混凝土课后答案第6,7章

6-1已知一钢筋混泥土矩形截面纯扭构件，截面尺寸b×h=150mm×300mm，作用于构件上的扭矩设计值T=3.60KN.m，采用C30混泥土，采用HPB235级钢筋。试计算其配筋量。『解』ft2fyv2fy2查表得：=1.43N/mm,=210N/mm=210N/mm取s=35mmC=25mmwtb2(3hb)1502(3300150)2.8125106mm366Acorbcor.hcor(15050)(30050)2.5104mm2ucor2(bcorhcor)2(100250)700mm取1.2由公式得：T0.35ftWt1.2fyvAstlAcors1043.61060.351.432.81251061.21.22102.5AstlsAstl3.61060.351.432.8125106s1.21.22.02.51040.3178取单肢8即Astl＝50.3s50.3158mm即0.3178取s＝150mmfyAstl.sfyvAstlucorAstl.fyvAstlucor1.221050.37002fy.s210150282mm由纵箍比可知：查附表11取410即Astl314mm2结构在剪扭作用下，结构受剪及受扭箍筋最小配筋率为：sv.minAsv.minbcor.s0.281.430.190050.30.3400210100150故满足最小配筋率。配筋图如下：4Φ108@1506-2已知一均布荷载作用下钢筋混泥土矩截面弯、剪、扭构件，截面尺寸为b×h=200mm×400mm。构件所承受的弯矩设计值M=50KN.m，剪力设计值V=52KN,扭矩设计值T=4KN.m。全部采用HPB235级钢筋，采用C20混泥土。试计算其配筋。『解』（1）受弯纵筋计算设钢筋按一排布置，取h040035365mm查表得：ft=1.1N/mm2fc=9.6N/mm2fy=210N/mm2M501060.1955s1.09.620036521fcbh021 1 2s 0.2196AS1fcbh01.09.60.2196200365733mm2fy210（2）受剪及受扭钢筋计算截面限制条件验算：h0hw365mm,hw/b365/2001.825VT5210341062bh0Wt2003650.86.71061.459N/mm0.25cfc0.251.09.62.4N/mm2故截面尺寸满足要求，VT1.459N/mm20.71.1N/mm20.251.09.60.77N/mm2又bh0Wt故需按计算配置受扭钢筋受剪钢筋计算由公式6-23可得：t1.51.50.9395V.Wt521036.7106110.50.5106200365Tbh04Acor

(200

50) (400

50)

52500mm2受剪箍筋

由公式

6-24可得：Asvl521030.7(1.50.9395)2003651.1sv1.252103650.2139受扭箍筋取1.3，由公式6-27可得：Astl41060.356.71060.93951.10.1045st1.21.32105.25104故得腹板单肢箍筋的需要量为Astl0.10450.2139s20.211取箍筋直径为8（Asvl50.3mm2）sAsvl50.30.211238取s=220mm0.211mm受扭纵筋：Astl1.350.32101000/210220297mm2故得腹板纵筋：2Φ10弯曲受压区纵筋总面积为As297/2149mm2选用210(AS157mm2)Φ8@2202Φ18弯曲受拉纵筋总面积为As733297/2882mm2选用220218(As883mm2)2Φ20配筋图如下：7-1

已知矩行截面柱 b=300mm，h=400mm。计算长度N=300KN，弯矩设计值 M=150KN.m，混泥土强度等级为

L0

为 3m，作用轴向力设计值C20，钢筋采用 HRB335级钢筋。设计纵向钢筋

As及

As/的数量。『解』查表得：fc=9.6N/mm2fy=300N/mm2取s=40mmh0has=360mme0M/N0.5mea20mmeieae0520mml0/h7.55要考虑10.5fcA/N0.59.6300400/3001031.921取11取 2 1.011(l0/h)21211(7.5)211.031400ei/h01400520/360ei1.03520535.60.3h00.3360属于大偏心受压eeih0/2as535.618040675.6mm取xbh0Ne1cf0h/2x)yfA(0hs)a由ab(xs得AsNea1fcbh02b(10.5b)/fy(h0as)560.9mm2Asmin0.02bh0.02300400240mm2AA选用414A615mm2ssminsNe20.5)yf01c0fa)由得sNefyAs(h0as)/a1fcbh020.385112s10.480.52h00.52360187.22as80mm由Na1fcbh0fyAsfyAs得Asa1fcbh0fyAs/fy9.63003600.53006152343mm2300选用525As2454mm2截面配筋图如下：5254142147-2 已知条件与题数量。『解』

