第三題
1.求有效深度 ( 拉或壓力鋼筋之其形心至壓力側距離 )
a.拉力側有效深度 d = 50cm ,
b.壓力側(如果有配壓力筋且壓力筋與最外層拉力筋同號數) d' = 60 - 53.5 = 6.5cm
2.檢核 T beam 尺寸
根據土木401-96之2.12.3 (page2-13)有獨立T beam之規定
a.有效長度 be 小於等於 4bw = 120 cm ,check NG
取 be=120 cm
b. hf 大於等於 bw / 2 = 15cm ,check OK
註:規定翼版厚度乃因其可發揮的有效抗壓影響範圍,所以不是翼版越長,抗壓面積越大
3.判是否可以視為矩形梁配筋
a. Mn = Cc * [ d - ( hf / 2 ) ] = 0.85 * fc' * be * hf * [ d - ( hf / 2 ) ]
=> Mn = 0.85*210*120*15*[50-(15/2)] = 136.5525 t-m
b. Md = f * Mn = 0.9 * 136.5525 = 122.90 t-m 小於 Mu = 125 t-m ,check NG
c. As = Asf + Asw = 57.375 + 20.75 = 78.12 cm2 ( 感謝 cance大大提供修正 )
d. check As,req 滿足最小鋼筋量Asmin否
Pmin = max { 14 / fy , 0.8 * fc'^0.5 / fy } = 0.00333
=> Asmin = Pmin * bw * d = 0.00333 * 30 *50 = 5 cm2 ,check OK
e. check As,req 滿足最大鋼筋量Asmax否
X0.004 = d * 3 / ( 3 + 4 ) = 50 * 3 / 7 = 21.43
Cc = 0.85 * fc' * bw * ( B1 * X0.004 ) = 0.85*210*30*(0.85*21.43) = 97.544 t
T = Asw,max * fy , 且 T = Cc
=> Asw,max = Cc / fy = 23.225 cm2
=> T beam 採單筋之Asmax = Asf + Asw,max = 80.6 cm2 ,check OK
f. check f = 0.9 OK否
Cc = 0.85 * fc' * bw * ( B1 * X )
Tw = Asw * fy ,且 Cc = Tw
X = 20.75 * 4200 / ( 0.85 * 210 * 30 * 0.85 ) = 19.15 cm
Es = Ecu * ( d - X ) / X = 0.004834 > 0.005 ,check NG !!
g. 採假設 f 介於 0.65 ~ 0.9 單筋設計
由 f = 0.65 + 0.25 * ( Es - 0.002 ) / ( 0.005 - 0.002 ) = 0.483 + 83.3 * Es ,
且 Es = Ecu * ( d - X ) / X
得 f = 0.483 + 0.25 * ( d - X ) / X = ( 12.5 + 0.233 * X ) / X
Mnw = Mn - Mnf = ( Mu / f ) - Mnf = ( 125 / f ) - 102.41 ( t-m )
Mnw = 125 * X / ( 12.5 + 0.233 * X ) - 102.41
Ccw = 0.85 * fc' * bw * ( B1 * X ) = 0.85*210*30*(0.85*X) = 4.552*X (t)
Mnw = Ccw * ( d - B1 * X / 2 ) = 4.552*X*( 50 - 0.85*X/2) / 100
Mnw = 2.276*X - 0.0193*X^2 ( t-m ) = 125 * X / ( 12.5 + 0.233 * X ) - 102.41 ( t-m )
由上可得 一元三次方程式 M(X) = 0 , 求 X
M(X) = 4.5*X^3 - 289.06*X^2 + 72688.47*X - 1280125 = 0
得 X = 18.59 cm
註:求一元三次可用牛頓定律
check f = ( 12.5 + 0.233 * X ) / X = 0.905 大於 0.9 ,check NG!!