冠梁及支撑配筋计算书

冠梁配筋计算书执行:《混凝土结构设计规范》(GB50010-2010),本文简称《混凝土规范》钢筋：d-HPB300;D-HRB335;E-HRB400;F-RRB400;G-HRB500;P-HRBF335;Q-HRBF400;R-HRBF500-----------------------------------------------------------------------1已知条件及计算要求：(1)已知条件：矩形柱b=800mm，h=1200mm计算长度L=8.50m砼强度等级C30，fc=14.30N/mm2ft=1.43N/mm2纵筋级别HRB400，fy=360N/mm2，fy'=360N/mm2箍筋级别HPB300，fy=270N/mm2轴力设计值N=1.00kN弯矩设计值Mx=775.00kN.m，My=0.00kN.m剪力设计值Vy=50.00kN，Vx=0.00kN(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-----------------------------------------------------------2受压计算2.1轴压比2Abh8001200960000mmN1.001030.000fA14.3960000c2.2偏压计算1(1)计算相对界限受压区高度ξ《混凝土规范》式6.2.7-1:b10.800.5176bf360.0y11E2000000.0033scu(2)计算轴向压力作用点至钢筋合力点距离e:hha1200351165mm0sM775.00e774.9999m774999.9mm0N1.00emax{20,h/30}40.0mmaeee774999.940.0775039.9mmi0ah1200eea775039.935775604.9mmi2s2(3)计算配筋e775039.9mm0.3h0.31165349.5mmi0Nfbh1.0014.380011650.51766898994Nb1c0b且N=1.00kN≤N=6898.99kN,按照大偏心受压构件计算,根据《混凝土规范》6.2.17:bN1.00103x0.1mmfb1.0014.38001c因x=0.1mm<2a=70mm,取x=70.0mmsxNefbxh1c02AA'ss''fhay0se0.5h0.5xNi''fhay0s1.00103775039.90.51200.00.570.01904mm2360.01165352.3轴压验算(1)计算稳定系数φl0850010.6b800根据《混凝土规范》表6.2.15:插值计算构件的稳定系数φ=0.971(2)计算配筋,根据《混凝土规范》公式6.2.15:N1.00103fA14.39600000.9c0.90.971A'38130mm2s'360.0fy取A=0mm2s偏压计算配筋:x方向A=1904mm2sx:y方向A=0mm2sy轴压计算配筋:x方向A=0mm2sx2:y方向A=0mm2sy计算配筋结果:x方向A=1904mm2sxy方向A=0mm2sy最终配筋面积:x方向单边:A=1904mm2≤ρ×A=0.0020×960000=1920mm2,取A=1920mm2sxminsxy方向单边:A=0mm2≤ρ×A=0.0020×960000=1920mm2,取A=1920mm2syminsy全截面:A=2×A+2×A=7680mm2>ρ×A=0.0055×960000=5280mm2ssxsymin3受剪计算3.1x方向受剪计算剪力为零,采用构造配筋:箍筋最小配筋率:0.40%由于箍筋不加密,故ρ=0.4%×0.5=0.2%vminAAbhsvxsvminmin0.002012008001.011mm2/mmssa10a1012002351080023510b2sh2s3.2y方向受剪计算hha1200351165mm0sM77500000013.30yVh5000011650λ=13.3>3.0,取λ=3.0yy(1)截面验算,根据《混凝土规范》式6.3.1:h/b=1.5≤4,受剪截面系数取0.25wV50.00kN0.25fbh0.251.0014.380011653331.90kNycc0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fbh0.07NAt0svy1sfhyv01.75500001.4380011650.071.001033.0011.695mm2/mm270.01165箍筋最小配筋率:0.40%由于箍筋不加密,故ρ=0.4%×0.5=0.2%vmin计算箍筋构造配筋A/s:svminAbhsvminmin0.002080012001.011mm2/mmsa10a1080023510120023510b2sh2sAAsvy1.695svmin1.0110.212%0.126%bs800bs800故箍筋配筋量:A/s=1.011mm2/mmsvy4配置钢筋3(1)上部纵筋:4E25(1963mm2ρ=0.20%)>As=1920mm2，配筋满足。