| 基础J-1(按《地基基础设计规范(GB5000-21)》进行设计) |
|
回目录 |
|
| 地基承载力特征值fak |
|
210 |
kpa |
|
| 承载力修正系数ηb |
|
0 |
|
| 承载力修正系数ηd |
|
1 |
|
| 基底以下土的重度γ |
|
20 |
kN/M^3 |
|
| 基底以上土的加权平均重度γm |
20 |
kN/M^3 |
|
| 基础埋深d(用于承载力修正) |
2000 |
mm |
|
| 基础根部高度H |
|
800 |
mm |
|
| 基础端部高度h1 |
|
400 |
mm |
|
| 柱宽bc' |
|
600 |
mm |
|
注意啦: |
|
| 柱高hc' |
|
600 |
mm |
|
轴心荷载pk |
通过 |
|
| Y向双柱形心距离cy |
|
1000 |
mm |
|
X向pkmaxX |
通过 |
|
| X向双柱形心距离cx |
|
1000 |
mm |
|
X向pkminX |
>0可以 |
|
| 覆土厚度 ds(用于计算基础自重) |
2000 |
mm |
|
Y向pkmaxY |
通过 |
|
| 永久荷载控制的荷载组合分项系数γz |
1.35 |
|
Y向pkminY |
>0可以 |
|
| 混凝土强度等级 |
|
25 |
|
X方向冲切验算 |
通过 |
|
| 钢筋强度fy |
|
300 |
N/mm^2 |
|
Y方向冲切验算 |
通过 |
|
| 保护层厚度as |
|
80 |
mm |
|
X方向剪切验算 |
通过 |
|
| 柱1竖向力Fk1 |
|
2540 |
kN |
柱2Fk2 |
0 |
kN |
|
Y方向剪切验算 |
通过 |
|
| 柱1基础顶面弯矩Mkx1' |
|
23 |
kN·M |
柱2Mkx2' |
0 |
kN·M |
|
柱下局部受压 |
通过 |
|
| 柱1基础顶面弯矩Mky1' |
|
-62 |
kN·M |
柱2Mky2' |
0 |
kN·M |
|
AsI= |
1336 |
mm^2/M |
|
| 柱1基础顶面剪力Vkx1 |
|
31 |
kN |
柱2Vkx2 |
0 |
kN |
|
AsⅡ= |
1273 |
mm^2/M |
|
| 柱1基础顶面剪力Vky1 |
|
80 |
kN |
柱2Vky2 |
0 |
kN |
|
Φ |
12 |
@85 |
|
| 基础长宽比(L/B) |
|
1 |
X向轴力点=Fk2*cx/(Fk1+Fk2)= |
0 |
mm |
Φ |
12 |
@89 |
|
| h0= |
720 |
mm |
Y向轴力点=Fk2*cy/(Fk1+Fk2)= |
0 |
mm |
|
| (双柱)柱根宽度bc |
1600 |
mm |
X向轴力偏心距ex0= |
|
-500 |
mm |
|
| (双柱)柱根长度hc |
1600 |
mm |
Y向轴力偏心距ey0= |
|
-500 |
mm |
|
| Fk= |
|
2540.00 |
kN |
fc= |
11.9 |
N/mm^2 |
|
| 竖向力F=γz*Fk= |
|
3429.00 |
kN |
ft= |
1.27 |
N/mm^2 |
|
| fa=fak+ηb*γ*(b-3)+ηd*γm*(d-0.5)= |
240 |
kpa |
|
| 轴心受压基底面积=(Fk+Gk)/(fa-γg*ds) |
12.70 |
M^2 |
(注:γg取20.0kN/M^3) |
|
| 计算基础长度b= |
|
3564 |
mm |
取基础长度b= |
5500 |
mm |
|
| 计算基础宽度L= |
|
3564 |
mm |
取基础宽度L= |
5100 |
mm |
|
| Mx=γz*{(Mkx1'+Mkx2')-(Vky1+Vky2)*H+Fk*ey0}= |
-1769.9 |
kN·M |
|
