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Trang 2
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Pearson® is a registered trademark of Pearson plc
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Instructors of classes using Fennimore, Sustainable Facility Management: Operational Strategies for Today, 1/e
may reproduce material from the instructor’s manual for classroom use
10 9 8 7 6 5 4 3 2 1
ISBN-13: 978-0-13-511393-6
Trang 3(d) Eq.(2-15): 2.64=50/(62.4Vs); v,=0.304 ft3
Vw = (62 - 50)/62.4 = 0.192 ft3
v ;=0.56 - 0.304 =0.256 ft3
Eq (2-7): e = 0.256/0.304 = 0.84(e) Eq (2-8): n=[(0.064 + 0.192)/0.56)(100) = 45.7%
v;= 1 - 0.59=0.41 ft3
Eq.(2-7) e = 0.4 /0.59 = 0.69(c) Ww = (0.18)(100.4) = 18.1 ib
Vw=18.1/62.4 =0.29ft3
Eq.(2-9): S = (0.29/0.41)(100) = 70.7%
(2-6) Work with 1 ft3 of specimen
Trang 53(a) Eq (2-7): 0.56 =Vv/Vs; v ; =0.S6Vs
(b) Eq (2-15): G,=79.4/[{0.46){62.4)] =2.77(2-8) Work with 1 ft3of specimen
(c) Eq (2-15): Gs== 15.80/[(0.581){9.81) = 2.77(2-9) Work with 1 ft3ofspecimen
v;+ Vs=1; 0.94SVs+v,=1; v ,= 0.514 ft3
v;= 1- 0.514 ==0.486 ft3
Ww=(62.4}{0.486) = 30.3 Ib
Eq (2-15): 2.70=W,/[{0.514)(62.4)]; Ws= 86.6 Ib
Trang 80.140, or 0.319 ft3, must be added to each cubic foot Thisis {0.319)(62.4}, or 19.9 Ib of water.
{2-22} (a) Work with 1 ft3 ofspecimen
Trang 9CHAPTER 3
(3-1) (a) Po= (10)(120)/2000 = 0.600 ton/ft''
Eq (3-1): CN=0.77log10 (20/0.600) =1.17Ncorrected=(1.17)(26) =30
(b) 10 ft =3.048 m and 120 Ib/ft3 =18.85 kN/m3
Po=(3.048}{18.85) =57.45 kN/m2
Eq (3-3): Ncorrected=(26)(100/57.45)1/2 =34(3-2) (a) Po= [(8)(120) + (2)(120 - 62.4)]/2000 =0.538 ton/ft''
Eq (3-1): CN= 0.77log10 (20/0.538) = 1.21Ncorrected=(1.21}{26) =31
(b) Po=1.08 klps/ft''=51 71 kN/m2
Eq (3-3): Ncorrected=(26)(100/51.71)1/2 =36(3-3) (a) Po=(7)(20.40) = 142.8 kN/m2
Eq (3-2): CN =0.77 IOg10(1915/142.8) =0.868Ncorrected= (0.868)(22) = 19
(b) Eq (3-3): Ncorrected=(22)(100/142.8)1/2 =18(3-4) (a) Po= (2)(20.40) + (5)(20.40 - 9.81) = 93.75 kN/m2
Eq (3-2): eN=O.77loglO (1915/93.75) =1.01Ncorrected=(1.01)(22) = 22
(b) Eq (3-3): Ncorrected=(22)(100/93.75)1/2 = 23(3-5) Eq,(3-4): c=61/{{rr)[{4/12)2(8/12}{1/2) + (4/12)3(1/6)]} =449 Ib/ft2
From Fig 3-17 with PI = 40%, Il = 0.85 Hence, ccorrected= (0.85)(449) = 3821b/ft2
(3-6) A plot of time versus distance is given on page 8
slopeline 1=0.083/80 =0.001038slopeline 2=(0.093 - 0.08675)/(180 -100) =0.00007813
VI=reciprocal of slopeline 1=1/0.001038 =963 ft/secV2=reciprocal of slopeline 2=1/0.00007813 =12,800 ft/sec
L=82 ft (from plot on page 8)
Eq (3-6): HI=(82/2)[(12,800 - 963)/(12,800 +963)]1/2=38 ftWith VI = 963 ft/sec, according to Table 3-3, the subsurface material in the first layer is estimated to benormal sand or loose sand above the water table With V2= 12,800 ft/sec, according to Table 3-3, thesubsurface material in the second layer is estimated to be hard limestone, basalt, granite, or unweatheredgneiss
(3-7) Electrode Resistance Resistivity Cumulative
Trang 119Electrode spacing (column 1 in the preceding table) gives the approximate depth of subsurface materialincluded in a given measurement Resistivity (column 3 in the table) is computed from Eq (3-7), where D
is electrode spacing (column 1) and R is resistance (column 2) Hence, for the first row in the table,
