Geotechnical Engineering

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Chapter 12

Text of Geotechnical Engineering

  • Solution 12.1

    The Mohr total stress circle for no-tension limit is as shown above.

    20z

    s satu=

    sat

    ucr

    s2zz ==

    su

  • Solution 12.2 From Eq. (12.28), ( )ax = su qs2z + If ( ) ax 0 then crus zs2q If no tension crack develops, crz = 0, then us s2q

  • Solution 12.3

    (a) ' ucr '4s 4 15H 5.9m

    (20 9.8)= = =

    (b) Derivation similar to derived equations (12.31) and (12.32) i.e. ' ucr '4sH =

    If the trench is filled with bentonite slurry of unit weight f then ' sat f = ' ucr

    sat f

    4sH =

    (c) If the trench is filled with water ' ucr water '4s 4 15(H ) 5.9m

    (20 9.8)= = =

    If the trench is filled with bentonite ' ucr bentonitesat f

    4s 4 15(H ) 6.7m(20 11)

    = = = Additional depth of excavation by using bentonite = 6.7 -5.9 = 0.8 m

  • Solution 12.4 Active Passive Depth ' ' Ka Lateral

    Pressure Water Depth Kp Lateral

    Pressure Water

    m kN/m3 kN/m3 kPa kPa m kPa kPa 0 0 4+ 17 17 20 0.49 .49 x 17 x 4 = 33.3 4- 0.33 .33 x 17 x 4 = 22.4 0 7+ 18 18 30 0.33 .33 x 18 x 3 + 22.4 =

    40.2 0

    7- 0.38 .38 x (17 x 4 + 18 x 3) = 46.4

    0 7 0 0

    12 17.8 8 27 0.38 .38 x 8 x 5 + 46.4 = 61.6

    49 12 2.66

    2.66 x 8 x 5 = 106.4

    49

  • Solution 12.5 All forces are per m length of wall. Determine aK

    Rankine: 31

    23045tan

    245tanK,0 2cs2aR =

    =

    ==

    Coulomb: 0,0,,20 cs ==== and from equation (12.16) From computer program utility on CD, aK = 0.3 Determine lateral forces Smooth wall: use Rankines method

    Rankine: 2 21 1 1 18 8 1922 2 3aR aR

    P K H kN= = =

    Surcharge kPa15qs = ; kPa53115qK saRx ===

    Surcharge force = kN40m8P xs == Total lateral force = 192 40 232aR aR sF P P kN= + = + = Rough wall: Use Coulombs method.

    Coulomb: 2 2ac ac1 1P = K H = 0.3 18 8 = 172.8 kN2 2

    Surcharge =sP 15 0.3 8 = 36kN Total lateral force = acF = 172.8 + 36 = 208.8 kN Horizontal component of ( )caxac F:F = acF cos= 208.8cos20=196.2kN Vertical component of ( )cazac F:F = acF sin=208.8sin20=71.4kN

  • Solution 12.6 All forces are per m length of wall.

    Use Coulombs method. Determine aCK using the computer program utility at www.wiley.com/college/budhu All forces are per m length of wall.

    48.0K aC = for cs = 30o, =15, =10, = 20 Before flood:

    2aC a

    1P = K H2

    = 21 0.48 17.5 5 105kN/m2

    = Horizontal component: PaC cos (20 + 10) = 90.9 kN After flood:

    Soil: 2aC a1P K H2

    = = 21 0.48 (17.5 9.8) 5 46.2kN2

    =

    Water: 2 2aC w1 1P H 9.8 5 122.5kN2 2

    = = = = Horizontal component from soil: PaC cos ( + ) = 46.2 cos (20 +10) = 40 kN Horizontal component from water: PaC cos () = 122.5 cos (10) = 120.6 kN Total horizontal component = 40 + 120.6 = 160.6 kN

    15o

    20o

    10o

    PaC (soil)

    10o

    Water pressure

  • Solution 12.7 No wall friction

    With wall friction

  • Solution 12.8

    (a) = 18 3mkN , = 30cs , = 15 , = = 0,0 From computer program utility: 3.0K aC = The horizontal component is KaC x cos (15) = 0.29 (b) =18 3mkN = 30cs , === 306090,15 , = 0 From computer program utility: aCK = 0.686 The horizontal component is KaC x cos (45) = 0.485 Therefore (b) gives the larger horizontal force

  • Solution 12.9

    Determine the stability of the concrete gravity wall Backfill: Coarse-grained soil:

