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1、4.14.24.34.44.54.64.74.84.94.104.114.124.134.145.1混凝土结构基本原理习题参考答案第4章 受弯构件正截面的性能与设计q = 19.4kN/mkh = 600 - 40 = 560mm, A = 875mm2 , 2 巫20 +118 (A =882mm2)0 s sh = 100 - 03 = 70mm, A = 177mm2, Q 6150 ( A =189mm2/m)0ssHRB400, C30, b X h = 200mmX500mm, A =450mm2, 3 密 14 ( A =462mm2)ssh = 450mm, h = 450 -
2、 40 = 410mm, A = 915mm2 0sh = 500mm, h = 500 - 40 = 460mm, A = 755mm20sh = 550mm, h = 550 - 40 = 510mm, A = 664mm2 0s随梁截面高度增加，受拉钢筋面积减小。b = 200mm, h = 500 - 40 = 460mm, A = 925mm2 0sb = 250mm, h = 500 - 40 = 460mm, A = 709mm 2 0sh = 300mm, h = 500 - 40 = 460mm, A = 578mm20s 随梁截面宽度增加，受拉钢筋面积减小。C20, h =
3、 500 - 40 = 460mm, A = 981mm20sC25, h = 500 - 40 = 460mm, A = 925mm20sC30, h = 500 - 40 = 460mm, A = 895mm20s随梁截面宽度增加，受拉钢筋面积减小。HRB400, h = 500 - 40 = 460mm, A = 925mm2 0sHRB500, h = 500 - 40 = 460mm, A = 765mm20s随受拉钢筋强度增加，受拉钢筋面积减小。(1) M 二 122.501 kNmu(2) M 二 128.777 kNmu(3) M = 131.126 kN :mu(4) M =
4、 131.126 kN :mua = 45mm , A = 878mm2，选配 3 20 (A = 942mm 2)sssa = a = 40mm , A = 1104mm2，选配 2 20+2 18 ( A = 1137mm2)s sss(1) M 二 121.882 kNmu(2) M 二 214.169 kNmu(1) A = 822mm 2，选配 2 20+218 ( A = 1137mm 2)ss(2) A = 2167mm2，选配 6亚22 ( A = 2281mm2)ssa = 60mm , A = 2178mm2，选配 622 (A = 2281mm2)sss第5章 受压构件f
5、 = 16.7N/mm2, f = 410N/mm2，取 b 二 400mm , h 二 400mm ,cy5.25.35.45.55.65.75.85.95.106.16.27.17.27.37.4A二 2718mm2，选配 8叮22。 sf = 19.lN/mm2 ,广二 360N/mm2 , A = 2818mm2，选配 8 “22。(10T22),cysf 二 16.7N/mm2, f = 410N/mm2, f = 435N/mm2，取A = 3800mm2 cyyvs螺旋箍筋为!Tl253, N = 4495.5kN 。uf = 16.7N/mm2, f 二435N/mm2 , f
6、 二410N/mm2, a = a = 40mm ,cyys s1)大偏心受压， A = 635mm2， A =555mm2；ss(2)大偏心受压，已知 A = 942mm2 , A = 374mm2 o ssf =19.1N/mm2， f =360N/mm2， f =360N/mm2，cyy小偏心受压，拟定 A = 603mm2, g hh，A = A = 915mm2 。fssf =16.7N/mm2，c对称配筋，排架柱，f = 435N/mm2， f = 410N/mm 2 ，yy小偏心受压，x (h - h )， A = A = 413mm2。 fss第6 章 受拉构件f = 16.7
7、N/mm2 ， f = 300N/mm2 ， f = 300N/mm2 ， cyy小偏心受拉， A =933mm2， A = 267mm2。ssf =14.3N/mm2， f = 360N/mm2 ， f = 360N/mm2 ， cyy大偏心受拉，A p bh，取 A = 308mm2，且x = 2a ， A = 2084mm2。s minss s第 7 章 构件斜截面受剪s = 205 mmAa 取 40mm, f = 0.785ssAsv = 0.675sa 取 40mms序号b X h (mm)混凝土强度等级-8计算（s值）实配（s值）1250x500C2594902250x500C3
8、01001003300x500C251041004250x 600C251281207.5支座边缘最大剪力截面取8250，并弯起1120 根。7.6 纵筋：跨中 A = 2556mm2，支座处 A 2370 mm2ssA腹筋：支座边缘箍筋 f 0.694,弯起近5 根s7.7 (此题纵筋类型不对，应改为HRB400级)a取35mms8200均布荷载设计值为69kN/m 丄10200均布荷载设计值为71kN/m (由正截面承载力控制)A7.8 a 取 40mm, f 0.783ssA7.9 a 取 35mm, f 0.586ss7.10 a 取 120mm，跨中截面 A 5382 mm2 1/4
9、 截面：A 3752 mm2sss可不用弯起筋则支座截面P 0.00695sv第8章 受扭构件8.1 保护层厚度 c = 20mm， a a 40mm mm， h h 400 40 360 m。s s w 0箍筋：双肢0 8200顶部纵筋：2娄12+1娄16( A = 427 mm2)s每侧面纵筋：2歩10( A = 157 mm2)s底部纵筋：2歩10 ( A = 157 mm2)s8.2 a a 40mm , h 500 40 460 mm, h h h 460 100 360 mm。s s0w 0 f腹板箍筋：双肢 8200腹板纵筋顶部纵筋：2 II 10( A = 157 mm2)s每
10、侧面纵筋：2娄10( A = 157 mm2)s底部纵筋：2亚 12+1 亚 18( As = 481 mm2)翼缘受扭箍筋：双肢 8200S翼缘受扭纵筋：4 II 10( A = 314 mm2)s第9章 正常使用阶段验算9.1箍筋直径6mm,c =26mm ，swmax 0.15mm9.2箍筋直径6mm,c =26mm ，w 0.21mmsmax9.3箍筋直径6mm,c =26mm ，w 0.16mm ,f 14.5mmsmaxM = 100.600kN-m, A = 780mm2，实配纵筋 2生 18+虫20 ( A = 823mm2)ss9.4 箍筋直径 6mm, c =26mm, w
11、 = 0.15mm , f = 13.7mmsmaxM = 100.600kN - m, A = 808mm2，实配纵筋 2生 18+空20 ( A = 823mm2)ss第10章预应力混凝土构件10.1 b = 40.625N / mm2 , c = 42.39N / mm2 , q = 117.75 N / mm2,11 1214c = 132.98 N / mm2,1510.2 c = 42.39N / mm2 , c = 40 N / mm2 , c = 48.825 N / mm212 1314c = 131.215 N / mm2, c = 156.9 N / mm2 , c =
12、288.115 N / mm211151c -c = -3.71N / mm2 f = 2.4N / mm2，满足要求ckpctc -c =-4.81N/mm2 460kN，满足要求u10.3预应力筋：6 S12.9低松弛钢绞线，A = 512.4mm2，曲线布置方程y = 0445x2 ; p4.52非预应力筋：受拉区、受压区各配HRB335级4 12, A = A = 452mm2 ;ssc = 355.62N；mm2 ;c -c = -2.6Nmm2 f = 2.4Nmm2 , 满足要求；ckpctkf = f - f = 12.23-11.82 = 0.41mm= 35mm，满足要求。1 p250如有侵权请联系告知删除，感谢你们的配合！
