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FOUNDATION ENGINEERING – CIVI 6501 Department of Building, Civil and Environmental Engineering Concordia University Major Assignment Dr. A. Hanna Winter 2016 ** Students should avoid submitting assignment’s solutions copied from ppt slides from TAs’ presentation or copying solutions from previous students who used to study this course. ** Students are strongly encouraged to use other theories or methods available in the literature to solve the problems. 1. A footing 6 ft. x 6 ft located at a depth of 5.0 ft below ground level is loaded with an axial load of 180 tons and Mx = 90 ft-tons; My = 60 ft-tons. The subsoil has a unit weight of 115 lbs/ft3; = 36°, C = 200 lbs/ft2 and the water table is at a depth of 30 ft below ground level. Calculate the factor of safety against shear failure of soil. y

180 Ton

ex ey x

Df=5 ft

B=L=6 ft γ=115 lb/ft3 =36 c=200 lb/ft2

qmin

Solution: = 36° C = 200 lbs/ft2 γ=115 lbs/ft3 d=30 ft

1

qmax

ex=

M y 60 B = =0.33 ft < =1 ft OK Q 180 6

e y=

M x 90 L = =0.50 ft < =1 ft OK Q 180 6

'

B =B−2 e x =6−2 ×0.33=5.34 ft L' =L−2 e y =6−2× 0.50=5.00 ft effetive width=B ' =5 ft effective length=L' =5.34 ft

q max=

6e 6e N 180 6 × 0.33 6 ×0.5 ton lb 1+ x + y = 1+ + =9.15 2 × 2240=20496 2 BL B B 6 ×6 6 6 ft ft

q min=

6e 6e N 180 6 × 0.33 6 ×0.5 ton lb 1− x − y = 1− − =0.85 2 × 2240=1904 2 BL B B 6 ×6 6 6 ft ft

(

(

)

)

(

)

(

)

Assume general shear failure occurs 1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 Meyerhof (1963): ϕ=36 yields N c =50.55, N q=37.7, N γ =44.4 →

S c =1.3 S q=1 S γ =0.8

Vertical Load :i c =i q=i γ =1 2

d=30 ft >B yields Water has no effect on theultimate bearing capacity →

1 q'u =1× S c c N c +1× S q γ D f N q+ × 1× S γ γ B ' N γ 2 1 q'u =1× 1.3× 200 ×50.55+1 ×1 ×115 ×5 × 37.7+ ×1 ×0.8 ×115 × 5 ×44.4=45033lb/ ft 2 2 qu (net) =qu −γD=45033−115 × 5=44458

FS=

lb 2 ft

q u(net ) 44458 = =2.2 q max 20496

* Students can solve the problem by using other theories such as Vesic,… 2. What will be the change in the factor of safety in Problem 1 if the load was inclined at 12° to the vertical? The other data being same as in Problem 1. Solution: 1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 Meyerhof (1963): ϕ=36 yields N c =50.55, N q=37.7, N γ =44.4 →

S c =1.3 S q=1 S γ =0.8

3

(

i c =i q= 1−

θ 2 12 2 = 1− =0.75 90 90

) (

)

θ 2 12 2 i γ = 1− = 1− =0.44 ϕ 36

( ) (

)

1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 1 q'u =0.75 ×1.3 ×200 ×50.55+ 0.75× 1× 115 ×5 ×37.7+ × 0.44 ×0.8 × 115 ×5 × 44.4=30609lb / ft 2 2 qu (net) =qu −γD=30609−115×5=30034

q max=

FS=

lb 2 ft

6e 6e N 180 ×cos 12 6× 0.33 6 × 0.5 ton lb 1+ x + y = 1+ + =8.95 2 ×2240=20048 2 BL B B 6×6 6 6 ft ft

(

)

(

)

q u(net ) 30034 = =1.5 q max 20048

* Students can solve the problem by using other theories such as Vesic,…

3. A 2.0m wide square footing is located at a depth of 1.5m in a layered sand deposit. The upper sand layer 2.5m thick have 1= 1.8 t /m3, 1 = 400 and C1 = 0 is followed by other sand layer having 2= 1.6 t/m3, 2 = 300 and C2 = 0. Using a factor of safety of 3 against shear failure of soil, calculate the safe load the footing can carry. Assume that the water table is at great depth.

γ1=1.8 t/m3 1=40 c=0

Df =1.5 m 2.5m

4

B=L=2 m

H=1m

γ2=1.6 t/m3 2=30 c=0

Solution: Dense sand layer over loose sand layer First Method: 2 Dcosα i s K s tanϕ quv =qbv +γ 1 H 2 1+ −γ 1 H ≤q tv H B

(

)

1 q bv=γ 1 ( D+ H ) N q 2 i q 2 S q 2+ γ 2 B N γ 2 i γ 2 S γ 2 2 1 q tv=γ 1 D N q 1 i q 1 Sq 1 + γ 1 B N γ 1 i γ 1 S γ 1 2 upper (top) layer : ϕ=40 yields N q 1=64.1 , N γ 1=93.6 →

lower ( bottom ) layer :ϕ =30 yields N q 2=18.4 , N γ 2=15.7 →

i q=i γ =1

S q 1=Sq 2=1 S γ 1=S γ 2=0.8

1 t q 2=qbv =1.8 × (1.5+ 1 ) ×18.4 ×1× 1+ ×1.6 ×2 ×15.7 ×1 × 0.8=102.9 2 2 m

5

1 t q1 =qtv =1.8 ×1.5 × 64.1×1 ×1+ ×1.8 × 2× 93.6 ×1 ×0.8=307.9 2 2 m

Vertical Load :i a=i s=1 qbv =0.33, ϕ 1=40 yields K s=5.5 qtv →

(

quv =102.9+1.8 × 12 1+

q allowable=

2 ×1.5 ×cos 0 1 ×5.5 × tan 40 −1.8 ×1=117.7 ≤ 307.9 1 2

)

qu 115.11 t = =38.4 2 FS 3 m

Qallowable=qallowable × A=38.4 × ( 2 ×2 ) =153.6 ton

Second Method: D 1.5 = =0.75 yields λ=4.545,θ=−9.69 B 2 →

ρ= λ

( HB )+θ=4.545 ×( 12 )−9.69=−7.42

μ=45+

ϕ1 40 =45+ =65 2 2 1 t q bv=1.8 × ( 1.5+1 ) × 18.4 ×1 ×1+ ×1.6 ×2 ×15.7 × 1× 0.8=102.9 2 2 m

