混凝土结构中册混凝土结构与砌体结构设计知识题目解析

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MuA:C20混凝土查得fc=9.6N/mm2316As=603mm2Asfyfb1c6033009.694mm200bh0MuAAsfy(h0 x)603300(465294)75.6KNm2按弹性分析：MaMuaqi212Mua1275.625.2kNm1262q125.2kN/m2、计算跨中极限抗弯承载力u1:216As=402mm40230063mm,M62006352.3kNm2总弯矩MMuAu175.652.3127.9kNmpul8M总8127

2、.9Pu28.4kN/m623、若均布荷载同为，按弹性计算的支座弯矩MAe则调幅系数127.985.3kNmMAeMAu85.375.60.11485.3MAe11.2解：As1=Asa=644mm2/m,fy=210N/mm2,h0=120-20=100mm644210hx14.1mmb08/101009.61000Mu644210(100,412.58kNm/m2M总2Mu25.2kNm/mpu8M总825.212.6kN/m2ln2142111.3解：塑性较线位置如图所示lA取出一块梯形板块为隔离体，对较支座AB取力矩平衡:232pul a l a a83l a pu l 2a24m(l

3、a)pu2mpulaala24824mpu12ala第12章影响线yiDmaxPminDminy3yiy2V40.450.0750.454.41.60.267610.0750.8080.9Pmaxyi0.9189.8109.8Pmin22.20.8080.2672.152.151150.9247.25222.5kN11522.2kNPmaxmax222.543kN水平荷载系数1Tk4Tmax,k12.2解:1150.120.123.94104.098222.59.87.93kN1151计算柱顶水平集中力Wk:柱顶标高处4.098kNz1.0,檐口处z1.07Wk31k2k0.80.52.10.

4、50.61.21.070.4561.32.10.121.070.457.54kNq1kq2kzW0B0.51.0剪力分配系数计算:nA2.1314.387.2nB19.51050.81.00.4560.1480.3690.2;0.4562.16kN/m1.35kN/mWkq1kSHUB TOC o 1-5 h z CoA332.86810.231110.0085.7570.148C0b32.96因只需相对值，故略去10910.23110.369H3H3H31uAEIC;c20AEc14.382.868Ec41.24H31H31UB；Ec19.52.96Ec57.72Ec3 41.2457.72

5、UAUBEc98.963H41.2457.7298.960.417,98.960.5834计算约束反力、Rb：RaC11A3 1 0.24110.1488 1 0.23113.02898.960.3620.148C11B3 10.2410.3698 10.2313.0088.110.3710.369Raq1HCha2.1610.50.3628.21kNRbq21.3510.50.3715.26kNR8.215.2613.47kN5剪力计算:Wk0.41713.477.540.41721.018.76kNWk0.58313.477.540.58321.0112.25kNA柱顶剪力VA=8.76-

6、8.21=0.55Kn1 q1H 2 Va H212.1610.520.5510.5124.85kNm21.3510.5710.5147.8kNmMa=124.85kNMB=147.8kNm12.3解：从前题知nA=0.148,nB=0.369,。1计算剪力分配系数：3.5110.318coa2.533110.3183110.148C3,2.840B3110.3181_Ec11.uAH314.382.5336.38-Ec1LuBH319.52.8455.380.369(36.3855.38)EcEcUAUB91.760H3空0.4,91.7655.380.62计算约束反力Ra、Rb：10.31

7、82C3A1.510.3180.14891.760.8991.51.1851.138C3B1.510.31821.278RaRbM203柱顶剪力:VaRaa10.318315.85VbRb4弯矩图:1.055（相对值）0.3691535111.13815.85kN4311115.2212.4解：fy=300N/mmAs3241032500.85300小于最小配筋量61.2780.63kN15.850.415.22kN0.6315.6kN15.220.60.6354.6kNm18.4kNm2,Fv=324kN,8004015.6kN98.6kNm54.6kNm40.6kNm76.4kNmFh=7

8、8kN,a=250mm1.2T8103418312730mm230012的面积，故按构造配筋12.5由于解答不唯一，故从略。第15章e M8.1 10 632 4mmN250 10315.1解：查得砌体抗压强度设计值f=1.5N/mm2, TOC o 1-5 h z H06800e32.410.97;0.052h620h620110.8461210.0015112120.73111120.052、1N 250kN120.846fA0.731.5490620332.66kN按轴压计算时0一16800490113.880.776承载力应会满足0.001513.88215.2解：抗压强度设计值f=1

9、.19N/mm2,翼墙间距s=6.8m,层高H=3.5m,2HsH,计算高度H00.4s02H0.46.80.23.53.42m0.00153.421.119.8,0.630.191010.001519.82底层轴力N=118+3.36*3.5=129.76Kn0fA0.631.191000190142.47129.7615 . 3 解：N l 180kN ,f 1.3N / mm 2N050kN,Ao25022402400.1752m2ao10600215mm240mm1.32A175200Alaobb21525053750mm;3.263Al53750010.35一3.2611.5262f

10、Ai0.7 1.526 1.353750 74.64 kNNi局部受压不满足要求。154解：首先计算中和轴位置yy=20225383600490500,24049074020.5360049024020.54905003600 240202mm惯性矩I:2I49074034907405387403600490240336004902402022401221221.65510101.023410100.358310100.501910103.53861010mm2A49050036002401.109106mm2回转半径i3.538610101.109106179mm折算厚度hT3.5i625m

11、m,高厚比H0820013.12hT62540106114mm350103hT1140.182,fA6250.74410.00213.220.403;查得f1.3N/mm21120.1820.4031.3承载力满足要求。解：f=1.5N/mma010.5001.5AofAl一1120.7441.109106581kNN350kN183mma240mm,Alaobm220023700.359.510.721.5局部承压承载力不满足要求O3702.02366002347800mm22.0取76.86kN2.085kN15.6（条件不足，无法算出灌孔砌体抗压强度设计值。1无

12、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）15.7解：f2.31N/mm2,eM1510683.3mmN18010350-50_5a100502500.2016fn2.311283.32450.0020164302.86N/mm24.2e83.38.57；查得0.51nfnA0.512.86490 壁柱间墙体验算:350.2kNN180kN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。65.9410 6(2)Q1带壁柱墙验算39030.5240039019020.5形心轴位置：y1390200 1902400534103 1235mm4 23390423902400 390 1903I390 123.5 2400190390 190 123.51221219.28 1087.776 10 8 11.489 108 3.102 108 41.647 108 mm4I41.647