Principles of Foundation Engineering 9th Edition by Braja M. Das and Nagaratnam SOLUTIONS MANUAL

SOLUTIONS MANUAL For

Principles of Foundation Engineering Ninth Edition

BRAJA M. DAS

NAGARATNAM SIVAKUGAN

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Contents Chapter 2............................................................................................................................. 1 Chapter 3........................................................................................................................... 11 Chapter 5........................................................................................................................... 23 Chapter 6........................................................................................................................... 31 Chapter 7........................................................................................................................... 49 Chapter 8........................................................................................................................... 59 Chapter 9........................................................................................................................... 69 Chapter 10......................................................................................................................... 83 Chapter 11......................................................................................................................... 89 Chapter 12......................................................................................................................... 91 Chapter 13....................................................................................................................... 113 Chapter 14....................................................................................................................... 123 Chapter 15....................................................................................................................... 125 Chapter 16....................................................................................................................... 131 Chapter 17....................................................................................................................... 141 Chapter 18....................................................................................................................... 153 Chapter 19....................................................................................................................... 171

Chapter 2 2.1

From Eq. (2.18),

Gs ρw 2450 2.80 × 1000 ; e = 0.0571 = = 1+ e 0.925 1+ e

ρd =

From Eq. (2.6), Porosity, n =

2.2

e 0.0571 = = 0.054 1 + e 1 + 0.0571

From Eq. (2.13), the dry density ρd =

ρ 2060 = = 1786.6 kg/m3 1 + w 1 + 0.153

From Eq. (2.18), ρd = e=

Gs ρw 1+ e

Gs ρw 2.70 × 1000 −1 = − 1 = 0.511 ρd 1786.6

Once saturated, from Eq. (2.19), ρsat =

2.3

(Gs + e ) ( 2.70 + 0.511) × 1000 = 2125.1 kg/m 3 ρw = (1 + e) (1 + 0.511)

Let’s consider a 1-m2 area in plan. The initial volume of this soil is V = 1 × 0.5 = 0.5 m3. Volume of the solids is Vs. e = 0 .9 =

0.5 − Vs ; Vs = 0.2632 m3 Vs

Ws = 0.2632 × 2.68 × 9.81 = 6.919 kN The new volume after compaction = 1 × 0.455 = 0.455 m3 The dry unit weight, γd = From Eq. (2.12), γd =

6.919 = 15.21 kN/m3 0.455

Gs γw 1+ e

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e=

Gs γ w 2.68 × 9.81 −1 = − 1 = 0.729 γd 15.21

From Eq. (2.13), γ = (15.21)(1 + 0.20) = 18.25 kN/m 3

2.4

At the compacted road base, the weight of solids, Ws = 120,000 × 19.5 = 2,340,000 kN At the borrow pit, the dry unit weight, γd = Volume of the pit, V =

γ 17.5 = = 16.13 kN/m 3 1 + w 1 + 0.085

2,340,000 = 145,080 m3 16.13

The moisture content has to be increased from 8.5% (at the borrow pit) to 14.0% (at the road base). The quantity of water to add, 2,340,000 × (0.14 ‒ 0.085) = 128,700 kN Volume of water to be added =

2.5

γd =

γ 110.4 = = 99.9 lb/ft 3 1 + w 1 + 0.105

γd =

Gs γ w 2.65 × 62.4 ; e= − 1 = 0.655 1+ e 99.9

From Eq. (2.23), Dr =

2.6

2.7

128,700 = 13,119.3 m 3 9.81

0.870 − 0.655 × 100 = 60.6% 0.870 − 0.515

a. A-1-a

c. A-3

b. A-1-b

d. A-7-6

Soil A: % of gravel = 50, % of sand = 13, % of fines = 37 D10 = 0.035 mm, D30 = 0.061 mm, D60 = 9.8 mm Cu = 280; Cc = 0.02 LL = 58, PL = 34, PI = 24 plots below the A-line; hence, silt The soil can be described as poorly (gap) graded sandy silty gravel with a group symbol of GM.

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Soil B: % of gravel = 24, % of sand = 69, % of fines = 7 D10 = 0.17 mm, D30 = 0.82 mm, D60 = 2.6 mm Cu = 15.3; Cc = 1.5 LL = 42, PL = 22, PI = 20 plots above the A-line; hence, clay The soil can be described as well graded clayey gravelly sand with a group symbol of SW-SM. Soil C: % of gravel = 1, % of sand = 99, % of fines = 0 D10 = 0.7 mm, D30 = 1.2 mm, D60 = 1.6 mm Cu = 2.3; Cc = 1.3 The soil can be described as poorly (uniformly) graded sand with a group symbol of SP. Soil D: % of gravel = 0, % of sand = 12, % of fines = 88 LL = 75, PL = 31, PI = 44 plots above the A-line; hence, clay.

The soil can be described as sandy clay of high plasticity with a group symbol of CH.

2.8

The head loss from the reservoir to the ditch, ∆h = 38 − 28 = 10.0 m The length of the sand seam in the direction of the flow, L = 200 cos10 = 203.1 m The hydraulic gradient, i = 10.0 203.1 = 0.0492

By Darcy’s law [Eq. (2.35)], v = ( 2.6 × 10 −5 m/s)( 0.0492 ) = 0.128 × 10 −5 m/s The cross section of the sand seam through which the flow takes place is

1.0 × 500.0 = 500.0 m2 The flow rate = (0.128 × 10 −5 m/s)(500.0 m 2 ) = 64.0 × 10 −5 m 3/s Volume of water flowing into the ditch per day: = (64.0 × 10 −5 m 3/s)(24 )( 3600) = 55.3 m 3

2.9

For the flow net shown in Figure P2.9, Nf = 3 and Nd = 10 Total head loss from right to left, hmax = 5.0 m The flow rate is given by [Eq. (2.46)] q = khmax

 3 = (1.5 × 10−5 )(5.0)  = 2.25 × 10−5 m 3/s/m length Nd  10 

Nf

= ( 2.25 × 10− 5 )(50.0)( 24)( 3600 m 3/day) = 97.2 m 3 /day

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2.10

On top of the soft clay layer (i.e., at 10 m depth), initially: σ ′ = 1 × 17.0 + 9( 20 − 9.81) = 108.7 kN/m 2

After the water table is lowered, σ ′ = 3 × 17.0 + 7( 20 − 9.81) = 122.3 kN/m 2

By lowering the water table, the effective stress has increased by (122.3 – 108.7) = 13.6 kN/m2

2.11

The soil below the water table can be assumed to be fully saturated (i.e., S = 1).

e = wGs = 0.25 × 2.70 = 0.675 The saturated unit weight can be computed as

γsat =

(Gs + e)γw (2.70 + 0.675) × 9.81 = = 19.8 kN/m3 1+ e 1 + 0.675

At a depth of 5 m into the sandy clay, σ = 4 × 9.81 + 5 × 19.8 = 138.3 kN/m2 u = 9 × 9.81 = 88.3 kN/m2 σ ′ = σ − u = 138.3 − 88.3 = 50 kN/m2

2.12

Refer to the figure.

a. The compression index Cc is given by [Eq. (2.53)],

Cc =

e1 − e2 1.10 − 0.85 = = 0.441 log σ 2′ − log σ1′ log 240 − log 65 SOURCE: Browsegrades.net

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b. Let the void ratio at 460 kN/m2 pressure be e3 .

 460  e1 − e3 = Cc (log 460 − log 65) = 0.441 × log  = 0.375  65  e3 = 1.10 − 0.375 = 0.725

2.13

a. The clay is below the water table and, hence, is saturated. The initial void ratio eo can be determined as eo = wG s = 0.225 × 2.72 = 0.612 The saturated unit weight is determined as

γsat =

(Gs + e)γw ( 2.72 + 0.612)(9.81) = 20.3 kN/m3 = 1+ e 1 + 0.612

The effective overburden stress at the middle of the clay is

σ o′ = 2 × 17.0 + 3( 20.2 − 9.81) + 1.5( 20.3 − 9.81) = 80.3 kN/m2 < 110.0 kN/m2 Since the preconsolidation pressure is greater than the current overburden pressure, the clay is overconsolidated. The overconsolidation ratio

OCR = 110.0 80.3 = 1.37 b. The 2-m-high compacted fill imposes a surcharge of 2 × 20 = 40 kN/m2 (i.e., ∆σ ′ = 40.0 kN/m2 , σo′ = 80.3 kN/m2 , and σ c′ = 110.0 kN/m2 ) Since σ o′ + ∆σ ′ > σ c′ , the consolidation settlement can be computed from Eq. (2.69) as Sp = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o ′ ′ 1 + eo σ 1 + e σ o o c     0.06 × 3000  110.0  0.52 × 3000  80.3 + 40.0  log log +  1 + 0.612 1 + 0.612 110  80.3   

= 52.9 mm

2.14

For U = 75%, Tv = 0.477 (Table 2.12)

Tv =

cv t H dr2

From two-way (doubly drained) to one-way (singly drained), Hdr is doubled.

For the same U and, hence, the same Tv, this would increase the time fourfold.

Therefore, it will take 4t years.

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2.15

a. The clay layer with one-way drainage has Hc of 6.0 m. After one year,

Tv =

cvt 0.0014 × 365 × 24 × 3600 = = 0.123 ; settlement, Sc(t) = 160 mm H2 6002

From Table 2.12, U = 39.6%

U=

Sc ( t ) Sc (max)

; Sc (max) = 160 0.396 = 404 mm

When t = 2 years, Tv = 0.246.

From Table 2.12, U = 55.8%.

Consolidation settlement during the first two years is 0.558 × 404 = 225.4 mm b. The initial effective overburden stress at the middle of the clay is σo′ = 1.5 × 17.0 + 0.5(18.5 − 9.81) + 3.0(19.0 − 9.81) = 57.4 kN/m2

∆σ ′ = 3 × 19 = 57 kN/m2 Sc =

 σ o′ + ∆σ ′  Cc H  log 1 + eo σ o′  

404 =

2.16

Cc × 6000  57.4 + 57.0  log ; Cc = 0.41 1 + 0.810 57.4  

a. For the clay layer, assuming S = 100% below the water table, eo = 0.45 × 2.70 = 1.215

γsat =

(Gs + e)γw (2.70 + 1.215) × 9.81 = 17.3 kN/m3 = 1+ e 1 + 1.215

The initial effective overburden stress at the middle of the clay layer is

σo′ = 1 × 16 + 1(19.0 − 9.81) + 1.5(17.3 − 9.81) = 36.4 kN/m2 With OCR = 1.5, the preconsolidation pressure σ c′ = 1.5 × 36.4 = 54.6 kN/m2 When the fill is placed, it imposes a surcharge of ∆σ ′ = 20 × 1.5 = 30.0 kN/m2 From Eq. (2.69), Sc = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o 1 + eo  σ o′  1 + eo  σ c′  0.08 × 3000  54.6  0.65 × 3000  66.4  log log +  = 93.9 mm 1 + 1.215 1 + 1.215  36.4   54.6  SOURCE: Browsegrades.net

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b. Including the fill load and the warehouse load, ∆σ ′ = 30 + 40 = 70 kN/m2

Sc = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o 1 + eo  σo′  1 + eo  σ c′  0.08 × 3000  54.6  0.65 × 3000  106.4  log log +  1 + 1.215 1 + 1.215  36.4   54.6 

= 274.2 mm Consolidation settlement due to the warehouse alone is 274.2 – 93.9 = 180.3 mm

2.17

The direct shear test data are plotted in the figure. From the failure envelope, c′ = 6.5 kN/m2 and ϕ' = tan‒1(0.4292) = 23.2°

2.18

a. σ3′ = 100 kN/m2 and ∆σ f = 260 kN/m2 Therefore, σ1′ = σ 3′ + ∆σ f = 360 kN/m 2

φ′  φ′    From Eq. (2.91), σ1′ = σ 3′ tan 2  45 +  + 2c′ tan  45 +  2 2   For normally consolidated clays, c′ = 0; hence,

φ′   360 = 100 tan 2  45 +  ; φ ′ = 34.4 ° 2 

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b. For the second specimen, σ3′ = 200 kN/m2

34.4   2 σ1′ = 200 tan 2  45 +  = 719.5 kN/m 2  

∆σ f = 719.5 − 200 = 519.5 kN / m 2

2.19

a. In normally consolidated clay, c′ = 0 For the first specimen (consolidated drained test), using Eq. (2.91),

φ′   260 + 150 = 410 = 150 tan 2  45 +  ; φ ′ = 27.7° 2  In the second specimen (consolidated undrained test), applying the same value of φ ′ in Eq. (2.91),

σ3 = 150 kN/m 2 and ∆σ f = 115 kN/m 2 σ1 = σ 3 + ∆σ f = 265 kN/m 2

σ3′ = 150 − u f and σ1′ = 265 − u f

φ′  27.7    σ1′ = σ 3′ tan 2  45 +  = σ 3′ tan 2  45 +  = 2.737σ 3′ 2 2    265 − u f = (150 − u f ) × 2.737; u f = 83.8 kN/m2

b. From Eq. (2.96), A f =

2.20

uf ∆σ f

=

83.8 = 0.73 115

At failure the pore water pressure is uf, σ3 = 100 kN/m2 and σ1 = 207 kN/m2.

σ3′ = 100 − u f and σ1′ = 207 − u f Substituting for σ3′ and σ1′ in Eq. (2.91),

φ′  φ′    σ1′ = σ 3′ tan 2  45 +  + 2c′ tan 45 +  2 2  

(207 − u f ) = (100 − u f )tan 2  45 + 262  + 2 × 10 tan 45 + 262  







uf = 51.9 kN/m2 SOURCE: Browsegrades.net

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2.21

The following table can be prepared from the given data, and the Mohr circles are plotted as shown in the figure.

Sample No.

(kN/m )

1 2 3

100 200 350

σ3 2

(∆σ d ) f

(∆ud ) f

2

2

(kN/m ) 88.2 138.5 232.1

(kN/m ) 57.4 123.7 208.8

σ3′

σ1

σ1′

(kN/m )

(kN/m )

(kN/m2)

188.2 338.5 582.1

42.6 76.3 141.2

130.8 214.8 373.3

2

2

The failure envelope is drawn tangent to the Mohr circles in the figure and, from measurements, c′ = 10.0 kN/m2 and ϕ' = 24.7°

2.22

The unconfined compressive strength of the clay specimen qu = 2cu = 90 kN/m2 Cross sectional area of the specimen =

π 4

× 752 = 4417.9 mm2

Maximum load the specimen can carry = 90 × 4417.9 × 10−6 = 0.398 kN = 398 N Weight of one steel plate = 1.5 × 9.81 N = 7.358 N Therefore, number of plates that can be stacked on the specimen = 398 7.358 = 54 With qu = 90 kN/m2 (see Table 2.14), it is a medium clay (consistency).

2.23

a. From Figure P2.7, D10 = 0.7 mm, D15 = 0.9 mm, D30 = 1.2 mm, D50 = 1.4 mm, D60 = 1.6 mm, and D85 = 2.05 mm

D60 1.6 D302 1 .2 2 = = 2.3 ; Cc = = = 1.3 D10 × D60 0.7 × 1.6 D10 0.7 From Eq. (2.87), φ ′ = 26 + (10 × 0.8) + (0.4 × 2.3) + 1.6 log(1.4) = 35.2° Cu =

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D   2.05  b. From Eq. (2.89), a = 2.101 + 0.097  85  = 2.101 + 0.097  = 2.322  0 .9   D15  From Eq. (2.90), b = 0.845 – 0.398a = 0.845 – 0.398×2.322 = ‒ 0.0792

From Eq. (2.88),

1  1   −1   = tan   = 36.8°  ae + b   2.322 × 0.61 − 0.0792 

φ ′ = tan −1 

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Chapter 3 3.1

Do = 3.0 in.; Di = 3.0 ‒ 2 × 0.065 = 2.870 in.

From Eq. (3.3), the area ratio is given by AR (%) =

3.2

Do2 − Di2 3.02 − 2.870 2 × 100 = × 100 = 9.3% Di2 2.870 2

Assume the wall thickness is t (mm). Therefore, the outer diameter is (75 + 2t) mm.

For the samples to be undisturbed, AR < 10%. From Eq. (3.3), (75 + 2t )2 − 752 AR = × 100 ≤ 10 752 2

2

 75 + 2t   75 + 2t    ≤ 1.1 ;   ≤ 1.0488; t ≤ 1.83 mm  75   75 

The maximum possible wall thickness is 1.83 mm.

3.3 Depth from ground surface (m)

cu (kN/m2)

OCR

N60

[Eq. (3.8b)]

σ′o (MN/m2)

[Eq. (3.9)]

3.0

5

76.5

1 [(1.5)(16.5) + (1.5)(19 − 9.81)] = 0.03854 1000

5.51

4.5

8

107.3

0.03854 +

1 (1.5)(16.8 − 9.81) = 0.0490 1000

6.46

6.0

8

107.3

0.0490 +

1 (1.5)(16.8 − 9.81) = 0.0595 1000

5.65

7.5

9

116.8

0.0595 +

1 (1.5)(16.8 − 9.81) = 0.07 1000

5.48

9.0

10

126.0

0.07 +

1 (1.5)(16.8 − 9.81) = 0.0805 1000

5.35

Average cu = 106.78 kN/m2

Average OCR = 5.69

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3.4 Depth from ground surface N60 (m) 3.0 5 4.5 8 6.0 8 7.5 9 9.0 10 a See Problem 3.3

(kN/m2) 38.54 49.0 59.5 70.0 80.5

Depth (m)

σ o′ (kN/m2 )

σ o′ a

OCR [Eq. (3.10)]

σ c′ (kN/m2 )

7.52 9.47 8.0 7.46 7.2

235 376 376 423 470

[Eq. (3.11)]

3.5 18 × 1.5 = 27 18 × 3.0 = 54 18 × 4.5 = 81 18 × 6.0 = 108 108 + (1.5)(20.2 – 9.81) = 123.6 123.6 + (1.5)(20.2 – 9.81) = 139.2

1.5 3.0 4.5 6.0 7.5 9.0

 1  Eq. (3.13) : CN =    (σ o′ pa ) Depth (m)

N60

σ o′ (kN/m 2 )

0.5

(N1 )60 a

CN

12 11 10 8 12 12

1.5 6 27 1.92 3.0 8 54 1.36 4.5 9 81 1.11 6.0 8 108 0.96 7.5 13 123.6 0.9 9.0 14 139.2 0.85 a Rounded off to nearest whole number

3.6

CN =

2 1 + (σ o′ pa )

Depth (m)

N60

σ′o a (kN/m2)

CN

(N1)60 b

1.5 3.0 4.5 6.0 7.5 9.0

6 8 9 8 13 14

27 54 81 108 123.6 139.2

1.57 1.3 1.1 0.96 0.89 0.84

9 10 10 8 12 12

a b

See Problem 3.5 Rounded off to nearest whole number

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3.7

From Problem 3.5, the average value of ( N1 )60 =

1 (12 + 11 + 10 + 8 + 12 + 12) = 10.83 ≈ 11 6

Eq. (3.31b): φ ′ = 15.4( N1 )60 + 20 = (15.4)(11) + 20 = 33°

3.8

3.9

From Problem 3.5, Depth (m)

σ′o (kN/m2)

pa (kN/m2)

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

φ ′ (deg) [Eq. (3.30)] 6 34.7 8 34.9 9 34.0 8 31.4 13 34.9 14 34.9 Average φ ′= 34.1° ≈ 34°° N60

Eq. (3.29): φ ′ = 27.1 + 0.3( N1 )60 − 0.00054[( N1 )60 ]2 Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0

N60

φ ′ (deg)

[Eq. (3.29)] 12 30.6 11 30.4 10 30.05 8 29.47 12 30.6 12 30.6 Average φ ′ = 29.93° ≈ 30.45°°

3.10 Depth (m)

σ′o (kN/m2)

pa (kN/m2)

N60

Dr (%) [Eq. (3.22)]

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

6 8 9 8 13 14

50.6 51.8 49.7 43.2 52.8 52.7

Average Dr = 50.13% ≈ 50%

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0.5

3.11

 1  Eq. (3.13): CN =   ; ( N1 )60 = C N N60  (σo′ pa ) 0 .5

Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0

σ′o (kN/m2) 27 54 81 108 123.6 139.2

pa (kN/m2) 100 100 100 100 100 100

N60

CN

1.08( N1 )60

6 8 9 8 13 14

1.92 1.36 1.11 0.96 0.9 0.85

12.44 11.75 10.78 8.29 12.64 12.85

 1.08( N1 ) 60  Dr (%) =   60   [Eq. (3.28)] 45.3 44.3 42.4 37.2 45.9 46.3 Average = 43.56% ≈ 44%

3.12 CN γavg Depth (m) (kN/m3) Liao & Whitman Skempton [Eq. (3.13)] [Eq. (3.14)] 3.5 18.0 1.27 1.23 8.8 18.0 0.80 0.78 12.4 18.5 0.66 0.62 18.5 18.5 0.54 0.46 23.6 19.0 0.48 0.37 26.9 19.0 0.45 0.33 a

3.13

(N1 )60 a

N60

Liao & Whitman Skempton [Eq. (3.13)] [Eq. (3.14)]

12 8 15 19 21 26

15 6 10 10 10 12

15 6 9 9 8 9

Rounded off to nearest whole number

Eq. (3.13):

 1  CN =    σo′ pa 

0.5

  1 =   (25 × 110) 2000

0.5

= 0.85

( N 1 ) 60 = C N N 60 = ( 0.85)( 28) ≈ 24 a. From Eq. (3.21):

  σ′  Dr = 12.2 + 0.75222 N 60 + 2311 − 711OCR − 779  o  − 50Cu2   pa   

0.5

  25 × 110  2 = 2.2 + 0.75222 × 28 + 2311 − 711 × 1 − 779  − (50)3.2   2000    = 71.4%

0.5

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b. From Eq. (3.22):     N 60   Dr =    17 + 24  σ o′    p     a   

0. 5

    28   =   17 + 24  25 × 110       2000     

0 .5

= 0.748 or 74.8%

c. From Eq. (3.23): 1 .7    0.06    N 60  0.23 +  D50   1     Dr =  ×  9 σ o′ pa      

0 .5

1 .7    0.06    28 0.23 +    1 1 .2     = ×   9  ( 25 × 110 ) 2000     

0 .5

= 0.510 or 51.0%

d. From Eq. (3.24): C P = 60 + 25 log D50 = 60 + 25 log 1.2 = 62 5000 = 1 .28 100

C A = 1 .2 + 0 .05 log

COCR = OCR1.8 = 1  ( N1 )60  Dr =    CPC ACOCR  e.

0.5

  24 =   (62)(1.28)(1) 

0.5

= 0.55 or 55%

From Eq. (3.28): (N )  Dr =  1 60   60 

0 .5

 24  =   60 

0 .5

= 0.632 or 63.2%

SOURCE: Browsegrades.net

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3.14 Friction angle (deg) CN σ′o Depth N60 (N1)60 (m) (kN/m2) Eq. (3.13) Eq. (3.29) Eq. (3.30) Eq. (3.31b) 2.0 37.0 17 1.644 27.9 35.1 43.6 40.7 3.5 64.8 23 1.243 28.6 35.2 44.1 41.0 5.0 92.5 26 1.040 27.0 34.8 43.3 40.4 6.5 120.3 28 0.912 25.5 34.4 42.4 39.8 8.0 148.0 29 0.822 23.8 33.9 41.3 39.2

3.15

a. At 25 ft depth, σo′ = (10)(111.5) + (15)(125.0 − 62.4) = 2054.0 lb/ft 2 From Eq. (3.13),

 1  CN =    σo′ pa 

0.5

  1 =   2054 2000 

0.5

= 0.99

(N1)60 = CN N60 = 0.99 × 26 ≈ 26

b. From Eq. (3.28), (N )  Dr =  1 60   60 

0 .5

 26  =   60 

0. 5

= 0.658 or 65.8%

c. From Eq. (3.30), −1 

 N 60 φ ′ = tan   12.2 + 20.3(σo′ pa )

0.34

−1 

 26 = tan   12.2 + 20.3(2054 2000)

0.34

= 42.7° d. From Eq. (3.32), Es = 10N 60 pa

Es = (10)(26)( 2000) = 520,000 lb/ft 2

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3.16

Eq. (3.34): cu =

Eq. (3.38) :

K = =

cu ( VST ) =

T K

πd 2  d

 d  + + 6h  12  cos iT cos iB 

π (0.0635)2  0.0635

0.0635   cos 45 + cos 45 + (6)(0.127) = 0.000994

12

51 = 51,308 N/m 2 ≈ 51.3 kN/m 2 0.000994

Eqs. (3.39) and (3.40a):

cu (corrected) = λcu (VST) = [1.7 − 0.54 log(PI)](51.3) = [1.7 − 0.54log(46 − 21)](51.3) = 48.48 kN / m 2

3.17

Eq. (3.42): OCR = β

cu ( field ) σ o′

σ′o = 59.5 kN/m2 (from Problem 3.3) Eq. (3.43): β = 22(PI)−0.48 = (22)(25)−0.48 = 4.69  51.3  OCR = ( 4.69)  = 4.04  59.5 

3.18

a.

h 4 = =2 d 2

From Eq. (3.36): K = cu ( VST ) =

7π d 3 (7)(π )(2)3 = = 29.32 in.3 6 6

T 230 = = 7.84 lb/in.2 ≈ 1129 lb/ft 2 K 29.32

b. From Eqs. (3.39) and (3.40a), cu (corrected) = [1.7 − 0.54 log(PI)](1129)] = [1.7 − 0.54 log(58 − 29)](1129) = 1027.7 lb / ft 2

3.19

Eq. (3.45): β =

1 1 = = 4.18 0.08 + 0.0055( PI) 0.08 + 0.0055(29)

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Eq. (3.42): OCR = β

3.20

cu ( field )  1129  = ( 4.18)  = 3.52 σ′o  1340 

From Eq. (3.35),

πd 2 

d  π × 1002  100  6 3 3 K= h +  =  200 +  = 3.665 × 10 mm = 0.003665 m 2  3 2 3   From Eq. (3.34), Peak undrained shear strength = 60/0.00366 = 16,393 N/m2 = 16.4 kN/m2 Residual undrained shear strength = 35/0.00366 = 9,563 N/m2 = 9.6 kN/m2

3.21

The vane is rotated quick enough to ensure that the test is carried out under undrained conditions. The vane shears a cylindrical failure surface shown in the figure where the shear stress at failure is assumed the same at the upper and lower horizontal circular areas and the vertical cylindrical surface. We calculate the torque resisted by the shear stresses along these surfaces.

Cylindrical surface: T1 = πd hτ f

d πd 2h = τf 2 2

One circular surface: T2 = ∫

r = 0.5d

r =0

τ f ( 2πrdr ) r = τ f 2π ∫

0.5d

0

r 2dr =

πd 3 τf 12

For equilibrium, T = T1 + 2T2 . So,

 πd 2h πd 3  T = τf  +  6   2 πd 2  d Since the test is undrained, τ f = cu , T = cu h +  2  3 2 T πd  d K = = h +  cu 2  3 SOURCE: Browsegrades.net

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3.22 Depth (m)

σ o′ (kN/m 2 )

qc (kN/m 2 )

φ ′ (deg) [Eq. (3.52)]

2.0 3.5 5.0 6.5 8.0

33.0 57.8 82.5 107.3 132.0

3.12 4.25 5.14 9.23 12.20

40.4 39.0 38.0 39.0 40.3

Depth (m)

σ o′ (kN/m 2 )

qc (kN/m 2 )

Dr (%) [Eq. (3.51)]

2.0 3.5 5.0 6.5 8.0

33.0 57.8 82.5 107.3 132.0

3.12 4.25 5.14 9.23 12.20

42.2 42.8 43.1 54.1 59.0

3.23

3.24

From Eq. (3.55), a = 0.26 and c = 5.44 (Table 3.9).

Thus,

3.25

(qc pa ) = 5.44 D 0.26 50

N 60

Depth (m)

σ o′ (kN/m 2 )

qc (kN/m 2 )

N60

2.0 3.5 5.0 6.5 8.0

33.0 57.8 82.5 107.3 132.0

3.12 4.25 5.14 9.23 12.20

6 8 10 18 24

σ o′ = 3 × 16 .5 + 5(19 .0 − 9 .81) = 95 .5 kN/m 2

σo = 3 × 16.5 + 5 × 19.0 = 144.5 kN/m2 Eq. (3.56): cu =

qc − σ o 750 − 144.5 = = 40.4 kN/m2 Nk 15

From Mayne and Kemper (1988) [Eq. (3.59)], σ c′ (MN/m 2 ) = 0.243( qc ) 0.96 = 0.243(0.75) 0.96 = 0.184 MN/m 2 = 184 kN / m 2

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1.01

 q − σo   From Eq. (3.60), OCR = 0.37 c  σo′ 

1.01

 750 − 144.5  = 0.37  95.5  

= 2.4

3.26

3.27

fs (kN/m2) 2.1 12.5 10.5 40.5 32.3 75.2

Depth (m) 5 10 15

qc (MN/m2) 0.21 1.05 3.23

qc/pa

20 25

9.66 12.50

96.6 125

60.7 62.5

Fr Soil type (from Figure 3.30) (%) 5.95 Organic soil – clay 3.86 Clay-silty clay to clay 2.33 Silt mixtures – clayey silt to silty clay 0.63 Sand – clean sand to silty sand 0.50 Sand – clean sand to silty sand

Eq. (3.61):

46 + 180  326.5 − 42.4   ∆p   E p = 2(1 + µs )(Vo + vm )  = ( 2)(1 + 0.5) 535 +   2  ∆v    180 − 46  = 4121.6 kN/m2

3.28

a. K D =

po − uo 280 − ( 9.81)(8 − 3) = = 2.43 σ′o 95

K  Eq. (3.68): Ko =  D   1 .5 

0.47

2.43  − 0.6 =    1 .5 

0.47

− 0.6 = 0.65

b. Eq. (3.69): OCR = (0.5KD)1.6 = (0.5 × 2.43)1.6 = 1.37 c. Eq. (3.72) : Es = (1 − µs2 ) E D = (1 − µs2 )( 34.7)( p1 − po ) = (1 − 0.352 )( 34.7)( 350 − 280) = 2131 kN/m 2

3.29

KD =

po − uo 260 − ( 4)( 9.81) = = 3 .2 σ o′ ( 2)(14.5) + ( 4)(19.8 − 9.81)

Eq. (3.74a): φ ′ = 31 +

KD 3.2 = 31 + = 38.2° 0.236 + 0.066 K D 0.236 + (0.066)( 3.2)

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3.30

105 =

3.31

Es 1 − µs ; µ = 0.32 (γ g ) (1 − 2µs )(1 + µs ) s

Eq. (3.79): v =

Es 0.68 × ; Es = 14.136 kN/m2 18 9.81 (0.36)(1.32)

A time-distance plot is shown on the next page.