7-1相同，但受压钢筋已配有

4 16的

HRB335

级的纵向钢筋。设计

As查表得：fc=9.6N/mm2fy=300N/mm2取ss=40mmh0has=360mme0M/N0.5mea20mmeieae0520mml0/h7.55要考虑10.5fcA/N0.59.6300400/3001031.921取11取 2 1.011(l0/h)21211(7.5)211.031400ei/h01400520/360ei1.03520535.60.3h00.3360属于大偏心受拉eeih0/2as535.618040675.6mmMfyAs(h0as)77.184106N.mmM1NeM131.184106N.mm取xbh0Ne1c0h/2x)yfA(0hs)a由afb(xs得AsNea1fcbh02b(10.5b)/fy(h0as)560.9mm2Asmin0.02bh0.02300400240mm2AA选用414A615mm2ssminsNea20.5)fA(ha)fbh(1yss由1c00得sM1/afbh2131.1841060.3511c09.63003602S0.7729AM/fysh21572mms110AsfyAs15723008043001032As1fy3001376mm选用325As1473mm2截面配筋图如下：325 4147-3已知条件同题7-1相同，但受压钢筋已配有4Φ16的HRB335级钢筋的纵向钢筋。设计As数量。已知矩行截面柱b=600mm，h=400mm。计算长度混泥土强度等级为C30，纵筋采用HRB400级钢筋。，柱上作用轴向力设计值N=2600KN,弯矩设计值M=78KN.m，混泥土强度等级为C30，钢筋采用HRB400级钢筋。设计纵向钢筋As及As/的数量，并验算垂直弯矩作用平面的抗压承载力。『解』查表得：fc=14.3N/mm2fy=300N/mm2fy=360N/mm2取ss=40mmh0has=560mme0M/N30mmea20mmeieae050mm15l0/h105要考虑10.5fcA/N0.514.3300600/2601034.951取11取21.011(l0/h)21211(10)211.81400ei/h0140050/560ei1.850900.3h00.3560168mm属于小偏心受压对AS合力中心取矩eh/2as(e0ea)3004010250mmAsNea1fcbh(h0h/2)/fy(h0as)260010314.3300600(560300)/360520As0取As0.02bh0.02300600360mm2M1NeM131.184106N.mm选用216As402mm2由asas2fyAs(1as/h0)/a1fcbh0(b1)fyAs(1as/h0)/a1fcbh0(b1)2Nefyh0h0a1fcbh020.7521b1.60.5181.082eeih0/2as1.85028040330mm由Nea1fcb2(1/2)fyA(h0as)得sAsNea1fcbh02(10.5)/fy(h0as)1214.6mm20.02bh选用420As1256mm2抗压承载力验算：l0/h10查表3-1得0.98A/bh01256/3005600.75030s00Nu0.9(fcAfyAs)0.90.98(14.33006003601256)2.67103kN2600kN所以平面外承载力满足要求。截面配筋图如下：2162Φ104207-4