6

1 t q tv=1.8 ×1.5 × 64.1× 1×1+ × 1.8× 2× 93.6 ×1 ×0.8=307.9 2 2 m

qbv =0.33 qtv

α =ρln

q bv + μ=−7.42× ln ( 0.33 ) +65=73.2 qtv

( )

[ ( ) ] [ ()

F=ln 1+2

]

H 1 tanα =ln 1+ 2 tan73.2 =1.5 B 2

(

upper layer : K p=tan2 45+

qu =q bv−γ 1 H +

γ 1 K p sinδ 2 Htanα −BF DF + tanα 2 tanα

qu =102.9−1.8 ×1+

q allowable=

[

ϕ 40 =tan 2 45+ =4.6 2 2

)

(

)

]

[

]

1.8 × 4.6× sin 36 2× 1× tan 73.2−2 ×1.5 t × 1.5 ×1.5+ = 101.26 2 tan73.2 2 × tan 73.2 m

qu 101.26 t = =33.8 2 FS 3 m

Qallowable=qallowable × A=33.8 × ( 2 ×2 )=135 ton 4. A 1.5m x 1.5m footing is located at a depth of 1.0m in a uniform deposit of clay 2.5m thick. The clay is normally loaded having liquid limit=35%, water content=30%, specific gravity of solids=2.7; saturated density of 1.9 g/cm3, and unconfined compressive strength of 1.0 kg/cm2. Assuming clay layer to be fully saturated, calculate the net safe load the footing can carry. Adopt a factor of safety of 3.0 against shear failure of soil. Calculate the settlement of footing under the net safe load.

7 Df =1 m B=L=1.5 m

LL=35% W=30% Gs=2.7 ρsat=1.9 g/cm3 cu=1 kg/cm2

2.5m

Solution: Bearing Capacity: 1 ' qu =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B N γ 2

ϕ=0 yields N c =5.14, N q=1, N γ =0 →

S c =1.3

S q=1

i c =i γ =1

γ sat =19

kN m3

q Unconfined compression strength 100 kN c u= = = =50 2 2 2 2 m

qu =1× 1.3× 50 ×5.14+1 ×1 ×19 ×1 ×1=353.1

8

kN 2 m

qu (net) =qu −γD=353.1−19 ×1=334.1

q allowable=

qu (net) FS

=

kN m2

334.1 kN =111.4 2 3 m

Qallowable=qallowable × A=111.4 × ( 1.5 ×1.5 ) =250.65 kN * Students can estimate the bearing capacity by using other theories such as Terzaghi, Vesic,… Settlement: Average effective pressure before load application: σ '0 =1.75× ( 19−10 ) =15.75

kN m2

(Or 2.5 kN σ '0 = × ( 19−10 )=11.25 2 2 m ) ' Predict the preconsolidation pressure σ c

based on soil parameters provided

S ×e=Gs × w

e 0=

2.7 ×0.3 =0.81 1

35 e L = ¿ G s= × 2.7=0.945 100 100

[ ]

From Das’s book: Principles of Geotechnical Engineering Nagaraj and Murthy (1985): e 1.112− 0 0.0463 σ '0 eL kN log σ 'c = =3.58 yields σ 'c =3802 2 0.188 → m

( )

9

kN ' can be estimated: (Hansbo (1957) σ c =α (VST ) Cu (VST) =6.34*50 = 317 m2

' Or σ c

Where: (VST) an empirical coefficient =

Cu(VST)

σ 'c >σ '0

'

Δσ =

'

undrained shear strength=50

= 6.34

kN 2 m

Overconsolidated soil

Q allowable 250.65 kN = =49.51 2 ( B+ z ) ( L+ z ) ( 1.5+0.75 ) ( 1.5+ 0.75 ) m

'

σ 0 + ∆ σ =11.25 + 49.51= 60.76 '

S C=

222 222 = ¿ 35

Cs Hc σ + Δσ log 0 ' 1+e 0 σ0

kN m2

kN < 3802 m2

'

Nagaraj and Murthy (1985):

CC =0.2343

[ ]

¿ G =0.2343 35 2.7=0.22 100 S 100

[ ]

1 C s= C c =0.044 5 Cs Hc σ '0 + Δσ ' 0.044 ×(2.5−1) 11.25 +49.51 S C= log = log =0.027 m=2.7 cm ' 1+e 0 1+ 0.81 1 1.25 σ0

From Das: Principles of Foundation Engineering Nagaraj and Murthy (1985): 1.122− log σ 'c =

e0 −0.0463 log σ '0 eL

( )

0.188

0.81 ( 0.945 )−0.0463 log 11.25 =1.15 yields σ =14.13 kN

1.122− =

0.188

10

→

' c

m2

σ 'c >σ '0

Normally consolidated soil

Nagaraj and Murthy (1985):

'

Δσ =

S C=

CC =0.2343

[ 100¿ ]G =0.2343 [ 100 ] 2.7=0.22 35

S

Q allowable 250.65 kN = =49.51 2 ( B+ z ) ( L+ z ) ( 1.5+0.75 ) ( 1.5+ 0.75 ) m

CC H C σ ' + Δ σ ' 0.22 ×1.5 15.75+ 49.51 log 0 ' = log =0.112 m 1+ e0 1+0.81 15.75 σ0

* Settlement can be estimated by assuming the clayey soil is normally consolidated, and apply the equation from the literature to estimate the settlement (safer because normally consolidated clay will experience more settlement than overconsolidated clay). 5- The reinforced concrete retaining wall shown in the following retains a dry sand backfill. On the surface of the fill, there is a 20 kPa surcharge. The angle of shearing resistance of the sand is 34°. The unit weight of the dry sand is 16.0 kN/m3. a

