Part-10 (Moments of Inertia)

7/30/2019 Part-10 (Moments of Inertia)

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R. GANESH NARAYANAN, IITG

XV = ( xc dv) YV = ( yc dv) ZV = ( zc dv)Centroid of volume:

XA = ( xc dA) YA = ( yc dA) ZA = ( zc dA)

Centroid of area:

Moments of inertia : The moment of inertia of an object about a givenaxis describes how difficult it is to change its angular motion about that axis.

First moment of volume

w.r.t. yz plane

Symmetry plane

Centroidof volume xc dv = 0


2/29


Consider a beam subjected to pure bending.

Internal forces vary linearly with distancefrom the neutral axis which passes through

the section centroid.

X-axis => neutral axis => centroid of sectionpasses (one side tension; other compression)

F = k y A vary linearly with distance y

momentsecond

momentfirst0

22 ==

====

=

dAydAykM

QdAydAykR

AkyF

x

r

MX = y F = k y2 A;

Moment of inertia of beam

section w.r.t x-axis, IX (+VE)


3/29


Second moments ormoments of inertia of an area

with respect to thex andy axes,

== dxIdAyI yx 22

For a rectangular area,

331

0

22bhbdyydAyI

h

x ===

IY = x2 dA = x2 h dx = 1/3 b3h0

b

Rectangular moment of inertia


4/29


Thepolar 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 rectangularmoments of inertia,

xy II

dAydAxdAyxdArJ

+=

+=+== 222220

Polar moment of inertia

This is polar MI w.r.t. O


5/29


Consider areaA with moment of inertiaIx. Imagine

that the area is concentrated in a thin strip parallel to

thex axis with equivalentIx.

IkAkI xxxx == 2

kx = radius of gyration with respect to thex axis

J

kAkJ

A

IkAkI

O

OOO

yyyy

==

==

2

2

222yxO kkk +=

Radius of gyration

y kx

C

b

h

Kx and y are different

y = h/2

Kx = h/3


6/29


Beer/Johnston

( )h

hh

x

yyh

h

b

dyyhyh

bdy

h

yhbydAyI

0

43

0

32

0

22

43

=

=

==

12

3

bhIx=

Determine the moment of inertia of atriangle with respect to its base.

For similar triangles,

dyh

yhbdAh

yhblh

yhbl ===

dA = l dy

Determination of MI by area of integration

About centroidal axis (X, Y): Ix = 1/12 bh3; Iy = 1/12 b3h


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yMI of rectangular area:

dA = bdy

xb

h

y

dy

Ix = y2 dA = y2 bdy = 1/3 bh3; Iy = 1/3 hb30

h

MI - Ix and Iy for elemental strip:

y dIx = 1/3 dx (y3) = 1/3 y3 dx

dIy = x2dA = x2y dx or 1/3 x3dy

x

YX

dA = Ydx From this, MI of whole area canbe calculated by integration

dx

dy

X

Y


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y

x

a

by = k x5/2

Find MI w.r.t Y axis

Beer/Johnston (9.1)


9/29


Triangle: bh3/12 (about base)

Circular area: /4 r4 (about dia)

Rectangular area: bh3/3 (about base)


10/29


Parallel axis theorem

Consider moment of inertiaIof an area A with

respect to the axisAA

= dAyI 2

The axisBBpasses through the area centroid

and is called a centroidal axis.