Slope of 0a =

v1 =

1 15.24 × 10 −3 = v1 7 .5

7.5 × 103 = 492 m/s (top layer) 15.24

v2 = slope of ab =

v3 = slope of bc =

10 × 103 ≈ 1390 m/s 7.19

20 × 103 = 3390 m/s 5.9

xc = 7.5 m

Z1 = =

1 v2 − v1 xc 2 v2 + v1 1 1390 − 492 (7.5) = 2.6 m 2 1390 + 492

2 Z1 v32 − v12   v3v2  1  ; Ti 2 ≈ 20 × 10−3 s Eq. (3.81): Z 2 = Ti 2 − 2 2 2 (v3 )( v1 )   v3 − v2     2 Z1 v32 − v12 ( 2)( 2.6) ( 3390) 2 − ( 492) 2 = = 0.0105 ( v3 )( v1 ) ( 3390)( 492) v3 v 2 v −v 2 3

2 2

So, Z 2 =

=

(3390)(1390) ( 3390) 2 − (1390) 2

= 1524

1 (0.02 − 0.0105)(1524) = 7.24 m 2 SOURCE: Browsegrades.net

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Chapter 5 5.1

For method C: Mold volume = 2124 cm3 Hammer mass = 4.54 kg Hammer drop = 0.4572 m No. of layers = 5 No. of blows per layer = 56 Energy per blow = (4.54 × 9.81)(0.4572) = 20.36 N⋅m Total energy imparted to the soil in the mold = 5 × 56 × 20.36 = 5701.5 N⋅m Therefore, energy per unit volume =

5.2

5701.5 N.m/m 3 = 2648 kN.m/m 3 ≈ 2700 kN.m/m3 −6 2124 × 10

a. γd =

Gs γw 1+ e

γd (max) =

Gs γw 2.65 × 9.81 = = 17.75 kN/m 3 1 + emin 1 + 0.465

γd (min) =

Gs γw 2.65 × 62.4 = 9.81 = 15.07 kN/m 3 1 + emax 1 + 0.725

 γ − γd (min)  γd (max) 16.46 − 15.07  17.75 Dr =  d = = 0.559 or 55.9%   17.75 − 15.07  16.46  γd (max) − γd (min)  γd

b. A =

γd (min) γd (max)

=

15.07 = 0.849 17.759

Relative compaction, RC =

5.3

A 0.849 = = 0.927 or 92.7% 1 − Dr (1 − A) 1 − 0.559(1 − 0.849 )

For standard Proctor, using Eqs. (5.9) and (5.10)

wopt (%) = [1.95 − 0.38(log E )]( PL) = [1.95 − 0.38(log 600)](25) = 22.4 SOURCE: Browsegrades.net

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γd (max) = 22.68exp(−0.0183wopt ) = 22.68exp(−0.0183 × 22.4) = 15.05 kN / m 3 For modified Proctor, using the same equations,

wopt (%) = [1.95 − 0.38(log E )]( PL) = [1.95 − 0.38(log 2700)](25) = 16.2 γd (max) = 22.68exp(−0.0183wopt ) = 22.68exp(−0.0183 ×16.2) = 16.86 kN / m 3 From standard to modified Proctor, wopt has decreased by 6.2% and γd (max) has increased by 1.81 kN/m3.

5.4

For standard Proctor: Using Eqs. (5.11) and (5.12),

wopt = (1.99 − 0.165 ln E )( PI) = (1.99 − 0.165 ln 600)(45 − 25) = 18.7% L = 14.34 + 1.195 ln 600 = 21.98 M = ‒ 0.19 + 0.073 ln 600 = 0.277

γd (max) = 21.98 − 0.277 × 18.7 = 16.80 kN/m3 For modified Proctor: Using Eqs. (5.11) and (5.12),

wopt = (1.99 − 0.165 ln E )( PI) = (1.99 − 0.165 ln 2700)(45 − 25) = 13.7% L = 14.34 + 1.195 ln 2700 = 23.78 M = ‒ 0.19 + 0.073 ln 2700 = 0.387

γd (max) = 23.78 − 0.387 ×13.7 = 18.48 kN / m 3 From standard to modified Proctor, wopt has decreased by 5.0% and γd (max) has

increased by 1.68 kN/m3

5.5

Borrow pit

Truck V = 4.8 m3 W = 72.7 kN

w = 15.0% = 19.1 kN/m3

Compacted fill w = 18.0% = 17.3 kN/m3 V = 38,500 m3

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a. The water content in the truck is the same as that in the borrow area. So, the weight of solids in the truck is Ws = 72.7 (1 + 0.15) = 63.2 kN

In the compacted fill, the weight of the solids, Ws = 17.3 × 38, 500 = 666, 050 kN

Hence, the number of truck loads required is 666,050/63.2 = 10,539 b. Weight of solids in the pit. Ws = 666,050 kN Moisture content = 15.0% The weight of soil including solids and water is 666,050 (1 + 0.15) = 765,957.5 kN Unit weight γ at the pit = 19.1 kN/m3 Hence, the volume of the pit = 765,957.5/19.1 = 40,102.5 m3 c. Weight of solids per truck load, Ws = 63.2 kN The moisture content has to be increased from 15% to 18%.

Thus, the quantity of water to be added can be calculated as: 63.2 × 0.03 = 1.896 kN = 1.896 9.81 = 0.1933 m3 or 193.3 liters d. Saturation water content of the fill In the compacted fill, γd =

Gs γw (270)(9.81) ;e= − 1 = 0.531 1+ e 17.3

From Se = wGs , assuming S = 1, w =

5.6

0.531 = 0.197 or 19.7% 2.70

Eq. (5.17):

   3  3 1 1 1 1 S N = 1.7  + + +  = (1.7)  + + + 2 2 2 2 2 2 ( D20 ) ( D10 )  (0.7) (0.65)   ( 2)  ( D50 ) = 3.86 Rating—Excellent (SN between 0 and 10)

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5.7 Soil A B C

D10 (mm) 0.034 0.15 0.7

D20 (mm) 0.046 0.47 1.05

D50 (mm) 4.8 1.8 1.35

SN [Eq. (5.17)] 62.2 12.0 3.6

Rating (Table 5.6) Unsuitable Good Excellent

A = Gap (poorly) graded sandy silty gravel, GM B = Well graded clayey gravelly sand, SW-SM C = Poorly (uniformly) graded sand, SP

5.8

a. Eq. (5.19):

Sc ( p ) =

σ ′ + ∆σ (′ p ) (0.54)(5) Cc H c 70.0 + 60.0 log o = log m = 464.9 mm 1 + eo σ o′ 1 + 0.95 70.0

b. At the end of time t2 = 1 year, Tv =

cvt 2 4.0 × 1 = = 0.64 H2 2.52

From Figure 5.19, the mid-plane degree of consolidation U = 73.6% ∆σ (′ p ) 60 = = 0.86 70 σ o′ From Figure 5.17, for

∆σ (′ f ) ∆σ (′ p ) = 0.54 = 0.86 and U = 73.6%, ∆σ (′ p ) σ o′

Therefore, ∆σ (′ f ) = 0.54 × 60 = 32.4 kN/m

2

The total preload required over one year is 60 + 32.4 = 92.4 kN/m 2

5.9

∆σ (′p ) = 60 kN/m 2 ; σ o′ = 70 kN/m2 After 6 months, Tv =

cv t2 4.0 × 0.5 = = 0.32 H2 2.52

From Figure 5.19, the mid-plane degree of consolidation U = 42.0%

∆σ (′ p ) 60 = = 0.86 σ o′ 70 From Eq. (5.22), for

∆σ (′ p ) = 0.86 and U = 42.0%, σ o′ SOURCE: Browsegrades.net

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0.42 =

log(1 + 0.86)   ∆σ (′ f )   log 1 + 0.86 1 +  ∆σ (′ p )    

∆σ (′ f ) = 2.93 ; ∆σ (′ f ) = 2.93 × 60 = 175.8 kN/m 2 ∆σ (′ p ) The total preload required over six months = 60 + 175.8 = 235.84 kN/m 2 Note:

5.10

This is a substantial increase in the magnitude of the temporary surcharge when the preload duration is shortened.

a. Sc ( p ) =

σ ′ + ∆σ (′ p ) (0.46)(3) Cc H c 105.0 + 50.0 log o log m = 123.5 mm = 1 + eo σ o′ 1 + 0.89 105.0

b. σ o′ = 105 kN/m 2 ; ∆σ (′p ) = 50 kN/m 2 ; ∆σ (′ f ) = 40 kN/m 2

∆σ (′ f ) 40 ∆σ (′ p ) 50 = = 0.8 = = 0.48 ; ∆σ (′ p ) 50 σ o′ 105 From Figure 5.17, the mid plane degree of consolidation required is given by

U = 63% From Figure 5.19, Tv = 0.50

Tv =

cv t2 3.5t2 ; t2 = 1.287 ≈ 1.29 ; 0.50 = 2 H 3.02

The required duration for the preload surcharging, t2 = 1.29 years

5.11

a. Eq. (5.28): n =

de 4 .5 = =9 2 rw ( 2)( 0.25)

Eq. (5.29): S =

rs 0.35 = = 1 .4 rw 0.25

Eq. (5.30): Tr =

cvr t2 ( 0.3)( 6) = = 0.089 d e2 ( 4.5) 2

SOURCE: Browsegrades.net

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Eq. (5.27) : m =

2 n2 kh  n 2 − S 2  n 3 S   ln S ln − + +   n 2 − S 2  S  4 4n 2 k s  n 2 

 92 − 1.42  92 1.42  9  3  ln(1.4) = 2 ln + ( 2) − + 2 9 − 1.42  1.4  4 (4)(92 ) 9   = 1.82  − 8Tr   − 8 × 0.089   = 1 − exp  = 0.347 = 32.4% 1.82  m   

Eq. (5.26): U r = 1 − exp

b. Tv =

cv t2 (0.3)( 6) = = 0.089 2 H dr2 9   2 2

Tv = 0.089 =

π  Uv %    ; Uv = 33.7% 4  100 

Eq. (5.25):

Uv,r = 1 – (1 – Ur)(1 – Uv) = 1 – (1 – 0.324)(1 – 0.337) = 0.552 = 55.2%

5.12

cv = cvr = 0.042 ft2/day = 15.33 ft2/year Vertical drainage: Tv =

cv t2 15.33t2 = = 0.613t2 2 H2  10    2

(a)

Radial drainage: Eq. (5.30): Tr =

cvr t2 15.33t2 = = 0.426t2 d e2 62

n=

de (6)(12) = = 4 .5 2rw ( 2)(8)

t2 (yr) 0.2 0.4 0.8 1.0

Tv [Eq. (a)] 0.123 0.245 0.49 0.613

Uv (Table 2.12) 0.395 0.555 0.76 0.82

(b)

Tr [Eq. (b)] 0.085 0.170 0.341 0.426

Ur [Eqs. (5.26) and (5.32)] 0.553 0.80 0.96 0.98

Ur,v [Eq. (5.25)] 0.73 0.91 0.99 0.997

SOURCE: Browsegrades.net

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5.13

cv tc (0.015)( 30) = = 0.015 H2 (5.5) 2

Eq. (5.39): Tc = Tv =

cv t2 (0.015)(50) = = 0.0248 H2 (5.5) 2

From Figure 5.23, Uv ≈ 15% For the sand drain, n = Eq. (5.35): Trc = Tr =

de 2 .5 = = 17.86 2 rw ( 2)( 0.07 )

cvr tc (0.015)( 30) = = 0.072 d e2 ( 2.5) 2

cvr t2 (0.015)(50) = = 0.12 d e2 ( 2.5) 2

Eq. (5.34): Ur = 1 − Eq. (5.36): A =

1 [exp( ATrc ) − 1] exp( − ATrc ) ATrc

2 m

m=

(3)(17.86) 2 − 1 n2 3n 2 − 1 17.862 ln( n ) − = ln( 17 .

86 ) − = 2.14 n2 − 1 4n 2 17.862 − 1 ( 4)(17.86) 2

A=

2 = 0.935 2.14

ATrc = 0.935 × 0.072 = 0.067 Ur = 1 −

1 [exp( 0.067) − 1] exp( −0.067 ) = 0.033 = 3.3% 0.067

Uv,r = 1 – (1 – Ur)(1 – Uv) = 1 – (1 – 0.033)(1 – 0.15) = 0.178 = 17.8% S( p ) =

Cc H σ′ + ∆σ′ ( 0.3)(5.5) 80 + 70  log o = log  = 0.256 m = 256 mm 1 + eo σ′o 1 + 0.76  80 

Settlement at 50 days = (256)(0.178) ≈ 45.6 mm

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Chapter 6 6.1

a. Eq. (6.10): qu = c′Nc + q Nq +

1 γ BNγ 2

φ ′ = 28°. Table 6.1: Nc = 31.61; N q = 17.81; N γ = 13.7 qall =

qu 1  1  = (400)(31.61) + (110)(3)(17.81) + (110)(3)(13.7)  FS 4  2 

= 5195 lb / ft 2 b. φ ′ = 35°. Table 6.1: Nc = 57.75; N q = 41.44; N γ = 45.41

qall =

qu 1  1  = (0)(57.75) + (17.8)(1.2)(41.44) + (17.8)(1.5)(45.41)  FS 4  2 

= 372.8 kN / m2 c. Table 6.1: φ ′ = 30°; N q = 22.46; N γ = 19.13 Eq. (6.19), with c′ = 0: qu 1 = ( q Nq + 0.4γ BNγ ) FS FS 1 = [( 2 × 16.5)( 22.46) + (0.4)(16.5)(3)(19.13)] = 280 kN/m2 4

qall =

6.2

For φ ′ = 25°, from Table 6.1, Nc = 25.13; N q = 12.72; N γ = 8.34 Eq. (6.19): qu = 1.3c′N c + qN q + 0.4γBN γ = (1.3)( 200)( 25.13) + (115 × 3)(12.72) + ( 0.4)(114 )(5)(8.34) = 12,840 lb/ft 2

qall =

qu 12,840 = = 4280.1 lb/ft 2 FS 3

Maximum allowable load: Qall = 4280.1 × 5 × 5 = 107,003 lb ≈ 107.0 kip

6.3

a. Eq. (6.28): qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi +

1 γ BNγ Fγs Fγd Fγi 2

Vertical load, so Fci , Fqi , and Fγi are equal to 1.

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Continuous foundation, so Fcs , Fqs , and Fγ s are equal to 1.

φ ′ = 28°. Table 6.2: Nc = 25.8; N q = 14.72; N γ = 16.72.

Table 6.3:

 3 = 1 + 2 tan 28(1 − sin 28)2   = 1.299 B  3 1 − Fqd 1 − 1.299 Fcd = Fqd − = 1.299 − = 1.321 N c tan φ ′ (25.8)(tan 28) Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df

Fγd = 1 qall = =

qu 1  1  =  c′N c Fcd + q Nq Fqd + γ BNγ Fγd  FS FS  2  1 1  ( 400 )( 25 .

8 )( 1 .

321 ) + ( 110 × 3 )( 14 .

72 )( 1 .

299 ) + (110)( 3)(16.72)(1)  4 2 

= 5675 lb/ft 2 b. c′ = 0 qall =

1  1   q Nq Fqd + γ B Nγ Fγd  FS  2 

φ ′ = 35°. Table 6.2: N q = 33.3; N γ = 48.03 Table 6.3: Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

 1 .2  = 1 + 2 tan 35(1 − sin 35)2   = 1 .2 B  1.5 

Df

Fγd = 1 qall =

1 1  (1.2 × 17.8)( 33.3)(1.2) + (17.8)(1.5)( 48.03)(1) = 373.7 kN/m2  4 2 

c. qu = q Nq Fqd Fqs +

1 γ BNγ Fγd Fγs 2

φ ′ = 30°. From Table 6.2: N q = 18.4; N γ = 22.4 Table 6.3: Fγ s = 0.6; Fγ d = 1; Fqs = 1 + tan φ ′ = 1.577

Fqd = 1 + 2 tan φ ′(1 − sin φ ′) 2

2 = 1 + 0.289  = 1.193 B  3

Df

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qall =

1 1  (2)(16.5)(18.4)(1.193)(1.577) + (16.5)(3)( 22.4)(0.6)(1)  4 2 

= 368.8 kN/m2

6.4

Eq. (6.28): qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi For φ ′ = 25°, from Table 6.2, Nc = 20.72; N q = 10.66; N γ = 10.88 Shape factors:  B  N   5  10.66  Fcs = 1 +   q  = 1 +    = 1.51  L  N c   5  20.72  B 5 Fqs = 1 +   tan φ ′ = 1 +   tan 25 = 1.47 L 5 B Fγs = 1 − 0.4  = 1 − 0.4(1) = 0.6 L Depth factors: Fci , Fqi , Fγi

 3 = 1 + 2 tan 25(1 − sin 25)2   = 1.19 B  5 1 − Fqd 1 − 1.19 Fcd = Fqd − = 1.19 − = 1.21 (20.72)(tan 25) N c tan φ ′ Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df

Fγd = 1 Since there is no inclination in the load, all three inclination factors ( Fci , Fqi , Fγi ) are 1. qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = ( 200)( 20.72)(1.51)(1.21)(1) + (3 × 115)(10.66)(1.47 )(1.19)(1) +0.5(115)(5)(10.88)( 0.6)(1)(1) = 7,571.5 + 6,433.4 + 1,875.8 = 15,880.7 lb/ft 2

qall =

qu 15,880.7 = = 5293.6 lb/ft 2 FS 3

Maximum allowable load, Qall = 5293.6 × 5 × 5 = 132,339 lb = 132.3 kip

6.5

1  1  2 ′  1  Qall =   B c N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + γ BNγ Fγs Fγd Fγi   2  FS    cos 20°  SOURCE: Browsegrades.net

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Note: Qall = inclined allowable load

φ ′ = 25°. Table 6.2: Nc = 20.72; N q = 10.66; N γ = 10.88.

Fcs = 1 +

B Nq  5  10.66  = 1 +    = 1.514 L Nc  5  20.72 

B  5 tan φ ′ = 1 +  (tan 25) = 1.466 L  5 B Fγs = 1 − 0.4 = 1 − 0.4(1) = 0.6 L Fqs = 1 +

 3 = 1 + 0.311  = 1.187 B  5 1 − Fqd 1 − 1.187 Fcd = Fqd − = 1.187 − = 1.206 ( 20.72)(tan 25) N c tan φ ′ Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df

Fγd = 1 2

20   Fci = Fqi = 1 −  = 0.605 90   2

2

 20  β  Fγi = 1 −  = 1 −  = 0.04 25    φ′ 

(600)( 20.72)(1.514)(1.206)( 0.605)  2  1   1  5   qall =   + (115 × 3)(10.66)(1.466)(1.187)(0.605) = 156.8 kip   cos 20°   3  1000    + (0.5)(115)(5)(10.88)(0.6)(1)(0.04) 

6.6

For φ ′ = 23°, from Table 6.2, Nc = 18.05; N q = 8.66; N γ = 8.20 B = 2.0 m, and Df = 1.5 m For strip foundation, L > B; hence, B/L ≈ 0 Shape factors:

Fcs = 1 +

B Nq ≈1 L Nc

Fqs = 1 +

B tan φ ′ ≈ 1 L

Fγs = 1 − 0.4

B ≈1 L

SOURCE: Browsegrades.net

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Depth factors:

 1.5  = 1 + 2 tan 23(1 − sin 23)2   = 1.24 B  2  1 − Fqd 1 − 1.24 Fcd = Fqd − = 1.24 − = 1.27 N c tan φ ′ (18.05)(tan 23) Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df

Fγd = 1 Since there is no inclination in the load, all three inclination factors ( Fci , Fqi , Fγi ) are 1.

From Eq. (6.28), the ultimate bearing capacity is given by: qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = (5)(18.05)(1)(1.27 )(1) + (1.5 × 19 )(8.66)(1.24 )(1) + 0.5(19 )( 2 )(8.20)(1)(1)(1) = 576.5 kN/m 2

The pressure applied on the soil is 350/2 = 175 kN/m2 FS =

6.7

576.5 = 3.29 (Using gross pressures) 175

For φ ′= 24°, from Table 6.2, Nc = 19.32; N q = 9.60; N γ = 9.44.

Shape factors:

Fcs = 1 +

B Nq 1.5 9.60 = 1+ × = 1.50 L Nc 1.5 19.32

Fqs = 1 +

B 1 .5 tan φ ' = 1 + tan 24 = 1.45 L 1 .5

Fγs = 1 − 0.4

Depth factors :

B 1 .5 = 1 − 0 .4 × = 0.60 L 1 .5

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df B

= 1 + 2 tan 24(1 − sin 24)2 × Fcd = Fqd −

1 − Fqd N c tan φ ′

= 1.21 −

1 = 1.21 1.5 1 − 1.21 = 1.23 19.32 tan 24

Fγd = 1

Inclination factors:

Fci = 1 ; Fqi = 1 ; Fγi = 1 SOURCE: Browsegrades.net

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qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = (10)(19.32)(1.5)(1.23)(1) + (1 × 19 )( 9.6)(1.45)(1.21)(1) +0.5(19)(1.5)( 9.44)( 0.60)(1)(1) = 757.2 kN/m 2

qall =

qu 757.2 = = 252.4 kN/m 2 FS 3

Qall = qall × BL = 252.4 × 1.5 × 1.5 = 576.8 kN

6.8

a. Terzaghi’s bearing capacity equation and factors For φ ′ = 34°, from Table 6.1: Nc = 52.64; N q = 36.50; N γ = 38.04.

qu = c′Nc + q Nq + 0.5γ BNγ = 0 + (1 × 19.5)(36.5) + 0.5(19.5)( 2)(38.04) = 1453.5 kN/m2 qall =

1453.5 = 484.5 kN/m 2 3

Qall = 484.5 × 2 = 969.0 kN/m b. Meyerhof’s modified bearing capacity equation For φ ′ = 34°, from Table 6.2: Nc = 42.16; N q = 29.44; N γ = 41.06.

B = 2.0 m, and Df = 1.0 m For strip foundation, L > B; hence, B/L ≈ 0 Shape factors:

Fcs = 1 +

B Nq ≈1 L Nc

Fqs = 1 +

B tan φ ′ ≈ 1 L

Fγs = 1 − 0.4

Depth factors:

B ≈1 L

Fqd = 1 + 2 tan ϕ ′(1 − sin ϕ ′) 2

Df B

= 1 + 2 tan 34(1 − sin 34) 2 × Fcd = Fqd −

1 − Fqd N c tan φ ′

= 1.21 −

1 = 1.13 2

1 − 1.21 = 1.23 19.32 tan 24

Fγd = 1 SOURCE: Browsegrades.net

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Since there is no inclination in the load, all three inclination factors are unity ( Fci , Fqi , and Fγi ).

From Eq. (6.28), the ultimate bearing capacity is given by qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = (1 × 19.5)( 29.44 )(1)(1.13)(1) + 0.5(19.5)( 2)( 41.06)(1)(1)(1) = 1449 .4 kN/m 2

qall =

1449.4 = 483.1 kN/m 2 3

Qall = 483.1 × 2.0 = 966.2 kN/m

6.9

From Eqs. (6.24) and (6.28): qall =

1  1  c′N c Fcs Fcd + q Nq Fqs Fqd + γ BNγ Fγs Fγd − q  FS  2 

Note: Fci = Fqi = Fγi = 1 Qall(net) = qall(net) BL

φ ′ = 25°. Table 6.2: Nc = 20.72; N q = 10.66; Nγ = 10.88.

Table 6.3:

 2  10.66   B  N  Fcs = 1 +   q  = 1 +    = 1.343  L  N c   3  20.72  B 2 Fqs = 1 +   tan φ ′ = 1 +  (tan 25) = 1.31  L  3 B 2 Fγs = 1 − 0.4  = 1 − 0.4  = 0.73  L  3  1 .5  = 1 + 2 tan 25(1 − sin 25) 2   = 1.233 B  2  1 − Fqd 1 − 1.233 = 1.233 − = 1.257 Fcd = Fqd − N c tan φ ′ (20.72)(tan 25) Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Df

Fγd = 1

γ′ = (19.5 – 9.81) = 9.69 kN/m3 q = (1 × 17) + (9.69)(0.5) = 21.85 kN/m2

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Qall(net) =

( 2)( 3) (70)( 20.72)(1.343)(1.257) + ( 21.85)(10.66)(1.31)(1.233)  3 + (0.5)(9.69)( 2)(10.88)( 0.73)(1) − 21.85 

= 5760 kN

6.10

From Eq. (6.43): I r =

Es 2(1 + µs )(c′ + q′ tan φ′)

B   1 q′ = γ D f +  = (17)1 +  = 25.5 kN/m2 2   2 Ir =

1400 = 6.38 2(1 + 0.35)(72 + 25.5 tan 20)

From Eq. (6.44): I r ( cr ) = =

1   B  φ ′   exp  3.3 − 0.45  cot  45 −   2   L  2   1   1  20   exp  3.3 − 0.45  cot  45 −   = 40.36 2   2  2  

Since Ir(cr) > Ir , use Eqs. (6.45) and (6.47)   (3.07 sin φ ′) log(2 I r )   B Fγc = Fqc = exp  − 4.4 + 0.6  tan φ ′ +    1 + sin φ ′ L      1  (3.07 sin 20) log(2 × 6.38)   = exp  − 4.4 + 0.6  tan 20 +    = 0.534 2 1 + sin 20    

Fcc = Fqc −

1 − Fqc N c tan φ ′

Fcc = 0.534 −

. For φ ′ = 20°, Nc = 14.83.

1 − 0.5347 = 0.448 14.83 tan 20

Now, from Eq. (6.42): qu = c′N c Fcs Fcd Fcc + q Nq Fqs Fqd Fqc +

1 γ BNγ Fγs Fγd Fγc 2

From Table 6.2, for φ ′ = 20°, Nc = 14.83; N q = 6.4; N γ = 5.39.

Table 6.3:  N  B   6 .4   1  Fcs = 1 +  q   = 1 +    = 1.216  14.83  2   N c  L  SOURCE: Browsegrades.net

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B 1 Fqs = 1 +   tan φ ′ = 1 +  (tan 20) = 1.182 L 2 B 1 Fγs = 1 − 0.4  = 1 − 0.4  = 0.8 L 2

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2 Fcd = Fqd −

1 − Fqd N c tan φ ′

 1 = 1 + 2 tan 20(1 − sin 20)2   = 1.315 B  1

Df

= 1.315 −

1 − 1.315 = 1.373 (14.83)(tan 20)

Fγd = 1

Thus, qu = ( 72 )(14.83)(1.216)(1.373)( 0.448) + (1 × 17 )( 6.4)(1.182 )(1.315)( 0.534 ) 1 + (17 )(1.0)(5.39)( 0.8)(1)( 0.534 ) = 909 kN/m 2 2

6.11

B′ = B – 2e = 1.5 – (2 × 0.15) = 1.2 m; L′ = 1.5 m Eq. (6.55): qu′ = q Nq Fqs Fqd +

1 γ BNγ Fγs Fγd 2

φ ′ = 36°. Table 6.2: N q = 37.75; Nγ = 56.31  B′   1.2  Fqs = 1 +   tan ϕ ′ = 1 +   (tan 36) = 1.58  L′   1.5  Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

 1  = 1 + 2 tan 36(1 − sin 36)2   = 1.165 B  1.5 

Df

 B′   1 .2  Fγs = 1 − 0.4  = 1 − 0.4  = 0.68  L′   1 .5  Fγd = 1 qu′ = (1 × 17)( 37.75)(1.58)(1.165) +

1 (17 )(1.2)(56.31)( 0.68)(1) 2

= 1571.9 kN/m 2

Qall =

6.12

qu′ B′L′ (1571.9)(1.2)(1.5) = = 707.3 kN FS 4

Eq. (6.59): Qall =

Qult BL  1  = q Nq( e ) Fqs( e ) + γB Nγ ( e ) Fγs ( e )   4 4  2  SOURCE: Browsegrades.net

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φ ′ = 36°.

e 0.15 = = 0.1 . From Figures 6.22 and 6.23, N q ( e ) ≈ 38; Nγ ( e ) ≈ 32 B 1 .5

Eqs. (6.61) and (6.62): Fqs ( e ) = 1

 2e B   3  e    B  Fγs ( e ) = 1 +  − 0.68  + 0.43 −     B L   2  B    L 

2

  1 .5   1 .5    3 = 1 + [( 2)(0.1) − 0.68]  + 0.43 −  (0.1)   1 .5   2   1 .5  = 1 − 0.48 + 0.28 = 0.8

Qall =

6.13

2

(1.5)(1.5)  1  ( 1 )( 17 )( 38 )( 1 ) + (17)(1.5)(32)(0.8) = 546.98 kN ≈ 547 kN  4 2  

c′ = 0; φ ′ = 35° qu = q Nq Fqs Fqd +

1 γ BNγ Fγs Fγd 2

Table 6.2: N q = 33.3; N γ = 48.03 Fqs = Fγs = 1 Fγd = 1

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

 0.9  = 1 + 2 tan 35(1 − sin 35)2   = 1.127 B  1.8 

Df

1 (16)(1.8)( 48.03)(1)(1) 2 = 540.4 + 691.6 = 1232 kN/m 2

qu = (16)( 0.9)( 33.3)(1)(1.127 ) +

Eq. (6.63): RK = 1 −

qu ( e ) qu (centric)

e Eq. (6.64): RK = a   B For

Df B

=

k

0.9 = 0.5 , a = 1.754; k = 0.8 1.8

1.754(0.12)0.8 = 1 −

qu ( e ) qu (centric)

= 1−

qu ( e ) 1232

. qu ( e) = 835.8 kN/m 2

Qu = B × qu(e) = (1.8)(835.8) = 1504.4 kN/m

SOURCE: Browsegrades.net

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6.14

Eccentricity, e =

M 300 = = 0.10 m Q 3000

B′ = B – 2e = 2.00 ‒ (2)(0.1) = 1.80 m For φ ′ = 32°, from Table 6.2: Nc = 35.49; N q = 23.18; N γ = 30.22 Shape factors:

Fcs = 1 +

B′ N q 1.8 23.18 = 1+ × = 1.39 L Nc 3 35.49

Fqs = 1 +

B′ 1 .8 tan 32 = 1.37 tan φ ′ = 1 + L 3 B′ 1 .8 = 1 − 0 .4 × = 0.76 L 3

Fγs = 1 − 0.4

Depth factors:

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2 Fcd = Fqd −

1 − Fqd N c tan φ ′

2 = 1 + 2 tan 32(1 − sin 32)2   = 1.28 B 2

Df

= 1.28 −

1 − 1.28 = 1.29 35.49 tan 32

Fγd = 1.00

Inclination factors: Fci = 1; Fqi = 1; Fγi = 1 From Eq. (6.55), the ultimate bearing capacity is given by qu′ = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = (5)( 35.49)(1.39)(1.29)(1) + ( 2 × 18.5)( 23.18)(1.37)(1.28)(1) +0.5(18.5 × 1.8)( 30.22 )( 0.76)(1)(1) = 2204.6 kN/m 2

Qult = qu′ × B′L = ( 2204.6)(1.8 × 3) = 11,904.7 kN FS =

11,904.7 = 3.97 3000

SOURCE: Browsegrades.net

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600 kN/m

6.15 50 kN/m

80 kN⋅m/m 0.5 m

GL 0.5 m

O e Q v 1.0 m

0.5 m

Qh

1.0 m

a. The line load passes through the center of the foundation base.

