1. Moment of Inertia

2. 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

3. Application : Design Steel ( Section properties)

4. Moment of Inertia of Areay y dx dy dy h h C x C x b b
Area of shaded element,
Area of shaded element,
Moment of inertia about x-axis
Moment of inertia about y-axis
Integration from h/2 to h/2
Integration from b/2 to b/2

5. J = polar moment of inertia
Table 6.2. Moment of inertia of simple shapes
Shape
J = polar moment of inertia
1. Triangle
2. Semicircle
3. Quarter circle
4. Rectangle
5. Circle b h y x y x r y x r b h y x x y r

6. 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.
From Table 6.1; Ix = and Iy =
The centroid is, ( , ) = (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 y x b h

7. Determine centroid of composite area
Example 6.2 y
140 mm
60 mm
160 mm
60mm x y
140 mm 1 3 2
60 mm x
60mm
160 mm
60 mm
Determine centroid of composite area
PART
AREA(mm2)
y(mm)
x(mm)
Ay (103)(mm3)
Ax (103) (mm3) 1
60(200) = 12000
200/2 = 100
60/2 = 30
1200
360 2
160(60)=9600
60 + [160/2] = 140
288
1344 3
/2= 250
3000 Σ:
Σ: 2688 x 103
Σ: 4704 x 103

8. Second moment inertia PART AREA (A)(mm2) Ix = bh3/12 (106) (mm4)
dy = |y-y|(mm)
Ady2(106)(mm4) 1
60(200) = 12000
60(2003)/12 = 40
|100– 80| = 20
4.8 2
160(60)=9600
160(603)/12 = 2.88
|30 – 80| = 50 24 3
|100 – 80| =20 Σ:
[Ix + Ady2]1 + [Ix + Ady2]2 +[Ix + Ady2]3
= [ ] x106
= x 106 mm4
PART
AREA(mm2)
Iy = b3h/12
(106) (mm4)
Sx=|x-x| (mm)
Asx2(106)(mm4) 1
60(200) = 12000
603(200)/12
=3.6
|30-140|=110
145.2 2
160(60)=9600
1603(60)/12
=20.48
|140 – 140 |= 0 3
|250 – 140|=110 Σ:
[Iy + As2]1 + [Iy + As2]2 +[Iy + As2]3
= [ ] x106
= x 106 mm4