地下室外墙计算书(纯手算)

工程号页码版本项目名称：图纸索引设计日期构件检查地下室外墙设计（非人防）(结构构件的重要性系数r0=)地下室层数N=3层(1~3层)计算简图室内外高差H1=mm地下水位距室外地面高度H2=mm土的重度gs=kN/m3地下水的重度gw=kN/m3土有效内摩擦角φ'=度地面堆载q0=kN/m2土的浮重度g0=kN/m3静止土压力系数Ko==混凝土标号为：则fc=N/mm2ftk=N/mm2ft=N/mm2钢筋型号为：则fy=N/mm2=fy'Es=N/mm2（1）外墙荷载及内力计算各荷载分项系数：水压力为，土压力为，地面堆载为则有：p0=r0q0Ko=1.2×1×35×0.5=kN/mp1=p0+r0gsH2Ko=21+1.2×1×18×0×0.5=kN/mp2=p1+r0g0(L1-H1-H2)K0+r0gw(L1-H1-H2)=21+1.2×1×8×(6.475-0-0)×0.5+1×1×10×(6.475-0-0)=kN/mp3=p1+r0g0(L1-H1-H2+L2)K0+r0gw(L1-H1-H2+L2)=21+1.2×1×8×(6.475-0-0+2.9)×0.5+1×1×10×(6.475-0-0+2.9)=kN/mp4=p1+r0g0(L1-H1-H2+L2+L3)K0+r0gw(L1-H1-H2+L2+L3)=21+1.2×1×8×(6.475-0-0+2.9+4.4)×0.5+1×1×10×(6.475-0-0+2.9+4.4)=kN/m外墙弯矩及剪力设计值为：+M1=kNm,+M2=kNm,+M3=kNm-MB=kNm,-MC=kNm,-MD=kNmVmax=kN,Vmax=kN,Vmax=kN荷载效应标准组合下的弯矩为：+M1K=kNm,+M2K=kNm,+M3K=kNm-MBK=kNm,-MCK=kNm,-MDK=kNmProgramBasementWallVersion1.0@ARUPChinaLtd.2007.06Page1DesignofBasementWall(ChinaCode)PrintedTime-20/Apr/204:59PM153.133.02151.1232.3186.8310.9248.1198330268.1351457.5159.7501.21.0224.870164.234.68160.51.221.0001.21.0116.8301.21.02000001.01.21.21.221.000C3516.72.21.57HRB33530010303581-sinφ'0.534400700001816475500229006001.0层号层高(mm)外墙厚(mm)工程号页码版本项目名称：图纸索引设计日期构件检查-（2）外墙配筋计算(b=mm考虑)a)地下室一层外墙（d1=m，a1=)①相对界限受压区高度xb=假设受拉钢筋按双层布置，则h0=(-)=mm单筋矩形截面所能承受的最大弯矩为：Mmax==kNm较大值(|M1|，|MB|)故不需配双筋，按单筋受弯构件计算。②计算混凝土受压区高度x混凝土保护层厚度c=mm，as=as'=c+=mm则截面有效高度为h0=d1-=mm代入公式，解得x=h0-√=-√×-2×248.1/(1×16700×1)=mxbh0=m满足规范要求③计算竖向受力钢筋的面积AsAs===mm2初选钢筋f20-(As=mm2)④验算配筋率=0.45ft/fy==%，取=%=xba1fc/fy==%===%===%配筋满足规范要求⑤计算水平分布钢筋的面积Ash===mm2单边实配钢筋为f20-(2As=mm2)配筋满足规范要求⑥抗剪验算：0.7bhftbh0=0.7×1×1.57×1000×440/1000=kNkN抗剪满足规范要求⑦外墙裂缝变形计算外墙最大裂缝宽度限值=mm抗裂需配钢筋为f22-(As=mm2)则截面有效高度为h0=d1--=mm荷载效应标准组合下的最大弯矩为：Mk=kNmρte==3801/(0.5×1000×500)=0.01，取ρte=σsk==×()=N/mm2ψ=1.1-=1.1-0.65×2.2/(0.015×160.018)=≥≤取ψ=则最大裂缝宽度为σskdeqEsρte=2.1×0.512×160.018×(1.9×50+0.08×22÷0.015)/200000=mm≤=mm裂缝宽度满足规范要求ProgramBasementWallVersion1.0@ARUPChinaLtd.2007.06Page2DesignofBasementWall(ChinaCode)PrintedTime)ωmax0.081.9c=2.1ψ20/Apr/204:59PM0.65ftk/ρteσsk0.5120.21.00.5120.182ωlim0.2(+232.3As/0.5bh0.0152040.015204Mk/0.87h0As232.3106/0.87×439×3801160.0175318483.56268.10.210038015011439Ashρminbh0.236%×1000×50011781504190ρminρAs/bh02095/(1000×440)0.476ρmaxρmax0.55×1×16.7/3003.062ρ1As/bh2095/(1000×500)0.419ρmin0.45×1.57/3000.2360.20%ρmin0.2360.0350.242a1fcbx/fy1×16.7×1000×35/3001949150209550106060440h02-2M/a1fcb0.440.440.440.55050060440a1fcbh02xb(1-0.5xb)1289.207取墙宽10000.51.00工程号页码版本项目名称：图纸索引设计日期构件检查-b)地下室二层外墙（d2=m，a1=)①相对界限受压区高度xb=假设受拉钢筋按双层布置，则h0=(-)=mm单筋矩形截面所能承受的最大弯矩为：Mmax==kNm较大值(|M2|，|MC|)故不需配双筋，按单筋受弯构件计算。