獨(dú)立基礎(chǔ)計(jì)算書例題

1、課程設(shè)計(jì):題一：有一框架結(jié)構(gòu)，4個(gè)框架柱均為800x800,如下圖所示?，F(xiàn)已知1號(hào)柱，軸力F=850kN 彎矩Mx=10kNM My=20kNM 剪力 Vx=30kN,Vy=50kN 2 號(hào)柱，軸力 F=500kN 彎矩 Mx=3kN M My=5kN-M 剪力 Vx=10kN,Vy=20kN 3 號(hào)柱，軸力 F=1100kN 彎矩 Mx=15kN -M My=25kN- M 剪力 Vx=2kN,Vy=3kN 4 號(hào)柱，軸力 F=1800kN 彎矩 Mx=20kN M, My=35kN- M,剪力 Vx=50kN,Vy=10kN 地質(zhì)報(bào)告如 附圖1-1 '所示，基礎(chǔ)持力層為第3層粘土層

2、，地質(zhì)參數(shù)如圖所示。 要求為框架柱設(shè)計(jì)獨(dú)立基礎(chǔ)，并繪制基礎(chǔ)平面圖和剖面配筋圖。提示:1.基礎(chǔ)進(jìn)入第3層500mm 2.基礎(chǔ)混凝土等級(jí)為C25,基礎(chǔ)鋼筋為 2級(jí)鋼；3. 土容重均為18kN/M; 3.由于第4層為軟弱下臥層，需驗(yàn) 算軟弱下臥層承載力，合理取用地基承載力，地基壓力擴(kuò)散角為5度 ,深度修正系數(shù)為1.0。題二：有一框架結(jié)構(gòu)，4個(gè)框架柱均為800x800 ,如下圖所示。現(xiàn)已知1號(hào)柱，軸力F=950kN 彎矩Mx=100kN M, My=200kN- M,剪 力 Vx=2kN,Vy=3kN 2 號(hào)柱，軸力 F=600kN 彎矩 Mx=3kN M, My=5kN-M 剪力 Vx=150kN,

3、Vy=120kN 3 號(hào)柱,軸力 F=2000kN,彎矩 Mx=5kN-M, My=2kN- M 剪力 Vx=210kN,Vy=35kN 4 號(hào)柱，軸力 F=1500kN ,彎矩 Mx=10kN M My=32kN- M,剪力 Vx=52kN,Vy=11kN 地質(zhì)報(bào)告 如附圖2-2 '所示，基礎(chǔ)持力層為第3層粘土層，地質(zhì)參數(shù)如圖所示。要求為框架柱設(shè)計(jì)獨(dú)立基礎(chǔ)，并繪制基礎(chǔ)平面圖和剖面配筋圖。提 示：1.基礎(chǔ)進(jìn)入第3層600mm 2.基礎(chǔ)混凝土等級(jí)為C25,基礎(chǔ)鋼筋為2級(jí)鋼；3. 土容重均為18kN/M; 3.由于第4層為軟弱下臥層，需驗(yàn)算軟弱下臥層承載力，合理取用地基承載力，地基壓力擴(kuò)散

4、角為10度，深度修正系數(shù)為1.1標(biāo)高(m)16.0014.0012.00工程地質(zhì)剖面圖1-1'ZCK1631.9132.00素填土1.50 (30.41)30.00028.00 -粘土26.0024.0022.0020.0018.0015.00 (16.91)8.40 (23.51)00 12.00 (19.91)0 fak=400kPa Es=15.0MPa坡殘積粘土fak=120kPa Eo=5.0MPa粉質(zhì)粘土fak=450kPa Es=25.0MPaLs 石灰?guī)rfa=2500kPa frk=65.2MPa20.00 (11.91)X3Ls工程地質(zhì)剖面圖2-2'標(biāo)高(m)

5、ZCK1631.9132.0030.0028.0026.0024.0022.0014.00-17' 0 粘土fak=400kPa Es=15.0MPa-20.00 (11.91)20.0018.0016.008.40 (23.51)0- -坡殘積粘土0 fak=80kPa Eo=3.0MPa012.00 (19.91)0粉質(zhì)粘土fak=450kPa Es=25.0MPa12.00-15.00 (16.91)1.50 (30.41)素填土 石灰?guī)rfa=2500kPa frk=65.2MPa計(jì)算示例：1軟弱下臥層驗(yàn)算:地面/ / 引/ /* '2 J丿1持力層/ Es1Fk+X X

