CIVI 6501-Major Assignment . Solution

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/ft}2 _{and the water table is at a depth of 30 ft below ground} level. Calculate the factor of safety against shear failure of soil.

Solution: = 36° C = 200 lbs/ft2 γ=115 lbs/ft3 d=30 ft 1 y 180 Ton ex ey x Df=5 ft B=L=6 ft γ=115 lb/ft3 =36 c=200 lb/ft2 qmin q_max

e_x=My Q = 60 180=0.33 ft < B 6=1 ft OK e_y=Mx Q = 90 180=0.50 ft < L 6=1 ft OK B'=B−2 ex=6−2 ×0.33=5.34 ft L' =L−2 ey=6−2× 0.50=5.00 ft effetive width=B'_{=5 ft} effectivelength=L'=5.34 ft q_max= N BL

(

1+ 6 e_x B + 6 e_y B

)

= 180 6 ×6

(

1+ 6 × 0.33 6 + 6 ×0.5 6

)

=9.15 ton ft2 × 2240=20496 lb ft2 q_min= N BL

(

1− 6 e_x B − 6 e_y B

)

= 180 6 ×6

(

1− 6 × 0.33 6 − 6 ×0.5 6

)

=0.85 ton ft2 × 2240=1904 lb ft2

Assume general shear failure occurs

q_u' =i_cS_cc N_c+i_qS_qγ D_fN_q+1 2iγSγγ B '_N γ Meyerhof (1963): ϕ=36 yields → Nc=50.55, Nq=37.7, Nγ=44.4 S_c=1.3 S_q=1 Sγ=0.8 Vertical Load :i_c=i_q=i_γ=1

d=30 ft >B yields

→ Water has no effect on theultimate bearing capacity

qu ' =1× Scc Nc+1× Sqγ DfNq+ 1 2× 1× Sγγ B ' Nγ q_u'=1× 1.3× 200 ×50.55+1 ×1 ×115×5 × 37.7+1 2×1 ×0.8 ×115× 5 ×44.4=45033lb/ ft 2 qu (net)=qu−γD=45033−115× 5=44458 lb ft2 FS=qu(net ) qmax =44458 20496=2.2

* 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: q_u' =i_cS_cc N_c+i_qS_qγ D_fN_q+1 2iγSγγ B '_N γ Meyerhof (1963): ϕ=36 yields → Nc=50.55, Nq=37.7, Nγ=44.4 S_c=1.3 S_q=1 S_γ=0.8 3

i_c=i_q=

(

1− θ 90

)

2 =

(

1−12 90

)

2 =0.75 i_γ=

(

1−θ ϕ

)

2 =

(

1−12 36

)

2 =0.44 q_u' =i_cS_cc N_c+i_qS_qγ D_fN_q+1 2iγSγγ B '_N γ qu ' =0.75 ×1.3 ×200 ×50.55+0.75× 1× 115×5 ×37.7+1 2× 0.44 ×0.8 × 115×5 × 44.4=30609lb/ ft 2 qu (net)=qu−γD=30609−115×5=30034 lb ft2 q_max= N BL

(

1+ 6 e_x B + 6 e_y B

)

= 180 ×cos 12 6 × 6

(

1+ 6× 0.33 6 + 6 × 0.5 6

)

=8.95 ton ft2 ×2240=20048 lb ft2 FS=qu(net ) qmax =30034 20048=1.5

* 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 = 40}0 _{and C1 = 0 is followed by other} sand layer having 2= 1.6 t/m3_{, 2 = 30}0 _{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.

Df =1.5 m

γ1=1.8 t/m3 1=40

c=0

Solution:

Dense sand layer over loose sand layer First Method: q_uv=q_bv+γ₁H2

(

1+2 Dcosα H

)

i_sK_stanϕ B −γ1H ≤qtv q_bv=γ₁(D+H ) N_{q 2}i_{q 2}S_{q 2}+1 2γ2B Nγ 2iγ 2Sγ 2 q_tv=γ₁D N_{q 1}i_{q 1}S_{q 1}+1 2γ1B Nγ 1iγ 1Sγ 1

upper (top)layer : ϕ=40 yields

→ Nq 1=64.1 , Nγ 1=93.6 lower(bottom)layer :ϕ =30 yields

→ Nq 2=18.4 , Nγ 2=15.7 i_q=i_γ=1 S_{q 1}=S_{q 2}=1 S_{γ 1}=S_{γ 2}=0.8 q2=qbv=1.8 × (1.5+1)×18.4 ×1× 1+ 1 2×1.6 ×2 ×15.7 ×1 × 0.8=102.9 t m2 5 H=1m B=L=2 m γ2=1.6 t/m3 2=30 c=0

q₁=q_tv=1.8 ×1.5 × 64.1×1 ×1+1 2×1.8 × 2× 93.6 ×1 ×0.8=307.9 t m2 Vertical Load :i_a=i_s=1 q_bv qtv =0.33, ϕ₁=40 yields → Ks=5.5 q_uv=102.9+1.8 × 12

