Homework _3 Solutions - 6.4 A column carrying a vertical...

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Unformatted text preview: 6.4 A column carrying a vertical downward dead load and live load of ISO k and l20 k, respectively, is to be supported ona 3-l‘t deep square spread footing. The soil beneath this footing is an undrained clay withs" = 3000 lb/ft‘ and y = 7‘lb/tt The groundwater table is below the bottom of the footing. Compute the width B reqmred to obtain a factor of safety 0 3 against a bearing capacity failure. Solution No:- sn . Na = llo (r1591: 3 qwr .,. l, 2,39,.) gm) —r can) gums :1 277:2; _ I'voo') meow-r 1' ‘o q- 5, 't , 3803) 2,700» ‘k— c v #413» E $ 17' 57‘ _, L1 0090 a 3 8 6.6 A llS-m wide. 2.5-m long, 0.5-m deep spread footing is underlain by a soil with c' = [0 kPa, ¢' = 32". y = l8.8 kN/m" The groundwater table is at a great depth. Compute the maximum load this footing can support while maintaining a factor of safety of 2.5 against a bearing capacity failure. Solution , V151 Ne 53M" “(K-13+ “‘6‘ “A Guy"— (Imflod’) " ‘lihkl’a I. ’ 1 .L ' —- - ‘ Sc, 2/1 C’:_J)(§%): )=|.3‘\ 2*” 1;)1N31- ‘ ['57 9’“ l'u‘l (35;)‘0-7é k 7. °'/:_ :0, At 3/71). (1(0.3JJ=(.[§ d1“. (Tl{"3}_)l‘r~37~ (lle3L3L=/'°7 d rvl has! ‘1‘” “033053)(l5‘Xl'91(“)W'WI’HXI-oqh 0&0 MM l-i‘) ( 30») (mo (.1 7 )Cpa . .1 )ls‘t 14 :5, :o (“511% wk =-(l.§)C1.y){u'y)11g) A.qu Pdth ( mt) ' ° 4 {Pent}ch Coduro — Foundation Design: Principles and Practices/2e Solutions Manual 6.14 Conduct a_bearing capacity analysis on the Fargo Grain Elevator (see sidebar) and back-calculate the average undrained shear strength of the soil. The groundwater table is at a depth of 6 ft below the ground surface. Soil strata A and B have unit weights ofl l0 lb/ft‘; stratum D has 95 lb/ft'. The unit weight ofstratum C is unknown. Assume that the load on the foundation acted through the centroid of the mat. Solution Use t-«m’ Wit-0 ?$>O (Uhdfuiwq) («9:65) qul 3%. M .. w (W) «4&7 w ?'4 V20 61% Q'HFto' ‘Q7rbi‘467‘o6u7 Wm— qwf -. 5., (5,7) 1. (C(67\</) ~, 51,7 7.4 A column that carries a vertical downward load of l20 k is supported on a 5-ft square, 2-ft deep spread footing. The soil below has a unit weight of l 24 lb/ft" above the groundwater table and l27 lb/ft' below. The groundwater table is at a depth of 8 ft bel0w the ground surface. a. Devel0p a plot of the initial vertical stress. om]. (i.e., the stress present before the footing was built) vs. depth from the ground surface to a depth of IS ft below the ground surface. b. Using the simplified method, develop a plot oan: vs. depth below the center ofthe bottom ofthe footing and plot it on the diagram developed in part a. c. Using Newmark‘s integration of Boussinesq's method, compute A0: at depths of2 ft and 5 ft below the center of the bottom of the footing and plot them on the diagram. 120 + 14’ onion V: f - " Slm “0’6..- SH wt: (s)(>')(t)(ln>)- ZSl— q 5‘ g Dep’fi‘ 53°, .3 {Ag—e— RT %;2. Dwuh lm}; fibu/ Quadmnb 3L1 LL ’HL’W‘ 310/11" ‘(H’ —— UK— 3»: 7.1’ IS: 0.100 A515: 4C0.Loo)(91w.— liq) —_-_ U’IFT‘ I PH“ 14' S’ BHUTW’L hr ELLVH‘ = l-ré ’ me am 14/ n» 0.0%! 059» (would) (QN- Dr?)= Coduto — Foundation Design: Principles and Practices/2e Solutions Manual 7.6 A l.0-m square, 0.5—m deep footing carries a downward load of 200 kN. It is underlain by an overconsolidated clay (OC case I) with the following engineering properties: C = 0.20, C, = 0.05, e = 0.7. and y = 15.0 kN/m‘ above the groundwater table and 16.0 kN/m" below. The groundwater table is at a depth of 1.0 m below the ground surface. The secondary compression settlement is negligible. Using the classical method with hand computations, determine the total settlement of this footing. Solution Wt: (1..)