Parallel axis theorem and their use on Moment Of Inertia

you can learn how to use parallel axis theorem to calculate the moment of inertia about centroidal axis and its transfer rule.

1. –h
2
h
X
X
y
–h
2
b h
X
X'
X
X'
y
Statement and Derivation of parallel axis theorem
Parallel axis theorem states that the moment of inertia of a plane area about any axis parallel to the centroidal axis of that area
is equal to the sum of moment of inertia about a parallel centroidal axis of that plane area and the product of the area and
square of the distance between the two axes.


[Ix'x' = I
xx +Ah
2
]

Proof:
Consider a lamina having area 'A', X – X be the centroidal axis of this lamina and x'- x' be the another axis which is parallel to the
centroidal x – x axis and at a distance of h from centroidal axis . Consider an elemental area dA at a distance 'y' from the
centroidal axis x – x.
Ixx = M.O.I of the lamina about x - x axis (centroidal axis)
we know, I
XX = y
2
dA
Here,
I
X'X' = moment of inertia of the lamina about x'-x' axis
so,
I
X'X' = (h+y)
2
dA [(h+y) is the distance of elemental area from x' – x' axis.]
=(h
2
+2hy +y
2
) dA.
=h
2
dA +2h.ydA+ y
2
dA
= Ah
2
+2h× 0 + I
XX
[Ix'x' = I
xx +Ah
2
]

Hence proved the theorem


Q. Determine moment of inertia about centroidal xx and yy axes of the plane figure shown in figure below.
Axis दिएको छैन भने जता मान्िा पनन हुन्छ !
तर सके सम्म object first quadrant मा पने गरर मान्नु !
Solution:
first divide the whole figure into known geometrical figure and taking left bottom corner as a origin.
For figure (i) A
1 = 0.14 × 1.5 = 0.21 m
2

x
1 =
0.14
2
= 0.07m
y
1 =
1.5
2
= 0.75 m
20cm
1.5m
14cm
1.2m 1.17m
1.5m
0.91m
0.13m
1.06m0.14m
20cm
1
2
3
14cm
(y.dA = 1
st
moment of area =
centroid = 0
i.e  ydA=


-h/2
h/2
y.xdy = 0 )

2. y1
y
h
x= 0.39m 1.17m
1.5m
0.91m
0.13m
1.06m0.14m
20cm
1
2
3
14cm
For figure (2)
A
2 = 1.06 × 0.13 = 0.1378m
2

x
2 =




0.14 +
1.06
2
= 0.67 m
y
2 =
0.13
2
= 0.065m
For figure (3)
A
3 =
1
2
× 0.91 × 1.17 = 0.532m
2

x
3 =




0.14 +
0.91
3
= 0.443m
y
3 =




0.13 +
1.17
3
= 0.52m

-
X =
A
1x
1 + A
2 x
2 + A
3x
3
(A
1 + A
2 + A
3)
=
0.21 × 0.07 + 0.1378 × 0.67 + 0.532 × 0.443
(0.21 + 0.1378 + 0.532)

= 0.39m


-
Y =
A
1y
1 + A
2 y
2 + A
3y
3
(A
1 + A
2 + A
3)
=
0.21 × 0.75 + 0.1387 × 0.065 + 0.532 × 0.52
(0.21 + 0.1378 + 0.532)

= 0.503m
M.O.I about centriodal x - x axis
Ixx = [ ]Ixx rectangle
1 + [ ]Ixx rectangle
2 + [ ]Ixx rectangle
3
M.O.I about centriodal axis is calculating by using parallel axis theorem.

(Note: i.e. M.O.I about centroidal axis of individual figure is transferred in to the centroidal axis of whole figure.)
Ixx =




b
1h
1
3

12
+ A
1 (y
1 –
–
y)
2
+




b
2h
2
3

12
+ A
2 (y
2 –
–
y)
2
+




b
3h
3
3

36
+ A
3 (y
3 –
–
y)
2

=




0.14 × 1.5
3
12
+ 0.21 (0.75 – 0.503)
2
+




1.06 × 0.13
3
12
+ 0.1378 (0.06575 – 0.503)
2

+




0.91 × 1.17
3
36
+ 0.532 (0.53 – 0.503)
2

= 0.0521 + 0.0266 + 0.0408
[Ixx= 0.11957 m
4
]



Iyy = [ ]Iyy rectangle
1 + [ ]Iyy rectangle
2 + [ ]Iyy rectangle
3
=




h
1b
1
3

12
+ A
1 (x
1 –
–
x)
2
+




h
2b
2
3

12
+ A
2 (x
2 –
–
x)
2
+




h
3b
3
3

36
+ A
3 (x
3 –
–
x)
2

=




1.5 × 0.14
3
12
+ 0.21 (0.07 – 0.39)
2
+




0.13 × 1.06
3
12
+ 0.1378 (0.67 – 0.39)
2
+




1.17 × 0.91
3
36
+ 0.532 (0.443 – 0.39)
2

= 0.02184 + 0.0237 + 0.0276
= 0.07317 m
4