6、 X X X-Gk地下水位3 軟弱下臥層Es2i.i基本資料i.i.i工程名稱：工程一1.1.2 基礎(chǔ)類型：矩形基礎(chǔ)，底邊寬度b = 2000mm,底邊長度I = 2000mm1.1.3 基礎(chǔ)高度 H = 400mm,基礎(chǔ)混凝土容重丫 c = 25.00kN/m1.1.4相應(yīng)于荷載效應(yīng)標(biāo)準(zhǔn)組合時(shí)，上部結(jié)構(gòu)傳至基礎(chǔ)頂面的軸力值Fk =1300kN1.1.5基礎(chǔ)埋置深度d二2500mm；第一層土（持力層上面）土層厚度di二 2000mm,土的重度 丫 1 = 18kN/m ; 持力層土的重度 丫 2 = 18kN/m1.1.6地基壓力擴(kuò)散線與垂直線的夾角9 = 30°1.1.7軟弱下臥層

7、地基承載力特征值fak = 100kPa,地基承載力修正系數(shù)n d = 0.51.1.8軟弱下臥層頂面以上土的加權(quán)平均重度丫 m = 20kN/m ,軟弱下臥層頂面埋置深度 dz = 8400mm1.2計(jì)算結(jié)果1.2.1軟弱下臥層頂面處經(jīng)深度修正后地基承載力特征值fazfaz = fak + n d * 丫 m * (d - 0.5) = 100+0.5*20*(8.4-0.5) = 179.0kPa 1.2.2 基礎(chǔ)自重、土重標(biāo)準(zhǔn)值 Gk基礎(chǔ)自重 Gk1 = 丫 c * Vc = 25*1.6 = 40.0kN基礎(chǔ)上的土重 Gk2 = 丫 * (I * b * d - Vc) = 18*(1

8、0-1.6) = 151.2kN Gk = Gk1 + Gk2 = 191.2kN1.2.3相應(yīng)于荷載效應(yīng)標(biāo)準(zhǔn)組合時(shí)，軟弱下臥層頂面處附加壓力值pz基礎(chǔ)底面至軟弱下臥層頂面的距離z = dz - d = 5900mmpk = (Fk + Gk) / A = (1300+191.2)/4 = 372.8kPapc = 丫 1 * d1 + 丫 2 * (d - d1) = 36+9 = 45.0kPa對(duì)于矩形基礎(chǔ)的pz值，可按基礎(chǔ)規(guī)范式5.2.7-3簡化計(jì)算：pz = I * b * (pk - pc) / (b + 2z * tan 0) * (I + 2z * tan 0 )=2*2*(37

9、2.8-45)/(8.813*8.813) = 16.9kPa1.2.4軟弱下臥層頂面處土的自重壓力值pcz丫 m' = (18*2+18*6.4)/8.4 = 18.00kN/mpcz = 丫 m' * dz = 18*8.4 = 151.2kPa1.2.5當(dāng)?shù)鼗芰臃秶鷥?nèi)有軟弱下臥層時(shí)，應(yīng)按下式驗(yàn)算：pz + pcz < faz(基礎(chǔ)規(guī)范式 5.2.7-1)pz + pcz = 16.9+151.2 = 168.1kPa < faz = 179.0kPa,滿足要求。2.獨(dú)立基礎(chǔ)計(jì)算:一、示意圖基礎(chǔ)類型：階梯柱基 平面：剖面:計(jì)算形式：驗(yàn)算截面尺寸二、基本參數(shù)1

10、 依據(jù)規(guī)范建筑地基基礎(chǔ)設(shè)計(jì)規(guī)范(GB 50007 2002)混凝土結(jié)構(gòu)設(shè)計(jì)規(guī)范(GB 50010 2002)簡明高層鋼筋混凝土結(jié)構(gòu)設(shè)計(jì)手冊(cè)(第二版)2. 幾何參數(shù)：已知尺寸：B1 =1000 mm,A1=1000 mmH1 =400 mm,H2=0 mmB=800 mm,A =800 mmB3 =1400 mm,A3=1400 mm無偏心：B2 =1000 mm,A2=1000 mm基礎(chǔ)埋深d = 1.50 m鋼筋合力重心到板底距離as = 50 mm3. 荷載值：(1)作用在基礎(chǔ)頂部的標(biāo)準(zhǔn)值荷載Fgk = 2000.00 kNMgxk = 25.00 kN m Mgyk = 60.00 kN

11、 m Vgxk = 100.00 kNVgyk = 20.00 kNFqk = 0.00 kNMqxk = 0.00 kN m Mqyk = 0.00 kN m Vqxk = 0.00 kNVqyk = 0.00 kN作用在基礎(chǔ)底部的彎矩標(biāo)準(zhǔn)值M xk :=M gxk + M qxk=25.00 + 0.00=25.00 kN-mM yk :=M gyk+ M qyk=60.00 + 0.00=60.00 kN-mVxk =:V gxk + V qxk =100.00 + 0.00=100.00 kN mVyk =:V gyk + V qyk =20.00 + 0.00 =20.00 kN m

