3.00(10 )mm 9 4