(2)下部纵筋:4E25(1963mm2ρ=0.20%)>As=1920mm2，配筋满足。(3)左右纵筋:5E25(2454mm2ρ=0.26%)分配As=2945mm2>As=1920mm2，配筋满足。(4)竖向箍筋:d8@190四肢箍(1058mm2/mρ=0.13%)>Asv/s=1011mm2/m，配筋满足。sv(5)水平箍筋:d8@190四肢箍(1058mm2/mρ=0.09%)>Asv/s=1011mm2/m，配筋满足。sv5裂缝计算5.1左右侧裂缝计算(1)根据《混凝土规范》第7.1.2注3),偏压计算时e/h=(0/150)/0.765=0.00<=0.55,不00需要验算裂缝。5.2上下侧裂缝计算(1)根据《混凝土规范》第7.1.2注3),偏压计算时e/h=(80/150)/1.165=0.46<=0.55,不00需要验算裂缝。5.3裂缝计算结果W=max{0.000,0.000}=0.000mmρ×A=0.0055×480000=2640mm2ssxsymin3受剪计算3.1x方向受剪计算剪力为零,采用构造配筋:箍筋最小配筋率:0.40%由于箍筋不加密,故ρ=0.4%×0.5=0.2%vminAAbhsvxsvminmin0.00208006000.738mm2/mmssa10a108002351060023510b2sh2s3.2y方向受剪计算hha80035765mm0sM3660000009.57yVh500007650λ=9.6>3.0,取λ=3.0yy(1)截面验算,根据《混凝土规范》式6.3.1:h/b=1.3≤4,受剪截面系数取0.25wV50.00kN0.25fbh0.251.0014.36007651640.93kNycc0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fbh0.07NAt0svy1sfhyv01.75500001.436007650.071132.001033.0011.532mm2/mm270.0765箍筋最小配筋率:0.40%由于箍筋不加密,故ρ=0.4%×0.5=0.2%vmin计算箍筋构造配筋A/s:svminAbhsvminmin0.00206008000.738mm2/mmsa10a106002351080023510b2sh2s7AAsvy1.532svmin0.7380.255%0.123%bs600bs600故箍筋配筋量:A/s=0.738mm2/mmsvy4配置钢筋(1)上部纵筋:5E20(1571mm2ρ=0.33%)>As=960mm2，配筋满足。(2)下部纵筋:5E20(1571mm2ρ=0.33%)>As=960mm2，配筋满足。(3)左右纵筋:3E20(942mm2ρ=0.20%)分配As=1257mm2>As=960mm2，配筋满足。(4)竖向箍筋:d8@200三肢箍(754mm2/mρ=0.13%)>Asv/s=738mm2/m，配筋满足。sv(5)水平箍筋:d8@200三肢箍(754mm2/mρ=0.09%)>Asv/s=738mm2/m，配筋满足。sv5裂缝计算5.1左右侧裂缝计算(1)根据《混凝土规范》第7.1.2注3),偏压计算时e/h=(0/150)/0.565=0.00<=0.55,不00需要验算裂缝。5.2上下侧裂缝计算(1)截面有效高度:hha80035765mm0s(2)受拉钢筋应力计算,根据《混凝土规范》式7.1.4-4:Mk80.00e0.533m533.3mm0N150.00k21l11130021011.0715se5338000h40004000h7650h800ya35365mms2s2eey1.0715533365936mms0s'0fh22z'0h765765604.3mm0.870.12100.870.1210.00fe936Nezk150000.093660452.4954N/mm2skAz1571604s(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:A0.5bh0.5600800240000mm2teAAsp157100.0065teA240000teρ=0.0065<0.01,取ρ=0.01tete(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7.1.2-2:0.65ftk0.652.011.11.11.38880.010052.4954tesψ=-1.3888<0.2,取ψ=0.28受拉区纵向钢筋的等效直径d:eqnd2iid20mmeqndiii根据《混凝土规范》表7.1.2-1构件受力特征系数α=1.9:cr(5)最大裂缝宽度计算,根据《混凝土规范》式7.1.2-1:σ=σssqdseq1.9c0.08maxcrEsste52.5020.01.90.20001.9250.080.021mm2000000.01005.3裂缝计算结果W=max{0.000,0.021}=0.021mm