(t»)(mt)(2s,s)= [Z lat-J q; 200:” -° = 2”. we I c514: 0;)(o.y) : 7 LP. Lat,“ i Ezs’ @i A61, (-33! (f/d‘f¢) “#1. i \ 0.5 H 015 l‘ié @011 0.01% 7, 1—0 .3 mm 47 as l 20 3 2.0 27 3"” Z, q: / 5L you” 7.8 Solve Problem 7.6 using Skempton and Bjerrum‘s method with 5,, = 200 kPa. Solution gm? (ammo) (o.q7)(o.7) 7 ZMM 6600” V” «M Coduto— Foundation Design: Principles and Practices/2e Solutions Manual Note: All depths for CPT and SPT data are measured from the ground surface. 7.14 A 250-k column load is to be supported on a 9 ft X 9 ft square footing embedded 2 ft below the ground surface. The underlying soil is a silty sand with an average N“, of32 and a unit weight of l 29 lb/ft‘. The groundwater table is at a depth of 35 ft. Using the simplified Schmertmann method, compute the settlement ofthis footing at l = 50 yr. Solution fie” 1‘40?» C 3-7?) ‘-' ’33 099 “’15:”- Gw'1 (1)024) : ZSX w/F—f" W = («)(«Mzmwc 29.3% w a; 1pIQGOTzH,]°° Cl. :Sjgg W161 c e , 0.! 3—51- - I l I (3386 _Lfl ,OKM) (.Lc. lfQL LamJ (2%); (.1, ¢ I cart. (613029" “8 ,___. Iep —- arm-(Y "NV'ZW 30.6” 5’?! 5, .loc‘tkrl(l.g.)(t) (33y; -—2m ( 0,5531: O.\]d’)(CU /3<:Np " °'13l '4’ ES - gzpcooJTr llowrélfi 93v goo HI: . lnnl _.| 1,,J' . u . ..... 1| H a) Y ’9 Home Insert Page Layout Formulas Data Review View Get Started Acrobat Schnter‘tntann.x|s [Read-Only] [CC-moat A "—" * Cut ‘ Arial v 10 v IA“ .t'l IE E EH? | j '.Ia|.1T-Ext aaCom' __ = = _ P _ P375“? jFormatPaimer IE I Q II_ II .2 .1 | I; g at [large e.-.enter | $ ' '-j.-'.;. Clipboard '7- Font Alignment Hun‘ =13 v fir A El C D E F G H 1 SETTLEMENTANALYSIS 0F SHALLOW FOUNDATIONS 2 Schmertmann Method 3 4 Date October 13 2003 5 Identification Problem? 1? B ? Input Results 8 Units E E or SI 3 Shape SO 30 CI (30 or RE = 3386 |lJ.-"ft"2 11] = 3 ft delta = l] 83 in 11 = ft 12 = 2 ft 13 F' = 250 k 14 Dw = 35 ft 15 gamma = 123 |b.-"fi"3 1B t= 50 yr 1? 13 l:-' 21] Depth to Soil Lager 21 Top Bottom ES 21 | epsilon strain delta 22 ijt': 18': [lb-"W2: 1ft: 1%: tin: 23 01] 2 C 24 21] 3E 434E££ 05 £188 £1?ISB [£212 25 3t] 4 C 43-—££C 15 C 238 [3168 [C388 28 4 fl 5 E 434E££ 2 5 E 431] E 45?1 [£543 2? 51] BE 434EEC 35 C581 E53?4 EC?1? 28 51] ?E 43-—CCC 45 C1334 C?380 [£886 23 ? fl 8 E 434E££ 5 5 [I342 [8833 [£820 3t] 81] 3C 43-—CCC 85 C 531 £828? CC?54 31 3 fl 10E 434E££ ? 5 E 533 £5?4U [£683 32 101] 11 E 434EEC 85 [488 C 5133 [0323 33 111] 12E 43-—CCC 35 C43? [4134? [£558 34 121] 13E 434E££ 105 E 385 [4100 [£432 35 13 i] 14 C 43-—CCC 115 C 334 £3553 [[4213 38 14 U 15 E 434E££ 12 5 E 283 £301]? [£361 3? 151] 18E 434EEC 135 C 231 [2460 [£235 38 1131] 1?E 43-—CCC 145 [181] C 1313 [£230 33 1?0 18E 434E££ 155 E 128 £136? [£164 41] 181] 13C 43-—CCC 185 (0?? £0820 [£1138 41 13 fl 20E 434E££ 1? 5 £028 £02?3 [£033 42 201] 21 E 434EEC 185 £001] £0000 [[1100 7.22 A steel frame otfice burlding with no diagonal bracing will be supported on spread footings founded in a natural clay. The computed total settlement of these footings is 20 mm. Compute the differential settlement. Solution 997 O~K(Z°)" 'E’MW‘ ...
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This note was uploaded on 11/29/2010 for the course CEE 452 taught by Professor Zapata during the Spring '10 term at University of Arizona- Tucson.

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Homework _3 Solutions - 6.4 A column carrying a vertical...

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