西南交通大学基础工程课程设计

Foundation Design 姓名姓名 马 德 林 学号学号 20100193 班级班级 2010 级土木茅 1 班 西南交通大学土木工程学院西南交通大学土木工程学院 2013 年年 5 月月 1 CONTENTS Problem 2 Design 4 Chapter one Unit conversion 4 Chapter two Design load calculation 4 Chapter three Geotechnical designing 4 Step 1 to Step 9 Allowable bearing pressure method 4 Step 10 Checking moment load 8 Chapter four Structural designing 8 Part A Determine required thickness based on a two way shear analysis 8 Check one way shear 9 Part B Design the flexural steel 9 Chapter five Sketch of the designed footing 11 Figure 2 Thickness and effective depth 11 Figure 3 Dimensions and reinforcing steel 11 Figure 4 Inner block and outer block 12 Figure 5 Structural show 12 2 Problem Design for Practice A proposed office building is to be constructed at the site with a geologic profile showed in figure 1 The ground table is at 5 5 ft The shallow strata are very soft The data for these strata maybe used in foundation design were obtained from a series in situ tests and laboratory tests and showed in table 1 In the table Su is undrained shear strength is preconsolidation stress m Medium Sand Dr 60 0 ft 12 ft 23 ft 27 ft GWT 5 5 ft High Plastic Clay CH Silty Clay CL Very Stiff Silty Clay CL Fig 1 geologic section for the construction site The design columns of the proposed office building will carry the following loads dead vertical load range 30 100 k live vertical load range 20 75 k and dead load moment range 0 50 ft k These columns are to be supported on spreading footings A sketch of an interior column and its spreading footing is given in figure 2 If such an interior carrying a 50k dead vertical load a 50k live vertical load and a dead load moment 20 ft k Try to determine the spreading footing of this column The design task should including followings 3 Table 1 data for different strata 1 Unit conversion Before beginning your design please convert the data in the figures and tables from English to SI and please use SI in your designing 2 Design load calculation There are two methods of expressing and working with design loads the allowable stress design ASD and resistance factor design LRFD Calculate both of them 3 Geotechnical designing Select a suitable type of the spreading footing determine the footing depth determine allowable bearing pressure and determine the required base dimensions for the footings of the column in Figure 2 Fig 2 A sketch of the typical interior column and its footing 4 Structural designing Determine the materials using in the designed footings determine the thickness of the footing and determine the reinforcing steel of the footing 5 Sketch of the designed footing Show your design in a sketch ft Depth Range Soil Description total pcf Su psf Cc 1 e0 Cr 1 e0 m psf 0 12 CH 105 1280 0 15 0 02 2600 12 23 CL 112 1800 0 11 0 015 3000 23 27 Med Sand 120 0 006 0 002 27 CL 118 2600 0 08 0 01 6000 4 A proposed design ChapterChapter oneone UnitUnit conversionconversion To ConvertTo Multiply by ftm0 3048 psfkPa0 04787 pcfkN m30 1571 Data for different strata SI Depth Range m Soil Description 3 total kN m u S kPa 0 1 C C e 0 1 r C e m kPa 0 3 657CH16 49561 2740 150 02124 462 3 657 7 010CL17 59586 1660 110 015143 610 7 010 8 229Med Sand18 852 0 0060 002 8 229CL18 538124 4620 080 01287 220 Tips to convenient English units will be used in the designing ChapterChapter twotwo DesignDesign loadload calculationcalculation The allowable stress design ASD 5050100Pkkk min 50Pk max 175Pk 20Mftk max 50Mftk Resistance factor design LRFD 1 4 501 7 50155 u Pkkk 1 4 2028 u Mftkftk ChapterChapter threethree GeotechnicalGeotechnical designingdesigning Step 1 per TABLE 8 1 5050100Pkkk Use an estimated D of 2 ft 24 in Step 2 The groundwater table is at 5 5 ft and is not a concern at this site Step 3 per Figure 6 11 Soil Type Clay Design F with Typical Range Use F 3 5 step 4 For high clay if saturated undrained conditions exist as same as the problem statement we may conduct a stress analysis with the shear strength defined as and In this case per TABLE 6 1 Tu cs 0 T 5 7 1 00 0 cqr NNandN Hence Using square foundation B L 1280cpsf 105pcf Using the BEARING XLS spreadsheet with the computed min 50Pk 5 allowable bearing pressure 2 2770 a qlbft Step 5 per TABLE 2 2 Typical commercial and residential buildings 1 500 a 1 500 20 12 0 48 DaaS in Per TABLE2 1 for office building use in order to control 0 95 a in differential settlement here Step 6 using TABLE 7 5 for clayey natural soil assuming the foundation is a rigid structure the design value of is 0 5 D Step 7 so the total settlement 0 48 0 95 0 50 475 DaaD requirement controls the settlement analysis Step 8 using classical method to compute total settlement of shallow foundation which is based on Terzaghi s theory of consolidation Because of the assumption that all of the soils are over consolidated