| My=γz*{(Mky1'+Mky2')+(Vkx1+Vky2)*H+Fk*ex0}= |
-1764.7 |
kN·M |
|
| Y 轴方向截面面积 Acb |
|
3.4 |
M^2 |
|
| X 轴方向截面面积 AcL |
|
3.64 |
M^2 |
|
| X轴基础顶面坡度 |
|
11.89 |
° |
|
| Y轴基础顶面坡度 |
|
13.24 |
° |
|
| 基础底面积A |
|
28.05 |
M^2 |
|
| X向Wx=l * b * b / 6 |
|
25.71 |
M^3 |
|
| Y向Wy = b * l * l / 6 |
|
23.84 |
M^3 |
|
| 基础及土自重标准值Gk=γg*A*ds= |
1122.00 |
kN |
|
| 基础及的土重设计值G=γz*Gk= |
1514.7 |
kN |
|
| 轴心荷载作用下pk = (Fk + Gk) / A |
130.55 |
< |
fa= |
240.0 |
kpa |
通过 |
|
| X向pkmaxX=(Fk+Gk)/A+|Mky|/Wx= |
181.39 |
< |
1.2*fa= |
288.0 |
kpa |
通过 |
|
| X向pkminX=(Fk+Gk)/A-|Mky|/Wx= |
79.71 |
> |
|
0.00 |
kpa |
>0可以 |
|
| X向偏心矩ex=Mky/(Fk+Gk)= |
-0.357 |
< |
b/6= |
0.92 |
m |
|
|
| Y向pkmaxY=(Fk+Gk)/A+|Mkx|/Wy= |
185.54 |
< |
1.2*fa= |
288.0 |
kpa |
通过 |
|
| Y向pkminY=(Fk+Gk)/A-|Mkx|/Wy= |
75.57 |
> |
|
0.00 |
kpa |
>0可以 |
|
| Y向偏心矩ey=Mkx/(Fk+Gk)= |
-0.358 |
< |
L/6= |
0.850 |
m |
|
中间结果 |
|
| pmaxX=γz*PkmaxX= |
244.88 |
kpa |
pjmaxX=pmaxX-G/A= |
190.9 |
kpa |
|
Alx=0.5*(L+bc+2*Ho)*(L-bc-2*Ho)/2+L*(b-hc-L+bc)/2= |
5212100 |
| pmaxY=γz*PkmaxY= |
250.48 |
kpa |
pjmaxY=pmaxY-G/A= |
196.5 |
kpa |
|
Alx=L*[0.5*(b-hc)-h0]= |
|
6273000 |
| X方向冲切验算 |
|
Alx=0.5*(b-hc+2*bc+2*Ho)*[(b-hc)/2-Ho]= |
|
5252100 |
|
因b - hc=
|
3900 |
> |
L - bc=
|
3500 |
mm |
|
| b= |
5500 |
> |
hc+2*Ho= |
3040 |
mm |
|
| L= |
5100 |
> |
bc+2*Ho= |
3040 |
mm |
|
Aly=0.5*(b+hc+2*Ho)*(b-hc-2*Ho)/2+b*(L-bc-b+hc)/2= |
4152100 |
| Alx=0.5*(L+bc+2*Ho)*(L-bc-2*Ho)/2+L*(b-hc-L+bc)/2= |
5212100 |
mm^2 |
|
Aly=b*[0.5*(L-bc)-h0]= |
|
5665000 |
| ab = Min{bc + 2 * Ho,l} = |
3040 |
mm |
|
Aly=0.5*(l-bc+2*hc+2*Ho)*[(l-bc)/2-Ho]= |
|
4192100 |
| amx = (bc + ab) / 2 = |
|
2320 |
mm |
|
| 0.7 * βhp * ft * amx * Ho = |
1336.49 |
> |
Flx=pjmaxX*Alx= |
994.88 |
通过 |
|
