Let us calculate the moment of inertia along the centroidal axis for the T-section as shown below.
Given data:
Width of flange = b= 120mm.
Depth of flange =d =25mm.
Depth of T - section = 150mm.
Web width = 25mm.
First, you have to find the area of the web & flange of the T-section separately. The center of gravity of the T-section should be calculated.
All these steps are already covered in a separate article as linked below.
👀. How to calculate the center of gravity of a T -section?
Area A1 of flange = 3000 mm2
Area A2 of web = 3125 mm2
CG of flange Y1 = 137.5mm.
CG of web Y2 = 62.5mm.
CG of T-section Ӯ = 99.23mm.
Note: All the Values are taken from the above-linked article.
1. Moment of inertia of T-section about X-axis
Iₓₓ = [Ixx1 + I xx2]
Where Ixx1 👉 Moment of inertia of flange about X-axis
Ixx2 👉 Moment of inertia of web about X-axis.
Let us draw the local axes X1X1 & X2X2 passing through the CG of flange & web respectively as shown below.
As these two axes are parallel to the X-axis, we shall use the parallel axis theorem to calculate MI.
Ixx1 = [IG1 + A1h1²]
Where IG1 is the MI of the flange part about its local X1-axis.
The formula for MI of rectangle = IG1= [ bd³ ÷12]
So, Ixx1 = [( bd³ ÷12 )+A1h1²]
Here, b = width of the flange = 120mm.
d= depth of flange = 25mm.
h1= [Y1 - Ӯ]
= [137.5mm. - 99.23mm.]
= 38.27mm.
Ixx1 = [( bd³ ÷12 )+A1h1²]
Ixx1 = [( 120×25³ ÷12 )+ {3000 × (38.27)²}]
= [(1,56,250) + {43,93,778.70}]
= 45,50,028.70 mm4
= 4.55× 10⁶mm4
Similarly,
Ixx2 = [IG2 + A2h2²]
Where IG2 is the MI of the web part about its local X2-axis.
The formula for MI of rectangle = IG2= [ bd³ ÷12]
So, Ixx2 = [( bd³ ÷12 )+A2h2²]
Here, b = width of the web = 25mm.
d= depth of web = 125mm.
h2= [ Ӯ - Y2]
= [99.23mm.- 62.5mm.]
= 36.73mm.
Ixx2 = [( bd³ ÷12 )+A2h2²]
Ixx2 = [( 25×125³ ÷12 )+ {3125 × (36.73)²}]
= [(40,69,010.42) + {42,15,915.31}]
= 82,84,925.77 mm⁴
= 8.285× 10⁶mm⁴
2. Moment of inertia of T-section about Y-axis
Iyy = [Iyy1 + I yy2]
Where Iyy1 👉 Moment of inertia of flange about Y-axis
Iyy2 👉 Moment of inertia of web about Y-axis.
As the CG of the web & flange lies over the centroidal Y-axis, there is no need to apply the parallel axis theorem.
Therefore,
Iyy1 = [ db³ ÷12]
Here, b = width of the flange = 120mm.
d= depth of flange = 25mm.
Iyy1 = [ 25×120³ ÷12]
= 36,00,000mm⁴
Iyy2 = [ db³ ÷12]
Here, b = width of the web = 25mm.
d= depth of web = 125mm.
Iyy2 = [ 125×25³ ÷12]
= 1,62,760.42mm⁴
