混凝土结构设计原理 第四版 程文.doc

思路岛教育网 整理提供第3章习题3.1解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 300 N mm2 ；C30 混凝土，一类环境，c=25mm，as=35mm，h0=500-35=465mm s =M 90 1061 f cbh0 1.0 14.3 250 4652= 0.116查表得： = 0.123 b = 0.55As =1 fcb h0f y=1.0 14.3 250 0.123 465300= 681.6mm2选 2 16+2 14 钢筋， As = 710mm2 (满足构造要求)。min = 0.45ftf y1.43300minbh = 0.0021 250 500 = 262.5mm2 As = 710mm2取 As = 710mm23.2 解： fc = 19.1 N mm2 ； ft = 1.71 N mm2 ； f y = 360 N mm2C40 混凝土，二类 a 环境，c=30mm，as=40mm，h0=450-40=410mm s =M 140 1061 fcbh0 119.1 200 4102查表得： = 0.249 b = 0.518As =1 f cb h0f y=1.0 19.1 200 0.249 410360= 1083.3mm2选 3 22 钢筋， As = 1140mm2 (满足构造要求)。 min = 0.45ftf y1.71360minbh = 0.0021 200 450 = 189mm2 0.002= = 0.2182= 0.45 = 0.0021 0.002C30 混凝土，一类环境，c=25mm，as=25+16/2=33mm，h0=450-33=417mmx =f y As1 fcb=360 8041.0 14.3 200= 101.2mm bh0 = 0.518 417 = 216mmmin = 0.45ftf y1.43360minbh = 0.002 200 450 = 180mm2 M = 84kN m3.5解： fc = 11.9 N mm2 ； f y = f y = 300 N mm2 ； C25 混凝土，一类环境，c=25mm，设受拉钢筋放两层，as=60mm，h0=h-as=500-60=440mmAs =M 1 fcbh0 b (1 0.5b )f y(h0 as )=260 106 111.9 200 4402 0.55 (1 0.5 0.55)300 (440 35)= 628mm2As =1 fcbbh0 + f y Af y=111.9 200 0.55 440 + 300 628300= 2548mm2As 选 3 25+3 22， As 2613mm2； As选 2 20， As628mm23.6 解：1） fc = 19.1 N mm2 ； ft = 1.71 N mm2 ； f y = 360 N mm2 ，C40 混凝土，二类 a 环境，c=30mm，设受拉钢筋放两层，as=65mm，h0=750-65=685mm1 fcb h (h0 ) = 1.0 19.1 550 100 (685 2 2)= 667.1106 N mm 500 106 N mm属于第一类 T 形截面 s =M 500 1061 fcb h0 119.1 550 6852查表得 = 0.107 b = 0.518（可不验算）As =1 fcb h0f y=1.0 19.1 550 0.107 685360= 2139mm2选 6 22 钢筋， As = 2281mm2 (满足构造要求)。 min = 0.45ftf y1.71360minbh = 0.0021 250 750 = 393.75mm2 500 106 N mm属于第一类 T 形截面 s =M 500 1061 fcb h0 0.98 27.5 550 6852= 0.072查表得 = 0.075 b = 0.499 （可不验算）As =1 fcb h0f y=0.98 27.5 550 0.075 685360= 2115mm2选 8 18 钢筋， As = 2036mm2 (满足构造要求)。 min = 0.45ftf y2.04360minbh = 0.00255 250 750 = 478.125mm2 As = 2036mm2取 As = 2036mm23.7解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 360 N mm2 ；C30 混凝土，一类环境，c=25mm，设受拉钢筋放两层，as=60mm，h0=h-as=500-60=440mm1 f cb h (h0 ) = 1.0 14.3 400 80 (440 2 2)= 183.04 106 N mm 300 106 N mm属于第二类 T 形截面Mu1 = 1 fc (b b)h (h0 hf2) = 1.014.3 (400 200) 80 (440 802)= 91.52106 N mmM u 2 = M M u1 = 300 106 91.52 106 =208.48106Nmm s =M u 2 208.48 1061 f cbh0 114.3 200 4402查表得 = 0.504 b = 0.518As 2 =1 f cb h0f y=1.0 14.3 200 0.504 440360= 1762mm2As1 =1 fc (bf b)hff y=114.3 (400 200) 80360= 636mm2As = As1 + As 2 = 1762 + 636 = 2398mm2选 5 25 钢筋， As = 2454mm2 (满足构造要求)。