已知矩行截面柱 b=600mm，h=300mm。计算长度筋，柱上作用轴向力设计值 N=780KN, 弯矩设计值

L0＝4m，受压区已配有 2 16的钢M=390KN.m，混泥土强度等级为C30，钢筋采用 HRB400级钢筋。设计配筋数量。『解』查表得：fc=14.3N/mm2fy=360N/mm2fy=360N/mm2取ss=40mmh0has=560mme0M/N500mmea20mmeieae0520mm15l0/h6.675要考虑10.5fcA/N0.514.3300600/7801031.651取11取21.011(l0/h)21211(6.67)211.0361400ei/h01400520/560ei1.036520538.720.3h00.3560168mm属于大偏心受压eeih/2as538.7230040798.72mmMfyAs(h0as)36040252075.2544106N.mmM1NeM780103538.7275.2544106344.9472106N.mmsM1/afbh2344.94721060.256s.max0.38381c014.33005602S0.8493AM/fysh22015mms110AsAs1fyAs20153604027801032minbh360mm2fy360250mm选用216As402mm2截面配筋图如下：216 2Φ10 2167-5 已知矩行截面柱 b=600mm，h=400mm。计算长度L0＝4.5m，受压区已配有 2 25的钢筋，柱上作用轴向力设计值 N=468KN, 弯矩设计值 M=234KN.m，混泥土强度等级为C30，钢筋采用 HRB400级钢筋。设计配筋数量。『解』查表得：fc=14.3N/mm2fy=360N/mm2fy=360N/mm2取ss=40mmh0has=560mme0M/N500mmea20mmeieae0520mm15l0/h7.55要考虑10.5fcA/N0.514.3400600/4681033.671取11取21.011(l0/h)21211(7.5)211.0431400ei/h01400520/560ei1.043520542.360.3h00.3560168mm属于大偏心受压eeih/2as542.3630040802.36mmMfyAs(h0as)3601964520367.6608106N.mmM1NeM468103802.36367.66081067.84368106N.mmsM1/a1fcbh0278436800.004s.max0.383814.34005602S0.998As1M1/fysh7843680/3600.99825m60m390fyAs360402780103250mm2minbh2AsAs12015fy360选用318As763mm2截面配筋图如下7-6已知条件同题7－1，设计对称配筋的钢筋数量。『解』查表得：fc=9.6N/mm2fy=300N/mm2取ss=40mmh0has=360mme0M/N0.5mea20mmeieae0520mml0/h7.55要考虑10.5fcA/N0.59.6300400/3001031.921取11取21.011(l0/h)21211(7.5)211.031400ei/h01400520/360ei1.03520535.60.3h00.3360属于大偏心受压e ei h0/2 as 535.6180 40 675.6mmxN/a310mm4.17由cfb30010/9.630012sxbh00.55360180mmAsAsNea1fcbx(h0x/2)300103675.69.6300104.17(36052.085)fy(h0as)300320AsAs1058.9mm2As.min240mm2选用420AsAs1256mm2420 4207-7 已知条件同题 7－3，设计对称配筋的钢筋数量。『解』查表得：fc=14.3N/mm2fy=300N/mm2fy=360N/mm2取ss=40mmh0has=560mme0M/N30mmea20mmeieae050mm15l0/h105要考虑10.5fcA/N0.514.3300600/2601034.951取11取21.011(l0/h)21211(10)211.81400ei/h0140050/560ei1.850900.3h00.3560168mm属于小偏心受压由N a1fcbh0 bNe0.43a1fcbh02ba1fcbh0(1b)(h0as)26001030.51814.33005600.5181.18326001033300.4314.3300560214.3400560(0.80.518)520AsAsNea1fcbh02(10.5)/fy(h0as)1097.6mm20.02bh选用420As1256mm2抗压承载力验算：As/bh01256/3005600.7500300Nu0.9(fcAfyAs)0.90.98(14.34006003601256)2.67103kN2600kN所以平面外承载力满足要求。