哈尔滨工业大学硕士毕业论文模板

《哈尔滨工业大学硕士毕业论文模板》由会员分享，可在线阅读，更多相关《哈尔滨工业大学硕士毕业论文模板（68页珍藏版）》请在装配图网上搜索。

1、 Midterm reportCourse: Advanced Foundation EngineeringTitle：Transcona Grain Elevator, Canada Member：Duofa Ji、Xici Han、Shiwang Zhao、Zhimin Zhang不要删除行尾的分节符，此行不会被打印- I -目录- 67 -参考文献 CONTENTSAbstract (Chinese)IAbstract (English)IIIChapter 1 Introduction1 1.1 Introduction of castellated and tapered membe

2、rs11.1.1 Introduction to H-section steel11.1.2 Summary of present research on lightweight portal frame31.1.3 Summary of present research on tapered members51.1.4 Summary of present research on castellated beams7 1.2 Present research on cyclic plasticity model9 1.3 Origin and significance of the rese

3、arch project11 1.4 Main contents of the dissertation11Chapter 2 Static Analysis of cantilever tapered castellated members13 2.1 Introduction13 2.2 Solution to finite element nonlinear equations13 2.3 Von-Mises yield criterion13 2.4 Selection of the Elment14 2.5 Effective length factors of cantilever

4、 tapered members under axial loads15 2.6 Stability analysis of axial compression cantilever tapered castellated members17 2.6.1 Distribution pattern of residual stress and member design202.6.2 Ultimate strength analysis of axial compression cantilever cantilever tapered castellated members232.6.3 In

5、fluence of steel strength312.6.4 Limitation of opening rate332.6.5 Stability design of tapered castellated members35 2.7 Flexural strength analysis of cantilever tapered castellated members37 Ultimate flexural strength and stiffness372.7.2 Stiffness analysis452.7.3 Stress and strain analysis of plas

6、tic zone48 2.8 Planar stability analysis of cantilever tapered castellated beam-columns552.8.1 Relative equations of beam-columns552.8.2 Analysis of cantilever tapered castellated beam-columns572.8.3 Plastic zone analysis63 2.9 Summary63Chapter 3 Hysteresis behavior of cantilever tapered castellated

7、 members under cyclic loading65 3.1 Introduction65 3.2 Steel stress-strain model under cyclic loading67 3.3 Hysteresis capability of tapered castellated beam-columns under cyclic loading693.3.1 Dimension of tapered castellated beam-columns693.3.2 Hysteresis behavior of the members71 3.4 Summary79Cha

8、pter 4 Experimental research for cantilever tapered castellated members under static and cyclic loading80 4.1 Introduction80 4.2 Testing process design80 4.3 Design of testing members814.3.1 Design of members814.3.2 Joins and load814.3.3 Locations and installation of measure instrument844.3.4 Materi

9、al properties85 4.4 Analysis of testing results854.4.1 Cantilever tapered castellated members under static load86 4.4.2 Cantilever tapered castellated members under cyclic load894.5 Summary91Chapter 5 Simplification of cantilever tapered castellated beam-columns under static load92 5.1 Introduction9

10、2 5.2 Elastic analysis of cantilever tapered castellated beam-columns92 5.2.1 Establishment of equilibrium differentiale quations935.2.2 Establishment of stiffness equation96 5.3 Influence of secondary moments985.3.1 Conception of secondary moments985.3.2 Deflection caused by secondary moments99 5.4

11、 Elastic plastic analysis of cantilever tapered castellated beam-columns102 5.4.1 Plastic hing model1025.4.2 The load-defection curves1045.4.3 Influence of tangent modulu and initial imperfection105 5.5 Result comparison of numerical and finite element method106 5.6 Summary108Conclusion109References

12、 111Papers published in the period of Ph.D. education119Statement of copyright120Acknowledgement121Resume122 Chapter 1 introduction1.1 Case DescriptionIn September 1913, the Canadian Pacific Railway Company completed construction of a million-bushel (about 36,400 m3) grain elevator at North Transcon

13、a, 11 km north-east of Winnipeg, Canada. The elevator was one of the most important structures and one of the largest gravity railroad yards in the world, which covered several square miles and was built on partly farmed, relatively flat prairie land. The structure consisted of a reinforced-concrete