    3concrete

    3cs mkN24,15,mkN18,32 ====

    Base soil: Coarse-grained soil:

    === 20,mkN19,30 b3cs Step 1: Determine aCK From computer program utility on CD: 32.0K aC = Step 2: Determine lateral forces

    kN1.4641832.021HK

    21P 22aCaC === ;

    aCP acts at an angle = 15 to the horizontal Horizontal component of ( ) kN5.4415cos1.46cosPP:P aCcaxaC === Vertical component of ( ) kN9.1115sin1.46sinPP:P accazaC === Surcharge Fx = KaCqsH cos = 0.32 10 4cos15 =12.4 kNFz = KaCqsH sin = 0.32 10 4sin15 = 3.3 kNTotal horizontal force

    Rx = 44.5 +12.4 = 56.9 kN

    Step 3: Determine wall stability

  • Solution 12.10

    Step 1: Determine the active lateral force per meter and its location. From computer program utility

    33.0K aC = Thickness of wall at top of base+ 0.5 + 6/20 = 0.8 m H = 6m + 1m + 2.5 tan 8 = 7.35m Soil mass:

    Lateral force from soil mass: Pac = 12 KaC satH2 = 1

    2 0.33 18.5 7.352 =164.9 kN

    Horizontal component: Pax = PaC cos =164.9 cos 20 = 155 kN Vertical component: Paz = PaC sin = 164.9sin 20 = 56.4 kN Step 2: Determine the resultant vertical for per unit length and its location. A table is useful to keep the calculation tidy and easy to check

  • Part Force moment arm

    from toe(m) moment + (kN/m)

    1 0.5 2.5 0.35 18.5 = 8.1 3.67 29.7 2 5.2775.180.65.2 = 3.25 901.9 3 72240.65.0 = 1.75 126 4 6.212463.05.0 = 1.4 30.2 5 108245.41 = 2.25 243 Paz 56.4 4.5 253.4 Rz = 543.6 (+) = 1584.2axP 155.0 2.45 (-) 379.8 Mo = 1204.4

    Location of resultant vertical component of force from toe is:

    x = 1204.4543.6

    = 2.22m Step 3: Determine eccentricity

    e = x2B = 4.5

    2 2.22 = 0.03m

    Step 4: Determine stability

    Rotation: e75.065.4

    6B >== ; therefore rotation is satisfactory

    Translation. T = Rz tan b = 543.6 tan 20 = 197.8 kN FS( )T = TPax =

    197.8

    155.0= 1.3 < 1.5; unsatisfactory

    Bearing capacity

    +=Be61

    AR

    )( zmaxz = 543.6

    4.5 1 1+6 0.03

    4.5

    = 125.6kPa

    Short term loading qult = 5.14su = 5.14 x 94 = 483.2kPa

    FS( )B = qult( z)max =483.4

    125.6= 3.8 > 3 ; therefore bearing capacity is satisfactory for short term loading.

    Long term loading For p = 30, N = 16.1 (assuming a layer of granular material will be placed between the base and the foundation soil) i = 0.37

    ult1 1q B N i 19 4.44 16.1 0.37 251 kPa2 2

    = = = FS( )B = qult( z)max =

    251

    125.64= 2 < 3; therefore bearing capacity is not satisfactory for long term loading.

  • A drainage system to prevent build up of pore water pressures behind the wall is shown.

  • Solution 12.11

    Step 1: Determine the active lateral force per meter and its location From computer program utility: 314.0K aC = Before rainfall Lateral force from soil mass

    kN3.7059.17314.021HK

    21P 22sataCac ===

    Horizontal component: kN2.6814cos3.70cosPF acax === Vertical component: az acF P sin 70.3sin14 17kN= = = Surcharge:

    kN6.714cos55314.0cosHqKF saCx === kN9.114sin55314.0sinHqKF saCz ===

    Vertical force from surcharge = 5 x 4.1 = 20.5 kN Paz = 17 + 1.9 = 18.9 kN Pax = 68.2+ 7.6 = 75.8 kN

    4

  • Step 2: Determine the resultant vertical force per unit length and its locations.

