地基基础课程设计——参考样式

1、2012级学生课程设计 存档编号： 土力学与地基基础 课程设计系 部： 机建学部 专 业： 建筑工程技术 姓 名： 程辉 学 号： 201203061133 指 导 老 师： 陈震 2014年 6 月 16 日目 录1 课程设计任务书······························

2、83;································11.1 设计资料················

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4、···················11.2 设计要求·····························

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6、83;·····21.3 设计内容···········································

7、··········································31.4 地基基础设计成果······

8、;··················································

9、;·················32 柱下钢筋混凝土独立基础设计······························&

10、#183;······················42.1 设计要求、柱号及相关参数的选择························

11、;·····························42.2 基础材料和类型的选择··················&#

12、183;················································42.3 确定基

13、础的埋置深度·················································

14、·····················42.4 地基承载力验算···························

15、·················································52.4.1

16、确定地基承载力·················································

17、;·························52.4.2初定基础尺寸·······················

18、;··················································

19、;····62.4.3计算作用在基础底部弯矩值···········································

20、;················62.4.4验算地基承载力·······························

21、3;··········································62.5 软弱下卧层验算·····

22、3;·················································

23、3;···················72.6 地基的变形验算····························

24、3;···············································82.7 基础冲切验算·

25、;··················································

26、;····························8计算基础底面反力设计值····················

27、;··········································8 验算柱边冲切······

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29、83;···················92.7.3 验算基础变阶处冲切····························

30、;······································102.8柱下基础的局部受压验算·········&

31、#183;·················································&

32、#183;···112.9 基础配筋计算············································&

33、#183;································112.9.1 各控制截面弯矩计算··············&#

34、183;·················································&#

35、183;·122.9.2 计算配筋··············································&#

36、183;··································122.10 基础配筋图·············&#

37、183;·················································&#

38、183;···············143 设计心得与体会································

39、83;·······································15参考文献·········

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41、83;··························161课程设计任务书 题目：柱下钢筋混凝土独立基础1.1 设计资料（1）地形：拟建建筑场地平整（2）资料：自上而下依次为： 杂填土：厚约0.6m，含部分建筑垃圾； 粉质粘土：厚1.5m，软塑，潮湿，承载力特征值=120 粘土：厚1.3m，可塑，稍湿，承载力特征值=197 全风化砂质

42、泥岩：厚2.3m,承载力特征值=230 淤泥质土：厚0.9m,承载力特征值=75 强风化砂质泥岩：厚3.0m,承载力特征值=305表1.1物理力学参数表地层代号土名天然地基土重度（）孔隙比（e）凝聚力（c）内摩擦角（）压缩系数(a1-2)压缩模量（Es）抗压强度（frk）承载力特征值（fak）KN/m³度杂填土18粉质粘土200.6534130.2010.0130粘土19.40.5825230.228.2210全风化砂质泥岩21223016.10.8240淤泥质土171.31272.24.285强风化砂质泥岩22202518.43.0300（3）水文资料为：地下水对混凝土无侵蚀性。地

43、下水位深度：位于地表下1.5m。（4）上部结构资料：上部结构为四层全现浇框架结构，丙级建筑，室外地坪标高同自然地面，室内外高差450mm，柱网布置见图1.1。图1.1 柱网布置图上部结构作用在柱底的荷载标准值见表1.2。表1.2 柱底荷载标准值轴号Fk (KN)Mk (KNm)Vk (KN)A轴B轴C轴A轴B轴C轴A轴B轴C轴9751548118714010019846484410321615125216412522155605210901730131219015024262665711501815137021017527171736712181843143323519329780837412

44、8218731496257218325869583上部结构作用在柱底的荷载效应基本组合设计值见表1.3。表1.3 柱底的荷载效应基本组合设计值轴号F (KN)M (KNm)V (KN)A轴B轴C轴A轴B轴C轴A轴B轴C轴12682012154418313025860625813422100162721416328872786714182250170624819531581867414962360178227422835393958815842397186330625138610411096166724351945334284423112124108（5）材料：混凝土等级25C30，钢筋HPB2

45、35（级）、HRB335（级）。1.2 设计要求每人设计一个阶梯形柱下独立基础。每人根据自己的学号，完成要求的浅基础设计。学号尾号为单号的选土层为持力层，双号的选土层为持力层。(学号201203061133尾数为3选土层为持力层按学号顺序依次选取轴位置处柱下独立基础设计。（学号201203061133尾数为3选轴位置）1.3 设计内容设计柱下独立基础包括确定基础埋深、基础底面尺寸，对基础进行结构的内力分析、强度计算，确定基础高度、进行配筋计算，并满足构造设计要求，编写设计计算书。绘制基础施工图包括基础平面布置图、基础大样图，并提出必要的技术说明。1.4 地基基础设计成果地基基础设计计算书（1）

