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Unformatted text preview: 60 L011 owl CE 467L Geotechnical Engineering
Spring 2010
Homework 4 Due: Tuesday, F ebruaty 23, 2010 Read Section 2.4 prior to completing these three problems. Show all your work clearly. I. For the darn shown below (k = 5 x 10'4 cm/s), calculate the seepage beneath the dam, q, per day,
and the pore water pressures at points F and G. 2. For the following flow conditions (k — 2 x 10'3 cm/s) ﬁnd:
a. The flow q per hour beneath the dam b. The pore water pressure at point C. 0. Assuming your answer for (b) represents the average pore water pressure across the base of
the dam, calculate the resultant upward force acting on the dam (per m out of the page).
Assuming mm = 24 kN/m3, approximate the downward force (weight) exerted by the concrete dam (per m of dam out of page). Is the dam is stable from uplift? w .0 t 3m“ ‘e' C“ ' ‘f I Meow.
‘llIlI’.
.IIIIIII‘ , . 3. Complete Problem 2.5 in Craig lHPEldIOU‘Za fra— 5x10"! W/S .— ﬁxzo‘bM/QC
ﬁj: ELL. ELF Alng
Nd :IZ
LN, : Gm
=5IO’5’V (Q, 34 )_ 5 3 3600
Cf 2‘ 5'20 M) [Z M ‘ 52740 m/S‘aC Y “in ﬁ24j1f
“ cw:
05 = 1.7% *2” W m 516 AW
Ah: I 5 T. LE... — 22: : "m
SS PM 9W dmp Ak w mama 05 (MP
19’“ 1“ Mdm = 2.5 $VOPS
L‘F; k ~ NANA/x = (m  25695) s 4.75 m
ha: ‘— "qlwv (Muma kw, 7 H.76»(4.2) = Ms M
W :XWWF ‘ q'glkfggﬁqs "‘5 “i 37.? “ﬁg,
m
Fe. 6 I, MdW—Lq‘g
1% = 6M — (15(05): 1.25%
has : “L‘IJZM’ hm : 1.1544(1) = 5.45 m {AG’[Ciszt’d‘t’gaéfﬁﬁwb = 535 RB J2. hv ZNOFSCM/ $244105 9‘75“ SEC
[Al—v" 26"" '7’“ =" N4: 5: 5
MA = {2
1"L. .. [M
k : —— ___ =
A Ma [LAVO‘VS ‘53 4MP
_ M  3 
as a5 M71:  in = WW v 3
WW “b dw~~>
5) Nam=55
h; —— 1m  MAMA/x  IOl  5.50.52) = 10.3! M
L321 " F[Om (dCJ‘ULh/x @ 6&0.)
hpc 3 [Calm  (4qu 7— 2054%
(A; = m 1A” s 4.441(203D = m LL?“
W
C) Fu?: uc Rim = = Emmi Awrmﬁwﬂ dam/v» ﬁd’v‘avx 06;. 0. 4171132108, '« VOL : {2(a) HMGOXD
3 1250 W3 ﬁgoml [2' H M (90 w»
PM i924 wrﬂﬂlwo m3) 7 201,520 m FW>FUP so down \‘g SEEM; ...
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 Spring '10
 Rechenmacher
 Geotechnical Engineering, Trigraph, Pore water pressure

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