    Part Force ( )mkN Moment arm from toe (m)

    Moment( )mkNm+ 1 6.3379.176.41.4 = 2.45 827.1 2 48244.05 = 0.2 9.6 3 4.39244.01.4 = 2.45 96.5 4 5 x 4.1 = 20.5 2.45 50.2

    Paz 18.9 4.5 85.1

    zR 464.4= Resisting moment=

    1068.7

    axP Soil: 68.2

    Surcharge: 7.6

    35

    25

    113.7

    19

    disturbing moment =

    132.7

    Location of resultant vertical component of force from toe is

    o

    z

    M 1068.5 132.7x 2 mR 464.4

    = = = Step 3: Determine eccentricty

    B 4.5e x 2 0.25z 2

    = = = m Step 4: Determine stability

    Rotation: ,e75.065.4

    6B >== therefore rotation is satisfactory

    Translation

    z bR tan 464.4 tan 20 169 kN = = =

    ( )ax

    169FS 2.2 1.5,P 75.8= = = > therefore, translation is satisfactory

    Bearing capacity

    zmax

    R 6e 464.4 6 0.251 1 138 kPaA B 4.5 4.5

    = + = + =

  • For = 32cs , N 22.5 =

    n 1 3ax

    z

    P 75.8i (1 ) (1 ) 0.59R 464.4

    + = = =

    B B 2e 4.5 2 0.25 4 = = = m ult

    1 1q B N i 19 4 22.5 0.59 504 kPa2 2

    = = =

    ( ) ultBmax

    q 504FS 3.7 3138

    = = = > , therefore, bearing capacity is satisfactory After rain fall: Step 1: Determine the active lateral force and its location H1 = 1.5m, H2 = 3.5m

    Lateral force from soil mass

    22aC21sataC

    21sataCac HK2

    1HHKHK21P ++= =

    kN4.516.155.293.6)5.3()8.99.17(314.0215.35.1)9.17(314.05.1)9.17(314.0

    21 22 =++=++

    Water force = kN2.60)5.3(8.921 2 =

    Horizontal component: kN5014cos4.51Fax == Vertical component: kN4.1214sin4.51Faz ==

    kN8.1176.72.6050FkN3.149.14.12F

    ax

    az

    =++==+=

    Step 2: Determine the resultant vertical force per unit length and its location The lateral stresses shown on the right of the figure above have been multiplied by cos (14) because we are using Coulombs analysis.

    4

    5

    6

    7

    8

    910

    water

    1.52 8.2 8.6 kPa

  • Part Force Moment arm from toe(m)

    Moment ( )mkNm 1 (4.1 x 1.5 x 17.9) + [4.1 x

    3.1 x (17.9 9.8)] = 213 2.45 521.9

    2 48 0.2 9.6 3 ( ) 3.238.9244.01.4 = * 2.45 57.1 4 5 x 4.1 = 20.5 2.25 50.2 5 0.5 x 9.8 x 12 = 4.9 1/3 16.3

    Paz 14.3 4.5 64.4 =zR 319.1 Resisting moment =

    735.8 6 1.52 x 5 = 7.6 2.5 19 7 0.5 x 8.2 x 1.5 = 6.2 4.25 26.4 8 8.2 x 3.5 = 28.9 1.75 50.6 9 0.5 x 8.6 x 3.5 =15 1.17 17.6

    10 0.5 x 9.8 x 3.52 = 60 1.17 70.2 Disturbing moment =

    =183.8

    *Although seepage effects are neglected, this part of the base in completely bouyant. Location of resultant vertical component of force from toe is

    o

    z

    M 735.8 183.8x 1.73R 319.1

    = = = Step 3: Determine eccentricity

    B 4.5e x 1.73 0.522 2

    = = = m Step 4: Determine stability

    Rotation: 75.065.4

    6B == > e, therefore rotation is satisfactory

    Translation

    z bR tan 319.1 tan 20 116kN = = = kN ( )

    ax

    T 116FS 1 1.5F (117.8 4.9)

    = = =

  • ( ) ultBmax

    q 97FS 0.8 3120

    = = =

  • Solution 12.12

    Coarse grained soil: 0,9.0S,mkN17,29 3cs ==== Determine embedment depth, d, and maxM using FSM

    156.19.0735.1

    735.17.2

    S

    Ge,

    e1SeG

    w

    wS

    ws =

    =

    =

    ++=

    3w

    ssat mkN53.178.9156.11

    156.17.2e1eG =

    ++=

    ++=

    3wsat mkN73.7)8.953.17( ===

    === 2.2325.1

    29Fs

    csdesign ;

    From Appendix C: 3.2K,45.0K xax Hydrostatic pressures cancel due to even water level

    Part Pressure Force Moment Arm Moment 1 65.711745. = 83.3165.75. = ( )od83.1 + (7.01+3.83 od )

