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# Geotechnical Eng II CEG 4012

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This 41 page Class Notes was uploaded by Arely McDermott on Monday October 12, 2015. The Class Notes belongs to CEG 4012 at Florida International University taught by Luis Prieto-Portar in Fall. Since its upload, it has received 31 views. For similar materials see /class/221744/ceg-4012-florida-international-university in Civil Engineering at Florida International University.

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Date Created: 10/12/15

The Design Procedure Determine the structural loads and member sizes at the foundation level Collect all the geotechnical data set the proposed footings on the geotechnical profile Determine the depth and location of all foundation elements Determine the bearing capacity Determine possible total and differential settlements check effects at 2B depths Select the concrete strength and possibly the mix Select the steel grade Determine the required footing dimensions Determine the footing thickness T or D in some textbooks 10 Determine the size number and spacing of the reinforcing bars 11 Design the connection between the superstructure and the foundation and 12 Check uplift and stability against sliding and overturning of the structuresoil system FPONQMer Nf The first studies performed on foundation structural failures were done by Professor Talbot at the University of Illinois in 1913 Advances in the next 50 years include Prof FE Richart s tests at the University of Michigan His results were synthesized into the methodology used today by a committee sponsored by the ACI and ASCE and published in 1962 Spread footings is still the most popular foundation around the world because they are more economical than piles adding weight to them does not affect any other member and their performance has been excellent Selection of materials Spread footings are usually designed to use 3 ksi ltfc lt 4 ksi whereas modern structural members frequently use concrete between the range of 4 ksi ltf39c lt 8 ksi A higher concrete strength helps reduction the member s size However the footing s design is govern by the bearing capacity and settlement That means that the strength of the soil might be limiting factor and a higher concrete strength would not be relevant Where a footing must carry a load greater than 500 kips an f39c 5 ksi might be justified Since exural stresses are usually small a grade 40 steel would usually be adequate although it is currently unavailable in the US The most common grade used for construction is Grade 60 steel which is almost universally used in the world today 4Eh 1 quot H l The typical details of a spread footing as sketched for drafting The standard thicknesses T are given in English Units as multiples of 3quot 12quot 15 quot 18quot etc A high precision in specifying the depth of excavation Df is unnecessary AC1 code specifies that at least 3 inches of concrete cover must be included from ground contact Which takes into consideration irregularities in the excavation and corrosion factors II1 a 7 I r u quot 139 KM quot quot911 g1 The same footing as built Note that forming is required in this site due to the presence of a clean gravelly sand that would not stand vertically at the sides of the excavation