chapter nine laith batarseh...chapter nine laith batarseh mass moment of inertia definition mass...

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Chapter nine

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moment of inertia

Inertia

Inertia is defined as the body resistance to motion.

the magnitude of inertia is related proportionally to the body mass.

The mathematical relation that relates the motion of a body to its

mass is Newton’s 2nd law of motion

There are two types of inertia: the translational (mass) and the

rotational (mass moment of inertia).

The translational inertia represents the resistance for a translational

motion while the mass moment of inertia represents the resistance for a

rotational motion

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moment of inertia

Moment of inertia

Moment of inertia is the

moment produced when an

area is subjected to a

distributed load that vary in

linear fashion as shown in the

fig.

The load (P) is a function in

the distance y: P= Cy. Where C

is a constant

Definition

x

y

z

dF dA

Load P

moment of inertia

Moment of inertia

To find the moment, we define the differential moment (dM) as:

Definition

dACyydFdM .. 2

To find the moment, we define the differential force (dF) as:

dACydAPdF ..

To find the moment, integrate the differential moment:

dAyCdACyM .. 22

moment of inertia of the area ,Ix, about the x-axis

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moment of inertia

Moment of inertia

Mathematically

The moment of inertia about x-axis, Ix : Ax dAyI .2

The moment of inertia about y-axis, Iy : Ay dAxI .2

x

y

x

y

dA

A

moment of inertia

Moment of inertia

Mathematically

The polar moment of inertia, Jo : yx

Ao IIdArJ .2

x

y

x

y

dA

A

r

O

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moment of inertia

Example 10.1: rectangular area

Determine the moment of inertia for the rectangular area shown in Fig. with respect to (a) the x’ axis, (b) y’ axis

x'

y'

h/2

h/2

b/2b/2

xb

C

moment of inertia


Solution

x'

y'

h/2

h/2

b/2b/2

xb

dy'

C

Differential element

x'

y'

h/2

h/2

b/2b/2

xb

dx'

C

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moment of inertia


Solution

Part a:

'.dybdA

3

2/

2/

22

'12

1'.'.' bhdybydAyI

h

hA

x

moment of inertia


Solution

Part b:

'.dxhdA

3

2/

2/

22

'12

1'.'.' hbdxhxdAxI

b

bA

y

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Chapter nine

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Parallel - Axis Theorem


This method is used to find the moment of inertia

about an axis not passing in the centroid of the

area.

The axis under study is parallel to axis passing

through the centroid which the moment of inertia is

known.

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Derivation

Ay

Ay

Ax

Ayx

dAddAyddAyI

dAdyI

22

2

'.2.

.'

22

'

2

0'.2

.

yA

y

Ay

xA

AddAd

dAyd

IdAy

2

' yxx AdII

2

' xyy AdII 2AdJJ Co

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Composite Areas

Main concept Basic concept

Composite area are a series of simple shaped areas

The moment of inertia of the composite area is found by finding the moment of inertia for each shape and then sum these moments

to find the total moment of inertia, we chose a certain axis locates at

the centroid of one of the simple areas. Then use the parallel axis

theorem to find the moment of inertia of other areas about the chosen

axis


Example [1]: Prob. 10:38-40

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Example [1]: Prob. 10-38

Solution


Example [1]: Prob. 10:38

Solution

Centroid

mmA

Ayy 170

300100200100

30010025020010050''

Moment of inertia

23

'1, 5017010020010020012

1xI

23

'2, 17025030010030010012

1xI

46

'2,'1,' 10722 mmIII xxx

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Solution

23

'1, 05.01.02.01.02.012

1xI

23

'2, 25.03.01.03.01.012

1xI

43

'2,'1,' 1017.2 mIII xxx



Solution

4633107.91100300

12

1200100

12

1mmI y

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Composite Areas


Determine the distance y’ to the centroid of the beam’s cross-sectional area; then find the moment of inertia about the x’ axis.

