基础课程设计样本独立基础文档格式.docx
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HRB400fy=360N/mm2
3.计算信息
结构重要性系数:
γo=1.0
基础埋深:
dh=2.000m
纵筋合力点至近边距离:
as=40mm
基础及其上覆土的平均容重:
γ=20.000kN/m3
最小配筋率:
ρmin=0.150%
4.作用在基础顶部荷载标准组合值
F=1759.100kN
Mx=11.500kN*m
My=12.300kN*m
Vx=0.700kN
Vy=0.700kN
ks=1.35
Fk=F/ks=1759.100/1.35=1303.037kN
Mxk=Mx/ks=11.500/1.35=8.519kN*m
Myk=My/ks=12.300/1.35=9.111kN*m
Vxk=Vx/ks=0.700/1.35=0.519kN
Vyk=Vy/ks=0.700/1.35=0.519kN
5.修正后的地基承载力特征值
fa=180.000kPa
四、计算参数
1.基础总长Bx=B1+B2=1.550+1.550=3.100m
2.基础总宽By=A1+A2=1.550+1.550=3.100m
3.基础总高H=h1+h2=0.250+0.250=0.500m
4.底板配筋计算高度ho=h1+h2-as=0.250+0.250-0.040=0.460m
5.基础底面积A=Bx*By=3.100*3.100=9.610m2
6.Gk=γ*Bx*By*dh=20.000*3.100*3.100*2.000=384.400kN
G=1.35*Gk=1.35*384.400=518.940kN
五、计算作用在基础底部弯矩值
Mdxk=Mxk-Vyk*H=8.519-0.519*0.500=8.259kN*m
Mdyk=Myk+Vxk*H=9.111+0.519*0.500=9.370kN*m
Mdx=Mx-Vy*H=11.500-0.700*0.500=11.150kN*m
Mdy=My+Vx*H=12.300+0.700*0.500=12.650kN*m
六、验算地基承载力
1.验算轴心荷载作用下地基承载力
pk=(Fk+Gk)/A=(1303.037+384.400)/9.610=175.592kPa【①5.2.2-2】
因γo*pk=1.0*175.592=175.592kPa≤fa=180.000kPa
轴心荷载作用下地基承载力满足要求
2.验算偏心荷载作用下的地基承载力
exk=Mdyk/(Fk+Gk)=9.370/(1303.037+384.400)=0.006m
因|exk|≤Bx/6=0.517mx方向小偏心,
由公式【①5.2.2-2】和【①5.2.2-3】推导
Pkmax_x=(Fk+Gk)/A+6*|Mdyk|/(Bx2*By)
=(1303.037+384.400)/9.610+6*|9.370|/(3.1002*3.100)
=177.479kPa
Pkmin_x=(Fk+Gk)/A-6*|Mdyk|/(Bx2*By)
=(1303.037+384.400)/9.610-6*|9.370|/(3.1002*3.100)
=173.705kPa
eyk=Mdxk/(Fk+Gk)=8.259/(1303.037+384.400)=0.005m
因|eyk|≤By/6=0.517my方向小偏心
Pkmax_y=(Fk+Gk)/A+6*|Mdxk|/(By2*Bx)
=(1303.037+384.400)/9.610+6*|8.259|/(3.1002*3.100)
=177.255kPa
Pkmin_y=(Fk+Gk)/A-6*|Mdxk|/(By2*Bx)
=(1303.037+384.400)/9.610-6*|8.259|/(3.1002*3.100)
=173.928kPa
3.确定基础底面反力设计值
Pkmax=(Pkmax_x-pk)+(Pkmax_y-pk)+pk
=(177.479-175.592)+(177.255-175.592)+175.592
=179.142kPa
γo*Pkmax=1.0*179.142=179.142kPa≤1.2*fa=1.2*180.000=216.000kPa
偏心荷载作用下地基承载力满足要求
七、基础冲切验算
1.计算基础底面反力设计值
1.1计算x方向基础底面反力设计值
ex=Mdy/(F+G)=12.650/(1759.100+518.940)=0.006m
因ex≤Bx/6.0=0.517mx方向小偏心
Pmax_x=(F+G)/A+6*|Mdy|/(Bx2*By)
=(1759.100+518.940)/9.610+6*|12.650|/(3.1002*3.100)
=239.597kPa
Pmin_x=(F+G)/A-6*|Mdy|/(Bx2*By)
=(1759.100+518.940)/9.610-6*|12.650|/(3.1002*3.100)
=234.501kPa
1.2计算y方向基础底面反力设计值
ey=Mdx/(F+G)=11.150/(1759.100+518.940)=0.005m
因ey≤By/6=0.517y方向小偏心
Pmax_y=(F+G)/A+6*|Mdx|/(By2*Bx)
=(1759.100+518.940)/9.610+6*|11.150|/(3.1002*3.100)
=239.295kPa
Pmin_y=(F+G)/A-6*|Mdx|/(By2*Bx)