Design Criteria 1 The qall and Q control the footing dimensions B X L footing area A 2 The designer controls the depth Df embedment of the foundation 3 Shear v controls the footing s thickness T d 3 the diameter of rebars a Diagonal tension punching shear for square footings and b Wide beam shear for rectangular footings that is When L B gt 12 Analysis The analysis of a square or rectangular footing may first be performed by assuming there is no steel in the member The depth d from the top of the footing to the tension axis is Z Fy 0 l Qu 2d Vc b d 2d Vc c d cdbd qo shear on 4 faces bottom face L SetQuzBqu Mill WM d24 Vc qo d 2 vc qob c BL cb qo 0 on 7 BLq 0 For the special case of a square column Where c b W b 7 7 d2Vc qo4 d vc q02w 32 W2 qo4 0 LCJ T wea l LOVE For the case of a round column With a 2 diameter 1 l l p v d2 vc q04 d vc q02 a BL Acol qo4 0 d Design Steps m Compute the footing area via B x L for a square footing 3x3 2 Q qall for a rectangular footing BxL Q qall Where Q is the critical load combination not Qu M Find the soil reaction under ultimate structural loads to check bearing capacity find the quotultimatequot contact bearing an 2 Qu BL and check that an S qu Step 3 Compute the shear in the concrete Vc Case a for a square footing check for diagonal tension punching shear Vall 4Q SQRT f39c Where Q 085 for shear For example for f39c 3000 psi Vall 186 psi Case b for a rectangular footing check for wide beam shear Vall S 2Q SQRTf39c Where Q 085 for shear For example for f39c 3000 psi Vall 931 psi Step 4 Find the effective footing depth d Note that use of d Via this method eliminates the need to use steel for shear which is used only for exure Use the appropriate equation from the Analysis Section Step 5 Com ate the re aired area 0 steel As each wa or bendin exure Bending moment unit Width M q L2 2 for a cantilever beam Mummy2 QAsfyol az Check p so that the maximum allowed percentage of steel is not exceeded Step 6 Compute bond length column bearingz and the steel area required or dowels Use as a minimum an As 2 0005 Acol usually With 4 equal bars Step 7 Draft the above information into a complete drawing showing all the details Example 1 Design a square reinforced concrete footing for the following conditions The column has a DL 2 100 kips a LL 2 120 kips and is a 15 X 15 With 4 8 bars The footing is upon a soil With qall 4 ksf with a FS25 use 1 3000 psi and fy 50 ksi Solution Step 1 Find the footing dimensions for serVice loads quiz 2 2 B 2 amp 742 ft UseB 75 t B qall 4 Step 2 Check the ultimate parameters that is the actual soil pressure 1 under Qu The ultimate contact pressure Q 14DL17LL1410017120 140204344 kips QM 344 q 132 752 61 ksf lt10 ksf for 5 0K Steg 3 Compute the allowable shear stress in the concrete Va 4 f 39C 40853000 186 psi 268 ksf Step 4 Find the effective footing depth d in this case punching shear governs a 2 4vc d vc w BZ wz 20 4 2 4 2 d2 4268 jd268E1 5 752 0 4 2 12 12 4 which yields two solutions d1 120 ft and d2 246 ft Choose d 120 t g B 75 gt39 15 d lt As a check use the modified equation for d which is 4d22bc d Bz w V 0 which yields two solutions d1 128 ft and d2 252 ft Choose d 128 ft a difference of only 7 Therefore use the largest d 128 154 in Use d 16 in It is not necessary to check for widebeam shear Steg 5 Compute the required exural steel area As