(

1+2 ×1.5 ×cos 0 1

)

1 ×5.5 × tan 40 2 −1.8 ×1=117.7≤ 307.9 q_allowable= qu FS= 115.11 3 =38.4 t m2

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

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

(

H B

)

+θ=4.545 ×

(

1 2

)

−9.69=−7.42 μ=45+ϕ1 2=45+ 40 2 =65 qbv=1.8 × (1.5+1)× 18.4 ×1 ×1+1₂×1.6 ×2 ×15.7 × 1× 0.8=102.9 t m2

q_tv=1.8 ×1.5 × 64.1× 1×1+1 2× 1.8× 2× 93.6 ×1 ×0.8=307.9 t m2 q_bv qtv =0.33 α=ρln

(

qbv q_tv

)

+μ=−7.42× ln (0.33)+65=73.2 F=ln

[

1+2

(

H B

)

tanα

]

=ln

[

1+2

(

1 2

)

tan73.2

]

=1.5

upper layer : K_p=tan2

(

45+ϕ 2

)

=tan 2

(

45+40 2

)

=4.6 q_u=q_bv−γ₁H +γ1Kpsinδ tanα

[

DF+ 2 Htanα −BF 2 tanα

]

q_u=102.9−1.8 ×1+1.8 × 4.6× sin 36 tan73.2 ×

[

1.5 ×1.5+ 2× 1× tan 73.2−2 ×1.5 2 × tan 73.2

]

=101.26 t m2 q_allowable= qu FS= 101.26 3 =33.8 t m2

Q_allowable=q_allowable× 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

Solution: Bearing Capacity: qu=icScc Nc+iqSqγ DfNq+1₂iγSγγ B ' Nγ ϕ=0 yields → Nc=5.14, Nq=1, Nγ=0 S_c=1.3 S_q=1 ic=iγ=1 γ_sat=19kN m3 c_u=q 2=

Unconfined compression strength

2 = 100 2 =50 kN m2 qu=1× 1.3× 50 ×5.14+1 ×1 ×19 ×1 ×1=353.1 kN m2 LL=35% W=30% Gs=2.7 ρsat=1.9 g/cm3 cu=1 kg/cm2 2.5m

q_{u (net)}=q_u−γD=353.1−19 ×1=334.1kN m2 qallowable= qu (net) FS = 334.1 3 =111.4 kN m2

Q_allowable=q_allowable× 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:

σ₀'_{=1.75× (19−10)=15.75}kN m2 (Or σ₀'=2.5 2 ×(19−10 )=11.25 kN m2 )

Predict the preconsolidation pressure σc' _{based on soil parameters provided} S ×e=G_s× w e₀=2.7 ×0.3 1 =0.81 e_L=

[

¿ 100

]

Gs= 35 100× 2.7=0.945

From Das’s book: Principles of Geotechnical Engineering Nagaraj and Murthy (1985):

log σ_c'= 1.112−

(

e0 e_L

)

0.0463 σ0 ' 0.188 =3.58 yields→ σc ' =3802kN m2 9

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

Where: (VST) an empirical coefficient = 222¿ =

222

35 = 6.34

Cu(VST) undrained shear strength=50

kN m2 σc'>σ0' Overconsolidated soil Δσ'= Qallowable (B+z ) (L+z )= 250.65 (1.5+0.75) (1.5+0.75)=49.51 kN m2 σ0 ' +∆ σ' _{=11.25 + 49.51= 60.76} kN m2 < 3802 kN m2 S_C=CsHc 1+e₀ log σ0 ' +Δσ' σ₀'

Nagaraj and Murthy (1985): CC=0.2343

[

₁₀₀¿

]

GS=0.2343

[

35 100

]

2.7=0.22 C_s=1 5Cc=0.044 S_C=CsHc 1+e0 logσ0 ' +Δσ' σ₀' = 0.044 ×(2.5−1) 1+0.81 log 11.25+49.51 1 1.25 =0.027 m=2.7 cm

From Das: Principles of Foundation Engineering Nagaraj and Murthy (1985):

log σ_c' = 1.122−

(

e0 e_L

)

−0.0463 log σ0 ' 0.188 = 1.122−

(

0.81 0.945

)

−0.0463 log 11.25 0.188 =1.15 yields→ σc '_=14.13kN m2

4.0 m

σc '

>σ0

'

Normally consolidated soil

Nagaraj and Murthy (1985): CC=0.2343

[

₁₀₀¿

]

GS=0.2343

[

35 100

]

