Let us calculate the moment of inertia about the centroidal axes of the L-section as shown below.
Given data:
Depth of L-section = 75mm.
Width of L-section = 75mm.
Thickness of L-section = 8mm.
First, we have to find the area of the rectangles ABCD & BEFG of the L-section separately. The center of gravity of the L-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 L-section?
Note: All the below-given values are taken from the above-linked article.
Area A1 of rectangle ABCD = 536 mm2
Area A2 of rectangle BEFG = 600 mm2
CG of rectangle ABCD from reference axis XX= Y1 = 41.5mm.
CG of rectangle BEFG from reference axis XX= Y2 = 4mm.
CG of rectangle ABCD from reference axis YY= X1 = 4mm.
CG of rectangle BEFG from reference axis YY= X2 = 37.5mm.
CG of L-section from reference axis XX = Ӯ = 21.69mm.
CG of L-section from reference axis YY = 艾= 21.69mm.
1. Moment of inertia of L-section about X-axis
Iₓₓ = [Ixx1 + Ixx2]
Where Ixx1 👉 Moment of inertia of rectangle ABCD about X-axis
Ixx2 👉 Moment of inertia of rectangle BEFG about X-axis.
Let us draw the local axes X1X1 & X2X2 passing through the CG of rectangles ABCD & BEFG respectively as shown below.
As these two 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 rectangle ABCD about its local X1-axis.
The formula for MI of rectangle = IG1= [ bd³ ÷12]
So, Ixx1 = [( bd³ ÷12 )+A1h1²]
Here, b = width of the rectangle ABCD = 8mm.
d= depth of rectangle ABCD = [75mm. - 8mm.] = 67mm.
h1= [Y1 - Ӯ]
= [37.5mm. - 21.69mm.]
= 15.81mm.
Ixx1 = [( bd³ ÷12 )+A1h1²]
Ixx1 = [( 8×67³ ÷12 )+ {536 × (15.81)²}]
= [(200,508.67) + {133,976.47}]
= 334,485.14 mm⁴
= 3.345× 10⁵mm⁴
Similarly,
b. Ixx2 = [IG2 + A2h2²]
Where IG2 is the MI of the rectangle BEFG about its local X2-axis.
The formula for MI of rectangle = IG2= [ bd³ ÷12]
So, Ixx2 = [( bd³ ÷12 )+A2h2²]
Here, b = width of the rectangle BEFG= 75mm.
d= depth of rectangle BEFG = 8mm.
h2= [ Ӯ -Y2]
= [21.69mm.- 4mm.]
= 17.69mm.
Ixx2 = [( bd³ ÷12 )+A2h2²]
Ixx2 = [( 75×8³ ÷12 )+ {600 × (17.69)²}]
= [(3200) + {187,761.66}]
= 190,961.66 mm⁴
= 1.91× 10⁵mm⁴
1. Moment of inertia of L-section about Y-axis
Iyy = [Iyy1 + Iyy2]
Where Iyy1 👉 Moment of inertia of rectangle ABCD about Y-axis
Iyy2 👉 Moment of inertia of rectangle BEFG about Y-axis.
Let us draw the local axes Y1Y1 & Y2Y2 passing through the CG of rectangles ABCD & BEFG respectively as shown below.
As these two axes are parallel to the Y-axis, we shall use the parallel axis theorem to calculate MI.
a. Iyy1 = [IG1 + A1h1²]
Where IG1 is the MI of the rectangle ABCD about its local Y1-axis.
The formula for MI of rectangle = IG1= [ db³ ÷12]
So, Iyy1 = [( db³ ÷12 )+A1h1²]
Here, b = width of the rectangle ABCD = 8mm.
d= depth of rectangle ABCD = [75mm. - 8mm.] = 67mm.
h1= [ 艾- X1 ]
= [ 21.69mm. - 4mm.]
= 17.69mm.
Iyy1 = [( db³ ÷12 )+A1h1²]
Iyy1 = [( 67×8³ ÷12 )+ {536 × (17.69)²}]
= [(2858.67) + {167,733.75}]
= 170,592.42 mm⁴
= 1.71× 10⁵mm⁴
Similarly,
b. Iyy2 = [IG2 + A2h2²]
Where IG2 is the MI of the rectangle BEFG about its local Y2-axis.
The formula for MI of rectangle = IG2= [ db³ ÷12]
So, Iyy2 = [( db³ ÷12 )+A2h2²]
Here, b = width of the rectangle BEFG= 75mm.
d= depth of rectangle BEFG = 8mm.
h2= [ X2 - 艾]
= [37.5mm. - 21.69mm.]
= 15.81mm.
Iyy2 = [( db³ ÷12 )+A2h2²]
Iyy2 = [( 8×75³ ÷12 )+ {600 × (15.81)²}]
= [(281,250) + {149,973.66}]
= 431,223.66 mm⁴
= 4.31× 10⁵mm⁴
Thank you for going through these calculation steps❤. Have a good day 😄.
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