The unit strip of the cantilever arm L is 15 B w 39 E 2 2 L39 3l3 ft k B75 gtl FT l gt gt gt gt gt gt The cantilever moment M is L 392 61313212 2359 in ki S 2 2 p M2610 where Mu goAsfy d j with q 09 for exure Asfy AS 50 a 085f 39C b 085312 in which yields 16As 081AS 797 0 whence As 050 in2 l 163AS but Total As 2 75 ft 05 inzft 375 in2 Check p 2 As bd 05 12 16 00026 gt 0002 minimum lt 0021 maximum OK InB 90 in 75 use 12 5 bars As 372 inz 75 CC or 7 7 bars As 420 inz 12 CC or 5 8 bars As 385 inz 18 cc Steg 6 Check the development length for bond Ld AC1 122 126 If Ab the area of each individual bar and db the diameter of bar 004Abfy 00406050000 d f 39C 73000 also check With Ld 00004 d bfy 00004 0875 in 50000 psi 2 175 inches 219 in gt12 in minimum Steg 7 Check the column bearing to determine need of dowels Area of the column Ag 2 W2 15 in2 225 in2 Effective area of footing A2 b 4d2 15 in 416 in2 6240 in2 Checking the contact stress between column and footing fc pf 1 where 02070 and 1 S 2 g g fc 0730002 3750 psi gt 3000 psi Need dowels The actual ultimate contact pressure at ultimate loads is fc Qu Ag 2 344 kips 225 in2 152 ksi lt 30 ksi OK However dowels are always required with at least As 2 0005 Ag As required 2 0005 225 in2 2 1125 in2 The diameter 1 of the dowels are i 1 column bars S 015 maximum diameter difference Use 4 7 bars or 4 X 060 in2 2 240 in2 2 As dowels whereas column has 8 s Check diameter difference 100 0875 0125 lt 015 OK Steg 8 Check the embedment length Ld for the dowels 02 d Ld A or 00004 fy db or 8 inches choose the largest of the three J7 L 02500000875 d 3000 Ld 0000450 0000875 18 inches gt 8 inches 0K 16 inches gt 8 inches 0K Note however that d 16 in need a 2 hook to reach 18 but minimum hook is 6 Steg 9 Sketch the results N 220 kips Qu 344 kips 3 clearance 7 7 12 cc each way Examgle 2 Design a rectangular footing to carry a moment induced by Wind With the following data DL 2 800 kN LL 800 kN M 800 kN m and qall 200 kNm2 With FS 2 15 Square columns With c 500 mm fc 21 MPa and fy 415 MPa Ste 1 Find the footin dimensions D 800 M I 8 0 0 L g 800 KIVN 13300 L500 This time perform a trial and error selection of B X L 05m Set BXL 32 and check the increase in the soil pressure due to Wind load F 825m 225m4 l 32 Q qau 1600 kN 200 kNm2 8 m2 B 282 m L Bra gt1 eMQ800160005 645 39L 2 605 3 m from 6eL qov IfL 3 m try afooting 25 mX 4m 866 qavg Q A 1600 kN 10 m2 160 kPa qmax Q BL 1 6eL 1600 101 6054 280 kPa Note that qmax exceeds qavg by 33 increase size to 275 m X 45 m qavg Q A 1600 27545 130 kPa qmax 1301 605 217 kPa lt 200 X 133 266 kPa Iterate one again and settle With B 3 m and L 5 m M Check the ultimate parameters that is the actual soil pressure 1 under Qu Pu 14DL 17LL 14800 17800 2480 kN Mu 14DL 17LL 14300 17500 1270 kN m e ZMPu 1270 2480 0152 qmax Pu A1 6eL 2480151 601525 266 kPa lt qu 300 kPa OK qmin Pu A1 6eL 2480151 601525 645 kPa qavg kPa lt qall kPa Step 3 Compute the allowable shear stress in the concrete The diagonal tension for f39c 21 MPa Vc 129 MPa Step 4 Find the effective footing depth d Using the simplified equation 4d22bcd Bquvc0 4d2 205m 05m d 15165 1290 0 Which yields d 050 m Now find the depth d for Wide beam 885 a X fromX0t0X225 d dd 1 dV 2 q dX 65 V I qu 266 402XdX 266x 40222 598 266d 201225 d2 Vcvc2d12902d l d 060 m 2 use the highest d 060 112 q 866 i 408x 866 