Determine the earth pressure acting on the wall.

b

Determine the factor of safety against overturning of the wall.

c

Determine the factor of safety against sliding.

d

Determine the maximum and the minimum stresses on the base of the foundation

Solution:

0.5 m

20kPa

4

11

4.0 m

3

2

1

7.50 m 0.25 m

0.75 m

Pa(surcharge) Pa(soil) 3.75 m

0.75 m

(

K a=tan 2 45−

Pp

2.5 m

0.5 m

5

0.25 m

ϕ 34 =tan 2 45− =0.28 2 2

)

(

)

1 1 Pa (soil )= γ H 2 K a = ×16 ×7.52 ×0.28=126 kN 2 2 Pa (surgharge )=qH K a =20× 7.5 ×0.28=42 kN Pa=Pa (soil) + P a(surgharge) =126+42=168

(

K p=tan 2 45+

ϕ 34 =tan 2 45+ =3.54 2 2

)

(

)

1 1 2 2 P p= γ D K p = × 16 ×0.75 ×3.54=15.93 kN 2 2

Determine the factor of safety against overturning:

Sectio n 1 2

Area 7×0.25=1.75 2.5×7/2=8.7 5

Weight/Unit Length of Wall (kN /m) 1.75×16=28

Moment arm measured from O (m) 3.875

Moment about O (kN.m/m) 108.5

8.75×16=140

2.92

408.8

12

2.5×7/2=8.7 5 0.5×7=3.5 0.5×4=2

3 4 5

8.75×24=210 3.5×24=84 2×24=48

2.08 1 2

Ʃ V=510

M o=Pa (soil) ×

FSoverturning =

436.8 84 96 Ʃ MR=1134.1

H H + P a(surgharge) × =126 ×2.5+ 42× 3.75=472.5 kN .m 3 2

∑ M R = 1134.1 =2.4> 2OK Mo

472.5

Determine the factor of safety against sliding: 1 2 2 δ ' =( ¿ ) ϕ yields δ ' =20o 3 →

FSsliding =

∑ V tan δ' + P p = 510× tan 20+15.93 =1.2 =0.67 2 6 q max=qtoe=

∑V

q min=qheel =

B

∑V B

(1+ 6Be )= 5104 ( 1+ 6 ×40.7 )=261.4 kN /m

(1− 6Be )= 5104 (1− 6 ×40.7 )=−6.4 kN /m

The minimum stress on the base of foundation is negative, which is the consequence of the amount of e that is more than B/6.

6

A 400 mm diameter concrete pile is driven into a cohesionless soil with angle of shearing resistance φ equal to 35°. The soil has a wet density of 1.9 t/m 3 and a submerged density of 0.9 t/m3. The water table is 3m below ground surface. Estimate the ultimate pile capacity (Assume δ = 22°).

3m

G.W.T. 14

L

Solution: ρ=1.9

t kN yields γ =19 3 3 m → m

ρ' =0.9

t kN yields γ ' =9 3 3 m → m

Based on Meyerhof Method: ¿ φ=35 yields N q=143 →

Q p= A p × q' × N ¿q=

2

π ×0.4 × ( 3 ×1.9+ ( L−3 ) × 0.9 ) ×143=53.91+ 16.17 L ton 4

Limiting point resistance: ql =0.5 Pa N ¿q tan φ' =0.5 ×10 × 143× tan 35=500

t m2

π × 0.42 ql × A p=500 × =62.83 ton 4 53.91+ 16.17 L=62.83 yields L=0.6 m →

Qs=∑ P × ∆ L× f Perimeter:

P=πD=0.4 π

¿ f =for z=0 L¿ ' :f =K σ '0 tan δ ' ¿ for z=L' ¿ L :f =f z= L ¿ '

L' ≈ 15 ¿ 20 D yields L' =15 D=15× 0.4=6 m →

15

High Displacement Driven :k ≈ k 0 ¿ 1.8 k 0 yields k =1.5 k 0=1.5 ( 1−sinφ )=1.5 ×(1−sin 35)=0.64 →

'

at z=0: σ 0=0 yields f =0 →

'

at z=3 m :σ 0 =3× 1.9=5.7

'

t t yields f =0.64 × 5.7 × tan22=1.5 2 2 m → m

at z=6 m:σ 0 =3 ×1.9+3 ×0.9=8.4

at z=6 m:Qs =

t t yields f =0.64 × 8.4 × tan 22=2.17 2 2 m → m

0+ 1.5 1.5+2.17 ×0.4 π ×3+ ×0.4 π ×3=9.75 t 2 2

Qs=9.75+f z=6 P ( L−6 )=9.75+2.17 ×0.4 π × ( L−6 )=2.73 L−6.61 at L>6 m :Qu=Q P +QS =( 53.91+16.17 L ) + ( 2.73 L−6.61 ) =47.3+18.9 L

at L=6 m:Q p=62.83 at L=6 m:Qu=Q P +QS =62.83+6.43=69.26 t

* Students can solve the problem by using other methods

7

Estimate the pile capacity of a circular concrete pile that is 23 m long and has a diameter of 450 mm. The pile is in an offshore environment and the 23 m is only the part embedded in the soil. The soil profile indicates an average unconfined compressive strength of 0.25 kg/cm2. The submerged density of soil is 0.9 t/m3.

