Mechanics of Material-Moments of Inertia

8/12/2019 Mechanics of Material-Moments of Inertia

1/20

1

Universiti Tun Hussein Onn MalaysiaUniversi ti Tun Hussein Onn Malaysia(UTHM)(UTHM)

Engineering Mechanics :Engineering Mechanics :MECHANICS OF MATERIALMECHANICS OF MATERIAL

Lecture 08

MOHD NO RIHAN B IBRAHIM JKM FKMP


2/20

2

Todays Objectives:Students will be able to:1. Define the moments of inertia (MoI) for an area.2. Determine the MoI for an area by integration.3. Apply the parallel-axis theorem.

4. Determine the moment of inertia (MoI) for a composite area.Topics:

MoI: Concept and Definition Parallel-Axis Theorem

MoI by IntegrationMethod for Composite Areas

MOMENTS OF INERTIAMOMENTS OF INERTIA


3/20

3

Reading QuizReading Quiz

1. The definition of the Moment of Inertia for an area involves anintegral of the form

A) x dA. B) x2 dA.

C) x2

dm. D) m dA.

2. Select the SI units for the Moment of Inertia for an area.

A) m 3

B) m4

C) kgm 2

D) kgm 3


4/20

4

10.0 Application

Many structural members like beams and columns have crosssectional shapes like I, H, C, etc..

Why do they usually not havesolid rectangular, square, or

circular cross sectional areas?What primary property of thesemembers influences designdecisions?

How can we calculate this property?


5/20

5

Many structural members are madeof tubes rather than solid squares orrounds.

Why?

What parameters of the cross sectionalarea influence the designersselection?

How can we determine the value ofthese parameters for a given area?

10.0 Application


6/20

6

Cross-sectional areas of structuralmembers are usually made ofsimple shapes or combination ofsimple shapes.

Is there a simpler method fordetermining the MoI of suchcross-sectional areas as comparedto the integration method?

If yes, can you describe the

method?

10.0 Application


7/20

7

This is another example of a structural member with a compositecross-area.

Design calculations typically require use of the MoI for thesecross-sectional areas.

Can you describe a simple method to calculate MoI?

10.0 Application


8/20


9/20

9

10.1 Definition of MOI

Consider a plate submerged in a liquid.The pressure of a liquid at a distance z

below the surface is given by p = z,where is the specific weight of theliquid.

The force on the area dA at that point is dF = p dA.The moment about the x-axis due to this force is z (dF).The total moment is A z dF = A z2 dA = A( z 2 dA).This sort of integral term also appears in solid mechanics when

determining stresses and deflection.This integral term is referred to as the moment of inertia of thearea of the plate about an axis.

Moments of Inertia for Areas (Moments of Inertia for Areas ( Cont.Cont. dd ))


10/20

10

For the differential area dA, shown in thefigure:d I x = y2 dA ,d I y = x 2 dA , and,d J O = r 2 dA , where J O is the polar

moment of inertia about the pole O or z axis.The moments of inertia for the entire area are obtained byintegration.

Ix = A y2 dA ; I y = A x2 dA

JO = A r 2 dA = A ( x 2 + y 2 ) dA = I x + Iy

The MoI is also referred to as the second moment of an area andhas units of length to the fourth power (m 4 or in 4).

Moments of Inertia for Areas (Moments of Inertia for Areas ( Cont.Cont. dd ))

10.1 Definition of MOI


11/20

11

10.2 Radius of Gyration

Radius of Gyration of an AreaRadius of Gyration of an Area

For a given area A and its MoI, I x ,imagine that the entire area is located atdistance k x from the x axis.

2Then, I x = k xA or k x = ( Ix / A). This

k x is called the radius of gyration of thearea about the x axis. Similarly;

k Y = ( Iy / A ) and k O = ( J O / A )

The radius of gyration has units of length and gives an indication

of the spread of the area from the axes. This characteristic isimportant when designing columns.

A

k x

x

y


12/20

12

ParallelParallel --Axis Theorem for an AreaAxis Theorem for an Area

This theorem relates the moment ofinertia (MoI) of an area about an axis

passing through the areas centroid tothe MoI of the area about acorresponding parallel axis. This

theorem has many practicalapplications, especially when workingwith composite areas.