46、设计计算书封面封面上应写明设计题目、学生姓名、专业、年级、指导教师姓名、完成日期（2）目录及正文格式目录要求包括章节标题及相应页码。纸张大小：A4复印纸页面设置：页边距：上、下2.54厘米；左、右3.18厘米；行间距：1.5倍行距；字体：宋体；字号：标题用“小三”加粗，正文用“小四”；文档格式：WORD文档。设计图纸设计要求：绘制比例为1:20、1:30或1:50，A3图纸打印或手绘。2 柱下钢筋混凝土独立基础的设计2.1 设计要求、柱号及相关参数的选择（1）根据课程设计任务书的设计要求，选择轴线处柱，基础持力层选择土层，即全风化砂质泥岩。（2）柱底荷载效应基本组合设计值的选择。其中按要求选择

47、F=1706KN、M=315KN·M、V=74KN。 2.2 基础材料和类型的选择 选择混凝土等级为C30，受力钢筋为HRB335级钢筋，基础类型为柱下的钢筋混凝土独立基础：阶梯形基础。 2.3 确定基础的埋置深度（1）持力层为全风化砂质泥岩层，厚2.7m, =240KPa。（2）确定基础的埋置深度：埋置深度是由室内地面标高算起。（3）由于按照以下四个方面的要求：与建筑物有关的条件，工程的地质条件，水文地质条件，地基的冻融条件，场地的环境条件。综合考虑埋置深度d=4m（因轴在轴网中部，埋深从室内地坪算起）。（4）基础的基本简图如图2.1和图2.2所示。 图2.1 基础受力及土层分布图

48、图2.2 基础示意图2.4 地基承载力验算2.4.1确定地基承载力（1）首先对基础进行深度进行修正。则 a=akdm（d0.5) 因为持力层为全风化的砂质泥岩，故 。 =14.3KN/ 故 =390.2KPa = =可取 Bx=2.9m，By=2.9m（基础底面积：A=Bx×By=2.9×2.9=8.410m2）。因为b<3m，故可不对进行宽度的修正。 2.4.2初定基础尺寸基础总长：Bx=b1+b2+b3+b4+bc=0.600+0.600+0.600+0.600+0.500=2.900m基础总宽：By= a1+a2+a3+a4+hc=0.600+0.600+0.6

49、00+0.600+0.500=2.900mA1=a1+a2+hc/2=0.600+0.600+0.500/2=1.450mA2=a3+a4+hc/2=0.600+0.600+0.500/2=1.450mB1=b1+b2+bc/2=0.600+0.600+0.500/2=1.450mB2=b3+b4+bc/2=0.600+0.600+0.500/2=1.450m基础总高：H=h1+h2=0.300+0.300=0.600m底板配筋计算高度：h0=h1+h2-as=0.300+0.300-0.040=0.560m基础底面积：A=Bx×By=2.900×2.900=8.410m2

50、2.4.3计算作用在基础底部弯矩值Gk=×Bx×By×dh=20.000×2.900×2.900×4=672.800KNG=1.35×Gk=1.35×672.800=908.280KNMdx=Mx-Vy×H=284.000+124.000×0.600=358.400KN·m2.4.4验算地基承载力（1）验算轴心荷载作用下地基承载力Pk=(Fk+Gk)/A=(2435.000+672.800)/8.410=369.536KPa因0×Pk =1.0×369.536=36

51、9.536KPafa=390.200KPa故 轴心荷载作用下地基承载力满足要求（2） 验算偏心荷载作用下的地基承载力ey=Mdx/(Fk+Gk)=358.400/(2435.000+672.800)=0.115m因 |ey|By/6=0.483m y方向小偏心Pkmax-y=(Fk+Gk)/A+6×|Mdx|/(By2×Bx)=(2435.000+672.800)/8.410+6×|358.400|/(2.9002×2.900)=457.707KPaPkmin-y=(Fk+Gk)/A-6×|Mdx|/(By2×Bx) =(2435.0

