Parallel Axis TheoremStevenVukazich

SanJoseStateUniversity

x

Recall the Definition of the Moment of Inertia of an Area About an Axis

𝑑𝐴 y'

𝐼$ = &𝑦(𝑑𝐴�

�

= & 𝑦* + 𝑑 (𝑑𝐴�

�

x'

y d

Consider an axis x’ that is parallel to the x axis and passes through the centroid of the area. The distance between the two parallel axes is d

y = y' + d

𝐶

x

Expand and Examine Terms

𝑑𝐴 y'

𝐼$ = & 𝑦* + 𝑑 (𝑑𝐴 =&𝑦′(�

�

𝑑𝐴�

�

+ 2𝑑&𝑦*𝑑𝐴 + 𝑑(&𝑑𝐴�

�

�

�

x'

y d𝐶

Moment of Inertia of the area about the x' axis

First moment of the area about the x' axis = 0

Area

x

Parallel Axis Theorem

𝑑𝐴 y'𝐼$̅* = &𝑦′(

�

�

𝑑𝐴x'

y d𝐶

Centroidal Moment of Inertia

𝐼$ = 𝐼$̅* + 𝑑(𝐴

General Form

𝐼 = 𝐼 ̅ + 𝑑(𝐴

Parallel Axis Theorem

If we know the moment of inertia of a body about an axis passing through its centroid, we can calculate the body’s moment of inertia about any parallel axis

x

y

Example Problem

𝑏

ℎ

Find the Moment of Inertia of the of the shaded area about the x and yaxes shown. Use the Parallel Axis Theorem.

Note that this we have already found Ix , Iy and the location of the centroid for this shape using integration.

Moment of Inertia About Centroidal Axes

x𝐶

ℎ

13ℎ

23𝑏

y

b

x'

y'

Use Tabulated Solution for 𝐼 ̅

𝐼$̅* =136𝑏ℎ5

𝐼6̅* =136ℎ𝑏5

Moment of Inertia About Centroidal Axes

x𝐶

ℎ

13ℎ = 𝑑$

23𝑏 = 𝑑6

y

b

x'

y'

𝐼$̅* =136𝑏ℎ5

𝐼6̅* =136ℎ𝑏5

𝐼 = 𝐼 ̅ + 𝑑(𝐴

𝐴 =12𝑏ℎ

𝐼$ = 𝐼$̅* + 𝑑$(𝐴 =

136𝑏ℎ5 +

13ℎ

( 12𝑏ℎ

Moment of Inertia About the xAxis

𝐼$ = 𝐼$̅* + 𝑑$(𝐴 =

136𝑏ℎ5 +

13ℎ

( 12𝑏ℎ

𝐼$ =136𝑏ℎ5 +

118𝑏ℎ5

𝐼$ =336𝑏ℎ5 =

112𝑏ℎ5

Agrees with both the tabulated solution and our result from integration

x

y

𝑏

ℎ

Moment of Inertia About the yAxis

𝐼6 = 𝐼6̅* + 𝑑6(𝐴 =

136ℎ𝑏5 +

23𝑏

( 12𝑏ℎ

𝐼6 =136ℎ𝑏5 +

418ℎ𝑏5

𝐼6 =936ℎ𝑏5 =

14ℎ𝑏5

Agrees with our resultfrom integration

x

y

𝑏

ℎ