the equation of the total settlement is c Case 1 zfc 00 log 1 zf r c z C rH e Case 2 0zczf 000 loglog 11 zf rcc c zc CC rHH ee Where rigidity factor per TABLE 7 1 for spread footings r0 85r initial vertical effective stress at midpoint of the soil layer 0z final vertical effective stress at midpoint of the soil layer zf and 0zfzz could be computed by simplified method equation z for square foundation 1 76 2 1 1 1 2 zzD f q B z In which bearing pressureq 6 depth from bottom of foundation to point f z vertical effective stress at a depth D below the ground surface zD in this case 105 2210 zD psf pre consolidation stress at midpoint of the soil layer and c 0cmz Now we can compute the consolidation settlement after dividing the soil beneath the footing into layers Figure 1 Dividing the soil beneath the footing into five layers Try with 12Bft max 175Pk 2 175000 150 21515 12 a qpsfazD qpsf At midpoint of soil layer Layer No H ft zf ft 0z psf z psf zf psf c psfCase 0 1 c C e 0 1 r C e c in 3 51 75289129015792889OC I0 150 020 53 6 56 7571683715533316OC I0 150 020 44 512 597839913773978OC I0 110 0150 11 618125122114724251OC I0 110 0150 06 4231515143165815150C II0 0060 0020 01 1 15 7 so the settlement criterion has not been satisfied 0 95 ca in Try with 14Bft max 175Pk 2 175000 150 21193 14 a qpsf 1193210983 azD qpsf At midpoint of soil layer Layer No H ft zf ft 0z psf z psf zf psf c psfCase 0 1 c C e 0 1 r C e c in 3 51 7528997612652889OC I0 150 020 46 6 56 7571671114273316OC I0 150 020 40 512 597837513533978OC I0 110 0150 11 618125121614674251OC I0 110 0150 06 4231515142165715150C II0 0060 0020 01 1 04 so the settlement criterion has not been satisfied 0 95 ca in Try with 15Bft max 175Pk 2 175000 150 21078 15 a qpsf 1078210868 azD qpsf At midpoint of soil layer Layer No H ft zf ft 0z psf z psf zf psf c psfCase 0 1 c C e 0 1 r C e c in 3 51 7528986311522889OC I0 150 020 43 6 56 7571665713733316OC I0 150 020 37 512 597836313413978OC I0 110 0150 10 618125121314644251OC I0 110 0150 06 4231515141165615150C II0 0060 0020 01 0 97 so the settlement criterion has been satisfied 0 95 ca in Tips using the SETTLEMENT XLS with and can also max 175Pk 0 95 a in produce 1078 a qpsf Step 9 1078 2770 so the settlement controls the design Rounding to a multiple of 500 psf gives 1000 a qpsf 8 For a 100k column load a 50k dead vertical load and a 50k live vertical load 22 150 2300 f WBB 2 100000300 11 95 10000 f aD PW B BAft qu use 12 144 Bftin Step 10 checking moment load 2 150 2 1243200 f Wlb Using Equation 5 5 20 0 139 10043 2 f Mftk eft PWkk 12 2 66 B ft so OK for eccentric loading 6 B e 2122 0 13911 86 B BBeft 2 10000043200 01018 11 86 f equivD PW qupsf A Since in practical engineering 5 error is allowed this 1000 equiva qqpsf design is satisfactory Hence a suitable type of the spreading footing may be chosen as The footing depth 2Dft Allowable bearing pressure 2 1000 a qlbft Base dimension for the footing square foundation 12 144 Bftin ChapterChapter fourfour StructuralStructural designingdesigning Now a 21 inch square reinforced concrete column c 21in carries a vertical dead load of 50k a vertical live load of 50k and a dead load moment 20 ft k that s to say 155 u Pk 28 u Mftk Because of the small applied load and inspection costs of high strength concrete we will use and 2 2000 c flb in 2 60 y fk in Part A Determine required thickness based on a two way shear analysis 9 Try T 12 in 1 bar diameter 312 1 38dTinin 2222 22 15500028000144 21 8 38 100 4421 8144 uu uc PMBcd Vlb cdB 0 21 829bcdin 0 44 29 8200041 500 ncc Vb dflb Not acceptable 38 1000 85 4150035 300 ucnc VVlb Try T 15 in 1 bar diameter 315 1 311dTinin 2222 22 15500028000144 21 11 37 700 4421 11144 uu uc PMBcd Vlb cdB 0 21 1132bcdin 0 44 32 11200063 000 ncc Vb dflb OK 37 7000 85 6300053 500 ucnc VVlb However moment load is present here one way shear needs to be checked too Check one way shear using Equation 9 8 22 2614421 2 116 28000 0 155000 0109 500 144144 u ucu BcdM VPlb BB Using Equation 9 9 22 2 144 112000283 400 ncwc Vb dflb OK 109 5000 85 283 400240 900 ucnc VVlb Therefore T 15 in d 11 in Part B Design the flexural steel Find the required steel area 14421 61 5 22 Bc lin 22 2155000 61 52 28000 61 5 2 059 500 22 144144 uu uc PlM l Minlb BB 10 2 2 535 1 176 cuc s yc f bM Add ff b 22 2000 1442 353 2059500 11113 61 1 176 600000 9 2000 144 in Check minimum steel for grade 60 steel 0 0018 15 1443 89 s A Use 3 89 s A Use 7 7 bars each way 2 4 21 s Ain Tips The final value of 7 bars is determined as a part of the flexural 0 875 b din analysis and is different from 1 in assumed before However this difference is small compared to the construction tolerances so there is no need to repeat the shear analysis with revised b d and Acceptable 4 21 0 00260 004 144 11 s A bd Clear space between bars 144 8 0 875 17 in 3T 45 in or 18 i