Composite Areas


Solution

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Composite Areas


Solution

Centroid

inA

Ayy 2.2

14226

1442261''

Moment of inertia

423

'1, 3.2112.2262612

1inI x

423

'2, 7.362.24414112

12 inI x

4

' 587.363.21 inI x

Composite Areas


Determine the moment of inertia of the beam’s cross-sectional area about the y axis.

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Composite Areas


Solution

Composite Areas


Solution

Moment of inertia

4623

'1, 10415.760153001530012

12 mmxI y

4623

'2, 103.40151201201512

12 mmxI y

46

' 103.45413.4 mmxI y

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Chapter nine

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mass moment of inertia

Definition

Mass moment of inertia is the body resistance to angular motion

It is important concept when dealing with rotational bodies such as: gears,

turbine blades and shafts.

The mathematical formula for the mass moment for a body that has a

mass (m) is:

mdmrI .2

Where:-

•dm is a differential element lies at the

boundary of the body

•r is the radial distance from dm to the axis of

rotation

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Procedures

Select a differential element. This element cans be:

Shell element

Disc element

represent the differential element as a function of

distance r.


Example [1]: cylinder

Find the mass moment of inertia about z-axis

Solution

Differential element dIz:

drhrdmrdI z .2. 32

y

z

x

y

z

r

drShell

R

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Example [1]: cylinder

Integrate:

hRdrhrdII

R

zz

4

0

3

2

1.2

But the mass (m) equal: m = ρ.volume, or

hRm 2

Substitute m in Iz:

2

2

1mRI z


Special cases

2mRIC

+R

2

2

1mRIC

+R

The following cases are for common bodies. The mass moment of inertia is

found about z-axis and the origin point of the coordinate system (0,0,0) is

located at the centroid

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Special cases

2

5

2mRIC

+R

Cone

R

2

10

3mRIC


Special cases

22

12

1bamIC

Thin plate

b

a

2

12

1mlIC

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Parallel axis theorem

2mrII Cz

To find the mass moment of inertia about an axis not located at the centroid, the parallel axis theorem is used

Where r is the distance between the parallel axes, Iz is the mass moment of

about z-axis and IC is the mass moment of inertia about axis passes through

the centroid .


Composite bodies

njIIn

j

j ,...,2,1;1

If the body is a combination of (n) simple bodies each one has a mass moment of inertia, the mass moment of inertia for this composite body is simply the summation of all individual mass moment of inertia

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Chapter nine

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Example [1]: problem 10-100

Determine the mass moment of inertia of the

pendulum about an axis perpendicular to the

page and passing through point O. The slender

rod has a mass of 10 kg and the sphere has a

mass of 15 kg.

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This body consisted from to bodies: A (rood) and B(sphere) and the mass

moment of inertia about an axis pass through point O. this means applying the

parallel axis theorem.

Solution

Body A

22 45.01012

1

12

1 mlIC

2

2

. 675.0

2/45.0;

mkgI

RmRII

O

CO



Solution

Body B

22 1.0155

2

5

2 mRIC

2

2

. 5975.4

55.0;

mkgI

RmRII

O

CO

Total

2

,,, . 273.55975.4675.0 mkgIII BOAOtotO

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Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg.m2.



This body consisted from to 5 bodies: the plate and 4holes the mass moment of

inertia about an axis pass through point O. this means applying the parallel axis

theorem.

Solution

Mass moment of inertia about the centroid

Solid plate mass:- kgAmp 2.3)4.0)(4.0(20.20

Solid plate mass amount of inertia :- 22

, 4.04.02.312

1PCI

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Solution

Hole mass:- kgAmh 157.005.020.20 2

Holes mass amount of inertia :-

22

, 15.0157.005.0157.02

14hCI

Total mass moment of inertia about the centroid :-

2

,, . 07.0 mkgIII hCPCC



Solution

Total mass moment of inertia about the point O using parallel axis theorem :-

2.dmII CO

Where d is the distance between the centroid and the rotating point:-

45sin4.0d

m is the total mass of the system = 3.6 – (4)(0.157) = 2.572 kg

22kg.m 276.045sin4.0572.207.0 OI

chapter nine laith batarseh...chapter nine laith batarseh mass moment of inertia definition mass...

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