    2 65.71745. = ( ) oo d65.748.11d5.165.7 +=+ ( )od5.1 + /2 (8.61+11.48 2oo d83.3d + ) 3 ( )od5.173.745. +

    =3.48 ( )od5.1 + ( ) =+ 2od5.148.35.0

    2oo d74.1d22.591.3 ++

    ( )3

    d5.1 o+ (1.96+3.913

    O2

    OO d58.0d61.2d ++ )

    Active force 2oo d74.1d87.1222.19 ++ Active moment = (17.58+19.22

    3o

    2oo d58.0d44.6d ++ )

    4 oo d78.17d73.73.2 = 0.5 x 17.18do = 8.89do

    3d o

    3od96.2

    Moment 058.17d22.19d44.6d38.2 o

    2o

    3o =

    1

    2 3 4

  • Solving for do we get do = 4.73m Design depth = 1.2 4.73 = 5.68m

    Average passive pressure = ( )oopx d1.11H1[K ++ = 2.3 [17 x 1+ 7.73 (2.5 - 1 + 1.1 4.73) = 158.3 kPa Average active pressure = oax d1.1K = 0.45 7.73 1.1 4.73 =18.1 kPa Net force = (158.3 - 18.1) 0.2 4.73 = 132.6kN Active force = 19.22 + 12.87(4.73) + 1.74(4.73) 2 =119 kN Passive force = 8.89(4.73) 2 =198.9 kN R =198.9 119 = 79.9 kN

  • Solution 12.13 (a) FSM

    Anchor force

  • Solution 12.13 (b) NPPM

  • Solution 12.14(a) FSM

  • Solution 12.14(b) NPPM

  • Solution 12.15 FMM Design depth = 8.98 m = 9 m Design anchor force = 396.3 kN/m = 400 kN/m

    Anchor force

  • Solution 12.16

    m8.5H,m5B,31 ocs ===

    Lateral pressure value =

    2

    45tanH65.0 cs = 0.65 17.5 5.8 x 0.32= 21 kPa Calculate the forces on struts at each level All loads are per meter length of wall. Level 1

    B1M 0 1.5A 21 2 1 = = A = 28 kN 0Fx = : A + 1B = 21 2 1B = 14 kN

    Level 2

    2 121 1.5B C

    2= = = 16 kN

    Level 3

    C22.3M 1.5D 21 2.32

    = D = 37 kN x 2F 0 : D C 21 2.3 = + = 2C = 11 kN

    Step 4: Calculate forces on each strut.

    A = 28 kN B = 21 BB + = 30 kN C = 21 CC + = 27 kN D = 37 kN

    A

    B1

    C1

    B2

    D

    C2

  • Solution 12.17 Fine-grained soil

    3sat mkN19= , = 25cs , kPa40s u = , m1.6H o = , = 6m

    Check the stability against bottom heave

    02.161.6

    BH o == , ( ) 22.702.12.16

    BH

    2.016N oC =+=

    += , 0q s =

    ( ) 49.21.619

    4022.7qH

    sNFS

    so

    uCheave ==+= > 1.5; therefore, excavation safe against bottom heave

    Determine the lateral pressure diagram

    u

    o

    sH

    = 90.240

    1.619 = < 4, Maximum lateral pressure: H4.0 = 0.4 1.619 = 46.4 kPa

    Step 3: Calculate the forces on struts at each level. All loads are per meter length of wall. Level 1

    1BM = 0 = 1.8A -

    +

    +

    2575.04.46575.0

    3525.1575.04.46525.1

    21 A = 25.6 kN

    :0Fx = A + 1.625.46575.04.46525.121B1 == 5.36B1 = kN

    Level 2

    A

    B1

    0.25H0 = 1.525 m

    C1

    B2

  • 8.414.468.121CB 12 === kN

    Level 3

    +

    +==

    2675.04.46675.0

    3525.1675.04.46525.1

    21D8.10M 2C D = 29.1 kN

    :0Fx = ==+ 4.46675.04.46525.121DC2 66.7 kN 2C = 37.6 kN

    Step 4: Calculate forces on each strut. A = 25.6 kN, B = 21 BB + = 78.3 kN, C = 1 2C C 41.8+ = + 37.6 = 79.4 kN, D = 29.1 kN

    D

    C2

  • Solution 12.18 All calculations are per m length. Calculate the allowable tensile strength of the geotextile. With IDFS = 1.5, CRFS = 2, CDFS = 1.3 and BDFS = 1.3