14、 work-house, which contained five rows of 13 bins. The bins were based on a concrete structure containing belt conveyors supported by a reinforced-concrete shallow raft foundation. After the structure was completed, the filling was begun and grain was distributed uniformly between the bins. The weig

15、ht of grains is 20000 tons, which is equivalent to 42.5% of the total weight when the grains are full. On October 18, 1913, after the elevator was loaded to 87.5% of its capacity, settlement of the bin-house was noted. Within an hour, the settlement had increased uniformly to about 30 cm following b

16、y a tilt towards the west (Fig. 3.1a,b), which continued for almost 24 hours until it reached an inclination of almost 27 degrees.1.2 ConstructionExcavation for the elevator foundations started in 1911. No borings were taken, but after the excavation reached its design depth of 3.7 m, field-bearing

17、capacity tests were carried out by loading a plate laid upon a prepared smooth clay surface. A 60 cm thick 23.5 59.5 m reinforced-concrete slab was built to serve as a foundation for the concrete framework of the underground conveyor belt tunnels supporting the bin-house. The construction of the bin

18、-house proceeded at a rapid rate during the autumn and winter of 1912, the concrete circular bins being raised at the rate of 1 m per day, until they reached the designed height of 31 m.Chapter2 Failure analysis2.1 Engineering parameters2.1.1 LoadDead：weight of the structure was 20,000 tonsLive: the

19、 weight of the grain in the bins (full) 27060 tons The weight of the grain in the bins (failure) 23677.5 tons2.1.2 Soil parameters1）Judgement of the category of the soilThrough refering to literatures, we know that the plastic index for the soil is, through checking literatures we can make classific

20、ation of soil, when plastic index, the soil is clay, and when, he soil is silt, so we can classify the soil layer as clay.2）After analying the related literatures, we can get some parameters of the soil, and as is shown in the table 2-1 Table 2-1numberH(m)name(%)(%)(%)12.2artificial fills-27.5gray s

21、lickensided clay054704510518.73534.9gray silty clay031705710518.73543.0limestone rock-Through the above parameters and by formulawe can get =0.1429 for second layer; and =0.3143 for the third layer. Reference:” Liu Hui, discuss of the relationship of SPT counts of the standard foundation penetration

22、 and the physical mechanical parameters” And we can get properties of some parameters of clay. The relations of SPT counts of the standard foundation penetration and the physical mechanical parameters. Table2-2 Foundation soil categoryregression equationclay From the formulas in table. we can get th

23、e SPT counts of the second soil layer:N=12.47, so we adopt N=13modulus of compression of the soil:=8.73 ，compressibility coefficient：=0.21SPT of the third layer of soil N=9.32, N=10 modulus of compression of soil =7.30，compressibility coefficient=0.28From the reference of litterature, as to clay, er

24、ror should be within 9%, results of calculation is close to the real value, this experience regression equation has applicability and reliability, so this project uses this experience formula to get the parameters is reasonable, and can meet the requirements of the error range.。 stucture dimension p

25、arameter Geometry - Width of the rectangular footing: B= 23.5 m;- Length of the rectangular footing: L= 59.5 m;- Depth of the foundation: D =3.7 m.Soil profile - Thickness of the upper clay layer below the footing D =6.0 m; - Undrained shear strength of the upper layer Cu1 =qu1/2=54kPa; - Undrained

26、shear strength of the lower layer Cu2 =qu2/2=31kPa;- Total unit weight of clay in both layers 18.7 kN/m3Specific shape of the barn, refer to figure 2-3 and figure 2-4 Figure 2-3Figure 2-42.2 calculation hypothesis1.）Because the project overtured in 24 hours time, and is almost instantaneous, so, unc

27、onfined compressive strength is considered when computing, that is , and the result is conservative.2.） In this design, it was assumed that the soil profile was homogeneous with the properties of the layer.2.2 foundation bottom pressure calculationThis project adopts the serviceability limit state t

28、o calculate loads, using the standard load combination is mainiy to consider the short-term effect. Foundation bottom pressure can be got from the following formula, Assumptions are based on the axis load is applied on the bottom of foundation. 1）When the barn is full, the overall vertical load is:F

29、 = (20000+27060) 9.8 1000=461188KN Stress on bottom of foundation. 2）When the barn is dumping, the vertical load is: = (20000+23677.5) 9.8 1000=428039.5kN the stress of foundation bottom： 3) determination of characteristic value of foundation bearing capacityWhen the ccentricity, according to shear

