area moment of inertia
TRANSCRIPT
1
x
yΔA
∫ ∫== ydAkkydAR
Consider a beam of uniform cross-section
1
ΔF=kyΔA ∫ ∫== dAykdAkyM 22
Moment of Inertia of an Area
•Geometrical property and depends on its reference axis.
the smallest value occurs at the axis passing
2
the smallest value occurs at the axis passing through the centroid.
•It measures the ability to resist bending.the larger the moment of inertia the less
bending will occur
z
Moment of Inertia of an Area
dAyIA
x ∫= 2
dAxIA
y ∫= 2
Moment of Inertia:
3
y
x
y
rx dArJ
Az ∫= 2
Polar Moment of Inertia:
dAyxJA
z ∫ += )( 22
yIxIzJ +=
y
x dydx
Moment of Inertia of an Area• by integration
y
xy
dy
dA=ydx
y
4
x
y
dAydI x2=
dAxdI y2=
x
dAxdI y2=
dx
dAydI x2=
x
y
dy
dA=(a-x)dy
x
a
Calculating Ix:
Find the moment of inertia about the x axis of the following figure:
5
bdydA =
bddI 2
Area Moment of Inertia (Rectangular Area)
y
6
bdyyxdI 2=
3312
0 bhdyyh bxI =∫=
b
x
dyh
y
2
Moment of Inertia (Circular Area)y
du
xO
ur
ududA π2=
dAudJO2=
7
∫ ∫ ∫=== r r duuuduuodJoJ0 0
32)2(2 ππ
ududA π2=
4
2rJO
π=
The distance kx is referred to as the radius of gyration of the area with respect to the x-axis.
2y A
Radius of Gyration
8
AIk
AkI
xx
xx
=
= 2
x
kx
y A
AkI 2
The distance ky is referred to as the radius of gyration of the area with respect to the y-axis.
Radius of Gyration
9
x
ky
AI
k
AkI
yy
yy
=
= 2 AkJ 2
The distance ko is referred to as the radius of gyration of the area with respect to the x-axis.
Radius of Gyration
y
A
10
AJk
AkJ
OO
OO
=
=
xko
Find the moment of inertia of the h d d
y
100 mm
y2 = 100x
Example A
11
shaded area about the x-axis.x
100 mm
dAyIA
x ∫= 2
dyxyIx )100(100100
2 −= ∫−
dyI )100(2
100 2
y
y
y2 = 100x
Solution
12
dyyyIx )100
100(100100
2 −= ∫−
461067.26 mmxIx =ANSWER:
x
3
Find the moment of inertia of the h d d
y
75 mm
Example B
13
shaded area about the y-axis.
x
x2 = 25y
dAxIA
y ∫= 2
dxyxIy )225(750
2 −= ∫
x 2
yx
Solution
14
dyxxIy )25
225(750
2 −= ∫
461066.12 mmxIy =ANSWER:x
x2 = 25y
Determine the moment of inertia of the
y
Example C
15
shaded area about the x-axis.
x
100 mm
y2 = 100 - x
dAyIA
x ∫= 2
dyyyIx
2100
2∫=
y
Solution
16
dyyIx ∫= 100
4
431020 mmxIx =ANSWER:x
yy2 = 100 - x
Find the moment of inertia of the shaded
y
y = 2 sin x
Assignment:
17
inertia of the shaded area about its vertical centroidal axis.
x0 π
Moments of Inertia of Common Areas
18
4
Parallel Axis Theoremy’
yx’
a
dAx
19
x’
x
b y’
C
d
r
z’
dAy
x
z
Inertia of the shaded area about the x’ axis isdAyI
AX ∫= 2
' )'(
Since y’ = y + b, thendAbyI
AX ∫ += 2
' )(
∫∫∫ ++=AAAdAbydAbdAy 22 2
20
2' AbII XX +=
In a similar manner, it can be shown that2
' AaII yy +=
Since thenIIJ yxz ,''' +=)()( 22
' baAIIJ yxz +++=2' AdJJz z +=
It follows the same concept used in composite figures for First Moments!!!
Moment of Inertia (Composite Area)
21
∑=+++=n
iiTotalAxisSpecified 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.
Find the moment of inertia of the
0.9 m
Example D
22
of inertia of the composite area about the x-axis.
x
1.5 m
1.8 m
Components:
21 xxx III +=2
iixixi dAII +=where0.9 m
Solution
23
22
222 dAII xx +=
12
111 dAII xx +=PARALLEL AXIS THEOREM1.5 m
1.8 m
423
1 66085.22
)8.1)(9.0)(8.1(36
)9.0)(8.1( mIx =+=
23
1 )8.1(236bhbhI x +=
21 xxx III +=Solution
24468585.4 mIx =ANSWER:
423
2 025.2)75.0)(5.1)(8.1(12
)5.1)(8.1( mIx =+=
23
2 )75.0(12
bhbhI x +=
5
Determine the radius of gyration of the shaded
y
100 mm
Example E
25
of the shaded area with respect to y-axis.
100 mm
100 mm
21 yyyT III +=2
iiyiyi dAII +=where100 mm
y
100 mm
Solution
26
Components:
623
1 1067.163
100236
xbhbhI y =⎟⎠⎞
⎜⎝⎛+=
PARALLEL AXIS THEOREM
64
2 1027.398
)100( xI y ==π
2111 dAII yy +=
yy II =2
1
2
21 yyyT III += 61094.55 xIyT =
21 AAAT +=T
yTAI
yk =Radius of Gyration
32
107125)100()100)(200( xA =+=π
Solution
27
1071.2522
xAT =+=
mmxxky 65.461071.251094.55
3
6==
ANSWER: mmky 65.46=
Obtain the moment of inertia of the composite area
y
75 mm
50 mm
150 mm
Example F
28
composite area about the y-axis100 mm
60 mm
321 yyyyT IIII −+=2
iiyiyi dAII +=where
PARALLEL AXIS THEOREM
y
75 mm
50 mm
100 mm
150 mm Solution
29
Components:
PARALLEL AXIS THEOREM
1
2
3
2111 dAII yy +=
yy II =2
2111
311
1 12dhbhbI y +=
2222
322
2 12dhbhbI y +=
23
24
3 28drrI y
ππ+=2
333 dAII yy +=
60 mm
623
1 1075.168)75)(150)(150(12
)150)(150( xIy =+=
623
2 101875.267)5.187)(100)(75(12
)75)(100( xIy =+=
Solution
30
6224
3 10447.25)60(2
)60(8
)60( xIy =+=ππ
46321 105.410 mmxIIII yyyyT =−+=
ANSWER:46105.410 mmxI yT =