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1、柱截面设计(ZJM1)项目名称构件编号日期设计校对审核执行规范:混凝土结构设计规范(GB50010-2010),本文简称混凝土规范建筑抗震设计规范(GB5(X)11-2010),本文简称抗震规范钢筋:d-HPB300;D-HRB335;E-HRB400;F-RRB400:G-HRB500;P-HRBF335;Q-HRBF400;R-HRBF5001已知条件及计算要求:(1)已知条件:矩形柱b=550mm,h=550nm计算长度1=1Oom位强度等级C30,fc=14.30Ni2ft=1.43Nmm2纵筋级别HRB400,fy=360Nmmfy,=360N2箍筋级别HRB400,fy=360Nm
2、m2轴力设计值N=100.OOkN弯矩设计值Mx=50.OOkN.m,My=50.OOkN.m剪力设计值Vy=20.OOkN,Vx=20.OOkN(2)计算要求:1 .正截面受压承载力计算2 .斜截面承载力计算3 .裂缝计算内力设计值简图2受压计算2.1轴压比2 Abh=550x550=302500mwN100.00IO3=0.023fcA14.33025002.2受压计算双向偏心受压采用混凝土规范式6.2.21-3验算:(NUX及NUy计算时,只考虑弯矩作用方向的对边纵筋,忽略侧边纵筋)经试算,取每侧全部纵筋面积:AsX=605mm;sy-605mm,四根角筋总面积近似按1018代入公式验算
3、如下:3 1N=Ioo.00X10111+NUXNUyNuo13=194.22101 11373371+3733714810465结论:满足!X方向单边:Asx=605mn2PniA=0.0020302500=605mm2,5XAsx=605mm2y方向单边:Asy=605mm2ninA=0.0020X302500=605mm2,Asy=605mm2全截面:As=2Asx+2Asy-4Asj=1402mm2PninXA=O.0055X302500=1664M为一根角筋的面积),As=1664mm2将纵筋按边长分配可得:AH416mm:Asy=416mm2验算一边最小配筋:Asx=605mm2P
4、ninA=O.0020302500=605mm2,Asx=605mm2此时:Asy=227m2Pnin=0.0020302500-605un取A)iy=605验算一边最小配筋:sy=605mm2Pnin=0.0020302500=605mm取人,二605加此时:8x=227m23.0,取入x=3.0(D截面验算,根据混凝土规范式6.3.1:hb=O.94,受剪截面系数取0.25VX=20.00ANO.25cfcbho=0.251.0014.3550515=1012.62女N截面尺寸满足要求。(2)配筋计算根据混凝土规范式6.3.12:1.75V-ftbho-O.O!Nsvxa+1sfyv01.
5、75320000-1.43550515-0.07100.00103.00+12=(-0.886mmmm36O.O515箍筋最小配筋率:0.40%由于箍筋不加密,故Pvhin=O.4%X0.5=0.2%计算箍筋构造配筋ACnin/s:svminOminbh0.00205505502(5-10)+力-2(。$-1。)asvx-0.886ASYmin=0.161%VhQ20000515y=4.93.0,取人=3.0(D截面验算,根据混凝土规范式6.3.1:hb=O.94,受剪截面系数取0.25Vy=20.00ZN025HCfCbhO=0.251.0014.3550515=1012.62ZN截面尺寸满
6、足要求。配筋计算根据混凝土规范式6.3.12:V-1754o-OO7Nhsvy4+1SfyvfiO1.75320000-1.43550515-0.07100.0010*3.00+1,、2=(-0.886mmmm360.0515箍筋最小配筋率:0.40%由于箍筋不加密,故PVnIn=O.4%X0.5=0.2%计算箍筋构造配筋AXVnin/s:AsvtninOminbh0.00205505502=、,、=,、=0.605mmnnsb-2(5-10)+-2(y-10)550-2(35-10)+550-2(35-10)0.605一550=0.110%svy0.886As,min=-0.161%bs55
7、0bs故箍筋配筋量:Ars=0.605mm2mm3.3Xy双向受剪计算(D截面验算,根据混凝土规范式6.3.16:计算剪力V的作用方向与X轴的夹角:(vy(20000=arctanII=arctanII=0.785VxJ20000)V=20.OoAN0.25zrccA0cos()=0.251.014.35505150.707=716.03ZNX向截面尺寸满足要求。Vy=20.00kNAs=605mm2,配筋满足。(2)下部纵筋:3E18(763mm2P=0.25%)As=605un%配筋满足。(3)左右纵筋:止12(113加=0.04%)全侧配筋人$=622咂2人5=605|1)1112,配筋
8、满足。(4)竖向箍筋:E8200三肢箍(7547nP*0.14%)Asvs=605mm7m,配筋满足。(5)水平箍筋:E8C200三肢箍(7542nPjir=O.14%)Asvs=605mm2m,配筋满足。5裂缝计算5.1左右侧裂缝计算(D根据混凝土规范第7.1.2注3),偏压计算时eh0=(0/150)/0.515=0.00=0.55,不需要验算裂缝。5.2上下侧裂缝计算(D截面有效高度:力O=力-as=550-35=515Mm(2)受拉钢筋应力计算,根据混凝土规范式7.1.4-4:Mk80.00eQ=0.533m=533.3mmUNk150.00zO1000=1.814h550取*=1.0
9、h550V _=-a_=-35=240/h/m“2S2ese0+ys=1.OOOO533+240=773V =0NKe-Z)_150000.0x(773-421Asz763421(515O.87-O.12(1-0.00)1773)=164.7465515=420.6mm(3)按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率,根据混凝土规范式7.1.2-4:2teAs+Ap=般_763+0151250=0.0050Afe=0.5bh=0.5550550=151250Pw=0.00500.01,取Pte=O.01(4)裂缝间纵向受拉钢筋应变不均匀系数,根据混凝土规范式7.1.2-2:65f1k0.
10、652.01=1.1-=1.1-=0.3070pteCS0.0100164.7465受拉区纵向钢筋的等效直径CU:nid1deq=18wrtniidi根据混凝土规范表7.1.2T构件受力特征系数cr=1.9:(5)最大裂缝宽度计算,根据混凝土规范式7.12T:164.75/18.0=1.90.30701.925+0.08=0.092mm2000000.0100)5.3裂缝计算结果Wnax=max0.000,0.092=0.092mmWIin=O.40Omm,满足。605,0.20%605,0.11%605,0.20%605,0.20%5503E18(763,025%,-254)5503E18(763,0.25%,-254)计算钢筋面积简图逊200(754,0.14%)1E12(113,0.04%,+254)配筋简图【理正结构设计工具箱软件7.0】计算日期:2023-08-0716:34:19