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by: Coty Von
Coty Von

GPA 3.94

Paul Mayne

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Paul Mayne
Class Notes
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This 0 page Class Notes was uploaded by Coty Von on Monday November 2, 2015. The Class Notes belongs to CEE 4803 at Georgia Institute of Technology - Main Campus taught by Paul Mayne in Fall. Since its upload, it has received 29 views. For similar materials see /class/234176/cee-4803-georgia-institute-of-technology-main-campus in Civil and Environmental Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15
problem 1 For a circular exible footing resting on a deep depositofsiltv mediumrcoarse sand determine ne h the foundation has a diameter of4 24 meters and supports an axial load of52 MN Plot these stresses with depth to a maximum zd 4 The givenvaluesare Diameter 424m 424 Radius m 212 m Q 52 MN 5200 kN Max 4 Using these valuesand knowing we have a circular footing the following can be foun Q 5200 7 7 A 7 n212Z q 35828 Z In order to find the stress in uence factor at the centerline 1 T 1 Zr5 a 1 2 where in thiscase since we know zd laradiusl 7 Nln 1 1 2 d Using zd 14 as a sample calculation 1 7 21 1 Hi 1 1 7 5 i Z 1 1 1 HQ d i17 1 1 1 Z 1 2 gt i 0130 The imposed vertical stresses can then be found as follows 02 i q kN oz 0130 36828 02 5 kPa Table 1 displavsthe data in order to calculate the stresseswith depth to a maximum zd 4 which are Vertimlstvess Aujq 000 0 50 100 problem 2 For a depth of 93 m use Boussinesq point load theorv to calculate the vertical stress increase for problem 1 above at the centerline and beneath the edge of the circular foundation when z 9 3 m R can be calculated as follows At the centerline R 22 r2 R 932 0Z R 93 m At the edge R vzZ r2 R 932 212Z R 954m At the centerline the vertical stress increase for the LilLuldl 39 1 L J follows 3023 A02 7 211Rg 35200 kN93 m3 Ml 2n93 m At the edge the vertical stress increase for the circular foundation in Problem 1 can be calculated asfoiiows A 3023 oz 7 211Rg 35200 kN93 m3 7 2n954 mS A02 2529 kPa A02 Comparing quot I to the elastic solution at a depth of9 3 m shown in Table 1 resultsin 2680 At the Centerline100 7 100 2680 2529 7 595 Difference problem 3 An octagonal footln restson the surface ofa 8rm thick clavev gravel deposit overlving verv ater table is01 m At the edge 100 7 100 v later calculations d 1 2 05 m Figure 3 s ows the octagon with a circle which demonstrates how the diameter isfound Now that the diameter iscalculated we can record normalized depth to diameter zd and normalized depth to radius za where za 2lzdl These values using the following equation A0 a Z 15 Z 1 7 1 q 2 h a w ere 7 z i a In order to nd q the area of the octagon must be found A 2nd where a 05 mand d 121m A 205 m121 m A 121 m Knowing q 523 kN q can be calculated Q 523 kN 7 7 X 7 121m2 kN q 43223 The horizontal stress changes can now be found knowing that since the water table isat 01 m deep the soil is undrained and Poisson39s ratio V 05 02 A0 1 o2 05 7 1 1 q 2121 zgt a2 705 12 1 z Theeiasticmoduiusisgivenas379barswhichmust be converted to 37900 kPa Now the incremental strain can be found as follows r A02 729mm The incremental displacementscan then be calculated at each depth A2 Incremental Displacements A r F I llddll I I I cumulative displacement in uence factor can be found A In 2012472 1 000701 e displacement ofthe octagonal footing can now be calculated v0 2because it isgravel 1 i 172 s p s p 43223 121m r 000701 r 1 e 022 7 37900 kPa 1000 m 1 m roblem 4 A central mat foundation for the Ritzr Carlton Hotel in Buckhead is constructed as a rectangular reinforced concrete with dimensions of 502 feet bv 572 feet The total load imposed on the mat is15028 kips The results ofa soil test boring show groundwater at 30 feet and the raw 5st are given below REFUSAL AT 801 feet Note Forverv high SPTs maximum count is100 lASTM u 1585i and thusrecorded as100 blows per so 5 p 929r 1075m 00929 mm with depth of the energvrcorrected Nevaluesat the site given that the ef ciencv rating is 40 for the hammer svstem ER 40 ER 60 N The N and N5 valuesare displaved inTable 3 42 Asthe Riterarlton iswithin the Piedmont geologic formation estimate the equivalent Young39s modulus from the corrected N values using the relationship presented for the First Am erican Bank ro39ect Gibson tvpe model and tan equation that describes the variation ofE39 with depth at this site Note the refusal at bedrock For a pure GibsonE 501 7 172 soil isdrained up to 30 feet because the groundwater table is located at30 feet while v 0 5 from 30 feetto 80 feet because the soil is undrained To convert a into tons per square feet tsf b Now thatall the data iscompiled in Table 3 a plotof E Egbars r 1044 variation of E with depth at this site Figure 4shows the plotalong with a trendline which displavsthe equation at best t Nole At elevationsgreater than 50 feet the points can be neglected in order to determine the line of best t The equation that best describes the variation is z depth 7 0 1154 E 8 67 7 z wherekE7867tcfin L L 39 M W feet s 44 The mat isconstructed