= = 0.3772斜截面受剪承载力计算例题1 12 22）复合截面尺寸hw=h0=h-c-25/2=500-30-12.5=457.5hwb=457.5200= 2.3 V = 124.6kN满足。3）验算是否按计算配置腹筋0.7 ftbh0 = 0.7 1.1 200 457.5 = 70.5kN V = 124.6kN应按计算配置腹筋4）计算腹筋数量只配箍筋由 V 0.7 ftbh0 + 1.25 f yvAsvsh0 得： = = 0.451 mm2/mms 1.25 f yv h0 1.25 210 457.5选双肢 8 箍筋 s nAsv1 2 50.30.451 0.451取 s=200mm验算最小配箍率 sv =nAsv1bs2 50.3 f 1.1200 200 f yv 210满足仅配箍筋时的用量为双肢 8200即配箍筋又配弯筋a. 先选弯筋，再算箍筋根据已配的 2 25+1 22 纵向钢筋，将 1 22 的纵筋以 45角弯起，则弯筋承担的剪力：Vsb = 0.8 f y Asb sin s = 0.8 380.1 300 22= 64.5kNnAsv1s = = 负值1.25 f yvh0 1.25 210 457.5按构造要求配置箍筋并满足最小配箍率要求选双肢6200 的箍筋， sv =nAsv1bs2 28.3 f 1.1200 200 f yv 210b. 先选箍筋，再算弯筋先按构造要求选双肢6200 的箍筋， sv =nAsv1bs2 28.3 f 1.1200 200 f yv 210满足要求。Asb V 0.7 ftbh0 1.25 f yv0.8 f y sin sAsvsh0=222 28.3200= 118.5mm225）验算弯起钢筋弯起点处的斜截面抗剪承载力弯起钢筋弯起点距支座边缘 500-30-22/2-30-20/2+50=469mm，该处剪力V1=124.6-0.46970=91.77kNVcs = 0.7 ftbh0 + 1.25 f yvAsvs2 28.32003故不需要弯起第二排钢筋或加大箍筋用量。42 解：1）剪力图见书2）复合截面尺寸hw=h0=h-c-20/2=600-30-10=560hwb=560200= 2.8 Vmax = VA = 180kN满足。3）验算是否按计算配置腹筋A 支座：V集V总=160180V 140V总 160梁的左右区段均应按集中荷载作用下的独立梁计算将梁分为 AC、CD、DE、EB 段来计算斜截面受剪承载力AC 段： =ah010005601.75 1.75 + 1.0 1.79 + 1应按计算配置腹筋= = 0.00142 sv,min = 0.24 t = 0.24 = 0.0013= = 0.00142 sv,min = 0.24 t = 0.24 = 0.0013124.6 103 70.5 103 1.25 210 457.50.8 300 弯起 1 22 的纵筋，Asb=380.1mmh0 = 70.5 103 + 1.25 210 457.5 = 104.5 103 N91.7710 N= 88% ；B 支座： 集 = = 87.5%= = 1.79ftbh0 = 1.1 200 560 = 77.3kN 3 ，取 = 3ftbh0 = 1.1 200 560 = 53.9kN VC=50kN= = 3.57 3 ，取 = 3ftbh0 = 1.1 200 560 = 53.9kN VE=70Kn= = 1.79ftbh0 = 1.1 200 560 = 77.3kN sv,min = 0.24 t = 0.24 = 0.00126(70 53.9) 103= = 734.3mm sv =nAsv1bs2 50.3 f 1.1200 250 f yv 210满足箍筋用量为双肢 8250EB 段：nAsv1sVB 1.75f bhf yv h0 210 560选双肢 8 箍筋 s nAsv1 2 50.30.703 0.703取 s=140mm验算最小配箍率 sv =nAsv1bs2 50.3 f 1.1200 140 f yv 210满足箍筋用量为双肢 814043 解：1）剪力图见书2）复合截面尺寸hw=h0-hf=h-60-200=700-60-200=440hwb=440250= 1.76 Vmax = VA = 343.75kN满足。3）验算是否按计算配置腹筋AC 段： =ah015006401.75 1.75 + 1.0 2.34 + 1应按计算配置腹筋CB 段： =ah025006401.75 1.75 + 1.0 3 + 1应按计算配置腹筋4）计算腹筋数量= = 0.00201 sv,min = 0.24 t = 0.24 = 0.00126(160 77.3) 103= = 143.1mm= = 0.00359 sv,min = 0.24 t = 0.24 = 0.00126= = 2.34ftbh0 = 1.43 250 640 = 119.88kN 3，取 = 3ftbh0 = 1.43 250 640 = 100.10kN VB=206.25kNAC 段：在上排 2 根25 纵筋中先弯起一根（Asb=490.9mm2）nAsv1sVA 1.75 + 1.0 t 0f yv h0=343.75 103 1.752.34 + 11.43 250 640 0.8 360 490.9 300 64022 = 0.645mm2 / mm选双肢 8 箍筋 s nAsv1 2 50.30.6