chapter 6
TRANSCRIPT
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-1
EML 3004C
Chapter 6: Center of Gravity and CentroidChapter 6: Center of Gravity and Centroid
Namas ChandraIntroduction to Mechanical engineering
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
Namas ChandraIntroduction to Mechanical engineering
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.
Namas ChandraIntroduction to Mechanical engineering
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
Namas ChandraIntroduction to Mechanical engineering
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-6
EML 3004C
Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-7
EML 3004C
Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
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
Namas ChandraIntroduction to Mechanical engineering
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
Namas ChandraIntroduction to Mechanical engineering
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-10
EML 3004C
Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-11
EML 3004C
Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-12
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-13
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-14
EML 3004C
Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-15
EML 3004C
Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-16
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-17
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)
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-18
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-19
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-20
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-21
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-22
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)
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-23
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-24
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-25
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-26
EML 3004C
Problem 6-87 (pg. 294, Sections 6.8-6.9)
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:
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-27
EML 3004C
Problem 6-92 (pg. 294, Sections 6.8-6.9)
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:
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-28
EML 3004C
Problem 6-8 (pg. 257, Sections 6.1-6.3)
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-29
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
Namas ChandraIntroduction to Mechanical engineering
HibbelerChapter 6-30
EML 3004C
Chapter 6: Concludes….Chapter 6: Concludes….