es11!10!2 student version
DESCRIPTION
ES 11TRANSCRIPT
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Lecture 10
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3
x
y A
F=kyA
AkyF
From mechanics of materials:
The resultant force is:
ydAkkydAR
The resultant moment is:
dAykdAkyM22
First moment of area
Second moment of area / moment of inertia
-
xdA
y
y
dAyI x2
x
dAxI y2
Moment of inertia about the x-axis:
Moment of inertia about the y-axis:
-
Geometric property and depends on its reference axis. the smallest value occurs at the axis passing through the centroid.
It measures the ability of cross-section to resist bending. the larger the moment of inertia the less bending will occur
x
Which has the largest moment of inertia about the x-axis?
Which is harder to rotate about the x-axis?
It measures the resistance to rotation about an axis
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Choosing a differential element
x
dy
x2
y
1x dx
y1
y2
= 2
= 2
METHOD 1
-
dyxxy
dAyI
y
y
x
12
2
2
2
1
Choosing a strip such that the distance of all parts of the strip to the axis is constant. (Strip is parallel to axis)
x
dy
x2
y
1x
y1
y2
dyxxxx
dAxI
x
x
ely
12
2
21
2
2
1 2
NOTE: x or y here is not the y-coordinate of the centroid! In this derivation, all x or y in the strip for the term x2 or y2 should be at the same distance from the axis
METHOD 2
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xdx 1y
y 2y
x1 x2
Similarly
dxyyx
dAxI
x
x
y
12
2
2
2
1
METHOD 2
-
y
b
x
h
Find the moment of inertia of a rectangle with respect to one of its base
dy
y
dAyI x2
dybyh
02
When using the form y2dA, all parts of the strip must have the same distance to the axis
3
3
1bh
-
x
dxydI x3
3
1
y
y dy
dx
Recall that the moment of inertia of a rectangle with respect to its base is (1/3)bh3
The moment of inertia of a differential strip perpendicular to and touching the axis is then
METHOD 3
OR, if the strip is not touching
dxyydI x ]3
1
3
1[ 3132
x
dx 1y
y 2y
x1 x2
-
xdA
y
y
r
dArJO2
xPolar moment of inertia wrt pole O
O
dAyxJO22
dAydAxJO22
yxO IIJ
It measures the resistance to rotation about a point
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xy
O
x
y
kx
Concentrating the area into a strip such that it has the same moment of inertia wrt x-axis as the original,
AkI xx2
A
Ik xx Radius of gyration wrt y-axis
-
xy
Ox
y
ky
Similarly, for the y-axis
Iy = ky2A ky =
Iy
ARadius of gyration wrt x-axis
-
xy
O
Similarly, concentrating the area into a ring such that it has the same polar moment of inertia as the original,
AkJ oO2
A
Jk Oo
ko
x
y
222
yxo kkk
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Find the polar moment of inertia of a circle wrt its center
r r duuuduuodJoJ 0 0
32)2(2
y
du
x O
u
r
ududA 2
dAudJO
2
4
2rJ
O
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Determine the moment of inertia with respect to the x and y axes.
0
0.5
1
1.5
2
2.5
0 1 2 3 4 5
x = ky2
y = 0.25 x
-
Find the moment of inertia of the shaded area about the x-axis.
x
y
100 mm
y2 = kx
100 mm
150 mm
-
dAyIA
x 2
461040 mmxI x
x
y
y
y2 = kx
y
x = 150
Find k.
At x = 150, y = 100, thus k = 200/3 and the equation of the curve is y2 = (200/3)x
Using horizontal strips, we can use the form:
Thus,
= 2 150
3
2002
100
100
= 2 2 1
100
100
-
x
y
y
y2 = (200/3)x
x
x = 150
Alternatively, the moment of inertia of the top half is the same as the moment of inertia of the bottom half. The moment of inertia dIx of the differential strip is:
Thus,
=1
3
200
3
1/23
150
0
=1
323
= 20106 4
Since we only got the moment of inertia of the top half, we multiply it to two:
= 40106 4
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Find the moment of
inertia of the shaded
area about its vertical
centroidal axis.
x
y
y = 2 sin x
0
-
21
Reference: Beer, F. B., Johnson, E. R., and Eisenberg, E. R., 2006. Vector Mechanics for Engineers: Statics. 9th Ed. McGraw-Hill.
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x
y
x
y
b y
x
a
C
d
r
z
dA y
x
z
Inertia of the shaded area about the x axis is
dAyIA
X 2
' )'(
Since y = y + b, then
dAbyIA
X 2
')(
AAA
dAbydAbdAy 22 2
2
'AbII XX
Assume x is a centroidal axis
Parallel-axis theorem always involves one centroidal axis!
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x
y
x
y
b y
x
a
C
d
r
z
dA y
x
z
Similarly
2
'AaII yy
Since
thenIIJ yxz ,'''
)()( 22'
baAIIJ yxz
JO ' = JO +Ad2
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Since integration is a summation of small areas, composite areas can be used
n
i
iTotalAxisSpecified IIIII ...321)(
The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.
