Civil Engineering Department - lecture-notes.tiu.edu.iq

Tishk International UniversityCivil Engineering Department

Engineering Mechanics Chapter 8 & 10

Centroid and moment of Inertia

Instructor: Twana A. Hussein

1Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Chapter Objectives

• To discuss the concept of the center of gravity, center

of mass, and the centroid.

• To show how to determine the location of the center of

gravity and centroid for a system of discrete particles

and a body of arbitrary shape

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CONCEPT OF CENTER OF GRAVITY (CG)

• A body is composed of an infinite number of particles,

and so if the body is located within a gravitational

field, then each of these particles will have a weight

dW.

• The center of gravity (CG) is a point, often shown as

G, which locates the resultant weight of a system of

particles or a solid body.

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CONCEPT OF CENTER OF GRAVITY (CG)

• The centroid, C, is a point which defines the geometric center of an object.

• The centroid coincides with the center of mass or the center of gravity only if

the material of the body is homogenous (density or specific weight is

constant throughout the body).

• Equations:-

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• Locate the centroid of the area shown in figure.

Example 8.1:-

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Solution[1]

• A differential element of thickness dx is selected. The element intersects the

curve at the arbitrary point(x,y), and so it has a height y.

• The area of the element dA=y dx, and its centroid is located at (x,y/2)

Example 8.1:-

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Solution[2]

• The differential element of thickness dy is

selected. The element intersects the curve at the

arbitrary point (x,y), and so it has a length (1-x).

• The area of the element is dA=(1-x)dy, and its

centroid is located

at(𝑥 + (1−2𝑥) = 1+2𝑥 , 𝑦)

Example 8.1:-

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Solution[2]

Example 8.1:-


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Composite Bodies

• Consists of a series of connected “simpler” shaped bodies, which may be

rectangular, triangular or semicircular.

• A body can be sectioned or divided into its composite parts.

• Provided the weight and location of the center of gravity of each of these

parts are known, the need for integration to determine the center of gravity

for the entire body can be neglected.


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Copyright © 2010 Pearson Education South Asia Pte Ltd 10





EXAMPLE 1




Example-8.3



Example-8.4



Example-8.5




Example-8.6






Homework


Homework


Homework


Homework


Moment of Inertia

• Moment of Inertia of a Rectangular Area


Moment of Inertia

• Moment of Inertia of a Rectangular Area


Moment of Inertia

• Moment of Inertia of a Triangular Area.


Moment of Inertia

• Radius of Gyration of an Area

• The radius of gyration of an areaA with respect to the x axis isdefined as the distance kx, whereIx= kx A. With similar definitions forthe radii of gyration of A withrespect to the y axis and withrespect to O, we have



Moments of Inertia for Composite Areas

• Composite area consist of a series of connected simpler parts or

shapes.

• Moment of inertia of the composite area = algebraic sum of the

moments of inertia of all its parts.

• Procedure for Analysis

Composite PartsDivide area into its composite parts and indicate the centroid of each part to the

reference axis.

Parallel Axis TheoremMoment of inertia of each part is determined about its centroidal axis.

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Moments of Inertia for Composite Areas

• Procedure for Analysis

Parallel Axis Theorem

• When centroidal axis does not coincide with the reference axis,

the parallel axis theorem is used.

Summation

• Moment of inertia of the entire area about the reference axis is

determined by summing the results of its composite parts.

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Example 10.1

• Compute the moment of inertia of the composite area shown

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Example 10.1 [Solution]

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Example 10.2

Determine the moments of inertia of the beam’s cross-sectional area

shown about the x and y centroidal axes

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Example 10.3

Determine the moments of inertia and the radius of gyration of the

shaded area with respect to the x and y axes.

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Example 10.4


shaded area with respect to the x and y axes.

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Example 10.5


shaded area with respect to the x and y axes and at the centroidal axes.

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Example 10.6

Locate the centroid x’ of the cross-sectional area for the angle.

Then find the moment of inertia about the ӯ centroidal axis.


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Example 10.6

Determine the moment of inertia of the beam’s cross-sectional area

about the x and y axis


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Example 10.7

Locate the centroid ӯ of the cross-sectional area for the angle. Then

find the moment of inertia about the x’ centroidal axis.


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Example 10.8

Determine the distance ӯ to the centroid of the beam’s cross-sectional

area; then determine the moment of inertia about the ẋ and y axis.


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Example 10.9

Determine the distance ӯ to the centroid of the beam’s cross-sectional area; then find

the moment of inertia about the x’ axis.


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Example 10.9


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Example 10.10

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


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Thank you,


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Civil Engineering Department - lecture-notes.tiu.edu.iq