Solution:

16

qu 0.25 kg 10 4 cm 2 kN −3 C = = =0.125 × 10× 10 × =12.5 2 u 2 2 Undrained Shear Strength: 2 2 cm m m 2

A p=

π ×0.45 =0.16 m2 4

P=0.45 π=1.41 m

γ ' =0.9

t kN × 103 × 10× 10−3=9 3 3 m m

embeded length : L=23 m Based on Meyerhof Method: Q p=N c Cu A p =9 Cu A p=9× 12.5× 0.16=18 kN

Based on λ Method:

L=23 m yields λ=0.159 →

A 0.5 × 9 ×232 kN ' ´ Average value of σ : σ 0= = =103.5 2 L 23 m ' 0

kN f av =λ ( σ´ '0+ 2C u )=0.159 ( 103.5+2× 12.5 )=20.4 2 m Qs=P × L × f av =1.41 ×23 ×20.4=661.6 kN Qu=Q P +QS =18+661.6=679.6 kN

Based on α Method: C u 12.5 = =0.125 yields α ≈ 1 P a 100 →

17

f =α C u=1 ×C u=12.5

kN 2 m

Qs=P × L × f =1.41 ×23 × 12.5=405.4 kN Qu=Q P +QS =18+405.4=423.4 kN

* Students can solve the problem by using other methods 9

A group of 25 piles arranged in a square pattern is to be proportioned in a deposit of soft clay. Assuming the piles to be square (30 cm side) and 10 m long, work out the spacing for 100% efficiency. Neglect bearing and assume an adhesion factor of 0.8 for shear mobilization around each pile.

0.3m 0.3m

L=10 m

B

η=1

n=25

B

η=1 yields Qu Grout =n ×Q u →

Qu Grout =4 BL∝Cu +9 B2 C u=4 B× 10 ×0.8 ×C u +9 B 2 C u=32 B C u+ 9 B 2 C u

18

2

Qu= A s ∝C u+ 9 A s Cu =4 × 0.3 ×10 ×0.8 ×C u +9 ×0.3 ×C u=9.6C u +0.81 Cu By neglecting bearing: Qu Grout =4 BL∝Cu=4 B ×10 ×0.8 ×C u=32 BC u Qu= A s ∝C u=4 ×0.3 ×10 × 0.8× Cu =9.6 Cu Qu Grout =n× Q u 32 B Cu =25 ×9.6 C u yields B=7.5 m →

Center to center spacing: 7.5−0.3 =1.8 m 4 * Students can solve the problem by using other methods 10 Proportion a pile group to carry a load of 250t including the weight of pile cap at a site where the subsoil consists of uniform clay up to a depth of 30 m underlain by rock. Average unconfined compressive strength of clay is 0.8 kg/cm 2. The clay is normally loaded having a liquid limit of 40% and an initial void ratio of 1.0. A factor of safety of 2 is required against shear failure of soil. Compute the settlement of the pile group assuming the load to be transferred at 2/3 point of length of piles. Followings are the information about the pile group: n1=n2=3 η=100

Pile cross section: 30 ×30 cm Center to Center spacing between the piles: d=120 cm

19

Q=250t

1

Qu=250 t qu =0.8

kg cm2

2/3L

L

z

qu kg C = =0.4 2 u Undrained Shear Strength: 2 cm

2

H

¿=40

FS=2

ROCK

e 0=1

This problem has been solved based on 2 different assumptions. Assumption 1: n1=n2=3

0.3m

η=100

0.3m

Pile cross section: 30 ×30 cm Center to Center spacing between the piles: d=120 cm for Clay :G s =2.7−2.9 yields G s=2.8 →

d=

ηp n1 n2−4 D 2 ( n1 +n2 −2 )

=

( 1× 4 × 30× 3× 3 )−4 ×30 =120 cm 2 ( 3+3−2 )

Qu (group )=η ∑ Qu=1 ×9 × Qu (single ) yields Q u( single )= →

20

250 =27.8 ton 9

30m

Pile capacity=FS ×Qu ( single)=2× 27.8=55.6 ton −3

Q p=N c Cu A p =9 Cu A p=9× 0.4 ×10 ×30 × 30=3.24 ton Qs=Q u−Q p=55.6−3.24=52.36 ton

Qs=P × L × f yields L= →

Qs Pf

f =α C u , α=1 yields f =C u=0.4 →

kg 2 cm

Q s 52.36 ×10 3 L= = =1090 cm=10.9 m≈ 11m Pf 4 ×30 ×0.4

Determining Settlement: Because the length of piles are 11m, the stress distribution starts at a depth of 7.33m below the top of the pile. B g=L g=2 d +30=2× 120+30=270 cm=2.7 m 2 2 30− L 30− × 11 3 3 z= = =11.3 m 2 2 ∆ σ '=

γ d=

Q 250 t = =1.28 2 ( B g+ z ) ( L g + z ) ( 2.7+ 11.3 ) ( 2.7+11.3 ) m

G s γ w 2.8× 1 t = =1.4 3 1+e 1+1 m

σ '0 =γd =1.4 ×18.63=26.1

t m2

21

Nagaraj and Murthy (1985):

[

CC =0.2343

[ 100¿ ]G =0.2343 [ 100 ] 2.8=0.26

]

40

S

Cc H σ '0+ ∆ σ ' 0.26 ×22.66 26.1+1.28 ∆ Sc = log = log =0.061 m=6.1 cm ' 1+e 0 1+ 1 26.1 σ0

[

]

Assumption 2: L=15 m B g=L g=3 m for Clay :Gs =2.7−2.9 yields Gs=2.8 →

Determining bearing capacity for checking the assumption: Q p=N c Cu A p =9 Cu A p=9× 0.8 ×32 ×10 4=648 ton 4

Qs=P × L ×C u=4 ×3 ×15 × 10 ×0.8=1440 ton Qu=Q P +QS =648+1440=2088 kN Q a=

Qu 2088 = =1044 ton> 250ton OK FS 2

Determining Settlement: Because the length of piles are 15m, the stress distribution starts at a depth of 10m below the top of the pile. 2 2 30− L 30− × 15 3 3 z= = =10 m 2 2 ∆ σ '=

Q 250 t = =1.48 2 ( B g+ z ) ( L g + z ) ( 3+10 )( 3+10 ) m

22

γ d=

G s γ w 2.8× 1 t = =1.4 3 1+e 1+1 m

σ '0 =1.4 ×20=28

t 2 m

Nagaraj and Murthy (1985): CC =0.2343

∆ Sc =

[

]

