XXX基础计算书

巧家县中寨乡基础计算书

场地基本条件：

1.本工程根据《巧家县中寨乡岩土工程勘察报告》由于○1层土为人工填土选用第○2层土（碎石混合土）作为基础持力层，fak=260 kPa,ηd=1.2，γ1=18，F1=600,F2=800基础选用柱下钢筋混凝土基础。经场地调整，场地标高以拟建建筑场地前方的水泥地面作为高程起算点，假设起算点100.00为0.000。无地下水，基础埋置深度为-2.1m(d=2.1m)。

1．地基承载力修正：

fa=fak+γ1ηd（d-0.5）

=260kpa+18x1.2x(2.1m-0.5m)

=295kpa

2．基础尺寸及地基强度验算：

A=F/(fa-γD)

A1=2.4 所以基础宽度为1.6m

A2=3.2 所以基础宽度为1.8m

3.[计算条件]

1）复核J-1基础抗冲切及配筋：

[计算条件]

独基类型:锥形现浇

独基尺寸(单位mm):

长 宽 高

一阶 1700 1700 300

二阶 550 550 300

基础底标高：-2.1m

基础移心：S方向:0mm B方向:0mm

底板配筋：

Y方向:12@150

X方向:12@150

单位面积的基础及覆土重：20.0kPa

柱截面信息：

柱截面高：450mm

柱截面宽：450mm

柱偏心x：0mm

柱偏心y：0mm

柱转角：0°

荷载信息

竖向荷载基本值： Nk= 600Kn

X方向弯矩基本值：Mx= 50Kn*m

Y方向弯矩基本值：My= 10Kn*m

-1.50000300112@150-2.10030212@1500100157505805505587112@1505001212@150100575550575100850850

[计算结果]

1、冲切验算

采用GB5007-2002建筑地基基础设计规范,公式如下： Fl0.7hpftmh0 m(tb)/2

FlpjAl 8.2.7-1

8.2.7-2

8.2.7-3

冲切力抗力计算:

X+方向,高度 H= 600

Fl = pj*Al = 219.82* 0.12= 26.79

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

X-方向,高度 H= 600

Fl = pj*Al = 196.48* 0.12= 23.95

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

Y+方向,高度 H= 600

Fl = pj*Al = 151.94* 0.12= 18.52

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 =489.31KN

本方向冲切验算满足

Y-方向,高度 H= 600

Fl = pj*Al = 268.67* 0.12= 32.74

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

四边冲切验算

H = 600.

Fl = N-pk*(bc+2*h0)*(hc+2*h0)

= 600.00- 207.6*(450.0+2*550.0)*(450.0+2*550.0)*1e-6

= 101.21Kn

Fr = 0.7*Bhp*ft*am*h0

= 0.7*1.00* 1270.9*(450.0+450.0+2*550.0)*550.0*1e-6

= 1957.25Kn

四边冲切验算满足

X+方向,高度 H= 600mm

Fl = pj*Al = 219.82* 0.04= 9.21

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 = 538.24KN

本方向冲切验算满足

X-方向,高度 H = 600mm

Fl = pj*Al = 195.76* 0.04= 8.20

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 = 538.24KN

本方向冲切验算满足

Y+方向,高度 H= 600mm

Fl = pj*Al = 148.35* 0.04= 6.21

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 =538.24KN

本方向冲切验算满足

Y-方向,高度 H= 600mm

Fl = pj*Al = 268.67* 0.04= 11.25

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 = 538.24KN

本方向冲切验算满足

2、配筋验算

采用GB5007-2002建筑地基基础设计规范,计算公式如下：

MⅠ121(2la')(pjmaxpj)(pjmaxpj)l12

 弯矩计算：

x方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(219824.95+210845.17)+(219824.95-210845.17)*1.70]/12

= 54.47KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(195399.95+204379.73)+(195399.95-204379.73)*1.70]/12

= 49.61KNm

y方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(146549.97+191448.86)+(146549.97-191448.86)*1.70]/12

= 39.88KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(268674.94+223776.05)+(268674.94-223776.05)*1.70]/12

= 64.20KNm

x方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(219824.95+210845.17)+(219824.95-210845.17)*1.70]/12

= 54.47KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(195399.95+204379.73)+(195399.95-204379.73)*1.70]/12

= 49.61KNm

y方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(146549.97+191448.86)+(146549.97-191448.86)*1.70]/12

= 39.88KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.62*0.62[(2*1.70+0.45)*(268674.94+223776.05)+(268674.94-223776.05)*1.70]/12

= 64.20KNm

配筋计算：

M1 = 54.471

AGx = M1/(0.9*h0*fy)

= 54470.891/(0.9*0.540*300.)

=373.600mm*mm

M2 = 64.201

AGy = M2/(0.9*h0*fy)

= 64201.320/(0.9*0.540*300.)

= 440.338mm*mm

M1 = 54.471

AGx = M1/(0.9*h0*fy)

= 54470.891/(0.9*0.540*300.)