( )

++=

+==

dAddAyddAy

dAdydAyI

22

22

2

2

d+=

dA

A A

y

CB B

d

y

C Centroid

BB Centroidal axis

MI of area withcentroidal axis

0

Parallel axis theorem

Jo = Jc + Ad2First moment ofarea w.r.t. BB


11/29


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.

x

y

It should be noted that the radius of gyration of a composite area isnot equal to sum of radii of gyration of the component areas


12/29


MI of some common geometric shapes


13/29


Moment of inertiaITof a circular area with respectto a tangent to the circle T,

445

224412

r

rrrAdIIT

=

+=+=

Application 1:

Application 2:Moment of inertia of a triangle with respect to acentroidal axis,

( )3

361

2

31

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA

===

+=

IDD = IBB + ad2 = 1/36 bh3 + 1/2bh (2/3h)2 = bh3


14/29


Find the centroid of the area of the un-equal Z section. Findthe moment of inertia of area about the centroidal axes

shames

Ai xi yi Aixi Aiyi

2x1=2 1 7.5 2 158x1=8 2.5 4 20 32

4x1=4 5 0.5 20 2

Ai = 14 Aixi = 42 Aiyi = 49

12

3

Xc = 42/14 = 3 in.; Yc = 49/14 = 3.5in

y

x

1

6

2 1 4

1

1

2

3

Xc, Yc; 3, 3.5

Ixcxc = [(1/12)(2)(13)+(2)(42)] + [(1/12)(1)(83)+(8)(1/2)2] +[(1/12)(4)(13)+(4)(32)] = 113.16 in4

Similarly, Iycyc = 32.67 in4


15/29


Beer/Johnston:

Determine the moment of inertia of the shadedarea with respect to thex axis.

Rectangle:

( )( ) 46313

31 mm102.138120240 === bhIx

3

Half-circle:

moment of inertia with respect toAA,

( ) 464814

81 mm1076.2590 === rIAA


16/29


( )( )

( )

23

2

212

21

mm1072.12

90

mm81.8a-120b

mm2.38

3

904

3

4

=

==

==

===

rA

ra

moment of inertia with respect tox,

46

362

mm1020.7

1072.121076.25

=== AaII AAx =25.76x106 (12.72x103) (38.2)2

moment of inertia with respect tox,

( )46

2362

mm103.92

8.811072.121020.7

=

+=+= AbII xx

46mm109.45 =xI

I = 46mm102.138 46mm103.92


17/29


Product of inertia, Ixy

= dAxyIxy[Similar to Ixx (or Ix), Iyy (or Iy)]

When thex axis, they axis, or both are an axis of

symmetry, the product of inertia is zero.

The contributions to Ixy of dA and dA will cancel out

Parallel axis theorem for products of inertia:

AyxIIxyxy

+=

Centroid C is defined by x, y


18/29


Moment of inertia, Product of inertia about rotated axes

Given

=

==

dAxyI

dAxIdAyI

xy

yx22

we wish to determine moments and product

of inertia with respect to new axesxandy

x, y rotated to x, y

2cos2sin2

2sin2cos22

2sin2cos

22

xy

yx

yx

xyyxyx

y

xyyxyx

x

I

II

I

IIIII

I

IIIII

I

+

=

+

+

=

++

=

The change of axes yields

Ix+Iy = Ix+Iy


19/29


y

xa

ImaxImin

Assume Ixx, Iyy, Ixy are known for the reference axes x, y

At what angle of , we have maximum and minimum I

Minimum angle will be at right angles to maximum angleThese axes are called Principal axes & MI are Principal MI

Principal axes & Principal MI

Imax, min = (Ix+Iy/2) (Ix-Iy/2)2 + Ixy2

tan 2 = 2Ixy / (Iy-Ix)


20/29


For the section shown, the moments of inertia with

respect to thex andy axes areIx = 10.38 cm4 andIy= 6.97 cm4.

Determine (a) the orientation of the principal axes

of the section about O, and (b) the values of the

principal moments of inertia about O.

Apply the parallel axis theorem to each rectangle,

+= AyxII yxxyNote that the product of inertia with respect to centroidal axes parallel to

thex, y axes is zero for each rectangle.

