混凝土结构中册习题答案

1、混凝土结构中册习题答案第11章11.1解：1、求支座截面出现塑性较时的均布荷载首先计算支座截面极限抗弯承载力MuA:C20混凝土查得fc=mm2,316As=603mm26033001fcb9.620094mmbh0MuAx94Asfy(h0-)603300(465万)按弹性分析：MAMuAql21212MuaqT1275.66225.2kNmq125.2kN/m2、计算跨中极限抗弯承载力Mg：216As=402mm24023009.620063mm,Mu1402300465636352.3kNm2总弯矩M总MuaMui75.652.3127.9kNm,2Pul/日得8Pu8M总8127.96

2、228.4kN/m3、假设均布荷载同为MAe2M总3,pu,按弹性计算的支座弯矩2127.985.3kNm3那么调幅系数AeMAuMAe85.375.60.11485.311.2解：Asi=AsA=644mm2/m,644210/x14.1mm9.61000fy=210N/mm2,h0=120-20=100mmMu64414210(100)bh08/1010012.58kNm/mM总2Mu25.2kNm/mPu8M总l218225.212.6kN/m2421取出一块梯形板块为隔离体,对较支座m(la)PulamPu24PuPu24m2al第12章12.1解:y30.45影响线ViViV2V4D

3、maxPminDmin60.450.0756160.26764.40.8080.0750.8080.9PmaxV0.9189.8109.80.2672.152.1511511522.2kNPmax水平荷载系数Dmax22.21150.12iax,k13l42AlAB取力矩平衡:la24Pul2a(la)20.9247.25222.5kN222.543kN0.123.94109.84.098kN4.098222.57.93kNQ计算柱顶水平集中力师：柱顶标高处z1.0,檐口处z1.07WkW1kW2k0.80.52.10.50.61.21.070.4561.32.10.121.070.4567.

4、54kN2q1ksZW0B0.81.00.4562.16kN/mq2k0.51.00.4561.35kN/m3剪力分配系数计算:2.13cnA0.14814.387.2nB0.36919.510.58.410.50.2;Cob310.2310.14833110.2310.3692.86810.0085.7572.96因只需相对值,故略去109UaUbH3H3H31EcI2c0AEc14.382.868Ec41.24H31H31Ec19.52.96Ec57.72;cc1Ua1Ub341.2457.7298.963HH41.2498.960.417,57.7298.960.5834计算约束反力Ra

5、、Rb：3.02898.960.362310.24-10.3693.00811B31810.2310.3698.110.371310.24100.148C11A；31810.2310.148Raq1HC11A2.1610.50.3628.21kNRbq2HC11B1.3510.50.3715.26kNR8.215.2613.47kN0.41721.018.76kN0.58321.0112.25kN5)剪力计算:ARWk0.41713.477.54bRWk0.58313.477.54A柱顶剪力Va=()Ma底Mb底B柱顶剪力Vb=()1 212qH2VaH2.1610.520.5510.5124

6、.85kNm2 21 “一1.3510.5710.5147.8kNmMa=Mb=12.3解：从前题知nA=,nB=,3.50.318111计算剪力分配系数:Coa3110.318310.1482.53UaEccob3110.3180.3692.84UbUaUb(36.3855.38)91.76EcH336.3891.7655.3891.760.62计算约束反力Ra、Rb：C3A1.510.3182C3B1.5310.31810.148一-210.31823110.31810.3691.51.50.8991.1850.8991.055RaMiRbHM2C3B153.21.138111311.27

7、81115.85kN15.22kN15.8515.220.63kN3柱顶剪力:VaRaR15.850.40.6315.6kNVbRbH314.382.53136.38119.52.84相对值1.1381.278R15.220.60.6315.6kN55.38M1M212.4解：fy=300N/mm2,Fv=324kN,Fh=78kN,a=250mm324103250As0.8530080040781031.23002418312730mm小于最小配筋量612的面积,故按构造配筋12.5由于解答不唯一,故从略.第15章15.1解：查得砌体抗压强度设计值f=N/mm2,M8.1106N250103

8、32.4mm;Hoh680010.97;62032.40.052;62010.00151120.846fA0.731120.0521120.8460.731.5490620332.66kNN250kN按轴压计算时陋13.88490100.77610.001513.882承载力应会满足15.2解：抗压强度设计值f=N/mm2,翼墙间距s=m,层高H=m,2H>s>H计算高度H00.4s02H0.46.80.23.53.42m3.421.10.19119.8,0.0015o210.001519.82底层轴力N=118+*=0.630fA0.631.191000190142.47129.

9、7615.3解：Nl180kN,No50kN,A0250224024020.1752m2-2f1.3N/mm,a010J600215mm240mm,1.3Ala0bb221525053750mm2;A0A17520003.2635375010.35,3.2611.526fA0.71.5261.35375074.64kNNl局部受压不满足要求.15.4解：首先计算中和轴位置y36004905003600240240500惯性矩I:I49074031.655121010740490740538-2_101.0234100.3583A49050036002401.109回转半径IA3.5386101

10、0.1.109106折算厚度hT3.5i625mm236004902403360049024020210106210mm179mm高厚比120.50191010_1023.538610mm224040106335010114mm,hT114625fA1120.182120.7440.4031.31.109106581kN承载力满足要求.HohT820062513.120.182,0.403;N350kN一一一210.00213.20.744一,一.2查得f1.3N/mm15.5解：f=N/mm2,500a010,183mma,1.5240mm,2Ala0bmA0200

11、23703702347800mm20.35.9.512.022.0取2.0fAi0.721.53660076.86kN85kN局部承压承载力不满足要求.15.6(条件缺乏,无法算出灌孔砌体抗压强度设计值.标准差,无法得标准值,从而得不到设计值;2不知Cb20混凝土、MU10砌块、Mb5砂浆的强度平均值)15.7解：f一一22.31N/mm,15.15.1510618010383.3mm12.650fn2.31查得5050100502500.2021283.32450.00202143022.86N/mmHoh4.20.490.51nfnA8略9解:8.57；83.34900.170.512.864902350.2kNN180kN平安(1)屋盖属二类,横墙间距40m,查表15-6可知属于刚弹性方案.(2)壁柱墙验算形心轴位置:323900.524003901900.565.94106y13902001902400534103123.5mmI390412239023902123.524003901903C“2400390190123.519019.281087.7761210811.4891083.10210841.647108mm44