筏板冲切计算

轴力Fl(kN)22521柱宽bc(mm)2000柱长hc(mm)2000筏板厚h(mm)2000不平衡弯矩筏板有效厚Munb(KN*混凝土抗震系h0(mm)m)ft(N/mm2)中柱数γRE1

190013111C

<[τmax]=

1.3167N/mm2

0.7611N/mm2τmax=Fl/(μm*h0)+αs*Munb*cAB/Is=

1.3167N/mm2[τmax]=0.7(0.4+1.2/βs)βhp*

ok

2中柱结果

15600mmμm=2c1+2c2=

Is=c1h0^3/6+c1^3h0/6+c2h0c1^2/2=

3900mmc1=hc+h0=

3900mmc2=bc+h0=

1950mmcAB=c1/2=

4.5E+12 +1.9E+13 +5.6E+13 =7.96E+13

0.75982 +0.00128 =0.7611N/mm2τmax=Fl/(μm*h0)+αs*Munb*cAB/Is=

1.3167N/mm2[τmax]=0.7(0.4+1.2/βs)βhp*ft=

s11213(c1/c2)0.4

βs=hc/bc=取[h]=1000

1.1βhp=

2

3边柱结果

9800mmμm=2c1+c2=

Is=c1h0^3/6+c1^3h0/6+2h0c1(c1/2-x)^2+c2h0x^2

=3.4E+12 +8.1E+12 +3.9E+12 +5.8E+12 =1.784E+13

c1=hc+h0/2=2950mm

3900mmc2=bc+h0=

2062mmcAB=c1-x=

888.01mmx=c1^2/(2c1+c

1.20951 +0.00556 =1.21506N/mm2τmax=Fl/(μm*h0)+αs*Munb*cAB/Is=

1.3167N/mm2[τmax]=0.7(0.4+1.2/βs)βhp*ft=

s11213(c1/c2)0.36701

βs=hc/bc=取[h]=1000

2

βhp=

1.1

4角柱结果

5900mmμm=c1+c2=

Is=c1h0^3/12+c1^3h0/12+h0c1(c1/2-x)^2+c2h0x^2

=1.7E+12 +4.1E+12 +3E+12 +3E+12 =1.016E+13

c1=hc+h0/2=2950mmc2=bc+h0/2=2950mm

cAB=c1-x=

2213mmx=c1^2/(c1+c2

737.5mmτmax=Fl/(μm*h0)+αs*Munb*cAB/Is=

[τmax]=0.7(0.4+1.2/βs)βhp*ft=

s11123(c1/c2)0.4

βs=hc/bc=2

取[h]=1000

βhp=

1.12.00901 +0.01141N/mm22.02042N/mm2 =1.3167混凝土应力ft(N/mm2)1.71C40E+13mm4

ok

E+13mm4

ok

E+13mm4

not ok