The inclination β of the load to the vertical is given by

 30  β = tan −1   = 3.4°  500  e=0 b. There is no inclination in the line load. The eccentricity is given by e=

50 = 0.1 m = 100 mm 500

c. The vertical and horizontal components of the reaction at the soil-foundation interface are Qv and Qh, with eccentricity of e.

From equilibrium considerations, Qh = 50 kN/m and Qv = 600 kN/m.

The inclination of the load to vertical is

 50   = 4.8° .  600 

β = tan−1

By taking the moment about O, 50 × 1.0 – 80 – 600 × e = 0 e = ‒ 0.05m = ‒ 50 mm (i.e., 50 mm to the left of the center line)

6.16

For φ ′ = 35°, from Table 6.2: N γ = 48.03 Table 6.3: Fγs = 1 − 0.4

B ≈1 L SOURCE: Browsegrades.net

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Fγd = 1.00

With no inclination in the load, Fγi = 1.00 From Eq. (6.28): qu = 0.5γ BNγ Fγs Fγd Fγi = 0.5(19)(1.5)( 48.03)(1)(1)(1) = 684.4 kN/m2 qall =

qu 684.4 = = 228.1 kN/m2 FS 3

Area of the foundation =

π

× 132 −

π

× 10 2 = 54.2 m2

4 4 Therefore, the maximum silo load that can be allowed on the ring beam is

54.2 × 228.1 = 12,363 kN or 12.36 MN

6.17

We consider the long-term (drained) and the short-term (undrained) stability separately and see which of the two is more critical.

Long-term (Drained) Stability For φ ′ = 26°, from Table 6.2: Nc = 22.25; N q = 11.85; N γ = 12.54 Shape factors:

Fcs = 1 +

B Nq 2.0 11.85 = 1+ × = 1.53 L Nc 2.0 22.25

Fqs = 1 +

B 2 .0 tan 26 = 1.49 tan φ ′ = 1 + L 2 .0 B 2 .0 = 1 − 0 .4 × = 0.60 L 2 .0

Fγs = 1 − 0.4

Depth factors:

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2 Fcd = Fqd −

1 − Fqd N c tan φ ′

 1  = 1 + 2 tan 26(1 − sin 26)2   = 1.15 B  2.0 

Df

= 1.15 −

1 − 1.15 = 1.16 22.25 tan 26

Fγd = 1.00

Inclination factors: Fci = 1; Fqi = 1; Fγi = 1 qu = c′N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γ BNγ Fγs Fγd Fγi = (1 × 19 )(11.85)(1.49 )(1.15)(1) + 0.5(19 )( 2 )(12.54 )( 0.6)(1)(1) = 528 .8 kN/m 2 SOURCE: Browsegrades.net

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qu 528.8 = = 176.3 kN/m 2 FS 3 Qall = qall × BL = (176.3)( 2 × 2) = 705.2 kN qall =

Short-term (Undrained) Stability For φ = 0, from Table 6.2, Nc = 5.14; N q = 1; N γ = 0 Shape factors:

Fcs = 1 +

B Nq 2 .0 1 = 1+ × = 1.195 L Nc 2.0 5.14

Fqs = 1 +

B 2 .0 tan 0 = 1.0 tan φ ′ = 1 + L 2 .0

B Fγs = 1 − 0.4  = 0.6 L

Depth factors:

Fcd = 1 + 0.4

Df B

= 1 + 0 .4 ×

1.0 = 1.20 2 .0

Fqd = Fγd = 1 qu = cu N c Fcs Fcd Fci + q Nq Fqs Fqd Fqi + 0.5γBNγ Fγs Fγd Fγi = (60)(5.14)(1.195)(1.2)(1) + (19 × 1)(1)(1)(1)(1) = 461.2 kN/m 2

qall =

461.2 = 153.7 kN/m2 3

Qall = qall × BL = 153.7 × 2.0 × 2.0 = 615.0 kN Short-term stability is more critical than long-term stability. So, the maximum load Q = 615.0 kN.

6.18

qu = q Nq Fqs Fqd +

1 γ BNγ Fγs Fγd 2

q = (2)(105) + (2)(118 – 62.4) = 321.1 lb/ft2 e=

5 − 2 = 0.5 ft.

2

Table 6.2: φ ′ = 35°; N q = 33.3; N γ = 48.03 Table 6.3: Fqs = Fγs = Fγd = 1

4 Fqd = 1 + 2 tan 35(1 − sin 35) 2   = 1.204 5 SOURCE: Browsegrades.net

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qu = (321.2)( 33.3)(1)(1.204 ) +

1 (118 − 62.4)(5)( 48.03)(1)(1) ≈ 19,554 lb/ft 2 2

e    0.5  2 Eq. (6.67): qu ( e ) = qu 1 − 2  = 19,554 1 − ( 2)  = 15,643 lb/ft B 5      Qu = B × qu ( e ) = (5)(15,643) = 78,215 lb/ft ≈ 78.2 kip/ft

6.19

e=

70 = 0.156 m; c′ = 0 450

1   Eq. (6.59): Qult = B 2  q Nq ( e ) Fqs ( e ) + γ ′BNγ ( e ) Fγs ( e )  2  

φ ′ = 30°;

e 0.156 = B B

Qult = (450)(FS) = (450)(6) = 2700 kN Fqs ( e ) = 1

B=L   2 × 0.156   0.156  2 Fγs ( e ) = 1 +  − 0.68 (1) + 0.43 − (1.5) (1) B    B   0.234   0.312   = 1+  − 0.68  +  0.43 −  B   B   0.312 0.234 = 0.75 + − B B

Hence,

 1 0.312 0.234    2700 = B 2 (1.2 × 16)[ N q ( e ) ](1) + (19 − 9.81)( B )[ N γ ( e ) ] 0.75 − −  2 B B    0.312 0.234    2700 = B 2 19.2 N q( e) + (4.595)( B ) N γ ( e)  0.7 − −  B B    TRIAL AND ERROR: Let B = 2 m;

(a)

e 0.156 = = 0.078 ; φ ′ = 30° B B

From Figures 6.22 and 6.23, N q ( e ) ≈ 20; Nγ ( e ) ≈ 14 Right-hand size of Eq. (a): 0.312 0.234   ( 2) 2 (19.2)( 20) + ( 4.595)( 2)(14 ) 0.75 − −  2 2    = 1781.4 kN < 2700 kN SOURCE: Browsegrades.net

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Let B = 2.5 m;

e 0.156 = = 0.0624; φ ′ = 30° B 2 .5

From Figures 6.22 and 6.23, N q ( e ) ≈ 17; Nγ ( e ) ≈ 16 Right-hand side of Eq. (a): 0.312 0.234    ( 2.5)2 (19.2)(17) + (4.595)( 2.5)(16) 0.75 − −  2 2    = 2588 kN < 2700 kN So, B ≈ 2.6 m

6.20

eB 0.12 0.36  1 e = = 0.08  < ; L = = 0.16 (< 0.5).

B 1.5  6  L 2.25 So, Case II; Figure 6.27 applies.

From Figure 6.27,

L1 L = 0.92; 2 = 0.33 L L

L1 = (0.92)(2.25) = 2.07 m; L2 = (0.33)(1.5) = 0.495 m Eq. (6.75): A′ =

1 1 ( L1 + L2 ) B = ( 2.07 + 0.495)(1.5) = 1.923 m 2 2 2

Eq. (6.76): B′ =

A′ 1.923 = = 0.929 m L1 2.07

Eq. (6.77): L′ = 2.07 m qu′ = q Nq Fqs Fqd +

1 γ B′Nγ Fγs Fγd 2

Table 6.2: For φ ′ = 35°; N q = 33.3; N γ = 48.03

 B′   0.929  Fqs = 1 +   tan φ′ = 1 +  (tan 35) = 1.314  L′   2.07   B′   0.929  Fγs = 1 − 0.4  = 1 − 0.4  = 0.82  L′   2.07   0.8  Fqd = 1 + 2 tan 35(1 − sin 35)2   = 1.316  1.5  Fγd = 1

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qu′ = (17 × 0.8)( 33.3)(1.314 )(1.316) +

1 (17)( 0.929 )( 48.03)( 0.82)(1) 2

= 1094.13 kN/m 2

Qall =

6.21

qu′ B′L′ (1094.13)( 0.929)( 2.07 ) = = 526 kN FS 4

e  β°   a. From Eq. (6.89): Qu ( ei ) = qu B 1 − 2 1 −  φ ′  B   qu = q Nq Fqd +

2−

Df B

1 γ BNγ Fγd 2

φ ′ = 38°; N q = 48.93; N γ = 78.03 (Table 6.2)  1  Fqd = 1 + 2 tan 38(1 − sin 38)2   = 1.165  1.4  Fγd = 1

qu = (17.5 × 1)( 48.93)(1.165) +

1 (17.5)(1.4)( 78.03)(1) = 1953.43 kN/m 2 2 1   2−  1.4 

  0.15    18   Qu ( ei ) = (1953 .43)(1.4) 1 − ( 2)  1 −    1.4    38   1.5 − 0.7

e  β°   b. Eq. (6.90) : Qu ( ei ) = qu B 1 − 2 1 −  φ ′  B  

≈ 941.4 kN/m

Df B

 1  1.5 − 0.7    1.4 

  0.15    18  = (1953.43)(1.4) 1 − ( 2)  1 −    1.4    38   = 1130.9 kN/m

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SOURCE: Browsegrades.net

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Chapter 7 7.1

B = 7.5 ft and H = 3.0 ft; H/B = 0.4 For φ ′ = 25° and H/B = 0.4, N c* = 55; N q* = 23; N γ* = 7.0 (Figures 7.3, 7.4, 7.5) Eq. (7.2):

qu = c′Nc* + qNq* + 0.5γBNγ* = (210.0)(55) + (115.0 × 3)( 23) + 0.5(115.0)(7.5)(7.0) = 22,504 lb/ft 2 = 22.5 kip/ft2

7.2

The depths beyond which the presence of a rigid stratum will not influence the ultimate bearing capacity (for φ ′ = 25°) are:

N c* → D = 1.4B (from Figure 7.3) N q* → D = 1.4B (from Figure 7.4) N γ* → D = 0.6B (from Figure 7.5)

Therefore, with H > 12.0 ft, there is no influence from the rigid stratum, and we can use the bearing capacity factors from Table 6.2.

At φ ′ = 25°, Nc* = 20.72; Nq* = 10.66; Nγ* = 10.88 Note that the lower bound Nγ* values in Figure 7.5 are not Vesic’s bearing capacity factors that are used in this book.

Eq. (7.2):

qu = c′Nc* + qNq* + 0.5γBNγ* = ( 210.0)( 20.22) + (115.0 × 3)(10.66) + 0.5(115.0)(7.5)(10.88) = 12,721 lb/ft 2 = 12.7 kip/ft 2

7.3

B = 1.5 m and H = 1.0 m, H/B = 0.67 For φ ′ = 40° and H/B = 0.67, m1 = 0.4 and m2 = 0.45 (Figure 7.6) For φ ′ = 40° and H/B = 0.67, N q* = 210, N γ* = 160 (Figures 7.4 and 7.5) From Eqs. (7.4) and (7.5), SOURCE: Browsegrades.net

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B  1.5  Fqs* ≈ 1 − m1   ≈ 1 − (0.4)  ≈ 0.70  L  2.0  B  1.5  Fγs* ≈ 1 − m1   ≈ 1 − (0.45)  ≈ 0.66 L  2.0  The ultimate bearing capacity is given by Eq. (7.3): qu = qNq* Fqs* + 0.5γBNγ* Fγs* = (1.0 ×18.5)(210)(0.70) + 0.5(18.5)(1.5)(160)(0.66) = 4, 185 kN / m 2

7.4

The depths beyond which the presence of a rigid stratum will not influence the ultimate bearing capacity (for φ ′ = 40°) are: N q* → D = 2.4 B (from Figure 7.4) N γ* → D = 1.2 B (from Figure 7.5)

Therefore, with H > 4.0 m, there is no influence from the rigid stratum, and we can use the bearing capacity factors from Table 6.2.

At φ ′ = 40°, N q* = 64.20; N γ* = 109.41 Note that the lower bound N γ* values in Figure 7.5 are not Vesic’s bearing capacity factors that are used in this book.

The ultimate bearing capacity is given by Eq. (7.3) as qu = qNq*Fqs* + 0.5γBNγ* Fγs* = (1.0 × 18.5)( 64.20)( 0.70) + 0.5(18.5)(1.5)(109.41)( 0.66) = 1,833 kN/m 2

7.5

Eq. (7.10):

  B  0.5 H  − 0.707    qu ( square ) = 5.14 1 + cu + q 5.14        2 .0   0.5 0.5  − 0.707    = 5.14 1 + (40.0) + (1 × 19.0) = 276.3 kN/m2 5.14    

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7.6

Referring to Figure 7.8a: B = 2.0 m; Df = 1.5 m; H = 0.5 m; hence, H/B = 0.25 cu ( 2 ) cu (1) = 60 40 = 1.5

From Figure 7.8b, N c = 6.7 Table 6.3: Fcs = 1.0

D   1. 5  Fcd = 1 + 0.4 f  = 1 + 0.4  = 1 .3  2 .0   B 

Eq. (7.11):

qu = cu (1) N c Fcs Fcd + q = (40)(6.7)(1.0)(1.3) + (18.5 ×1.5) = 376.2 kN/m2 qall =

qu 376.2 = = 125.4 kN/m2 FS 3

Maximum wall load that can be allowed is qall × B = 251 kN/m

7.7

Table 7.2. For H/B = 0.25 and cu (1) cu ( 2 ) = 0.67, m = 1.033 N c Fcs = 5.14  Df  1 .5   = 1 + 0.4 Fcd = 1 + 0.4  = 1 .3  2 .0   B 

Eq. (7.12):

qu = cu (1) mNc Fcs Fcd + q = (40)(1.033)(5.14)(1.3) + (18.5 ×1.5) = 303.9 kN/m 2 qall =

qu 303.9 = = 101.3 kN/m2 FS 3

Maximum wall load that can be allowed is qall × B = 203 kN/m

7.8

B = 2.0 m; H = 0.5 m; Df = 1.0 m For φ ′ = 38°, from Table 6.2, N γ (1) = 78.03, N q (1) = 48.93

c N q2 (25.0)(5.14) = u ( 2) c( 2) = = 0.092 q1 0.5γ1BN γ (1) (0.5)(18.0)(2.0)(78.03) From Figure 7.10, Ks ≈ 3 Eq. (7.27) [for a continuous foundation]: SOURCE: Browsegrades.net

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2 D f  K s tan φ1′  qu = 5.14 cu ( 2 ) + γ1H 2  1 +  + γ1D f H  B  2 × 1  3 × tan 38  = 5.14 × 25 + (18)( 0.52 ) 1 + + 18 × 1  0 .5  2  = 128.5 + 26.4 + 18 = 172.9 kN/m 2 ← ← USE Eq. (7.26) [for a continuous foundation]:

qt = γ1D f N q(1) + 0.5γ1BNγ (1) = (18.0)(1)(48.93) + (0.5)(18.0)(2)(78.03) = 2285.3 kN/m2 qall =

qu 172.9 = = 57.6 kN/m2 FS 3

Qall = 57.6 × 2 = 115.2 kN/m

7.9

B = L = 2.0 m; Df = 1.5m; H = 1.5 m

q2 cu ( 2) 30 = = = 0.5 q1 cu (1) 60 Figure 7.11: ca cu (1) = 0.85 Eq. (7.32):

B  2c   B    qu = 1 + 0.2 (5.14)cu ( 2 ) + 1 +  a H  + γ1D f L  B  L        2  2 × 0.85 × 60  2   × 1.5 (19)(1.5) = 1 + 0.2 (5.14)(30) + 1 +   2.0  2      2  = 185 + 153 + 28.5 = 366.5 kN/m2

← ← USE

Eq. (7.33):

B  qt = 1 + 0.2 (5.14)cu (1) + γ1D f L  2  = 1 + 0.2 (5.14)(60) + (19.0)(1.5) = 398.6 kN/m2 2  With FS = 3, qall = 122.2 kN/m2 Allowable column load is: 122.2 × 2 × 2 = 489 kN

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7.10

B = 1.5 m; Df = 1.0 m; H = 1.5 m. D ≈ 1.5 m For a continuous foundation, shape factors are unity. γ (kN/m3)

φ ′ (deg)

Nq

Nγ

17.0 18.0

32 40

23.18 64.20

30.22 109.41

Upper layer Lower layer

Eq. (7.36): qt = c1′ Nc (1) Fcs (1) + γ1 D f Nq (1) Fqs (1) + 0.5γ1 B Nγ (1) Fγs (1) = (17 )(1)( 23.18)(1) + (0.5)(17 )(1.5)( 30.22 )(1) = 779 .37 kN/m 2

Eq. (7.37): qb = c2′ Nc ( 2 ) Fcs ( 2 ) + γ2 D f Nq ( 2 ) Fqs ( 2 ) + 0.5γ2 B Nγ ( 2 ) Fγs ( 2 ) = (18 )(1)( 64 .2 )(1) + ( 0.5)(18 )(1.5)(109 .41)(1) = 2632.64 kN Eq. (7.35): 2

2

H  1.5  2 qu = qt + ( qb − qt )  = 779.37 + (2632.64 − 779.37)  = 1821.8 kN/m D  2.0 

7.11

a. Assuming undrained conditions (φ = 0); Table 6.2: N c = 5.14 Nq = 1; Nγ = 0 Shape factors: Fcs = Fqs = Fγs = 1 Depth factors:

Fcd = 1 + 0.4

Df B

= 1 + 0 .4 ×

1 .5 = 1 .3 2 .0

Fqd = Fγd = 1

qu = cu Nc Fcs Fcd + q Nq Fqs Fqd + 0.5γB Nγ Fγs Fγd = (40)(5.14)(1)(1.3) + (18 × 1.5)(1)(1)(1) + 0 = 297.3 kN/m2 b. With the trench in place: W = 2.0 m; H = 4.0 m; φ ′ = 42° Figures 7.14, 7.15, and 7.16: N c ( T ) = 17; N q ( T ) = 7; Nγ ( T ) = 10 Eq. (7.38): qu = c2′ Nc ( T ) + D f γ2 Nq ( T ) + 0.5γ2 B Nγ ( T ) = ( 40 )(17 ) + (1.5)(18 .0)( 7.0) + 0.5(18.0)( 2.0)(10.0) = 1049 kN/m 2

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7.12

Eq. (7.40): qu = qNq ζ q +

1 γB Nγ ζ γ 2

φ ′ = 35°. Table 6.1: Nq = 41.44; Nγ = 45.41. x = 2 = 1.67 .

B

1 .2

For φ ′ = 35°, from Figure 7.18, ζ q = 1.38; ζ γ = 2.39 qu = (1 × 16.8)( 41.44 )(1.38) +

1 (16.8)(1.2)( 45.41)( 2.39) = 2054.72 kN/m 2 2

qnet = qu − γD f = 2054.72 − (1 × 16.8) = 2037.92 kN/m 2 Net allowable bearing capacity =

7.13

2037.92 ≈ 509.5 kN/m2 4

a. B = 1 m; H = 4 m. Since B < H, Ns = 0.

Df B

=

1 = 1 .0 .

1

Eq. (7.43): qu = cu Ncq Figure 7.21: qall =

cu Ncq FS

b 2 = = 2; β = 60°; N cq ≈ 6.5 B 1 =

(68)( 6.5) = 147.3 kN/m2 3

b.

B (m)

b/B

N cq [Figure 7.21]

qu = cu Ncq (kN/m2)

0 1 1 1 2 1 3 1 4 1 5 1 6 1 Note: Df /B = 1

0 1 2 3 4 5 6

4.2 5.5 6.5 7.0 7.0 7.0 7.0

285.6 374 442 476 476 476 476

b (m)

7.14

Eq. (7.42): qu =

Df B

=

1 γB Nγ q 2

Df b 6 b 4 = 1; = = 1.5. From Figure 7.20 for = 1 and = 1.5, ϕ ′ = 40° 4 B 4 B B

and β = 30°, the value of Nγq ≈ 135.

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qu = 0.5γB Nγq = (0.5)(110)( 4)(135) = 29,700 lb/ft 2

7.15

  q  Eq. (3.52): φ ′ = tan −1 0.1 + 0.38 log c   σ o′   Depth (m) 2 4 6 8 10 15

qc (MN/m2) 1.73 3.6 4.9 6.8 8.7 13

σ o′ (MN/m2) φ ′ (deg) [Eq. (3.52)] 0.033 0.066 0.099 0.132 0.165 0.248

37 37.2 36.6 36.9 37 37 Average φ ′ ≈ 37°

From Figure 7.28, for φ ′ = 37°, N q ≈ 43; N γ ≈ 60 tan θ =

kh 0 .2 = = 0 .2 1 − kv 1 − 0

From Figure 7.29, for tan θ = 0.2 and φ ′ = 37°,

N γE Nγ

= 0.39 and

N qE Nq

= 0.58

N γE = (0.39)(60) = 23.4; N qE = (0.58)(43) = 24.9

Eq. (7.48): quE = q NqE +

7.16

Df B

=

1 1 γ BNγE = (16.5 × 1)( 24.9) + (16.5)(1.5)( 23.4) = 700.5 kN/m2 2 2

1 = 0.67 . For φ ′ = 37°, FS = 4.

1 .5

Using Figure 7.30 and interpolating, kh* ≈ 0.32.

From Figure 7.31 for kh* ≈ 0.32 and φ ′ = 37°, the value of tan α AE ≈ 1.1.

Eq. (7.49): −4

−4 V2   0.352  0.32  = 0.174  tan α AE  (1.1) Ag 0 .

30 0 .

3 × 9 .

81     = 0.0062 m = 6.2 mm

k* SEq = 0.174 h A

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7.17

φ′   Eq. (7.53): quc = 2c′ tan  45 +  2  35   50 = 2c′ tan 45 +  ; c′ = 26.0 MN/m2 2 

7.18

From Problem 7.17, φ ′ = 35° and c′ = 26.0 MN/m2 From Eq. (6.19): qu = 1.3c′Nc + qNq + 0.3γBNγ 35  φ′    where N c = 5 tan 4  45 +  = 5 tan 4  45 +  = 68.1 2 2  

φ′  35    N q = tan 6  45 +  = tan 6  45 +  = 50.3 2 2   N γ = Nq + 1 = 51.3

qu = (1.3)( 26 × 1000)(68.1) + (1.5 × 18.0)(50.3) + 0.3(26.0)( 2.0)(51.3) = 2303 × 103 + 1358 + 800.3 kN/m 2 = 2305 MN/m 2 Allowing for the rock joints, qu ( modified) = qu ( RQD )2 = 2305 × 0.702 = 1129 MN/m 2

qall =

qu 1129 = = 376 MN/m2 > 30 MN/m2 ( f c′) FS 3

Use qall as 30 MN/m2.

The maximum allowable load = 30 × 2 × 2 = 120 MN

7.19

Df B

=

D  5 = 1.25; φ ′ = 35°. Table 7.5 :  f  ≈ 5 . So it is a shallow foundation.

4  B cr

m = 0.25; K u = 0.936 Eq. (7.59):

  D  D  Fq = 1 + 2 1 + m f  f  Ku tan φ ′  B  B   = 1 + 2[1 + (0.25)(1.25)](1.25)(0.936)(tan 35) = 3.15 Qu = Fq A γ Df = ( 3.15)( 4 × 4)(112 )(5) = 28.224 lb ≈ 28.2 kip SOURCE: Browsegrades.net

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7.20

Df B

=

2 = 1.67 1.2

Eq. (7.63):

 Df    = 0.107 cu + 2.6 = (0.107)(74 ) + 2.5 = 10.4 > 7  B cr -square

D  So,  f  = 7 .  B sq

D  D    L  Eq. (7.64) :  f  =  f  0.73 + 0.27    B   B cr - rectangular  B cr -square   Df    2.4  = 7 0.73 + 0.27   < 1.55  1.2    B cr -square  Df    = 8.89. Hence, this is a shallow foundation.  B cr

Df 1.57 B Eq. (7.65): α ′ = = = 0.188 8.89  Df     B cr

B 1.2  Eq. (7.67): Fc*−rectangular = 7.56 + 1.44  = 7.56 + 1.44  = 8.28  L  2.4  From Figure 7.39, with α' = 0.188, the value of β' = 0.275.

Eq. (7.69): Qu = A( β ′Fc*cu + γD f ) = (1.2 × 2.4)[( 0.275)(8.28)(74) + (18)( 2)] ≈ 589 kN

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Chapter 8 8.1

a. Point O: The four point loads are at equal distance from O and will contribute equally to ∆σ. So, we calculate the vertical stress increase due to one load and multiply by four.

Due to one of the four point loads,

r = OA = (0.5b)2 + (0.5b)2 = 0.707b ; z = 0.5b; r/z = 1.414.

Eq. (8.1): ∆σ =

3P P = 0.1226 2 2 5/ 2 2π (0.5b) [1 + 1.414 ] b 2

The vertical stress increase due to the four point loads is 0.49

P b2

Point A: Due to the load at B (or D), r = b; z = 0.5b. Hence, r/z = 2.0.

Eq. (8.1): ∆σ =

3P

[

]

2 5/ 2

2π (0.5b)2 1 + 2.0

P = 0.0342 2 b

Due to the load at C, r = 1.414b; z = 0.5b. Hence r/z = 2.828 Eq. (8.1): ∆σ =

3P

[

]

2 5/ 2

2π (0.5b) 1 + 2.828 2

P = 0.0079 2 b

Due to the load at A, r = 0; z = 0.5b. Hence r/z = 0 Eq. (8.1): ∆σ =

3P P = 1.9099 2 5/ 2 2 b 2π (0.5b) [1 + 0]

Stress increase under A: (1.9099 + 2 × 0.0342 + 0.0079)

P P = 1,986 2 2 b b

b. For P = 500 kN and b = 5 m: ∆σ under O = 0.49

P = 9.8 kN/m2 b2

∆σ under A = 1.986

P = 39.7 kN/m2 2 b

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∆σ values were computed from Eq. (8.1) and plotted for the three depths in the following figure. It can be seen that ∆σ decays with depth and lateral distance.

8.3

Substituting P = 1000 kN in Eq. (8.1), values for ∆σ can be generated for r = 1 m, 2 m, and 4 m at different depths. They are plotted in the figure below.

∆σ

8.2

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8.4

a. Under the center [Eq. (8.3)]: z (ft)

∆σ (lb/ft2)

0.50 1.25 2.50 3.75 5.00 6.25 7.50 8.75 10.00 11.25 12.50 13.75 15.00 16.25 17.50 18.75 20.00

1198.8 1182.9 1092.7 940.8 775.7 628.6 508.8 414.6 341.3 284.3 239.5 204.0 175.4 152.3 133.3 117.5 104.3

b. Under the edge:

2z/B 0.1 0.3 0.5 1.0 1.5 2.0 2.5 3.0 4.0

∆σ /qo (Table 8.1) 0.484 0.451 0.417 0.332 0.256 0.196 0.151 0.118 0.075

z (ft)

∆σ (lb/ft2)

0.5 1.5 2.5 5.0 7.5 10.0 12.5 15.0 20.0

580.8 541.2 500.4 398.4 307.2 235.2 181.2 141.6 90.0

The following are the plots of ∆σ with z.

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8.5

z/(B/2) = 2.5/5 = 0.5. From Table 8.1, the ∆σ values are computed as follows.

r/(B/2) 0 0.2 0.4 0.6 0.8 1.0

∆σ /qo (Table 8.1) 0.911 0.902 0.869 0.796 0.646 0.417

r (ft)

∆σ (lb/ft2)

0 1 2 3 4 5

1093.2 1082.4 1042.8 955.2 775.2 500.4

The plot is shown below.

8.6

We compute the vertical stress increases due to the two line loads separately and add them.

