Architecture
°ÇÃà Architecture


Áß°ø ¿øÇü ºÎÀçÀÇ µÎ²² °áÁ¤(thickness of hollow circular cross section)

ÀÛ¼ºÀÚ Uploader : ¸¶Áß¹° ÀÛ¼ºÀÏ Upload Date: 2022-06-09º¯°æÀÏ Update Date: 2022-06-09Á¶È¸¼ö View : 265

¿Ü°æ Do (mm) Áß°ø ¿øÇüºÎÀçÀÇ Çã¿ëÀÀ·ÂÀÌ ¥òa (GPa) ÀÏ ¶§, Çã¿ëÇÏÁßÀÌ P (kN) ÀÌ µÇ±â À§ÇÑ µÎ²² t (mm) ¸¦ ±¸ÇÑ´Ù.

³»°æÀ» Di (mm) ¶ó Çϸé,

Di = Do-2t

°¡ µÈ´Ù.

ÈûÀ» ¹Þ´Â ¸éÀû A (mm^2) ´Â,

A = (¥ð/4)(Do^2-Di^2)

ÀÌ°í, ÀÀ·Â ¥ò (kN/mm^2, GPa) ´Â,

¥ò = P/A = 4P/(¥ð*(Do^2-Di^2))

ÀÌ µÈ´Ù. ´ÜÀ§¸¦ ÀÏÄ¡½ÃÅ°¸é ¥ò ¡Â ¥òa/1000 À̹ǷÎ,

4P/(¥ð*(Do^2-Di^2)) = ¥òa/1000

Do^2-Di^2 = 4000*P/(¥ð*¥òa)

Di = sqr(Do^2-4000*P/(¥ð*¥òa))

Do-2t = sqr(Do^2-4000*P/(¥ð*¥òa))

t = (Do-sqr(Do^2-4000*P/(¥ð*¥òa)))/2

ÀÌ µÈ´Ù.

*** Âü°í¹®Çå[References] ***

t = (Do-sqr(Do^2-4000*P/(¥ð*¥òa)))/2
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