Steg 5 Compute the required longitudinal exural steel area As A fy 415AS a AS 085f cb 085211 225 225 Mu Iva I266x 40222dx597 kN m O O Mu 597 kN m 0Asfyd AS2 00515AS 00137 2 0 from whence AS 282 cm2 m use 17 25 mm rebars Checkp As bd 000282l06 00047 gt 0002 minimum lt 0021 maximum OK Steg 6 Compute the required transverse exural steel area As Using a high average q qavg qmax 165 266 215 kPa Mu W122 qb 05222 2152153 05222 168 m KN and again Asd a2 Muny A82 00515As 0000386 2 0 As 761 cm2m Checkp As bd 761 X 10394 m2 106 000126 lt 0002 less than minimum 2 use minimum As 2 0002bd 0002l06 00012 mZm 12cm2m Therefore the longitudinal steel 282 cmZm X 3111 846 cm2 X l 111210000cm2 846 X 10393 m2 gt 17 25 mm bars placed at 176 cm cc The transverse steel 12 cmZm X 5m 2 60 X 10393 m2 3 13 25mm bars placed at 38 cm cc mm I 848ULltN Pu I 1870 m RN Steg 7 Draft a sketch of the results 05m I 885m gem 4 T 06m 888 cmEm 070m 39 l Iooooooooooooooooooooooooo iiyg grim BARS T QUESm 0075 U39l 3 A W II N 0 3 V Examgle 3 Design continuous wall footing for a warehouse building given DL 2 3 kft LL 2 12 kft the gall 2 ksf with FS 2 2 75011 2 110 pcf and f39c 3000 psi fy 60 ksi Yooncrete 150 pcf Steg 1 Assume a footing thickness D 1 ft By ACI 7731 the minimum cover In soils is 3 Also assume using 4 bars Q 12quot 1 12quot 3quot 05quot2 875 RDDF MASDNRY DVERHEAD CRANE WALL FLDDR SLAB W Twoogvewge Boom Stwlp Continuous or strip or wall footings are reinforced transversely to the direction of the wall The footing acts in essence as two small cantilever beams projecting out from under the wall and perpendicular to it in both directions The reinforcement is placed at the bottom where the stresses are in tension exure Although the main reinforcement is transverse to the axis of the wall there is also a requirement for longitudinal reinforcement to control temperature shrinkage and concrete creep Steg 2 Find the soil pressure 1 under ultimate loads Estimate the footing Width B Q qall 42 k 2 ksf 21 ft assume B 3 feet AC1 92 required strength U 14D 17L U 1431712 42 20 62 kips The soil pressure at ultimate loads qo U B1 62 kips 3ft1ft 21ksf lt qu 4 ksf D1ITB T l J olj Steg 3 Check the shear strength of the concrete The critical section for shear occurs at a distance d from the face of the wall AC1 155 and 11 11 1 2 12 T V I I I m I I I j CRITICAL SECTIDN QSIEIRSF q 81 R510 FDR MDMENT CRITICAL SECTIDN V18 ogtqs FDR SHEAR the ultimate shear Va 2 12quot 875quot 1ft ft 12 in X 21 kft2 057 kips ft of wall Check the concrete strength vu S Q vc 2Q f c bd 0852300012 875 98 kipsft of wall OK Since vu ltlt 120 can reduce D thickness of the footing to say 085 ft 10 d 10quot 3quot 052 2 E gt 6quot dmm AC1157 rechecking Vu lt Vc 2Q 70 bd 20853000 12in650in 73 kipsft of wall OK Steg 4 Compute the required transverse exural steel area As From ACI 154 Mu qo l2 2 where l 12 21 ksf21ft2 105 k ft ft of wall W W F r buta Asfy 085f c b 60 ksi As 0853 ksi12 in 196 As inches b 12quot of wall and Mn 2 Asfy d a2 2 As 60 kin265quot a2 but Mu Q Mn 09 Mn 105 kipsin in 0960 kinz As 65quot 098As 53As2 351As 105 0 which yields A51 2 66 in2 per ft of wall As2 0003 in2 per ft of wall p1 As1Bd 66 in2 12 in65 in 0085 p2 As2 Bd 2 0003 in2 12in65in 00004 lt 00018 The maximum steel percentage allowed pmax 075 pb where pb 085f cfy 8700087000 fy 085360 085 8700087000 fy 0021 