Let us calculate the moment of inertia along the centroidal axis for the I-section as shown below.
Given data:
Width of top flange = 75mm.
Thickness of top flange = 16mm.
Depth of web = 75mm.
Width of web = 16 mm.
Width of bottom flange = 100mm.
Thickness of bottom flange = 20mm.
First, you have to find the area of the web & flange of the I-section separately. The center of gravity of the I-section should be calculated.
All these steps are already covered in a separate article as linked below.
👀. How to find the center of gravity of an I -section?
Area A1 of top flange = 1200 mm2
Area A2 of web = 1200 mm2
Area A3 of bottom flange =2000 mm2
CG of top flange Y1 = 103mm.
CG of web Y2 = 57.5mm.
CG of bottom flange Y3 = 10mm.
CG of I-section Ӯ = 48.32mm.
Note: All the Values are taken from the above-linked article.
1. Moment of inertia of I-section about X-axis
Iₓₓ = [Ixx1 + Ixx2 + Ixx3]
Where Ixx1 👉 Moment of inertia of top flange about X-axis
Ixx2 👉 Moment of inertia of web about X-axis.
Ixx3 👉 Moment of inertia of bottom flange about X-axis
Let us draw the local axes X1X1, X2X2, & X3X3 passing through the CG of web & flange as shown below.
As all these three axes are parallel to the X-axis, we shall use the parallel axis theorem to calculate MI.
a. Ixx1 = [IG1 + A1h1²]
Where IG1 is the MI of the top 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 top flange = 75mm.
d= depth of top flange = 16mm.
h1= [Y1 - Ӯ]
= [103mm. - 48.32mm.]
= 54.68mm.
Ixx1 = [( bd³ ÷12 )+A1h1²]
Ixx1 = [( 75×16³ ÷12 )+ {1200 × (54.68)²}]
= [(25,600) + {3,587,882.88}]
= 3,613,482.88 mm4
= 3.613× 10⁶mm⁴
Similarly,
b. 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 = 16mm.
d= depth of web = 75mm.
h2= [ Y2 - Ӯ]
= [57.5mm.- 48.32mm.]
= 9.18mm.
Ixx2 = [( bd³ ÷12 )+A2h2²]
Ixx2 = [( 16×75³ ÷12 )+ {1200 × (9.18)²}]
= [(562,500) + {101,126.88}]
= 663,626.88 mm⁴
= 0.664× 10⁶mm⁴
c. Ixx3 = [IG3 + A3h3²]
Where IG3 is the MI of the bottom flange part about its local X3-axis.
The formula for MI of rectangle = IG3= [ bd³ ÷12]
So, Ixx3 = [( bd³ ÷12 )+A3h3²]
Here, b = width of the bottom flange = 100mm.
d= depth of bottom flange = 20mm.
h3= [ Ӯ - Y3]
= [ 48.32mm. - 10mm.]
= 38.32mm.
Ixx3 = [( bd³ ÷12 )+A3h3²]
Ixx3 = [( 100×20³ ÷12 )+ {2000 × (38.32)²}]
= [(66,666.67) + {2,936,844.80}]
= 3,003,511.47 mm⁴
= 3.0× 10⁶mm⁴
2. Moment of inertia of I-section about Y-axis
Iyy = [Iyy1 + [Iyy2 + I yy3]
Where Iyy1 👉 Moment of inertia of top flange about Y-axis
Iyy2 👉 Moment of inertia of web about Y-axis.
Iyy3 👉 Moment of inertia of bottom flange about Y-axis
As the CG of web & flange lies over the centroidal Y-axis, there is no need to apply the parallel axis theorem.
Therefore,
a. Iyy1 = [ db³ ÷12]
Here, b = width of the top flange = 75mm.
d= depth of top flange = 16mm.
Iyy1 = [ 16×75³ ÷12]
= 562,500mm⁴
b. Iyy2 = [ db³ ÷12]
Here, b = width of the web = 16mm.
d= depth of web = 75mm.
Iyy2 = [ 75×16³ ÷12]
= 25,600mm⁴
c. Iyy3 = [ db³ ÷12]
b = width of the bottom flange = 100mm.
d= depth of bottom flange = 20mm.
Iyy3 = [ 20 ×100³ ÷12]
= 1,666,666.67mm⁴
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