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第三題My=Mpy<=1.5My要檢核,應該是1.5My控制
現在的時間是 2025 6月 18 (週三) 4:40 am
版主: 紫煌
CCP1942 寫:(1) "Cm formulation for LRFD design" appears collect. Please see 102年結構技師鋼構第四題 given formulations for Cm, B1 and B2 calculation.
CCP1942 寫:(2) 題目沒給邊界條件, but Kx = 1.79 is given. " Using K=1.0" is not conservative. Because you get much higher capacity.
CCP1942 寫:Thank your for your detailed decription.
(1) I am not familiar with 內政部營建署頒布之"鋼構造建築物鋼結構設計技術規範-鋼結構極限設計法規範. I just answered the problem per real "AISC-LFRD" code as given in test. I can add your remark to my answer.
CCP1942 寫:After a second thought, I do not agree with your comment in (2). Just as you said when K decreases for 1.79 to 1.
如果k愈小--->Pe1 愈大
如果Pe1愈大--->Pu/Pe1 愈小
如果Pu/Pe1愈小--->(1-Pu/Pe1) 愈大
如果(1-Pu/Pe1)愈大---->B1=Cm/(1-Pu/Pe1) 愈小
B1xMnt愈小---->(Not Conservative)
in this case,
K = 1.0 愈小
KL/rx = 1 x 3,750 / 156 = 24.04 愈小
Pel = 3.1416^2 x 200,000 x 17,097 / 24.04^2 = 58,396 KN 愈大
Pu/Pel = 2,050 / 58,396 = 0.035 愈小
( 1 - Pu/Pel ) = 0.965 愈大
B1 = Cm / 0.965 = 1.037 x Cm 愈小 ( was 1.127 x Cm )
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