30、strength parameters of the soil to determin the foundation characteristic value of subgrade bearing capacity, and to canculate it according to the following formular. table2- bearing capacity factor 土的内摩擦角标准值 02468101200.030.060.100.140.180.231.001.121.251.391.551.731.943.143.323.513.713.934.174.42A

31、ccording to the existing reference and table 2. We know that .by refering to the table 2, we know that. So characteristic value of bearing strengh for the soil of foundation bottom .According to table 2-4, we find that in the place below 6 m of the foundation bottom , there is a kind of relatively w

32、eak soil, That is soft substratum, the characteristic value of bearing strengh: 4）correction of depth and width of characteristic value bearing capacity for the foundation.）when the width of foundation is greater than 3m or buried depth is greater 0.5m ,the bearing capacity should be modified by the

33、 following equation. 表2- 承载力修正系数Soil category artificial fills. Clay of which the e orIL is greater than 0.85.01.0 Clay of which the e orIL is smaller than 0.85.0.31.6）through the above table we know that the correction coefficient of characteristic value of bearing capacity for foundation bottom is

34、: ，so，Foundation failureThe correction coefficient of characteristic value of bearing capacity for weak underlying stratum is:，so，5）the calculation of strengh of weak underlying stratumFrom the table 2-4. we can find that there exists a relatively weak layer at the depth of 6m, that is weak underlyi

35、ng stratum According to the regulations, if there exists weak underlying stratum below bearing stratum of foundation, the calculation of weak underlying stratum should be carried out according the following equation. For rectangular foundation, refer to the litterature () z/b 0.25 0.53510when z/b0.2

36、5, set; in this project z/b=3.7/2 3.5=0.1574 Foundation failure2.4 The limit load of the foundationThis project adopts three kinds of methods for calculating the limit load of the foundation. Namely methods of Vesic/K.Terzaghi and Hansen J.B. since the cohesion c of every layer is so different, So t

37、he calculation of ultimate load is carried out by using stratified calculation and average soil layer for calculation. As to equation of Vesic, which is related to safety coefficient, this project reffers tolitterature() and adopts 1.22.4.1 Vesic formula calculation1）limit load of the second layer o

38、f the the soil.2）. limit load of the third layer of the the soil.2.4.2 K.Terzagh formula computationfrom the litterature () we know that the equation to canculate strip foundation is Table 2- K.Terzaghi formula bearing capacity factors001.005.71when=0， So K.Terzaghi formula changes into 1）limit load

39、 of the second layer of the the soil.2）limit load of the third layer of the the soil. 2.4.3 Hansen J.B formula computationfrom the litterature() we know that the equation to calculate strip foundation isfrom the parameters we know that when =0 reffering to literature ()table 2-001.005.14when=0， Hans

40、en J.B formula changes into 1) limit load of the second layer of the the soil.2.) limit load of the third layer of the the soil.2.4.4 analysis of the calculation resultWe can see that , the three methods have similar results. As the load is,which is the failure load. And it proves that our calculati

41、on methods and assumptions are correct and credibility.2.5 settlement of the foundation According to the litterature () we can know the curve of this layertable2-5from the figure we can find the Pre-consolidation pressure and effective stress of the layer h=29.6ft=9m is so，Overconsolidation margin A

42、dopt the middle point （2.55,1.4） and point（4,1.24）, According to the litterature (),we know:1）The second layer =8.73 ，=0.21 =0.8So 2）The third layer=7.30，=0.28 reffering to litterature=1.05 Adopt the middle point （0.8,1.62） and point（1.1,1.60）, According to the litterature (),we know 1）The second la

43、yer=8.73 ，=0.21 reffering to litterature，=0.8 so2) The third laye=7.30，=0.28 reffering to litterature，=1.05so，计算谷仓空载时的沉降To calculat the total settlement of this project, we adopt two methods, and they are excel table and delamination summation method with checking computations, then compare the resu

44、lts of the calculation. Observe the calculation to sure whether the results of the gap between the two is within permitted error range. If it is within permitted error range, the calculation of the settlement method is right and credible, If it is not within the permitted scope of the error, the cau

