推荐回答(1个)
一. 参数信息
QTZ-315塔吊天然基础的计算书
塔吊型号:QTZ315, 自重(包括压重)F1=250.00kN, 最大起重荷载F2=30.00kN,
塔吊倾覆力距M=315.40kN.m, 塔吊起重高度H=28.00m, 塔身宽度B=1.40m,
混凝土强度等级:C35, 基础埋深D=1.30m, 基础最小厚度h=1.30m,
基础最小宽度Bc=5.00m, 二. 基础最小尺寸计算基础的最小厚度取:H=1.30m
基础的最小宽度取:Bc=5.00m三. 塔吊基础承载力计算 依据《建筑地基基础设计规范》(GB50007-2002)第5.2条承载力计算。
计算简图:
当不考虑附着时的基础设计值计算公式:
当考虑附着时的基础设计值计算公式:
当考虑偏心距较大时的基础设计值计算公式:
式中 F──塔吊作用于基础的竖向力,它包括塔吊自重,压重和最大起重荷载,F=1.2×280=336.00kN;
G──基础自重与基础上面的土的自重,G=1.2×(25.0×Bc×Bc×Hc+20.0×Bc×Bc×D) =1275.00kN;
Bc──基础底面的宽度,取Bc=5.00m;
W──基础底面的抵抗矩,W=Bc×Bc×Bc/6=20.83m3;
M──倾覆力矩,包括风荷载产生的力距和最大起重力距,M=1.4×315.40=441.56kN.m;
a──合力作用点至基础底面最大压力边缘距离(m),按下式计算:
a=5.00/2-441.56/(336.00+1275.00)=2.23m。
经过计算得到:
无附着的最大压力设计值 Pmax=(336.00+1275.00)/5.002+441.56/20.83=85.63kPa
无附着的最小压力设计值 Pmin=(336.00+1275.00)/5.002-441.56/20.83=43.25kPa
有附着的压力设计值 P=(336.00+1275.00)/5.002=64.44kPa
偏心距较大时压力设计值 Pkmax=2×(336.00+1275.00)/(3×5.00×2.23)=96.50kPa四. 地基基础承载力验算 地基基础承载力特征值计算依据《建筑地基基础设计规范》GB 50007-2002第5.2.3条。
计算公式如下:
其中 fa──修正后的地基承载力特征值(kN/m2);
fak──地基承载力特征值,取85.00kN/m2;
b──基础宽度地基承载力修正系数,取0.30;
d──基础埋深地基承载力修正系数,取1.60;
──基础底面以下土的重度,取19.00kN/m3;
γm──基础底面以上土的重度,取19.00kN/m3;
b──基础底面宽度,取5.00m;
d──基础埋深度,取0.50m。
解得地基承载力设计值 fa=96.40kPa
实际计算取的地基承载力设计值为:fa=96.40kPa
地基承载力特征值fa大于最大压力设计值Pmax=85.63kPa,满足要求!
地基承载力特征值1.2×fa大于偏心距较大时的压力设计值Pkmax=96.5kPa,满足要求!五. 受冲切承载力验算
依据《建筑地基基础设计规范》GB 50007-2002第8.2.7条。
验算公式如下:
式中 hp──受冲切承载力截面高度影响系数,取 hp=0.96;
ft──混凝土轴心抗拉强度设计值,取 ft=1.57kPa;
am──冲切破坏锥体最不利一侧计算长度:
am=[1.40+(1.40 +2×1.30)]/2=2.70m;
h0──承台的有效高度,取 h0=1.25m;
Pj──最大压力设计值,取 Pj=96.50kPa;
Fl──实际冲切承载力:
Fl=96.50×(5.00+4.00)×0.50/2=217.12kN。
允许冲切力:
0.7×0.96×1.57×2700×1250=3560760.00N=3560.76kN
实际冲切力不大于允许冲切力设计值,所以能满足要求!
六. 承台配筋计算
依据《建筑地基基础设计规范》GB 50007-2002第8.2.7条。
1.抗弯计算,计算公式如下:
式中 a1──截面I-I至基底边缘的距离,取 a1=1.80m;
P──截面I-I处的基底反力:
P=96.50×(3×1.40-1.80)/(3×1.40)=55.14kPa;
a“──截面I-I在基底的投影长度,取 a“=1.40m。
经过计算得 M=1.802×[(2×5.00+1.40)×(96.50+55.14-2×1275.00/5.002)+(96.50-55.14)×5.00]/12
=208.63kN.m。
2.配筋面积计算,公式如下:
依据《建筑地基基础设计规范》GB 50007-2002第7.2条。
式中 1──系数,当混凝土强度不超过C50时, 1取为1.0,当混凝土强度等级为C80时,
1取为0.94,期间按线性内插法确定;
fc──混凝土抗压强度设计值;
h0──承台的计算高度。
经过计算得 s=208.63×106/(1.00×16.70×5.00×103×12502)=0.002
=1-(1-2×0.002)0.5=0.002
s=1-0.002/2=0.999
As=208.63×106/(0.999×1250×210.00)=795.43mm2。
由于最小配筋率为0.15%,所以最小配筋面积为:9750mm2。
故取 As=9750mm2。
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