p = (2)(n)(10)(12.73) = 800 ohm-ftValues in column 4 are cumulative resistivity values A plot of electrode spacing versus cumulativeresistivity is shown on page 10 From this plot, the thickness of the first soil layer is determined to beapproximately 63 ft Because the resistivity of the upper layer is in the range from 50 to 500 ohm-ft
according to Table 3-4, the subsurface material in this layer is estimated to be moist to dry silty and sandysoils Because the resistivity of the lower layer is in the range from 500 to 1000 ohm-ft according to Table3-4, the subsurface material in this layer is estimated to be well-fractured to slightly fractured bedrockwith moist-soil-filled cracks
Trang 13Chapter 4(4-1) Eq (2-11): 'y=[(3815 - 2050}/453.6J/(1/30) == 116.7Ib/fe
Eq.(4-1): 'Y d=116.7/(1 + 0.0 1) =107.0 Ib/fe
(4-2) : ':: : ': : : :-;';: ';'.: ;':.';: ;f(: ' : ':i":';:' : :: ! ' : _ _
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M ois t re Co n tent ( I)(4-3) 95% of the maximum dry unit weight =(0.95)(112.6) =107.0 Ib/fe
From the compaction curve of Problem 4-2, the range of water content most lkely to attain 95% or more
of the maximum dry u it weight is 9%to 19%
(4-4) From Table 4-3, with maximum dry unit weight =104.8 Iblfeand optimum moisture content =20.7%, the
possible type of soil for this sample is estimated to be A-7-6 clay or A-6 silty clay
(4-5) Weight of soil used in test hole =845 - 323 = 522 g
Volume of test hole ={522/453.6}/100 =0.0115 fe
Eq (2-11): 'Y ={648/453.6)/0.0115 = 124.2Ib/fe
Eq (4-1): 'Yd=124.2/(1 +0.16} =107.1Ib/fe
(4-6) Percent compactio achieved = (107.1/112.6)(100) =95.1%
(4-7) let subscript lib" denote soil in b rrow pit and subscript "f 'denote soil in the fill
O.4S{Vsl!+{Vsl =2500; (Vs}f=1724 m3Because (Vs)b={Vs)f,0.5952Vb =1724
(4-8) Eq (4-3): D = (0.5)[(20)(10)]1/2 = 7.1 m
11
Trang 16(6-1) Eq (6-2): j5=(20.72)(2.5) +(20.72 - 9.81)(6) +(19.69 - 9.81)(7) =186 kN/m2
Eq (6-3): ~w =(9.81)(6 + 7)=128 kN/m2
Eq (6-4): Plotal= 186+128 = 314 kN/m2(6-2) Eq (6-7): P=(3)(200,000)/{(2)(n)(15)2[1 +(0/15)2]s/2} = 4241b/ft2(6-3) Eq (6-7): P = (3)(200,000)/{(2)(n)(15)2[1 +(1O/15)2]s/2} = 1691b/ft2
P=(0.424)(4500) = 1908 Ib/ft2(b) z/a= 18/12 =1.5; ria =6/12 =0.5; Influence coefficient =0.374
P=(0.374)(4500) = 1683 Ib/ft2
(6-7) (a) z/a =3/(3/2) =2.0; ria =(3/2)/(3/2) =1.0; Influence coefficient =0.194
P=(0.194)(250) = 48.50 kN/m2(b) Overburden pressure at 3-m depth = (3)(16.38) = 49.14 kN/m2Total vertical pressure =48.50 +49.14 =97.64 kN/m2
(6-8) mz=12; z=15; m=12/15 =0.80
nz = 8; z = 15; n=8/15 =0.533Influence coefficient ==0.115
P = (0.115)(6000) = 690 Ib/ft2
(6-9) mz=6; z=25; m=6/25 =0.24
nz=6; z=25; n=6/25 =0.24Influence coefficient = 0.026
P=(4)(0.026)(5000) = 520 Ib/ft2(6-10) Divide the footing area into four equal parts, each 1 m by 1m For each part,
mz = 1; z= 4; m = 1/4 = 0.25
nz= 1; z=4; n=1/4=0.25Influence coefficient =0.027Net vertical pressure increment at base of footing =1000/[(2)(2)]- (16.80)(1.8) = 219.8 kN/m2
p=(4)(0.027)(219.8) =23.74 kN/m2(6-11) (a) Divide the area into three rectangular parts:
Part 1, upper left area, 16ft by 12 ftPart 2,lower left area, 16 ft by 12 ftPart 3, lower right area, 20 ft by 12 ftPart 1: mz=16; z = 24; m= 16/24 =0.667
nz= 12; z=24; n= 12/24 =0.500
Trang 17(b) Divide the area into three rectangular parts :
Trang 18(b) Below point J:
, 1'
,I
Influence coefficient =0.062
Influence coefficient =0.022
Area JHDE: same as Area FCHJ;
Trang 19The "e-log p" curve is shown on page 18.