56.6

28.375.125.15.1

0005.1

28.375.125.15.1cm,cm,cm,cmArea,Rectangle

42

=

+

+

Ayx

III

II

IAyxyx


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( )

=

+=

=

=

255.4and4.752

85.397.638.10

56.6222tan

m

yx

xym

II

I

== 7.127and7.37 mm

( )22

22

minmax,

56.6

2

97.638.10

2

97.638.10

22

+

+

=

+

+= xy

yxyxI

IIIII

4min

4max

cm897.1

cm45.15

==

==

II

II

b

a


22/29


Cables With Concentrated Loads

Cables are applied as structural elements

in suspension bridges, transmission lines,

aerial tramways, guy wires for high

towers, etc. For analysis, assume:

a) concentrated vertical loads on given

vertical lines,

b) weight of cable is negligible,c) cable is flexible, i.e., resistance to

bending is small,

d) portions of cable between successive

loads may be treated as two force

members

Wish to determine shape of cable, i.e.,

vertical distance from supportA to each

load point.


23/29

R. GANESH NARAYANAN, IITG constantcos ===

xATT

Four unknowns are involved and threeequations of equilibrium are not sufficient to

determine the reactions.

Consider entire cable as free-body. Slopes of

cable atA andB are not known - two reaction

components required at each support.

For other points on cable,2yields02 yMC =

yxyx TTFF ,yield0,0 ==

Additional equation is obtained by

considering equilibrium of portion of cableAD and assuming that coordinates of pointD

on the cable are known. The additional

equation is .0 =DM


24/29


Cables With Distributed Loads

For cable carrying a distributed load:

a) cable hangs in shape of a curve

b) internal force is a tension force directed along

tangent to curve.

Consider free-body for portion of cable extending

from lowest point Cto given pointD. Forces are

horizontal force T0

at C and tangential force TatD.

From force triangle:

0

220

0

tan

sincos

T

WWTT

WTTT

=+=

==

Horizontal component ofT is uniform over cable. Vertical component ofT is equal to magnitude ofW

measured from lowest point.

Tension is minimum at lowest point and maximum

atA andB.


25/29


Parabolic Cable

Consider a cable supporting a uniform, horizontally

distributed load, e.g., support cables for a

suspension bridge.

With loading on cable from lowest point Cto a

pointD given by internal tension force

magnitude and direction are,wxW =

0

222

0 tan T

wx

xwTT =+=

Summing moments aboutD,

02

:0 0 == yTx

wxMD

0

2

2Twxy =

or

The cable forms a parabolic curve with

vertical axis and its vertex at the origin of

coordinates.


26/29


The cableAEsupports three vertical loads from the

points indicated. If point Cis 5 m below the leftsupport, determine (a) the elevation of pointsB and

D, and (b) the maximum slope and maximum

tension in the cable.

Determine reaction force components at

A from solution of two equations formed

from taking entire cable as free-body

and summing moments aboutE, and

from taking cable portionABCas a free-body and summing moments about C.

Calculate elevation ofBby considering

AB as a free-body and summing

momentsB. Similarly, calculate

elevation ofD usingABCD as a free-

body.

Evaluate maximum slope and

maximum tension which occur inDE.


27/29


Determine two reaction force components atA

from solution of two equations formed from

taking entire cable as a free-body and summing

moments aboutE,

( ) ( ) ( )

06606020

041512306406020

:0

=+

=+++=

yx

yx

E

AA

AA

and from taking cable portionABCas a

free-body and summing moments about C.

( ) 0610305

:0

=+

=yx

C

AA

M

Solving simultaneously,

kN5kN18 == yx AA


28/29


Calculate elevation ofBby consideringAB as

a free-body and summing momentsB.

( ) ( ) 020518:0 == BB yM

m56.5=y

Similarly, calculate elevation ofD using

ABCD as a free-body.

( ) ( ) ( ) ( ) 0121562554518

:0

=++

=Dy

M

m83.5=D

y


29/29


Evaluate maximum slope and

maximum tension which occur inDE.

15

7.14

tan = = 4.43

cos

kN18

max

=T kN8.24max =T

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Part-10 (Moments of Inertia)