We use the parallel axis theorem to get the moment of inertia about a non-centroidal axis
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Find the moment of inertia of the composite area about the x-axis.
x
0.9 m
1.5 m
1.8 m
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Components:
21 xxx III 2
iixixi dAII where
2 2
222 dAII xx
1 2
111 dAII xx
PARALLEL AXIS THEOREM
0.9 m
1.5 m
1.8 m
Solution
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468585.4 mI x ANSWER:
423
2 025.2)75.0)(5.1)(8.1(12
)5.1)(8.1(mI x
423
1 66085.22
)8.1)(9.0)(8.1(
36
)9.0)(8.1(mI x
23
1 )8.1(236
bhbhI x
23
2 )75.0(12
bhbh
I x
21 xxx III Solution
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Determine the moment of inertia of the area with respect to the x-axis.
y 75 mm
50 mm
100 mm
150 mm
60 mm
x
-
The moment of inertia of the figure is the moment of inertia of the large rectangle minus the moment of inertia of the small triangle and the semicircle.
y 75 mm
50 mm
100 mm
150 mm
60 mm
x
d2
d1
1
2
3
= 1 2 3
1 =1
3225 1503 = 253.125 1064
2 =1
375 503 = 3.125 1064
To get I3, the form Ix1 = r4/8 is the moment of
inertia about x1 and is not a centroidal axis of the semicircle, thus we cannot use this in the parallel axis theorem to get Ix of the semicircle directly. We can, however find the moment of inertia with respect to the centroid, Ix,centroid first
1 = + 12
60 4
8= +
60 2
2
4 60
3
2
y 75 mm
50 mm
100 mm
150 mm
r = 60 mm x1
Xcentroid of semicircle
x
d2
d1 2
-
= 1.42245 106
Using the parallel axis theorem again to get the moment of inertia of the semicircle wrt x.
3 = + 22
3 = 1.42245 106 + 60 2
2150
4 60
3
2
3 = 89.12388 1064
= 253.125 106 3.125 1064
89.12388 1064
Thus, the total moment of inertia is
= 160.88 1064
y 75 mm
50 mm
100 mm
150 mm
60 mm
x
d2
d1
1
2
3
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1. Beer, F. B., Johnson, E. R., and Eisenberg, E. R., 2006. Vector Mechanics for Engineers: Statics. 9th Ed. McGraw-Hill.
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Determine the moment of inertia of the shaded area about the x-axis.
x
y
100 mm
y2 = 100 - x
-
dAyIA
x
2
Ix = y2(
0
10
100- (100- y2 ))dy
dyyIx
10
0
4
431020 mmxIxANSWER:
x
y
y
y2 = 100 - x
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Determine the radius of gyration of the shaded area with respect to y-axis.
100 mm
y
100 mm
-
Components:
21 yyyT III
2
iiyiyi dAII where
623
1 1067.163
100
236x
bhbhI y
PARALLEL AXIS THEOREM
64
2 1027.398
)100(xI y
2
111 dAII yy
yy II 2
1
2
100 mm
y
100 mm
Solution
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21 yyyT III
61094.55 xI yT
21 AAAT T
yT
A
I
yk Radius of Gyration
32
1071.252
)100(
2
)100)(200(xAT
mmx
xky 65.46
1071.25
1094.553
6
ANSWER: mmky 65.46
Solution
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Obtain the moment of inertia of the composite area about the y-axis
y
75 mm
50 mm
100 mm
150 mm
60 mm
-
Components:
321 yyyyT IIII
2
iiyiyi dAII where
PARALLEL AXIS THEOREM
1
2
3
2111 dAII yy
yy II 2
2111
311
112
dhbhb
I y
2222
322
212
dhbhb
I y
23
24
328
drr
I y
2
333 dAII yy
y
75 mm
50 mm
100 mm
150 mm
60 mm
-
623
1 1075.168)75)(150)(150(12
)150)(150(xI y
623
2 101875.267)5.187)(100)(75(12
)75)(100(xI y
6224
3 10447.25)60(2
)60(
8
)60(xI y
46321 105.410 mmxIIII yyyyT
ANSWER: 46105.410 mmxI yT
Solution
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Find the moment of inertia of the shaded area about: a) y-axis b) x-axis
x
y
75 mm
x2 = 25y
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Find the moment of inertia of the shaded area about: a) y-axis b) x-axis
x
y
75 mm
x2 = 25y
-
x
x2 = 25y
y
x x=75, y=225
Using horizontal strips:
a) Moment of inertia about the x-axis, Ix
x
x2 = 25y
y
y
x=75, y=225 =
2
= 2 25 1/2 0
225
0
Alternatively, we can use vertical strips:
= 1
323 1
313
=1
3 2253
2
25
3
75
0
= 244.06106 4
= 244.06106 4
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x
x2 = 25y
y
x x=75, y=225
Using vertical strips:
b) Moment of inertia about the y-axis, Iy
= 2
= 2 225
2
25
75
0
Alternatively, we can use horizontal strips:
= 1
323 1
313
=1
3 25 1/2
3 03
225
0
= 12.66106 4
x
x2 = 25y
y
y
x=75, y=225
= 12.66106 4