混泥土结构(第四版)第五章答案

1、第5章受压构件的截面承载力已知某多层四跨现浇框架结构的第二层内柱，轴心压力设计值N1100kN,楼层高H6m,混凝土强度等级为C20,采用H级钢筋(HRB335ffi筋)。柱截面尺寸为350mm350mm,求所需纵筋面积。解：根据混凝土结构设计规范规定：l01.25H1.2567.5m则：r端021.43查表61得:0.714由N0.9fcAfyAs得:fy0.9fcA30031100109.63500.90.71423501786mm选取42521964mmAsA1.6%350350min0.6%,可以截面每一侧配筋率：0.2% ,可以19640.5TZTTZT0.8%min350350已知

2、圆形截面现浇钢筋混凝土柱,直径不超过350mm,承受轴心压力设计值N1900kN,计算长度I。4m,混凝土强度等级为C25,柱中纵筋采用HRB400钢筋，箍筋用I级钢筋。试设计该柱截面。解：柱子直径取为350mm(1)先按普通纵筋和箍筋柱计算104ml0d400035011.43,查表得:0.9311 Nf 0.9y12122A-d2-35mm244由N0.9fcAfyAs得：,119000002fcA11.996162.53120mm3600.90.931-3.2%A96162.5选取822As3041mm2(2)按采用螺旋箍筋柱计算，假定3%得：As2A2885m

3、m选取6125As22945mm混凝土的保护层取30mm,则:dcor30235060290mmd2dcor66018.5mm20.9fcAcor2fyAssofyAs混凝土强度等级为C25,1.0A-N0.9fcAcorfyAs2fy19000000.911.966018.53602368mm其中：AssodcorAss1取螺旋箍筋直径2d10mm,则Assi78.5mmdA-corss1sAsso3.1429078.5194mm368取s50mm,以满足不小于40mm,并不大于80mm及0.2dcor的要求。所以：柱中纵筋选取625,As22945mm箍筋采用螺旋箍筋,s50mm,d10m

4、m验算：AssodcorAss13.1429078.521429.6mm50Asso0.25As20.252945736.25mm2,可以N0.9fcAcor2fyAssofyAs0.911.966018.521.03601429.636029452587.6kNN 0.9fcAfy As0.9 0.93111.9 96162.5 360 30411876.1kN1.5 1876.1kN 2814.2kN2587.6kN所以该柱的轴心受压承载力设计值为2587.6kN 1900kN ,故满足要求。已知柱的轴向力设计值N 800kN ,弯矩M 160kNm ；截面尺寸b 300mm ,h 500

5、mm, as as45mm ;混凝土强度等级为C30,采用HRB400级车冈筋;计算长度l0 3.5m 0求钢筋截面面积As 及 As。解：e0M160 1033N800 100.2m 200mm20mm , h0as 455mme0ea20020 220mmfcA0.5 c 0.514.3 300 5008001031.341,ei3.50.515,取1.01400ei/h0 h11400 220455235001 1 1.075001.07 220 235.4mm0.3h0136.5mm可以按大偏心受压计算:he ei-asi2 s235.4 等 45440.4mmNe 1 fcbx h0

6、fy As hoas,取b 0.5182Nefcbhcb10.5bfyhgas800103440.41.014.330045520.51810.50.51836045545rr 277mm minbh-20.002 300 500 300mm取As300mm21 fcbx fy AsfyAsAs1fcb bh0 NfyAsfy3601.014.33000.5184558001033002886mm受拉钢筋选用2 25, As2982mm ；受压钢筋选用 2 16, As2402mm。N fy AsfyAsx 1 fcb800 103 360 402 360 982 ” 235mm1.0 14.

7、3300x 2350.517h04550.518故假定为大偏心受压是正确的。 垂直于弯矩作用平面的验算：I03535 11.67,查表得0.30.955Nu0.9fcAfyAsfyAs0.9 0.955 14.3300500 360 402360 9822271864 N 800103 N满足要求。已知柱的轴向力设计值N 550kN ,弯矩 M450kN m ;截面尺寸 b 300mm ,h600mm,aa45mm；混凝土强度等级为C35,采用in级钢筋；计算长度I。7.2m 0求钢筋截面面积As及As。帼e0M45U030.818m818mme20mm,h0ha555mm81820838mm

8、fcA0.5c-N0.516.73006005501032.7311.0l0h7.20.612151.01400ei/h02l0h1838140055572001.068838895mm0.3h0166.5mm600111.068可以按大偏心受压计算:Nehei21 fcbxh。89560045 1150mmfyAs h0b 0.518Ne1 fcbh： b0.5 bfyh0as550 1 03 1150 1.0 1 6.7 30025552 0.5181 0.5 0.518360555452263mm min bh0.2% 300 600 360mm2取As2360mm1 fcbx fy A

