Area moment of_intertia

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Area Moment of InertiaEngineering Mechanics- Distributed Forces

• Encounter in the engineering problems involving deflection, bending, buckling in statics

Moment of Inertia of an Area• Consider distributed forces F whose magnitudes are proportional to the

elemental areas on which they act and also vary linearly with the distance of from a given axis.

AA

2M y dA• The integral is already familiar from our study of centroids.

• The integral is known as the area moment of inertia, or more precisely, the second moment of the area.

y dA

y2 dA

Engineering Mechanics- Distributed Forces

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Moment of Inertia of an Area by Integration• Second moments or moments of inertia of

an area with respect to the x and y axes,

dAxIdAyI yx22

• Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes.

• For a rectangular area,3

31

0

22 bhbdyydAyIh

x

• The formula for rectangular areas may also be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

31


Polar Moment of Inertia

• The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs.

dArJ 20

• The polar moment of inertia is related to the rectangular moments of inertia,

xy II

dAydAxdAyxdArJ

222220


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Radius of Gyration of an Area• Consider area A with moment of inertia

Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix.

AIkAkI x

xxx 2

kx = radius of gyration with respect to the x axis

• Similarly,

AJkAkJ

AI

kAkI

OOOO

yyyy

2

2

222yxO kkk


Sample Problem

Determine the moment of inertia of a triangle with respect to its base.

SOLUTION:

• A differential strip parallel to the x axis is chosen for dA.dIx y2dA dA l dy

• For similar triangles,

lb h y

hl b h y

hdA b h y

hdy

• Integrating dIx from y = 0 to y = h,

h

hh

x

yyhhb

dyyhyhbdy

hyhbydAyI

0

43

0

32

0

22

43

12

3bhI x

Could a vertical strip have been chosen for the calculation? What is the disadvantage to that choice?


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Sample Problem

a) Determine the centroidal polar moment of inertia of a circular area by direct integration.

b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter of the area.

SOLUTION:

• An annular differential area element is chosen,

dJ O u 2dA dA 2 u du

JO dJ O u 2 2 u du 0

r 2 u 3du

0

r

42

rJO

• From symmetry, Ix = Iy,

JO Ix Iy 2Ix2

r 4 2I x

44

rII xdiameter

Why was an annular differential area chosen? Would a rectangular area have worked?


Sample ProblemEngineering Mechanics- Distributed Forces

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Determine the moment of inertia of the area under the parabola about the x-axis


Sample Problem

Parallel Axis Theorem

• Consider moment of inertia I of an area Awith respect to the axis AA’

dAyI 2

• The axis BB’ passes through the area centroid and is called a centroidal axis.

dAddAyddAy

dAdydAyI22

22

2

2AdII parallel axis theorem


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Parallel Axis Theorem• Moment of inertia IT of a circular area with

respect to a tangent to the circle,

4

45

224412

r

rrrAdIIT

• Moment of inertia of a triangle with respect to a centroidal axis,

IA A I B B Ad 2

I B B IA A Ad 2 112 bh3 1

2 bh 13 h 2

136 bh3


Sample ProblemEngineering Mechanics- Distributed Forces

Determine the moment of inertia of the half circle with respect to the x-axis

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Moments of Inertia of Composite Areas• The moment of inertia of a composite area A about a given axis is

obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.


Sample Problem

Determine the moment of inertia of the shaded area with respect to the x axis.

SOLUTION:

• Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.

• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.


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Sample ProblemSOLUTION:• Compute the moments of inertia of the bounding

rectangle and half-circle with respect to the x axis.

Rectangle: 463

313

31 mm102138120240 .bhI x

Half-circle: Moment of inertia with respect to AA’,

464814

81 mm10762590 .rI AA

23

2212

21

mm1072.12

90

mm 81.8a-120b

mm 2.383

90434

rA

ra

Moment of inertia with respect to x’,

46

2362

mm10207)238(107212107625

....AaII AAx

Moment of inertia with respect to x,

46

2362

mm10392

88110721210207

....AbII xx


Sample Problem • The moment of inertia of the shaded area is obtained by

subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.

46 mm109.45 xI

xI 46 mm102.138 46 mm103.92
