2022年混凝土结构与砌体结构设计方案课后习题答案中册 .pdf
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=9.6N/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 .1273.526.751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=100mm 01 .1410006.9210644hmmxb, mkNmMu/58.12)214100(210644mkNmMMu/2 .252总222/6.12142.25818mkNlMpnu总113解：塑性铰线位置如图所示。A B 1 8/10100 8/10100 A B l a a 2alA B 4al2)(31al精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页，共 8 页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 .2211528.9108.9185.22225.2479.011515.29 .09 .015.2267.0808.0075.01maxmaxminminminmaxmax水平荷载系数12.0kNTkNTkk93.75 .222115098.4098.48.91094.312.041max,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. 7645.007. 112.01.23. 1645. 007.12.16.05. 01. 25.08. 0212mkNBwqzsk/16.2645. 00.18 .001mkNqk/35. 1645. 00. 15.020.45 4.4 1.15 4.4 1.6 y1y2y3y4精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页，共 8 页3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. 238.143333023ccBccAcAEHEHuEHEHCIEHu3396.9872.5724.4111HEHEuuccBA417.096.9824.41A，583.096.9872.57B4计算约束反力AR、RB：371.011.8008.31369.012 .0181369.012 .013362.096.98028.31148.012 .0181148.012 .01334113411BACCkNHCqRkNHCqRBBAA26.5371. 05.1035.121.8362. 05.1016.2112111kNR47.1326. 521.8A B kWq1kq2k精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 8 页5剪力计算：kNWRkNWRkBkA25.1201.21583.054.747.13583.076.801.21417.054.747.13417.0A 柱顶剪力 VA=8.76-8.21=0.55Kn () B 柱顶剪力 VB=12.25-5.26=7kN () 6弯矩图：kNmMkNmHVHqMBAA8.1475 .1075.1035.12185.1245 .1055.05.1016.22121221底底123 解：从前题知 nA=0.148, nB=0.369, 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：MA=124.85kNMB=147.8kNm M1精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 8 页278.1055.1899.05. 11369.01318. 01318. 015. 1138.1185.1899.05. 11148.01318. 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：124 解： fy=300 N/mm2, FV=324 kN, Fh=78 kN, a=250 mm 23373031241830010782 .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=1.5 N/mm2, mmNMe4.3210250101.836。97.1062068000hH。052.06204.32he。73. 01846.01121052.01211846.0110015.011112220kNNkNfA25066.3326204905.173.0M2A B 54.6kNm 98.6kNm 18.4kNm 54.6kNm 76.4kNm 40.6kNm 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 5 页，共 8 页按轴压计算时88.134906800776.088.130015.01120承载力应会满足。152 解：抗压强度设计值 f=1.19 N/mm2, 翼墙间距s=6.8 m, 层高 H=3.5 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+3.36*3.5=129.76Kn 76.12947.142190100019. 163.00fA153 解：,180kNNl,500kNN201752. 02402402250mA2/3. 1mmNf, mmmma2402153.16001002053750250215mmbaAbl。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y mmy20224036005004905.024049036005.074049022惯性矩 I: 210101010102323105386. 3105019. 0103583. 0100234. 110655.122402022404903600122404903600274053874049012740490mmI2610109.12403600500490mmA490 3600 538 y=2022 500 240 114 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页，共 8 页回转半径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。155 解： f=1.5 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。15 6 （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：2/31.2mmNf，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。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页，共 8 页（2）1 带壁柱墙验算形心轴位置：mmy5.123105341094.6524001902003905.019039024005 .0390362314888882322410647.4110102.310489.1110776.71028.1921905.12319039024001219039024005.123239039012390mmImmAIi3.881053410647.4138；mmihT3095. 3s=40 m, H=6.5 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=6.5 m, sH。msH4 .246. 06 .0016.2063.1219.04 .221满足要求。390 2400 y2y1200 190 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 8 页，共 8 页
