Second moment of area / Moment of Inertia
Definition:
The second moment of area about axes x-x and y-y are given by: y
dA
I xx = ∑ y2dA or I xx = ∫ y2dA
y
I yy = ∑ x 2dA or I yy = ∫ x 2dA x
x
For the basic geometries
Rectangle Circle Triangle
y
y
x x R
d x x h
x’ x’ x
x
b b
y
y
Second moment of area
4 3
I xx = 12
1
bd 3 I xx = πR4 = I yy I xx = bh
12
HOME WORK: show that the
I x ' x ' = 13 bd 3 Second Polar moment of area moment of inetia about the centroidal
3
I yy = 12
1
db3 J = I xx + I yy = πR 4 axis in this case is: I x = bh
36
2
The Parallel axis theorem
If I is the second moment of area of an area about the centroidal axis x-x,
then the second moment Ixx about an axis parallel to x-x and
perpendicular distance d is given by:
I xx = I + Ad 2
1
y
Find the second moment of area of the given section T- 50
x’
section about its centroidal axes Ix and Iy 5 1
G x
We can start with Ix 50
2
Use, table as shown below:
Part Area A (cm2) yi (cm) Ayi (cm3) Ayi2 (cm4) Ii (cm4)
1 50 × 5 = 250 2.5 625 1562.5 520.83
2 45 × 5 = 225 27.5 6187.50 170156.25 37968.75
Σ 475 6812.50 171718.75 38489.58
yi = distance of the centroid of the part from axis x’-x’
Ii = second moment of area of the section/part about its centroidal axis
If Ix’ is the second moment of area of the composite area about x’-x’ then,
Ix’ = Σ Ii + Σ Ayi2 = 38489.58 + 171718.75 = 210,208.33 cm4 (from the parallele axis theorem)
If y is the centroid, then y =
∑ Ai yi = 6812.50 = 14.34 cm
∑A 475
By using the parallel axis theorem: Ix = Ix’ −y2 ΣA = 210,208.33 − (14.34)2(475)
= 112531.42 cm4
To find Iy , you can use addition of the moments of area of the two sections, since their centroidal
axes coincides. Thus,
Iy = I1y’ + I2y = 1
12
(5)(50)3 + 12
1
(45)(5)3 = 52552.088 cm4
2
Find the second moment of area of the given section chennel- y Web
section about its centroidal axes Ix and Iy 60
x’
Start with Ix 2
50 x
G
1 3
5
Flanges
Units are in cm
Element Area A yi (cm) Ayi (cm3) Ayi2 (cm4) Ii (cm4)
1 250 25 6250 156250 52,083.33
2 250 5 1250 6250 520.83
3 250 25 6250 156250 52,083.33
750 13,750 318,750 104,687.49
If Ix’ is the second moment of area of the composite area about the axis x’-x’ then,
Ix’ = Ayi2 + Σ Ii = 318,750 + 104,687.49 = 423437.49 cm4
If y is the centroid, then y = ∑ Ai yi = 13, 750 = 18.33 cm
∑A 750
I x can be determined using the parallele axis theorem, thus, I x
Ix = Ix’ −y2 ΣA
= 423437.49 − (18.33)2(750) = 171,445.815 cm4
I can be obtained from:
I y = 12
1
(50)(60)3 − 12
1
(45)(50)3 = 431,250 cm4
3
Find the second moment of area of the given section I-section y
y’
about its centroidal axes Ix and Iy (UNITS in cm)
1
Start with Iy 2
x x
60
y’
50
Element Area A yi (cm) Ayi (cm3) Ayi2 (cm4) Ii (cm4)
1 250 25 6250 31250 625000
2 250 25 6250 31250 250
3 250 25 6250 31250 625000
Σ 750 18,750 93,750 1250250
Centroid y = ∑ Ai yi = 18, 750 = 25
∑A 750
Moment of area about axis y’- y’ is Iy’ = ΣAyi2 + ΣIi = 1344,000 cm4
From parallel axis theorem: Iy = Iy’ −y2ΣA = 1344,000 − 252(750)
= 875,250 cm4
1 1 45
Moment of area about Ix is obtained from: Ix = (50)(60)3 − 2 × ( )(50)3
12 12 2
= 431,250 cm4
4
Tutorial Part_3: 2nd Moment of Area
1. Determine the Ixx for the thin walled sections given on the figures below, whose wall thickness
is 0.01 m. (All units are in m)
x 0.2
x
x x 0.15 x x 0.2
0.1
0.15
0.15
0.15
x y x
0.2 x 0.2
x
y
0.25
2. Find the moment of inertia of the given sections about their centroidal axes (Ix and Iy). C is the
centroid. y
Ix’ = 22.9(106) mm4
6 4
Ix’ = 2.51(10 ) mm Iy’ = 3.60(106) mm4
Iy’ = 29.8(106) mm4
Ix’ = 124(106) mm4
Iy’ = 1.21(109) mm4 y = 334 mm, Ix’ = 3.83 (109) mm4
Iy’ = 914(106) mm4
5
y
5 cm
Square 10 cm
60 cm
x
20 cm
y = 22.5 mm, Ix’ = 34.4 (106) mm4
Iy’ = 122(106) mm4
3. Find the moment of inertia of the following beams about their centroidal axes
Ix = 52.7 (106) mm4
Iy = 2.51(106) mm4
