How to find the center of gravity of a C-section?/ Calculating centroid of a channel section.

 Let us consider a channel-section or C-section as shown below.

Given data:

Depth of C-section = 250mm.

Width of C-section = 80mm.

Thickness of C-section = 10mm.

Calculation:

The given C- section is symmetrical to X-axis 

What does that mean?

Let us observe the drawing as shown below.

When we divide the C-section into two equal halves by drawing  X-axis,  two parts on either side are identical having the same mass. The same is not true for the Y-axis.

Therefore the center of gravity lies on the X-axis.

Let us split the C-section into three rectangles ABCD, BEFG & EHIJ having the line YY as the reference axis.

The Axis of reference is the edge line that we take on the drawing. As CG lies on X-axis, in this case, line YY becomes the reference axis.

Area of rectangle ABCD 

A1= [width × thickness]

    = [80mm. × 10mm.]

 A1= 800 mm2

Co-ordinate X1 

  = [ width of rectangle ABCD ÷ 2]

 = [ 80mm.÷ 2)]

X1 = 40mm.

Area of rectangle BEFG 

A2 = [thickness × depth]

    = [10mm. ×  230mm.]

A2= 2300 mm2

Co-ordinate X2

= [ thickness of rectangle BEFG  ÷ 2)]

= [ 10mm.  ÷ 2)]

X2= 5mm.

Area of rectangle EHIJ

A3= [width × thickness]

    = [80mm. × 10mm.]

 A3= 800 mm2

Co-ordinate X3 

  = [ width of rectangle EHIJ ÷ 2]

 = [ 80mm.÷ 2)]

X3 = 40mm.

Now, the center of gravity of the given C-section/channel section

 艾= [{(A1×X1) + ( A2 ×X2) + (A3 × X3)} ÷ ( A1 + A2 + A3)]

    = [{(800× 40) + (2300 ×5) + (800 × 40)} ÷ ( 800 + 2300 + 800)]

   = [75,500  ÷ 3900 ]

   艾 = 19.36mm.

Note: 

1. Distance of X1, X2, X3 & center of gravity 艾, is measured from the reference axis - YY.

Also, go through 👇

👀. How to find the center of gravity of an L-section?

Thank you for going through these calculation steps❤. Have a good day 😄.

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