Engineering Mechanics: Engineering Mechanics: Statics Statics

Appendix A: Area Moments of Inertia

Moment of InertiaMoment of Inertia When forces are distributed continuously over an area, it is

often necessary to calculate moment of these forces about some axis (in or perpendicular to the plane of area)

Frequently, intensity of the distributed force is proportional to the distance of the line of action from the moment axis, p = ky

dM = y(pdA) = ky2dA

I is a function of geometry only!

Hydrostatic pressure

Bending moment in beam

Torsion in shaft

2M k y dA

Moment of inertia of area/ Second moment of area (I )

Rectangular moment of inertia

Polar moment of inertia

DefinitionsDefinitions

2

2

x

y

I y dA

I x dA

-- Moment of inertia about x-axis

2 x yzI r dA I I

• Notice that Ix, Iy, Iz involve the square of the distance from the inertia axis -- always positive!

• dimensions = L4 (ex. m4 or mm4)

Sample Problem A/1Sample Problem A/1 Determine the moments of inertia of the rectangular area about the

centroidal x0- and y0-axes, the centroidal polar axis z0 through C, the x-axis, and the polar axis z through O.

-- Must remember!: for a rectangular area,

: for a circular area, - see sample problem A/3

3 3,

12 12x ybh hbI I

4

4x yrI I

For an area A with moment of inertia Ix and Iy

Visualize it as concentrated into a long narrow strip of area A a distance kx from the x-axis. The moment of inertia about x-axis is Ix. Therefore,

The distance kx = radius of gyration of the area about x-axis

Radius of GyrationRadius of Gyration

2x xk A I

x xk I A

Similarly,

Do not confused with centroid C!

Radius of GyrationRadius of Gyration

y yk I A

z zk I A 2 2 2z x yk k k

Transfer of AxesTransfer of Axes Moment of inertia of an area about a noncentroidal axis

The axis between which the transfer is made must be parallel

One of the axes must pass through the centroid of the area

2 20( )x xdAdI y y d dA

2 20 0 2x x x xd dA d dA dI I y y dA

2x x xI I Ad

0Ay 0 0y and with the centroid on x0-axis

2

2

2

x x x

y y y

z z

I I Ad

I I Ad

I I AdParallel-axis theorems

Composite AreasComposite Areas Centroid of composite areas:

i i

i

A xx

A

Part Area, A

Sum A

x y AyAx

Ax Ay

i i

i

A yy

A

100 mm

400 mm

400 mm

100 mm

Composite AreasComposite Areas The moment of inertia of a composite area about a particular

axis is the sum of the moments of inertia of its component parts about the same axis.

I = I + Ad2

o The radius of gyration for the composite area cannot be added, k = I/A

xIPart Area, A dx dy Adx2 Ady

2yI

Sum A Adx2 Ady

2

xI yI

Example A/7Example A/7

Calculate the moment of inertia and radius of gyration about the x-axis for the shaded area shown

Products of InertiaProducts of Inertia Unsymmetrical cross section

Ixy = xydA

  may be positive, negative or zero Ixy = 0 when either the reference axes is an axis of symmetry

because x(-y)dA cancel x(+y)dA

Transfer of Axes 

Ixy = (x0+dy)(y0+dx)dA

Ixy = Ixy + Adx dy

  

Sample Problem A/8 & A/10Sample Problem A/8 & A/10 Determine the product of inertia of the area shown with respect

to the x-y axes. 

Rotation of AxesRotation of Axes To calculate the moment of inertia of an area

about an inclined axes 

Ix’ = y’2 dA = (ycos – xsin )2 dA

Iy’ = x’2 dA = (ysin – xcos )2 dA

  -- expand & substitute sin2 = (1- cos 2)/2

cos2 = (1+ cos 2)/2

 

'

'

' '

cos2 sin22 2

cos2 sin22 2

sin2 cos22

x y x yx xy

x y x yy xy

x yx y xy

I I I II I

I I I II I

I II I

Rotation of AxesRotation of Axes The angle which makes Ix’ and Iy’ either max or

min

dIx’/d = (Iy - Ix)sin 2 - 2Ixycos 2 = 0

The critical angle :

2tan2 xy

y x

I

I I

This equation gives two value of 2 [tan 2 = tan (2+) ] obtain two values for (differ by /2)

axis of minimum moment of inertia axis of maximum moment of inertia

called “Principal Axes of Inertia”

2 2max

2 2min

1( ) 4

2 21

( ) 42 2

x yx y xy

x yx y xy

I II I I I

I II I I I

Mohr’s Circle of InertiaMohr’s Circle of Inertia

1. Draw x-axis as I and y-axis as Ixy

2. Plot point A at (Ix, Ixy) and B at (Iy, -

Ixy)

3. Find the center of the circle at O2 2R OS AS

R

S ImaxImin

6. Imax = O + R and Imin = O - R

tan2ASOS

5. Angle 2 is found from AS and OS as

4. Radius of the circle is OA or OB

Sample Problem A/11Sample Problem A/11

Determine the orientation of the principle axes of inertia through the centroid of the angle section and determine the corresponding maximum and minimum moments of inertia.