12、繞 X 軸彎矩：M0xk = Mxk Vyk (H1+ H2) = 25.00 20.00 X 0.40 = 17.00 kN m 繞 Y 軸彎矩：M0yk = Myk + Vxk (H1+ H2) = 60.00 + 100.00X 0.40 = 100.00 kN (3)作用在基礎(chǔ)頂部的基本組合荷載不變荷載分項(xiàng)系數(shù)rg = 1.00活荷載分項(xiàng)系數(shù)rq = 1.00F = rg Fgk + rq Fqk = 2000.00 kNMx = rg Mgxk + rq Mqxk = 25.00 kN m My = rg Mgyk + rq Mqyk = 60.00 kN m Vx = rg Vgx

13、k + rq V qxk = 100.00 kNVy = rg V gyk + rq V qyk = 20.00 kN(4)作用在基礎(chǔ)底部的彎矩設(shè)計(jì)值繞 X 軸彎矩：M0x = Mx Vy (Hi+ H2) = 25.00 20.00X 0.40 = 17.00 kN m繞 Y 軸彎矩：M0y = My+ Vx (H1+ H2) = 60.00 + 100.00X 0.40 = 100.00 kN m4. 材料信息：混凝土： C25鋼筋： HRB335(20M nSi)5. 基礎(chǔ)幾何特性：底面積：S = (A1 + A2)(B 1+ B2) = 2.00 2X0 = 4.00 m2繞 X 軸抵

14、抗矩： Wx = (1/6)(B 1+B2XA1+A2)2 = (1/6) 2.00 X.002 = 1.33 m3繞 Y 軸抵抗矩： Wy = (1/6)(A 1+A2)(B1+B2)2 = (1/6) 2.00 X.002 = 1.33 m3三、計(jì)算過程1. 修正地基承載力修正后的地基承載力特征值fa = 600.00 kPa2軸心荷載作用下地基承載力驗(yàn)算計(jì)算公式：按建筑地基基礎(chǔ)設(shè)計(jì)規(guī)范(GB 50007 2002)下列公式驗(yàn)算：pk = (Fk + Gk)/A(5.2.2 1)Fk = Fgk+ Fqk = 2000.00 + 0.00 = 2000.00 kNGk = 20S d -

15、= 20 4.00 X.50 = 120.00 kNpk = (Fk+ Gk)/S = (2000.00 + 120.00)/4.00 = 530.00 kPa < fa,滿足要求。3. 偏心荷載作用下地基承載力驗(yàn)算計(jì)算公式：按建筑地基基礎(chǔ)設(shè)計(jì)規(guī)范(GB 50007 2002)下列公式驗(yàn)算：當(dāng) ew b/6 時(shí)，pkmax = (Fk+ Gk)/A + M k/W(5.2.2 2)pkmin = (Fk + Gk)/A Mk/W(5.2.2 3)當(dāng) e> b/6 時(shí)，pkmax = 2(Fk + Gk)/3la(5.2.2 4)X、Y方向同時(shí)受彎。偏心距 exk = M 0yk/(

16、Fk+ Gk) = 100.00/(2000.00 + 120.00) = 0.05 me = exk = 0.05 m w (B1+ B2)/6 = 2.00/6 = 0.33 mpkmaxX = (Fk+ Gk)/S+ M 0yk/Wy=(2000.00 + 120.00)/4.00 + 100.00/1.33 = 605.00 kPa偏心距 eyk = M 0xk/(Fk+ Gk) = 17.00/(2000.00 + 120.00) = 0.01 me = eyk = 0.01 m w (A 1 + A2)/6 = 2.00/6 = 0.33 mpkmaxY = (F k+ Gk)/S

17、+ M 0xk/Wx=(2000.00 + 120.00)/4.00 + 17.00/1.33 = 542.75 kPapkmax = pkmaxX + pkmaxY (Fk + Gk)/S = 605.00 + 542.75 530.00 = 617.75 kPaw 1.2X fa = 1.2 X 600.00 = 720.00 kPa，滿足要求。4. 基礎(chǔ)抗沖切驗(yàn)算計(jì)算公式：按建筑地基基礎(chǔ)設(shè)計(jì)規(guī)范(GB 50007 2002)下列公式驗(yàn)算：FlW 0.7 f hp ft am 110(8.2.7 1)Fl :=pj Al(8.2.7 3)am=(at + ab)/2(8.2.7 2)pma

18、x,x =F/S + M 0y/W y =2000.00/4.00 + 100.00/1.33=575.00 kPapmin,x =F/S M 0y/W y =2000.00/4.00 100.00/1.33 :=425.00 kPapmax,y =F/S + M 0x/W x =2000.00/4.00 + 17.00/1.33 =512.75 kPapjmin,y =F/S M0x/Wx =2000.00/4.00 17.00/1.33 =487.25 kPapj = pjmax,x + pjmax,y F/S；=575.00 + 512.75 500.00=587.75 kPa(1)柱對(duì)