=(1759.100+518.940)/9.610-6*|11.150|/(3.1002*3.100)
=234.803kPa
1.3因Mdx≠0Mdy≠0
Pmax=Pmax_x+Pmax_y-(F+G)/A
=239.597+239.295-(1759.100+518.940)/9.610
=241.842kPa
1.4计算地基净反力极值
Pjmax=Pmax-G/A=241.842-518.940/9.610=187.842kPa
Pjmax_x=Pmax_x-G/A=239.597-518.940/9.610=185.597kPa
Pjmax_y=Pmax_y-G/A=239.295-518.940/9.610=185.295kPa
2.柱对基础的冲切验算
2.1因(H≤800)βhp=1.0
2.2x方向柱对基础的冲切验算
x冲切面积Alx=max((A1-hc/2-ho)*(bc+2*ho)+(A1-hc/2-ho)2,(A2-hc/2-ho)*(bc+2*ho)+(A2-hc/2-ho)2
=max((1.550-0.400/2-0.460)*(0.400+2*0.460)+(1.550-0.400/2-0.460)2,(1.550-0.400/2-0.460)*(0.400+2*0.460)+(1.550-0.400/2-0.460)2)
=max(1.967,1.967)
=1.967m2
x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=1.967*187.842=369.467kN
γo*Flx=1.0*369.467=369.47kN
因γo*Flx≤0.7*βhp*ft_b*bm*ho(6.5.5-1)
=0.7*1.000*1.43*860*460
=396.00kN
x方向柱对基础的冲切满足规范要求
2.3y方向柱对基础的冲切验算
y冲切面积Aly=max((B1-bc/2-ho)*(hc+2*ho)+(B1-bc/2-ho)2,(B2-bc/2-ho)*(hc+2*ho)+(B2-bc/2-ho)2)
=max((1.550-0.400/2-0.460)*(0.400+2*0.460)+(1.550-0.400-0.460)2/2,(1.550-0.400/2-0.460)*(0.400+2*0.460)+(1.550-0.400-0.460)2/2)
y冲切截面上的地基净反力设计值Fly=Aly*Pjmax=1.967*187.842=369.467kN
γo*Fly=1.0*369.467=369.47kN
因γo*Fly≤0.7*βhp*ft_b*am*ho(6.5.5-1)
=0.7*1.000*1.43*860.000*460
=396.00kN
y方向柱对基础的冲切满足规范要求
八、基础受剪承载力验算
1.计算剪力
pk=(Fk+Gk)/A=(1303.0+384.4)/9.6=175.6kPa
A'
=(A1+A2)*(max(B1,B2)-0.5*bc)
=(1550+1550)*(max(1550,1550)-0.5*400)
=4.18m2
Vs=A'
*pk=4.18*175.6=734.9kN
基础底面短边尺寸大于柱宽加两倍基础有效高度,不需验算受剪承载力!
九、柱下基础的局部受压验算
因为基础的混凝土强度等级大于等于柱的混凝土强度等级,所以不用验算柱下扩展基础顶面的局部受压承载力。
十、基础受弯计算
双向受弯
因ex≤Bx/6=0.517mx方向小偏心
a=(Bx-bc)/2=(3.100-0.400)/2=1.350m
Pj1=(Bx-a)*(Pmax_x-Pmin_x)/Bx+Pmin_x-G/A
=(3.100-1.350)*(239.597-234.501)/3.100+234.501-518.940/9.610
=183.378kPa
因ey≤By/6=0.517my方向小偏心
a=(By-hc)/2=(3.100-0.400)/2=1.350m
Pj2=(By-a)*(Pmax_y-Pmin_y)/By+Pmin_y-G/A
=(3.100-1.350)*(239.295-234.803)/3.100+234.803-518.940/9.610
=183.339kPa
βx=1.025
βy=1.025
MI=1/48*βx*(Bx-bc2*(2*By+hc)*(Pj1+Pjmax_x)
=1/48*1.025*(3.100-0.400)2*(2*3.100+0.400)*(183.378+185.597)
=378.99kN*m
MII=1/48*βy*(By-hc)2*(2*Bx+bc)*(Pj2+Pjmax_y)
=1/48*1.025*(3.100-0.400)2*(2*3.100+0.400)*(183.339+185.295)
=378.64kN*m
十一、计算配筋
1.计算基础底板x方向钢筋
Asx=γo*MI/(0.9*ho*fy)
=1.0*378.99*106/(0.9*460.000*360)
=2542.9mm2
Asx1=Asx/By=2542.9/3.100=820mm2/m
Asx1=max(Asx1,ρmin*H*1000)
=max(820,0.150%*500*1000)
=820mm2/m
选择钢筋14@180,实配面积为855mm2/m。
2.计算基础底板y方向钢筋
Asy=γo*MII/(0.9*ho*fy)
=1.0*378.64*106/(0.9*460.000*360)
=2540.5mm2
Asy1=Asy/Bx=2540.5/3.100=820mm2/m
