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1、精品word 可编辑资料 - - - - - - - - - - - - -混凝土构造中册习题答案第 11 章11 1 解： 1.求支座截面出现塑性铰时的均布荷载q1首先计算支座截面极限抗弯承载力M uA ：C20 混凝土查得f c=9.6N/mm 2 .316As=603mm 2As f yx1 f cb6039.6300200x94mmb h094A1BM uAAs f y (h0)6032300(465)75.6KNm2按弹性分析：M AM uAql 21212 MuA. ql 21275.66 225.2kNm2q125.2kN / m2.计算跨中极限抗弯承载力M u 1 ： 216A
2、s=402mmx4029.630020063mm .M u140230046563252.3kNm总弯矩 M 总M uAM u 175.652.3127.9kNmp2由 Mu l得p8M 总8127.928.4kN / m总8ul 2623.假设均布荷载同为pu .按弹性计算的支座弯矩2M AeM 总32127.9385.3kNm那么调幅系数M AeM Au85.375.60.114M Ae85.311 2解： As1=A sA=644mm 2 /m ,f y=210N/mm 2. h0 =120-20=100mm644x9.6210100014.1mmb h0.8/10100M u64421
3、0(10014)212.58kNm / mM 总2M u25.2kNm/ m8M 总p825.212.6kN / m2A8/10100Bl2nu1421-word.zl.第 1 页,共 9 页 - - - - - - - - - -精品word 可编辑资料 - - - - - - - - - - - - -11 3解：塑性铰线位置如下图；B1(la) 32laa4ala2lA取出一块梯形板块为隔离体,对铰支座AB 取力矩平衡：m (la)2pla1lau232laala 242lamp u24a la 8lalapua83lapu l2a2424mpul2 ala第 12 章12 1 解：y3
4、y110.4560.075y 2y3y1y 4影响线0.45y264.40.8084y1.660.2670.454.41.154.41.6yi10.0750.8080.2672.15Dmax0.9 Pmaxyi0.92.151150.9247.25222.5kNPmin189.81029.811522.2kNDminPmin PmaxD max22.2115222.543kN水平荷载系数0.121Tk0.1243.94109.84.098kNTmax. k4.098115222.57.93kN-word.zl.第 2 页,共 9 页 - - - - - - - - - -精品word 可编辑资
5、料 - - - - - - - - - - - - -12 2 解： 1 计算柱顶水平集中力Wk ：柱顶标高处z1.0.檐口处z1.07WkW1kW2k0.80.52.10.50.61.21.070.4561.32.10.121.070.4567.54kN 2 q1ksz w0 B0.81.00.4562.16kN / mq2 k0.51.00.4561.35kN / mW 3剪力分配系数计算：kn2.13A14.387.20.148q1kq2knB19.50.36910.58.40.2;C0 A10.531310.008 5.757AB2.868C0 B10.2310.2310.148311
6、0.3692.96 因只需相对值,故略去109uH 3H 3AH 31;Ec I 2 C0 AEcH 31uB14.38H 32.868Ec1;41.24Ec19.52.96Ec57.7211u AuBEc41.24H 357.7298.96 EcH 341.24A98.960.417 ,57.72B98.960.583 4计算约束反力RA .RB ：- word.zl.第 3 页,共 9 页 - - - - - - - - - -精品word 可编辑资料 - - - - - - - - - - - - -C11 AC11B3 10.248 10.233 10.248 10.23110.148
7、110.148110.369110.3693.02898.963.0088.110.3620.3712.1610.50.3628.21kN1.3510.50.3715.26kNRAq1 HC11ARBq2 HC11BR8.215.2613.47kN 5剪力计算：RWk0.41713.477.540.41721.018.76kNRWk0.58313.477.540.58321.0112.25kNABA柱顶剪力V A =8.76-8.21=0.55Kn(B柱顶剪力V B=12.25-5.26=7kN() 6弯矩图：M A 底121qHVH1A222.1610.520.5510.5124.85kNm
8、M B 底11.35210.5710.5147.8kNmM A =124.85kNM B=147.8kNm- word.zl.第 4 页,共 9 页 - - - - - - - - - -精品word 可编辑资料 - - - - - - - - - - - - -12 3 解：从前题知nA=0.148.nB=0.369. 1计算剪力分配系数：3.5110.318C0 A310.318312.531uEc110.148AH 314.382.5336.38C32.84uEc110 B10.3183110.369BH 319.52.8455.3811u AuB36.38A91.76(36.380.4
9、 ,55.38)BEc H 355.3891.7691.76 EcH 30.6相对值 2计算约束反力RA .RB：2C3 A1.510.3181.50.8991.138M 1M 210.3183110.1481.185C3 B1.510.31821.50.8991.27810.3183110.3691.055ABRM 1A C3 AH153.2 111.13815.85kNRCM 2B 3 BH131111.27815.22kNR15.8515.220.63kN 3柱顶剪力：V ARAV BRBAR15.85BR15.220.40.60.630.6315.6kN15.6 kN 4弯矩图：98.6kNm54.6kNm54.6kNm76.4kNm18.4kNm40.6kNm-word.zl.第 5 页,共 9 页 - - - - - - - - - -精品word 可编辑资料 - - - -