7-8已知矩行截面偏心受压构件，b=300mm，h=500mm，as=as/＝35mm，l0＝4.0m，采用对称配筋As＝As＝804mm2(416)，混泥土强度等级为C30，纵筋采用HRB400级钢筋。设轴向力沿长边方向的偏心矩e0=120mm，求此柱的受压承载力设计值。『解』查表得：fc=14.3N/mm2fy=360N/mm2取ss=35mmh0has=465mm取ea20mmeieae012020140mm0.3h0139.5按大偏心受压计算15l0/h85要考虑取11，21.011(l0/h)21211(8)211.1521400ei/h01400140/465eeih0/2as376.28mmN1由acfbxyfsAyfsA14.3300x。。。。。。。。。。。（1）Ne1c0h/2x)yfA(0hs)aafb(xsN376.2814.3300(465x/2)360804430（2）由（1）（2）解得：N1550.883kNx315.2mm2所以此柱的受压承载力设计值为1550.883KN7-9已知矩形截面偏心受压构件 b× h=400mm×600mm，L0＝6.0m，截面配筋： As/＝1256mm2(420)，As＝1964mm2(425)。混泥土强度等级为C30，纵筋为HRB400级钢筋。as=as/＝40mm，轴向力沿长边方向偏心矩e0=100mm。求该受压构件的受压承载力设计值N。『解』查表得：fc2fy2=14.3N/mm=360N/mm取ss=40mmh0has=560mm取ea20mmeieae010020120mm0.3h0168mm按小偏心受压计算15l0/h105要考虑10.22.7e/h00.77921.0i，11(l0/h)21211(10)210.7791.261400ei/h01400140/465eeih0/2as411.2mmNafbxyfsAsA由1cs.（1）Ne1cf0h/2x)yfA(0hs)aab(xs（2）由（1）（2）解得：x1112mm(舍去)x2358.mm4N2470.306kNa1fcbh14.34006003432kN所以此柱的受压承载力设计值为2470.306KN7-10 已知矩形截面偏心受压构件 b×h=400mm×600mm，L0＝6.0m，在截面上作用一偏心力N＝75KN，其偏心矩分别为 e0=20mm，40mm，60mm，80mm，120mm，150mm，180mm，200mm。试分别求设计以上 8种情况下截面的配筋数量。采用对称配筋（ AsAs/），混泥土强度等级为C30，纵筋采用HRB400级钢筋。并绘出用钢量随偏心矩变化的关系图。『解』查表得：fc=14.3N/mm2fy=360N/mm2取ss=40mmh0has=560mm取ea20mmeieae020204mm0同理可得e040mm,60mm,80mm,120mm,150mm,180mm,200mm,时ei 60mm,80mm,100mm,140mm,170mm,200mm,220mm并分别取为ei1,ei2,ei3,ei4,ei5,ei6,ei7,ei815l0/h105要考虑10.5fcA/N0.514.3240000/7510322.881，取11.0,21.011(l0/h)21211(10)2121400ei/h0140040/465同理可得：21.667,31.5,41.4,51.286,61.236,71.2,81.1821ei240800.3bh0168mm同理可求得前4组ei0.3bh0后4组的ei0.3bh0同理可得：e2360.02mm,e3380mm,e4400mm,e5440.04mm,e6470.12mm,e7500mm,e8520.04mm由于是对称配筋计算NbNba1fcbh0b14.34000.5185601659.2kNN75kN属于大偏心受压.xN/a1fcb75103/14.34001.013.112as80mm取x80mmAs1As1Ne7510328073mm22fy(h0as)360520minbh480mm同理可得：As2As264mm2As3As357mm2As4As449mm2As5As533mm2As6As611mm2As7As74mm2As8As81mm2都小于mbhi480mmn2选用316AsAs603mm27-11知工字形截面柱尺寸如图弯矩设计值M=226.2KN.m

7－36所示，计算长度 L0＝6.0m，轴向力设计值 N=650KN,，混泥土强度等级为 C25，钢筋采用 HRB335级钢筋。设计对称配筋的数量。736fc=1.9N/mm2fy=360N/mm2ss=40mmh0has=660mmea20mm或eah/3023.3mmea23.3mmeieae034823.3371.3mm15l0/h10510.5fcA/N0.511.9118720/6501031.09111.0,21.011(l0/h)21211(8.57)211.0931400ei/h01400371.3/660eeih0