    3.13.125.145

    FSFSFSFST

    TBDCDCRID

    ulta == = 8.88 kN/m

    Calculate the vertical spacing

    aRK =

    =

    2

    2945tan2

    45tan 2cs2 = 0.35

    Lateral stress due to surcharge = 1535.0qK saR = = 5.2 kPa. 2.5z5.1735.02.5z35.0qKK saRzaRx +=+=+= = 6.13z +5.2 ( ) =+=+= 2.5613.62.5H13.6maxx 42 kPa

    From equation (12.73) with ( )spFS = 1.3, we get ( ) ( )( ) 3.142

    88.8FSqK

    TS

    spzaR

    aminz =+= = 0.163m = 163mm

    Check spacing requirement at mid-height (z=3) 2.5313.6x += = 23.6 kPa

    mm290m29.03.16.23

    88.8Sz === Use zS = 150mm for the bottom half of the wall and 300mm for the top half of the wall. Determine length of reinforcement required at the base for translation. From computer program utility: ( )=== 20,2931.0K csaC and ( ) 29.020cos3.0cosKK aCxaC ===

    ( ) ( ) 5.117b1529.0b5.1729.021HqKHK

    21P 2saC

    2aCax =+=+= kN

    34725.0s5.0s uw === kPa < 50 kPa, therefore use ws = 34 kPa Eq. (12.70):

    ( ) 117.5 1.5 5.234

    ax Tb

    w

    P FSL m

    s= = =

    Eq. (12.71): ( ) ( )

    ( )150.5 0.29 0.5 6 1.5

    17.5tan tan 16

    saC x T

    bb

    qK H FSL

    + + = = = 5.85m Use bL = 6m

    For top layer: ( )

    =2

    45tanzHL csr = ( )

    2

    2945tan3.06 = 3.36m

    ( ) 0.35 0.3 1.3 0.192 tan 2 tan 20

    aR z te

    i

    K S FSL m

    = = = L = 3.36 + 0.19 = 3.55m

  • Determine the total length of reinforcement at each level for internal stability

    z(m) Sz LR Le L L used (m)0.30 0.15 3.36 0.09 3.45 40.45 0.15 3.27 0.09 3.36 40.60 0.15 3.18 0.09 3.27 40.75 0.15 3.09 0.09 3.19 40.90 0.15 3.00 0.09 3.10 41.05 0.15 2.92 0.09 3.01 41.20 0.15 2.83 0.09 2.92 41.35 0.15 2.74 0.09 2.83 41.50 0.15 2.65 0.09 2.74 41.65 0.15 2.56 0.09 2.66 41.80 0.15 2.47 0.09 2.57 41.95 0.15 1.16 0.09 1.26 42.10 0.15 1.12 0.09 1.21 42.25 0.15 1.08 0.09 1.17 42.40 0.15 1.03 0.09 1.13 42.55 0.15 0.99 0.09 1.08 42.70 0.15 0.95 0.09 1.04 42.85 0.15 0.90 0.09 1.00 43.00 0.15 0.86 0.09 0.95 43.30 0.30 0.77 0.19 0.96 43.60 0.30 0.69 0.19 0.88 43.90 0.30 0.60 0.19 0.79 44.20 0.30 0.52 0.19 0.70 54.50 0.30 0.43 0.19 0.62 54.80 0.30 0.34 0.19 0.53 55.10 0.30 0.26 0.19 0.45 65.40 0.30 0.17 0.19 0.36 65.70 0.30 0.09 0.19 0.27 66.00 0.30 0.00 0.19 0.19 6

    Check external stability Stability against translation is already satisfied. Check bearing capacity ( ) 65.17H omaxz == = 105 kPa Short term

    ultq = 5.14 us 5.14 72= = 370 kPa

  • ( ) ( )ultB xq 370FS 3 5

    105max

    .= = = > 3: okay Long term: For 28 = , N = 11.5

    3117.5(1 ) 0.54105 6

    1 1 18 6 11.5 0.54 = 335kPa2 2

    = == = ult

    i

    q BN i

    ( ) 335 3.2105

    = =B

    FS > 3 ; Okay

  • Solution 12.19 Assume zs = 0.5m, ys = 1m

    aRK = 35.0245tan cs2 =

    52.029sin1sin1K cso === At base: K = aRK = 0.35

    ( ) ( ) ( )5

    0.35 17.5 6 15 0.5 1 30.075 2.5 10

    s z y trr

    y

    K H q s s FSt

    wf + + = = = m10336

    5 = 3.36mm mm25.150025.0t corrosion ==

    25.136.3t design += = 4.61mm Use t = 5mm Determine length of reinforcement required at base.

