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Radius of gyration of area, hollow circular section

ÀÛ¼ºÀÚ Uploader : a.c.e. ÀÛ¼ºÀÏ Upload Date: 2023-06-10º¯°æÀÏ Update Date: 2023-06-10Á¶È¸¼ö View : 80

±×¸²°ú °°ÀÌ Á÷°æ D (cm), µÎ²² t (cm) ÀÎ Áß°ø¿øÇü ´Ü¸éÀÇ ´Ü¸é 2Â÷ ¹Ý°æ r (cm) ¸¦ ±¸ÇÑ´Ù.

As shown in the figure, find the radius of gyration of area r (cm) of the cross section of a hollow circular with diameter D (cm) and thickness t (cm).

Since r = sqr(I/A), I is the second moment of area and A is the area,

I = (¥ð/64)*(D^4 - (D-2*t)^4)

A = (¥ð/4)*(D^2 - (D-2*t)^2)

Thus,

r = sqr( (1/16) * (D^4 - (D-2*t)^4) / (D^2 - (D-2*t)^2) )

 = (1/4)*sqr( D^2 + (D-2*t)^2 )

*** Âü°í¹®Çå[References] ***

r = (1/4)*sqr( D^2 + (D-2*t)^2 )
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