【這是我的計算過程，不知道是否有錯誤，請各位前輩指教。】
[2009/07/22訂正]
Asb = [0.85fc’(ab)×b] / fy + [As’(fy- 0.85fc’)/fy]
因為fy = 4200 kgf/cm^2，所以εy = 0.002
Xb = 0.003×d / (0.003+εy) = 0.003×55 / (0.003+0.002) = 33 cm
(ab) = β1×Xb = 0.85×33 = 28.05 cm
Asb = [0.85×280×(28.05×25+15×50)] / 4200 + 18×(1- 0.85×280 / 4200) = 99.22 cm^2
Asy = [0.85fc’(ay)×b] / fy + [As’(fy- 0.85fc’)/fy]
Xy = 0.003×d’ / (0.003-εy) = 0.003×(6) / (0.003-0.002) = 18 cm
(ay) = β1×Xy = 0.85×18 = 15.3 cm
Asy = [0.85×280×(15.3×25+15×50)] / 4200 + 18×(1- 0.85×280 / 4200) = 81.155 cm^2
As = 70 cm^2 ＜ Asb = 99.22 cm^2
As = 70 cm^2 ＜ Asy = 81.155 cm^2
所以拉力筋降服，壓力筋未降服。
判斷壓力區塊深度是否大於15cm
As15 = [0.85fc’(a15)×b] / fy + [As’(f's- 0.85fc’)/fy]
a = 15cm，X = 15 / 0.85 = 17.65cm
ε's = 0.003×(X - d') / X = 0.003×(17.65 - 6) / 17.65 = 0.00198 ＜εy = 0.002
所以 f's = ε's×Es = 0.00198×2.04×10^6 = 4039.2 kg/cm^2
As15 = [0.85×280×15×75] / 4200 + [18×(4039.2- 0.85×280)/4200] = 80.05cm^2
As = 70 cm^2 ＜ As15 = 80.05 cm^2
所以 a ＜ 15cm
Ts = Cc + Cs
As×fy = 0.85fc’β1×X×b + As’×(fs’-0.85fc’)
70×4200 = 0.85×280×085×X×75 + 18×{[0.003×(X-6)/X]×2.04×10^(6) – 0.85×280}
15172.5X^(2) – 188124X – 660960 = 0
解方程式得X =15.255
(a) = β1×X = 0.85×15.255 = 12.97 cm
Mn = Ts×(d – a/2) = Cc×(d – a/2) + Cs×(d – d’)
= 0.85fc’ab×(d – a/2) + As’×(fs’-0.85fc’)×(d – d’)
=0.85×280×12.97×75×(55 - 12.97/2)
+18×{[0.003×(15.255-6)/15.255]×2.04×10^(6) – 0.85×280}×(55-6)
= 142.97 tf - m