9、sfyAsAs1fcbbh0N受拉钢筋选用232+228,As22841mm；受压钢筋选用216,As2402mm。N fy AsfyAs1 fcb550103fy1.0 16.7 300 0.518 555 550 1032360 2833mm4023602841237.5mm1.016.7360b 0.518x237.5八0.428h0555故假定为大偏心受压是正确的。垂直于弯矩作用平面的验算：b 0.4l0b72-7224,查表得0.30.65Nu0.9fcAfyAsfyAs0.90.6516.730060036040236028413170kN,弯矩 M 83.6kN45mm ;混凝土

10、强度等级为m；截面尺寸 C35,采用田32441485.8N55010N满足要求。已知荷载作用下柱的轴向力设计值b400mm,h600mm,a级钢筋；计算长度106.0m0求钢筋截面面积As及As。解：e00.0264 m 26.4 mm3M83.610N317010320mm,h。555mmeie0ea26.42046.4mmfcA0.5c-N0.516.740060031701030.632l0h6.00.61015,1.01400e/h01.46.41400-555260000.63260011.544ei1.54446.471.4mm0.3h0166.5mm初步判定为小偏心受压。0.8

11、,b0.518minbh_一20.002400600480mmfyhe-eia30071.445183.6mm2heeia71.430045326.4mm2Neifcbx-asAha,23X317010183.61.016.7400x4520.80.5180.8360解得x461mmxNe1fcbxh02fyAsh0aNe1fcbxh00.5xfyh0a46155523_317010326.41.016.7400461e。ea1fcbhh0fyh0as31701033004526.4201.016.740060055530036055545负数不会发生反向破坏。受拉钢筋选用218,As509m

12、m360 555 452192 mm min bh 480mm为了防止反向破坏，对As进行验算。；受压钢筋选用218,As509mm2。由N1fcbxfyAssAs和sLfyb1得x444mmbh00.518555287.5mm所以按小偏心受压计算是正确的。验算垂直于弯矩作用方向的承载力：l06015,查表得0.895Nu0.9fcAfyAsfyAs0.90.89516.740060036050936050933523643.64N317010N满足要求。已知轴向力设计值N7500kN,弯矩M1800kNm；截面尺寸b800mm,h1000mm,aa45mm；混凝土强度等级为C30,采用II级

13、钢筋；计算长度I。6m,采用对称配筋（AsAs）;求钢筋截面面积AsAs。解：N1fcbxfyAsfyAsN 1fcb37500 101.0 14.3 800655.6mm xbbh0 525.25mm按小偏心受压计算。e0M 1800 0.24m 240mmN100075003033mm,h0 h as 955mme0ea24033 273mml0hCL fcA 0.5 N6000 人610000.514.3 800 10007500 1030.763ei15 ，取1400e /h01.068 273292mmei2按简化方法计算:292 5000.81.012731400 9550.3h0

14、286.5mm45 747mm2600010000.763 1 1.06837500 103 0.55 1 14.3 800 955327500 10 747 0.43 1 14.3 800 9550.8 0.55955 451 14.3 800 9550.55Ne0.43ifcbho0.8bhoa1.0 9.6 40033100 101.0 9.6 400x 360所以 x 610.5mm h取x h 600mm进行计算。x1 fcbx h0fy As h0 ;2e N1.0 9.6 400 600 555 30033100 1019643600.8555-555 360 6030.518

15、0.81964555 450.645xh00.645955616mmNe1fcbh(210.5；Asfy%a_3_2_750010747114.3800955064510.50.645305.8mme-a305.86004550.8mm22取1.0则向50.8mme0eiea50.82030.8mmMuNe0 310010330.8 95.48kN m M 85kN m经计算1.35与假定的1.0相矛盾重新选取1.4,则ei36.3mme0eiea36.32016.3mmM u Ne033100 1016.3 50.53kN m经过计算1.4,可以。因MuM85kNm,故截面不安全。已知某单层

16、工业厂房的形截面边柱，下柱高5.7m,柱截面控制内力N870kN,M420kNm;截面尺寸b80mm,h700mm,bfbf350mm,hfhf112mm,aa45mm;混豌土强度等级为C35,采用HRB400钢筋；对称配筋。求钢筋截面面积。解：e0420 1033870 100.483m 483mm7003023mm , h0h a 655mmeie0ea48323506mmCL 0.50.516.7 1164803870 101.17 1 ,取 11.0l01.0H5.7ml05700c.，8.14h7001400ei/h0l0h1c5061400-655570070021.01.01.06先按大偏心受压计算。8701031fcbf1.0 16.7350149mm hf此时中和轴在腹板内N 1 fchf bf b8703101.0 16.7 112350 801%b1.016.780273mm xb 0.55 655360mm按大偏心受压公式计算钢筋。ei1.06 506536mmheei2841mm由 Ne 1fc cbxbfb hfhffy Ash0 aNe1fcbxbf bhfhehffyha112112 655 23273870108411.016.780273655350802300655452109