[email protected] ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed

[email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp10:Moment of

Inertia



Introduction Previously we considered distributed forces

which were proportional to the area or volume over which they act; i.e., Centroids• The resultant Force was obtained by summing or

integrating over the areas or volumes• The moment of the resultant about any axis was

determined by Calculating the FIRST MOMENTS of the areas or volumes about that axis– FIRST MOMENT = An Area/Volume/Mass INCREMENT

(or INTENSITY) times its LEVER ARM (kind of like F•d for Torque) Lever Arm = the ┴ Distance to Axis of Interest



Introduction cont. Next consider forces which are proportional

to the Area or Volume over which they act, but also VARY LINEARLY WITH DISTANCE from a given axis• It will be shown that the magnitude of the

resultant depends on the FIRST MOMENT of the force distribution with respect to the axis of interest

• The Equivalent point of application of the resultant depends on the SECOND moment of the Force distributionwith respect to the given axis



Area Moment of Inertia Consider distributed forces, F ,

whose magnitudes are proportional to the elemental areas, A, on which they act, and also vary LINEARLY with the Distance of A from a given axis.

Example: Consider a beam subjected to pure bending. A Mechanics-of-Materials Analysis Shows that forces vary linearly with distance from the NEUTRAL AXIS which passes through the section centroid. In This Case

Moment) (Resultantmoment second

Force) (Resultantmoment first 0

CONST Matl/Geom

22

dAydAykM

QdAydAykR

k

FyMAkyF

y



Example → 2nd Moment of Area Find the Resultant hydrostatic force on a submerged

CIRCULAR gate, and its Point of Application

y

k

Ay

I

dAy

dAyydAyRyMdM

ydAyydFdMAydAyRdF

ydApdAdFdAAAyApF

xxPPxx

xA

22

2

also taking

Let

• Recall That the gage Pressure, p, is Proportional to the Depth, y, by p = γy

• Consider an Increment of Area, A• Then The Associated Force Increment

F = pA• Locate Point of Resultant Application, yP, by

Equating Moments about the x-Axis

More onthis Later



Second Moment

The Areal Moments of Inertia Take these forms

In both the Integral and the Sum Observe that since u is a distance, so the expressions are of the form

Thus “moments of Inertia” are the SECOND Moment of Area or Mass

iivv AuIdAuI 22

IntensityArmLever 2 vI



Moment of Inertia by Integration SECOND MOMENTS or MOMENTS OF

INERTIA of an area with respect to the x and y axes: dAxIdAyI yx

22

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

ydxxdAxI

dyxaydAyI

y

x

22

22



Base Case → Rectangle Find Moment of Inertia for a Rectangular

Area

Apply This Basic formula to strip-like rectangular areas That are Most conveniently Parallel to the Axes

Consider a Horizontal Strip• dIx: b∙y → [a-x]∙y

• dIy → Subtract strips: b+ → a; b− → x

3

0

22

3

1bhbdyydAyI

h

x

dyxadyxdyadI

dIdIdI

dyxaydAydI

y

yyy

x

3333

22

3

1

3

1

3

1

Length x of StripaLength of Strip



Use BaseLine Case for Iy





Polar Moment of Inertia The polar moment of inertia is an important parameter in

problems involving torsion of cylindrical shafts, Torsion in Welded Joints, and the rotation of slabs

In Torsion Problems, Define a Moment of Inertia Relative to the Pivot-Point, or “Pole”, at O

dArJO2

Relate JO to Ix & Iy Using The Pythagorean Theorem

xy

O

II

dAydAxdAyxdArJ

22222



Radius of Gyration (k↔r) 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, Then Define the RADIUS OF GYRATION, r

A

IrArI yyyy 2

Similarly in the Polar Case the Radius of Gyration

222

2

yxO

OOOO

rrr

A

JrArJ

xr A

IrArI xxxx 2

yrOr



Example 1

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

SOLUTION:• Choose dA as a differential

strip parallel to the x axis

dyldAdAydI x 2

By Similar Triangles See that l in linear in y

dyh

yhbdA

h

yhbl

h

yh

b

l

Integrating dIx from y = 0 to y = h

h

hh

x

yyh

h

b

dyyhyh

bdy

h

yhbydAyI

0

43

0

32

0

22

4312

3bhI x



Example 2

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.

