Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 6-1 EML 3004C Chapter 6: Center of Gravity and Centroid

Namas ChandraIntroduction to Mechanical engineering

HibbelerChapter 6-1

EML 3004C

Chapter 6: Center of Gravity and CentroidChapter 6: Center of Gravity and Centroid


HibbelerChapter 6-2

EML 3004C

Chapter 6.1 C.G and Center of MassChapter 6.1 C.G and Center of Mass

R iW W

1 1 2 2

Taking moments about point O

...

i i

Wx W x W x

W xx

W

We can replace W by

VolumeW=

specific GravityV At


HibbelerChapter 6-3

EML 3004C

Chapter 6.1 C.G and Center of Mass..2Chapter 6.1 C.G and Center of Mass..2

i ii iW xx

W

A x

A

i ii iW yy

W

A y

A

The same way you can find the centroid of the line and the volume.


HibbelerChapter 6-4

EML 3004C

Example 6-2 (pg.251, sections 6.1-6.3)

Locate the centroid of the area shown in figure

Solution I

Differential Element. A differential element of thickness dx is shown in the Figure. The element intersects the curve at the arbitrary point (x,y), and so it has a height of y.

Area and Moment Arms. The area of the element is dA=y dx, and its centroid is located at ,

2

yx x y


HibbelerChapter 6-5

EML 3004C

Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)

Integrations. Applying Equations 6-5 and integrating with respect to x yields

1 1 3

0 01 1 2

0 0

1 1 2 2

0 01 1 2

0 0

0.2500.75

0.333

( / 2) ( / 2) 0.1000.3m

0.333

A

A

A

A

xy dx x dxx dAx

dA y dx x dx

y y dx x x dxy dAy

dA y dx x dx


HibbelerChapter 6-6

EML 3004C


Solution II

Differential Element. The differential element of thickness dy is shown in Figure. The element intersects the curve at the arbitrary point (x,y) and so it has a length (1-x).

Area and Moment Arms. The area of the element is dA = (1-x) dy, and its centroid is located at

1 1,

2 2

x xx x y y


HibbelerChapter 6-7

EML 3004C


Integrations. Applying Equations 6-5 and integrating with respect to y, we obtain

11

001 1

0 0

1 1 3/ 2

0 01 1

0 0

1(1 )(1 ) / 2 (1 ) 0.2502 0.75m

0.333(1 ) (1 )

(1 ) ( ) 0.100= 0.3m

0.333(1 ) (1 )

A

A

A

A

y dyx x dyx dAx

dA x dy y dy

y x dy y y dyy dAy

dA x dy y dy


HibbelerChapter 6-8

EML 3004C

Chapter 6: In-class exerciseChapter 6: In-class exercise

Assume the following:

h=6m b=3m. Find x and y


HibbelerChapter 6-9

EML 3004C

Example 6-6 (pg. 261, Sections 6.1-6.3)

Locate the centroid C of the cross-sectional area for the T-beam

Solution I The y axis is placed along the axis of symmetry so that To obtain we will establish the x axis (reference axis) thought the base of the area.l The area is segmented into two rectangles and the centroidal location for each is established.

y

y

5in. (10 in.) (2 in.) + 11.5 in (3 in.)(8 in.)

(10 in.) (2 in.) + (3 in.)(8 in.)

8.55in.

yAy

A

0x


HibbelerChapter 6-10

EML 3004C


Solution II

Using the same two segments, the x axis can be located at the top of the area. Here 1.5in. (3 in.) (8 in.) + -8 in (10 in.)(2 in.)

(3 in.) (8 in.) + (10 in.)(2 in.)

4.45in.

yAy

A

The negative sign indicates that C is located below the origin, which is to be expected. Also note that from the two answers 8.55 in + 4.45 in =13.0 in., which is the depth of the beam as expected



EML 3004C


Solution III

It is also possible to consider the cross-sectional area to be one large rectangle less two small rectangles. Hence we have

6.5in. (13 in.) (8 in.) - 2 5 in (10 in.)(3 in.)

(13 in.) (8 in.) 2(10 in.)(3 in.)

8.55in.

yAy

A



EML 3004C

Problem 6-30 (pg. 264, Sections 6.1-6.3)

Determine the distance to the centroid of the shaded area. y

2 3

3 6

Segment (mm ) (mm) (mm )

1 300(600)=180(10) 300 54(10)

1 2 (300)(600) =

2

A y yA

3 6

2 3 6

2 3

90(10) 200 18(10)

1 3 ( )(300) =141.37(10) -127.32 -18(10)

2

4 -( )(100) 31.71(10) 0 0

3 6

6

3

379.96(10) 54(10)

54(10) = =142mm

379.96(10)

yAy

A

Solution



EML 3004C

6.6 Moments of Inertia For Areas

By definition moments of inertia with respect to any axis (i.e. x and y) are

2 2,x y

A A

I y dA I x dA

Polar moment of inertia

yxAA

zoIIdAyxdArJorJ )( 222

Always positive value

Units: 44 in,m



EML 3004C

Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)



EML 3004C

Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)



EML 3004C

Con’t 6.6 Moments of Inertia For Areas

Example 6.14 Lets proof for rectangular areaSolution Part (a)

3

3

2/

2/

22/

2/

22

121

(b)Part 121

)(

hbI

bh

ydybydbydAyI

y

h

hA

h

hx

yxII

and



EML 3004C

Example 6-15 (pg. 287 Sections 6.8-6.9)

Determine the moment of inertia of the shaded area about the x axis.