②计算混凝土受压区高度x混凝土保护层厚度c=mm，as=as'=c+=mm则截面有效高度为h0=d2-=mm代入公式，解得x=h0-√=-√×-2×198/(1×16700×1)=mxbh0=m满足规范要求③计算竖向受力钢筋的面积AsAs===mm2初选钢筋f20-(As=mm2)④验算配筋率=0.45ft/fy==%，取=%=xba1fc/fy==%===%===%配筋满足规范要求⑤计算水平分布钢筋的面积Ash===mm2单边实配钢筋为f20-(2As=mm2)配筋满足规范要求⑥抗剪验算：0.7bhftbh0=0.7×1×1.57×1000×540/1000=kNkN抗剪满足规范要求⑦外墙裂缝变形计算外墙最大裂缝宽度限值=mm抗裂需配钢筋为f22-(As=mm2)则截面有效高度为h0=d2--=mm荷载效应标准组合下的最大弯矩为：Mk=kNmρte==3801/(0.5×1000×600)=0.01，取ρte=σsk==×()=N/mm2ψ=1.1-=1.1-0.65×2.2/(0.013×104.802)=≤取ψ=则最大裂缝宽度为σskdeqEsρte=2.1×0.2×104.802×(1.9×50+0.08×22÷0.013)/200000=mm≤=mm裂缝宽度满足规范要求ProgramBasementWallVersion1.0@ARUPChinaLtd.2007.06Page3DesignofBasementWall(ChinaCode)PrintedTimeωmax0.081.9c)=2.1ψ20/Apr/204:59PM0.65ftk/ρteσsk0.0230.21.00.20.052ωlim0.2(+186.8As/0.5bh0.012670.01267Mk/0.87h0As186.8106/0.87×539×3801104.8023412593.463510.210038015011539Ashρminbh0.236%×1000×60014131504190ρ1As/bh2095/(1000×600)0.349ρminρAs/bh02095/(1000×540)0.388ρmax0.20%ρmin0.236ρmax0.55×1×16.7/3003.062a1fcbx/fy1×16.7×1000×22/30012251502095ρmin0.45×1.57/3000.236h02-2M/a1fcb0.540.540.540.0220.297a1fcbh02xb(1-0.5xb)1941.801501060605400.61.000.55060060540工程号页码版本项目名称：图纸索引设计日期构件检查-c)地下室三层外墙（d3=m，a1=)①相对界限受压区高度xb=假设受拉钢筋按双层布置，则h0=(-)=mm单筋矩形截面所能承受的最大弯矩为：Mmax==kNm较大值(|M3|，|MD|)故不需配双筋，按单筋受弯构件计算。②计算混凝土受压区高度x混凝土保护层厚度c=mm，as=as'=c+=mm则截面有效高度为h0=d3-=mm代入公式，解得x=h0-√=-√×-2×330/(1×16700×1)=mxbh0=m满足规范要求③计算竖向受力钢筋的面积AsAs===mm2初选钢筋f20-(As=mm2)④验算配筋率=0.45ft/fy==%，取=%=xba1fc/fy==%===%===%配筋满足规范要求⑤计算水平分布钢筋的面积Ash===mm2单边实配钢筋为f20-(2As=mm2)配筋满足规范要求⑥抗剪验算：0.7bhftbh0=0.7×1×1.57×1000×640/1000=kNkN抗剪满足规范要求⑦外墙裂缝变形计算外墙最大裂缝宽度限值=mm抗裂需配钢筋为f22-(As=mm2)则截面有效高度为h0=d3--=mm荷载效应标准组合下的最大弯矩为：Mk=kNmρte==3801/(0.5×1000×700)=0.01，取ρte=σsk==×()=N/mm2ψ=1.1-=1.1-0.65×2.2/(0.011×147.131)=≥≤取ψ=则最大裂缝宽度为σskdeqEsρte=2.1×0.205×147.131×(1.9×50+0.08×22÷0.011)/200000=mm≤=mm裂缝宽度满足规范要求ProgramBasementWallVersion1.0@ARUPChinaLtd.2007.06Page4DesignofBasementWall(ChinaCode)PrintedTime+)ωmax0.081.9c20/Apr/204:59PM=2.10.65ftk/ρteσsk0.2050.21.00.2050.082ωlim0.2ψ(310.9As/0.5bh0.010860.01086Mk/0.87h0As310.9106/0.87×639×3801147.1305104703.36457.50.210038015011639Ashρminbh0.236%×1000×70016491504190ρ1As/bh2095/(1000×700)0.299ρminρAs/bh02095/(1000×640)0.327ρmax0.20%ρmin0.236ρmax0.55×1×16.7/3003.062a1fcbx/fy1×16.7×1000×32/30017821502095ρmin0.45×1.57/3000.236h02-2M/a1fcb0.640.640.640.0320.352a1fcbh02xb(1-0.5xb)2727.578501060606400.71.000.55070060640