[ ]

¿ G =0.2343 40 2.8=0.26 100 S 100

[ ]

Cc H σ ' + ∆ σ ' 0.26 ×20 28+ 1.48 log 0 ' = log =0.058 m=5.8 cm 1+e 0 1+1 28 σ0

[

]

* Students can solve the problem by using other assumptions

23

View more...
180 Ton

ex ey x

Df=5 ft

B=L=6 ft γ=115 lb/ft3 =36 c=200 lb/ft2

qmin

Solution: = 36° C = 200 lbs/ft2 γ=115 lbs/ft3 d=30 ft

1

qmax

ex=

M y 60 B = =0.33 ft < =1 ft OK Q 180 6

e y=

M x 90 L = =0.50 ft < =1 ft OK Q 180 6

'

B =B−2 e x =6−2 ×0.33=5.34 ft L' =L−2 e y =6−2× 0.50=5.00 ft effetive width=B ' =5 ft effective length=L' =5.34 ft

q max=

6e 6e N 180 6 × 0.33 6 ×0.5 ton lb 1+ x + y = 1+ + =9.15 2 × 2240=20496 2 BL B B 6 ×6 6 6 ft ft

q min=

6e 6e N 180 6 × 0.33 6 ×0.5 ton lb 1− x − y = 1− − =0.85 2 × 2240=1904 2 BL B B 6 ×6 6 6 ft ft

(

(

)

)

(

)

(

)

Assume general shear failure occurs 1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 Meyerhof (1963): ϕ=36 yields N c =50.55, N q=37.7, N γ =44.4 →

S c =1.3 S q=1 S γ =0.8

Vertical Load :i c =i q=i γ =1 2

d=30 ft >B yields Water has no effect on theultimate bearing capacity →

1 q'u =1× S c c N c +1× S q γ D f N q+ × 1× S γ γ B ' N γ 2 1 q'u =1× 1.3× 200 ×50.55+1 ×1 ×115 ×5 × 37.7+ ×1 ×0.8 ×115 × 5 ×44.4=45033lb/ ft 2 2 qu (net) =qu −γD=45033−115 × 5=44458

FS=

lb 2 ft

q u(net ) 44458 = =2.2 q max 20496

* Students can solve the problem by using other theories such as Vesic,… 2. What will be the change in the factor of safety in Problem 1 if the load was inclined at 12° to the vertical? The other data being same as in Problem 1. Solution: 1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 Meyerhof (1963): ϕ=36 yields N c =50.55, N q=37.7, N γ =44.4 →

S c =1.3 S q=1 S γ =0.8

3

(

i c =i q= 1−

θ 2 12 2 = 1− =0.75 90 90

) (

)

θ 2 12 2 i γ = 1− = 1− =0.44 ϕ 36

( ) (

)

1 q'u =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B' N γ 2 1 q'u =0.75 ×1.3 ×200 ×50.55+ 0.75× 1× 115 ×5 ×37.7+ × 0.44 ×0.8 × 115 ×5 × 44.4=30609lb / ft 2 2 qu (net) =qu −γD=30609−115×5=30034

q max=

FS=

lb 2 ft

6e 6e N 180 ×cos 12 6× 0.33 6 × 0.5 ton lb 1+ x + y = 1+ + =8.95 2 ×2240=20048 2 BL B B 6×6 6 6 ft ft

(

)

(

)

q u(net ) 30034 = =1.5 q max 20048

* Students can solve the problem by using other theories such as Vesic,…

3. A 2.0m wide square footing is located at a depth of 1.5m in a layered sand deposit. The upper sand layer 2.5m thick have 1= 1.8 t /m3, 1 = 400 and C1 = 0 is followed by other sand layer having 2= 1.6 t/m3, 2 = 300 and C2 = 0. Using a factor of safety of 3 against shear failure of soil, calculate the safe load the footing can carry. Assume that the water table is at great depth.

γ1=1.8 t/m3 1=40 c=0

Df =1.5 m 2.5m

4

B=L=2 m

H=1m

γ2=1.6 t/m3 2=30 c=0

Solution: Dense sand layer over loose sand layer First Method: 2 Dcosα i s K s tanϕ quv =qbv +γ 1 H 2 1+ −γ 1 H ≤q tv H B

(

)

1 q bv=γ 1 ( D+ H ) N q 2 i q 2 S q 2+ γ 2 B N γ 2 i γ 2 S γ 2 2 1 q tv=γ 1 D N q 1 i q 1 Sq 1 + γ 1 B N γ 1 i γ 1 S γ 1 2 upper (top) layer : ϕ=40 yields N q 1=64.1 , N γ 1=93.6 →

lower ( bottom ) layer :ϕ =30 yields N q 2=18.4 , N γ 2=15.7 →

i q=i γ =1

S q 1=Sq 2=1 S γ 1=S γ 2=0.8

1 t q 2=qbv =1.8 × (1.5+ 1 ) ×18.4 ×1× 1+ ×1.6 ×2 ×15.7 ×1 × 0.8=102.9 2 2 m

5

1 t q1 =qtv =1.8 ×1.5 × 64.1×1 ×1+ ×1.8 × 2× 93.6 ×1 ×0.8=307.9 2 2 m

Vertical Load :i a=i s=1 qbv =0.33, ϕ 1=40 yields K s=5.5 qtv →

(

quv =102.9+1.8 × 12 1+

q allowable=

2 ×1.5 ×cos 0 1 ×5.5 × tan 40 −1.8 ×1=117.7 ≤ 307.9 1 2

)

qu 115.11 t = =38.4 2 FS 3 m

Qallowable=qallowable × A=38.4 × ( 2 ×2 ) =153.6 ton

Second Method: D 1.5 = =0.75 yields λ=4.545,θ=−9.69 B 2 →

ρ= λ

( HB )+θ=4.545 ×( 12 )−9.69=−7.42

μ=45+

ϕ1 40 =45+ =65 2 2 1 t q bv=1.8 × ( 1.5+1 ) × 18.4 ×1 ×1+ ×1.6 ×2 ×15.7 × 1× 0.8=102.9 2 2 m