SOLUTION:• Choose dA as a differential

annulus of width du

By symmetry, Ix = Iy, so

rr

OO

O

duuduuudJJ

duudAdAudJ

0

3

0

2

2

22

2

4

2rJO

xxyxO IrIIIJ 22

2 4

4

4rII xdiameter



Parallel Axis Theorem With Respect to The

OffSet Axis AA’

dAddAyddAy

dAdydAyI AA22

22'

2

Now Consider the Middle Integral

BB' Moment WRT1st 22 ddAyd

But this is a 1st Moment about an Axis Thru the Centroid, Which By Definition = 0

Consider the moment of inertia I of an area A with respect to the axis AA’

Next Consider Axis BB’ That Passes thru The Area Centriod



Parallel Axis Theorem cont.

Next Consider the Last Integral

II

AdII

dAddAyddAyI

BB

BBAA

AA

'

2''

22'

so Axis, Centroidal a is BB'but

0

2

Thus The Formal Statement of the Parallel Axis Theorem

AdII 2• Note: This Theorem

Applies ONLY to a CENTROIDAL Axis and an Axis PARALLEL to it

AddAd 22 The Relation for I w.r.t.

the Centroidal Axis and a parallel Axis:



Parallel Axis Theorem Exmpls Find the Moment of inertia IT of a

circular area with respect to a tangent to the circle

445224

412 rrrrAdIIT

Find the Moment of inertia of a triangle with respect to a centroidal axis

3

361

2

31

213

1212

22

bhI

hbhbhAdII

AdIAdII

AA

BBAA



Composite Moments of Inertia 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.• This Analogous to Finding the Centroid of a Composite



Standard Shapes are Tabulated

44622 in 385mm 102.160in 11.20mm 7230



Example 3

The strength of a W14x38 rolled steel beam is increased by welding a ¾” plate to its upper flange.

Determine the Moment Of Inertia and Radius Of Gyration with respect to an axis which is parallel to the plate and passes through the centroid of the NEW section.

SOLUTION PLAN• Determine location of the

Centroid for the Composite section with respect to a coordinate system with origin at the centroid of the BEAM section.

• Apply the PARALLEL AXIS THEOREM to determine moments of inertia of the beam section and plate with respect to COMPOSITE SECTION Centroidal Axis.

• Calculate the Radius of Gyration from the Moment of Inertia of the Composite Section

4

2

in 385

in 11.20

beam

beam

I

A





Example 3 cont. Determine location of the CENTROID

of the COMPOSITE SECTION with respect to a coordinate system with origin at the CENTROID of the BEAM section. The First Moment Tabulation

12.5095.17

0011.20Section Beam

12.50425.76.75Plate

in ,in. ,in ,Section 32

AyA

AyyA

in. 792.2in 17.95

in 12.502

3

A

AyYAyAY

Calc the Centroidin. 792.2



Example 3 cont.2 Apply the parallel axis theorem to

determine moments of inertia of beam section and plate with respect to COMPOSITE SECTION centroidal axis.

Then the Composite Moment Of Inertia About the Composite Centroidal Axis, at x’

4

23

43

1212

plate,

4

22sectionbeam,

in 2.145

792.2425.775.69

in3.472

792.220.11385

AdII

YAII

xx

xx

4plate,section beam, in5.6172.145 3.472 xxxx IIII

in. 792.2



Example 3 cont.3 Finally Calculate the Radius of

Gyration for the Composite From Quantities Previously Determined

So Finally

2

4

in 17.95

in 5.617

A

Ik xx

in. 87.5xk

in. 792.2



WhiteBoard Work

Let’s WorkThis NiceProblem

Find the Area Moment of Inertia of the Parallelogram about its Centroidal axis y’

http://www.mupad.de/

http://www.mupad.de/



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhT

µs

T

µx

dx

dy





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[email protected] ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE Licensed