= 373.600mm*mm

M2 = 64.201

AGy = M2/(0.9*h0*fy)

= 64201.320/(0.9*0.540*300.)

= 440.338mm*mm

X方向配筋 Y方向配筋

373.600 440.338

原钢筋X方向配筋量满足

原钢筋Y方向配筋量满足

计算的配筋方案为：

AGx:12@150 AGy:12@150

2）复核J-2基础抗冲切及配筋：

[计算条件]

独基类型:锥形现浇

独基尺寸(单位mm):

长 宽 高

一阶 2000 2000 300

二阶 550 550 300

基础底标高：-2.1m

基础移心：S方向:0mm B方向:0mm

底板配筋：

Y方向:12@130

X方向:12@130

单位面积的基础及覆土重：20.0kPa

柱截面信息：

柱截面高：450mm

柱截面宽：450mm

柱偏心x：0mm

柱偏心y：0mm

柱转角：0°

荷载信息

竖向荷载基本值： Nk= 800Kn

X方向弯矩基本值：Mx= 50Kn*m

Y方向弯矩基本值：My= 10Kn*m

-1.50000300112@130-2.10030212@1300100152070010550050112@130217001212@13010072555072510010001000

[计算结果]

1、冲切验算

采用GB5007-2002建筑地基基础设计规范,公式如下： Fl0.7hpftmh0 m(tb)/2

FlpjAl 8.2.7-1

8.2.7-2

8.2.7-3

冲切力抗力计算:

X+方向,高度 H= 600

Fl = pj*Al = 207.50* 0.40= 82.87

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

X-方向,高度 H= 600

Fl = pj*Al = 194.19* 0.40= 77.55

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

Y+方向,高度 H= 600

Fl = pj*Al = 170.94* 0.40= 68.27

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 = 489.31KN

本方向冲切验算满足

Y-方向,高度 H= 600

Fl = pj*Al = 237.50* 0.40= 94.85

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.45+1.55)*0.55/2 =489.31KN

本方向冲切验算满足

四边冲切验算

H = 600.

Fl = N-pk*(bc+2*h0)*(hc+2*h0)

= 800.00- 200.0*(450.0+2*550.0)*(450.0+2*550.0)*1e-6

= 319.50Kn

Fr = 0.7*Bhp*ft*am*h0

= 0.7*1.00* 1270.9*(450.0+450.0+2*550.0)*550.0*1e-6

= 1957.25Kn

四边冲切验算满足

X+方向,高度 H= 600mm

Fl = pj*Al = 207.50* 0.32= 66.27

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 =538.24KN

本方向冲切验算满足

X-方向,高度 H = 600mm

Fl = pj*Al = 193.81* 0.32= 61.90

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 = 538.24KN

本方向冲切验算满足

Y+方向,高度 H= 600mm

Fl = pj*Al = 169.06* 0.32= 53.99

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 =538.24KN

本方向冲切验算满足

Y-方向,高度 H= 600mm

Fl = pj*Al = 237.50* 0.32= 75.85

0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1270.94*(0.55+1.65)*0.55/2 =538.24KN

本方向冲切验算满足

2、配筋验算

采用GB5007-2002建筑地基基础设计规范,计算公式如下：

121(2la')(pjmaxpj)(pjmaxpj)l12

MⅠ 弯矩计算：

x方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(207500.00+201687.50)+(207500.00-201687.50)*2.00]/12

= 91.72KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(192500.00+198312.50)+(192500.00-198312.50)*2.00]/12

= 86.46KNm

y方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(162500.00+191562.50)+(162500.00-191562.50)*2.00]/12

= 75.95KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(237500.00+208437.50)+(237500.00-208437.50)*2.00]/12

= 102.23KNm

x方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(207500.00+201687.50)+(207500.00-201687.50)*2.00]/12

= 91.72KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(192500.00+198312.50)+(192500.00-198312.50)*2.00]/12

= 86.46KNm

y方向,h0 = 540mm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(162500.00+191562.50)+(162500.00-191562.50)*2.00]/12

= 75.95KNm

M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]

=

0.78*0.78[(2*2.00+0.45)*(237500.00+208437.50)+(237500.00-208437.50)*2.00]/12

= 102.23KNm

配筋计算：

M1 = 91.721

AGx = M1/(0.9*h0*fy)

= 91720.922/(0.9*0.540*300.)

= 629.087mm*mm

M2 = 102.234

AGy = M2/(0.9*h0*fy)

=102233.742/(0.9*0.540*300.)

= 701.192mm*mm

M1 = 91.721

AGx = M1/(0.9*h0*fy)

= 91720.922/(0.9*0.540*300.)

= 629.087mm*mm

M2 = 102.234

AGy = M2/(0.9*h0*fy)

= 102233.742/(0.9*0.540*300.)

= 701.192mm*mm

X方向配筋 Y方向配筋

629.087 701.192

原钢筋X方向配筋量满足

原钢筋Y方向配筋量满足

计算的配筋方案为：

AGx:12@130 AGy:12@130