Due to q1 = 30 kN/m, x = 6 m, z = 3 m. From Eq. (8.5):

∆σ 2 ∆σ 2 = = = ; ∆σ = 0.255 kN/m 2 2 2 q1 / z π[( x / z ) + 1] 30 / 3 π[(6 / 3) 2 + 1]2 Due to q2 = 40 kN/m, x = 3 m, z = 3 m:

∆σ 2 ∆σ 2 = = = ; ∆σ = 2.102 kN/m2 2 2 2 2 q2 z π[( x / z ) + 1] 40 / 3 π[(3 / 3) + 1] Total vertical increase at A due to the two line loads is 0.255 + 2.120 = 2.38 kN/m2 SOURCE: Browsegrades.net

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8.7

B = 9.0 ft, qo = 800.0 lb/ft2 Point A:

x = 6.0 ft, z = 3.0 ft Table 8.3: For 2z/B = 0.667 and 2x/B = 1.33, ∆σ/qo = 0.222 ∆σ = 0.222 × 800 = 177.6 lb/ft2

Point B:

x = 4.5 ft, z = 3.0 ft Table 8.3: For 2z/B = 0.667 and 2x/B = 1.0, ∆σ/qo = 0.493 ∆σ = 0.493 × 800 = 394.4 lb/ft2

Point C:

x = 0.0 ft, z = 3.0 ft Table 8.3: For 2z/B = 0.667 and 2x/B = 0.0, ∆σ/qo = 0.919 ∆σ = 0.919 × 800 = 735.2 lb/ft2

8.8

Point A:

Considering the whole raft, B = 8 m, L = 16 m, and z = 4 m. m = B/z = 8/4 = 2.0; n = L/z = 16/4 = 4.0 From Table 8.5, I = 0.2391 ∆σ = 0.2391 × 80 = 19.1 kN/m2

Point B:

Considering the left half of the raft, B = L = 8 m; z = 4 m. m = n = 8/4 = 2 From Table 8.5, I = 0.2325 ∆σ = 2 × 0.2325 × 80 = 37.2 kN/m2

Point C:

Considering a quarter of the raft, B = 4 m, L = 8 m; z = 4 m. m = 4/4 = 1.0 and n = 8/4 = 2.0 From Table 8.5, I = 0.1999 ∆σ = 4 × 0.1999 × 80 = 64.0 kN/m2

Point D:

Considering the upper half of the raft, B = 4 m, L = 16 m, and z = 4 m. m = 4/4 = 1.0 and n = 16/4 = 4.0 From Table 8.5, I = 0.2042 ∆σ = 2 × 0.2042 × 80 = 32.7 kN/m2

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8.9

Use Eq. (8.10) with I values derived from Table 8.5, noting that they are valid only below corners.

We label some points in the raft as follows.

At 4 m below point A: (Note: m, n values in parentheses) I = IAKDE + IAJBK + IAFCJ = I (1.0, 1.5) + I (1.5, 2.0) + I (2.0, 2.0) = 0.1935 + 0.2234 + 0.2325 = 0.6404 ∆σ = 0.6404 × 60 = 38.4 kN/m2 At 4 m below point B: I = IBDEJ + IBKFC – IBKAJ = I (3.0, 1.5) + I(3.5, 2.0) ‒ I(1.5, 2.0) = 0.2280 + 0.2385 – 0.2234 = 0.2431 ∆σ = 0.2431 × 60 = 14.6 kN/m2 At 4 m below point C: I = ICBDH + ICJAF – ICJEH = I (3.0, 3.5) + I(2.0, 2.0) ‒ I(3.5, 2.0) = 0.2447 + 0.2325 – 0.2385 = 0.2387 ∆σ = 0.2387 × 60 = 14.3 kN/m2

8.10

a. Considering a quarter of the foundation of B/2 × B/2, with z = B/2, m = n = 1.

Hence, I = 0.17522 ∆σ = 4 × 0.17522 × qo = 0.701qo SOURCE: Browsegrades.net

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b. From Eq. (8.15),

∆σ =

qo BL qo ( B)( B) = = 0.444qo ( B + z )( L + z ) ( B + B / 2)( B + B / 2)

c. From Figure 8.11, at z = 0.5B, ∆σ = 0.7qo (approximately)

8.11

a. Referring to Figure P8.11 and Eq. (8.25):

 H 2 I a ( H 2 ) − H1I a ( H1 )  ∆σ av = 4qo   H 2 − H1   50 × 2000 lb qo = = 4000 lb/ft 2 5×5

B L

H1 = 3 ft; H2 = 13 ft

For I a ( H 2 ) : m2 = n2 =

(5 2 ) = 0.192 B = H2 13 (5 2 ) = 0.192 L = H2 13

Figure 8.13: I a ( H 2 ) ≈ 0.09 For I a ( H1 ) : m2 =

(5 2 ) = 0.83 B = H2 3

n2 =

(5 2 ) = 0.83 L = H1 3

Figure 8.13: I a ( H1 ) ≈ 0.212

 (13)(0.09) − (3)(0.212)  ∆σ av = (4)(4000)  ≈ 854 lb/ft 2  13 − 3   b. As in Part a, consider a quarter of the foundation and multiply by four. qo = 4000 lb/ft2 z (ft) 3.0 8.0 13.0

B (ft) 2.5 2.5 2.5

L I m n (ft) (Table 8.5) 2.5 0.833 0.833 0.1512 2.5 0.313 0.313 0.0400 2.5 0.192 0.192 0.0165

∆σ = 4 qo I (lb/ft2) 2419 640 264

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Eq. (8.26):

∆σav ( H 2 / H1 ) =

∆σ t + 4∆σ m + ∆σ b 2419 + 4 × 640 + 264 = = 874 lb/ft2 6 6

c. Eq. (8.15): ∆σ =

qo BL where B = L = 5 ft and qo = 4000 lb/ft2. ( B + z )( L + z )

For z = 3 ft, ∆σ = 1562.5 lb/ft2 For z = 8 ft, ∆σ = 591.7 lb/ft2 For z = 13 ft, ∆σ = 308.6 lb/ft2 Eq. (8.26):

∆σav ( H 2 / H1 ) =

8.12

∆σ t + 4∆σ m + ∆σb 1562.5 + 4 × 591.7 + 308.6 = 706 lb/ft2 = 6 6

Point A:

B1 3 B 20 = = 0.6; 2 = = 4. Figure 8.20 : I ′ ≈ 0.472 z 5 z 5

∆ σ = (2)(10 × 17)(0.472) = 160.5 kN/m2 Point B:

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For 1:

B1 0 B 20 = = 0; 2 = = 4; I ′ ≈ 0.415 z 5 z 5

For 2:

B1 6 B 20 = = 1.2; 2 = = 4; I ′ ≈ 0.485 z 5 z 5

∆σ = (10 × 17)(0.415 + 0.485) = 153 kN/m2 Point C:

For 1:

B1 26 B 20 = = 5.2; 2 = = 4; I ′ ≈ 0.5 z 5 z 5

For 2:

B1 0 B 20 = = 0; 2 = = 4; I ′ ≈ 0.415 z 5 z 5

∆σ = (10 × 17)(0.5 − 0.415) = 14.45 kN/m2

8.13

For the Westergaard solution, µs = 0.2. Therefore [Eq. (8.32)]:

η=

1 − 2µ s 1 − 2 × 0 .2 = = 0.612 2 − 2µ s 2 − 2 × 0 .2

The computed values are shown in the table. b

a

∆σW ∆σ B z r r z 2 (ton/ft ) (ft) (ft) (ton/ft2) A 6 0 0.0 0.66 0.59 B 6 6 1.0 0.12 0.08 C 6 12 2.0 0.01 0.01 D 12 0 0.0 0.17 0.15 E 12 6 0.5 0.09 0.07 F 12 12 1.0 0.03 0.02 a ∆σ B is from Eq. (8.1); b ∆σW is from Eq. (8.31) Point

∆σ B is generally larger than ∆σ W .

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8.14

The values computed are given in the table and are plotted in the figure below. z (ft) 0.25 0.50 0.75 1.00 1.50 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 15.00

B 2z 12.000 6.000 4.000 3.000 2.000 1.500 1.000 0.750 0.600 0.500 0.429 0.375 0.333 0.300 0.273 0.250 0.231 0.214 0.200

∆σ B (ton/ft2) 2.00 1.99 1.97 1.94 1.82 1.66 1.29 0.98 0.74 0.57 0.45 0.36 0.29 0.24 0.20 0.17 0.15 0.13 0.11

∆σ W

(ton/ft2) 1.90 1.80 1.70 1.60 1.41 1.24 0.96 0.74 0.57 0.45 0.36 0.29 0.24 0.20 0.17 0.15 0.13 0.11 0.10

Note: ∆ σB is generally larger than ∆ σW . ∆σB from Eq. (8.3); ∆σW from Eq. (8.35).

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Chapter 9 9.1

Table 9.1. PI = 45; OCR = 2; β ≈ 400 E s = β cu = ( 400 )( 60 ) = 24,000 kN/m 2

L 3 H 4 = = 1.5; = = 2; A1 = 0.65 B 2 B 2 Df 1.5 = = 0.75; A2 = 0.93 B 2.0 Eq. (9.1): Se = A1 A2

9.2

qo B 60 × 2 = 0.65 × 0.93 × m = 3.02 mm Es 24,000

Bulk modulus K is defined as

K=

Mean normal stress Volumetric strain

When there is no volume change, volumetric strain is zero and, hence, K = ∞.

K ∞ 1 = =∞= E E 3(1 − 2 µs ) Thus, µs = 0.50

9.3

B = 6 ft; L = 12.0 ft; m′ = L/B = 2.0.

Figure 9.4 [or Eq. (9.6)]; Icenter = 1.53 Eq. (9.3): qo B (2.4 × 103 )(6.0) 2 Se = (1 − µ s ) I = (1 − 0.12 )(1.53)ft = 0.87 in.

3 Es (300.0 × 10 )

9.4

For one of the four quarters of the foundation, B = 3.0 ft, L = 6.0 ft; H = 9.0 ft m′ = L/B = 2.0; n′ = H/B = 9/3 = 3 Tables 9.3 and 9.4: F1 = 0.402; F2 = 0.084 Eq. (9.11) with µs = 0.1,

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 1 − 2 µs   1 − 2 × 0.1   F2 = 0.402 +  I s = F1 +  (0.084) = 0.477  1 − 0.1   1 − µs  From Eq. (9.3), for only a quarter of the foundation, Se =

qo B (2.4 ×103 )(3.0) (1 − µs2 ) I = (1 − 0.12 )(0.477)ft = 0.136 in.

3 Es (300.0 × 10 )

Considering the four quarters, settlement beneath the center is 4 × 0.136 = 0.54 in.

9.5

µs = 0.1; L/B = 2.0; Df /B = 3/6 = 1/2 (using B = 6 ft for the entire foundation) Figure 9.5b: If ≈ 0.76 Eq. (9.22):

Se = 4 ×

qo B ( 2.4 × 103 )(3.0) (1 − µs2 ) I s I f = 4 × (1 − 0.12 )(0.477)(0.76)ft 3 Es (300 × 10 )

= 0.41 in.

9.6

Consider a quarter of the foundation and multiply Is by four.

B = 1.0 m and L = 2.0 m; m′ = L/B = 2.0 H = 4.0 m (surface foundation); so, n′ = H/B = 4.0 Tables 9.3 and 9.4: F1 = 0.476; F2 = 0.069  1 − 2 µs   1 − 2 × 0 .2   F2 = 0.476 +  Eq. (9.11): I s = F1 +  ( 0.069 ) = 0.528  1 − 0 .2   1 − µs 

µs = 0.2; L/B = 2.0; Df /B = 1/2 (using B = 2 m for the entire foundation) Figure 9.5b: If ≈ 0.79 Considering the four quarters, from Eq. (9.22):

Se = 4 ×

qo B 150 × 103 × 2 (1 − µs2 ) I s I f = 4 × (1 − 0.22 )(0.528)(0.79)m 6 Es 16 × 10

= 30.0 mm

9.7

The equivalent foundation diameter Be is given by [Eq. (9.24)]: 4 BL 4×2×3 Be = = = 2.764 m

π

π

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β=

12 × 106 Eo = = 10.85 kBe 400 × 103 × 2.764

H 3 = = 1.085 Be 2.764 Figure 9.8: IG = 0.61 Eq. (9.28): I F =

π + 4

=

1   3   2t  Ef 4.6 + 10  B    Eo + e k  Be  2   1

π + 4

  9    (2)(0.25)  3 25 × 10  4.6 + 10   12 × 106 +  2.764  (400 × 103 )   2.764       2  = 0.794 1

Eq. (9.29): I E = 1 −

B  3.5 exp (1.22 µs − 0.4) e + 1.6   Df    1 = 1− = 0.924  2.764  3.5exp[(1.22)(0.4) − 0.4] + 1 .6   1 

Eq. (9.27):

qo Be I G I F I E (125 × 103 )(2.764)(0.61)(0.794)(0.924) 2 Se = (1 − µs ) = (1 − 0.22 )m Eo 9000 = 12.4 mm

9.8

Eq. (9.27): Se =

qo Be I G I F I E (1 − µs2 ) Eo

qo = 150 kN/m2 Be =

(4)( 3)2 4B2 = = 3.385 m π π

µs = 0.3; Eo = 16,000 kN/m2

β =

Eo 16,000 = = 11.82 kBe (400)(3.385)

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H 20 = = 5.91 Be 3.385 From Figure 9.8, IG ≈ 0.89.

Eq. (9.28) :

IF =

π + 4

=

π + 4

1      2t  3 Ef   4.6 + 10  E + Be k  Be  o 2   1   6    ( 2)(0.25)  3 15 × 10 4.6 + 10    16,000 +  3.385 (400)   3.385     2 

= 0.815

Eq. (9.29) :

IE = 1 −

= 1−

1 B  3.5 exp(1.22 µs − 0.4) e + 1.6  D   f  1  3.385  + 1.6  3.5 exp[(1.22)(0.3) − 0.4]  1.5 

= 0.923

 1 − 0.32  Se = (150)(3.385)( 0.89)(0.815)( 0.923)  = 0.0193 m = 19.3 mm  16,000 

9.9

For the square foundation, the I z − z diagram is shown in the following figure.

Here, z1 = 0.5B = 1.0 m

1.5 m

0.1

2.0 m 1.0

Iz Iz(m)

2.0 Sand z1 = 0.5B = 1.0 m 4.0 = (1.5 + 1.0)(18.5) = 46.3 z (m)

2

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I z ( m ) = 0.5 + 0.1

q−q 140 − 1.5 × 18.5 = 0.5 + 0.1 = 0.656 q′z (1) 46.3

The area of the influence factor diagram is given by

 0.1 + 0.656  (1.0) + 0.5(3.0)(0.656) = 1.362 m 2 

∑ I ∆z =  z

Eq. (9.35): Es = 2.5qc = 2.5 × 10 = 25 MN/m2

C1 = 1 − 0.5

q 1.5 ×18.5 = 1 − 0.5 = 0.88 q −q 140 − 1.5 ×18.5

C 2 = 1 + 0.2 log

25 = 1.48 0.1

From Eq. (9.30): 4 m I z ∆z

Se = C1C2 ( q − q ) ∑0

Es

 1.362  = ( 0.88)(1.48)(140 − 1.5 × 18.5) m 3  25 × 10 

= 8.0 mm

With C2 = 1.0, the settlement would be 4 m I z ∆z

Se = C1C2 ( q − q ) ∑0

Es

 1.362  = (0.88)(1.0)(140 − 1.5 × 18.5) m 3  25 × 10 

= 5.4 mm

Settlement expected due to creep during the 25 years: (8.0 – 5.4) = 2.6 mm

9.10

Eq. (9.31): I z ( m ) = 0.5 + 0.1

q −q q′z (1)

q − γDf = (18)(1.5) = 27 kN/m 2 qo = q − q = 195 − 27 = 168 kN/m 2

q′z (1) = ( Df + z1 )γ = (1.5 + B )γ = (4)(18) = 72 kN/m 2

I z ( m ) = 0.5 + 0.1

168 = 0.653 72

See the strain influence factor diagram on the next page.

Note: [Es = 3.5qc – Eq. (9.36)].

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Depth (m) ∆z (m) Es (kN/m2) Average Iz 0 –2 2.0 6,125 0.3812 2 – 2.5 0.5 12,075 0.6077 2.5 – 8 5.5 12,075 0.4135 8 – 10 2.0 10,150 0.087

(I z Es )(∆z ) 0.000124 0.000025 0.000188 0.000017 ∑0.000354

 q  27   10  = 1.4 C1 = 1 − 0.5  = 1 − 0.5  = 0.92; C2 = 1 + 0.2 log   195 − 27   0 .1  q −q Se = C1C2 ( q − q ) ∑

Iz ( ∆z ) = ( 0.92 )(1.4)(195 − 27 )( 0.000354 ) Es

= 0.0766 m = 76.6 mm

9.11

The strain influence diagram is shown on the next page.

Also, Es = 1.4Esquare = (1.4)(3.5qc) = 4.9qc

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(I z Es )(∆z )

Depth (m)

∆z (m)

Es (kN/m2)

Iz

0 – 1.25 1.25 – 2 2–8 8 – 10

1.2 0.75 6 2

8,575 8,575 16,905 14,210

0.4 0.5745 0.343 0.0685

Df B

=

0.000056 0.00005025 0.0001217 0.00000964 ∑0.0002376

1 .5 = 0.6 ; Table 9.6: Cd ≈ 0.84 2 .5

Eq. (9.39):

Iz ( ∆z ) = (0.84)(168)(0.0002376) = 0.0335 m = 33.5 mm 0 Es

z2

Se = Cd ( q − q) ∑

Post construction settlement:

Eq. (9.39):

   0 .1   t days  0.02    z2 log  1 day   ∑ (qc ∆z )  z2  

Note: qc is in MN/m2.

∑ (qc ∆z ) (1.75)( 2) + (3.45)(6) + ( 2.9)(2) = = 3 MN/m2 z2 10  0.1   365 × 10  Settlement = (0.02) (10) log  = 0.0237 m = 23.7 mm 1  3    Total settlement = 33.5 + 23.7 = 57.2 mm SOURCE: Browsegrades.net

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9.12

For net applied pressure = 150 kN/m2 and N60 = 20, Se,plate = 6 mm (Figure 9.16) From Eq. (9.43), 2 2  2B   1 D f   2 × 2.0   1 1.0    Se,footing = Se, plate  1 − × = ( 6 )     1 − ×    B + 0.3   3 B   2.0 + 0.3   3 2.0  = 15.1 mm

9.13

From Eq. (9.47): 2

qnet =

9.14

2

N 60  B + 1  25  9 + 1   1 6  2   Fd Se =   1 + × (1.2) = 11.3 kip/ft 4  B  4  9   3 9

 B  z′ = 1.4  Eq. (9.56): BR  BR 

0.75

 1  ; z ′ = (1.4)( 0.3)   0 .3 

0.75

= 1.036

2

L   0.7 1.25     B   q′  Se B       = α1α2α3  Eq. (9.57): L   BR   pa  BR  0.25 +  B  Normally consolidated sand: α1 = 0.14; α2 =

1.71 1.71 = = 0.053 1.4 ( N 60 ) (12)1.4

Table 9.7: α3 = 1. q′ = 153 kN/m2 2

 2  1.25 1    1  0.7  153  Se   = (0.14)(0.053)(1)      0 .3  0.25 + 2   0.3   100   1  Se ≈ 0.00879 m = 8.79 mm

9.15

The applied pressure at the footing level = From Eq. (9.63), Dr =

260 = 130 kN/m2 2

( N1 ) 60 32 = = 0.730 or 73.0% 60 60

The influence zone extends to 2B below the foundation. Therefore, the in situ effective overburden stress at the centre of the influence zone is computed as

σ o′ = 3.0 × 19 = 57.0 kN/m 2 SOURCE: Browsegrades.net

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The vertical stress increase at this depth can be estimated as [2:1 method; see Eq. (8.15)]

∆σ o′ =

130 × 2 = 65.0 kN/m2 ( 2 + 2)

The influence depth Z = 2B = 4.0 m (below the foundation).

From Figure 9.18, for Z/B = H/B = 2 and B/L = 0.1, (1 − µ s2 ) I ≈ 1.06; pa ≈ 100 kN/m2

 σ o′ + 0.5∆σ o′    pa  

0.5

 57.0 + 0.5 × 65.0  =  100  

0.5

= 0.946

For Dr = 73.0% and Z = 2B, from Eq. (9.62), KE, 0.1% = 759. From Eq. (9.60),

 σ ′ + 0.5∆σ o′   E0.1% = K E , 0.1% pa  o pa  

0.5

= 759 × 100 × 0.946 = 71,801 kN/m2

Substituting these values into Eq. (9.66),

 Se    B

0.3

=

125 × 130 × 1.06 = 0.240 ; Se/B = 0.00859 71,801

Settlement = 0.00859 × 2000 = 17.2 mm 9.16

The applied pressure at the footing level = 250 kN/m2 From Eq. (9.63); Dr =

( N1 ) 60 25 = = 0.646 or 64.6% 60 60

The influence zone extends to B below the foundation. Therefore, the in situ effective overburden stress at the center of the influence zone is computed as

σ o′ = 2.0 × 19 = 38.0 kN/m 2 The vertical stress increase at this depth can be estimated as [2:1 method; see Eq. (8.15)]

∆σ o′ =

250 × 2 × 2 = 111.1 kN/m2 (2 + 1)(2 + 1)

The influence depth Z = B = 2.0 m (below the footing) From Figure 9.18, for Z/B = H/B = 1 and B/L = 0.1, (1 − µ s2 ) I ≈ 1.06; pa ≈ 100 kN/m2

 σ o′ + 0.5∆σ o′    p a  

0.5

 38.0 + 0.5 × 111.1  =  100  

0.5

= 0.967

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For Dr = 64.6% and Z = B, KE, 0.1% = 680 [Eq. (9.61)] Eq. (9.60):

 σ ′ + 0.5∆σ o′   E0.1% = K E , 0.1% pa  o pa  

qo =

0.5

= 680 × 100 × 0.967 = 65,756 kN/m2

1000 = 250 kN/m2 2×2

Substituting these values in Eq. (9.66),

 Se    B

0.3

=

125 × 250 × 0.56 = 0.266 ; Se/B = 0.0121 65,756

Settlement = 0.0121 × 2000 = 24.2 mm

9.17

B = 5 ft; L = 10 ft; d = 5 ft; e = 0.5 ft B 5 Eq. (9.71): f L / B = 0.8 + 0.2  = 0.8 + (0.2)  = 0.9 L  10  e 0 .5  Eq. (9.72): f e (center) = 1 − 0.33  = 1 − (0.33)  = 0.967 B  5  2

2

δ 10 Eq. (9.74): f δ (center) = 1 −   = 1 −   = 0.988  90   90 

β = 26.6°, so it is 2H:1V slope. d  Eq. (9.77): f β ,d = 0.71 +  B 

0.15

5  = 0 .7  1 +  5 

0.15

= 0.777

f = fL/B fe fδ fβ,d = (0.9)(0.967)(0.988)(0.777) = 0.668 The table on the following page can now be prepared.

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 ∆R    R o  

(lb/in.2)

Se B

0.002 0.004 0.008 0.012 0.024 0.05 0.08 0.10 0.20

7.2 24.2 32.6 42.4 68.9 126.1 177.65 210.5 369.6

0.00048 0.00096 0.00192 0.00288 0.00576 0.012 0.0192 0.024 0.048

pp(m)

1

Se 2 (in.)

Γ3

qo 4 (lb/in.2)

Qo 5 (kip)

0.0288 0.0576 0.1152 0.1728 0.3456 0.72 1.152 1.44 2.88

2.27 2.17 2.0 1.86 1.7 1.35 1.1 1.07 0.9

10.92 35.08 43.55 52.68 78.24 113.72 130.54 150.46 222.2

78.62 252.58 313.56 379.3 563.33 818.78 939.89 1083.31 1599.84

S e  ∆R  ( 0.24) = B  Ro 

1

2

S  S  Se =  e ( B ) =  e (60) in. B B

3 Figure 9.24 4

f Γ pp ( m ) = 0.668Γ pp ( m ) = qo

5

Qo (kip) = qo BL =

( qo lb/in.2 )( 60 in. × 120 in.)

1000

The plot of Se versus Qo is shown.

From the plot, Qo at 1 in. settlement is 900 kip.

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9.18

Eq. (9.31): I z ( m ) = 0.5 + 0.1

q−q 200 − (18.5)(1.5) = 0 .5 + 0 .1 = 0.68 q′z (1) (18.5)( 3.0)

The strain influence diagram is shown:

0.68  0.1 + 0.68  At =  × 4.5 = 2.115 m  ×1.5 + 2 2   z = 0, Aw = At = 2.115 m z = 0.75 m, Aw = 2.115 – 0.5(0.1+0.39)(0.75) = 1.931 m z = 1.5 m, Aw = 0.5(4.5)(0.68) = 1.530 m z = 2.0 m, Aw = 0.5(4.0)(0.604) = 1.209 m z = 3.0 m, Aw = 0.5(3.0)(0.453) = 0.680 m z = 4.0 m, Aw = 0.5(2.0)(0.302) = 0.302 m z = 1.0 m, Aw = 0.5(1.0)(0.151) = 0.076 m The additional settlements due to water table rise computed from Eq. (9.87) are given in the table on the next page and are plotted against the water table depth in the figure.

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z (m) 0.0 0.75 1.5 2.0 3.0 4.0 5.0 6.0

Aw

Se,additional

(m) 2.115 1.931 1.530 1.209 0.680 0.302 0.075 0

(mm) 14.00 12.78 10.13 8.00 4.50 2.00 0.50 0.00

Some observations about the rate of settlement increase are as follows.

•

The rate of increase in settlement is not linear.

•

The rate of increase is very low at the beginning when the water table is beginning to affect the settlements.

•

The rate of settlement increases with the water table rise up to a depth of B/2 below the foundation.

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Chapter 10 10.1

X=

150 × 13.5 = 8.1 ft (100 + 150)

L = 2( L2 + X ) = 2(3.0 + 8.1) = 22.2 ft Therefore, L1 = L − ( L2 + L3 ) = 22.2 − (3.0 + 13.5) = 5.7 ft Required plan area of the combined footing is

10.2

A=

Q1 + Q2 (100 + 150) × 1000 = = 100.0 ft 2 qall(net) 2500

B=

100.0 = 4.50 ft 22.2

L = L1 + L2 + L3 = 3.0 + 4.5 + 12.0 = 19.5 ft Required area of the footing is

A=

Q1 + Q2 (200 + 100)(1000) = = 115.4 ft 2 qall(net) 2600

 B + B2   B1 + B2  A= 1 L =  (19.5) = 115.4  2   2  B1 + B2 = 11.83 ft Eq. (10.8): X =

(a)

Q2 L3 100 × 12.0 = = 4.0 ft Q1 + Q2 200 + 100

 B + 2 B2  L    Eq. (10.9): X + L2 =  1  B1 + B2  3   B + 2 B2  19.5   4.0 + 4.5 =  1  B + B  1 2  3 

B1 + 2 B2 = 1.308 B1 + B2 B1 = 2.247B2

(b)

Solving Eqs. (a) and (b): B1 = 8.19 ft and B2 = 3.64 ft SOURCE: Browsegrades.net

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10.3

From Eq. (10.12), Df  0.195 B   qu ( net ) = 5.14 cu 1 +   1 + 0.4 L  B   24  9   = 5.14 (1200 )1 + 0.195 ×   1 + 0 .4 ×  30  24   = 8200.0 lb/ft 2

10.4

From Eq. (9.51): 2 2 D f  Se  N 60  B + 0.3   23  6 + 0.3   2  75  qnet =   =   1 + 0.33   1 + 0.33   0.08  B   B  25  0.08  6   6  25 

= 1055 kN/m2 From Eq. (10.14):

qnet ≈

10.5

 D f   Se  N 60  23   2  75    ≈ 1 + 0.33   ≈ 957.4 kN/m2 1 + 0.33  0.08   6  25   B  25  0.08 

  D f   Se Eq. (10.15): qnet ≈ 0.25N 60 1 + 0.33  B  

  9  5.2 ≈ 0.25 × 201 + 0.33  Se ; Se = 0.94 in.  27   10.6

a. The gross pressure applied by the building on the underlying soil is 8 × 15 = 120 kN/m2 qnet = qgross − γ D f

For qnet = 0, qgross = γ D f So, Eq. (10.21): Df =

120 = 6.32 m 19.0

b. With D f = 3 m , the net applied pressure qnet = 120 − 3 × 19.0 = 63.0 kN/m 2 .

The net ultimate bearing capacity from Eq. (10.12) is

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Df  0.195 B   qu ( net ) = 5.14 cu 1 +  1 + 0.4 L  B   15  3  = 5.14 (50) 1 + 0.195 ×  1 + 0.4 ×  = 318 .2 kN/m 2 20  15  

Factor of safety is

10.7

318.2 = 5.1 63.0

a. The net applied pressure

Q 36 × 1000 − γ Df = = 120 kN/m2 A 15 × 20

b. The initial effective overburden stress at the middle of the clay is

σ o′ = 9 × 16 .5 + 2.5(19.0 − 9.81) = 171 .5 kN/m 2 σ c′ = 210 kN/m 2 The clay is slightly overconsolidated. c. We determine the average stress increase, ∆σ′av , within the clay layer under the center by first considering a quarter of the raft where B = 7.5 m and L = 10 m.

The depth to the top and bottom of the clay are: H1 = 6 m and H2 = 11 m.

For the top of the layer, m2 =

B 7.5 L 10 = = 1.25 and n2 = = = 1.67 H1 6 H1 6

From Figure 8.13, I a ( H1 ) = 0.23 For the bottom of the layer, m2 =

B 7.5 L 10 = = 0.68 and n2 = = = 0.91 H 2 11 H 2 11

From Figure 8.13, I a ( H 2 ) = 0.20 From Eq. (8.20), − H 1 I a ( H1 )  H I  (11)( 0.20) − ( 6)( 0.23)  ′ = qo  2 a ( H 2 ) ∆σ av = (120 )    H 2 − H1 11 − 6    = 19.7 kN/m 2 ′ = 4 × 19 .7 = 78 .8 kN/m 2 For the four quarters, ∆σ av

From Eq. (2.69),

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Sc =

Cs H c σ′ C H σ ′ + ∆σ ′ log c + c c log o 1 + eo σo′ 1 + eo σ c′

(0.06)(5.0) 210 (0.35)(5.0) 171.5 + 78.8 + log log 1 + 0.82 171.5 1 + 0.82 210 = 0.0878 m = 87.8 mm

=

10.8

For B = 15 m and L = 20 m, with H1 = 6 m and H2 = 11 m, For the top of the layer, m2 =

B 15 L 20 = = 2.5 , and n2 = = = 3.33 H1 6 H1 6

From Figure 8.13, I a ( H1 ) = 0.245 For the bottom of the layer, m2 =

B 15 L 20 = = 1.36 , and n2 = = = 1.82 H 2 11 H 2 11

From Figure 8.13, I a ( H 2 ) = 0.235 From Eq. (8.20), − H 1 I a ( H1 )  H I  (11)( 0.235) − ( 6)( 0.245)  ′ = q0  2 a ( H 2 ) ∆σ av = (120 )  = 26.8 kN/m 2   H 2 − H1 11 − 6    

From Eq. (2.67),

Sc =

10.9

Cs H c σ ′ + ∆σ ′ (0.06)(5.0) 171.5 + 26.8 log c = log = 0.0104 m = 10.4 mm 1 + eo σo′ 1 + 0.82 171.5

For a 7.5-ft-wide square foundation, 2

2

 B + 1  7 .5 + 1  3 k = k1   = (350)  = 112.4 lb/in.