2.7=0.22 Δσ' = Qallowable (B+z ) (L+z )= 250.65 (1.5+0.75) (1.5+0.75)=49.51 kN m2 S_C=CCHC 1+e₀ log σ0'+Δ σ' σ₀' = 0.22 ×1.5 1+0.81 log 15.75+49.51 15.75 =0.112m

* 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: 4 3 2 1 11 0.5 m 20kPa

5 K_a=tan2

(

45−ϕ 2

)

=tan 2

(

45−34 2

)

=0.28 P_{a (soil )}=1 2γ H 2 K_a=1 2×16 ×7.5 2 ×0.28=126 kN P_{a (surgharge )}=qH K_a=20× 7.5 ×0.28=42 kN P_a=P_{a (soil)}+P_a(surgharge)=126+42=168 K_p=tan2

(

45+ϕ 2

)

=tan 2

(

45+34 2

)

=3.54 P_p=1 2γ D 2 K_p=1 2× 16 ×0.75 2 ×3.54=15.93 kN

Determine the factor of safety against overturning:

Sectio n Area Weight/Unit Length of Wall (kN /m) Moment arm measured from O (m) Moment about O (kN.m/m) 1 7×0.25=1.75 1.75×16=28 3.875 108.5 2 2.5×7/2=8.75 8.75×16=140 2.92 408.8 Pa(surcharge) 7.50 m 0.75 m 0.25 m Pa(soil) 3.75 m 0.75 m Pp 2.5 m 0.5 m 0.25 m

qtoe=qmax qheel=qmin 3 2.5×7/2=8.75 8.75×24=210 2.08 436.8 4 0.5×7=3.5 3.5×24=84 1 84 5 0.5×4=2 2×24=48 2 96 Ʃ V=510 _MR=1134.1Ʃ Mo=Pa (soil)× H 3+Pa(surgharge)× H 2=126 ×2.5+42× 3.75=472.5 kN .m FS_overturning=

∑

MR M_o = 1134.1 472.5 =2.4>2OK

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

∑

V tan δ ' +Pp P_a = 510× tan 20+15.93 168 =1.2<1.5

The factor of safety against sliding is less than 1.5, so, the width of retaining wall should be increased or a base key should be added to increase the factor of safety.

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

13

G.W.T. 3m M_net=

∑

M_R−

∑

M_o=1134.1−472.5=661.6 kN . m ´X = Mnet

∑

V= 661.6 510 =1.3 m e=B 2− ´X=2−1.3=0.7 m> B 6=0.67 qmax=qtoe=

∑

V B

(

1+ 6 e B

)

= 510 4

(

1+ 6 × 0.7 4

)

=261.4 kN /m qmin=qheel=

∑

V B

(

1− 6 e B

)

= 510 4

(

1− 6 × 0.7 4

)

=−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/m3 _{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°).

Solution: ρ=1.9 t m3 yields → γ=19 kN m3 ρ'=0.9 t m3 yields → γ ' =9kN m3

Based on Meyerhof Method:

φ=35 yields → Nq ¿₌₁₄₃ Q_p=A_p× q'_{× N} q ¿ =π ×0.4 2 4 ×

(

3 ×1.9+(L−3)× 0.9

)

×143=53.91+16.17 L ton

Limiting point resistance:

ql=0.5 PaNq ¿ tan φ'=0.5 ×10 × 143× tan 35=500 t m2 q_l× A_p=500 ×π × 0.4 2 4 =62.83 ton 53.91+16.17 L=62.83 yields → L=0.6 m Q_s=

∑

P × ∆ L× f Perimeter: P=πD=0.4 π ¿ f =for z=0 L¿':f =K σ₀' _{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 ≈ k₀¿1.8 k₀yields → k =1.5 k0=1.5(1−sinφ)=1.5 ×(1−sin 35)=0.64 at z=0: σ0 ' =0 yields → f =0 at z=3 m :σ0 ' =3× 1.9=5.7 t m2 yields_→ f =0.64 × 5.7 × tan22=1.5 t m2 at z=6 m:σ₀'_{=3 ×1.9+3 ×0.9=8.4} t m2 yields→ f =0.64 × 8.4 × tan 22=2.17 t m2 at z=6 m:Q_s=0+1.5 2 ×0.4 π ×3+ 1.5+2.17 2 ×0.4 π ×3=9.75 t Q_s=9.75+f_z=6P (L−6)=9.75+2.17 ×0.4 π ×( L−6)=2.73 L−6.61 at L>6 m :Q_u=Q_P+Q_S=(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:Q_u=Q_P+Q_S=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/m}3_.