39pmax 075 pb 0750021 00016 Now note that p1 0085 gt pmaX 0016 therefore use pmin 00018 As pmin bd 0001812 in65 in 014 in2 per ft of wall use 1 4 every foot of wall As 2 020 in2 m Check the reinforcement development length Ld AC1 122 Ld 004 Ab fy fquotc but not less than 00004 d b fy Ld 004 020 in260000 psi 3000psi 88 inches Check Ld 0004 d b fy 0004 65 in60000 12 inches 12 controls design The depth is thus 12quot 3 cover 9 lt 12quot thus are missing 3quot on each side for Ld must increase footing to B 35 feet Minimum steel in longitudinal direction to offset shrinkage and temperature effects as per AC1 712 As 00018 b d 0001842 in65 in 049 in2 39 Provide 3 4 bars at 12quot As 2 060 in2 Steg 6 Sketch the finished design 3 10 Cantilever sign are more economical th n the overhead bridge type They are built from hotdipped galvanized steel pipes to a maximum truss span of 44 ft 134 m The singlecolumn steel post is bolted to a drilled shaft foundation Design follows AASHTO s Standard Specifications for Structural Supports for Highway Signs Luminaires and Traffic Signals published in 1985 This NJDOT research structure tended to move vertically and horizontally With the passing of trucks underneath Anchor bolts loosened after nine months in service The ratio of stiffness to mass gave the signs a loW natural frequency of 1 Hz Q muss EXISTING CAISSDN ELEVATION In addition to low natural frequencies of about 1 Hz these structures have extremely low damping ratios typically 02 to 05 1 percent Cantilever support structures are susceptible to large amplitude vibration and fatigue cracking caused by wind loading Report 412 of the National Research Program NCHRP Fatigue Resistance Design of Cantilevered Signal Sign and Light Supports studies three phenomena 1 buffeting by natural wind gusts 2 buffeting by trucks and 3 galloping also known as Den Hartog instability Galloping is an aeroelastic phenomenon caused by wind generated aerodynamic forces Maximum response is from truck gusts as an equivalent static pressure of 37 psf 1770 Pa times a drag coefficient of 145 for vertical movement and 17 for wind in the horizontal direction Full strength from truck induced gusts occurred at 18 ft heights and decreased to zero at 30 ft heights CE Sept 2000 Homework the billboard Sign Design a spread footing using FBC and ASCE 7 02 Ignore the effect of the water table Jr 239 J t 3 t 6 I I I I 20 I I I I I P10k 24H STEEL COLUMN 40 39 HIGH A WIIH 1 THICK WALLS 4 Z L Xgt X gtgtgt STEP 1 Find the Wind load as per ASCE 793 assuming an Exposure C Cat IF qz Gh Cf AfT he force Whereqz 000256 kz IV2 qz 34 psfandKz 0981 105 V 110 mph GH 126CF 12 The sign shape factor is MN 3220 16 F 34 psf 126 12 32ft X 20ft 1000 324 kipsCalculate loads on footing STEP 2Weight of steel column 2 gs LA 2 049 MX 10 kips X 15 150 kftMy 324 k X 30 972 k fth 324 k X 15 486 k ftTotal normal load N 10 k 5 k 15 kips STEP 3Calculate the footing s bearing capacity using Hansen s formulaqult c Nc sc dc Ic q Nq sq dq Iq 05 g B Ng sg dg IgWherec cohesion 0150 ksfq ng embedment pressure 2 0130 ksf3 ft 039 ksz footing Width initial assumptions 2 5 ftL footing length initial assumptions 2 15 fth bearing cap factor for embedment at f 20 2 ep tan f tan245f2 640Nc bearing cap factor for cohesion at f 20 2 Nq 1 cot f 1483Ng bearing cap factor for Width at f 20 2 15 Nq 1 tan