19、基礎(chǔ)的沖切驗(yàn)算：Ho = H i + H2 as = 0.40 + 0.00-0.05 = 0.35 mX方向：Aix = 1/4 (A + 2H0+ Ai + A2)(Bi + B2 B 2H0)=(1/4) X (0.80+ 2X 0.35+ 2.00)(2.00 0.80 2X 0.35)=0.44 m2Fix = pj A ix = 587.75 0X4 = 257.14 kNab = minA + 2H°, A1 + A2 = min0.80 + 2 X 0.35, 2.00 = 1.50 mamx = (at + ab)/2 = (A + ab)/2 = (0.80 +

20、1.50)/2 = 1.15 mFlx w 0.7 hp ft amx H0 = 0.7 1X0 X270.00 X150 X.350=357.82 kN，滿足要求。Y方向：Aly = 1/4 (B + 2H0+ B1+ B2)(A 1 + A2 A 2H°)=(1/4) X (0.80+ 2X 0.35+ 2.00)(2.00 0.80 2X 0.35)=0.44 m2Fly = pj Aly = 587.75 0X4 = 257.14 kNab = minB + 2H0, B1 + B2 = min0.80 + 2X 0.35, 2.00 = 1.50 mamy = (at +

21、ab)/2 = (B + ab)/2 = (0.80 + 1.50)/2 = 1.15 mFly w 0.7 hp ft amy H0 = 0.7 1X0 X270.00 X150 X.350=357.82 kN，滿足要求。(2)變階處基礎(chǔ)的沖切驗(yàn)算：5. 基礎(chǔ)受壓驗(yàn)算計(jì)算公式：混凝土結(jié)構(gòu)設(shè)計(jì)規(guī)范(GB 50010 2002)Fl w 1.35c t0c Aln(7.8.1-1)局部荷載設(shè)計(jì)值：Fl = 2000.00 kN24.00 m混凝土局部受壓面積：Aln = Al = B X = 0.80 0.80 = 0.64 m2混凝土受壓時(shí)計(jì)算底面積：Ab = minB + 2A, B 1+

22、B2 Xmin3A, A 1+ A2混凝土受壓時(shí)強(qiáng)度提高系數(shù)：B l = sq.(A b/Al) = sq.(4.00/0.64) = 2.501.35l fc A ln=1.35 X00 X.50 X1900.00 X64=25704.00 kN > Fl = 2000.00 kN，滿足要求。6. 基礎(chǔ)受彎計(jì)算計(jì)算公式：中下列公式驗(yàn)算:(11.4 7)(11.4 8)按簡明高層鋼筋混凝土結(jié)構(gòu)設(shè)計(jì)手冊(cè)(第二版)M I = f /48 (L a)2(2B + b)(pjmax+ pjnx)M n = f /48 (B b)2(2L + a)(pjmax+ pjny)(1) 柱根部受彎計(jì)算：

23、G = 1.35Gk = 1.35 120.00 = 162.00kNI -1截面處彎矩設(shè)計(jì)值：pnx = pjmin,x + (pjmax,x pjmin,x)(B 1 + B2+ B)/2/(B 1 + B2)=425.00 + (575.00 425.00) X (2.00 + 0.80)/2/2.00 =530.00 kPaM 1 = 3 /48 1半昭2 B) 22(A 1 + A2)+ A(p jmax,x + pjnx)=1.0000/48 X (2.00 0.80)2x (2 X 2.00+ 0.80) X (575.00 + 530.00)=159.12 kN m -n -n

24、截面處彎矩設(shè)計(jì)值：pny = pjmin,y + (pjmax,y pjmin,y )(A 1 + A2+ A)/2/(A 1 + A 2)=487.25 + (512.75 487.25) X (2.00 + 0.80)/2/2.00 =505.10 kPaM n = 3 /48 1 -|(AA2 A) 22(B 1 + B2)+ B(p jmax,y + pjny)=1.0000/48 X (2.00 0.80)2X (2 X 2.00+ 0.80) X (512.75 + 505.10)=146.57 kN m -I -I截面受彎計(jì)算：相對(duì)受壓區(qū)高度： Z = 0.056154配筋率：p = 0.002227計(jì)算面積：779.60 mm2/mn -n截面受彎計(jì)算：相對(duì)受壓區(qū)高度： Z = 0.051604配筋率：p = 0.002047計(jì)算面積：716.44 mm2/m(2) 變階處受彎計(jì)算：I -1截面處彎矩設(shè)計(jì)值：pjnx = pjmin,x