    aCK = 0.29, ( )xaCK = 0.27 Eq. (12.70) ( ) 117.5 1.5 5.2

    34ax T

    bw

    P FSL ms

    = = =

    Eq. (12.71):

    150.27 0.5 6 1.517.5 5.45

    tan 20bL m

    = =

    Eq. (12.68): ==

    20tan075.023.115.035.0

    Le 4.17m

    Since bL > Le, use bL = 5.5m Calculate length of reinforcement required.

    z (m) Sz(m) K Le Lr L (m) Lused(m)0.50 0.5 0.50 5.97 1.42 7.4 7.5 1.00 0.5 0.49 5.80 1.33 7.1 7.5 1.50 0.5 0.47 5.63 1.25 6.9 7.5 2.00 0.5 0.46 5.47 1.17 6.6 6.5 2.50 0.5 0.45 5.30 1.08 6.4 6.5 3.00 0.5 0.43 5.13 1.00 6.1 6.5 3.50 0.5 0.42 4.97 0.92 5.9 6.5 4.00 0.5 0.40 4.80 0.50 5.3 5.5 4.50 0.5 0.39 4.63 0.25 4.9 5.5 5.00 0.5 0.38 4.46 0.00 4.5 5.5 5.50 0.5 0.36 4.30 -0.25 4.0 5.5 6.00 0.5 0.35 4.13 -0.50 3.6 5.5

    Check external stability Bearing capacity ( ) 10565.17Homaxz === kPa

  • Short term

    ultq 5 14 72.= = 370 kPa ( ) ( )ultB x

    q 370FS105

    max

    = = = 3.5 > 3; okay Long term Long term: For 28 = , N = 11.5

    3117.5(1 ) 0.54105 6

    1 1 18 6 11.5 0.54 = 335kPa2 2

    = == = ult

    i

    q BN i

    ( ) 335 3.2105

    = =BFS > 3 ; Okay

  • Solution 12.20 Assume spacing and width of ties.

    m1Sy = w = 300mm zS = 1m Calculate required thickness of reinforcement

    52.029sin1sin1K cso === 35.0

    22945tan

    245tanK 2cs2aR =

    =

    =

    Eq. (12.75): At base, K = 0.35

    +64152.0

    64 = 0.41

    ( ) ( ) ( )5

    y

    tryzSr

    105.23.03111241741.0

    wfFSSSqHK

    t =+= = 131.2 mm3.1m10 5 =

    corrisiont = annual corrosion rate design life = 0.025 50 = 1.25mm designt = calculated thickness + corrosion thickness = 1.3 + 1.25 = 2.55mm

    Use t = 3mm Determine the length of reinforcement required at base From computer program utility: aCK = 0.29, = 29cs , = 20 ( ) 27.020cos29.0cosKK aCxaC ===

    From Eq. (12.71) m02.620tan

    3171245.027.0

    L b =

    +

    = For internal stability the effective length at the wall base is:

    Eq. (12.68): ( )

    08.220tan3.02

    3.11135.0tanw2

    FSSSKL

    i

    tyzaRe =

    == m < 6.02 m Use a length of 8m and tie into each wall face. Pax = 0.5 x 0.35 x 17 x 42 = 47.6 kN/m Check external stability Check bearing capacity ( ) === 417Hmax 68 kPa For = 29cs , N = 22.5

    347.6(1 ) 0.7668 8

    1 1 17 8 22.5 0.76 = 1163kPa2 2

    = == = ult

    i

    q BN i

    ( ) ( )ultB xq 1163FS 17

    68= = =

    max

    > 3; therefore bearing capacity is satisfactory.

  • Solution 12.21 See spreadsheet solution on the next page for NPPM. z is the moment arm. ESA is used for sand and TSA is used for the clay. Design depth = 1.2 od Determination of net available force

    pavP = Average passive pressure = ( ) ( )o2uo1pxoospx d1.1S2HKd1.1HqK ++++ aavP = Average active pressure = 5z = od1.15

    Net force = ( ) oaavpav d2.0PP Bending moment Assume maximum bending moment occurs below excavation level. Let z be the location of the point of maximum bending moment (zero shear force) from the excavation level.