2 B 2 × 7 .

5     For a 7.5-ft × 10.5-ft rectangular foundation, using Eq. (10.44), B 7 .5    k( B × B ) 1 + 0.5  (112.4)1 + 0.5 ×  L 10.5    k = = = 101.7 lb/in.3 1 .5 1 .5 10.10 a. The coefficient of subgrade reaction for the 300 × 450 mm plate is 250 0.005 kN/m3, or 50 MN/m3.

From Eq. (10.44),

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B  k ( B× B ) 1 + 0.5  L  k= 1 .5 0 .3   k 0.3 1 + 0.5  0.450   50 = ; k0.3 = 56.3 MN/m3 1 .5

b. For a 2-m-wide foundation, 2

2

 B + 0 .3   2 .0 + 0 .3  3 k( 2×2 ) = k0.3   = (56.3)  = 18.6 MN/m  2B   2 × 2 .0  From Eq. (10.44), B 2   k( B × B ) 1 + 0.5  (18.6)1 + 0.5  L 3   k = = = 16.5 MN/m 3 1 .5 1 .5

10.11 I F =

1 (2.0)(0.4) 3 = 0.0107 m 4 12

From Eq. (10.45): k = 0.65 12

Es B 4 Es ( 30 × 10 6 )( 2.0) 4 30 × 10 6 12 = 0 .

65 EF iF B (1 − µs2 ) ( 30 × 10 9 )(). 0107 ) ( 2)(1 − 0.252 )

= 10 .7 × 10 6 N/m 3 = 10.7 MN/m 3

From Eq. (10.46):

Es 30 × 106 k= = = 16.0 × 106 N/m3 = 16.0 MN/m3 2 2 B(1 − µs ) (2)(1 − 0.25 )

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Chapter 11 11.1

For φ ′ = 26°, from Table 6.1, Nc = 27.09, Nq = 14.21; Nγ = 9.84 From Eq. (6.10), qu = c′N c + q Nq + 0.5γ BNγ = (10)( 27.09) + (19)(14.21) + (0.5)(19)( B )( 9.84) = 93.48 B + 540.89 kN/m2

qall =

93.48B + 540.89 = 31.2 B + 180.3 kN/m2 3

The design load = 600 + 400 = 1000 kN/m

1000 ≤ 31.2 B + 180.3 B

31.2 B 2 + 180.3B − 1000 ≥ 0 ; B ≥ 3.47 m ≈ 3.5 m Use B as 3.5 m.

11.2

Design load = 1.25 × 600 + 1.5 ×400 = 1350 kN/m Therefore, the applied pressure =

1350 kN/m2 ← S* B

Factored cohesion c′ = 0.5 × 10 = 5 kN/m2 Factored φ ′ = tan −1 (0.8 × tan 26 ) = 21.3° For φ ′ = 21.3°, from Table 6.1, Nc = 19.33, Nq = 8.54; Nγ = 4.54 From Eq. (6.10), qu = c′N c + q Nq + 0.5γ BNγ = (5)(19.33) + (19)(8.54) + (0.5)(19)( B )( 4.54) = 43.13B + 248.91 kN/m 2 ← R * From Eq. (11.6), S* ≤ R* SOURCE: Browsegrades.net

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1350 ≤ 43.13B + 258.91 B 43.13B2 + 258.91B – 1350 ≥ 0; B ≥ 3.36 m ≈ 3.4 m Use B as 3.4 m.

11.3

Total load = 1.25 × 600 + 1.75 × 400 = 1450 kN/m The pressure applied to the soil is

1450 1450 kN/m2; i.e., S* = kN/m2 B B

For φ ′ = 26°, from Table 6.1, Nc = 27.09, Nq = 14.21; Nγ = 9.84 From Eq. (6.10),

qu = c′N c + q Nq + 0.5γ BNγ = (10)( 27.09) + (19)(14.21) + (0.5)(19)( B )(9.84) = 93.48B + 540.89 kN/m2 From Eq. (11.7), R * = ΦR = R* = 0.5[93.48B + 540.89] = 46.74 B + 270.45 From Eq. (11.6), S* ≤ R*

1450 ≤ 46.74 B +270.45 B 46.74B2 + 270.45B – 1450 ≥ 0; B ≥ 3.38 m ≈ 3.4 m Use B as 3.4 m.

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Chapter 12 12.1

a. With uniform f(z) in Soils 1 and 2, the pile load Q(z) should decrease linearly within the two soils.

Perimeter of the pile, p =   0.5 = 1.571 m From Eq. (12.9): dQ( z )   p f( z ) dz Shaft load within the Soil 1 = 1.571  10  40 = 628.4 kN Therefore, Q(z = 10 m) = 1500 ‒ 628.4 = 871.6 kN Shaft load within the Soil 2 = 1.571  5  70 = 549.9 kN Therefore, Q(z = 15 m) = 871.6 ‒ 549.9 = 321.7 kN The variation of Q(z) with depth is shown in Part a of the figure below. b. Noting that the pile load Q(z) varies linearly within Soils 1 and 2, it is evident that f(z) is constant within the two soils.

Eq. (12.9): For Soil 1, f ( z )   For Soil 2, f ( z )  

1 dQ( z ) 1 1500  800    44.6 kN/m2 p dz 1.571 10 1 dQ( z ) 1 800  300    63.7 kN/m2 p dz 1.571 5

The variation of f(z) with depth is shown in Part b of the figure. f(z) (kN/m2)

Q(z) (kN) 1500

10 m

Soil 1

44.6

Soil 1

871.6 5m

Soil 2 321.7

63.7

Soil 2

z (m)

z (m)

Part a

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Part b

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12.2

a. Eq. (12.18): Q p  Ap qN q*  Ap ql Given 2  42 . From Table 12.6, N q* = 525 Ap qN q*  (0.46  0.46)(18.6  20)(525)  41,325.5 kN Eq. (12.19): Ap ql  Ap ( 0.5 pa N q* tan 2 )  ( 0.46  0.46)[( 0.5)(100)(525)(tan 42)]  5001.3 kN

Hence, Qp = 5001.3 kN b. Eq. (12.28): Es  m pa  (600)(100)  60,000 kN/m2     25   42  25  Eq. (12.29): s  0.1  0.3   0.1  0.3   0.355  20   20 

Eq. (12.30):

42  25  18.6  20      25  q   Δ  0.0051     0.0051     0.00279 20  pa  20  100    Eq. (12.27): Ir 

Es 60,000   66.1 2(1   s ) q tan 2 2(1  0.355)(18.6  20)(tan 42)

Eq. (12.26): I rr 

Ir 66.1   55.8 1  I r  1  (66.1)( 0.00279 )

Eq. (12.21): Q p  Ap o N σ*  1  2(1  sin 42)  o   (18.6  20)  206.06 kN/m2  3  

N σ*  120 (Table 12.8) Q p  (0.46  0.46)( 206.06)(120)  5232.3 kN

c. Eq. (12.38): Q p  qN q* Ap L 20   43.48. From Figure 12.20, N q*  82 D 0.46 Q p  (18.6  20)(82)(0.46  0.46)  6454.6 kN

2  42 ,

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12.3

Ap 



D2 



4 4 Point load Qp:

 0.52  0.196 m 2

From Table 12.6, N q* = 81 qN q*  (20  18.5)(81)  29,970 kN/m2 ql  0.5 pa N q* tan    (0.5)(100)(81)(tan 32)  2531 kN/m2 Using the smaller of the above two values, Qp = 0.196  2531 = 496.0 kN Shaft load Qs: Perimeter, p = D = 1.571 m Ko  1  sin    0.470

   0.7   22.4 K = 1.5  0.470 = 0.705 Critical depth, L = 15D = 7.5 m We apply Eq. (12.43) for computing the skin friction as f  K o tan   At z = 0, f = 0 At z = 7.5 m, f = (0.705)(7.5  18.5)(tan 22.4) = 40.3 kN/m2 At z = 20 m, f = 40.3 kN/m2  0  40.3  Qs   (7.5)(1.571)  ( 40.3)( 20  7.5)(1.571) = 1028.8 kN 2  

Ultimate load, Qu = 477.7 + 1028.8 = 1506.5 kN Allowable load, Qall 

12.4

1506.5  502 kN 3

Point load Qp: L 20   40 ;    32 D 0 .5

From Figure 12.20, N q* = 30 From Eq. (12.38),

Qp  qNq* Ap  (20 18.5)(30)(0.196)  2176 kN Shaft load Qs: SOURCE: Browsegrades.net

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L 20   40 ;    32 D 0 .5

From Figure 12.22, K ≈ 0.45 From Eq. (12.46), Qs  K o tan(0.8 ' ) pl  0.45(10  18.5) tan(0.8  32)(  0.5  20)  1253.1 kN Ultimate load, Qu = Qp + Qs = 2176 + 1253 = 3429 kN 3429  1143 kN 3

Allowable load, Qall 

12.5

Ap  0.4  0.4  0.16 m 2 Point load Qp: From Table 12.6, N q* = 96 qN q*  ( 20  18.5)( 96)  25,920 kN/m2 From Eq. (12.19), ql  0.5 pa N q* tan    (0.5)(100)( 96)(tan 33)  3117.2 kN/m2 Using the smaller of the above two values, Qp = 0.16  3117.2 = 498.7 kN Shaft load Qs: Ko  1  sin    0.455

   23.1 K = 1.4  0.455 = 0.637 Perimeter, p = 4D = 1.6 m Critical depth, L = 15D = 6.0 m From Eq. (12.43) for computing the skin friction as f  K o tan   At z = 0, f = 0 At z = 6.0 m, f = (0.637)(6.0  18.0)(tan 23.1) = 29.3 kN/m2 At z = 20 m, f = 29.3 kN/m2  0  29.3  Qs   ( 6.0)(1.6)  ( 29.3)(15.0  6.0)(1.6) = 562.6 kN 2  

Ultimate load, Qu = 498.7 + 562.6 = 1061.3 kN SOURCE: Browsegrades.net

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Allowable load, Qall 

12.6

1061.3  353.8 kN 3

Point load Qp: From Table 12.6, for  = 33, N*q = 96

q p  qNq*  [(8)(17.0)  (10)(19.0  9.81)](96)  21,878 kN/m2 ql  0.5 pa N q* tan    (0.5)(100)( 96)(tan 33)  3117.2 kN/m2 Using the smaller of the two values,   Qp = (3117.2)  0.450 2   495.8 kN 4 

Shaft load Qs (Upper layer): Critical depth, L = 15D = 6.75 m For the upper layer, Ko  1  sin 31  0.485; K  1.5Ko  0.727;    20.2 For the lower layer, Ko  1  sin 33  0.455; K  1.5Ko  0.683;    21.5 Applying Eq. (12.43), f  K o tan   for the upper layer, At z = 0, f = 0 At z = 6.75 m, f = (0.727)(6.75  17)(tan 20.2) = 30.7 kN/m2 At z = 8.0 m, f = 30.7 kN/m2  0  30.7   Qs, upper layer = (0.450)  (6.75)  (30.7)(8.00  6.75) = 200.7 kN 2    For the lower layer, the skin friction remains the same at 30.7 kN/m 2.

Qs, lower layer = (  0.450)(10)(30.7) = 434.0 kN Qu = 495.8 + 200.7 + 434.0 = 1130.5 kN Qall 

12.7

130.5  376.8 kN 3

a. Eq. (12.19) will apply: q p  ql  0.5 pa N q* tan 

  = 40; N q* = 346 (Table 12.6). So, qp = (0.5)(100)(346)(tan 40) = 14,516 kN/m2 SOURCE: Browsegrades.net

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  2 Q p  Ap q p   0.381 (14,516)  1655 kN 4

b.   = 40; Irr = 50; N σ*  93 (Table 12.8). Q p  Ap o N σ*  1  2 Ko   σ o    q ; Ko = 1 – sin 40 = 0.357  3 

q = (3)(15.7) + (3)(18.2 – 9.81) + (15)(19.2 – 9.81) = 213.12 kN/m 2  1  ( 2)( 0.357)  o   ( 213.12)  121.76 kN/m 2  3  

   2 Q p   0.381 (121.76)(93)  1291 kN  4   c. Q p 

1655  1291  1473 kN (estimated ) 2

d. L  15D  (15)0.381  5.72 m. Let us assume L  6 m.

Qs  p L fav

 15.7  3  Qs ( z  0 to 3 m )  (  0.381)(3)( 1.4) (tan 0.6  32)  41.23 kN 2       K



 o (av)

 ( 3  15.7)  ( 3)(18.2  9.81)  o ( between z 3 to 6 m )    36.14 kN/m 2  2   Qs ( z 3 to 6 m )  (   0.318)(3)(1.4)(36.14)(tan 0.6  32)  52.81 kN

Qs ( z  6 to 21 m )  (  0.318)(15)(1.4)[( 3  15.7)  ( 3)(18.2  9.81)]  (tan 0.6  40)  675 kN

Qs = 41.23 + 52.81 + 675 = 769.04 kN e. Qall 

12.8

Q p  Qs FS



1473  769.04  560.51 kN 4

a. Point load Qp: From Eq. (12.20), SOURCE: Browsegrades.net

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  Q p  9 Ap cu  (9)  0.5 2 (60) =106.0 kN 4 

Shaft load Qs: For cu = 60 kN/m2, from Table 12.11,  = 0.62 Qs = (0.518.0)(0.62)(60) = 1051.8 kN Qu = 106.0 + 1051.8 = 1157.8 kN Qall 

1157.8  386 kN 3

b. Percentage of ultimate load carried by the pile shaft =

1051.8  100 = 91% 1157.8

This is a friction pile, where most of the load is carried by the shaft.

12.9

Point load Qp: From Eq. (12.20),   Q p  9 Ap cu  (9)  0.450 2 (80) =114.5 kN 4  

Shaft load Qs: For cu = 40 kN/m2, from Table 12.11,  = 0.74 For cu = 60 kN/m2, from Table 12.11,  = 0.62 For cu = 80 kN/m2, from Table 12.11,  = 0.54 Qs = (  0.450)[5  40  0.74 + 6  60  0.62 + 8  80  0.54] Qall 

1127.9  376 kN 3

c  1600  12.10 Eqs. (12.37) and (12.35): I rr  I r  347 u   33  (347)   33  244.6  2000   pa  Eq. (12.34): N c* 

4  4 (ln I rr  1)   1  (ln 244.6  1)  2.57  11.24 3 2 3 2

15 1600 Q p  Ap cu N c*     kip/ft 2 (11.24)  28.1 kip  12   1000  Eq. (12.61):

Qs   cu p L SOURCE: Browsegrades.net

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cu 1600   0.8 . Table 12.11:  = 0.54. pa 2000

1600 15 Qs  (0.54) kip/ft2  4  (50)  216 kip 12   1000  Qall 

Q p  Qs FS



28.1  216  81.4 kip 3 2

1600  15  12.11 Eq. (12.20): Q p  9cu Ap  (9)    22.5 kip  1000  12  Qs  pL fav Eq. (12.53): f av   ( o  2cu ) o 

(121)(50)  3025 lb/ft 2  3.025 kip/ft 2 2

L = 50 ft = 15.24 m. Table 12.10:  = 0.2

15 Qs   4  (50)(0.2)[3.025  (2)(1.6)]  311.25 kip 12   Qall 

22.5  311.25  111.25 kip 3

12.12 L = 20 m;  = 0.173 o 

( 20  18)  1805 kN/m 2 2

Qs  pL( o  2cu )  (4  0.46)( 20)(0.173)(180  2  80)  2164.58 kN Eq. (12.20): Q p  9cu Ap  (9)(80)(0.46)2  152.35 kN Qall 

2164.58  152.35  772.31 kN 3

16 12.13 a. Qs  p[L11cu (1)  L2 2cu ( 2 ) ]. p  ( 4)   5.33 ft  12  z = 0 to 20 ft: o 

(118)( 20)  1180 lb/ft 2 2

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cu (1) pa cu ( 2 ) pa



700  0.35 . Table 12.11: 1  0.78 2000



1500  0.75 . Table 12.11: 2  0.56 2000

Qs = 5.33[(20)(0.78)(700) + (40)(0.56)(1500)] = 237,292 lb  237 kip b.

cu ( av ) 

(700)( 20)  (1500)( 40)  1233.3 lb/ft 2 60

1 1 ( 20)( 2360)  ( 40)( 2360  4760) 2 o ( av )  2  2766.7 lb/ft 2 60

Length of pile = 60 ft  18.3 m;   0.18 Qs  pL [ o( av)  2cu ( av ) ] 

1  16   4  (60)(0.18)[2766.7  (2)(1233.3)] 1000  12 

 301.4 kip  301 kip

1  c. f av(1)  (1  sin R ) tan R  o(av)  (1  sin 20)(tan 20)  118  20  2   282.6 lb/ft 2 SOURCE: Browsegrades.net

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2360  4760  2 f av(2)  (1  sin 20)(tan 20)   852.6 lb/ft 2   16   1  Qs  p[L1 f av (1)  L2 f av ( 2 ) ]    4  [( 20)( 282.6)  (40)(852.6)] 12   1000   212 kip

12.14 Top Layer: o 

20 (118)  1180 lb/ft 2 ; cu = 700 lb/ft2 2

f av(1)  0.5(cu o )0.5  0.5[(700)(1180)]0.5  454.42 lb/ft 2 f av(1)  0.5(cu )0.75 ( o )0.25  0.5[(700)0.75 ][(1180)0.25 ]  398.8 lb/ft 2 Use f av(1)  454.42 lb/ft 2 Bottom Layer:  40  2 2 o  118  20  (122.4  62.4)   2360  1200  3560 lb/ft ; cu = 1500 lb/ft  2 

f av(2)  0.5(cu o )0.5  0.5[(1500)(3560)]0.5  1155.4 lb/ft 2 f av(2)  0.5(cu )0.75 ( o )0.25  0.5[(1500)0.75 ][(3560)0.25 ]  930.85 lb/ft 2 Use f av(2)  1155.4 lb/ft 2 Qs  p[L1 f av (1)  L2 f av ( 2 ) ]

p = 5.33 ft (See Problem 12.13, Part a) Qs  5.33[( 20)( 454.42)  (40)(1155.4)]  294,772 lb  294.8 kip 12.15 Top Layer: cu = 700 lb/ft2; o  1180 lb/ft 2 cu 700   0.593 o 1180

Given: PI = 15;   0.5 (Figure 12.26) Bottom Layer: cu = 1500 lb/ft2; o  3560 lb/ft 2 SOURCE: Browsegrades.net

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cu 1500   0.42 o 3560

Given: PI = 20;   0.605 Qs  p[L1 f av (1)  L2 f av ( 2 ) ]  5.33[( 20)(0.5)( 700)  (40)(0.605)(1500)]  230,789 lb  230.8 kip

12.16 Eq. (12.70): f   o  = 15.5 kN/m3;   10.72 zm1.3 ; zm 

10  5;   10.72(5) 1.3  1.323 2

 10  The average value of o  (15.5)   77.5 kN/m 2  2 

f av   o  (1.323)(77.5)  102.53 kN/m2 Qs   f av pL  (102.53)(  0.45)(10)  1449.48 kN   10  13  15  13 Average N 60  1 pile diameter above and  3 pile diameters below the tip   3

Eq. (12.74a): qp = 57.5N60 Qp 



  D 2 q p   (0.45) 2 [(57.5)(13)]  118.88 kN 4 4

Qu  Qs  Q p  1449.48  118.88  1568.36 kN  1568 kN

12.17 Eq. (12.76):  

56.2 cu

 56.2  ( cu )  56.2 kN/m2 f av  cu   c  u  Qs  D fav L  ( )(0.6)(56.2)(10)  1059.3 kN

12.18 Eq. (12.79): qu (design) 

qu (lab) 5



80  16 MN/m 2 5

  36    N   tan 2  45    tan 2  45    3.852 2 2   

SOURCE: Browsegrades.net

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Eq. (12.80): Q p (all) 

qu (design) ( N   1) Ap FS

For HP 310  125, Ap = 15.9  103 m2 Q p (all) 

(16)(3.852  1)(15.9  103 )  411.45  103 MN  411.5 kN 3

12.19 a. The load vs. total settlement plot is shown in the figure below.

4

d 350  4  6.9 mm 120 120

L 20  m/N  0.0083 mm/kN Ap Ep   0.352  25  109 4 Draw a straight line with a slope of 0.0083 mm/kN with an intercept of 6.9 mm on the settlement axis. The intersection of this line (dashed) with the settlement curve gives the ultimate load as 1000 kN. With FS = 2.5, the allowable pile load is 400 kN. b. At Q = 1250 kN:

st  39.6 mm se  (0.0083)(1250)  10.375 mm snet  st  se  39.6  10.375  29.225 mm SOURCE: Browsegrades.net

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12.20 Assume the pile is incompressible and apply Eq. (12.83).

H = ; L = 25 m; D = 0.6 m; and s = 0.2.

L L  42 and 0 D H

Therefore,

From Figure 12.36, I = 1.85 Q 1200  103 I   1.85 m  3.6 mm LEs 25(25  106 )

st  snet 

The average pile load on the cross-section can be taken as 0.5(1200 + 300) = 750 kN From Eq. (12.84),

se 

Qavg L Ap E p



(750  103 )(25) m  2.2 mm  2 9   0.6 (30  10 ) 4 

Settlement of the pile head = 3.6 + 2.2 = 5.8 mm

12.21 From Eq. (12.88),



Ap 

4

se (1) 

 0.6 2 = 0.283 m2

(Qwp  Qws ) L Ap E p

(300  0.5  900)(103 )(25)  m = 2.2 mm (0.283)(30  109 )

From Eq. (12.89),

se ( 2 )  qwp 

se ( 2) 

qwp D Es Qwp Ap



qwp D Es

(1   s2 ) I wp 300  103  1061.1  103 N/m 2 , and Iwp = 0.85 0.283

(1   s2 ) I wp 

(1061  103 )(0.6) (1  0.2 2 )(0.85) m  20.8 mm 6 (25  10 )

From Eq. (12.91), Q  D se ( 3)   ws  (1  μs2 ) I ws  pL  E s  900  103  0.6    (1  0.22 )( 2  0.35 25 / 0.6 ) m  1.9 mm 6  π  0.6  25  25  10 SOURCE: Browsegrades.net

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12.22 Total elastic settlement, se = 2.2 + 20.8 + 1.9 = 24.9 mm Q

Q

z z Q(z) dz

Q(z)

(b)

(c)

kQ

(1 – k)Q

(a) We take the pile length as L and the cross section as Ap. The modulus of elasticity of the pile material is Ep. Load Q is applied at the top (head or butt) of the pile. As shown in part a of the figure, it is assumed that kQ is transferred to the soil through the skin friction along the shaft and (1 - k)Q is transferred through the pile tip, where k < 1.

We consider an infinitesimal pile element of thickness dz, located at depth z.

Since the skin friction is uniform, the shaft load in part b is proportioned as z kQ . From equilibrium of forces along the vertical direction, L

z  Q( z )  1  k Q L  This is the axial load that acts on the pile element in part c.

Therefore, the elastic shortening of the pile element is dse (1) 

Q( z ) dz Ap E p



Q  z 1  k dz Ap E p  L SOURCE: Browsegrades.net

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Integrating the above expression for the entire pile length L, zL

zL

Q z Q  z  1  k dz 1  k dz   Ap E p  L Ap E p z  0  L z 0

se (1)   se (1) 

QL  k  1   Ap E p  2 

(a)

se(1) can also be expressed as se (1) 

(Qwp  Qws ) L Ap E p

Substituting Qwp = (1 – k)Q and Qws = kQ in the above equation, se (1) 

(Qwp  Qws ) L Ap E p



[(1  k )Q  kQ]L Ap E p

(b)

Equating the expressions on the right-hand sides of Eqs. (a) and (b),

 k 1    (1  k )  k ;  = 0.5  2

12.23 I p 

1 3 1 bh  ( 0.305) 4  0.0007 m 4 ; E p  21 106 kN/m2 12 12

Eq. (12.105): T  5

EpI p nh

5

(21  106 )(0.0007)  1.098 m 9200

L 30   5 — So, long pile. Mg = 0.

T 1.098 From Eq. (12.100): x z ( z )  Ax

Qg T 3 EpI p

; Qg 

x z ( z ) Ep I p AxT 3

At z = 0, xz = 12 mm = 0.012 m; Ax = 2.435 (Table 12.15) Qg 

( 0.12)( 21  106 )( 0.0007)  54.7 kN ( 2.435)(1.098) 3

Check for Moment Capacity: M z (max)  FY

Qg 

Ip  AmQg T ; d 2

2 FY I p d Am T SOURCE: Browsegrades.net

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From Table 12.15, the maximum value of Am is 0.772. So

Qg 

( 2)( 21,000)(0.0007)  113.7 kN (0.305)(0.772)(1.098)

Use Qg = 54.7 kN 12.24 Check for Bending Failure:

   I 0.0007  M y  FY p  ( 21,000)    96.39 kN - m d  0.305          2   2 My 4

D γK p



My

   D 4γ tan 2  45   2 

From Figure 12.42(a), with



96.39  232 (0.305) (16) tan 2 (45  15) 4

Q e  0, u ( g3)  50 D KpD 

Qu(g) = 50 tan2 (60)(0.305)3(16) = 68 kN Check for Pile Head Deflection:

nh 9200 5  0.91 m 1 6 EpI p (21  10 )(0.0007)

Eq. (12.117):   5

L = (0.91)(30) = 27.3. From Figure 12.43(a) for L = 27.3, e  0.

L

3

2

xz ( z 0 ) ( E p I p ) 5 (nh ) 5 Qg L 3

Qg 

 0.15 2

x z ( z  0 ) ( E p I p ) 5 ( nh ) 5 (0.15) L

3

2

(0.012)[( 21  106 )(0.0007)] 5 (9200) 5   32.5 kN (0.15)( 30)

Use Qg = 32.5 kN 12.25 Eq. (12.122): Qu 

EH E S C

HE = 40 kip-ft = 40 12 kip-in.; E = 0.85 SOURCE: Browsegrades.net

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Qu 

(0.85)( 40  12)  2040 kip 1  0.1 10

2040  340 kip 6

Qall 

E WR h WR  n Wp 12.26 Qu  S  C WR  Wp 2

Wp  weight of (pile  cap) 

90  100  2.4  11.4 kip 1000

   (0.85)( 40  12)   12  (0.35)2 (11.4)  Qu     1167.9 kip  1 12  11 .

4    0 .1   10  

1167  292 kip 4

Qall 

E HE ; E HE L S 2 Ap Ep

12.27 Qu 

E HE L  2 Ap Ep

Qu 

(0.85)( 40  12)  680 kip 0 .1  0 .5

Qall 

680  227 kip 3

12.28 Wp = Wpile + Wcap = S

 4

(0.85)( 40  12)(90  12)  0.5 ( 2)( 29.4)(30  103 )

 0.4 2  20  24 + 5.0 = 65.3 kN

6  0.24 blows/mm 25

2 EWR h WR  n W p (0.8)(35.6)(1000) 35.6  (0.4 2 )(65.3) Qu     = 4675.4 kN S C WR  W p 0.24  2.54 35.6  65.3

Qall 

4675.4  1169 kN 4

12.29 Wp = Wpile + Wcap = 0.4  0.4  15  24  4.0  61.6 kN SOURCE: Browsegrades.net

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SF  4 , Qu  4  400  1600 kN 2 EWR h WR  n Wp Qu   S C WR  W p

(0.85)(30.0)(1000) 12.2  0.352  61.6 1600   S  2.54 12.2  61.6 S = 1.724 mm (approximately 15 blows per 25 mm).

