混凝土结构中册习题66366

1、混凝土构造中册习题答案66366混凝土构造中册习题答案6636610/10混凝土构造中册习题答案66366混凝土构造中册习题答案第11章111解：1、求支座截面出现塑性铰时的均布荷载q1第一计算支座截面极限抗弯承载力MuA：C20混凝土查得fc=mm2,316As=603mm21BAAsfy603300bh0 x9.620094mm1fcbMuAAsfy(h0 x)603300(46594)75.6KNm22按弹性解析：MAMuAql2q12MuA1275.6,l26225.2kNm12q125.2kN/m、计算跨中极限抗弯承载力Mu1：216As=402mm22402300Mu1402300

2、46563x20063mm,52.3kNm9.62总弯矩M总MuAMu1127.9kNmpul2得pu8M总8127.928.4kN/m由M总l26283、若均布荷载同为pu,按弹性计算的支座弯矩MAe2285.3kNmM总127.933则调幅系数MAeMAu0.114MAe85.3112解：As1=AsA=644mm2/m，fy=210N/mm2,h0=120-20=100mm644210bh0,x14.1mm9.610008/101008/10100BAMu644210(10014)12.58kNm/m2M总2Mu25.2kNm/mpu8M总825.212.6kN/m2ln21421113

3、解：塑性铰线地址以下列图。B(la)132laa4ala2lA取出一块梯形板块为隔断体，对铰支座AB取力矩平衡：l21lalalam(la)apu2322a42lapul2ampulaalapulalaa2488324pu24m2alal第12章121解：y2y3y1y4y31y10.450.0756影响线y20.454.40.8086y41.60.2676yi10.0750.8080.2672.15Dmax0.9PmaxyiPmin189.8109.811522.2kN2PminDminDmax22.2222.543kN115Pmax水平荷载系数0.12Tk14Tmax,k4.0987.93

4、kN222.5115122解：计算柱顶水平集中力Wk：柱顶标高处z1.0,檐口处z1.071WkW1kW2k0.80.52.161.32.10.121.070.4567.54kNq1kzw0B0.81.02sq2k0.51.00.4561.35kN/m剪力分配系数计算：Wk3nA2.130.14814.38q1kqnB0.3692k19.510.58.40.2;10.5ABC0A332.8680.2311110.148C0B32.96因只需相对值，故略去10910.23110.369uAH3H3H31EcI2C0AEc14.382.868Ec;41.24uBH31H31;Ec19.52.96E

5、c57.7211Ec41.2457.72EcuAuB398.963HH41.2457.720.583A98.960.417，B98.964计算拘束反力RA、RB：310.24113.028C11A0.1480.362198.96810.2310.148310.24113.008C11B0.3690.37118.11810.2310.369RAq1HC11A2.1610.50.3628.21kNRBq2HC11B1.3510.50.3715.26kNR8.21剪力计算：5ARWk0.41713.477.540.41721.018.76kNBRWk0.58313.477.540.58321.011

6、2.25kNA柱顶剪力VA=()B柱顶剪力V=()B弯矩图：6MA底1q1H2VAH12.1610.5222MB底11.3510.5710.5147.8kNm2M=MB=A123解：从前题知nA=,nB=,3.50.318111计算剪力分配系数：CC32.530A31Ec1110.3181uA0.148H314.382.5336.380B32.84uBEc111H319.52.8455.380.3183110.36911(36.3855.38)EcEc（相对值）91.76A36.380.4，B55.380.691.7691.762计算拘束反力RA、RB：C3A1.510.31820.899M1

7、M211.51.138131.1850.3180.1481C3B1.510.31821.50.8991.27810.3183111.0550.369RAM1C3A153.21.13815.85kNH11RBM2C3B1311.27815.22kNH11AB0.63kN3柱顶剪力：VARAAVBRBB4弯矩图：124解：fy=300N/mm2,FV=324kN,Fh=78kN,a=250mmAs324103250401.278103418312730mm20.85300800300小于最小配筋量612的面积，故按构造配筋125由于解答不唯一，故从略。第15章151解：查得砌体抗压强度设计值f=N

8、/mm2,M8.110632.4mm;H06800;e32.40.052;eN250103h62010.97h6200110.8461210.001511210.731121120.0521120.846fA0.731.5490620332.66kNN250kN按轴压计算时680013.88149000.776承载力应会满足。10.001513.882152解：抗压强度设计值f=N/mm2,翼墙间距s=m,层高H=m,2HsH,计算高度H00.4s02H3.421.119.8,0.00150.19010.630.001519.821基层轴力N=118+*=0fA0.631.1910001901

9、42.47129.76153解：Nl180kN,N050kN,A025022402400.1752m2f1.3N/mm2,a010600215mm240mm1.3Ala0bb21525053750mm2;A0175200Al3.26353750010.353.2611.5262fAl0.75375074.64kNNl局部受压不满足要求。154解：第一计算中和轴地址y3600y=2022402114538500490y49074020.5360049024020.54905003600240202mm惯性矩I:49074037402490240327405383600240I1249021236

10、0049024020221.65510101.023410100.358310100.501910103.53861010mm2A49050036002401.109106mm2辗转半径iI3.53861010mmA1.109106179折算厚度hT3.5i625mm，高厚比M40106,eN350103114mm0113.220.7440.0021H08200hT13.12625e114,hT0.182625120.403;查得f1.3N/mm21111120.1820.74412fA106581kNN350kN承载力满足要求。155解：f=N/mm2,a010500183mma240mm,

11、Ala0bb21.5A2002370370347800mm2010.359.512.022.0取2.0fAl0.721.53660076.86kN85kN局部承压承载力不满足要求。156（条件不足，无法算出灌孔砌体抗压强度设计值。1无标准差，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）157解：f2.31N/mm2，eM1510683.3mmN18010312.650501000.20165050250fn2.3121283.30.0020164302.86N/mm2245H04.28.57；e83.30.17h0.49h490查得0.51nfnA0.512.864902350.2N180kN安全kN158略159解：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）1带壁柱墙验算形心轴位置：39030.5240039019020.565.94106y13902001902400534103123.5mm39043902390190322123.524002400190I390212390190123.512219.281087.77610811.4891083.10210841.64710