萬科第五園一期建筑施工圖基礎(chǔ)j04a

1、錐型柱基計算書項(xiàng)目名稱_J4A_日 期_設(shè) 計 者_(dá)校 對 者_(dá)一、示意圖基礎(chǔ)類型：錐型柱基計算形式：驗(yàn)算截面尺寸平面:剖面:二、基本參數(shù)1依據(jù)規(guī)范建筑地基基礎(chǔ)設(shè)計規(guī)范（GB 50007-2002）混凝土結(jié)構(gòu)設(shè)計規(guī)范（GB 50010-2002）2幾何參數(shù)：已知尺寸：B1 = 1550(mm), A1 = 2400(mm)H1 = 300(mm), H2 = 300(mm)B = 200(mm), A = 1850(mm)無偏心：B2 = 1550(mm), A2 = 2400(mm)基礎(chǔ)埋深d = 2000(mm), 鋼筋合力重心到板底距離 = 50(mm), 3荷載值：(1)作用在基礎(chǔ)頂部

2、的基本組合荷載豎向荷載F = 2598.00kNMx = 0.00kN*mMy = 4.00kN*mVx = 11.00kNVy = 2.00kN(2)作用在基礎(chǔ)底部的彎矩設(shè)計值繞X軸彎矩: M0 x = Mx - Vy * d = 0.00 - (2.00)*1.00 = -2.00kN*m繞y軸彎矩: M0y = My + Vx * d = 4.00 + (11.00)*1.00 = 15.00kN*m4材料信息：混凝土： C25鋼筋： HRB335(20MnSi)三、計算過程1修正地基承載力修正后的地基承載力特征值 fa = 160.00 kPa2軸心荷載作用下地基承載力驗(yàn)算計算公式：按

3、建筑地基基礎(chǔ)設(shè)計規(guī)范（GB 50007-2002）下列公式驗(yàn)算：pk = (FkGk)/A (式5.2.41)其中：A = 4.80 * 3.10 = 14.88Fk = F/Ks = 2598.00/1.25 = 2078.40kN Gk = A*d*20 = 14.88*1.00*20 = 297.60kN pk = (FkGk)/A = ( 2078.40 297.60 ) / 14.88 = 159.68 kPa fa, 滿足要求3偏心荷載作用下地基承載力驗(yàn)算計算公式：按建筑地基基礎(chǔ)設(shè)計規(guī)范（GB 50007-2002）下列公式驗(yàn)算：當(dāng)eb/6時,pkmax=3(FkGk)/3la (

4、式5.2.24)X、Y方向同時受彎:Mxk = M0 x/Ks = 2.00/1.25 = 1.60kN*mMyk = M0y/Ks = 15.00/1.25 = 12.00kN*mFk = F/Ks = 2598.00/1.25 = 2078.40kNGk = G/1.35 = 401.76/1.35 = 297.60kN pkmax = 161.37 kPa 1.2 * fa = 1.2 * 160.00 = 192.00 kPa, 滿足要求4基礎(chǔ)抗沖切驗(yàn)算計算公式：按建筑地基基礎(chǔ)設(shè)計規(guī)范（GB 50007-2002）下列公式驗(yàn)算：Fl 0.7 * hp * ft * am * h0 (8

5、.2.71) Fl = pj * Al (8.2.72) am = (at ab)/2 (8.2.73) (1)柱子對基礎(chǔ)的沖切驗(yàn)算：X方向(a方向)：pj = pjmax = 176.72 kPaAlx = 3.47 mFlx = pj * Ax= 176.72 * 3.47= 612.32 kNab = Minbc2*ho，b= Min 1.85 2 * 0.55 ，4.80= 2.95 mamx = (atab)/2= (1.85 2.95)/2 = 2.40 mFlx 0.7 * hp * ft * amx * ho = 0.7 * 1.00 * 1270.00 * 2.400 * 0

6、.550= 1173.48 kN, 滿足要求Y方向(b方向)：pj = pjmax = 176.72 kPaAly = 2.06 mFly = pj * Ay= 176.72 * 2.06= 363.59 kNab = Minac2*ho，a= Min 0.20 2 * 0.55 ，3.10= 1.30 mamy = (atab)/2= (0.20 1.30)/2 = 0.75 mFly 0.7 * hp * ft * amy * ho = 0.7 * 1.00 * 1270.00 * 0.750 * 0.550= 366.71 kN, 滿足要求5基礎(chǔ)受壓驗(yàn)算計算公式：混凝土結(jié)構(gòu)設(shè)計規(guī)范（7.

7、8.1-1）Fl 1.35*c*l* fc * Aln局部荷載設(shè)計值：Fl = 2598.00 kN混凝土局部受壓面積：Al = bc * hc = 0.20 * 1.85 = 0.37m混凝土受壓時計算底面積：Ab = 2.88m混凝土受壓時強(qiáng)度提高系數(shù)：= sqr(Ab/Al) = sqr(2.88/0.37) = 2.79 1.35 * c * * fc * Al = 1.35 * 1.00 * 2.79 * 11900.00 * 0.37 = 16583.56 kN Fl = 2598.00 kN 滿足要求！6基礎(chǔ)受彎計算計算公式：按建筑地基基礎(chǔ)設(shè)計規(guī)范(GB 50007-2002)下

8、列公式驗(yàn)算：M=a1*2*(2*l+a)*(pmax+p-2*G/A+(pmax-p)*l)/12 (8.2.74)M=(la)*2 * (2*bb)*(pmax+pmin-2*G/A)/48 (8.2.75)(1)柱根部受彎計算：X方向受彎截面基底反力設(shè)計值：pnx=pminx(pmaxxpminx)*(l/2.0lo/2.0)/l=199.65(203.55199.65)*(3.100.20)/3.10/2=201.47 kPaY方向受彎截面基底反力設(shè)計值：pny=pminy(pmaxypminy)*(b/2.0bo/2.0)/b=201.43(201.76201.43)*(4.801.8

9、5)/4.80/2=201.53 kPa-截面處彎矩設(shè)計值：M= (l/2-l0/2)*(l/2-l0/2)*(2*b+bo)*(pmaxx+pnx-2*G/A)+(pmaxx-pnx)*b)/12= (3.10/2-0.20/2)*(3.10/2-0.20/2)*(2*4.80+1.85) *(203.55+201.47-2*401.76/14.88)+(203.55-201.47)*4.80)/12= 705.94 kN.m-截面處彎矩設(shè)計值：M= (b/2-bo/2)*(b/2-bo/2)*(2*l+lo)*(pmaxy+pny-2*G/A)+(pmaxy-pny)*l)/12= (4.80/2-1.85/2)*(4.80/2-1.85/2)*(2*3.10+0.20) *(201.76+201.53-2*401.76/14.88)+(201.76-201.53)*3.10)/12= 405.43 kN.m-截面受彎計算：相對受壓區(qū)高度： = 0.041726 配筋率： = 0.001517計算面積：4369.57 mm-截面受彎計算：相對受壓區(qū)高度： = 0.037017 配筋率： = 0.001346 min = 0.0015