    = z( ) ( )

    3zz

    21

    2zzs2

    3zzz5

    21

    2zzHK

    21z

    3H

    HK21

    2zH

    zHqK 2uo1axo2

    o1axo

    osax +

    +

    +

    +++

    ( ) 22u2o1ax2o1axosaz z21zs2z

    25zHK

    21HK

    21zHqK:0

    z++++=

    = 0

    Use a spreadsheet to solve for z. In Excel, use Tools Goal Seek to find z. See next page for the solution for d d = 12.26 m Design d = 1.2 x 12.26 = 14.7 m

  • qs 20 Above

    excavation H 2.5 m ' 7.2 kN/m3 Ka 0.4 cs 27 deg js 0.0 kN/m3 Kp 2.5 a 0 deg aj 7.2 kN/m3 Below excavation b 0 deg pj 7.2 kN/m3 Ka 0.4 h 0 m uc 0.0 kPa Kp 2.5

    Depth to water 10 m Pw 0.0 kN

    17 kN/m3 y 8.17 m a 0 m

    d 12.26 m clay below excavation 19 ' 9.2 kN/m3 cs 27 aj 9.2 kN/m3 a 0 deg pj 9.2 kN/m3 b 0 deg sw 25 active su 50 kPa sw 25 passive Drained

    Part Pressures Force la M 1 8.0 20.0 1.25 25.0 2 68.0 340.0 6.67 2266.7 3 68.0 0.0 10.00 0.0 4 0.0 0.0 10.00 0.0 5 76.0 931.8 8.63 8041.8

    water 0.0 0.0 6.59 0.0 Sum 1291.8 10333.5

    6 236.9 1452.1 10.67 15499.6 (FS)r 1.50

    h

    anchor

    3

    4

    5 6

    d

    H

  • Solution 12.22

    3861.028cos10cos10cos28cos10cos10cos

    k1k

    22

    22

    pRaR =

    +==

    o1 91.162.0tan =

    = , o1p 85.46sin

    sinsin21

    224=

    ++=

    ( ) 4377.2sinsincoscos cossin2sin1)cos(k 222 p2

    pR =++=

    1

    3

    2

    10

    0.7m

    6m Granular backfill

    3m 0.7m 0.8m

    0.5m

    Compacted sand 1.4m

    WS2

    WS1

    W2 W1

    W3

  • 12

    3

    s1

    s2

    W 0.5 6 23.5 70.5kN at1.25m from toeW 0.5 0.2 6 23.5 14.1kN at 0.93m from toeW (0.8 0.7 3) 0.7 23.5 74kN at 2.25m from toeW 3 6 18 324kN at 3.0m from toeW 0.5 3 0.529 18 14.3kN at 3.5m from toe

    = == == + + == == =

    2

    aR

    ax

    az

    zo

    cb

    T

    1P (0.38)(18)(6 0.7 3 tan10) 178.8 kN2

    P 178.8cos(10) 176kNP 178.8sin(10) 31kNR 70.5 14.1 74 324 14.3 31 527.9kN

    20 , Base resis tan ce T 527.9 tan 20 192kN192(FS) 1.1176

    = + + == == == + + + + + =

    = = == =

    Unsatisfactory in translation

    oM 70.5 1.25 14.1 0.93 74 2.25 324 3 14.3 3.5 31 4.51176 7.233

    1005kN m

    = + + + + +

    =

    Mo 1005x 1.9Rz 527.9

    = = =

    B 4.5e x 1.9 0.352 2

    = = = B 4.5 0.75 0.35 safe6 6

    = = > Unlikely to fail by rotation. Bearing Capacity:

    Zmax

    R 6e 527.9 6 0.351 1 172kPaA B 4.5 1 4.5

    = + = + = , 35, N 37.1 = =

    n 1 2 1

    n

    H 176i 1 1 0.3V 527.9

    + +

    = = =

    , 'B 4.5 2 0.35 3.8m= =

    ult1q B N i 0.5 18 3.8 37.1 0.3 381kPa2

    = = =

    BC381(FS) 2.2172

    = = < 3.0 Unsatisfactory in bearing capacity.