Trang 20tie - log p " Cur.v e for Problem (1''. ' ')
Trang 21[p= 160 kN/m2] > [Po'=125 kN/m2]; Use Eq (7-22)
s, =(0.06)[3.8/(1 + 0.70)] log (125/108) + (0.30)[3.8/(1 + 0.70)] log (160/125) '" 0.080 rn, or 80 mm(7-9) Because the compressible clay layer is underlain by permeable sand and gravel, this is a double drainage
for clay layer, with H = (85 - 76)/2 = 4.5 ft
(1) When U = 10%, Tv = 0.0077 (from Fig 7-17)
Eq (7-29): tlO = [0.0077/(2.18 x 10-3)](4.5 x 12)2 = 10,300 min, or 0.020 yr
(2) When U=20%, Tv '" 0.0314
t20= [0 0314/(2.18 x 10-3)](4.5 x 12)2=42,001 min, or 0.080 yr
(3) When U = 30%, Tv = 0.0707
t30 = [0.0707/(2.18 x 10-3)](4.5 x 12)2=94,569 min, or0.180 yr(4) When U=40%, Tv=0.126
t40 = [0_126/(2.18 x 10-3)](4.5 x 12)2= 168,539 min, or 0.321 yr(5) When U=50%, Tv=0.196
t50 = [0.196/(2.18 x 1O-3)]{4.5 x 12)2 = 262,172 min, or 0.499 yr(6) When U = 60%, Tv = 0.286
t60 = [0.286/(2.18 x 1O-3}](4.5 x 12}2 = 382,558 min, or 0.728 yr(7) When U = 70%, Tv = 0.403
t70 = [0.403/(2.18 x 10-3)](4.5 x 12)2 = 539,059 min, or 1.026 yr(8) When U = 80%, Tv = 0.567
t80 = [0.567/(2.18 x 10-3)](4.5 x 1d = 758,428 min, or 1.443 yr(9) When U = 90%, Tv = 0.848
t90 = [0.848/(2.18 x 10-3)](4.5 x 12)2= 1,134,297 min, or 2.158 yr
U (fraction of total settlement)
(%)102030405060708090100
Settlement(in.)
0.370.73
1.10
1.46
1.83
2.202.562.933.293.66
The time-versus-settlement curve is given on page 20
Time(yr)0.0200.080
0.180
0.321
0.4990.728
1.026
1.4432.15800
(7-10) P =Po+ t.p =108 + 52=160 kN/m2
[p = 160 kN/m2] < [Po'= 185 kN/m2]; Use Eq (7-21)
Sc=(0.06)[3.8/(1 + 0.70)] log (160/108) =0.023 m, or 23 mm(7-11) (1) When U = 90%, Tv = 0.848 (from Fig 7-17)
H = 12 ft (single drainage)
Eq (7-29): t90 = [0.848/(9.04 x10-4)](12 x 12)2 = 19,451,469 min, or 37 yr(2) 1yr = 525,600 min
Eq (7-29): 525,600 =[Tv/(9.04 xlO-4)](12 x 12)2
Tv = 0.023From Fig 7-17, with Tv=0.023, U=15%
Settlement at 1 yr = (0_15)(4.60) = 0.69 in
Trang 22A
-I 0
Trang 23For a depth of 5 ft, avera e corrected N-value =28.
For a depth of 7.5 ft, average corrected N-value = {28 + 27)/2 = 28
For a depth of 10 ft, average corrected N-value = {28 + 27 + 30)/3 =28
For a depth of 12.5 ft, averag corrected N-value ={28 +27 +30+28)/4 =28
For a depth of 15 ft, average correcte N-value =(28 +27+30 +28+23)/5 =27
be multiplied byXB. Therefore, in this problem, q = 3.06 tons/ft" will produce (1)(1.062) in of settlement
Therefore, 3.06/1.062 =qallowableforloin.settiemenJ1qallowableforloin.settlemen=t2.88 tons/ft''
(7-17) For Eq (7-33),
Ap =800/[(2}(2)] - (17.5)(1) =182.5 kN/m2
Cl=1- (0.5)(17.5/182.5) =0.952(2 = 1+0.2 log [(10)(4)] = 1.32
table on page 22)
Eq (7-33): St(4yr)=(0.952){1.32)(182.5)(lo.5 x 10.5) = 0.024 m, or 24 mm
Trang 24*Column headings are:
A layer thickness (Az), m
B Depth from base of footing to center of layer, m
C Modulus of elasticity of sand layer (Es),kN/m2
D Strain influencefactor for soilzonezdepth below foundation (obtained from graph
above) (lz), dimensionless
E (lJEs)(Az), m3/kN
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