12.30 Eq (12.131) : Qn 

 ) γf H 2f tan   p (1  sin fill

2 (  0.305)(1  sin 25)(17.5)(32 ) tan( 0.5  25)  2  9.66 kN

12.31 Qn 

 ) γf H 2f tan   p (1  sin fill

2 (  0.305)(1  sin 25)(19.8  9.81)(32 ) tan( 0.5  25)  2  5.51 kN

12.32 Eq. (12.132): L  H f  L  H f γ f H f  2γf H f   L1  2 γ   γ (18  3.5) 18  3.5 (17)( 3.5)  ( 2)(17)( 3.5)     2 L1 19.8  9.81 19.8  9.81 191.55 L1   11.91 ; L1  9 m L1 L1 

Eq. (12.134):  ) γf H f tan   L1  Qn  p (1  sin clay

 ) γ tan   L12 p (1  sin clay

2    0.406 (1  sin 20)(17)( 3.5) tan( 0.6  20)(9) (9)2   0.406 (1  sin 20)(19.8  9.81) tan(0.6  20) 2  167.71 kN 

SOURCE: Browsegrades.net

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12.33 n1 = 4; n2 = 6; D = 0.4 m; d = 1.2 m a. From the Converse-Labarre equation (Table 12.18):  ( n1  1)n2  ( n2  1)n1  1 D  tan d 90 n n 1 2  

  1 

 ( 4  1)6  (6  1)4  1 0.4  1   tan 1.2  0.68 or 68% (90)( 4)(6)  

b. From the Los Angeles Group Action equation:

  1  1

D [n ( n  1)  n2 ( n1  1)  2 ( n1  1)( n2  1)] dn1n2 1 2 0 .4 [4( 6  1)  6( 4  1)  2 ( 4  1)( 6  1)  0.74 or 74%  (1.2)( 4)( 6)

12.34 Piles Acting Individually: Eq. (12.138):  Qu  n1n2 [9 Ap cu ( p )   p cu L ] cu 90   0.9 ; Table 12.11:   0.51 pa 100









 Qu  (3)(3) (9) 4  0.3162 (90)  (0.51)(  0.316)(90)(20)   (9)( 63.53  911.34)  8773.83 kN

Qall 



Qu 8773.83   2925 kN FS 3

Piles Acting as A Group: Eq. (12.139): Qu = Lg Bg cu(p) N c* + 2(Lg + Bg)cu L (2)(0.316) D Lg  Bg  ( n  1) d  2    (2)(0.6)   1.516 m 2 2

Lg Bg

 1;

L 20   13.19. From Figure 12.55, N c* = 9. So Bg 1.516

Qu = (1.516)(1.516)(90)(9) + (2)(1.516 + 1.516)(90)(20) = 1861.59 + 10,915.2 = 12,776.79 kN SOURCE: Browsegrades.net

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Qall 

12, 776.79  4258.93 kN 3

So, Qall = 2925 kN

12.35 Piles Acting Individually: Eq. (12.138):  Qu  n1n2 [9 Ap cu ( p )   p cu L ] cu 860   0.43 ; Table 12.11:   0.72 pa 2000

    12  2   12    Qu  (3)(3) (9)  4  12  (860)  (0.72)  12 (860)(45)      (9)( 6078  87,537)lb  842.5 kip

Piles Acting as A Group: Eq. (12.139): Qu = Lg Bg cu(p) N c* + 2(Lg + Bg)cu L D Lg  Bg  ( n  1) d  2   ( 2)( 2.5 ft )  1 ft  6 ft  2

Lg

 1;

L 45   7.5. From Figure 12.55, N c* = 9. So Bg 6

 Qu 

(6)(6)(860 )(9)  (2)(6  6)(860)(45 )  1207.4 kip 1000

Bg

Qall 

842.5  280.8  281 kip 3

12.36 L = 12.0 m Ap = (/4)(0.402) = 0.126 m2 p = D = 1.257 m From Table 12.11, for cu = 60 kN/m2,  = 0.62; and for cu = 90 kN/m2,  = 0.51 From Eq. (12.138),

 Qu  n1n2 [9 Ap cu  ( cu )( pΔL)]  (3  5)[(9)(0.126)(90)  (0.62  60)(1.257  5.0)  (0.51  90)(1.257  7)]  11,096.0 kN SOURCE: Browsegrades.net

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Bg  1200  2  400  2800 mm ; Lg  4  1200  400  5200 mm

L L 12 5 .2   4.29; g   1.86 B g 2 .8 B g 2 .8

From Figure 12.55, N c* = 8.5 From Eq. (12.139),

 Qu  Lg Bg cu N c*  2 ( Lg  Bg )cu ΔL  (5.2)( 2.8)(90)(8.5)  (2)(5.2  2.8)(5.0)(60)  (2)(5.2  2.8)(7.0)( 90)  26,018.4 kN Using the lesser of the two values, the allowable load on the pile group is given as Qall 

1096.0  3700 kN 3

12.37 The pressure distribution diagram is shown.

 o  Δ  (kN/m2)

Layer

 o (kN/m )

1

(15.72)(3) + (18.55 – 9.81)(3) + (13)(19.18 – 9.81) = 47.16 + 26.22 + 121.81 = 195.19

1335  22.23 (2.75  5) 2

217.42

2

195.19 + (5)(19.18 – 9.81) + (2.5)(18.08 – 9.81) = 262.72

1335  5.74 (2.75  12.5) 2

268.46

2

Δ  (kN/m ) 2

SOURCE: Browsegrades.net

111

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1335  3.6 (2.75  16.5) 2

3

262.72 + (2.5)(18.08 – 9.81) + (1.5)(19.5 – 9.81) = 297.93

Δsc (1) 

(0.8)(10)  217.42  log   0.208 m 1  0 .8  195.19 

Δsc ( 2 ) 

( 0.31)(5)  268.46  log   0.0073 m 11  262.72 

Δsc ( 3) 

( 0.26)( 3)  301.53  log   0.0024 m 1  0 .7  297.93 

301.53

 sc  217.7 mm

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Chapter 13 13.1

Point load Qp(net): From Eq. (13.18): Q p( net ) = Ap q′(ωN q* − 1) Ap =

π 4

× 5.0 2 = 19 .64 ft 2

q′ = 24 × 110 + 3 × 115 = 2985.0 lb/ft2 L = 27 ft; Db = 5 ft; L1 = 24.0 ft; Ds = 2.5 ft L 27 = = 5 .4 Db 5

From Table 13.3, ω = 0.83 From Eq. (13.19), N q* = 0.21e 0.17φ ′ = 0.21 × e 0.17 ×38 = 134 .2

Qp(net) = (19.64)(2985.0)(0.83×134.2 - 1) = 6471.4×103 lb = 6471.4 kip Shaft load Qs:

φ ′ = 31° and δ ′ = 18.6° Ko = 1 − sin φ ′ = 0.485 p = πDs = 7.85 ft Critical depth is 15Ds = 37.5 ft, which is more than L1. Therefore, the frictional resistance f ( = Kσo′ tan δ′ ) along the pile shaft will increase over the entire length of the shaft.

At z = 0, f = 0 At z = 24.0 ft, f = (0.485)(24 × 110)(tan 18.6) = 430.9 lb/ft2  0 + 430.9  Qs =   (7.85 × 24.0) = 40,591.0 lb = 40.6 kip 2   Qall =

6471 .4 + 40.6 = 1628.0 kip 4

Percentage of the load carried by the shaft =

40 .6 × 100 = 0.62% 40 .6 + 6471 .4

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13.2

a. Qp from Chen and Kulhawy method L = 15.0 m, Db = 1.0 m Base area A p =

π 4

× 1.0 2 = 0.785 m 2

Perimeter, p = πDs = 3.142 m For φ ′ = 34°, from Table 13.1, Nq = 29.44; Fqs = 1.675; C = 0.262; Icr = 106.37; µs = 0.235; n = 0.00275 From Eq. (13.7),

 L   15  Fqd = 1 + C tan −1   = 1 + 0.262 tan −1   = 1.39 1  Db  At the bottom of the shaft, q′ = 15 × 18.0 = 270.0 kN/m2 For φ ′ = 34°, take m as 200.

Es = 200 × 100 = 20,000 kN/m2 From Eq. (13.11), Ir =

Es 20,000 = = 44.5 2(1 + µs ) q′ tan φ ′ 2(1 + 0.235)( 270) tan 34

From Eq. (13.14), ∆=n

q′ 270 = 0.00275 × = 0.00743 pa 100

From Eq. (13.10), I rr =

Ir 44.5 = = 33.4 1 + I r ∆ 1 + ( 44.5)(0.00743)

Since Irr < Icr, From Eq. (13.17),   (3.07 sin φ ′)(log 2 I rr )   Fqc = exp  − 3.8 tan φ ′ +   1 + sin φ '      ( 3.07 sin 34 )(log( 2 × 33.4))   = exp  − 3.8 tan 34 +    = 0.575 1 + sin 34   

Therefore, from Eq. (13.5) Q p ( net ) = Ap q ′( N q − 1) Fqs Fqd Fqc ] = ( 0.785 )[ 270 ( 29 .44 − 1)(1.675 )(1.39 )( 0.575 )] = 8070 kN SOURCE: Browsegrades.net

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b. Qp using Berezantzev et al. method For

L = 15 and φ ′ = 34°, N q* = 67.99 (Table 13.2); ω = 0.680 (Table 13.3) Db

From Eq. (13.18),

Qp(net) = Ap q′(ω Nq* − 1) = (0.785)(270)(0.680 × 67.99 − 1) = 9587 kN c. Load carrying capacity of the shaft Qs: Critical depth is 15Ds = 15.0 m.

Ko tan δ′ = (1 − sin 34) tan(0.9 × 34) = 0.26 At z = 0 m, σo′ = 0; hence, f = Kσo′ tan δ′ = 0 At z = 15.0 m, σo′ = 15 × 18.0 = 270 kN/m2; f = (270)(0.26) = 70.2 kN/m2 The skin friction increases linearly to the entire depth of 15 m.  0 + 70.2  Qs = ( 3.142 ) (15) = 1654 kN 2  

d. Maximum allowable load Let’s use an average value for Q p ( net ) from the two methods as Q p (net ) =

8070 + 9587 = 8829 kN 2

Qu = 8829 + 1654 = 10,483 kN Qall =

13.3

10,483 = 3494 kN 3

a. Point load Q p ( net ) From Eq. (13.18), Q p( net ) = Ap q′(ωN q* − 1) Ap =

π 4

× 1.52 = 1.77 m 2

q′ = 10 × 17.0 + 0.75 × 19.0 = 184.3 kN/m2 L = 10.75 m; Db = 1.5 m; L1 = 10.0 m; Ds = 0.75 m L 10.75 = = 7.2 ; ω = 0.82 (Table 13.3) Db 1 .5 SOURCE: Browsegrades.net

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From Eq. (13.19),

N q* = 0.21e0.17φ ′ = 0.21 × e0.17×40 = 188.5 Q p ( net ) = (1.77)(184.35)(0.82 × 188.5 – 1) = 50,109.8 kN

Shaft load Qs:

φ ′ = 32° and δ ′ = 16.0° Ko = 1 − sin φ ′ = 0.470 p = πDs = 2.36 m Critical depth is 15Ds = 11.3 m which is more than L1. Therefore, the frictional resistance f ( = Kσo′ tan δ′ ) along the pile shaft will increase over the entire length of the shaft.

At z = 0, f = 0 At z = 10.0 m, f = (0.470)(10.0 × 17)(tan 15) = 21.49 kN/m2  0 + 21.4  Qs =  ( 2.36 × 10.0) = 252.2 kN 2   Qall =

50,109.6 + 252.5 = 12, 590 kN 4

b. Point load Qp(all): Allowable settlement 15.0 = = 0.01 or 1.00% Db 1500

Q p ( all )

From Figure 13.11,

Q p ( net )

= 0.32

Therefore, Qp(all) = 0.32 × 50,109.8 = 16,035.1 kN Shaft load Qs(all): Allowable settlement 15.0 = = 0.02 or 2.00% Ds 750

From Figure 13.12a,

Qs ( all ) Qs

= 0.92

Therefore, Qs(all) = 0.92 × 252.5 = 232.3 kN Qall = Qp(all) + Qs(all) = 16,035.1 + 232.3 = 16,267 kN

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13.4

In computing the load carrying capacity of the shaft Qs, we divide the sand into four layers.

Layer No. 1 2 3 4

β1 = 1.5 − 0.244 zi0.5

Depth to mid-layer, zi (m) 1.0 3.0 5.0 7.0

1.256 (Use 1.2) 1.077 0.954 0.854

′ = γzi σ ozi (kN/m2) 16.5 49.5 82.5 115.5

′ f i = β1 σ ozi (kN/m2) 19.8 53.3 78.7 98.6

The ultimate shaft load Qs is given by Qs = p

∑ f ∆L = (π × 0.75)[(19.8 × 2) + (53.3 × 2) + (78.7 × 2) + (98.6 × 2)] 4

1 i

i

= 1180.0kN

The point bearing capacity qp is given by [Eq. (13.28)],

qp = 57.5N60 = 57.5× 25 = 1437 kN/m2 (less than 4310 kN/m2) Since the base diameter is larger than 1.27 m, let’s reduce this as [Eq. (10.29a)] qpr =

1.27 × 1437 = 1217 kN/m 2 1.50

The ultimate point load, Q p =

π 4

× 1.5 2 × 1217 = 2150.6 kN

Therefore, the ultimate load carrying capacity Qu = Qp + Qs = 2150.6 + 1180.0 = 3331 kN

13.5

Settlement 10 × 100 = × 100 = 1.33% Ds 750 From Figure 13.12(a), the fraction of the ultimate shaft load that is mobilized is 0.97. Therefore, the shaft load = 0.97×1180 = 1144.6 kN Settlement 10 × 100 = × 100 = 0.67% Db 1500

From Figure 13.11, the fraction of the ultimate tip load that is mobilized is 0.21.

Therefore, the shaft load at 10 mm settlement = 0.21×2150.6 = 451.6 kN Therefore, the total load that can be allowed on the drilled shaft is Q = 1144.6 + 451.6 = 1596.2 kN

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13.6

Point load Qp(net): From Eq. (13.33), Qp( net) = Apcu Nc

*

*

From Table 13.4, for cu = 150 kN/m2, Es/3cu = 285 and Nc = 8.85 π  Q p ( net ) =  × 1.0 2 (150)(8.85) = 1042 .6 kN 4 

Shaft load Qs: From Eq. (13.37), Qs = Σ α *cu p ∆ L From Eq. (13.38),

p   100  * For cu = 80 kN/m2, α = 0.21 + 0.25 a  = 0.21 + 0.25  = 0.52  80   cu  p   100  * For cu = 150 kN/m2, α = 0.21 + 0.25 a  = 0.21 + 0.25  = 0.38  150   cu  p = π × 1.0 = 3.14 m Qs = (3.14)[0.52×5×80 + 0.38×10×150] = 2442.9 kN Qall =

13.7

1042 .6 + 2442 .9 = 1162 kN 3

Within the sand layer: The effective overburden stress σ o′ increases linearly within the sand, from top to bottom.

At the bottom of the sand layer, σ o′ = 18×115 = 2,070.0 lb/ft2 Ko = 1 – sin 34 = 0.44 From Eq. (13.23),  0 + 2070 .0  Qs ,sand = πDs K o tan δ ′ σ o′ dz = π ( 3.0)( 0.44 )(tan 0.8 × 34 ) × 18.0  2   = 39,704.6 lb = 39.7 kip

∫

For the clay layer, from Eq. (13.38),  100   = 0.71  50 

α * = 0.21 + 0.25

Qs,clay = π Ds α*cu ∆L = π (3.0)(0.71 × 1000)(12.0) = 80,299 lb = 80.3 kip cu/pa = 0.5 SOURCE: Browsegrades.net

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*

From Table 13.4, Nc = 7.95 π  Q p (net) = Ap cu N c* =  × 32 (1000)( 7.95) = 56,195.2 lb = 56.2 kip 4  Qu ( net ) = Q p ( net ) + Qs ,sand + Qs ,clay = 56.2 + 39.7 + 80.3 = 176.2 kip

With FS = 3, Qall = 176.2 3 = 58.7 kip

13.8

se(1) =

(Qwp + ξQws ) L Ap E p

=

(60 × 103 + 0.65 × 160 × 103 )(30.0 × 12) in. = 0.023 in. π 2 6  × 30 (3.6 × 10 ) 4 

From Eq. (12.89)

60 × 103 × 30 qwp D (π / 4)( 302 ) 2 se ( 2 ) = (1 − µs ) I wp = (1 − 0.52 )(0.85) in. = 0.451 in.

Es 3600 From Eq. (12.90), se ( 2 ) =

Qwp C p Dq p

=

(60 × 10 3 )(0.04) ft = 0.533 in. ( 2.5)( 2.4 × 10 3 × 9)

From Eq. (12.91),

 Q  D  se ( 3) =  ws  (1 − µs2 )(2 + 0.35 L / D )  πDL  Es   160 × 103  2.5  2  =  (1 − 0.5 )(2 + 0.35 30 / 2.5 ) ft = 0.095 in.  π × 2.5 × 30  3600 × 144  From Eq. (12.93),

QwsCs (160 × 103 )[0.93 + 0.16 30 / 2.5] (0.04) se( 3) = = ft = 0.176 in.

Lq p (30)(2400 × 9) From the elastic analysis equations, se = 0.023 + 0.451 + 0.095 = 0.57 in.

From the semi-empirical equations for se(2) and se(3), se = 0.023 + 0.533 + 0.176 = 0.73 in.

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13.9

π  Eq. (13.60): Q p = Ap [4.83(qu )0.51 ] =  (1.5) 2  [(4.83)(1.81)0.51 ] = 11.55 MN 4  Eq. (13.62): Qs = 0.8(qu )0.5 (π DL) = (0.8)(1.81)0.5 [(π )(1.5)(8)] = 40.58 MN Qall =

13.10 a.

11.55 + 40.58 = 13.03 MN 4

′ φ′

 γ Ds K p   Eq. (13.45): Qc = 1.57 Ds2 ( EP RI )  EP RI 

0.57

Ep = 22 × 106 kN/m2; RI = 1

Ip =

πD 4 ( π)(1.25)4 = = 0.1198 m4 64 64

φ′  35    K p = tan 2  45 +  = tan 2  45 +  = 3.69 2 2    (17.5)(1.25)(35)(3.69)  Qc = (1.57)(1.25) (22 × 10 )(1)   ( 22 × 106 )(1)   2

Qg Qc

=

0.57

6

= 326,630 kN

260 = 0.0008 326,630

From Figure 13.20, xo/Ds = 0.0025 xo = (0.0025)(1.25) = 0.00313 m = 3.13 mm

b. Eq. (10.47): ′ φ′

 γ Ds K p   M c = 1.33Ds3 ( E P RI )  E P RI 

0 .4

 (17.5)(1.25)( 35)( 3.69 )  = (1.33)(1.25)3 ( 22 × 106 )(1)   (22 × 106 )(1)   . = 1,586,544 kN m

From Figure 10.23, for

Qg Qc

= 0.0008,

0.4

M max ≈ 0.000375 Mc

Mmax = (0.000375)(1,586,544) = 594.9 kN·m

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D   1.25  M max  s  (594.9)   2  2    = = 3104 kN / m 2 c. σ tensile = Ip 0.1198 d.

E P RI ( 22 × 106 )(1) = = 7787.2 γ′Ds φ ′K p (17.5)(1.25)(35)(3.69)  L    ≈ 6 ; Lmin = (6)(1.25) = 7.5 m  Ds  min

13.16 From Eq. (10.51b): f = 6.564qu0.5 ≤ 0.15qu Since qu(concrete) < qu(rock), use qu(concrete) in Eq. (10.51b) qu(concrete) = 28,000 kN/m2

f = (6.564)(28,000)0.5 = 1098.4 kN/m2 Check: f = 0.15qu = (0.15)(28,000) = 4200 kN/m 2 > 1098.4 kN/m2 So, use f = 1098.4 kN/m2 From Eq. (10.52): Qu = π Ds L f = ( π)(1.5)(8)(1098.4) = 41,409 kN From Eqs. (10.53), (10.54), and (10.55): se =

Qu I f Qu L + Ac Ec Ds Emass

For RQD ≈ 75%, from Eq. (10.56): Emass = 0.266( RQD ) − 1.55 = (0.0266 )( 75) − 1.66 = 0.335 Ecore

Emass = Ecore = (0.335)(12.1 × 106 ) = 4.05 × 106 kN/m2 Ec 22 × 106 = ≈ 5.43 Emass 4.05 × 106

L 8 = = 5.33 Ds 1.5 From Table 10.6 for

Ec L = 5.43 and = 5.33, the magnitude of I f is about 0.42.

Emass Db

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se =

(41,409 kN)(8)

π

+

( 41,409 kN)(0.42) (1.5)( 22 × 106 kN/m 2 )

( 22 × 106 kN/m 2 ) 4 = 0.0114 m = 11.4 mm > 10 mm

So, from Eq. (10.78):

    c 500   3+ s   3+ Ds  q = (3) π × 1.52   1500 (28,000) Qu = 3 Ap    0.5  u 0.5   4   3      101 + 300 δ   101 + 300    500     cs     = 29,567 kN

Qall =

Qu 29,567 = ≈ 9856 kN FS 3

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Chapter 14 14.1

From Eq. (14.6): K r (1 − α rp ) Pr K r (1 − 0.85) 0.15 K r = X = = = Pt K p + K r (1 − 2α rp ) K p + K r (1 − 2 × 0.85) K p − 0.7 K r Pp Pt

= 1− X = 1−

K − 0.85 K r 0.15K r = p K p − 0 .7 K r K p − 0.7 K r

From Eq. (14.7),

K pr =

14.2

K p + K r (1 − 2α rp ) K p + K r (1 − 2 × 0.85) K p − 0.7 K r = = K 2 Kr 2 Kr 1 − α rp 1 − 0.85 1 − 0.72 r Kp Kp Kp

The equivalent raft area per pile = 1.60 × 1.63 = 2.608 m2 The equivalent cap area for a single pile ro can be determined by: πrc2 = 2.608 ; rc = 0.91 m

Assuming uniform soil conditions with depth, ξ = 1 and ρ = 1.

Eq. (14.10),

rm = {0.25 + ξ [2.5 ρ(1 − µs ) − 0.25]}L = {0.25 + 1[2.5 × 1(1 − 0.1) − 0.25]}× 13 = 29.3 m

 29.3  Eq. (14.9), ζ = ln  = 4.87  0.225  From Eq. (14.8), r   0.91  ln  c  ln   r 0.225  α rp = 1 −  o  = 1 −  = 0.71 ζ 4.87

14.3

From Eq. (14.7),

K pr =

K p + K r (1 − 2α rp ) 5300 + 3250(1 − 2 × 0.71) = = 5697 MN/m 3250 2 Kr 2 1 − α rp 1 − 0.71 × Kp 5300 SOURCE: Browsegrades.net

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From Eq. (14.6),

X =

14.4

K r (1 − α rp ) K p + K r (1 − 2α rp )

345 = 61 mm 5697 m

From Eq. (14.5), P1 =

Pup 1− X

=

500 = 657.9 MN 1 − 0.24

Thus, settlement =

14.6

3250(1 − 0.71) = 0.24 5300 + 3250(1 − 2 × 0.71)

Kpr = 5697 MN/m Settlement =

14.5

=

657.9 m ≈ 116 mm 5697

Solution: (only brief details given in bullet points)

• • • • • •

• • • •

•

Total load of the superstructure = 8,800 MN Height = 1000+ m Footprint = 3720 m2 Average pressure on the subgrade below the raft = 2.37 MPa The raft consists of three wings and the center core.

226 1.5 m diameter and 44 1.8 m diameter bored piles – all connected to the continuous raft covering the entire footprint. Piles are 45 m long at the wings and 105 m long at the center of the tower.

Pile spacing = 2.5 times the pile diameter Raft thickness = 4.5 m at the center and increases to 5.0 m at the ends of the wings.

Located in Jeddah Site investigation included 7 boreholes 120-200 m deep. The top 1-3 m of silty sand was underlain by coralline limestone rock, followed by successive layers of gravel and weak conglomerate and decomposed sandstone.

Expected settlement = 83-107 mm

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Chapter 15 15.1

From Eq. (11.5): γd =

2.72 × 9.81 Gs γ w = LL LL 1+ Gs 1 × × 2.72 100 100

By substituting different values of LL, the following plot can be obtained.

Since the point lies below the envelope, collapse will occur.

15.2

σ′c = 100.0 kN/m2; σ′o = 70.0 kN/m2

σ c′ 100 = = 1.33 σ o′ 75

(See Step 7 of Section 15.4)

The soil is normally consolidated.

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15.3

Eq. (2.63): ∆H =

∆e Ho 1 + eo

where eo and Ho are the initial void ratio and layer thickness, respectively.

The change in height, or the settlement, can be calculated as ∆H =

15.4

(0.81 − 0.68) ( 4000) = 287 mm 1 + 0.82

LL = 50; wn = 20%. From Figure 11.13, sw(free) ≈ 3.0% Eq. (15.10): ∆S F = 0.0033Zsw (free) = ( 0.0033)( 3000 )( 3.0) = 29.7 mm

15.5

The plot of sw(1) vs. depth is shown in the figure below.

The total likely swell ∆S is given by the area of the figure; hence, ∆S =

15.6

1 1 1 1  (1)( 2.1) + (1)( 2.1 + 3.7) + (1)( 3.7 + 5.2) = 0.8040 m = 84 mm  100  2 2 2 

From the plot of sw(1) vs. depth shown in Problem 15.5: At 4.5 m depth, total swell, ∆S = 0 At 3.5 m depth, total swell, ∆S =

1 1  (1)( 2.1) = 0.0105 m  100  2 

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At 2.5 m depth, total swell, ∆S =

1 1 1  (1)( 2.1) + (1)( 2.1 + 3.7) = 0.0395 m  100  2 2 

At 1.5 m depth, total swell, ∆S =

1 1 1 1  ( 1 )( 2 .

1 ) + ( 1 )( 2 .

1 + 3 .

7 ) + (1)( 3.7 + 5.2) = 0.8040 m  100  2 2 2 

The total swell ∆S (mm) at four different depths is summarized as follows: Depth, z (m) 1.5 2.5 3.5 4.5

∆S (mm) 84 39.5 10.5 0

The plot of ∆S versus depth is shown below in the figure. From the figure, the 25mm swell is reached at a depth of 2.9 m. Hence, the depth of 2.9 – 1.5 = 1.4 m below the foundation is the undercut, requiring non–swelling replacement soil.

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15.7

a. The plot of sw(1) vs. z is shown.

The total likely swell ∆S is given by the area of the figure shown.

∆S =

1 1 1 1 1  + + + + + ( 1 )( 1 .

2 ) ( 1 )( 1 .

2 2 .

5 ) ( 1 )( 2 .

5 3 .

7 ) (1)(3.7 + 4.6)  100  2 2 2 2 

= 0.1020 m = 102 mm b. From the plot shown in Part a: At 5.0 m depth, total swell, ∆S = 0 At 4.0 m depth, total swell, ∆S =

1 1  (1)(1.2) = 0.0060 m  100  2 

At 3.0 m depth, total swell, ∆S =

1 1 1  ( 1 )( 1 .

2 ) + (1)(1.2 + 2.5) = 0.0245 m  100  2 2 

At 2.0 m depth, total swell, ∆S =

1 1 1 1  (1)(1.2) + (1)(1.2 + 2.5) + (1)( 2.5 + 3.7) = 0.0555 m  100  2 2 2 

At 1.0 m depth, total swell, ∆S =

1 1 1 1 1  (1)(1.2) + (1)(1.2 + 2.5) + (1)( 2.5 + 3.7) + (1)(3.7 + 4.6)  100  2 2 2 2 

= 0.1020 m The total swell ∆S (mm) at five different depths is summarized as follows: SOURCE: Browsegrades.net

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∆S (mm) 102.0 55.5 24.5 6.0 0.0

Depth, z (m) 1.0 2.0 3.0 4.0 5.0

The plot of ∆S versus depth is shown in the figure on the next page. It can be seen that the 20 mm swell is reached at a depth of 3.2 m. Hence, the undercut will be 3.2 – 1.0 = 2.2 m below the bottom of the foundation.

15.8

′ tan φ ps ′ Eq. (15.14): U = π Ds Z σ sw

Ds = 1 m; Z = 9 m U = ( π)(1)( 9 )( 600)(tan 20) = 6174.6 kN

Eq. (15.17): U =

cu N c π 2 ( Db − Ds2 ) FS 4

 (150)( 6.14 )  π  2 2 2 6174.6 =   4 ( Db − 1 ) = 241.1( Db − 1) 3 

Or, 25.61 + 1 = Db2 Db = 5.16 m

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15.9

π (cu N c ) ( Db2 − Ds2 ) 4  Eq. (15.17): FS = U −D

π (150)(6.14) ( Db2 − 12 ) 723.35( Db2 − 1) 4 4= = (6174.6) − (1500) 4674.6 Or, 25.85 = Db2 − 1 Db = 5.2 m

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Chapter 16 16.1

From Eqs. (16.1) and (16.4),

K o = (1 − sin 33)1.5sin 33 = 0.568 Po = 0.5Ko γH 2 = 0.5(0.568)(115.0(15)2 = 7348.5 lb/ft The resultant Po is acting at H 3 = 5 ft above the bottom

of the wall

16.2

When the entire soil is submerged, σ h consists of σ h′ and u. They all increase linearly from zero at the top to the maximum value at the bottom of the wall.

At the bottom of the wall,

σ o′ = 15(127.0 – 62.4) = 969.0 lb/ft2 K o = 0.568

σ h′ = 0.568 × 969 = 550.4 lb/ft2 u = 15 × 62.4 = 936.0 lb/ft2

σ h = 550.4 + 936.0 = 1486.4 lb/ft2 Po = 0.5(1486.4)(15) = 11,148.0 lb/ft The resultant Po will act at 15 3 = 5 ft above the bottom of the wall 16.3

At A: σ o′ = 2 × 16.5 = 33.0 kN/m2

σ h′ = 0.45 × 33.0 = 14.85 kN/m2 u=0

σ h = 14.85 + 0 = 14.85 kN/m2

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At B: σ o′ = 3 × 16.5 + 4(19.0 – 9.81) = 86.26 kN/m2

σ h′ = 0.55 × 86.26 = 47.44 kN/m2 u = 4 × 9.81 = 39.24 kN/m2

σ h = 47.44 + 39.24 = 86.68 kN/m2 16.4

Effect of a surcharge q at the ground level is to increase the effective vertical stress at any point within the soil by q. This in turn would increase the effective horizontal stress by Koq. At A: σ h = 14.85 + 0.45×20 = 23.85 kN/m2 At B: σ h = 86.68 + 0.45×20 = 97.68 kN/m2

16.5

Ko, sand1 = 1 – sin 32 = 0.470; Ko, sand2 = 1 – sin 36 = 0.412 At top of sand layer 1: The vertical effective pressure σ o′ = 0; hence σ h′ = 0.

At the bottom of sand layer 1: σ o′ = 2×17.5 = 35.0 kN/m2

σ h′ = 0.470 × 35.0 = 16.5 kN/m2 At the top of sand layer 2:

σ o′ = 2 × 17.5 = 35.0 kN/m2 σ h′ = 0.412 × 35.0 = 14.4 kN/m2

There is a break in the σ h′ value at 2 m depth due to different Ko values.

At the bottom of sand layer 2: σ o′ = 2 × 17.5 + 3 × 18.5 = 90.5 kN/m2

σ h′ = 0.412 × 90.5 = 37.3 kN/m2 The variation of σ h′ with depth is shown in the figure on the next page. The lateral force on the wall per unit length is the area of the diagram in the figure.

