Section Properties.pdf

7/28/2019 Section Properties.pdf

1/18

Hong Kong Institute of Vocational Education (Tsing Yi)Higher Diploma in Civil Engineering Structural Mechanics

Chapter 2 SECTION PROPERTIES

Page 1 of 18

Section Properties

Centroid

The centroid of an area is the point about which the area could be balanced if itwas supported from that point. The word is derived from the word center, and itcan be though of as the geometrical center of an area. For three-dimensionalbodies, the term center of gravity, or center of mass, is used to define a similarpoint.

Centroid of Simple Areas

For simple areas, the location of the centroid is easy to visualize.

B

HH/2

B/2

y

xH/3

H

B

xCC

C

D/2

D

CS/2

S/2


2/18



Page 2 of 18

Centroid of Complex Areas

Most complex shapes can be considered to be made up by combining several

simple shapes together.

If the area has an axis of symmetry, the controid will be on that axis. Somecomplex shapes have two axes of symmetry, and therefore the centroid is at theintersection of these two axes.

C C C

CC

C


3/18



Page 3 of 18

Where two axes of symmetry do not occur, the method of composite areas can beused to locate the centroid. For example, consider the following shape. It has avertical axis of symmetry but not a horizontal axis of symmetry. Such areas can

be considered to be a composite of two or more simple areas for which thecentroid can be found by applying the following principle:The product of the total area times the distance to the centroid of the total area isequal to the sum of the products of the area of each component part times thedistance to its centroid, with the distances measured from the same reference axis.

Moment of area

= ydA =(Aiyi)

This can be stated mathematical as

ATY =(Aiyi) Y =(Aiyi)/AT

Where AT =total area of the composite shapeY =distance to the centroid of the composite shape measured from some

reference axisAi=area of one component part of the shapeY i =distance to the centroid of the component part from the reference

axis.


4/18



Page 4 of 18

Example

Find the location of the centroid of following composite area.

Solution

=31.9 mm

T

2

1 ii

A

yAY

=

10605040

551060255040 xxxxY

+

+=


5/18



Page 5 of 18

Example

Find the location of centroid of the following T-section.

Solution

Y =(100*10*60+100*10*5) / (100*10+100*10)

=32.5 mm


6/18



Page 6 of 18


7/18



Page 7 of 18


8/18



Page 8 of 18


9/18



Page 9 of 18

Moment of Inertia (2nd moment of area)

In the study of strength of materials, the property of moment of inertia is an

indication of the stiffness of a particular shape. That is, a shape having a highermoment of inertia would deflect less when subject to bending moments than onehaving a lower moment of inertia.Moment of inertia of simple shapes

Where y =distance from an element of area to the reference axis

A =Area of an element

Example

Determine the moment of inertia, I, of the following area with respect to itscentroidal axis.

dAyI =2

12

)*(

*

3

2/

2/

2

2

bhI

dybyI

dybdA

dAyI

h

h

=

=

=

=

Moment of Inertia, I

In static, it refers to Area moment of Inertia

In dynamic, it refers to Mass moment of Inertia

where I = y2dm

Moment of inertia of a rectangle about centroidal axis:

I =12bd3


10/18



Page 10 of 18

Moment of inertia of complex shapes

If the component parts of a composite area all have thesame centroidal axis, thetotal moment of inertia can be found by adding or subtracting the moments ofinertia of the components parts with respect to the centroidal axis.

Example

The cross-section of a beam shown in the following figure has its centroid at theintersection of its axes of symmetry. Compute the moment of inertia of thesection with respect to the horizontal axis x-x.

Ix =I1+I2 +I3I2 = 30*80

3/12 =1.28 x 106 mm4I1 =I3 =30*40

3/12 =0.16 x 106 mm4

Then,

Ix =I1+I2 +I3

=1.6 x 106 mm4

Ix =I1+I2 +I3


11/18



Page 11 of 18

Example

Compute the moment of inertia for the following section with respect to the axis

x-x.

Ix =I1- I2I1

= 504/12 =520.8 x 103 mm4

I2 =*354/64 =73.7 x 103 mm4

Then,

Ix =I1- I2 =447.1 x 103 mm4


12/18



Page 12 of 18

Polar Moment of Inertia (Perpendicular axis theorem)

The moment of inertia for an area relative to a line or axis perpendicular to theplane of the area is called the polar moment of inertia and is denoted by thesymbol J.

The moment of inertia of an area in the xy plane with respect to the z axis is

JZ = r2dA

= (x2

+ y2

)

dA

= x2dA + y2dA

Therefore, (Perpendicular axis theorem)

JZ = Ix + Iy

Expressed in words, this equation states that the polar moment of inertia for anarea with respect to an axis perpendicular to its plane is equal to the sum of themoments of inertia about any two mutually perpendicular axes in its plane that

interest on the polar axis.


13/18



Page 13 of 18

Parallel axis theorem

The theorem states that the second moment of an area with respect to any axis is

equal to the second moment of the area with respect to a parallel axis through thecentroid of the area added to the product of the area and the square of the distancebetween the two axes.

The second moment of the area in the following figure about the b axis is

I b = (y + d)2 dA

= y2

dA +2d y dA +d2

dA

=Ic + 2d y dA +d2 A

Where,

2d y dA =0

Therefore, Parallel axis theorem

I b = Ic +Ad2

Where Ic is the moment of inertia about centroidal axis

Id is the distance of Ib from Ic


14/18



Page 14 of 18

Example

Compute the moment of inertia for the following inverted T-section with respect

to its centroidal axis.

Solution

Y =(100*10*60+100*10*5)/(100*10+100*10)=32.5 mm

By using the Parallel Axis Theorem,

I =10*1003/12 +100*10*(60-32.5)2+100*103/12 +100*10*(32.5-5)2

=2.35 x 106

mm4


15/18



Page 15 of 18

Example

Compute the moment of inertia for the following section with respect to its

centroidal axis.

Solution

Method 1,

We divide the section into 3 small sections and apply the Parallel AxisTheorem,

I =15*3003/12 +2*[215*153/12 +215*15*(322.5-165)2]

=1.94 x 108

mm4


16/18



Page 16 of 18

Method 2,

Moment Inertia of the I-section,

I = Iarea 1 Iarea 2 Iarea 3

= 215*3303/12 100*3003/12 100*3003/12

=1.94 x 108 mm4


17/18



Page 17 of 18

Example

Compute the moment of inertia for the following section with respect to itscentroidal axis.

Solution

The x-x centroidal axis is shown. It is the axis of symmetry. Only one transferdistance is required (part 1).

Ixx for parts 2 and 4 =50*2003/12

=33.3 x 106 mm4

Ixx for part 3 =150*503/12

=1.56 x 106 mm4

Ixx for parts 1 and 5 =250*503/12 +250*50*(150-25)2

= 197.9 x 106 mm4

Total Ixx =2*33.3 x 106+1.56 x 106 +2*197.9 x 106

=464 x 106 mm4


18/18



Page 18 of 18

Documents

Section Properties.pdf