Undrained Shear Strength: Cu= q_u 2= 0.25 2 =0.125 kg cm2× 10× 10 −3_×104cm2 m2 =12.5 kN m2 A_p=π ×0.45 2 4 =0.16 m 2 P=0.45 π=1.41 m γ'=0.9 t m3× 10 3 × 10× 10−3=9kN m3 embeded length : L=23 m

Based on Meyerhof Method:

Q_p=N_cC_uA_p=9 C_uA_p=9× 12.5× 0.16=18 kN Based on λ Method: L=23 m yields → λ=0.159 Average value of σ₀' : ´σ'₀=A L= 0.5 × 9 ×232 23 =103.5 kN m2 fav=λ

(

σ´0 ' +2Cu

)

=0.159 (103.5+2× 12.5)=20.4 kN m2 Q_s=P × L × f_av=1.41 ×23 ×20.4=661.6 kN Qu=QP+QS=18+661.6=679.6 kN Based on α Method: C_u Pa =12.5 100=0.125 yields→ α ≈ 1 17

f =α C_u=1 ×C_u=12.5kN

m2

Q_s=P × L × f =1.41 ×23 × 12.5=405.4 kN

Q_u=Q_P+Q_S=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.

L=10 m η=1 n=25 η=1 yields → Qu Grout =n ×Q_u Qu Grout=4 BL∝Cu+9 B2Cu=4 B× 10 ×0.8 ×Cu+9 B2Cu=32 B Cu+9 B2Cu 0.3m 0.3m B B

Qu=As∝Cu+9 AsCu=4 × 0.3 ×10 ×0.8 ×Cu+9 ×0.3 2 ×Cu=9.6Cu+0.81 Cu By neglecting bearing: Q_{u Grout}=4 BL∝C_u=4 B ×10 ×0.8 ×C_u=32 BC_u Q_u=A_s∝Cu=4 ×0.3 ×10 × 0.8× Cu=9.6 Cu Q_{u Grout}=n× Q_u 32 B C_u=25 ×9.6 C_uyields → B=7.5 m

Center to center spacing: 7.5−0.3

4 =1.8 m

* 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/cm2_{. 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

2 Q=250t ROCK L z 30m 2/3L H 1 Q_u=250 t q_u=0.8 kg cm2

Undrained Shear Strength: Cu= qu 2=0.4 kg cm2 ¿=40 FS=2 e₀=1

This problem has been solved based on 2 different assumptions.

Assumption 1:

n₁=n₂=3

η=100

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

→ Gs=2.8

d=ηp n1n2−4 D

2

(

n₁+n₂−2

)

=

(1× 4 × 30× 3× 3)−4 ×30

2(3+3−2) =120 cm

Q_u(group)=η

_∑

Q_u=1 ×9 × Q_u(single)yields

→ Qu(single)=

250

9 =27.8 ton

0.3m 0.3m

Pile capacity=FS ×Q_u(single)=2× 27.8=55.6 ton Qp=NcCuAp=9 CuAp=9× 0.4 ×10−3×30 × 30=3.24 ton Q_s=Q_u−Q_p=55.6−3.24=52.36 ton Q_s=P × L × f yields → L= Q_s Pf f =α C_u, α=1 yields → f =Cu=0.4 kg cm2 L=Qs Pf= 52.36 ×103 4 ×30 ×0.4=1090 cm=10.9 m≈ 11m 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 z= 30−2 3L 2 = 30−2 3× 11 2 =11.3m ∆ σ' = Q

(

B_g+z

) (

L_g+z

)

= 250 (2.7+11.3) (2.7+11.3 )=1.28 t m2 γ_d=Gsγw 1+e = 2.8× 1 1+1 =1.4 t m3 σ₀'=γd=1.4 ×18.63=26.1 t m2 21

Nagaraj and Murthy (1985): CC=0.2343

[

₁₀₀¿

]

GS=0.2343

[

40 100

]

2.8=0.26 ∆ Sc= C_cH 1+e₀log

[

σ'₀+∆ σ' σ0'

]

=0.26 ×22.66 1+1 log

[

26.1+1.28 26.1

]

=0.061 m=6.1 cm 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:

Qp=NcCuAp=9 CuAp=9× 0.8 ×3 2 ×104=648 ton Qs=P × L ×Cu=4 ×3 ×15 × 104×0.8=1440 ton Q_u=Q_P+Q_S=648+1440=2088 kN Q_a=Qu FS= 2088 2 =1044 ton>250ton OK Determining Settlement:

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

(

B_g+z

) (

L_g+z

)

= 250 (3+10 )(3+10 )=1.48 t m2

γ_d=Gsγw 1+e = 2.8× 1 1+1 =1.4 t m3 σ₀'_{=1.4 ×20=28} t m2

Nagaraj and Murthy (1985):

C_C=0.2343

[

¿ 100

]

GS=0.2343

[

40 100

]

2.8=0.26 ∆ Sc= C_cH 1+e₀log

[

σ'₀ +∆ σ' σ0'

]

=0.26 ×20 1+1 log

[

28+1.48 28

]

=0.058 m=5.8 cm * Students can solve the problem by using other assumptions