土木人

沒看到有人貼RC解答
我貼我的好嚕 不過這是我試後再算一次的
(靠 跟我考試時算的不一樣 )

對了 我是古人 所以我用 土木401-86a 解
不過以下我還是會算一下新軌範的答案

有錯記得批評指教 並幫我更正 謝謝

第三題
h=75，b=50
1. 求斷面性質
a. d = 75-4-1.27-3.22/2=68.12 cm & d'=4+1.27+2.54/2=6.54 cm
b. As = 4*3.22 = 32.572cm2 & As’ = 2*5.067 = 10.134 cm2
2. 計算拉力筋降伏時，Asb
a. Cb = {ecu/(ecu+ey)}*d = 40.4 cm => ab=0.85*Cb =34.34 cm
b. Asb*fy = 0.85*fc’*(b-20)*10 + 0.85*fc'*20*(ab-10) + As'*fy
Asb = 54.72cm2 > As (此時壓力筋為降伏，check我省略)
表示實際斷面之 es > ey ， fs = fy
3. 計算壓力筋降伏時，As,lim
a. C,lim = {ecu/(ecu-ey)}*d’ = 20.85 cm => a,lim=0.85*C,lim =17.72 cm
b. As,lim*fy = 0.85*fc’*(b-20)*10 + 0.85*fc'*20*(a,lim-10) + As'*fy
As,lim = 35.88cm2 > As
表示實際斷面之es’ < ey ， fs’ < fy
4. 假設中性軸距拉力側為x (且x < 10/0.85 =11.76 cm)
a. Cc = 0.85fc’*b*(0.85*x) = 10.115x (我就是此步錯了(0.85*x)我忘了*0.85 )
b. Cs = As’*fs = 62.02*(x-d’/x)
c. T = As*fy = 136.8 t
d. Cc + Cs = T 得 10.1x2 - 74.78x - 405.61 = 0
得 x = 11.03 < 10/0.85 = 11.76 (OK)
Cc = 111.57t & Cs = 25.25t
Check Cc + Cs = 136.82 = T (OK)
5. 求Mu (土木401-86a)
a. Mn = Cc*(d – a/2) + Cs*(d-d’) = 86.31 t-m
b. Mu = 0.9 * Mn = 77.68 t-m
6. 求Mu(新規範)
a. Mn = Cc*(d – a/2) + Cs*(d-d’) = 86.31 t-m
b. et = (d-x)/x*ecu = 0.002514 界在0.002~5之間，過渡斷面
Ft = 0.483 + 0.833*et = 0.692
c. Mu = 0.69 * Mn = 59.73 t-m
第四題 稍後在PO