6

1 t q tv=1.8 ×1.5 × 64.1× 1×1+ × 1.8× 2× 93.6 ×1 ×0.8=307.9 2 2 m

qbv =0.33 qtv

α =ρln

q bv + μ=−7.42× ln ( 0.33 ) +65=73.2 qtv

( )

[ ( ) ] [ ()

F=ln 1+2

]

H 1 tanα =ln 1+ 2 tan73.2 =1.5 B 2

(

upper layer : K p=tan2 45+

qu =q bv−γ 1 H +

γ 1 K p sinδ 2 Htanα −BF DF + tanα 2 tanα

qu =102.9−1.8 ×1+

q allowable=

[

ϕ 40 =tan 2 45+ =4.6 2 2

)

(

)

]

[

]

1.8 × 4.6× sin 36 2× 1× tan 73.2−2 ×1.5 t × 1.5 ×1.5+ = 101.26 2 tan73.2 2 × tan 73.2 m

qu 101.26 t = =33.8 2 FS 3 m

Qallowable=qallowable × A=33.8 × ( 2 ×2 )=135 ton 4. A 1.5m x 1.5m footing is located at a depth of 1.0m in a uniform deposit of clay 2.5m thick. The clay is normally loaded having liquid limit=35%, water content=30%, specific gravity of solids=2.7; saturated density of 1.9 g/cm3, and unconfined compressive strength of 1.0 kg/cm2. Assuming clay layer to be fully saturated, calculate the net safe load the footing can carry. Adopt a factor of safety of 3.0 against shear failure of soil. Calculate the settlement of footing under the net safe load.

7 Df =1 m B=L=1.5 m

LL=35% W=30% Gs=2.7 ρsat=1.9 g/cm3 cu=1 kg/cm2

2.5m

Solution: Bearing Capacity: 1 ' qu =i c Sc c N c +i q S q γ Df N q + i γ S γ γ B N γ 2

ϕ=0 yields N c =5.14, N q=1, N γ =0 →

S c =1.3

S q=1

i c =i γ =1

γ sat =19

kN m3

q Unconfined compression strength 100 kN c u= = = =50 2 2 2 2 m

qu =1× 1.3× 50 ×5.14+1 ×1 ×19 ×1 ×1=353.1

8

kN 2 m

qu (net) =qu −γD=353.1−19 ×1=334.1

q allowable=

qu (net) FS

=

kN m2

334.1 kN =111.4 2 3 m

Qallowable=qallowable × A=111.4 × ( 1.5 ×1.5 ) =250.65 kN * Students can estimate the bearing capacity by using other theories such as Terzaghi, Vesic,… Settlement: Average effective pressure before load application: σ '0 =1.75× ( 19−10 ) =15.75

kN m2

(Or 2.5 kN σ '0 = × ( 19−10 )=11.25 2 2 m ) ' Predict the preconsolidation pressure σ c

based on soil parameters provided

S ×e=Gs × w

e 0=

2.7 ×0.3 =0.81 1

35 e L = ¿ G s= × 2.7=0.945 100 100

[ ]

From Das’s book: Principles of Geotechnical Engineering Nagaraj and Murthy (1985): e 1.112− 0 0.0463 σ '0 eL kN log σ 'c = =3.58 yields σ 'c =3802 2 0.188 → m

( )

9

kN ' can be estimated: (Hansbo (1957) σ c =α (VST ) Cu (VST) =6.34*50 = 317 m2

' Or σ c

Where: (VST) an empirical coefficient =

Cu(VST)

σ 'c >σ '0

'

Δσ =

'

undrained shear strength=50

= 6.34

kN 2 m

Overconsolidated soil

Q allowable 250.65 kN = =49.51 2 ( B+ z ) ( L+ z ) ( 1.5+0.75 ) ( 1.5+ 0.75 ) m

'

σ 0 + ∆ σ =11.25 + 49.51= 60.76 '

S C=

222 222 = ¿ 35

Cs Hc σ + Δσ log 0 ' 1+e 0 σ0

kN m2

kN < 3802 m2

'

Nagaraj and Murthy (1985):

CC =0.2343

[ ]

¿ G =0.2343 35 2.7=0.22 100 S 100

[ ]

1 C s= C c =0.044 5 Cs Hc σ '0 + Δσ ' 0.044 ×(2.5−1) 11.25 +49.51 S C= log = log =0.027 m=2.7 cm ' 1+e 0 1+ 0.81 1 1.25 σ0

From Das: Principles of Foundation Engineering Nagaraj and Murthy (1985): 1.122− log σ 'c =

e0 −0.0463 log σ '0 eL

( )

0.188

0.81 ( 0.945 )−0.0463 log 11.25 =1.15 yields σ =14.13 kN

1.122− =

0.188

10

→

' c

m2

σ 'c >σ '0

Normally consolidated soil

Nagaraj and Murthy (1985):

'

Δσ =

S C=

CC =0.2343

[ 100¿ ]G =0.2343 [ 100 ] 2.7=0.22 35

S

Q allowable 250.65 kN = =49.51 2 ( B+ z ) ( L+ z ) ( 1.5+0.75 ) ( 1.5+ 0.75 ) m

CC H C σ ' + Δ σ ' 0.22 ×1.5 15.75+ 49.51 log 0 ' = log =0.112 m 1+ e0 1+0.81 15.75 σ0

* Settlement can be estimated by assuming the clayey soil is normally consolidated, and apply the equation from the literature to estimate the settlement (safer because normally consolidated clay will experience more settlement than overconsolidated clay). 5- The reinforced concrete retaining wall shown in the following retains a dry sand backfill. On the surface of the fill, there is a 20 kPa surcharge. The angle of shearing resistance of the sand is 34°. The unit weight of the dry sand is 16.0 kN/m3. a

Determine the earth pressure acting on the wall.

b

Determine the factor of safety against overturning of the wall.

c

Determine the factor of safety against sliding.

