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钱时的均布荷载q1首先计算支座截面极限抗弯承载力Ma：C20混凝土查得fc=mm 316 A s=603mA1Asfy 603 300 x1 fcb 9.6 20094mm bh0MuaAsfy(h0 |) 603300(46594)75.6KNm 2按弹性分析：M a M uAql21212Mua q -守 25.2kNm8/10100Bq125.2kN/m2、计算跨中极限抗弯承载力Mui ： 2216 As=402mm402 300x 63mm,9.6 200M u1 40263300 465 52.3kNm2总弯矩M 总 M

2、 uaMui75.6 52.3127.9kNmPu8M总 8-l 一127 9 28.4kN /m 623、若均布荷载同为pu,按弹性计算的支座弯矩22MAe-M总一127.985.3kNm33则调幅系数MAeMAu85.375.60114MAe85.311.2解:A1=AA=644mmm,fy=210N/mrmh0=120-20=100mm644210x14.1mmbh08/101009.61000Mu64414210(100)12.58kNm/m2Mu25.2kNm/mPu8M总l21产12.6kN/m24111.3解：塑性较线位置如图所示。lAl13a4la2(la)2取出一块梯形板块为

3、隔离体，对较支座AB取力矩平衡:m(la)Pula2mPu24Pula24Pul2a24mPul2ala第12章12.1解:影响线yiDmaxPminDminy3yiV2V410.0750.9Pmax0.450.07560.454.40.8086160.26760.808yi0.9189.8109.8Dmax22.21150.2672.152.151150.9247.2511522.2kN222.543kN222.5kN水平荷载系数0.1210.123.94109.84.098kN44.098Tmax,k-222.57.93kN11512.2解:1计算柱顶水平集中力Wk：柱顶标高处z1.0,0

4、7WkW1kW2k0.80.52.10.50.61.21.070.451.32.10.121.070.457.54kNqkszW0B0.81.00.4562.16kN/m0.51.00.4561.35kN/m3剪力分配系数计算:nA2.1314.380.1487.2nB0.36919.50.2;2kq10.5 8.410.5COACOB310.2310.14833110.2310.3692.86810.0085.7572.96因只需相对值，故略去109H3H3H31Ua-EcI2c0AEc14.382.868Ec41.24H31H31Ub-；Ec19.52.96Ec57.72cc1Ua1UbE

5、c41.2457.7298.96Ec41.24A98.960.417,57.7298.960.583计算约束反力Ra、RB:3 1 0.2410.148C11A318 1 0.2310.148413 1 0.21-0.3693.02898.960.36211B)318 1 0.23 10.3693.0088.110.371Ra5HC11A2.1610.50.3628.21kNRbqzHCnB1.3510.50.3715.26kNR8.215.2613.47kN5剪力计算:A R WkB R WkA柱顶剪力0.417 13.47 7.540.583 13.47 7.54Va=()0.417 21

6、.01 8.76kN0.583 21.01 12.25kNB柱顶剪力Vb=(6）弯矩图:Ma底1q1H2211.352VaH10.5MA=12.162710.512.3解：从前题知na=,10.520.5510.5124.85kNm147.8kNmMB=1计算剪力分配系数:C0AC0B3110.31810.1483313110.31830.369B=,3.5110.3182.532.84UaUbEch3Ech314.382.5319.52.8436.3855.38UaUb(36.3855.38)EcH3Ec91.76H3（相对值）36.38A91.760.455.3891.760.62）计算约

7、束反力R、R:C3A1.50.3182C3B1.53110.31810.148210.31821.50.8991.1851.138M2RaMiRbHM23110.318310.369C3AC3B15.853柱顶剪力:VaRaaVbRbb0弯矩图:1.50.8991.0551.278153.21.13811-1311.2781115.220.63kN15.850.415.85kN15.22kN0.6315.6kN15.220.60.6315.6kN12.4解：fy=300N/mn2,Fv=324kN,Fh=78kN,a=250mmAs3241032500.8530080040c781031.23

8、002418312730mm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,M8.1106”，332.4mmN2501033243240.052；620210.00151120.8460.731120.052fA0.731.5按轴压计算时120.846490620332.66kN鬻13.880210.001513.8820.77615.2解：抗压强度设计值2HsH,计算高度H3.421.10.19119.8,0210.001519.820.63H。680010.97620250kN承载力应会满足f=N/mm2,翼墙

9、间距s=0.4s02H0.46.80.20.0015m,层高H=m,3.53.42m底层轴力N=118+*=0fA0.631.191000190142.47129.7615.3解：N1180kN,N。50kN,A025022402400.1752m2-2f1.3N/mm,a。10600215mm240mmAla0bb21525053750mm2;AoA175200537503.263010.35,3.2611.5262NlfAl0.71.5261.35375074.64kN局部受压不满足要求15.4解：首先计算中和轴位置y360049049074020.5360049024020.549050

10、03600240惯性矩I：240500202mm_3_2_3490740374036004902403490740538-1221236004902402022401.65510101.023410100.358310100.501910103.53861010mm2A49050036002401.109106mm2回转半径i3.53861010,A1.10910179mm折算厚度hT3.5i625mm,高厚比H08200013.12hT625hT1140.1826256M4010e114mmN3501030210.00213.220.7440.403；查得f-21.3N/mm21120.18

11、2,1210.744fA0.4031.31.109106581kNN350kN承载力满足要求。15. 5解：f= N/mm2,a0101.5240mm,A a0bb183 20036600mm2Ao200 2 370370 2347800 mm1 0.35.9.5 12.022.0 取 2.0fAl0.7 2 1.53660076.86kN 85kN（。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局部承压承载力不满足要求。15.6（条件不足，无法算出灌孔砌体抗压强度设计值。砌块、Mb5砂浆的强度平均值）15.7解：f2.31N/mm2,6M1510八e-三83.3m

12、mN18010312.6505050502501000.2016fn2.31283.32450.0020164302.86N/mm24.20.498.57;e83.30.17h490查得0.51nfnA0.512.864902350.2kNN180kN安全15.8略15.9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)0)带壁柱墙验算39030.5240039019020.5y1390200190240065.941063-123.5mm5341039042390I3902123.512224003901903122400390190123.519019.281087.77610811.4891083.10210841.647108mm44