center of mass and centroids :: guidelines notes/me101-lecture17-kd.pdf · center of mass and...

Center of Mass and Centroids :: Guidelines

Centroids of Lines, Areas, and Volumes

1.Order of Element Selected for Integration

2.Continuity

3.Discarding Higher Order Terms

4.Choice of Coordinates

5.Centroidal Coordinate of Differential Elements

A

dAzz

A

dAyy

A

dAxx

ccc

V

dVzz

V

dVyy

V

dVxx

ccc

L

zdLz

L

ydLy

L

xdLx

1ME101 - Division III Kaustubh Dasgupta

Center of Mass and Centroids

Theorem of Pappus: Area of Revolution- method for calculating surface area generated by revolving a plane curve

about a non-intersecting axis in the plane of the curve

Surface Area

Theorem of Pappus can also be used to determine centroid of plane

curves if area created by revolving these figures @ a non-intersecting

axis is known

Area of the ring element: circumference times dL

dA = 2πy dL

Total area,

or

is the y-coordinate of the centroid C for the line

of length L

Generated area is the same as the lateral area of a

right circular cylinder of length L and radius

dLyA 2

dLyLy LyA 2

y

y

LyA

If area is revolved

through an angle θ<2π

θ in radians


Center of Mass and Centroids

Theorem of Pappus: Volume of Revolution- method for calculating volume generated by revolving an area about a

non-intersecting axis in the plane of the area

Volume

Theorems of Pappus can also be used to determine centroid of plane areas if

volume created by revolving these figures @ a non-intersecting axis is known

Volume of the ring element: circumference times dA

dV = 2πy dA

Total Volume,

or

is the y-coordinate of the centroid C of the

revolved area A

Generated volume is obtained by multiplying the

generating area by the circumference of the circular

path described by its centroid.

dAyV 2

dAyAy AyV 2

y

AyV

If area is revolved

through an angle θ<2π

θ in radians


Area Moments of Inertia

• Previously considered distributed forces which were proportional

to the area or volume over which they act.

- The resultant was obtained by summing or integrating over

the areas or volumes

- The moment of the resultant about any axis was determined

by computing the first moments of the areas or volumes

about that axis

• Will now consider forces which are proportional to the area or

volume over which they act but also vary linearly with distance

from a given axis.

- the magnitude of the resultant depends on the first moment of

the force distribution with respect to the axis.

- The point of application of the resultant depends on the

second moment of the distribution with respect to the axis.



• Example: Consider a beam subjected to pure bending.

Internal forces vary linearly with distance from the

neutral axis which passes through the section centroid.

moment second

momentfirst 022

dAydAykM

QdAydAykR

AkyF

x

• Consider distributed forces whose magnitudes are

proportional to the elemental areas on which they

act and also vary linearly with the distance of

from a given axis.

F

A

A

First Moment of the whole section about the x-axis =

A = 0 since the centroid of the section lies on the x-axis.

Second Moment or the Moment of Inertia of the beam

section about x-axis is denoted by Ix and has units of

(length)4 (never –ve)

y


Area Moments of Inertia 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


Area Moments of Inertia by Integration

• For a rectangular area,

331

0

22 bhbdyydAyIh

x

• The formula for rectangular areas may also

be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

31



• The polar moment of inertia is an important

parameter in problems involving torsion of

cylindrical shafts and rotations of slabs

dArIJ z

2

0

• The polar moment of inertia is related to

the rectangular moments of inertia,

xy

z

II

dAydAxdAyxdArIJ

22222

0

Polar Moment of Inertia

Moment of Inertia of an area is purely a mathematical

property of the area and in itself has no physical significance.


Area Moments of InertiaRadius 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.

A

IkAkI x

xxx 2

kx = radius of gyration with respect

to the x axis

• Similarly,

A

JkkAkAkIJ

A

IkAkI

OzOzOzO

y

yyy

22

2

2222

yxzO kkkk

Radius of Gyration, k is a measure of distribution of area from a reference axis

Radius of Gyration is different from centroidal distances


Area Moments of InertiaExample: 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.

dyldAdAydIx 2

• For similar triangles,

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

43

12

3bhI x


Area Moments of InertiaExample: 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:

• An annular differential area element is chosen,

rr

OO

O

duuduuudJJ

duudAdAudJ

0

3

0

2

2

22

2

4

2rJO

• From symmetry, Ix = Iy,

xxyxO IrIIIJ 22

2 4

4

4rII xdiameter


Area Moments of InertiaParallel Axis Theorem

• Consider moment of inertia I of an area A

with respect to the axis AA’

dAyI 2

• The axis BB’ passes through the area centroid

and is called a centroidal axis.

• Second term = 0 since centroid lies on BB’

(∫y’dA = ycA, and yc = 0

dAddAyddAy

dAdydAyI

22

22

2

2AdII Parallel Axis theorem

Parallel Axis theorem:

MI @ any axis =

MI @ centroidal axis + Ad2

The two axes should be

parallel to each other.


dAddAyddAy

dAdydAyI

22

22

2

Area Moments of InertiaParallel 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,

3

361

2

31

213

1212

2

bh

hbhbhAdII

AdII

AABB

BBAA

• 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.


Area Moments of Inertia: Standard MIs


Area Moments of InertiaExample:

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.


Area Moments of InertiaExample: Solution

SOLUTION:

• Compute the moments of inertia of the bounding

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

Rectangle:

463

3

13

3

1 102.138120240 mmbhI x

Half-circle:

moment of inertia with respect to AA’,

464

814

81 mm1076.2590 rI AA

2mm

rA

mm 81.8a-120b

mm r

a

3

2

2

12

2

1

1072.12

90

2.383

904

3

4

Moment of inertia with respect to x’,

4

AAx

mm

AaII

6

2362

1020.7

2.381072.121076.25

moment of inertia with respect to x,

4

xx

mm

AbII

6

2362

103.92

8.811072.121020.7


Area Moments of InertiaExample: Solution

• 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.

46mm109.45 xI

xI 46mm102.138 46mm103.92


Area Moments of InertiaProducts of Inertia: for problems involving unsymmetrical cross-sections

and in calculation of MI about rotated axes. It may be +ve, -ve, or zero

• Product of Inertia of area A w.r.t. x-y axes:

x and y are the coordinates of the element of area dA=xy

dAxyI xy

• When the x axis, the y axis, or both are an

axis of symmetry, the product of inertia is

zero.

• Parallel axis theorem for products of inertia:

AyxII xyxy

+ Ixy

+ Ixy - Ixy

- Ixy

Quadrants


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