45、se of excessive produce error should be analysed, and the methods should be improved.1） the first method firstly consider the settlement of non-load barn. Figure2-6 total settlement of non-load barn (2) the second method: delamination summation-method (considering the settlement of non-load barn whe

46、n)1)Draw profile between foundation and soil as shown in figure 2-7 Figure 2-7 2）Calculation of gravity stress of foundation soilat the bottom around the underground water 3）contact pressure of foundation bottom Set the average soil gravity of foundation and backfill soil above 4）Subsidiary stress a

47、t the bottom of foundation. 5）The shape of the foundation bottom is rectangle, to calculat with vertex point method. subsidiary stress , of which the stress state parameter can be got by refering to the related table. The calculation process is as follows. table2-8 Table 2-8 subsidiary stress calcul

48、ation depth z/ml/bz/bStress coefficient subsidiary stress 0.01.22.43.64.86.08.010.92.52.52.52.52.52.52.52.500.100.200.310.410.510.680.930.25000.24960.24920.24650.24400.23850.22530.2088144.80144.57144.34142.77141.32138.14130.49120.946）Foundation settlement calculation， as is shown in table 2-9，Table

49、2-9 Foundation settlement calculationnumbera /mm12345671.21.21.21.21.222.90.210.210.210.210.210.280.280.80.80.80.80.81.051.05144.69144.46143.56142.05139.73134.32125.7220.2620.2220.1019.8919.5636.6949.807）Total settlement compare the two settlement calculation plans, We found when using excel softwar

50、e to calculate, the total settlement of the empty barn is S = 192.52 mm. When using delamination summation method to calculate, the settlement is S = 193.10 mm.The error of results by two methods is within the scope of the allowable range, therefore, the settlement result is right and credible2.6 su

51、mmaryFrom the above the calculation result, it is intensity failure for this foundation , the bearing capacity of the soil is too small, and could not support the upper structure, which eventually led to the destruction. It gaves our designer a enlightenment: strict exploration and careful checking

52、is necessary in the process of design. Chapter3 Redesign of the Foundation 3.1Scheme SelectionAs to the the geological conditions of the foundation is not that complex. the desigh scheme of this foundation should be considered in two aspects, one is shallow foundation, the other is deep foundation.3

53、.2 Shallow foundation designThis design adopt scheme of raft foundation Embedded depth of foundation d should be less than groundwater depth, usually should be buried at least 0.5 m depth, embedded depth of foundation should not be less than times of Building height for the sake of seismic requireme

54、nts, we choose embedded depth of foundation 2.2 m in depth . Bearing capacity calculation of bottom of foundation1）According to the literature () ,Wheneccentricity：According to the existed geological material, we know, by referring to table2, we know that at the present so characteristic value of be

55、aring capacity at the foundation bottom is. and characteristic value of bearing capacity of the weak underlying stratum at the 7.5m depth below bottom of foundation2）when the width of the foundation is greater than 3m or the embedded depth is greater than 0.5m. the bearing capacity of foundation sho

56、uld be modified according by the following formula. 通过表2-，我们可以知道，基础底面地基承载力特征值修正系数 ，所以，correction factor of characteristic value of bearing capacity of the weak underlying stratum is:，所以，2） To determine the size of foundation bottomFor the axially loaded foundation, the basic size of the foundation b

57、ottom could be got by取L=70，B=30，A=2100 can meet the requirement.3.3基础底面压力计算 This project uses Serviceability Limit state to calculate the load. In consideration of Short-term effect, we use the standard load combination for calculation. The foundation bottom pressure can be calculated by the followi

58、ng formula. Assume that foundation is axially loaded.When the barn is full, the vertical load is:F = (20000+27060) 9.8 1000=461188KN Foundation bottom stress is: 3.4 strength calculation of underlying weak underlying stratum.According to the regulations, if there exists soft soil layer under bearing

59、 stratum, the fomula of calculating of the strength should be as follows. For rectangular foundation, refer to literature():当z/b0.25时取=，for this project z/b=3.7/23.5=0.1574 foundation failer3.5 redesign This design is based on the bearing capacity of weak underlying stratum to calculate the area of

60、foundation. The depth of foundation D=2.2m. According to the previous calculation we know characteristic value of bearing capacity of weak underlying stratum at the depth of 7.5m below the foundation.1) Characteristic value of bearing capacity of weak underlying stratum2) according to the previous calculation,we can get: 所