f 295sq shape factor for embedment 10 BL sin f 111sc shape factor for cohesion 10 NqNc BL 114sg shape factor for Width 2 10 04 BL 867dq depth factor for embedment 1 2 tan f 1 sin f2 DfB 119dc depth factor for cohesion 10 04 DfB 124dg depth factor for width 2 10 ic inclination factor 05 l 1 H Af Ca Where ca 2 06 to 10 ciq 1 05 H V Af Ca cot fd Where 2 S d S 5ig 1 07 H V Af Ca cot fa Assume FS 2 3Theref0re qult 015 ksf 1483 114 124 10 03964011111910 05 0130 ksf 5 295 867 10 727 and qa11 qult FS 2 242 3 AssumeB 10 SC 10 0431 X 02 109L 50 DC 10 04310112D 3 Q 130X3 390BL 02SQ 210 02 sin 20 107FS 308 210 0402 092DQ 1 0315310 109 QULT 150 1483 109 112 390 64 107 109 05 130 10 29 092 10 7361 psf Qa qULT FS 2 2453 psf 25 ksf QMAXMIN PBL 6peyb21 6PeXBL2 P 15 kips 3X10X50X0150 FTG WT 240 kips EX 2 MyP 972240 2 405 Ey MXP 150240 2 0625 QmaXmin 240kips500 62400625 102 X 50 624040510X502 089 ksf lt 25 0067 OK 6X40550 6X062510 086 lt 10 0k 6 in middle 13 3 CHECK OT LONG DIRECTION MOT 324 KIPS 303 1069 KIP FT MR 2 5 KIPS X 25 225 KIPS X 25 5750 KIP FT ES 2 57501069 2 54 gtgt 15 OK CHECK SLIDING RSSVtanfCB 240 tan f 15010 1587 KIPS gtgt 324 KIPS 3 LOAD COMBINATION 075 14D 17L 17W 105D 1275W FACTORED LOADS PU 105 X 240 KIPS 252 KIPS MUX 2105 X 150 KIP FT 2 158 KIP FT MUY 1275 X 972 KIP FT 2 1239 KIP FT THEREFORE QMAXMIN 33610 X 50 i 6 X 336 X 0625 102 X 50 i 6 X 336 X 4910 X 502 0989 KSF 0019 KSF CHECK BEAM SHEAR D 36 4 32 VU 0019X50 2133 01550 21330989 0019 144 KFT FVC 085 X 2 X 3000 X 12 X 32 1000 36 KIPS gt144 KIPS OK PUNCHING SHEAR WILL NOT GOVERN BY OBSERVATION 3 DESIGN FOR FLEXURE IN LONG DIRECTION F C 3000 PSI FY 60000 PSI MU 0019 X 26 X 26 2 05 X 26 X 0989 0019 X 263 2 116 KIP FT AASXFY085F CB ASMUX12FFYD A2 A 2164 AS 083 IN2 R 2 AS BD 2 083 12 X 32 00022 gt 00018 OK USE 7 8 OC AS 090 IN2 OK THEREFORE MU 09 X 60 132 164212 126 KIP FT gt116 KIP FT OK USE FOOTING 10 X 50 X 3 W 7 8 0C TampB EW NOTE IN LIEU OF SUCH A LARGE EXPENSIVE FOOTING A DRILLED SHAFT MIGHT BE RECOMMENDED IN PRACTICE Shallow foundations are the most commonly used type of foundations primarily because they are simple and very economical to build Similarly their design is simple albeit tedious Fortunately the average price of software for the design and analysis of shallow foundations is about 400 An example of an even simpler software is the use of Microsoft s Excel spreadsheet Given the loading conditions the soil properties and the footing s material properties a spreadsheet can give the dimensions of the footing and the maximum moment and shear acting upon it InQut Required a Physical Layout 1 the desired ratio of length to width and 2 the estimated thickness of the footing b Material Properties 1 Unit Weight of Concrete and 2 Allowable Soil Bearing Capacity 0 Loading Conditions 1 Dead Load 2 Live Load 3 Moments about the xaxis and y axis and 4 Dead Load Imposed on Footing E SFODTING ex1s iEile Edit Eiew Epreachnnting induw 454 MHHWEI le EEEEEEEEEEEE in g G 39OLII39Id E f Re ady 39 I References Braja M Das Principles of Foundation Engineering 4th Edition Joseph E Bowles Foundation Analysis and Design 4th Edition Arthur H Nilson Design of Concrete Structures 12th Edition Edward G Nawy Reinforced Concrete Leonard Spiegel amp George F Limbrunner Reinforced Concrete Design 4th Edition

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