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1、筏板冲切抗剪计算项目名称 构件编号 日期设计者校对者一、设计资料1. 筏板信息筏板类型：平板式筏基筏板厚度:d=600 mm混凝土强度等级：C55筏板计算用保护层厚度破=30 mm2. 地基信息w w1地基土平均净反力设计值：Fl = 300.00 KPa3. 柱信息柱截面高：h=900 mm柱截面宽：w=500 mm柱x方向弯矩：Mx=0.00 Kn*m柱y方向弯矩： 的=800.00 Kn*m柱轴力设计值：N=1000.00 Kn柱冲切类型：边柱柱边类型：上边4. 柱墩信息柱墩截面宽：w1= 1300 mm柱墩截面高：h1= 1500 mm柱墩高：d1= 100 mm5. 设计时执行的规范
2、：建筑地基基础设计规范(GB 50007-2002)以下简称基础规范二、计算结果1. Y方向筏板抗剪切验算Y方向H = 700h0 = 670 mma计算冲且临界截面周长及极惯性矩根据基础规范附录P.0.2,采用以下公式计算c1 = hc + h0/2 = 900+670/2 = 1235.0mmc2 = bc + h0 = 500+670 = 1170.0mmum = 2c1 + c2 = 2X 1235.0+ 1170.0 = 3640.0mmx = c12/(2c1+c2) = 1235.02/(2 X 1235.0 + X 1170.0) = 419.0mm cAB = c1 - x
3、= 1235.0 - 419.0 = 816.0mmI = ch 3/6 + c 3h /6 + 2hc (c /2 - X)2 + chX2 s 1 01 00 1 1/2 0=1235 X 6703/6 + 12353 X 670/6 + 2 X 670X1235X(1235/2 - 419)2 + 1170 X 670X 4192=475077345280mm4以s = 1-2= 1-2 .= 0.41s 1 + m(C/c2)1 + 01.24/1.17)b计算作用在冲且临界截面重心上的弯矩P = areaXFl = 1.445 X 300.000 = 433.485ep = 0.25
4、2eN = 0.419Munb = Ne.Pe -Mc = 300.000X 0.419-433.485 X 0.252-0.000 = 16.68Kn*mc筏板冲切验算P C基础规范8.4.7条，采用以下公式计算Tmax = Fl/Umh0 + asMUnbcAB/Is=566.52/3.64X0.67 + 0.41X16.68X0.82/0.48=243.9 kn*m2Ps = 2.00Vh 不大于 800mm,B hp= 1.0f = 1960.00 KPa t0.7(0.4+1.2/p s)p hpf t = 0.7(0.4+1.2/2.00)1.00X 1960.00 = 1372.
5、00.SmaxWW+Sp s)P hpf t.筏板抗冲切验算满足2. X方向筏板抗剪切验算X方向H = 700a计算冲且临界截面周长及极惯性矩根据基础规范附录P.0.2,采用以下公式计算c1 = hc + h0/2 = 500+670/2 = 1235.0mmc2 = bc + h0 = 900+670 = 1170.0mmum = 2c1 + c2 = 2X 1235.0+ 1170.0 = 3640.0mmx = c12/(2c1+c2) = 1170.02/(2 X 1170.0 + X 1235.0) = 382.9mmcAB = c1 - x = 1170.0 - 382.9 = 7
6、87.1mmI = ch 3/6 + c 3hJ6 + 2hc (c. /2 - X)2 + chX2 s 1 01 00 1 12 0=1170X6703/6 + 11703 X 670/6 + 2X670X1170X(1170/2 - 383)2 + 1235 X 670 X 3832=422845874176mm4以s = 1-2= 1-2 .= 0.39s 1 + |1(c1/c2)1 + 3(1.17/1.24)b计算作用在冲且临界截面重心上的弯矩P = areaX F = 1.445 X 300.000 = 433.485eP = 0eN = 0Munb = NeN-Pep-Mc
7、= 300.000 X 0.000-433.485 X 0.000-800.000 = -800.00Kn*mC筏板冲切验算基础规范847条，采用以下公式计算t = Flu h+OLM c !I max 1 m 0 s unb AB s=566.52/3.64X0.67 + 0.39X800.00X0.79/0.42=818.3 kn*m2P =2.00 sh 不大于 800mm, p = 1.0hpf = 1960.00 KPa0.7(0.4+1.2/ps)Phi/t = 0.7(0.4+1.2/2.00)1.00X 1960.00 = 1372.00Vt 0.7(0.4+1.2/p )p fmaxs hp t. 筏板抗冲切验算满足
