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1、完美WORD格式5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铰时的均布荷载qi首先计算支座截面极限抗弯承载力Ma：zz/z22C20 混凝土查得 fc=9.6N/mm, 316 A s=603mm603 3009.6 200二 94mm ： bh0Asfyx 二：1 fcbx94Mua 二 Asfy(h。)=603 300(465) = 75.6KNm2专业知识分享A按弹性分析:M A 二 M UAql12M ua 12 75.61262二 25.2kNm.q1 = 25.2kN / m2、计算跨中极限抗弯承载力Mu1 : 2216 As=402mmX 二丝呱63mm,
2、9.6 200Mu402 沢 300465L52.3kNmI 2丿总弯矩 M 总=M uA Mu1 =75.652.3 =127.9kNm由M总总、Pul28Pu8M总623、若均布荷载同为 pu,按弹性计算的支座弯矩2 2M Ae M 总 127.9 =85.3kNm3 3则调幅系数M Ae - M AuM Ae85.3 -75.685.3= 0.11411. 2 解：As1 =A5A=644mrr/m,2y=210N/mm, h 0=12O-2O=1OOmm644 210x =9.6 1000=14.1mm ： bh0,)0M 总=2M u = 25.2kNm/m8M总2In 18 25.
3、2242 1= 12.6kN/m211. 3 解：塑性铰线位置如图所示。=12.58kNm/mZ/Z%XV%Xzz厂二z zaXXlaia2完美WORD格式1. l -a322l-a224Pu取出一块梯形板块为隔离体，对铰支座AB取力矩平衡:m (l - a) = pu专业知识分享24ml 2a l -a第12章12. 1 解：影响线y3 = 10.45y10.07560.45 4.4 y20.80861.6 y40.2676*7：3E?E0.454.41.154.46x yi =10.0750.8080.267 = 2.15Dmax =0.9Pmax=0.92.15 115=0.9 247.
4、25 = 222.5kN18 98 10 9.815=22.2kNDminPmax水平荷载系数Dmax222.5 =43kN115:-=0.121 Tk0.12 3.94 10 9.8 =4.098kN44.098Tmax,k222.5 =7.93kN11512. 2 解：计算柱顶水平集中力 Wk :柱顶标高处Jz :1.0, 檐口处Jz : 1.07Wk 二 W1k W, 0.8 0.5 2.10.5 - 0.6 1.2】 1.07 0.45 6=1.3 2.1 -0.121.07 0.45 6 = 7.54kN- Jzw0B=0.8 1.0 0.45 6 =2.16kN/mq2k = -0
5、.5 1.0 0.45 6 = T.35kN/m剪力分配系数计算:WknA2.1314.38= 0.1487.2 nB0.36919.510.58.410.5=02CAC0BUaUb1 0.2331 0.2H3q1kq2k777TB丄一10.14810.008 5.757=2.868- =2.96因只需相对值，故略去109-1 I0.369H3H3Ecl2C0AEc 14.38 2.868 Ec 41.24H3H3Ec41.2498.964计算约束反力C11AC11B剪力计算:19.5 2.96 Ec 57.72Ec 41.2457.72 =98.96 EcH3H3= 0.417Ra、FB:3
6、 1 0.248 1 0.233 1 0.248 1 0.23眶二 0.58398.96-12.148 丿2.148 丿0.3690.3691叱 0.36298.96迴8 3718.11Ra =qHC11A =2.16 10.5 0.362 =8.21kNRB =q2HC11B =1.35 10.5 0.371 = 5.26kN R =8.21 5.26 =13.47kNA RWki=0.41713.477.54 =0.41721.01 =8.76kNb RWk =0.58313.477.54 = 0.58321.01 = 12.25kNA柱顶剪力 Va=8.76-8.2仁0.55Kn (；、
7、B柱顶剪力 VB=12.25-5.26=7kN(；1 2 1 2MA底q1HVaH2.16 10.50.55 10.5 =124.85kNm2 21MB底1.35 10.5 7 10.5 =147.8kNm2MA=124.85kN12. 3 解：从前题知 na=0.148, nb=0.369.3.511-0.318计算剪力分配系数:3C0A2.53311 +0.31831 I0.148丿3Cb2.84311+0.3183 -1 10.369丿UbEcH3Ec114.38 2.53119.5 2.84136.38155.38丄丄-UA - u B（相对值）= (36.38 55.38)全=91.
8、76 昱H 3H 336.3891.76= 0.4，55.3891.76= 0.62计算约束反力RA、 RB:M1C3A-1.521-0.318310.3183C3B= 1.510.1481 -0.3182M2310.318丄一10.3690.899= 1.51.2781.055M1153.2,RaC3 a1.138 = 15.85 kN 1.1M 2-131RB 2 C3B1.278 - -15.22kN - R =15.85 -15.22 =0.63kN柱顶剪力:VA 二 RA - a R =15.85 -0.4 0.63 =15.6kN Vb =RB - b R =15.22 -0.6
9、0.63 - -15.6kN，：-2212. 4 解：fy=300 N/mm, F v=324 kN, Fh=78 kN, a=250 mm3a32410 750丄，A1.20.85 300800 -40378 10300= 418 312 =730mm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2f=1.5 N/mm ,8.1 1063 = 32.4mm；250 10一也二迎ra;h 620二_11 210.0015 112=0.84632.46200.052 ;1+10.052 +2 -0.7312 sH,.计算高
10、度H0 =0.4s 02H= 0.4 6.80.2 3.5 = 3.42m3.42 1.10.19= 19.8,11 0.0015 19.82:-=0.0015= 0.632专业知识分享底层轴力 N=118+3.36*3.5=129.76Kn0fA =0.63 1.19 1000 190 =142.47129.76215. 3 解：M =180kN, No =50kN,A。=250 2 240 240 =0.1752m2f =1.3N/mm2,a =10600= 215mm c 240mm.1.3A =a0bb =215 250 =53750mm2；血 =175200 =3.26 3Al53750- : =0孑=1 0.35.3.26 一1 =1.526 490 汉 740I =+ 490 沃 740 538740 f +(3600-490/ 2403 +2 12(3600-490240 2021210 10 10 10 10 2=1.655 101.0234 100.3583 100.5019 10= 3.5386 10 mm240 丫62A =490 5003600 240 =1.109 10 mm完美WORD格式专业知识分享回转半径=179mmT _ 35386 OO101 A =