  • Solution 12.23 (a) Refer excel spreadsheet Soln 12.23.xls for computations 12.23 (a) Geotextile mechanical stabilized earth wall Help Kac 0.3 KaR 0.3333FS 3

    phi 30degrees Ko 0.5t (use standard size) 4.00 20degrees tr 1.43 mm 18 fy 450 Mpa qs 15kPa corrosion rate 0.025 mm/yrH 6m design life 75 yr w 50mm tcorrosion 1.9 mm Sz 0.50m t design 3.30

    Sy 1m z(m) Sz K LR Le L 0.50 0.50 0.486 1.72 17.81 19.52 1.00 0.50 0.472 1.63 17.30 18.93 1.50 0.50 0.458 1.55 16.79 18.34 2.00 0.50 0.444 1.47 16.28 17.75 2.50 0.25 0.431 1.38 15.77 17.16 3.00 0.25 0.417 1.30 15.26 16.56 3.50 0.25 0.403 1.22 14.75 15.97 4.00 0.25 0.389 1.00 14.25 15.25 4.50 0.25 0.375 0.75 13.74 14.49 5.00 0.25 0.361 0.50 13.23 13.73 5.50 0.25 0.347 0.25 12.72 12.97 6.00 0.25 0.333 0.00 12.21 12.21

  • (b) Refer excel spreadsheet Soln 12.23.xls for computations 12.23 (b) Geotextile mechanical stabilized earth wall Help Kac 0.3 KaR 0.33FS 3 phi 30 degrees sp 1.3 20 degrees ID 1.5 18 CR 2 qs 15 kPa CD 1.3 H 6 m BD 1.3 Pax 116.7098 kN Tult 58.5 kN Sz 219 mm Tall 11.5 kN Sy 390 mm

    z(m) Sz LR Le L 0.50 0.50 3.18 0.30 3.47 1.00 0.50 2.89 0.30 3.18 1.50 0.50 2.60 0.30 2.90 2.00 0.50 2.31 0.30 2.61 2.25 0.25 2.17 0.15 2.31 2.50 0.25 2.02 0.15 2.17 2.75 0.25 1.88 0.15 2.03 3.00 0.25 1.73 0.15 1.88 3.25 0.25 1.59 0.15 1.74 3.50 0.25 1.44 0.15 1.59 3.75 0.25 1.30 0.15 1.45 4.00 0.25 0.54 0.15 0.68 4.25 0.25 1.01 0.15 1.16 4.50 0.25 0.40 0.15 0.55 4.75 0.25 0.72 0.15 0.87 5.00 0.25 0.58 0.15 0.73 5.25 0.25 0.20 0.15 0.35 5.50 0.25 0.29 0.15 0.44 5.75 0.25 0.07 0.15 0.22 6.00 0.25 0.00 0.15 0.15

  • Solution 12.24 Wall rotation is limited to 0.005H0 Properties: Assume unit weight of concrete is 23.5 kN/m3 Since wall rotation is limited to 0.0005H0 . Use at rest condition, '0 csK 1 sin 0.53= = Lateral pressure distribution on the wall is shown below (a) due to surcharge (b) due to soil

    0.75

    4.5Granular backfill

    3.5m 0.75

    0.5m

    9 kN/m Masonry wall

    4m

    1.25m

    10 kPa load from home site

    5.25m

    2.625 m

    1.75 m

    127.9 kN

    27.8kN

    0.53x17.5x5.25= 48.7 0.53x10= 5.3 kPa

  • Weight of concrete (Wc) = 142.7 KN Weight of soil (Ws) = 286.7 KN Total force in vertical direction Rz = 142.7+286.6+ 9 +10 x 3.5= 473.3 kN Total force in horizontal direction Rx = 27.8+127.9=155.7 kN Overturning moment (M0) about toe= 296.8 kN.m Resisting moment about toe = 962.4 kN.m Resultant moment = 665.6 kN.m Base resistance T =473.3 x tan (25) = 220.7 kN FS against translation = 220.7/155.7 = 1.4 < 1.5. Not satisfactory . Distance from toe to centroid = 665.6/473.3 = 1.41 e = 4.25/2 1.41 = 0.72 B/6 = 0.71 < 0.72 . Centroid lies approximately at middle third. Max. vertical stress = (473.3/4.25) x (1 +6 x 0.72/4.25) = 224.6 i = (1-155.7/473.3)^3 = 0.36 qult = 0.5 x 17.5 x (4.25 2 x 0.72) x 61.4 x 0.36 = 543.5 FS against bearing capacity = 543.5/224.6 = 2.4 Unsatisfactory in bearing capacity. Redesign. By changing the width of the footing base from 4.25 to 4.75 m using the spreadsheet program retwall.xls, we get the following factors of safety Translation = 1.6, bearing capacity = 3.2 Note: You can trick the spreadsheet to use Ko by adjusting the friction angle of the backfill so that Ko = 0.53. A value of friction angle of 17.8 deg. will give a Ko of 0.53. \\