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# EXAM#2 - CEE 3413 — SOIL MECHANICS SUMMER 2009 Department...

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Unformatted text preview: CEE 3413 — SOIL MECHANICS, SUMMER 2009 Department of Civil and Environmental Engineering Mississippi State University HOMEWORK 3 — Due Thursday, July 23 1. The plan ofa mat foundation is shown below. The uniformly distributed load on the foundation is 1800 psf. Determine the increase of vertical stress (A0) at depths of 5, 10, and 15 feet below the ground surface beneath points A, B, and C. I 10ft l 2. If the building in problem #1 was constructed above the soil profile shown below, calculate the total amount of consolidation settlement of the clay layers. Calculate the total settlement beneath points A and B. Clay yt = 110 pcf CEC = 0.34, CBC: 0.033 Normaliy Consolidated 5ft Clay ' vsat = 120 pcf I CE: = 0.40, C39: 0.043 At a depth of 10 ft: Pp = 840 psf 15 ft At a depth of 15 ft: F'p = 1300 psf CE 3413 Homework 8 continued 3. A consolidation test is performed on an undisturbed sample of soil and the following results are obtained: Effective Void Ratio stress (psf) Initial 0.800 100 0.796 250 0.787 500 0.78 1000 0.775 2000 , 0.728 4000 0.622 8000 0.528 16000 0.440 32000 0.355 64000 0.267 (a) Plot the curve of void ratio vs. effective stress on semi—log paper. (y—axis: linear; x-axis: log scale) (b) Compute the compression index (Cc) and the recompression index (Cr). Compute the compression ratio (CBC) and the recompression ratio (Car). (c) Estimate the preconsolidation pressure of the soil sample. (d) lfthe soil stratum in the field is 10—ft thick and the initial effective stress at the center ofthe layer is 900 psf, how much settlement will occur if a wide embankment is constructed that produces an increase in stress of 2500 psf? 4. Problem 10.8 in your textbook (page 368). 5. Problem 10.9 in your textbook (page 369). TREADWELL & ROLLO, INC. SHEELL 0F 4 Consulting Engineers and Scientists - JOB NO. ' DATE PROJECT W3 “d“ﬂﬂkﬂjm, . 13:39:66 COMPUTED BY SUBJECT CHECKED BY as .,m.m.tm.ewww_mnwwr C49 1% (D ,7 ex 6 4' 5 eﬁ’ 5523:“) mm (2m (gs-em? % AW“: mew (mad? as 22155 [mfg + Air-=2 WA F39 + Ag“? [44% WP + TREADWELL & ROLLO, INC. SHEET£0F L Consulting Engineers and Scientists JOB NO. ‘ DATE PROJECT M COMPUTED BY SUBJECT M CHECKED BY 1: 3’2. m? m n 5.35 .44. K7 2; W E: )0 Je. 1’0 1;; to 3 TE: «a: 3 g; (g M 33 E rE r} 3% H ' F TREADWELL & ROLLO, INC, SHEET Z; 0F 4 Consulting Engineers and Scientists JOB NO. DATE PROJECT COMPUTED BY SUBJECT CHECKED BY } r "3"” 0-» ZDM AW?“ (0,2344%)U50032 “‘5‘ ”Rb/”7‘12“ V53 «m (minus? } ”7",“: 0.451; MT Calaﬁpmﬁmﬁ «x- 2%? £1,917 SC, 1 g) [0/1 {3}” "E- QE’)C310+¢¢"§:> +65 >020 =(02»/~{ 17‘9”???)1’ 5%55 3%:- l M0 1 @ ,5”: (33’? 6%? gym? w’ra’omwbaém “3 HM EMF. Mfg/ﬂ \_ . Ema?» mwﬂafmfx EWWM% WIN“; A “‘1’ 8 Gm g A . WWW‘.~54.mmm”A,mn,,imamlpmmkwmw”4“,“.W.H,..,-_..,.‘..,,u.,,mw§, ﬂ A --- = ,r’ 5 Mam am , 7 7 , \$qu «Qmwﬁ QM”: m9? @Erxz sﬂﬁ V \$me va GE T E Eh @mv W Hug \Wermmxlmﬁ \$qu Lita i5 1 E,“ 23.5, W": ‘5; 1- ‘2) 3 a Z" ' 31: "1‘ 9’2”? AG"? (7,2,4 (WW? 11‘) "\$431,. F543 M": 2.23: “243 i I 5' + 3 ﬁw7E> “’"ﬁﬁgiﬂM"? x SD (745' AG”? 2,779 F333;. 3 1/5 3 :39; “-3 I 253 l "7 __:_, ’ZJrL—u— / f: m: 0 m.“ 2+t75’ W)": + (7,207 3‘ i {”73 3 1:7 ﬂngéﬁ [SW-"'3 115”. 2/ 393 3:“ 3 Wmmummmmm Aim KHZ-{é “a"; ﬂﬁﬂvgv\$v5+~ ﬂit-41.; +3 3:: “\$4? ﬂ-Iﬂsﬁi + 0.2:,ng A: 3%?! V33: Aw? W20 {:0 ,075+ﬂ.0§r7+ a0”??? + 0.04%?) 43¢“? 54%.4} FE]: /\ ﬁmmw mmmhwm mﬁwwmﬁm 889 809 89 “Na “\$.35 R m. p g E ﬂ“ Q n f J6 m f? F f? C700 +£5095) I!!! l l BMWMW.H_.._N _w._,..ug_ir.,ﬁ,, .. .M £1“ .. .. 2. rm...“ W...W....mw 55»; [M 5.5: 2,5 / . Q: A"? CV? i551) éikgﬁai‘WfCﬁéTjExJ-CC/ '3 (f)i’§’}>(i% ‘1? /F‘/;§5 "“5" (32.55" (/34?) (:3 }‘ C5" 3 3.... 2 56 5/5/ § 5755/0565 Liz/59w? 555.43 7% Mil/{EE/ fo/(‘jFT-"QLAEZX A735 35%;?” 81555-4 7" W 9/" C’ "Chirp-"f EC- " IVA CCCMUCH4’ " “:jmmmmmmg { V1,] ﬂashy? Elf ~ w l l’4C"'Cﬂz/5 : 551,657?! 55%.» \ ‘ "w-‘V'hwvu. gm} 3;; ‘ - ‘v ./ a; mﬂpcv“l(4£“=~2>wﬂzw\ \ u, (:5 m 029’?) 5 ca «w ~ ----- Wm man/3;, _ - 4— 5563.7 5 5 5 .71., \ Ami}. ‘x r. r / __,. 7--—— ‘. . / W' __ \— _/7. .. a . 5 .. "L" (I 5351/sz CC” 5W"£‘)§ C“ ffiﬁ‘J .m 4.22.55 m m?» ,5 Aaémegp _ 5+€Meylmwwx1 .. , 13 f1 v / “‘ Z) r’m 3 (j p w“ < .14: DCIWQ> “i“ CW >Cﬂﬂlzrﬂ7lfﬂ> "Jr () f/ 1:53p} :7 c» .. J r ”mint” h .1 1.;25; I} m... 0’ y/f’D4 4 “:3“ I “magi/7 ...
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