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1、混凝土结构中册习题答案第 11 章111 解： 1、求支座截面出现塑性铰时的均布荷载q1首先计算支座截面极限抗弯承载力MuA：C20混凝土查得fc=mm2, 316 As=603mm2KNmxhfAMhmmbffAxysuAbcys6 .75)294465(300603)2(942006.9300603001按弹性分析：,122qlMMuAAkNmlMquA2.2566.75121222mkNq/2.2512、计算跨中极限抗弯承载力1uM：216 As=402mm2mmx632006.9300402, kNmMu3 .522634653004021总弯矩kNmMMMuuA9 .1273.526
2、.751总由82lpMu总得mkNlMpu/4.2869 .1278822总3、若均布荷载同为pu, 按弹性计算的支座弯矩kNmMMAe3.859 .1273232总则调幅系数114.03.856 .753.85AeAuAeMMM112解： As1=AsA=644mm2/m， fy=210N/mm2, h0=120-20=100mm01.1410006.9210644hmmxb, mkNmMu/58.12)214100(210644mkNmMMu/2 .252总222/6 .12142 .25818mkNlMpnu总AB18/101008/10100AB精品资料 - - - 欢迎下载 - -
3、- - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 1 页，共 9 页 - - - - - - - - - - 113解：塑性铰线位置如图所示。取出一块梯形板块为隔离体，对铰支座AB取力矩平衡：422312)(2alaalalalpalmualalmpalpalaalalpalaalpmuuuu224224388242第 12 章121 解：影响线267.066.1808.064.445.0075.0645.014213yyyykNDPPDkNPkNyPDyii435.2221152.222.2211528 .9108.9185.22225.247
4、9.011515.29 .09 .015.2267.0808.0075.01maxmaxminminminmaxmax水平荷载系数12. 0kNTkNTkk93.75 .222115098.4098.48.91094.312.041max,laa2alAB4al2)(31aly1y2y3y4精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 2 页，共 9 页 - - - - - - - - - - 122 解：1 计算柱顶水平集中力kW：柱顶标高处,0.1z檐口处07.1zkNWWWkkk54. 7645.
5、007.112.01.23. 1645.007.12. 16. 05. 01. 25 . 08. 0212mkNBwqzsk/16.2645.00.18 .001mkNqk/35.1645.00 .15. 023剪力分配系数计算：;2.05 .104.85.10369.05.192 .7148.038.1413.2BAnn930301096.21369.012.013868.2757.5008.0131148.012.013因只需相对值，故略去BACC;72.57196.25.191;24.411868.238.143333023ccBccAcAEHEHuEHEHCIEHu3396.9872.
6、5724.4111HEHEuuccBA417.096.9824.41A，583. 096.9872.57B4计算约束反力AR、RB：ABkWq1kq2k精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 3 页，共 9 页 - - - - - - - - - - 371.011.8008. 31369.012 .0181369.012 .013362.096.98028. 31148.012 .0181148.012 .01334113411BACCkNHCqRkNHCqRBBAA26.5371.05 .10
7、35.121.8362.05.1016.2112111kNR47.1326.521. 85剪力计算：kNWRkNWRkBkA25.1201.21583.054.747.13583.076.801.21417.054.747.13417.0A柱顶剪力 VA= (?)B柱顶剪力 VB= (?)6弯矩图：kNmMkNmHVHqMBAA8 .1475.1075.1035.12185.1245.1055.05 .1016.22121221底底MA=MB=精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 4 页，共 9
8、 页 - - - - - - - - - - 123 解：从前题知 nA=, nB=, 318. 0115. 31计算剪力分配系数：84.21369.01318.01353.21148.01318.0133030BACC38.55184.25 .19138.36153.238.14133HEuHEucBcA3376.91)38.5538.36(11HEHEuuccBA（相对值）4 .076.9138.36A，6.076.9138.55B2计算约束反力RA、RB：278.1055.1899.05.11369.01318.01318.015.1138.1185.1899.05 .11148.013
9、18.01318.015.1323323BACCkNCHMRkNCHMRBBAA22.15278.11113185.15138.1112.1533231kNR63. 022.1585.153柱顶剪力：kNRRVkNRRVBBBAAA6 .1563.06.022.156.1563.04.085.154弯矩图：M1M2AB精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 5 页，共 9 页 - - - - - - - - - - 124 解： fy=300 N/mm2, FV=324 kN, Fh=78 kN,
10、 a=250 mm23373031241830010782 .14080030085.025010324mmAs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/mm2,mmNMe4.3210250101 .836; 97.1062068000hH; 052.06204 .32he;73.01846.01121052.01211846.0110015.011112220kNNkNfA25066.3326204905 .173.0按轴压计算时88.134906800776.088.130015.01120承载力应会满
11、足。152 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mHsH42.35.32 .08.64.0024.008.1919. 01 .142.3, 0015.063.08 .190015.01120底层轴力 N=118+*=76.12947.142190100019.163.00fA153 解：,180kNNl,500kNN201752.02402402250mA2/3 .1mmNf, mmmma2402153.1600100精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - -
12、- - - - -第 6 页，共 9 页 - - - - - - - - - - 2053750250215mmbaAbl; 326.3537501752000lAA02526.1126.335.01llNkNfA64.74537503.1526.17 .0局部受压不满足要求。154 解：首先计算中和轴位置ymmy20224036005004905 .024049036005.074049022惯性矩 I:210101010102323105386.3105019. 0103583. 0100234. 110655.122402022404903600122404903600274053874
13、049012740490mmI2610109.12403600500490mmA回转半径mmAIi17910109.1105386. 3610折算厚度mmihT6255 .3，高厚比12.1362582000ThHmmNMe11410350104036, 182.0625114The, 744.02 .13002.01120403.01744.01121182. 012112; 查得2/3 .1mmNfkNNkNfA35058110109.13 .1403.06承载力满足要求。4903600538y=2022500240114精品资料 - - - 欢迎下载 - - - - - - - - -
14、- - 欢迎下载 名师归纳 - - - - - - - - - -第 7 页，共 9 页 - - - - - - - - - - 155 解： f= N/mm2,mmamma2401835.1500100, 2036600200183mmbaAbl203478003703702200mmA0 .20 .202.215.935.01取kNkNfAl8586.76366005.127 .0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：2/31.
15、2mmNf，mmNMe3.83101801015362016.0100250505050506.122/86.2430002016.02453 .8321231.2mmNfn57.849.02 .40hH；17.04903.83he查得51.0kNNkNAfnn1802.35049086.251.02安全158 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：mmy5.123105341094.6524001902003905 .019039024005.0390362314888882322410647.4110102.3
16、10489.1110776.71028.1921905.12319039024001219039024005.123239039012390mmImmAIi3.881053410647.4138；mmihT3095 .3精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 8 页，共 9 页 - - - - - - - - - - s=40 m, H= m查表 15-8 得mHH8.75 .62.12.1024.25309.08. 7, 11, 84. 046 . 14 .012, 2416.202484.0121不满足要求。2壁柱间墙体验算：s=4 m, H= m, sH; msH4.246.06 .0016.2063.1219.04.221满足要求。3902400y2y1200190精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 9 页，共 9 页 - - - - - - - - - -