d

Determine the maximum and the minimum stresses on the base of the foundation

Solution:

0.5 m

20kPa

4

11

4.0 m

3

2

1

7.50 m 0.25 m

0.75 m

Pa(surcharge) Pa(soil) 3.75 m

0.75 m

(

K a=tan 2 45−

Pp

2.5 m

0.5 m

5

0.25 m

ϕ 34 =tan 2 45− =0.28 2 2

)

(

)

1 1 Pa (soil )= γ H 2 K a = ×16 ×7.52 ×0.28=126 kN 2 2 Pa (surgharge )=qH K a =20× 7.5 ×0.28=42 kN Pa=Pa (soil) + P a(surgharge) =126+42=168

(

K p=tan 2 45+

ϕ 34 =tan 2 45+ =3.54 2 2

)

(

)

1 1 2 2 P p= γ D K p = × 16 ×0.75 ×3.54=15.93 kN 2 2

Determine the factor of safety against overturning:

Sectio n 1 2

Area 7×0.25=1.75 2.5×7/2=8.7 5

Weight/Unit Length of Wall (kN /m) 1.75×16=28

Moment arm measured from O (m) 3.875

Moment about O (kN.m/m) 108.5

8.75×16=140

2.92

408.8

12

2.5×7/2=8.7 5 0.5×7=3.5 0.5×4=2

3 4 5

8.75×24=210 3.5×24=84 2×24=48

2.08 1 2

Ʃ V=510

M o=Pa (soil) ×

FSoverturning =

436.8 84 96 Ʃ MR=1134.1

H H + P a(surgharge) × =126 ×2.5+ 42× 3.75=472.5 kN .m 3 2

∑ M R = 1134.1 =2.4> 2OK Mo

472.5

Determine the factor of safety against sliding: 1 2 2 δ ' =( ¿ ) ϕ yields δ ' =20o 3 →

FSsliding =

∑ V tan δ' + P p = 510× tan 20+15.93 =1.2 =0.67 2 6 q max=qtoe=

∑V

q min=qheel =

B

∑V B

(1+ 6Be )= 5104 ( 1+ 6 ×40.7 )=261.4 kN /m

(1− 6Be )= 5104 (1− 6 ×40.7 )=−6.4 kN /m

The minimum stress on the base of foundation is negative, which is the consequence of the amount of e that is more than B/6.

6

A 400 mm diameter concrete pile is driven into a cohesionless soil with angle of shearing resistance φ equal to 35°. The soil has a wet density of 1.9 t/m 3 and a submerged density of 0.9 t/m3. The water table is 3m below ground surface. Estimate the ultimate pile capacity (Assume δ = 22°).

3m

G.W.T. 14

L

Solution: ρ=1.9

t kN yields γ =19 3 3 m → m

ρ' =0.9

t kN yields γ ' =9 3 3 m → m

Based on Meyerhof Method: ¿ φ=35 yields N q=143 →

Q p= A p × q' × N ¿q=

2

π ×0.4 × ( 3 ×1.9+ ( L−3 ) × 0.9 ) ×143=53.91+ 16.17 L ton 4

Limiting point resistance: ql =0.5 Pa N ¿q tan φ' =0.5 ×10 × 143× tan 35=500

t m2

π × 0.42 ql × A p=500 × =62.83 ton 4 53.91+ 16.17 L=62.83 yields L=0.6 m →

Qs=∑ P × ∆ L× f Perimeter:

P=πD=0.4 π

¿ f =for z=0 L¿ ' :f =K σ '0 tan δ ' ¿ for z=L' ¿ L :f =f z= L ¿ '

L' ≈ 15 ¿ 20 D yields L' =15 D=15× 0.4=6 m →

15

High Displacement Driven :k ≈ k 0 ¿ 1.8 k 0 yields k =1.5 k 0=1.5 ( 1−sinφ )=1.5 ×(1−sin 35)=0.64 →

'

at z=0: σ 0=0 yields f =0 →

'

at z=3 m :σ 0 =3× 1.9=5.7

'

t t yields f =0.64 × 5.7 × tan22=1.5 2 2 m → m

at z=6 m:σ 0 =3 ×1.9+3 ×0.9=8.4

at z=6 m:Qs =

t t yields f =0.64 × 8.4 × tan 22=2.17 2 2 m → m

0+ 1.5 1.5+2.17 ×0.4 π ×3+ ×0.4 π ×3=9.75 t 2 2

Qs=9.75+f z=6 P ( L−6 )=9.75+2.17 ×0.4 π × ( L−6 )=2.73 L−6.61 at L>6 m :Qu=Q P +QS =( 53.91+16.17 L ) + ( 2.73 L−6.61 ) =47.3+18.9 L

at L=6 m:Q p=62.83 at L=6 m:Qu=Q P +QS =62.83+6.43=69.26 t

* Students can solve the problem by using other methods

7

Estimate the pile capacity of a circular concrete pile that is 23 m long and has a diameter of 450 mm. The pile is in an offshore environment and the 23 m is only the part embedded in the soil. The soil profile indicates an average unconfined compressive strength of 0.25 kg/cm2. The submerged density of soil is 0.9 t/m3.

Solution:

16

qu 0.25 kg 10 4 cm 2 kN −3 C = = =0.125 × 10× 10 × =12.5 2 u 2 2 Undrained Shear Strength: 2 2 cm m m 2

A p=

π ×0.45 =0.16 m2 4

P=0.45 π=1.41 m

γ ' =0.9

t kN × 103 × 10× 10−3=9 3 3 m m

embeded length : L=23 m Based on Meyerhof Method: Q p=N c Cu A p =9 Cu A p=9× 12.5× 0.16=18 kN

Based on λ Method:

L=23 m yields λ=0.159 →

A 0.5 × 9 ×232 kN ' ´ Average value of σ : σ 0= = =103.5 2 L 23 m ' 0

kN f av =λ ( σ´ '0+ 2C u )=0.159 ( 103.5+2× 12.5 )=20.4 2 m Qs=P × L × f av =1.41 ×23 ×20.4=661.6 kN Qu=Q P +QS =18+661.6=679.6 kN

Based on α Method: C u 12.5 = =0.125 yields α ≈ 1 P a 100 →

17

f =α C u=1 ×C u=12.5

kN 2 m

Qs=P × L × f =1.41 ×23 × 12.5=405.4 kN Qu=Q P +QS =18+405.4=423.4 kN

* Students can solve the problem by using other methods 9

A group of 25 piles arranged in a square pattern is to be proportioned in a deposit of soft clay. Assuming the piles to be square (30 cm side) and 10 m long, work out the spacing for 100% efficiency. Neglect bearing and assume an adhesion factor of 0.8 for shear mobilization around each pile.