Zone Horizontal load (kN/m) Col. 1 Col. 2 1 0.5 × 2 × 16.5 = 16.5 2 3 × 14.4 = 43.2 3 0.5 × 3 × (37.3 – 14.4) = 34.4 94.1 ∑

Centroidal height from base (m) Moment (kN⋅m/m) Col. 3 Col. 2 × Col. 3 3.67 60.6 1.5 64.8 1.0 34.4 159.8

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The lateral load per unit length Po is 94.1 kN/m. It acts horizontally at a height of 159.8 ≈ 1.7 m above the base of the wall 94.1

16.6

φ′  25    a. K a = tan 2  45 −  = tan 2  45 −  = 0.406 2 2   The maximum tensile stress occurs at the top of the clay, and is given by [Eq. (16.11)]

σ a′ = −2c′ K a = −2 × 10 × 0.406 = −12.7 kN/m2 b. The depth of the tensile crack is given by [Eq. (16.12)] zc =

2c′ 2 × 10 = = 1.65 m γ K a 19.0 0.406

c. At the bottom of the wall,

σ a′ = (5 × 19.0)( 0.406 ) − 2 × 10 × 0.406 = 25.87 kN/m 2 Pa = 0.5( 25.87)(5.0 − 1.65) = 43.33 kN/m It acts at a height of

5.0 − 1.65 = 1.12 m above the base 3

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16.7

34 For the sand, K a = tan 2  45 −  = 0.283 2  

36   For the silty sand, Ka = tan 2  45 −  = 0.260 2  At z = 0 m:

σ′a = 0

At z = 4– m:

σ a′ = (0.283)( 4 × 16.5) = 18.68 kN/m 2

At z = 4+ m:

σ a′ = (0.260 )( 4 × 16.5) = 17.16 kN/m 2

At z = 7 m:

σ a′ = (0.260 )( 7 × 16.5) = 30.03 kN/m 2

Horizontal load (kN/m) Zone Col. 2 Col. 1 1 0.5 × 4 × 18.68 = 37.36 2 3 × 17.16 = 51.48 3 0.5 × 3 × (30.3 – 17.6) = 19.31 108.15 ∑

Centroidal height from base (m) Moment (kN⋅m/m) Col. 3 Col. 2 × Col. 3 4.33 161.77 1.5 77.22 1.0 19.31 258.30

The plot of σ a′ against depth is shown in the figure below.

The active thrust Pa = 108.15 kN/m It acts at a height of

158.30 = 1.46 m above the bottom 108.15 SOURCE: Browsegrades.net

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16.8

Ka, sand = 0.283 and Ka, clay = 0.406 At top of the sand: σ o′ = 0; σ a′ = 0.

At bottom of the sand layer: σ o′ = 4 ×18 = 72 kN/m 2

σ a′ = Ka σ o′ = 0.283 × 72.0 = 20.4 kN/m 2 At top of the clay layer: σ o′ = 72.0 kN/m2

σ a′ = Ka σ o′ − 2c′ K a = 0.406 × 72.0 − 2 × 5 × 0.406 = 22.9 kN/m2 At bottom of the clay layer,

σ o′ = 4 × 18 + 2(19.0 − 9.81) = 90.4 kN/m 2

σ a′ = Ka σ o′ − 2c′ Ka = 0.406 × 90.4 − 2 × 5 × 0.406 = 30.3 kN/m2 The pore water pressure increases linearly within the clay layer.

At the bottom of the layer, it is 2 × 9.81 = 19.6 kN/m2 The variation of σ a′ and u with depth is shown in the figure below.

Zone Horizontal load Centroidal height from Moment (kN⋅m/m) Col. 1 (kN/m) -- Col. 2 base (m) -- Col. 3 Col. 2 × Col. 3 1 3.33 135.9 0.5 × 4 × 20.4 = 40.8 2 1.0 45.8 2 × 22.9 = 45.8 3 0.67 4.7 0.5 × 2 × 7.4 = 7.4 4 0.67 13.1 0.5 × 2 × 19.6 = 19.6 113.6 199.5 ∑

Pa = 113.6 kN/m acting at a height of 199.5 113.6 = 1.76 m above the bottom of the wall SOURCE: Browsegrades.net

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16.9

a.

α φ′ θ sin α cos α sin φ ′ cos φ ′ cos θ (deg) (deg) (deg)

ψa (deg)

sin ψa cos ψa

β ′a (deg)

cos(α–θ )

Ka(R)

ηa (deg)

0

34

0

0.000 1.000 0.559

0.829 1.000

0.000

0.000

1.000

0.000

1.000

0.283 62.000

5

34

0

0.087 0.996 0.559

0.829 1.000

3.967

0.069

0.998

5.000

0.996

0.286 60.017

10

34

0

0.174 0.985 0.559

0.829 1.000

8.091

0.141

0.990

10.000

0.985

0.294 57.954

15

34

0

0.259 0.966 0.559

0.829 1.000 12.571 0.218

0.976

15.000

0.966

0.311 55.715

20

34

0

0.342 0.940 0.559

0.829 1.000 17.708 0.304

0.953

20.000

0.940

0.338 53.146

0

34

10

0.000 1.000 0.559

0.829 0.985 20.000 0.342

0.940

21.951

0.985

0.333 62.000

10

34

10

0.174 0.985 0.559

0.829 0.985 28.091 0.471

0.882

27.460

1.000

0.388 57.954

20

34

10

0.342 0.940 0.559

0.829 0.985 37.708 0.612

0.791

31.524

0.985

0.481 53.146

0

34

20

0.000 1.000 0.559

0.829 0.940 40.000 0.643

0.766

32.162

0.940

0.461 62.000

10

34

20

0.174 0.985 0.559

0.829 0.940 48.091 0.744

0.668

33.595

0.985

0.553 57.954

20

34

20

0.342 0.940 0.559

0.829 0.940 57.708 0.845

0.534

33.983

1.000

0.693 53.146

b. The values are the same. c. → For θ = 0 and α = 0 (i.e., vertical wall and horizontal backfill), as expected, Pa acts horizontally, with β′a = 0; Ka(R) is the same as the value determined from Ka = tan2 (45 – φ′/2); The slope of the failure wedge ηa is the same as (45 + φ′/2). → For a vertical wall (θ = 0), as expected, Pa acts parallel to the backfill slope (see the first five rows). → For a given wall, when α increases, Pa and β also increase. → For a given backfill slope (say α = 20°), when θ increases, the magnitude of Pa increases. Here, the slope of the failure wedge ηa is independent of θ, as seen from Eq. (16.21).

16.10 The values are the same as those in Problem 16.9. α (deg) Ka [Eq. (16.22)] 0 5 10 15 20

0.283 0.286 0.294 0.311 0.338

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16.11 a. Eq. (16.34): zc =

2c′ 1 + sin φ ′ 2 × 10.5 1 + sin 15 = = 1.56 m γ 1 − sin φ ′ 17.5 1 − sin 15

b. z (m)

c′ γz

Ka′

σ a′ (kN/m2)

2.0 3.0 4.0 5.0 6.0

0.30 0.20 0.15 0.12 0.10

0.1302 0.2867 0.3655 0.4130 0.4449

4.540 14.994 25.488 36.000 46.537

c. Pa = 0.5 × 46.54 × 4.44 = 103.3 kN/m, acting at 5°° to the horizontal at a height of 4.44/3 = 1.48 m above the base 16.12 From Table 16.6, for α = 10°, φ ′= 36°, and β = 85°, Ka = 0.3072 Pa = 0.5 K a γH 2 = 0.5( 0.3072 )(17.0)( 6.02 ) = 94.0 kN/m

It acts at a height of 2.0 m above the base of the wall, inclined at an angle of 24°° to the normal to the wall surface.

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16.13 Eq. (16.38):

 sin β  2q   sin 85  2 × 25  3 γeq = γ +    = 17 +    = 25.3 kN/m    sin(β + α )  H   sin(85 + 10)  6.0  Pa = 0.5 K a γH 2 = 0.5( 0.3072 )( 25.3)( 6.0 2 ) = 139.9 kN/m

16.14 Since a < 0.4, use Eq. (16.41). z (ft)

b

σ h (lb/ft3)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.000 0.067 0.133 0.200 0.267 0.333 0.400 0.467 0.533 0.600 0.667 0.733 0.800 0.867 0.933 1.000

0.000 20.018 34.256 40.600 40.540 36.825 31.719 26.552 21.924 18.018 14.817 12.230 10.150 8.477 7.128 6.034

16.15 From Eqs. (16.45) and (16.46),

 b′  −1  2.0   = tan   = 18.43° H   6.0 

θ1 = tan −1 

 a′ + b′  −1  1.5 + 2.0   = tan   = 30.26°  H   6.0 

θ2 = tan −1 

The resultant horizontal load on the wall is given by [Eq. (16.44)]

P=

q [H (θ 2 − θ1 )] = 40 [6(30.26 − 18.43)] = 31.5 kN/m 90 90

From Eqs. (16.48) and (16.49), R = (1.5 + 2.0)2 × (90 – 30.26) = 731.82 Q = 2.02 × (90 – 18.43) = 286.28 From Eq. (16.47),  6.02 (30.26 − 18.43) + (731.82 − 286.28) − 57.3(1.5)( 6.0)  z = 6 .0 −   = 3.49 m 2(6.0)( 30.26 − 18.43)   SOURCE: Browsegrades.net

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The horizontal load P = 31.5 kN/m acts at a height of 3.49 m above the bottom of the wall.

16.16 Eq. (16.50): Pae = For kh = 0.3,

1 2 γH (1 − kv ) K ae 2

δ ′ 22 2 = = ; φ ′ = 33°; α = 10°; β = 80° φ ′ 33 3

 kh  −1  0.15  Eq. (16.29): θ ′ = tan −1   = 8 .5 °  = tan  1− 0   (1 − k v ) 

From Eq. (16.51) for β = 80°; α = 10°; δ ′ = 22º; θ ′ = 8.5°; and φ ′ = 33º, Kae = 0.5567 From Eq. (16.50),

Pae =

1 2 1 γH (1 − kv ) K ae = (18.5)(6.02 )(1 − 0)(0.5567) = 185.4 kN/m 2 2

16.17 For kh = 0.3,

δ ′ 22 2 = = ; φ ′ = 33°; α = 10°; β = 80° φ ′ 33 3

 kh  −1  0.15  Eq. (16.29): θ ′ = tan −1   = 8 .5 °  = tan  1− 0   (1 − k v ) 

Eq. (16.53): β ′ = β − θ ′ = 80 − 8.5 = 71.5° Eq. (16.54): α ′ = θ ′ + α = 8.5 + 10 = 18 .5° Eq. (16.52): Pae =

 sin 2 β ′  1  γH 2 (1 − k v )[ K a ( β ′, α ′)] 2 2  cos θ ′ ⋅ sin β 

From Eq. (16.36) for β ′ = 71.5°; α ′ = 18.5°; δ ′ = 22º; and φ ′ = 33º, the value of K a ( β ′, α′) = 0.5937   sin 2 71.5 1  = 185.4 kN/m Pae =  (18.5)( 6) 2 (1 − 0)( 0.5937 ) 2 2  cos 8.5 × sin 80 

Eq. (16.35): Pa = For

1 ( Ka γH 2 ) 2

δ′ 2 = : φ ′ = 33°; α = 10°; β = 80°; Ka = 0.3953 (Table 16.6) φ′ 3

Pa = (1 2)(0.3953)(18.5)( 6)2 = 131.6 kN/m

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∆Pae = 185.4 – 131.6 = 53.8 kN/m H 6 0.6 H ( ∆Pae ) +   Pa (0.6)(6)(53.8) +  (131.6)  3 = 3 Eq. (16.56): z = = 2.46 m Pae 185.4

16.18 For c* =

c′ 100 = = 0.058 ; kh = 0.3, from Figure 16.20c, Pae* = 0.24 γH 115 × 15

Pae = Pae* γH 2 = 0.24 (115)(152 ) = 6210 lb/ft

16.19 We apply Rankine’s passive earth pressure equation in terms of total stresses For φ = 0, Kp = 1.0

σ p = Kp σ o + 2c K p = (1.0)(115.0) z + 2(300 ) = 115 z + 600 lb/ft 2

The plot of passive pressure with depth is shown.

Zone Horizontal load (lb/ft) Col. 1 Col. 2 1 15 × 600 = 9,000 2 0.5 × 1725 × 15 = 12.937.5 21,937.5 ∑

Centroidal height from base (ft) Moment (lb⋅ft/ft) Col. 3 Col. 2 × Col. 3 7.5 67,500.0 5.0 64,687.5 132,187.5

The horizontal passive load Pp = 21,937.5 lb/ft acts at a height of 132,187.5 21,937.5 = 6.02 ft

16.20 Table 16.9 Table 16.10 φ′ [or Eq. (16.72)] (deg) [or Eq. (16.70)] 30 2.5900 cos 15 = 2.502 2.502 40 4.1168 cos 15 = 3.977 3.977

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Chapter 17 17.1

The retaining wall and the soil regions of interest are divided into rectangles and triangles as shown in the figure below for ease of computation of the weights and moment arms

For φ ′ = 35° and α = 15°, Ka = 0.2968 (Table 16.3)

Pa =

1 1 γH ′2 Ka = (18.5)(6.54)2 (0.2968) = 117.4 kN/m 2 2

Ph = Pa cos α = 113.4 kN/m Pv = Pa sin α = 30.4 kN/m Section Weight Moment arm Moment about C (kN/m) from C (m) (kN⋅m/m) 1 (0.5)(1.5)(6.0)(24.0) = 108.0 1.0 108.0 2 (0.5)(6.0)(24.0) = 72.0 1.75 126.0 3 (0.5)(2.0)(6.0)(24.0) = 144.0 2.67 384.5 4 (0.5)(2.0)(6.0)(18.5) = 111.0 3.33 369.6 5 (0.5)(2.0)(0.54)(18.5) = 10.0 3.33 33.3 Pv = 30.4 4.0 121.6 ΣV = 475.4 ΣMR = 1143.0 SOURCE: Browsegrades.net

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Stability with respect to overturning

Overturning moment M O = Ph

H′ = 113.4 × 2.18 = 247.2 kN.m/m 3

Resisting moment ∑ M R = 1143.0 kN.m/m FS(overturning) =

ΣM R 1143.0 = = 4.62 MO 247.2

Stability with respect to sliding

φ′   For φ ′ = 35°, from Eq. (16.64), K p = tan 2  45 +  = 3.690 2  Pp =

1 2 1 γD K p = (18.5)(1.0)2 (3.690) = 34.1 kN/m 2 2

Taking δ ′ as 2 3 φ ′, δ ′ = 23.3° FS( sliding) =

( ΣV ) tan δ ′ + Pp Ph

=

475 .4 × tan 23.3 + 34.1 = 2.11 113 .4

Stability with respect to bearing capacity failure

e=

B ΣMR − ΣM O 4.0 1143.0 − 247.2 B 4  − = − = 0.116 m <  = = 0.67 m  2 ΣV 2 475.4 6 6 

From Eq. (17.18), qmax = qtoe =

ΣV  6e  475.4  (6)( 0.116  2 1 +  = 1 +  = 139.5 kN/m B  B 4 .0  4 .0 

In granular soil, Eq. (17.20) becomes qu = q Nq Fqd Fqi + 0.5γ B′Nγ Fγd Fγi

where B′ = B – 2e = 3.768 m and q = γD = (18.5)(1.0) = 18.5 kN/m2 For φ ′ = 35°, from Table 6.2: N q = 33.30; N γ = 48.03

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

1 = 1 + 2 tan 35(1 − sin 35)2   = 1.07 B 4

Df

Fγd = 1  P cos α  −1  113.4   = tan   = 13.4°  ΣV   475.4 

ψ = tan −1  a

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2

2

2

2

 ψ   13.4  Fqi = 1 −  = 1 −  = 0.72 90   90    ψ 15   Fγi = 1 −  = 1 −  = 0.33 35  φ′    qu = (18.5)(33.30)(1.07)(0.72) + (0.5)(18.5)(3.768)(48.03)(1)(0.33) = 1027.0 kN/m 2

FS( bearing capacity ) =

17.2

qu 1027.0 = = 7.4 qmax 139.5

The passive resistance Pp was not considered in computing FS(overturning) which remains unchanged at 4.62. However, neglecting Pp will reduce FS(sliding), which becomes FS( sliding ) =

( ΣV ) tan δ ′ + Pp Ph

=

475 .4 × tan 23.3 + 0.0 = 1.81 113 .4

FS(bearing capacity) remains the same at 7.4

17.3

Refer to the figure below. For φ ′ = 34° and α = 10°, Ka = 0.2944 (Table 16.3)

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Pa =

1 1 γH ′2 Ka = (18.0)(6.54)2 (0.2944) = 116.5 kN/m 2 2

Ph = Pa cos α = 114.7 kN/m Pv = Pa sin α = 20.2 kN/m Section

Weight (kN/m)

1 2 3 4 5

(0.5)(1.5)(6.0)(24.0) = 108.0 (0.5)(6.0)(24.0) = 72.0 (0.5)(2.0)(6.0)(24.0) = 144.0 (0.5)(2.0)(6.0)(18.0) = 108.0 (0.5)(2.0)(0.54)(18.0) = 9.7 Pv = 20.2 ΣV = 461.9

Moment arm from C (m) 1.0 1.75 2.67 3.33 3.33 4.0

Moment about C (kN⋅m/m) 108.0 126.0 384.5 359.6 32.4 80.8 ΣMR = 1091.3

Stability with respect to overturning

Overturning moment M O = Ph

H′ = 114.7 × 2.18 = 250.0 kN.m/m 3

Resisting moment ∑ M R = 1091.3 kN.m/m Therefore, FS(overturning) =

ΣM R 1091.3 = = 4.37 MO 250.0

Stability with respect to sliding

φ′   For φ ′ = 24°, from Eq. 16.64, K p = tan 2  45 +  = 2.37 2  1 1 Pp = γD 2 K p = (18.5)(1.0)2 (2.37) = 21.9 kN/m 2 2 Taking δ′ = 23 φ ′ , δ ′ = 16.0°, and ca′ as 2 3 c′, ca′ = 6.7 kN/m2 FS( sliding ) =

( ΣV ) tan δ ′ + Bca′ + Pp Ph

=

461 .9 × tan 16 + 4 × 6.7 + 21 .9 = 1.58 114 .7

Stability with respect to bearing capacity failure

e=

B ΣM R − ΣM O 4.0 1091.3 − 250.0 B 4  − = − = 0.179 m <  = = 0.67 m  2 ΣV 2 461.9 6 6 

From Eq. (17.18), SOURCE: Browsegrades.net

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q max = q toe =

ΣV  6e  461.9  (6)(0.179  2 1 +  = 1 +  = 146.5 kN/m B  B 4 .0  4 .0 

qu = c′N c Fcd Fci + q Nq Fqd Fqi + 0.5γ B′Nγ Fγd Fγi

where B′ = B – 2e = 3.642 m; q = γD = (19.0)(1.0) = 19.0 kN/m2 For φ ′ = 24°, from Table 6.2: Nc = 19.32; N q = 9.60; N γ = 9.44

Fqd = 1 + 2 tan φ ′(1 − sin φ ′)2

Fcd = Fqd −

1 − Fqd N c tan φ ′

1 = 1 + 2 tan 24(1 − sin 24)2   = 1.08 B 4

Df

= 1.08 −

1 − 1.08 = 1.09 19.32 tan 24

Fγd = 1  P cos α  −1  114.7   = tan   = 13.9°  ΣV   461.9 

ψ = tan −1  a

2

2

ψ    13.9  Fci = Fqi = 1 −  = 1 −  = 0.72 90  90    2

2

 ψ  13.9  Fγi = 1 −  = 1 −  = 0.36 φ′  35    qu = (10)(19.32 )(1.09 )( 0.72 ) + (19.0)( 9.60)(1.08)( 0.72 ) + (0.5)(19.0)( 3.642 )( 9.44 )(1.0)( 0.36) = 411.0 kN/m 2

FS( bearing capacity ) =

17.4

qu 411.0 = = 2.81 qmax 146.5

φ′   For φ ′ = 33°, Rankine’s K a = tan 2  45 −  = 0.295 2  Pa =

1 1 γH ′2 K a = (18.0)(6.0)2 (0.295) = 95.6 kN/m 2 2

Section 1 2

Moment arm Moment about C from C (m) (kN⋅m/m) (0.5)(6.0)(24.0) = 72 2.75 198.0 (0.5)(2.5)(6.0)(24.0) = 180 1.67 300.6 Pv = 0 0 ΣV = 252.0 ΣMR = 498.6 Weight (kN/m)

SOURCE: Browsegrades.net

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Stability with respect to overturning

Overturning moment M O = Ph

H′ = 95.6 × 2.00 = 191.2 kN.m/m 3

Resisting moment ∑ M R = 498.6 kN.m/m FS( overturning ) =

ΣM R 498.6 = = 2.61 MO 191.2

Stability with respect to sliding

Taking δ ′ as 2 3 φ ′, δ ′ = 22° FS( sliding ) =

( ΣV ) tan δ ′ + Pp Ph

=

252 .0 × tan 22 + 0 = 1.07 95.6

Eccentricity at the base

e=

B ΣM R − ΣM O 3.0 498.6 − 191.2 − = − = 0.280 m 2 ΣV 2 252.0

The eccentricity is within B/6 (= 0.5 m).

17.5

qmax = qtoe =

ΣV  6e  252.0  ( 6)( 0.280)  1+ = 131.0 kN/m 2 1 +  =   B  B 3 .0  3 .0 

qmin = qheel =

ΣV  6e  252.0  ( 6)( 0.280)  1− = 38.6 kN/m2 1 −  =   B  B 3 .0  3 .0 

Refer to the figure.

From Table 16.6, for φ ′ = 33°,

δ ′= 22°, Ka = 0.2645 Pa = 0.5γH ′2 Ka = 0.5(18.0)(6.0)2 (0.2645) = 85.7 kN/m Ph = Pa cos 22 = 79.5 kN/m Pv = Pa sin 22 = 32.1 kN/m

SOURCE: Browsegrades.net

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Section

Weight (kN/m)

1 2

(0.5)(6.0)(24.0) = 72 (0.5)(2.5)(6.0)(24.0) = 180 Pv = 32.1 ΣV = 284.1

Moment arm from C (m) 2.75 1.67 3.00

Moment about C (kN⋅m/m) 198.0 300.6 96.3 ΣMR = 594.9

Stability with respect to overturning

Overturning moment M O = Ph

H′ = 79.5 × 2.00 = 159.0 kN.m/m 3

Resisting moment ∑ M R = 594.9 kN.m/m FS( overturning ) =

ΣM R 594.9 = = 3.74 MO 159.2

Stability with respect to sliding

Taking δ ′ as 2 3 φ ′, δ ′ = 22° FS( sliding ) =

( ΣV ) tan δ ′ + Pp Ph

=

284 .1 × tan 22 + 0 = 1.44 79 .5

Eccentricity at the base

e=

B ΣM R − ΣM O 3.0 594.9 − 159.0 − = − = −0.034 m 2 ΣV 2 284.1

The eccentricity is within B/6 (= 0.5 m).

17.6

qmax = qheel =

ΣV  6e  284.1  ( 6)( 0.034)  1+ = 101.1 kN/m 2 1 +  =   B  B 3 .0  3 .0 

qmin = qtoel =

ΣV  6e  284.1  ( 6)( 0.034 )  2 1− 1 −  =  = 88.3 kN/m B  B 3.0  3 .0

a. Based on Coulomb’s active earth pressures: Refer to the figure on the next page. From Table 16.6: α = 10°, φ ′ = 36°;

δ ′ = 24°. Coulomb’s active earth pressure coefficient: Ka = 0.2633 Pa =

1 1 γH ′2 K a = (18.5)(6.0)2 (0.2633) = 87.7 kN/m 2 2

SOURCE: Browsegrades.net

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Ph = Pa cos 24 = 80.1 kN/m Pv = Pa sin 24 = 35.7 kN/m

Section 1 2 3

Weight (kN/m) (0.5)(0.5)(5.5)(24.0) = 33.0 (0.5)(5.5)(24.0) = 66.0 (0.5)(3.5)(24.0) = 42.0 Pv = 35.7 ΣV = 176.7

Moment arm from C (m) 1.83 2.25 1.75 2.5

Moment about C (kN⋅m/m) 60.4 148.5 73.5 89.3 ΣMR = 371.7

Stability with respect to overturning

Overturning moment M O = Ph

H = 80.1 × 2.0 = 160.2 kN.m/m 3

Resisting moment ∑ M R = 371.7 kN.m/m FS( overturning) =

ΣM R 371.7 = = 2.32 MO 160.2

Stability with respect to sliding

φ′   For φ ′ = 36°, K p = tan 2  45 +  = 3.852 2  Pp =

1 2 1 γD K p = (18.5)(1.0)2 (3.852) = 35.6 kN/m 2 2 SOURCE: Browsegrades.net

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FS( sliding ) =

( ΣV ) tan δ ′ + Pp Ph

=

176 .7 × tan 24 + 35.6 = 1.43 80.1

b. Based on Rankine’s active earth pressures.

Refer to the figure.

From Table 16.3, for α =10° and φ ′ = 36°, Rankine’s active earth pressure coefficient is given by Ka = 0.2696

Pa = 0.5γH ′2 K a = 0.5(18.5)(6.18)2 (0.2696) = 95.2 kN/m Ph = Pa cos 24 = 87.0 kN/m Pv = Pa sin 24 = 38.7 kN/m

Section

Weight (kN/m)

1 2 3 4 5

(0.5)(0.5)(5.5)(24.0) = 33.0 (0.5)(5.5)(24.0) = 66.0 (0.5)(3.5)(24.0) = 42.0 (1.0)(5.5)(18.5) = 101.8 (0.5)(1.0)(0.18)(18.5) = 1.67 Pv = 35.7 ΣV = 280.2

Moment arm from C (m) 1.83 2.25 1.75 3.00 3.17 2.5

Moment about C (kN⋅m/m) 60.4 148.5 73.5 305.4 6.1 89.3 ΣMR = 683.2

Stability with respect to overturning

Overturning moment M O = Ph

H = 87.0 × 2.06 = 179.2 kN.m/m 3

SOURCE: Browsegrades.net

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Resisting moment ∑ M R = 683.2 kN.m/m FS( overturning ) =

ΣM R 683.2 = = 3.81 MO 179.2

Stability with respect to sliding

φ′   For φ ′ = 36°, K p = tan 2  45 +  = 3.852 2  Pp =

1 2 1 γD K p = (18.5)(1.0) 2 (3.852) = 35.6 kN/m 2 2

Taking δ ′ as 2 3 φ ′, δ ′ = 24° FS( sliding ) =

17.7

a.

( ΣV ) tan δ ′ + Pp Ph

=

280 .27 × tan 24 + 35.6 = 1.84 87.0

φ′  34    K a = tan 2  45 − 1  = tan 2  45 −  = 0.2827 2 2    Eq. (17.46):

t=

(γ1HK a SV S H )[ FS( B ) ] w fy

=

(16)(10)( 0.2827)(1)(1.25)(3) (0.12)( 260,000)

= 0.005437 m ≈ 5.44 mm b. Eq. (17.45). At z = 0,

φ ′  FS S S ( K γz )  L = H tan 45 − 1  + ( P ) V H a 2 (2 w tan φµ′ )( γz )  = 10 tan( 45 − 17) +

17.8

(3)(0.2827)(1)(1.25) = 14.79 m (2)(0.12)(tan 25)

a. Check for overturning: Pa =

1 1 γ1H 2 K a = (16)(10) 2 (0.2827 ) = 226.16 kN/m 2 2

 10  M O = Pa z ′ = ( 226.16)  = 753.87 kN .m/m ; L = 14.79 m  3

Eq. (17.48): SOURCE: Browsegrades.net

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 L2  (10)(16)(14.79)2  L M R = ( HL)( γ1 )  = Hγ1   = ≈ 17,500 kN.m/m 2 2 2 FS(overturning) =

17,500 = 23.2 753.87

b. Check for sliding: Eq. (17.50):

 2   2  γ1HL tan  φ1′  (16)(10)(14.79) tan  (34 ) 3  =  3   = 4.37 FS(sliding) = Pa 226.16 c. Check for bearing capacity:

φ2′ = 25°. From Table 6.2, N c = 20.72; N γ = 10.88 1 1 Eq. (17.51): qult = c2′ Nc + γ2 LNγ = (30)(20.72) + (15.5)(14.79)(10.88) 2 2 2 = 621.6 + 1247.1 ≈ 1869 kN/m

σ v ( H ) = γ1H = (16)(10) = 160 kN/m 2 FS(bearing) =

17.9

1869 = 11.68 160

φ1′ = 30°. K a = tan 2  45 − 

Eq. (17.56): SV =

Tall γ1zKa [ FS( B ) ]

At z = H = 6 m: SV = L=

30   = 0.333 .

2

16 = 0.336 m (15.9)(6)(0.333)(1.5)

S K [ FS( B ) ] H −z + V a φ′  2 tan φF′  tan  45 + 1  2 

At z = 0 (maximum length):

L=

(0.336)(0.333)(1.5) 6 + = 3.47 + 0.23 = 3.7 m tan (45 + 15) 2    2 tan  (30)  3   SOURCE: Browsegrades.net

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Eq. (17.60): ll =

SV Ka [ FS( P ) ] (0.336)(0.333)(1.5) = = 0.115 m - - - Use 1 m 4 tan φF′  2   4 tan  (30)  3  

17.10 Check for overturning: Pa =

1 (15.9)( 6) 2 (0.333) = 95.3 kN/m 2

M O = Pa

6 = 190.6 kN .m/m 3

Use L = 3.7 m for all depths.

Hγ1 L2 (6)(15.9)(3.7) 2 MR = = = 653 kN .m/m 2 2 FS(overturning) =

653 = 3.43 190.63

2  γ1HL tan  φ1′   3  = (15.9)( 6)( 3.7)(tan 20) = 1.35 Check for sliding: FS(sliding) = Pa 95.3 Check for bearing capacity:

For φ2′ = 20°, from Table 6.2, Nc = 14.83; N γ = 5.39 qult = c2′ Nc +

FS(bearing) =

1 1 γ2 L Nγ = (55)(14.83) + (16.8)( 3.7)(5.39) = 983.17 kN/m 2 2 2

983.17 = 10.31 (15.9)(6)

Note: FS against sliding is below acceptable values, so increase L to about 8 m.

SOURCE: Browsegrades.net

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Chapter 18 18.1

Refer to the figure.

