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单筋矩形截面梁正截面受弯承载力计算例题1.钢筋混凝土简支梁,计算跨度l=5.4m,承受均布荷载,恒载标准值gk=10kN/m,活载标准值qk=16kN/m,恒载和活载的分项系数分别为γG=1.2,γQ=1.4。试确定该梁截面尺寸,并求抗弯所需的纵向受拉钢筋As。解:⑴选用材料混凝土C30,2cN/mm3.14f,2tN/mm43.1f;HRB400钢筋,2yN/mm360f,518.0b⑵确定截面尺寸mm675~450540081~12181~121lh,取mm500hmm250~16750021~3121~31hb,取mm200b⑶内力计算荷载设计值kN/m4.34164.1102.1kQkGqgq跨中弯矩设计值mkN4.1254.54.34818122qlM⑷配筋计算布置一排受拉钢筋,取mm40sa,则mm46040500s0ahh将已知值代入20c1xhbxfM,得24602003.140.1104.1256xx整理为0876929202xx解得mm238460518.0mm1080bhx,满足适筋梁要求由基本公式,得2yc1smm8583601082003.140.1fbxfA002.000179.036043.145.045.0ytff,002.0min2min2smm200500200002.0mm858bhA,满足最小配筋率要求选3Φ20,2smm942A2.已知钢筋混凝土梁截面尺寸为b×h=250mm×550mm,混凝土强度等级C30,已配纵向受拉钢筋为6根直径20mm的HRB400级钢筋,承受的弯矩设计值M=250kN·m。验算此梁受弯承载力是否满足要求。解:由已知条件可得,2smm1884A,mm48565550s0ahh根据基本公式,得mm251485518.0mm7.1892503.140.118843600bc1syhbfAfx,满足适筋梁条件mkN250m264.6kNmmN106.26427.1894857.1892503.140.1260c1uMxhbxfM受弯承载力满足要求3.条件同题1,试用系数法求该梁抗弯所需的纵向受拉钢筋。解:2072.04602003.140.1104.1252620c1sbhfM8826.02072.02112121121ss260sysmm8584608826.0360104.125hfMA双筋矩形截面梁正截面受弯承载力计算例题1.已知钢筋混凝土梁截面尺寸为b×h=200mm×500mm,混凝土C30,HRB400级钢筋,承受的弯矩设计值M=260kN·m。试计算此梁受弯所需的纵向受力钢筋。解:查表得2cN/mm3.14f,2yyN/mm360ff布置两排受拉钢筋,取mm65sa,则mm43565500s0ahh384.0480.04352003.140.110260maxs,2620c1sbhfM若按单筋截面设计,将成为超筋梁,所以应按双筋截面设计。取mm40sa为使总用钢量最小,取0bhx,则226s0y20c1maxs,smm367404353604352003.140.1384.010260ahfbhfMA2syb0c1ysmm2157367360518.04352003.140.136011AfbhffA选受拉钢筋6Φ22(2smm2281A),受压钢筋2Φ16(2smm402A)2.已知数据同题1,但已配受压钢筋3Φ18,2smm763A,试求受拉钢筋用量sA。解:mmN105.151404357633601026066s0sy1ahAfMM2799.04352003.140.1105.1512620c11sbhfM8317.02799.02112121121ss260sy1s1mm11634358317.0360105.151hfMA2syys1smm19267631163AffAA选用2Φ22+4Φ20(2smm20161256760A)3.某钢筋混凝土梁截面尺寸b×h=250×600mm,混凝土C30,HRB400钢筋,已配6根直径22mm的受拉钢筋(As=2281mm2),3根直径20mm的受压钢筋(sA=942mm2),(取as=65mm,sa=40mm)。该梁承受弯矩设计值M=340kN∙m。验算该梁的正截面受弯承载力是否满足要求。解:53565600s0ahhmmmm8.1342503.140.19423602281360c1sysybfAfAfxmm804022mm277535518.0s0bahmkN340mkN393mmN103934053594236028.1345358.1342503.140.126s0sy0c1uMahAfxhbxfM满足承载力要求钢筋混凝土受弯构件斜截面受剪承载力计算例题1.钢筋混凝土简支梁,截面尺寸b×h=250×550mm,h0=485mm,净跨ln=5.76m,均布荷载作用,荷载设计值q=75kN(包括自重),混凝土C30,箍筋HPB300,试计算斜截面受剪所需的箍筋用量。解:查表得2cN/mm3.14f,2tN/mm43.1f,2N/mm270yvf⑴求剪力设计值支座边缘处截面剪力值kN21676.5752121nqlV⑵验算截面尺寸216kNVkN5.433N4334694852503.1425.025.00cbhf截面尺寸满足要求⑶验算是否需要按计算配箍筋kN216121.4kNN12137148525043.17.07.00Vbhft需按计算配置箍筋⑷计算受剪所需的箍筋mmmm7226.04852701213712160007.020yv0t1svhfbhfVsnA选双肢8箍筋,mm1397226.03.502s,取mm125s实配双肢箍筋125@8偏心受压构件正截面受压承载力计算例题1.钢筋混凝土柱截面b×h=300×400mm,as=sa=40mm,弯矩设计值M=220kN∙m,轴向压力设计值N=400kN,混凝土C30(fc=14.3N/mm2),HRB400钢筋(2yyN/mm360ff),不考虑P-δ效应的附加影响,求所需的纵向受力钢筋。解:mm36040400s0ahh550mmm55.04002200NMe因20mmmm3.133040030h,所以取mm20aemm57020550a0ieee因mm1083603.00.3mm5700ihe,所以按大偏心受压情况计算mm7304024005702siahee取0bhx,则得2min223s0ybb20c1smm240400300002.0mm68240360360518.05.01518.03603003.140.1730104005.01bhahfbhfNeA23syb0c1ysmm179310400682360518.03603003.140.136011NAfbhffA选受拉钢筋5Φ22(2smm1900A),受压钢筋3Φ18(2smm763A)2.已知数据同题1,但已配受压钢筋3Φ22,2smm1140A,试求受拉钢筋用量sA。解:mmN107.1604036011403607301040063s0sy1ahAfNeM384.02890.03603003.140.1107.160maxs,2620c11sbhfM即满足0bhx的条件8889.036040118248.02890.021121211210sssha即满足s2ax的条件260sy1s1mm15033608248.0360107.160hfMA23ysyys1smm15323601040011401503fNAffAA选受拉钢筋4Φ22(2smm1520A)3.钢筋混凝土柱截面b×h=350×500mm,as=sa=40mm,弯矩设计值M=420kN∙m,轴向压力设计值N=1100kN,混凝土C35(fc=16.7N/mm2),HRB400钢筋,不考虑P-δ效应的附加影响,对称配筋,求所需的纵向受力钢筋。解:mm46040500s0ahh382mmm382.011004200NMe7.163050030hmm<20mm,20aemm402203820aieeemmmm6124025004022siaheemm2.1883507.160.111000001bfNxcmm804022mm238460518.0s0bah2001mm17904046036022.1884602.1883507.160.161211000002sycssahfxhbxfNeAA每侧选用钢筋5Φ22(2ssmm1900AA)
本文标题:混凝土结构计算例题
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