0.3m 0.3m

L=10 m

B

η=1

n=25

B

η=1 yields Qu Grout =n ×Q u →

Qu Grout =4 BL∝Cu +9 B2 C u=4 B× 10 ×0.8 ×C u +9 B 2 C u=32 B C u+ 9 B 2 C u

18

2

Qu= A s ∝C u+ 9 A s Cu =4 × 0.3 ×10 ×0.8 ×C u +9 ×0.3 ×C u=9.6C u +0.81 Cu By neglecting bearing: Qu Grout =4 BL∝Cu=4 B ×10 ×0.8 ×C u=32 BC u Qu= A s ∝C u=4 ×0.3 ×10 × 0.8× Cu =9.6 Cu Qu Grout =n× Q u 32 B Cu =25 ×9.6 C u yields B=7.5 m →

Center to center spacing: 7.5−0.3 =1.8 m 4 * Students can solve the problem by using other methods 10 Proportion a pile group to carry a load of 250t including the weight of pile cap at a site where the subsoil consists of uniform clay up to a depth of 30 m underlain by rock. Average unconfined compressive strength of clay is 0.8 kg/cm 2. The clay is normally loaded having a liquid limit of 40% and an initial void ratio of 1.0. A factor of safety of 2 is required against shear failure of soil. Compute the settlement of the pile group assuming the load to be transferred at 2/3 point of length of piles. Followings are the information about the pile group: n1=n2=3 η=100

Pile cross section: 30 ×30 cm Center to Center spacing between the piles: d=120 cm

19

Q=250t

1

Qu=250 t qu =0.8

kg cm2

2/3L

L

z

qu kg C = =0.4 2 u Undrained Shear Strength: 2 cm

2

H

¿=40

FS=2

ROCK

e 0=1

This problem has been solved based on 2 different assumptions. Assumption 1: n1=n2=3

0.3m

η=100

0.3m

Pile cross section: 30 ×30 cm Center to Center spacing between the piles: d=120 cm for Clay :G s =2.7−2.9 yields G s=2.8 →

d=

ηp n1 n2−4 D 2 ( n1 +n2 −2 )

=

( 1× 4 × 30× 3× 3 )−4 ×30 =120 cm 2 ( 3+3−2 )

Qu (group )=η ∑ Qu=1 ×9 × Qu (single ) yields Q u( single )= →

20

250 =27.8 ton 9

30m

Pile capacity=FS ×Qu ( single)=2× 27.8=55.6 ton −3

Q p=N c Cu A p =9 Cu A p=9× 0.4 ×10 ×30 × 30=3.24 ton Qs=Q u−Q p=55.6−3.24=52.36 ton

Qs=P × L × f yields L= →

Qs Pf

f =α C u , α=1 yields f =C u=0.4 →

kg 2 cm

Q s 52.36 ×10 3 L= = =1090 cm=10.9 m≈ 11m Pf 4 ×30 ×0.4

Determining Settlement: Because the length of piles are 11m, the stress distribution starts at a depth of 7.33m below the top of the pile. B g=L g=2 d +30=2× 120+30=270 cm=2.7 m 2 2 30− L 30− × 11 3 3 z= = =11.3 m 2 2 ∆ σ '=

γ d=

Q 250 t = =1.28 2 ( B g+ z ) ( L g + z ) ( 2.7+ 11.3 ) ( 2.7+11.3 ) m

G s γ w 2.8× 1 t = =1.4 3 1+e 1+1 m

σ '0 =γd =1.4 ×18.63=26.1

t m2

21

Nagaraj and Murthy (1985):

[

CC =0.2343

[ 100¿ ]G =0.2343 [ 100 ] 2.8=0.26

]

40

S

Cc H σ '0+ ∆ σ ' 0.26 ×22.66 26.1+1.28 ∆ Sc = log = log =0.061 m=6.1 cm ' 1+e 0 1+ 1 26.1 σ0

[

]

Assumption 2: L=15 m B g=L g=3 m for Clay :Gs =2.7−2.9 yields Gs=2.8 →

Determining bearing capacity for checking the assumption: Q p=N c Cu A p =9 Cu A p=9× 0.8 ×32 ×10 4=648 ton 4

Qs=P × L ×C u=4 ×3 ×15 × 10 ×0.8=1440 ton Qu=Q P +QS =648+1440=2088 kN Q a=

Qu 2088 = =1044 ton> 250ton OK FS 2

Determining Settlement: Because the length of piles are 15m, the stress distribution starts at a depth of 10m below the top of the pile. 2 2 30− L 30− × 15 3 3 z= = =10 m 2 2 ∆ σ '=

Q 250 t = =1.48 2 ( B g+ z ) ( L g + z ) ( 3+10 )( 3+10 ) m

22

γ d=

G s γ w 2.8× 1 t = =1.4 3 1+e 1+1 m

σ '0 =1.4 ×20=28

t 2 m

Nagaraj and Murthy (1985): CC =0.2343

∆ Sc =

[

]

[ ]

¿ G =0.2343 40 2.8=0.26 100 S 100

[ ]

Cc H σ ' + ∆ σ ' 0.26 ×20 28+ 1.48 log 0 ' = log =0.058 m=5.8 cm 1+e 0 1+1 28 σ0

[

]

* Students can solve the problem by using other assumptions

23

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