# Product of Inertia and Mohr's Circle

**Topics:**Second moment of area, Perpendicular, Euclidean geometry

**Pages:**27 (3351 words)

**Published:**June 27, 2012

Lecture 1

LECTURE 1 TOPICS

I. Product of Inertia for An Area

Definition Parallel Axis Theorem on Product of Inertia Moments of Inertia About an Inclined Axes Principal Moments of Inertia Mohr’s Circle for Second Moment of Areas

II. Unsymmetrical Bending II Unsymmetrical Bending

Unsymmetrical Bending about the Horizontal and Vertical Axes of the Cross Section Unsymmetrical Bending about the Principal Axes

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Lecture 1, Part 1

Product of Inertia for an Area

Consider the figure shown below

y x A dA y x Product of Inertia of A wrt x and y axis: Product of Inertia of Element dA:

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Product of Inertia for an Area

Consider the figure shown below

y x A dA y x Unit: length4 – m4, mm4, ft4, in4 g NOTE: 1. Ixy can be positive, negative or zero. 2. The product of inertia of an area wrt any two orthogonal axes is zero when either of the axes is an axis of symmetry. Product of Inertia of A wrt x and y axis:

Product of Inertia for an Area

Parallel Axis Theorem Parallel‐Axis Theorem

y’ y x’ Product of Inertia of A wrt x and y axis: dA C dy x dx y’ x’ ’ Product of Inertia of Element dA:

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Product of Inertia for an Area

Parallel Axis Theorem Parallel‐Axis Theorem

y’ y x’

dA C dy

y’ x’ ’

x dx

The product of inertia of an area wrt any two perpendicular axes x and y is equal to the product of inertia of the area wrt a pair of centroidal axes parallel to the x and y axes added to the product of the area and the two centroidal distances from the x and y axes.

Product of Inertia for an Area

Example 1

y b

Determine the following: a) Product of Inertia, Ixy

h

x

b) Product of Inertia, Ix’y’ , with ) , xy respect to a pair of centroidal axes x’ and y’ parallel to the given axes x and y

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Product of Inertia for an Area

Example 1

y x dA h y x dy b

Solution: a) Product of Inertia, Ixy • Consider the strip ⎛ x⎞ dI xy = ⎜ ⎟ ydA ⎝ 2⎠ • The area, dA, is equal to q dA = x dy • Substituting gives x2 ⎛ x⎞ dI xy = ⎜ ⎟ y ( x dy ) = ydy 2 ⎝ 2⎠

Product of Inertia for an Area

Example 1

y x dA h y x dy b

Solution: a) Product of Inertia, Ixy • But x is a function of y, and using similar triangles b x b ⇒ x= y = h y h • Substitute x to dIxy gives b2 x2 = 2 y 3dy dI xy = ydy 2h 2

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Product of Inertia for an Area

Example 1

y x dA h y x dy b

Solution: a) Product of Inertia, Ixy • Integrating I xy = ∫ I xy = I xy = h 0

b2 3 y dy 2h 2

b2 h 3 yd dy 2h 2 ∫0 b2h2 8

Product of Inertia for an Area

Example 1

y’ y b/3

Solution: b) Product of Inertia, Ix’y’ • Parallel Axes Theorem I xy = I x ' y ' + Ad x d y

x’

I x ' y ' = I xy − Ad x d y

I x'y ' = b2h2 8 ⎛ 1 ⎞⎛ b ⎞⎛ 2h ⎞ − ⎜ bh ⎟⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠⎝ 3 ⎠

C 2h/3 x

I x'y' =

b2h2 72

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Moments of Inertia about Inclined Axes

Transformation Equations:

y y’’

dA

y

θ

θ

y’

y cosθ

Moments and Product of Inertia of dA wrt x’ and y’ axes:

x’

x sinθ

x

θ y sinθ

x

x cosθ

x’

Moments of Inertia about Inclined Axes

Expanding and integrating each expression and realizing that

Gives

These equations may be simplified using the trigonometric identities

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Moments of Inertia about Inclined Axes

Which then gives

y y’

A θ

x’

θ

x Moments of Inertia of An Area About an Inclined Axes x’ and y’ in terms of Ix, Iy, Ixy and θ

Moments of Inertia about Inclined Axes

Adding the first and second equations

y y’ y’’

A

x’

o

x x’’

The sum (also called the polar moment of inertia) Ix’ + Iy’ is a constant. Since sum is constant, Ix’ will be maximum and the corresponding Iy’ will be minimum for one particular value of θ.

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Principal Moments of Inertia

First Objective: Determine the orientation of the axes or angle θ which the moments Ix’ and Iy’ are maximum or minimum. y’

y

DEFINITION: 1)...

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