of Moment inertia of an area
For linearly-varying distributed loads,
Resultant force
R=
ky dA = k
A
y dA = k Q x
A
The first moment of the area about the x -axis
Qx =
y dA
A
The first moment of the area about the y -axis
Qy =
A
x dA
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Moment of inertia of an area (cont.)
For linearly-varying distributed loads,
Resultant moment about the x -axis
M=
ky 2 dA = k
A
y 2 dA = k I x
A
The second moment of the area about the x axis.
Ix =
A
y 2 dA
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Moment of inertia of an area (cont.)
The second moment, or moment of inertia, of an area about the x -axis.
Ix =
y 2 dA
A
The moment of inertia of the area about the y -axis.
Iy =
A
x 2 dA
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Moment of inertia of an area (cont.)
Direct integration
Ix =
y 2 dA
Ix =
dI x
=
y 2 ( a x ) dy
A
Iy =
A
x 2 dA
y
Iy =
dI y
=
x 2 y dx
x
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Moment of inertia of an area (cont.)
Moment of inertia of a rectangular section
Ix =
y 2 dA
A
=
h
y 2 b dy
0
1 3 13
I x = bh = h b
3
3
1 3 13
I y = hb = b h
3
3
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Moment of inertia of an area (cont.)
Computing I x and I y from the same elemental strip
For the elemental strip (rectangular) shown,
13
13
dI x = h b = y dx
3
3
Ix =
dI x =
13
y dx
3
x 2 y dx
x
dI y = x 2 y dx
Iy =
dI y =
x
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Moment of inertia of an area (cont.)
For the elemental strip (rectangular) shown,
dI x = y 2 dA = y 2 ( a x ) dy
Ix =
dI x =
y 2 ( a x ) dy
y
13
13
dI y = h2 b h1 b
3
3
13
13
= a dy x dy
3
3
13
= ( a x 3 ) dy
3
Iy =
dI y =
y
13
( a x 3 ) dy
3
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Polar moment of inertia of an area
The magnitude of a force per unit area varies
linearly with the distance from the z -axis,
dF = kr dA
The resultant moment about the z -axis
Mz =
dM =
Mz = k
r dF =
kr 2 dA
A
r 2 dA = k J O
A
The polar moment of inertia of the area
about the z -axis is
JO =
A
r 2 dA
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Polar moment of inertia of an area (cont.)
JO =
r 2 dA
A
r 2 = y2 + x 2
JO =
( y 2 + x 2 ) dA
y 2 dA +
A
=
A
JO = I x + I y
A
x 2 dA
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Radius of gyration of an area
Ix =
y 2 dA
Iy =
A
x 2 =
k2
x
Iy dA
A
Ix =
k2
y
JO =
2
kO
A
A
A
kx =
Ix
A
ky =
Iy
A
kO =
JO
A
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Radius of gyration of an area (cont.)
JO = I x + I y
2
kO A = k 2 A + k 2 A
x
y
2
kO = k 2 + k 2
x
y
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Moments of inertia
Example 57
Example 58
CE 297
13
Parallel axis theorem
Given: The moment of inertia about centroidal axis BB
Ix =
y
x
x
dy
y 2 dA
dA
y
C
Find: The moment of inertia about a
parallel non-centroidal axis AA
Ix =
y = y + d x
y 2 dA
y
d
dx
O
Ix =
y 2 dA
=
( y + d x )2 dA
=
y 2 dA + 2 d x
x
y dA + d 2
x
dA
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Parallel axis theorem (cont.)
y
y 2 dA = I x
x
x
dy
y dA = y A = 0
dA
dA = A
C
y
d
dx
O
Ix =
y 2 dA + 2 d x
2
k2 A = k x A + d2 A
x
x
y dA + d 2
x
dA
y
x
Ix = Ix + d2 A
x
2
k2 = k x + d2
x
x
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15
Parallel axis theorem (cont.)
Ix = Ix + d2 A
x
y
x
x
dy
2
k2 = k x + d2
x
x
dA
C
Similarly,
Iy =
( x + d y )
d
2
dA
O
Iy = Iy + d2 A
y
2
k2 = k y + d2
y
y
y
y
dx
x
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Parallel axis theorem (cont.)
For the polar moment of inertia,
y
JO = I x + I y
= Ix +
x
dy
d2 A
x
+ Iy +
dA
d2 A
y
C
= ( I x + I y ) + (d 2 + d 2 ) A
x
y
JO = JC + d A
2
2
x
2
kO = kC + d 2
d
O
y
y
dx
x
Moments of inertia of common geometric shapes
CE 297
18
Moments of inertia of rolled-steel shapes (U.S.)
CE 297
19
Moments of inertia of rolled-steel shapes (S.I.)
CE 297
20
Composite areas
If an area A is made up of several simple areas, A1 , A2 , ..., an integral over
A may be divided into integrals over A1 , A2 , ....
Ix =
y 2 dA
y 2 dA +
A
=
A1
y 2 dA + . . .
A2
= I x 1 + d 2 1 A1 + I x 2 + d 2 2 A2 + . . .
x
x
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Composite areas
I x = I x 1 + d 2 1 A1 + I x 2 + d 2 2 A2 + . . .
x
x
Thus, the moment of inertia may be found by
1.
For each simple area, compute the moment of inertia about a centroidal axis that is parallel to the desired axis.
2.
Using the parallel axis theorem, determine the moment of inertia
for the desired axis.
3.
Add (or subtract) the calculated moments of inertia.
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Parallel axis theorem (cont.)
Example 59
Example 60