Let us consider an I - section as shown below.
Given data:
Width of top flange = 75mm.
The thickness of top flange = 16mm.
Depth of web = 75mm.
Width of web = 16 mm.
Width of bottom flange = 100mm.
The thickness of bottom flange = 20mm.
Calculation:
The given I- section is symmetrical to Y-axis
What does that mean?
Let us observe the below drawing.
When we divide the I-section into two equal halves by drawing the central Y-axis, two parts on either side are identical having the same mass. The same is not true for the X-axis.
Therefore the center of gravity lies on the central Y-axis.
Now, let us calculate the area of the top flange, web, & bottom flange, & CG of respective rectangles, having the line XX as the reference axis.
The Axis of reference is the lowest line that we take on the drawing. As CG lies on Y-axis, in this case, line XX becomes the reference axis.
1. Area of the top flange
A1= [width × thickness]
= [75mm. × 16mm.]
A1= 1200 mm2
Co-ordinate Y1 of the top flange with reference to the line XX
= [ thickness of bottom flange + depth of web + (top flange thickness ÷ 2)]
= [ 20mm. + 75mm. + ( 16mm. ÷ 2)]
Y1 = 103mm.
2. Area of the web
A2 = [width × depth]
= [16mm. × 75mm.]
A2= 1200 mm2
Co-ordinate Y2 of the web with reference to the line XX
= [ thickness of bottom flange + (depth of web ÷ 2)]
= [ 20mm. + ( 75mm. ÷ 2)]
Y2= 57.5mm.
3. Area of the bottom flange
A3= [width × thickness]
= [100mm. × 20mm.]
A3= 2000 mm2
Co-ordinate Y3 of the bottom flange with reference to the line XX
= [bottom flange thickness ÷ 2]
= [20mm. ÷ 2]
Y3 = 10mm.
Now, the center of gravity of given I-section
Ӯ = [(A1×Y1) + (A2 ×Y2) +(A3 ×Y3 ) ÷ ( A1 + A2 +A3)]
= [(1200× 103) + (1200 ×57.5 ) + (2000 ×10 ) ÷ ( 1200 + 1200 +2000)]
= [2,12,600 ÷ 4400 ]
Ӯ = 48.32mm.
Note:
1. Distance of Y1, Y2, Y3, & center of gravity Ӯ, is measured from the reference axis - XX.
Also, go through 👇
👀. How to find the center of gravity of a C-section?
Thank you for going through these calculation steps❤. Have a good day 😄.
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