to bear 16 feetbelow rade Determine the modulusvalue at thisbearing elevation Egg for the settlement calculation is 7 8 z E 7 867 ref 7 16 ft is 7 13872 c f 45 Evaluate the dimensionless Gibson parameter Be for the matgg 7 ES where d 7 shortest side of as the rectangular footing 7 502feet G 13872 tsf 67 ref 7 502 ft Ea 867 ref 7 502 ft 6 7 0318 45 Determine the displacement in uence factor iGH 1 6H 056 0235 W 1 Ea E d where h 7 80 7 16 7 64 feet d 7 502 feet as 7 03187 1 lay r u 56 u 215 n m llt 5II 1gt 16 0357 lflhe mat is 225 feet thick what is the foundation rigiditv factor Evaluate the modi er Ir Assuming 6 reinforcing steel Ema 0944000 ksi 00629000 ksi 5 0 0 ksi 396000 tsf Es taken ata z 7 a 502 E 7 867 tsf 2 tsf To nd the foundation rigiditv factor 3 RF 52 2 s where l thickness 25 m and a radius d2 3 K 7 131 The modi er N can now be found asfollows n F 4 4610KF n 1 1 7777 F 4 4610131 1F7 842 8 Perlheembedmenlglvenin44aboveevaluate the factor IE IE 7 17 d 35 exp122v 7 04 7 16 where v 7 02 because it is above the water table d 502 feel 2 16 feet 1 35 exp12202 7 04 7 16 1571 E 7 0912 49 Estimate the magnitude ofcenterpoint settlement ConvertE 7 13872 tsf 7 2 7 27744 ksf The magnitude of 39 39 s p 445 ksf 7 502 ft 7 0357 7 0842 7 0912 7 17 022 in s 7 p 7 0212 feet 7 12f t p 7 255 in 50 Estimate the displacementsat the foundation corners lf 39 1 E255 in 7 1273 in Problem Given the below recordingsfrom a seismic piezocone test evaluate the nonlinear loa 7 displacement7capacitv response ofa square footing that hasa width ofB 7 22 m The groundwater table is11 m deep and the footing rests0 3 m below grade The footing is 0 35 m thick and is constructed of reinforced concrete 1 Plot e SCPTu sounding using 4 s39de7bv7s39de graphs for each ofthe four readings with depth whattvpe of soil servesasthe foundation bearing material for this footing Erom evaluation ofthe graphs it can be assumed that the soil type is sand Thisconclusion isformed bv the qT graph showing valuesgreater than 5 MPa ateach velocitv must be used Because it is being found within a depth range of1SB the shear wave velocities between 0 and 1 SB 1522 33 m Now the unit weight can be found as follows y 7 ym 7 8631ong 7 1611ogz where z 7 3 3 m and the average shear wave velocities V5 7 171 06 ms y 7 ym 7 8631og171067 1611og33 kN VT 7541 1844 3 m To nd the friction angle the following equation must be used qr oz 7 176 11 log am To solve this equation the total vertical stress porewater pressure and effective vertical stress averagesfor the depth up to 158 7 3 3 m must be found The total vertical loverburdenl stress can be calculated asfol ows 0u0 VTdZ At 33 m where VT 7 1844 lltNm3 as found above avg 7 1844 w33 m avg 7 8 kPa The hvdrostatic porewater pressure can then be found sfollows no 7 z 2wyw where z 73 3 m 2w 7 11 m andvw 7 98 kNm 110 7 33 m 7 11 m98 w 7 2156 kPa The effective vertical stress can be calculated at a depth of33 m 0v0 0u0 140 ova 7 6084 7 2156Pa 928 kPa Now the friction angle can be calculated 4 o 7 176 111og qd6HlFlE1VZ sp E where 15028 kips 0027quot h q39 to33mwhichis597 MPa oam 7 1013 kPa and o39v 7 3928 We 597MPa 7 1000 kP Mpa V 175 nlog q 7 393 4 Use limit plasticitv to calculate the drained bearing capacity in order to nd the operational unit weight Vt for when the footing base depth isgreater than where the groundwater level lies use the following equation V PV 7541 Vw where the operational unit weight isnow setequal to the effective unit weight Referring to Table 1 the operational unit weight can e found at depth of33 m as shown 77 VSAT Vw 18 44 9 8 W V 7 V 7 ma kN y 77 y 7 864 3 To calculate the bearing capacity the bearing factor must rst be found in order to nd the bearing factor the rst step isto calculate the overburden bearing capacity Na Na a l anz45 3933 N 2 q entantaqaa anz455 q Next substitute Nq into the friction bearing capacity quation 7 w39 N 7 25856 1 tan3933 N 7 7 Now he bearing capacity factor equation isas follows N ngdev where B y 7 17 04 7 since we know we have a square footing where B 7 22 m 87A Therefore 9 7 1 7 04 y 7 06 also yd 1 Back to the bearing factorequation and substituting the above va ues N 7 yazyaNy N 7 0 1N V Where NV9762 N 7 0619762 Nquot 7 5857 Now that the bearing factor iscalculated the drained bearing capacity can be found asfollows quiz 537W 1 kN out 7 E 22 m 864 5857 qua 7 55648 kPa 5 Evaluate the initial small7strain stiffness Emax from the shear wave velocity data lagain within 158 depth range The shear wave velocity constant in the sand up to 155 733 m isequal to V5 717106 ms Thismust be converted to the equivalent low7slrain shear modulus 6mm pr where p 7 mass density ofthe material 7Vrg 5min g V 1844k A31 6 7 Tquot 17106 987 S s m 7 Now Gmax can be converted to equivalent small7strain elastic modulus EW 72 m 1v where u 7 Poisson sratio 7 02 for drained Em 7 2550480 kPa1 02 Em 7 1321160 kPa 6 Using the nonlinear hvbrid elasticitv 7 plasticitv solution estimate the load 7displacement response up to 80 mm e ection The equation for de ection for the hybrid elasticitv7 plasticitv approach is as follows


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