52、00+672.800)/8.410-6×|358.400|/(2.9002×2.900) =281.365KPa>0（3） 确定基础底面反力设计值0×Pkmax-y =1.0×457.707=457.707KPa1.2×fa =1.2×390.200=468.240KPa由以上验算可知，地基承载力满足要求。2.5 软弱下卧层验算（1）软弱下卧层顶面处经深度修正后地基承载力特征值 faz地面至软弱下卧层顶面总深度d=6.350 mfaz = fak +d×m×(d-0.5)=85.000+3.0×13

53、.055×(6.350-0.5)=314.117 KPa（2）计算基础底面处土的自重压力值Pc=i×ti=18×0.95+20×1.2+19.4×1.5+21×(6.350-4.0)=57.050 KPa（3）计算地基压力扩散角上层土压缩模量 Es1=16.100 Mpa 下层土压缩模量 Es2=4.200 MpaEs1/Es2=16.100/4.200=3.833 z/b=2.350/2.900=0.810查基础规范 表5.2.7,地基压力扩散角=24°（4）计算相应于荷载效应标准组合时，软弱下卧层顶面处附加压力值 Pz矩

54、形基础：Pz=l×b×(Pk-Pc)/(b+2×z×tan)×(l+2×z×tan)=2.900×2.900×(369.536-57.050)/(2.900+2×2.350×0.442)×(2.900+2×2.350×0.442)=106.128 KPa（5）计算软弱下卧层顶面处土的自重压力值PczPcz=i×hi=18×0.95+20×1.2+19.4×1.5+21×6.350=82.900 KPa（6）

55、软弱下卧层验算Pz+Pcz=106.128+82.900=189.028< faz =314.117 KPa 软弱下卧层承载力满足要求2.6 地基的变形验算由于，建筑物层数，且为丙级建筑，所以可以不做地基的变形验算。2.7基础冲切验算 计算基础底面反力设计值（1）计算x方向基础底面反力设计值ex=Mdy/(F+G)=0.000/(2435.000+908.280)=0.000m因 ex Bx/6.0=0.483m x方向小偏心Pmax-x=(F+G)/A+6×|Mdy|/(Bx2×By)=(2435.000+908.280)/8.410+6×|0.000|/

56、(2.9002×2.900)=397.536KPaPmin-x =(F+G)/A-6×|Mdy|/(Bx2×By) =(2435.000+908.280)/8.410-6×|0.000|/(2.9002×2.900)=397.536KPa（2）计算y方向基础底面反力设计值ey=Mdx/(F+G)=358.400/(2435.000+908.280)=0.107m因 eyBy/6=0.483 y方向小偏心Pmax-y=(F+G)/A+6×|Mdx|/(By2×Bx) =(2435.000+908.280)/8.410+6

57、15;|358.400|/(2.9002×2.900)=485.707KPaPmin-y=(F+G)/A-6×|Mdx|/(By2×Bx)=(2435.000+908.280)/8.410-6×|358.400|/(2.9002×2.900)=309.365KPa（3）计算地基净反力极值因 Mdx0 并且 Mdy=0Pmax=Pmax-y=485.707KPaPmin=Pmin-y=309.365KPaPjmax=Pmax 验算柱边冲切YH=h1+h2=0.600m, YB=bc=0.500m, YL=hc=0.500mYB1=B1=1.450

58、m, YB2=B2=1.450m, YL1=A1=1.450m, YL2=A2=1.450mYHo=YH-as=0.560m因 (YH800) hp=1.0（1）x方向柱对基础的冲切验算x冲切位置斜截面上边长 bt=YB=0.500mx冲切位置斜截面下边长 bb=YB+2×YHo=1.620mx冲切不利位置 bm=(bt+bb)/2=(0.500+1.620)/2=1.060mx冲切面积 Alx=max(YL1-YL/2-ho)×(YB+2×ho)+(YL1-YL/2-ho)2,(YL2-YL/2-ho)×(YB+2×ho)+ （YL2-YL/

59、2-ho)2 =max(1.450-0.500/2-0.560)×(0.500+2×0.560)+(1.450-0.500/2-0.560)2, (1.450-0.500/2-0.560)×(0.500+2×0.560)+(1.450-0.500/2-0.560)2)=max(1.446,1.446)=1.446m2x冲切截面上的地基净反力设计值 Flx=Alx×Pjmax=1.446×377.707=546.316KNo×Flx=1.0×546.316=546.32KN0.7×hp×ftb&#

60、215;bm×YHo=0.7×1.000×1.43×1060×560=594.19KNx方向柱对基础的冲切满足规范要求（2）y方向柱对基础的冲切验算y冲切位置斜截面上边长 at=YL=0.500my冲切位置斜截面下边长 ab=YL+2×YHo=1.620my冲切面积 Aly=max(YB1-YB/2-ho)×(YL+2×ho)+(YB1-YB/2-ho)2,(YB2-YB/2-ho)×(YL+2×ho)+(YB2-YB/2-ho)2) =max(1.450-0.500/2-0.560)×