| Y方向冲切验算 |
|
| Aly=0.5*(l-bc+2*hc+2*Ho)*[(l-bc)/2-Ho]= |
|
4192100 |
mm^2 |
|
| ab = Min{hc + 2 * Ho,b} |
3040 |
mm |
|
| amy = (hc + ab) / 2 |
|
2320 |
mm |
|
| 0.7 * βhp * ft * amY * Ho = |
1336.49 |
> |
Fly=pjmaxY*Aly= |
823.65 |
通过 |
|
| X 方向(b 方向)剪切验算 |
|
| 计算宽度Lo={1.0-0.5*[1.0-(bc+2*50)/L]*(Ho-h1)/Ho}*L= |
|
4344.44 |
mm |
|
| Vx=pj*Ax=pj*(b-hc)*L/2= |
1898.29 |
< |
0.7*βh*ft*Lo*Ho= |
2780.79 |
通过 |
|
| Y 方向(l 方向)剪切验算 |
|
| 计算宽度bo={1.0-0.5*[1.0-(hc+2*50)/b]*(Ho-h1)/Ho}*b= |
|
4685.19 |
mm |
|
| Vy=pj*Ay=pj*(l-bc)*b/2= |
|
1837.21 |
< |
0.7*βh*ft*bo*Ho= |
2998.89 |
通过 |
|
| X 方向(b 方向)柱边(绕 Y 轴)抗弯计算 |
|
| pmaxX=γz*PkmaxX= |
244.88 |
kpa |
|
| pminX=γz*PkminX= |
107.61 |
kpa |
|
| pX=pminX+(pmaxX-pminX)*(b+hc)/b/2= |
196.21 |
kpa |
|
| MIx=(b-hc)^2*[(2*L+bc)*(pmaxX+pX-2*G/A)+(pmaxX-pX)*L]/48= |
|
1324.1 |
kN·M |
|
| MⅡx=(L-bc)^2*(2*b+hc)*(pmaxX+pminX-2*G/A)/48= |
|
786.2 |
kN·M |
|
| Y 方向(l 方向)柱边(绕 X 轴)抗弯计算 |
|
| pmaxY=γz*PkmaxY= |
250.48 |
kpa |
|
| pminY=γz*PkminY= |
102.02 |
kpa |
|
| pY=pminY+(pmaxY-pminY)*(L+bc)/L/2= |
199.53 |
kpa |
|
| MIy=(b-hc)^2*[(2*L+bc)*(pmaxY+pY-2*G/A)+(pmaxY-pY)*L]/48= |
|
1361.1 |
kN·M |
|
| MⅡy=(L-bc)^2*(2*b+hc)*(pmaxY+pminY-2*G/A)/48= |
|
786.2 |
kN·M |
|
| MⅠ = Max{MⅠx,MⅡy} = |
1324.12 |
kN·M |
|
| AsⅠ=MⅠ/0.9*h0*fy*L= |
|
1336 |
mm^2/M |
Φ |
12 |
@85 |
|
| MⅡ = Max{MⅡx,MⅠy} = |
1361.15 |
kN·M |
|
| AsⅡ=MⅡ/0.9*h0*fy*B= |
|
1273 |
mm^2/M |
Φ |
12 |
@89 |
|
| 柱下局部受压承载力计算 |
|
| 混凝土局部受压面积 Al = bc * hc = |
2560000 |
mm^2 |
|
| Ab = (bx + 2 * c) * (by + 2 * c)= |
2890000 |
mm^2 |
|
| βl = Sqr(Ab / Al)= |
1.06 |
|
| 1.35 * βc * βl * fc * Al = |
43696.80 |
> |
F = |
3429.0 |
kN |
通过 |
|
|
|
|
回目录 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|