土木在线论坛 \ 建筑结构 \ PKPM \ 请问JCCAD中工具箱的计算结果怎么看,短柱的配筋怎么看

请问JCCAD中工具箱的计算结果怎么看,短柱的配筋怎么看

发布于:2010-03-21 10:58:21 来自:建筑结构/PKPM [复制转发]
主要是基础短柱的配筋怎么看. 希望高手指点一下。谢谢

*************************************************************************************************
计算项目: 独立基础-5
*************************************************************************************************
[计算条件]
独基类型:阶形现浇
独基尺寸(单位mm):
长 宽 高
一阶 1500 1500 400
二阶 1000 1000 400
三阶 500 500 700
基础底标高:-1.5m
基础移心:S方向:0mm B方向:0mm
底板配筋:
Y方向:10@200
X方向:10@200

单位面积的基础及覆土重:18.0kPa
柱截面信息:
柱截面高:500mm
柱截面宽:500mm
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= 200Kn
X方向弯矩基本值:Mx= 0Kn*m
Y方向弯矩基本值:My= 0Kn*m




[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:
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= 1500
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*0.94*1101.28*(0.50+1.50)*1.45/2 = 1052.60KN
◎◎◎本方向冲切验算满足◎◎◎
X-方向,高度 H= 1500
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*0.94*1101.28*(0.50+1.50)*1.45/2 = 1052.60KN
◎◎◎本方向冲切验算满足◎◎◎
Y+方向,高度 H= 1500
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*0.94*1101.28*(0.50+1.50)*1.45/2 = 1052.60KN
◎◎◎本方向冲切验算满足◎◎◎
Y-方向,高度 H= 1500
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*0.94*1101.28*(0.50+1.50)*1.45/2 = 1052.60KN
◎◎◎本方向冲切验算满足◎◎◎

四边冲切验算
H = 1500.
Fl = N-pk*(bc+2*h0)*(hc+2*h0)
= 200.00- 88.9*(500.0+2******)*(500.0+2******)*1e-6
= -827.56Kn
Fr = 0.7*Bhp*ft*am*h0
= 0.7*0.94* 1101.3*(500.0+500.0+2******)*******1e-6
= 8210.24Kn
◎◎◎四边冲切验算满足◎◎◎

X+方向,高度 H= 800mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(0.50+1.50)*0.75/2 = 578.17KN
◎◎◎本方向冲切验算满足◎◎◎
X-方向,高度 H = 800mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(0.50+1.50)*0.75/2 = 578.17KN
◎◎◎本方向冲切验算满足◎◎◎
Y+方向,高度 H= 800mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(0.50+1.50)*0.75/2 = 578.17KN
◎◎◎本方向冲切验算满足◎◎◎
Y-方向,高度 H= 800mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(0.50+1.50)*0.75/2 = 578.17KN
◎◎◎本方向冲切验算满足◎◎◎

X+方向,高度 H= 400mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(1.00+1.50)*0.35/2 = 337.27KN
◎◎◎本方向冲切验算满足◎◎◎
X-方向,高度 H = 400mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(1.00+1.50)*0.35/2 = 337.27KN
◎◎◎本方向冲切验算满足◎◎◎
Y+方向,高度 H= 400mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(1.00+1.50)*0.35/2 = 337.27KN
◎◎◎本方向冲切验算满足◎◎◎
Y-方向,高度 H= 400mm
Fl = pj*Al = 88.89* 0.00= 0.00
0.7*βhp*ft*(at+ab)*ho/2 = 0.7*1.00*1101.28*(1.00+1.50)*0.35/2 = 337.27KN
◎◎◎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:

弯矩计算:
x方向,h0 = 340mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.25*0.25[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 3.24KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.25*0.25[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 3.24KNm

y方向,h0 = 340mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.25*0.25[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 3.24KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.25*0.25[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 3.24KNm

x方向,h0 = 740mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm

y方向,h0 = 740mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm

x方向,h0 = 1440mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm

y方向,h0 = 1440mm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm
M = 1/12*a1*a1*[(2l+a`)*1(Pjmax+Pj)+(Pjmax-Pj)*l]
= 0.50*0.50[(2*1.50+0.50)*( 88888.89+ 88888.89)+( 88888.89- 88888.89)*1.50]/12
= 12.96KNm

配筋计算:
M1 = 3.241
AGx = M1/(0.9*h0*fy) = 3240.741/(0.9*0.340*210.) = 50.432mm*mm
M2 = 3.241
AGy = M2/(0.9*h0*fy) = 3240.741/(0.9*0.340*210.) = 50.432mm*mm

M1 = 12.963
AGx = M1/(0.9*h0*fy) = 12962.964/(0.9*0.740*210.) = 92.685mm*mm
M2 = 12.963
AGy = M2/(0.9*h0*fy) = 12962.964/(0.9*0.740*210.) = 92.685mm*mm

M1 = 12.963
AGx = M1/(0.9*h0*fy) = 12962.964/(0.9*1.440*210.) = 47.630mm*mm
M2 = 12.963
AGy = M2/(0.9*h0*fy) = 12962.964/(0.9*1.440*210.) = 47.630mm*mm

X方向配筋 Y方向配筋
92.685 92.685

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:10@200 AGy:10@200

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全部回复(4 )

只看楼主 我来说两句
  • 结构抛物线
    都是这样的。、
    2010-12-13 13:34:13

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  • 一个海贼
    一个海贼 板凳
    没人讲的更详细点吗,我来学习的。
    2010-12-13 08:52:13

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