61、;(0.500+0.560)+(1.450-0.500/2-0.560)2,(1.450-0.500/2-0.560)×(0.500+0.560)+(1.450-0.500/2-0.560)2)=max(1.446,1.446)=1.446m2y冲切截面上的地基净反力设计值Fly=Aly×Pjmax=1.446×377.707=546.316KNo×Fly=1.0×546.316=546.32KN0.7×hp×ftb×am×YHo=0.7×1.000×1.43×1060

62、15;560=594.19KNy方向柱对基础的冲切满足规范要求验算基础变阶处冲切YH=h2=0.300mYB=bc+b2+b4=1.700mYL=hc+a2+a4=1.700mYB1=B1=1.450m, YB2=B2=1.450m, YL1=A1=1.450m, YL2=A2=1.450mYHo=YH-as=0.300-0.040=0.260m因 (YH800mm) hp=1.0（1）x方向变阶处对基础的冲切验算x冲切位置斜截面上边长 bt=YB=1.700mx冲切位置斜截面下边长 bb=YB+2×YHo=2.220mx冲切不利位置 bm=(bt+bb)/2=(1.700+2.22

63、0)/2=1.960mx冲切面积 Alx=max(YL1-YL/2-ho)×(YB+2×ho)+(YL1-YL/2-ho)2,(YL2-YL/2-ho)×(YB+2×ho)+ (YL2-YL/2-ho)2=max(1.450-1.700/2-0.560)×(1.700+2×0.560)+(1.450-1.700/2-0.560)2,(1.450-1.700/2-0.560)×(1.700+2×0.560)+(1.450-1.700/2-0.560)2)=max(0.114,0.114)=0.114m2x冲切截面上的

64、地基净反力设计值 Flx=Alx×Pjmax=0.114×377.707=43.210KNo×Flx=1.0×43.210=43.21KN0.7×hp×ftb×bm×YHo=0.7×1.000×1.43×1960×260=510.11KNx方向变阶处对基础的冲切满足规范要求（2）y方向变阶处对基础的冲切验算y冲切位置斜截面上边长 at=YL=1.700my冲切位置斜截面下边长 ab=YL+2×YHo=2.220my冲切面积 Aly=max(YB1-YB/2-ho)&

65、#215;(YL+2×ho)+(YB1-YB/2-ho)2,(YB2-YB/2-ho)×(YL+2×ho)+(YB2-YB/2-ho)2) =max(1.450-1.700/2-0.560)×(1.700+0.560)+(1.450-1.700/2-0.560)2, (1.450-1.700/2-0.560)×(1.700+0.560)+(1.450-1.700/2-0.560)2)=max(0.114,0.114)=0.114m2y冲切截面上的地基净反力设计值Fly=Aly×Pjmax =0.114×377.707=43.

66、210KNo×Fly=1.0×43.210=43.21KN0.7×hp×ftb×am×YHo=0.7×1.000×1.43×1960×260=510.11KNy方向变阶处对基础的冲切满足规范要求故：基础各部分的冲切验算均满足要求，所定基础各部分高度合理。2.8柱下基础的局部受压验算因为基础的混凝土强度等级大于等于柱的混凝土强度等级，所以不用验算柱下扩展基础顶面的局部受压承载力。2.9 基础配筋计算2.9.1 各控制截面弯矩计算因Mdx=0 Mdy0 并且 exBx/6=0.483m x方向单向受

67、压且小偏心a=(Bx-bc)/2=(2.900-0.500)/2=1.200mP=(Bx-a)×(Pmax-Pmin)/Bx+Pmin=(2.900-1.200)×(485.707-309.365)/2.900+309.365=412.738KPaMI-1=1/12×a2×(2×By+hc)×(Pmax+P-2×G/A)+(Pmax-P)×By) =1/12×1.2002×(2×2.900+0.500)×(485.707+412.738-2×908.280/8.410)+(485.707-412.738)×2.900)=541.32KNMMII-1=1/48×(By-hc)2×(2×Bx+bc)×(Pmax+Pmin-2×G/A) =1/48×(2.900-0.500)2×(2×2.900+0.500)×(485.707+309.365-2×908.280/8.410)=437.78KNMa=(Bx-bc-b2-b4)/2=(2.900-0.500-0.6