A differential element of area that is parallel to the x axis is chose for integration.

dA =(100-x) dy.

Limits of integration wrt y, y=0 to y=200 mm

2 2

2200 200 2002 2 4

0 0 0

6 4

(100 )

1(100 ) 100

400 400

107(10 ) mm

x A AI y dA y x dy

yy dy y dy y dy

Solution I (CASE I)



EML 3004C

6-7 Parallel-Axis Theorem

If we know Moment of Inertia of a given axis, we can compute M.I about another parallel axis

22 2

31 2

2

( ) 2

1. Moment of inertia with respect to x axis (centroidal axis)

2. is zero, since 0

3.

x y y y

A A A A

x

y

I y d dA y dA d y dA d dA

I

y y dA y dA

Ad



EML 3004C

Con’t 6-7 Parallel-Axis Theorem

As a result:2

2

20

2

similarly

and

For composite body:

x x y

y y x

c

x x i i i

I I Ad

I I Ad

J J Ad

I I A d



EML 3004C

Con’t 6-7 Parallel-Axis Theorem

iA

yx

iII

yyxxd

A

yx

II

i

ii

iyix

yx

segment of area :

) and( axis centroidal

respect tobody with in the ofsegment of inertia ofmoment :,

) and or and (i.e. axes parallel obetween tw distance :

body of area total:

) and(

axis centroidal respect tobody with a of inertia ofmoment :,

2

2

xyy

yxx

AdII

AdII



EML 3004C

Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)

Solution II (CASE 2)A differential element parallel to the y axis is chosen for integration.

Use the parallel-axis theorem to determine the moment of inertia of the element with respect to this axis.

For a rectangle having a base b and height h, the moment of inertia about its centroidal axis is

31

12xI bh



EML 3004C

For the differential element , b = dx and h = y, and thus

Since the centroid of the element is at from the x axis, the moment of inertia of the element about this axis is

31

12xdI dx y

/ 2y y

22 3 31 1

12 2 3x x

ydI dI dA y dx y y dx y dx

Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)

1003 3/ 2

0

6 4

1 1(400 )

3 3

107(10 ) mm

x x AI dI y dx x dx

Solution II (CASE 2)



EML 3004C

Example 6-18

49999

321

49232

3

493

2

49

232

1

mm109.210425.11005.010425.1 Total

mm10425.1 )200)(300)(100()300)(100(121

D Rectangle

mm1005.0)100)(600(121

B Rectangle

mm10425.1

)200)(300)(100()300)(100(121

A Rectangle

.1

:Solution

xxxx

yxx

x

yxx

x

IIII

AdII

I

AdII

I



EML 3004C

Con’t Example 6-18

49999

321

49232

3

493

2

49

232

1

mm106.51090.11080.11090.1 Total

mm1090.1 )250)(300)(100()100)(300(121

D Rectangle

mm1080.1)600)(100(121

B Rectangle

mm1090.1

)250)(300)(100()100)(300(121

A Rectangle

.2

yyyy

xyy

y

xyy

y

IIII

AdII

I

AdII

I



EML 3004C

Example 6-17 (pg. 291, Sections 6.8-6.9)

Determine the moment of inertia of the cross-sectional area of the T-beam about the centroidal axis.x

In class workout



EML 3004C


Determine the moment of inertia of the shaded area with respect to a horizontal axis passing through the centroid of the section

3 2 3 2 4

1(2)(6) 5(6)(1)2.333in

2(6) 1(6)

1 1(6)(2) 2(6)(2.333 1) (1)(6) 1(6)(5 2.333) 86in

12 12xx

yAy

A

I

Solution:



EML 3004C


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

3 3 2 41 1(2)(6) 2 (4)(1) 1(4)(1.5) 54.7 in

12 12yI

Solution:



EML 3004C


Determine the location of the centroid of the quarter elliptical plate.

( , )x y

Area of the differential element

22 2

2

21 2 2and , .22 2

bdA y dx b x dx

a

y bx x y b x

a

Solution



EML 3004C

Con’t Problem 6-8 (pg. 257, Sections 6.1-6.3)

22 2

0 2 4

322 2

0 2

2 21 2 2 2 20 2 22

4

322 2

0 2

ba x b x dxx dA aAx adAA ba b x dx

a

b ba b x b x dxa ay dAAy a

dAA ba b x dxa

Con’t Solution



EML 3004C

Chapter 6: Concludes….Chapter 6: Concludes….

Documents

Namas Chandra Introduction to Mechanical engineering Hibbeler Chapter 6-1 EML 3004C Chapter 6: Center of Gravity and Centroid