Consider an area with centroid C. The x' and y' axes pass throughC. The MoI about the x-axis, which is parallel to, and distance d yfrom the x ' axis, is found by using the parallel-axis theorem.

10.4 Parallel-Axis Theorem


13/20

13

ParallelParallel --Axis Theorem for an Area (Axis Theorem for an Area ( Cont.Cont. dd ))

Using the definition of the centroid:

y' = ( A y' dA) / ( A dA) . Now

since C is at the origin of the x' y' axes,

y' = 0 , and hence A y' dA = 0 .

Thus IX

= IX

' + A d y

2

IX = A y 2 dA = A (y' + d y)2 dA

= A y' 2 dA + 2 d y A y' dA + d y 2 A dA

y = y' + d y

Similarly, I Y = IY' + A d X 2 and

JO = J C + A d 2

10.4 Parallel-Axis Theorem


14/20

14

10.3 Calculation Integration Method

For simplicity, the area element used has adifferential size in only one direction(dx or dy). This results in a single integrationand is usually simpler than doing a doubleintegration with two differentials, dxdy.

The step-by-step procedure is:

1. Choose the element dA: There are two choices: a vertical strip or ahorizontal strip. Some considerations about this choice are:a) The element parallel to the axis about which the MoI is to be

determined usually results in an easier solution. For example,we typically choose a horizontal strip for determining I x and avertical strip for determining I y.

Moments of Inertia for an Area by IntegrationMoments of Inertia for an Area by Integration


15/20

15

b) If y is easily expressed in terms of x (e.g.,y = x 2 + 1), then choosing a vertical stripwith a differential element dx wide may

be advantageous.2. Integrate to find the MoI. For example, given the element shown in

the figure above:Iy = x

2 dA = x2 y dx and

Ix = d Ix = (1 / 3) y 3 dx (using the information for arectangle about its base from the inside back cover of the textbook).

Since in this case the differential element is dx, y needs to beexpressed in terms of x and the integral limit must also be in terms ofx. As you can see, choosing the element and integrating can bechallenging. It may require a trial and error approach plus experience.

Moments of Inertia for an Area by Integration (Moments of Inertia for an Area by Integration ( Cont.Cont. dd ))

10.3 Calculation Integration Method


16/20

16

Moment of Inertia for a Composite AreaMoment of Inertia for a Composite Area

A composite area is made by adding orsubtracting a series of simple shapedareas like rectangles, triangles, andcircles.

For example, the area on the left can bemade from a rectangle minus a triangleand circle.

The MoI of these simpler shaped areas about their centroidalaxes are found in most engineering handbooks as well as the

inside back cover of the textbook.Using these data and the parallel-axis theorem, the MoI for acomposite area can easily be calculated.

10.5 Calculation Composite Method


17/20


18/20

18

Given : The beams cross-sectional area.

Find : The moment of inertia of the areaabout the x-axis

Plan : Follow the steps for analysis.

ExampleExample



19/20

19

TutorialTutorial

Given: The shaded area as shown in thefigure.

Find: The moment of inertia for the areaabout the x-axis

Plan: Follow the steps for analysis.

Solution

1. The given area can be obtained by subtracting both the circle (b) andtriangle (c) from the rectangle (a).

2. Information about the centroids of the simple shapes can beobtained from the inside back cover of the book. The perpendiculardistances of the centroids from the x-axis are: d a = 5 in , d b =4 in, and d c = 8 in.

(a) (b) (c)



20/20

20

Solution (Solution ( Cont.Cont. dd ))

3. I Xa = (1/12) 6 (10)3

+ 6 (10)(5)2

= 2000 in 4

IXb = (1/4) (2) 4 + (2) 2 (4) 2

= 213 .6 in 4

IXc = (1 /36) (3) (6) 3 +

() (3) (6) (8) 2 = 594 in 4

IX = IXa IXb IXc = 1190 in 4

k X = ( IX / A )

A = 10 ( 6 ) (2) 2 () (3) (6) = 38 .43 in 2

k X = (1192 / 38 .43) = 5 .57 in.

(a) (b) (c)


Documents

Mechanics of Material-Moments of Inertia