L1 = 2 m; L2 = (6 – 2) = 4 m

  = 40. Ka = 0.217; Kp = 4.599; Kp – Ka = 4.382 γ  19 – 9.81 = 9.19 kN/m3

At the water table level, to the right of the sheet pile, σ1  0.217 ( 2  17 .5)  7.60 kN/m 2

At the excavation level, to the right of the sheet pile, σ 2  0.217(2  17.5  4  9.19)  15.57 kN/m 2

From Eq. (18.6), L3 

σ 2 15.57   0.387 m ( K p  K a )γ 4.382  9.19

Compute P and z : P  (0.5  7.60  2)  (7.60  4)  (0.5  7.97  4)  (0.5 15.57  0.387)  7.60  30.40  15.94  3.01  56.95 kN/m

SOURCE: Browsegrades.net

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Pz  7.60(0.67  4  0.387)  30.40(2  0.387)  15.94(1.33  0.387)  (3.01 0.258)  139.137 kN m/m 139.137 z  2.44 m 56.95

From Eq. (18.11), σ 5  K p ( γm L1  γL2 )  ( K p  K a ) γL3  4.599(17.5  2  9.19  4)  4.382  9.19  0.387  345.6 kN/m 2

A1 

σ 5 345.6   8.68 m γ( K p  K a ) 9.19  4.328

A2 

8P 8  56.95   11.31 m 2 γ( K p  K a ) 9.19  4.382

A3 

6 P[2 zγ( K p  K a )  σ 5 ] γ2 ( K p  K a ) 2

 A4 

6  56.95  (2  2.44  9.19  4.382  345.6)  114.2 m3 9.192  4.3822 P  6 z  5  4 P  56.95(6  2.44  345.6  4  56.95)   185.68 m 4 2 2 2 2 γ ( K p  K a ) 9.19  4.382

Substituting in Eq. (18.16): L 44  8.68 L 34  11.31L 24  114.2 L 4  185.68  0 Solving the above equation by trial and error, L4 = 4.1 m D = L4 + L3 = 4.1 + 0.387 = 4.487 m Increasing D by 30%, L + D = 6 + (1.3)(4.487) ≈ 11.83 m 18.2

Refer to the figure on the next page.

The maximum bending moment will occur between E and F , at depth z below E. This depth is given by [Eq. (18.21)]

z 

2P  ( K p  K a )γ 

2(66.37)  1.82 m ( 4.382)9.19

The maximum bending moment is given by [Eq. 18.22)]

SOURCE: Browsegrades.net

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1 γ( K p  K a ) z3 6 1  66.37( 2.39  1.82)  (9.19)( 4.382)1.823  238.96 kN.m/m 6

M max  P( z  z) 

The section modulus of the required section is given by S

M max 238.96  103   1.406  10  3 m 3/m of the wall 6 σ all 170  10

Section PZC-17 would be suitable.

18.3

Sheet pile depth:

 = 32. Ka = 0.307; Kp = 3.255; Kp – Ka = 2.95

 2  γLK a  17.5  5  0.307  26.9 kN/m 2 ( K p  K a ) γ  2.95  17.5  51.6 kN/m 2 /m depth

L3 

 2 K p  Ka

L

0.307  5  0.52 m 2.95

P  ( 0.5  26.9  5)  (0.5  26.9  0.52)  67.25  7.00  74.25 kN/m

SOURCE: Browsegrades.net

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z

67.25  2.187  7.00  0.347  2.02 m 74.25

Note: z can be approximated as [Eq. (18.30)]

z  L3 

(Use the exact value of 2.02 m L 5.0  0.52   2.19 m determined above) 3 3

 5  K p γL  ( K p  K a )γL3  3.255  17 .5  5  2.95  17.5  0.52  311.7 kN/m 2 A1 

A2 

A3   A4 

 5 γ( Kp  Ka )



311.7  6.04 m 17.5  2.95

8P 8  74.25   11.51 m2 γ ( Kp  K a ) 17.5  2.95

6 P[2 z γ ( K p  K a )   5 ] γ 2 ( Kp  K a )2 6  74.25  [2  2.02  17.5  2.95  311.7]  86.97 m3 2 2 17.5  2.95 P 6 z  5  4 P  74.25(6  2.02  311 .7  4  74.25)   113.52 m 4 2 2 2 2 γ (Kp  Ka ) 17.5  2.95 SOURCE: Browsegrades.net

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Eq. (18.31) becomes

f ( z2 )  L 44  6.04 L 34  11.51L 24  86.97L 4  113.52  0 By trial and error, f (4.01) = 0 gives L4 = 4.01 m D = L4 + L3 = 4.01 + 0.52 = 4.53 m Increase the embedment length D by 30% to 5.89 m.

Thus, the sheet pile has to be driven to a depth of 5 + 5.89 = 10.89 m Sheet pile section: Zero shear (i.e., maximum moment) occurs at z given by

z 

2  74.25  1.70 m 2.95  17.5

2P  ( Kp  K a )γ

1 γ ( K p  Ka ) z3 6 1  74.25( 2.02  1.70)  (17.5)(2.95)1.703  233.94 kN.m/m 6

M max  P( z  z) 

The required section modulus of the sheet pile is given by S

M max

 all



233.94  103  1.231  10 3 m3/m 6 190  10

Use section PZC-13 (from Table 18.1).

18.4

Original ground level L1 Granular soil

L L2

P1

z1 Excavation level

z1

z L3 D

Cohesive soil G

J

L4 I

B SOURCE: Browsegrades.net

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For the sand, Ka = 0.295

 1  1.5  16.5  0.295  7.30 kN/m2  2  (1.5  16.5  3.0  9.19)(0.295)  15.43 kN/m2 P1  (0.5  7.30  1.5)  (7.30  3.0)  (0.5  8.13  3.0)  5.48  21.90  12.20  39.58 kN/m

z1 

5.48  3.5  21.90  1.5  12.20  1.0  1.62 m 39.58

From Eq. (18.48),

4c  (γL1  γL2 )D 2  2 P1D  P1 ( P1  12cz1 )  0 (γL1  γL2 )  2c

4  50  (16.5  1.5  9.19  3)D2  2  39.58  D 

39.58(39.58  12  50  1.62) 0 (16.5  1.5  9.19  3)  2  50

147.68 D 2  79.16 D  262.86  0; D  1.63 m Increase D by 50% to 2.45 m The depth of the sheet pile must be 4.5 + 2.45 = 6.95 m 18.5

From Eq. (18.42):

 6  4c  ( γL1  γL2 )  4  50  [(16.5)(1.5)  (9.19  3)]  147.68 kN/m2 Eq. (18.49): z  

P1

6



39.58  0.268 m 147.68

Eq. (18.50): M max  P1 ( z   z1 ) 

 6 z2 2

 39.58(0.268  1.62) 

147.68  0.2682 2

 74.73  5.30  69.43 kN.m/m

The required section modulus is given by: S

18.6

M max

 all



69.43  103  0.365  10  3 m3 /m 6 190  10

  = 35. Ka = 0.271; Kp = 3.69;

Kp FS



3.69  2.46 ; γ   20  9.81  1 .5

2

10.19 kN/m

SOURCE: Browsegrades.net

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The right side of the sheet pile is in active state and the left is in passive state. The lateral earth pressure variation with depth is shown in the figure. There is also a horizontal reaction R at the bottom of the sheet pile (see figure) for equilibrium.

At C, the active earth pressure is (0.271)(6 ×18) = 29.27 kN/m2 At C, the passive earth pressure is (2.46)(2×18) = 88.56 kN/m 2 At the bottom of the sheet pile O, active earth pressure is (0.271)[(6 ×18) + 10.19D] = 29.27 + 2.76D, and passive earth pressure is (2.46)[(2×18) + 10.19D] = 88.56 + 25.07D The lateral earth pressure diagram is shown in the figure. We divide the earth pressure loads into six separate components as shown in the figure.

Height above O (m)

Moment about O (kNm/m)

Pa1 = 0.5  6  29.27 = 87.81

D+2

87.81D + 175.62

Pa2 = D  29.27 = 29.27D

0.5D

14.64D2

Pa3 = 0.5  2.76D  D = 1.38D2

0.33D

0.455D3

Pp1 = 0.5 × 2 × 88.56 = 88.56

D + 0.67

88.56D + 59.33

Pp2 = D × 88.56 = 88.56D

0.5D

44.28D2

Pp3 = 0.5 × 25.07D × D = 12.535D2

0.33D

4.136D3

Horizontal load (kN/m)

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 M O  87.81D  175.62  14.64 D 2  0.455D 3 88.56 D  59.33  44.28 D 2  4.136 D 3  0

3.681D3  29.64 D 2  0.75D  116.29  0 ; D  1.8 m The sheet pile should be driven to a depth of 4 + 2 + 1.8 = 7.8 m

18.7

a.

Kp

4.023  2.682 FS 1 .5 The right side of the sheet pile is in active state and the left is in passive state.

The lateral earth pressure variation with depth is shown in the figure. There is also a horizontal reaction R at the bottom of the sheet pile for equilibrium.

  = 37. Ka = 0.249; Kp = 4.023;



At B, the active earth pressure is (0.249)(5 ×17) = 21.2 kN/m2 At the bottom of the sheet pile O: active earth pressure is (0.249)[5 ×17 + 10.19D] = 21.2 + 2.54D, and passive earth pressure is (2.682)(10.19D) = 27.33D The lateral earth pressure diagram is shown in the figure. Divide the earth pressure loads into four separate components as shown.

Height above O (m)

Moment about O (kNm/m)

Pa1 = 0.5  5  21.2 = 53.0

D + 1.67

53.0D + 88.5

Pa2 = D  21.2 = 21.2D

0.5D

10.6D2

Pa3 = 0.5  2.54D  D = 1.27D2

0.33D

0.42D3

Pp1 = 0.5  D  27.33D = 13.67D2

0.33D

4.56D3

Horizontal load (kN/m)

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 M O  53.0 D  88.5  10.6 D 2  0.42 D 3  4.56 D 3  0 4.14 D 3  10.6 D 2  53.0 D  88.5  0; D  5.6 m The sheet pile should be driven to a depth of 5.0 + 5.6 = 10.6 m b. Let the maximum bending moment Mmax be at a depth of z below the excavation level, where the shear force is zero.

For equilibrium, Pa1 + Pa2 + Pa3 – Pp1 – V = 0

V = Pa1 + Pa2 + Pa3 – Pp1 = 0 53.0 + 21.2z + 1.27z2 – 13.67z2 = 0 12.4z2 – 21.2z – 53 = 0 z = 2.92 m

M max  (53.0  2.92  88.5)  (10.6  2.922 )  (0.42  2.923 )  ( 4.56  2.923 )  230.7 kN.m/m The required section modulus S is S

M max

 all



230.7  103  121.4  10  5 m3 /m 190  106

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18.8

a.   = 35. Ka = 0.271; Kp = 3.690;

Kp FS



3.960  2.46 1 .5

1  0.271  (3  17)  13.8 kN/m2  2  2.46  10.19D  25.1D kN/m2  3  (0.271)[3  17  (4  D)  10.19]  24.9  2.76D kN/m2 Horizontal load (kN/m) Pa1 = 0.5 × 3 × 13.8 = 20.7

Distance below O (m) 1

Moment about O (kNm/m) 20.7

Pa2 = (D + 4) × 13.8 = 55.2 + 13.8D

4 + 0.5D

220.8 + 82.8D + 6.9D2

Pa3 = 0.5 × (D + 4) × (2.67D +11.1) = 1.38D2 + 11.07D + 22.2

4.67 + 0.67D

0.92D3 + 13.9D2 + 66.6D + 103.7

Pp1 = 0.5 × 25.1D × D = 12.55D2

6 + 0.67D

75.3D2 + 8.4D3

 M O  20.7  (220.8  82.8 D  6.9 D 2 )  (0.92 D 3  13.9 D 2  66.6 D  103.7)  75.3D 2  8.4 D 3  0 7.48 D 3  54.5 D 2  149.4 D  345.2  0; D  3.25 m The sheet pile should be driven to a depth of 7 + 3.25 = 10.25 m SOURCE: Browsegrades.net

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b. For equilibrium, F  Pa1  Pa 2  Pa 3  Pp1  20.7  (55.2  13.8D )  (1.38 D 2  11.07 D  22.2)  (12.55D 2 )  0 Substituting D = 3.25 m, F = 61.0 kN/m.

If the tie rods are spaced at 3-m intervals, load per tie rod is: 61.0  3 = 183.0 kN 18.9

a. Sheet pile depth For   = 35, Ka = 0.271; Kp = 3.690; Kp – Ka = 3.419

σ1  0.271  (3  16.5)  13.41 kN/m2 σ2  (3  16.5  6  9.19)(0.271)  28.36 kN/m2 L3 

σ 2 28.36   0.903 m γ ( K p  K a ) 9.19  3.419

1 1 1 σ1L1  σ 2 L2  (σ 2  σ1 ) L2  σ 2 L3 2 2 2 1 1  1     13.41  3  (13.41  6)  ( 28.36  13.41)(6)    28.36  0.903 2 2  2   20.12  80.46  44.85  12.81  158.24 kN/m

P

Pz  ( 20.12  7.903)  (80.46  3.903)  ( 44.85  2.903)  (12.81  0.602)  610.95 kN.m/m

Therefore, z  3.861 m From Eq. (18.72),

L34  1.5(l2  L2  L3 ) L24 

3P( L1  L2  L3 )  ( z  l1 ) 0 γ( K p  K a )

L34  1.5(1.5  6  0.903) L24 

3  158.24(3  6  0.903)  (3.861  1.5) 0 9.19  3.419

L34  12.60 L24  68.62  0 ; L4 = 2.16 m

Dtheoretical = L3 + L4 = 0.903 + 2.16 = 3.06 m We increase this by 30% to 4.0 m. Thus, the required depth of the sheet pile is 6 + 3 + 4 = 13.0 m

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b: Anchor force

F  P

1 1 γ( K p  K a ) L24  158.24   9.19  3.419  2.162 2 2

 84.94 kN/m With 3 m of horizontal spacing between the tie rods, the force per tie rod is 3  84.94 = 254.8 kN. c. Maximum bending moment From Eq. (18.74), Mmax occurs at z given by

1 1  1 L1  F  1( z  L1 )  Ka γ( z  L1 ) 2  0 2 2 1 1  13.41  3  84.94  13.41( z  3)   0.271  9.19  ( z  3)2  0 2 2 Let z – 3 = x

 64.83  12.41x  1.245x 2  0 ; x 2  10.77 x  64.83  0 x = z – 3 = 4.3 m; z = 7.3 m Taking the moment about the point of zero shear,

1 L1  x2 1  x M max   σ1L1  x    σ1  Ka γx 2    F (l2  x ) 2 3 2 2   3 1 3 4.32     13.41  3   4.3    13.41  2 3 2  1 4.33   0.271  9.19   84.94  1.5  4.3 2 3  106.61  123.98  33.0  492.65  229.06 kN.m/m 18.10 a.   = 35; l1 = 1.5 m; L1 = 3 m; L2 = 6 m. So, l1 L1  L2  0.167 From Figures 18.24, 18.25, and 18.26, the generalized non-dimensional values are: GD = 0.18; GF = 0.06; GM = 0.019

L1 ( L1  L2 )  0.333 and l1 ( L1  L2 )  0.167 .

From Figures 18.27, 18.28, and 18.29: CDL1 = 1.145; CFL1 = 1.072; CML1 = 1.031 From Eq. (18.75), D  ( L1  L2 )(GD )(CDL1 )  (3  6)(0.18)(1.145)  1.55 m SOURCE: Browsegrades.net

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Depth of sheet pile = 6 + (1.3)(1.55) ≈ 11 m b. From Eq. (18.78):

16.5  32  9.19  62  2  16.5  3  6 γa   13.25 kN/m3 2 (3  6) From Eq. (18.76):

F  γa ( L1  L2 )2 (GF )(CFL1 )  13.25(3  6)2 (0.06)(1.072)  69.03 kN/m With 3-m horizontal spacing of tie rods, the force per tie rod = 69.03  3 = 207.1 kN c. From Eq. (18.77),

M max  γa ( L1  L2 )3 (GM )(CML1 )  13.25(3  6)3 (0.019)(1.031)  189.22 kN.m/m 18.11 H  = L1 + L2 + Dactual = 3 + 6 + 4 = 13.0 m Mmax = 229.06 kNm/m I (m4/m)

Section

H   (m3/MN) Log  (m)

S (m3/m)

Md = Sall Md /Mmax (kNm/m)

PZC-12 192.06  10-6 13.0 7.73  10-4 –3.11 120.42  10-5

252.88

1.10

116.2  10-6 13.0 12.77  10-4 –2.89 98.92  10-5

207.23

0.90

PZ-22

The sand will be classified as medium dense sand. We have tried the two small sections in Table 18.1 and, with Mmax being relatively small, both sections are adequate. They both plot in the safe section region in Figure 18.30. Use PZ-22.

Note: The Hagerty and Nofal (1992) chart used in Problem 18.10 also suggests that the moment estimated in Problem 18.9 is high.

18.12 Depth of sheet pile

  γav

16.5  3  9.19  6  11.6 kN/m3 36

Ka = 0.271

 L  0.72  0.271  11.6  9  20.4 kN/m2 σ a  CKa γav SOURCE: Browsegrades.net

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σ p  Rσ a  0.6  20.4  12.2 kN/m 2

l1 1.5   0.167 L 9.0 From Eq. (18.84),  92  (1  2  0.167 )  0 D 2  2 D(9)1  0.167     0 .6 

D2 + 15D – 90 = 0; D  4.6 m Check this R using Eq. (18.84):

R

L( L  2l1 ) 9(9  2  1.5) 54    0.6 (O.K.) D(2 L  D  2l1 ) 4.6(2  9  4.6  2  1.5) 90.16

The required depth of sheet pile = 6 + 3 + 4.6 = 13.6 m Anchor force

From Eq. (18.85), F  σ a ( L  RD)  20.4(9  0.6  4.6)  127.3 kN / m Maximum moment

RD 0.6  4.6 2l 2  1.5   0.307 ; 1   0.33 L 9 L 9 From Eq. (18.86),





M max  0.5  20.4  92  (1  0.307)2  0.33  (1  0.307)  205.9 kN.m/m 18.13   = 34 gives Ka = 0.283

σ1  (0.283)( 4  17)  19.2 kN/m2 σ2  (0.283)(4  17  6  9.19)  34.8 kN/m2 σ 6  4  40  (4  17  6  9.19)  36.9 kN/m2

P1  0.5  19.2  4  19.2  6  0.5  15.6  6  38.4  115.2  46.8  200.4 kN/m P1z1  38.4  7.333  115.2  3.0  46.8  2.0  720.80 kN.m/m z1 

720 .80  3.60 m 200 .4

From Eq. (18.88) SOURCE: Browsegrades.net

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 6 D 2  2 6 D( L1  L2  l1 )  2 P1 ( L1  L2  l1  z1 )  0 36.9 D 2  2  36.9 D(4  6  2)  2  200.4  (4  6  2  3.60)  0 36.9 D 2  590.4 D  1763.5  0; D = 2.573 m From Eq. (18.87), F  P1   6 D  200.4  35.9  2.573  105.5 kN/m 18.14 From Eq. (18.91): FS  T 

1 K p  K a γH 2 s 2

1 3.69  0.271  17  H 2  3 2 H = 1.77 m (say 1.8 m) 1.5  183 

Take the anchor beam height as 1.6 m so that h > 0.5H (see Figure 18.42)

18.15 Step 1: Basic Case – Strip Anchor of Height H For   = 32, from Figure 18.44a, Ka = 0.28 Weight of the anchor plate W = (2.0 m)(0.1 m)(24.0 kN/m3) = 4.8 kN/m From Eq. (18.93), SOURCE: Browsegrades.net

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K p sin δ 

W  0.5γH 2 Ka sin   4.8  0.5(17.5)(22 )(0.28) sin 32   0.286 0.5γH 2 0.5(17.5)(22 )

From Figure 18.44b, K p cos δ  4.7 From Eq. (18.92),

  0.5(17.5)(22 )(3.7  0.28  cos 32)  121.2 kN/m Pult Step 2: Strip Anchor of Height h

From Eq. (18.94) with Cov = 14,      C 1   14  1  ov   Pus    Pult (121.2)  116.5 kN/m H   C    14   2    ov  h     1.25   Step 3: The Actual Plate Anchor

S   B 2.5  1.5   0.308 H  h 2.0  1.25 From Figure 18.46,

Be  B  0.18 H h Be = 0.18(2.0 + 1.25) + 1.5 = 2.09 m From Eq. (18.95), Pult  Pus Be  116.5  2.09  243.5 kN

18.16 H = 1.2 m; h = 0.75 m; B = 0.9 m;   = 32 H/h = 1.60 and B/h = 1.2 From Fig. 18.48: M γq  7.0 From Fig. 18.49: Fs = 1.45 From Eq. (18.97): Pult  M γq ( γh 2 ) BFs  7.0(17.5  0.752 )( 0.9)(1.45)  89.9 kN

SOURCE: Browsegrades.net

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5.4  H 2    18.17 Pult  tan    A 

0.28

5.4  1.22    γAH  tan 32  0.75  0.9 

0.28

(17.5)(0.75  0.9)(1.2)

 151.4 kN From Eq. (18.99), P u  2 .2   Pult H

0 .3

100  u   2.2  151.4  1 .2 

0. 3

The anchor displacement, u = 0.0217 m = 21.7 mm

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Chapter 19 19.1

a.

γH 18.5 × 7 = 6.5 > 4 = c 20 We treat the soil as soft to medium clay and use the envelope shown in Figure 19.8. The maximum lateral earth pressure is given by the larger of [Eq. (19.6)] σ a = γH − 4c = 18.5 × 7 − 4 × 20 = 49.5 kN/m 2 σ a = 0.3γH = 0.3 × 18.5 × 7 = 38.9 kN/m 2

Use σa = 49.5 kN/m2

From the equilibrium considerations of the upper block,

1.75   A(3.0) = (49.5)(2.25)(1.125) + (0.5)(49.5)(1.75) 2.25 +  = 248.02 3   A = 82.67 kN/m A + B1 = (49.5)(2.25) + (0.5)(49.5)(1.75) = 154.69 kN/m; B1 = 72.02 kN/m For the bottom block, C(2.0) = (49.5)(3)(1.5); hence, C = 111.38 kN/m C + B2 = (49.5)(3) = 148.5 kN/m; B2 = 37.12 kN/m SOURCE: Browsegrades.net

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The strut loads are as follows.

PA = As = 82.67 × 5 = 413.4 kN

PB = Bs = ( B1 + B2 ) s = (72.02 + 37.12)(5) = 545.7 kN PC = Cs = 111.38 × 5 = 556.9 kN b. In the upper block, the maximum moment occurs between A and B, at distance z1 above B1, where the shear force is zero. This can be determined as follows.

B1 = 72.02 = (49.5)(z1); z1 = 1.455 m

 1.455  M max = 72.02(1.455) − (49.5)(1.455)  = 52.39 kN.m/m 2   In the lower block, from the bending moment diagram, it can be seen that the maximum moment occurs at C. This can be determined as Mmax = (49.5)(1)(0.5) = 24.75 kN⋅m/m.

Use the larger of the two values: 52.39 kN·m/m.

Thus, the required section modulus is given by

S=

M max 52.39 = = 30.8 × 10-5 m3 /m of the wall 3 σall 170 × 10

From Table 18.1, PZ-22 sheet pile sections can be used. c. For the wales at B,

M max =

( B1 + B2 )s 2 (72.02 + 37.12)(52 ) = = 341.06 kN.m 8 8

For the wales at C,

Cs 2 (111.38)(52 ) M max = = = 348.06 kN⋅⋅m 8 8

19.2

a. K a = tan 2 ( 45 − φ ′ / 2 ) = 0.249 σ a = 0.65γHK a = 34.96 kN/m 2

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Leaving the top and bottom struts, the middle two struts meet the sheet pile at a hinge. As shown in the figure, the struts are labelled A, B, C and D. The same notation is given to the strut load per meter width of excavation. The loads B and C are split into two components.

From the equilibrium considerations of the top block,

A(3.0) = (34.96)(5)(2.5) = 437.0 ; A = 145.67 kN/m A + B1 = 34.96 × 5 = 174.8 kN/m; B1 = 29.13 kN/m In the middle block, by symmetry, B2 = C1 = 0.5 × 34.96 × 3 = 52.44 kN/m2 For the bottom block, D (3.0) = (34.96)(4.0)(2.0); D = 93.22 kN/m D + C2 = (34.96)(4.0) = 139.84 kN/m; C2 = 46.62 kN/m PA = As = 145.67 × 2.5 = 364.2 kN

PB = Bs = ( B1 + B2 ) s = (29.13 + 52.44)(2.5) = 203.9 kN PC = Cs = (C1 + C2 ) s = (52.44 + 46.62)(2.5) = 247.7 kN PD = Ds = 93.22 × 2.5 = 233.1 kN b. Find the maximum moments within the three sections of the sheet pile shown in the figure.

For the top block, At distance z1 above B1 (and below A), the shear force is zero.

B1 = 29.13 = 34.96 z1; z1 = 0.833 m At this location, bending moment = (34.96)(0.833)(0.833/2) = 12.13 kN⋅m/m SOURCE: Browsegrades.net

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MA = (34.96)(2)(1) = 69.92 kN⋅m For the middle block,

(34.96)(32 ) M max = = 39.33 kN/m 8 For the bottom block, At distance z2 below C2, the shear force is zero.

34.96 z2 = C2; z2 = 46.62/34.96 = 1.333 m At this location, M = (46.62)(1.333)-(34.96)(1.333)(1.333/2) = 31.08 kN⋅m/m MD = (34.96)(1.0)(0.5) = 17.48 kN⋅m/m From the different values of bending moments calculated, it can be seen that the maximum (69.92 kN⋅m/m) occurs at A. Thus, the required section modulus is given by

S=

M max

σ all

=

69.92 = 41.3 × 10−5 m3 /m of the wall 3 170 × 10

From Table 18.1, PZ-22 sheet pile sections can be used.

c. For the wales at B,

M max =

( B1 + B2 ) s 2 (29.13 + 52.44)(2.52 ) = = 63.73 kN⋅⋅m 8 8

For the wale at C,

M max =

19.3

(C1 + C2 ) s 2 (52.44 + 46.62)(2.52 ) = = 77.39 kN⋅⋅m 8 8

γH 17.65 × 9 = = 5.3 > 4 c 30 We will treat the soil as soft to medium clay and use the envelope shown in Figure 19.8. The maximum lateral earth pressure is given by the larger of [Eq. (19.6)] σ a = γH − 4c = 17.65 × 9 − 4 × 30 = 38.9 kN/m 2 σ a = 0.3γH = 0.3 × 17.65 × 9 = 47.7 kN/m 2 SOURCE: Browsegrades.net

174 © 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Use σa = 47.7 kN/m2 Leaving the top and bottom struts, the middle three struts meet the sheet pile at a hinge. As shown in the figure below, the struts are labelled A, B, C, D, and E. The same notation is given to the strut load per meter width of excavation.

From equilibrium consideration for the top block, find the loads A and B1.

By taking moment about hinge B1, 2.25   0.25   A( 2.0) = ( 47.7 × 0.25)  + (0.5 × 47.7 × 2.25) 0.25 +  = 55.15 3   2  

Thus, A = 27.58 kN/m

A + B1 = (47.7 × 0.25) + (0.5 × 47.7 × 2.25) = 65.59 kN/m ; B1 = 38.01 kN/m.

For the second block, by symmetry, B2 = C1 = 0.5 × 47.7 × 2.0 = 47.7 kN/m Similarly, C2 = D1 = 0.5 × 47.7 × 2.0 = 47.7 kN/m For the bottom block, taking moment about the hinge D,

E (2.0) = (47.7 × 2.5)(1.25) ; E = 74.53 kN/m E + D2 = 47.7 × 2.5 = 119.25 kN/m ; D2 = 44.72 kN/m The strut loads are as follows.

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PA = As = 27.58 × 3 = 82.7 kN

PB = Bs = ( B1 + B2 ) s = (38.01 + 47.70)(3) = 257.1 kN PC = Cs = (C1 + C2 ) s = (47.70 + 47.70)(3) = 286.2 kN

PD = Ds = ( D1 + D2 ) s = (47.70 + 44.72)(3) = 277.3 kN PE = Es = 74.53 × 3 = 223.6 kN

19.4

33   a. For the sand, K a = tan 2  45 −  = 0.295 2  Unconfined compressive strength of the clay = 2c = 70 kN/m2 From Eq. (19.8), cav = =

γav =

b.

1 [γs K s H s2 tan φs′ + ( H − H s )n ′qn ] 2H 1 [(17.0)( 0.295)( 4 2 ) tan 33 + (10 − 4 )( 0.75)( 70)] = 18.35 kN/m 2 2 × 10

1 [γs H s + ( H − H s )γc ] = 1 [(17.0)( 4) + (18.5)(10 − 4)] = 17.9 kN/m3 H 10

γav H 17.9 × 10 = = 9.75 > 4 , so we treat the soil as soft to medium clay. cav 18.35 The lateral earth pressure envelope will be as shown in Figure 19.8.

From Eq. (19.6), σ a = γav − 4Cav = (17.9 × 10 )( 4)(18.35) = 105 .6 kN/m 2 , or

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σ a = 0.3 × 17.9 × 10 = 53.7 kN/m 2

Use σ a = 105 .6 kN/m 2

19.5

From Eq. (19.10), cav =

1 [5 × 600 + 8 × 1200 + 12 × 850] = 912 lb/ft 2 25

From Eq. (19.11), γav =

19.6

1 [5 × 110 + 8 × 120 + 12 × 105] = 110.8 lb/ft 3 25

L = ∞; T = ∞; H = 7 m; B = 5 m B′ = B

2 = 3.535 m

B′′ = 2 B′ = 5.0 m c = 20 kN/m2 and γ = 18.5 kN/m3

0.2(5.0)  ( 20)( 7) 0.2 B′′  cH   5.14(20)1 + 5.14c1 + + + ∞  (3.535) L  B′   FS = = = 1.10 γH + q (18.5)( 7) + 0

19.7

0 .2 B ′  0.2(3.535)    cN c 1 +  (20)(5.14)1 +  L  ∞    = 1.14 FS = = q c 0 20     − H 7 γ + 18.5 + − ′ H B 7 3 .

535    

19.8

H = 9.0 m; T = ∞; L = 25 m; q = 0; c = 30 kN/m2; and γ = 17.65 kN/m3 B′ = B

2 = 2.121 m B′ = B / 2 =2.121 m

0 .2 B ′  0.2( 2.121)    cN c 1 +  (30)(5.14)1 +  L  25    = 4.97 FS = = q c 0 30     − H γ + 17.65 + −  ( 9 .0 ) ′ H B 7 2 .

121    

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177 © 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SOURCE: Browsegrades.net

178 © 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.