Asy1=max(Asy1,ρmin*H*1000)
JC-2锥形基础计算
JC-2计算类型:
验算截面尺寸
基础端部高度h1=300mm
基础根部高度h2=250mm
基础长度B1=1650mmB2=1650mm
基础宽度A1=1650mmA2=1650mm
F=2043.400kN
Mx=15.300kN*m
My=16.500kN*m
Vx=0.000kN
Vy=0.000kN
Fk=F/ks=2043.400/1.35=1513.630kN
Mxk=Mx/ks=15.300/1.35=11.333kN*m
Myk=My/ks=16.500/1.35=12.222kN*m
Vxk=Vx/ks=0.000/1.35=0.000kN
Vyk=Vy/ks=0.000/1.35=0.000kN
1.基础总长Bx=B1+B2=1.650+1.650=3.300m
2.基础总宽By=A1+A2=1.650+1.650=3.300m
3.基础总高H=h1+h2=0.300+0.250=0.550m
4.底板配筋计算高度ho=h1+h2-as=0.300+0.250-0.040=0.510m
5.基础底面积A=Bx*By=3.300*3.300=10.890m2
6.Gk=γ*Bx*By*dh=20.000*3.300*3.300*2.000=435.600kN
G=1.35*Gk=1.35*435.600=588.060kN
Mdxk=Mxk-Vyk*H=11.333-0.000*0.550=11.333kN*m
Mdyk=Myk+Vxk*H=12.222+0.000*0.550=12.222kN*m
Mdx=Mx-Vy*H=15.300-0.000*0.550=15.300kN*m
Mdy=My+Vx*H=16.500+0.000*0.550=16.500kN*m
pk=(Fk+Gk)/A=(1513.630+435.600)/10.890=178.993kPa【①5.2.2-2】
因γo*pk=1.0*178.993=178.993kPa≤fa=180.000kPa
exk=Mdyk/(Fk+Gk)=12.222/(1513.630+435.600)=0.006m
因|exk|≤Bx/6=0.550mx方向小偏心,
=(1513.630+435.600)/10.890+6*|12.222|/(3.3002*3.300)
=181.033kPa
=(1513.630+435.600)/10.890-6*|12.222|/(3.3002*3.300)
=176.952kPa
eyk=Mdxk/(Fk+Gk)=11.333/(1513.630+435.600)=0.006m
因|eyk|≤By/6=0.550my方向小偏心
=(1513.630+435.600)/10.890+6*|11.333|/(3.3002*3.300)
=180.885kPa
=(1513.630+435.600)/10.890-6*|11.333|/(3.3002*3.300)
=177.100kPa
=(181.033-178.993)+(180.885-178.993)+178.993
=182.925kPa
γo*Pkmax=1.0*182.925=182.925kPa≤1.2*fa=1.2*180.000=216.000kPa
ex=Mdy/(F+G)=16.500/(2043.400+588.060)=0.006m
因ex≤Bx/6.0=0.550mx方向小偏心
=(2043.400+588.060)/10.890+6*|16.500|/(3.3002*3.300)
=244.395kPa
=(2043.400+588.060)/10.890-6*|16.500|/(3.3002*3.300)
=238.885kPa
ey=Mdx/(F+G)=15.300/(2043.400+588.060)=0.006m
因ey≤By/6=0.550y方向小偏心
=(2043.400+588.060)/10.890+6*|15.300|/(3.3002*3.300)
=244.195kPa
=(2043.400+588.060)/10.890-6*|15.300|/(3.3002*3.300)
=239.086kPa
=244.395+244.195-(2043.400+588.060)/10.890
=246.949kPa
Pjmax=Pmax-G/A=246.949-588.060/10.890=192.949kPa
Pjmax_x=Pmax_x-G/A=244.395-588.060/10.890=190.395kPa
Pjmax_y=Pmax_y-G/A=244.195-588.060/10.890=190.195kPa
=max((1.650-0.400/2-0.510)*(0.400+2*0.510)+(1.650-0.400/2-0.510)2,(1.650-0.400/2-0.510)*(0.400+2*0.510)+(1.650-0.400/2-0.510)2)
=max(2.218,2.218)
=2.218m2
x冲切截面上的地基净反力设计值Flx=Alx*Pjmax=2.218*192.949=428.039kN
γo*Flx=1.0*428.039=428.04kN
=0.7*1.000*1.43*910*510
=464.56kN
y冲切面积Aly=max((B1-bc/2-ho)*(hc+2*ho)+(B1-bc/2-ho)2,(B2-bc/2-ho)*(