#### Transcript Moment of Inertia

```MOMENT OF INERTIA





Type of moment of inertia
 Moment of inertia of Area
 Moment of inertia of mass
Also known as second moment
Why need to calculate the moment of Inertia?
 To measures the effect of the cross sectional shape of a
beam on the beam resistance to a bending moment
Application
 Determination of stresses in beams and columns
Symbol
 I – symbol of area of inertia
 Ix, Iy and Iz

Application : Design Steel ( Section properties)
Moment of Inertia of Area
y
y
dx
dy
h
C
dy
x
h
C
b
x
b
Area of shaded element, dA  bdy
Area of shaded element, dA  hdx
Moment of inertia about x-axis
Moment of inertia about y-axis
I y   x 2dA   hx 2dx
I x   y 2dA   by 2dy
Integration from h/2 to h/2
Integration from b/2 to b/2
h2
Ix 
 by
b2
2
dy
h 2
Iy 
 hx dx
2
b 2
h2
by 3 
bh3


3  h 2 12
b2
hx 3 
b3h


3  b 2 12
Table 6.2. Moment of inertia of simple shapes
Shape
bh 3
36
b3h
36
1 4
r
8
1 4
r
8
1 4
r
4
1 4
r
16
1 4
r
16
1 4
r
8
y
bh3
12
b3h
12
1
bhb2  h2 
12
x
1 4
r
4
1 4
r
4
1 4
r
2
x
h
y
x
b
y
2. Semicircle
r
3. Quarter circle
Iy
y
1. Triangle
y
x
y
r
y
x
yx
4. Rectangle
x
h
5. Circle
y
b
x
J = polar moment
Ix
r
of inertia
PARALLEL - AXIS THEOREM

There is relationship between the moment of inertia about two parallel axes
which is not passes through the centroid of the area.
y
h
x
y
x
b

From Table 6.1; Ix =

The centroid is, ( x ,


bh3
12
y
and Iy =
b 3h
12
) = (b/2, h/2)
Moment of inertia about x-x axis, Ixx = Ix + Ady2
 where dy is distance at centroid y
Moment of inertia about y-y axis, Iyy = Iy + Asx2
 where
sx is distance at centroid x
Example 6.2
y
y
140 mm
140 mm
60 mm
60 mm
1
3
2
x
x
60mm 160 mm 60 mm
60mm 160 mm 60 mm
Determine centroid of composite area
PART
AREA(mm2)
y(mm)
x(mm)
1
60(200) =
12000
160(60)=9600
200/2 =
100
60/2 = 30
60(200) =
12000
200/2 =
100
2
3
Σ: 33 600
Ay (103)(mm3)
Ax (103) (mm3)
60/2 = 30
1200
360
60 +
[160/2] =
140
220 +60/2=
250
288
1344
1200
3000
Σ: 2688 x 103
Σ: 4704 x 103

Ax
 140 mm
A

Ay
y
 80 mm
A
x
Second moment inertia
PART
1
2
3
AREA (A)(mm2)
60(200) = 12000
160(60)=9600
60(200) = 12000
Ix = bh3/12
(106) (mm4)
dy = |y-y|(mm)
60(2003)/12 = 40
|100– 80| = 20
4.8
|30 – 80| = 50
24
|100 – 80| =20
4.8
160(603)/12 = 2.88
60(2003)/12 = 40
Σ: 33 600
 I xx 
=
=
PART
1
2
3
[Ix + Ady2]1 + [Ix + Ady2]2 +[Ix + Ady2]3
[44.8 + 26.88 + 44.8] x106
116.48 x 106 mm4
AREA(mm2)
60(200) = 12000
160(60)=9600
60(200) = 12000
Iy = b3h/12
(106) (mm4)
603(200)/12
=3.6
1603(60)/12
=20.48
603(200)/12
=3.6
Σ: 33 600
 I yy 
[Iy + As2]1 + [Iy + As2]2 +[Iy + As2]3
= [148.8 + 20.48 + 148.8] x106
= 318.08 x 106 mm4
Sx=|x-x| (mm)
|30-140|=110
Asx2(106)(mm4)
145.2
|140 – 140 |= 0
0
|250 – 140|=110
145.2
```