Fundamentals of Geotechnical Engineering (MindTap Course List)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
Question
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Chapter 15, Problem 15.4P
To determine

Find the factors of safety with respect to overturning, sliding, and bearing capacity failure.

Expert Solution & Answer
Check Mark

Answer to Problem 15.4P

The factor of safety with respect to overturning is 4.37_.

The factor of safety with respect to sliding is 1.58_.

The factor of safety with respect to bearing capacity failure is 2.81_.

Explanation of Solution

Given information:

The cohesion (c1) of backfill soil is 0.

The unit weight (γ1) of the backfill soil is 18.0kN/m3.

The friction angle (ϕ1) of the backfill soil is 34°.

The unit weight (γc) of the concrete is 24.0kN/m3.

The cohesion (c2) of soil in front of the heel and under the base is 10kN/m2.

The unit weight of soil (γ2) in front of the heel and under the base is 19.0kN/m3.

The friction angle (ϕ2) of the soil in front of the heel and under the base is 24°.

The backfill angle (α) is 10°.

Calculation:

Check stability with respect to overturning.

Consider point C as the left end of the toe base as named as C.

Divide the retaining wall into section as in Figure 1.

Sketch the section of the retaining wall as shown in Figure 1.

Fundamentals of Geotechnical Engineering (MindTap Course List), Chapter 15, Problem 15.4P

Here, P0 is the resultant thrust on the retaining wall.

Refer Table 14.2, “Values of Ka” in the textbook.

Take the value of active earth pressure coefficient (Ka) with corresponding inclination of backfill (α) of 10° and effective soil friction angle (ϕ) of 34° is 0.2944.

Refer Figure 1.

Find the height of the inclined portion of backfill (H1), using the tangent rule:

tanα=H1x3H1=x3tanα

Substitute 2 m for x3 and 10° for α.

H1=2×tan10°=0.353m

Find the total height of the inclined backfill (H).

H=H+D+H1

Here, H is the height of retaining wall and D is the depth to the bottom of the base slab.

Substitute 5.0 m for H, 1.0 m for D, and 0.353 m H1.

H=5+1.0+0.353=6.353m

Find the active earth pressure (Pa) on the retaining wall due to the backfill using the equation:

Pa=12γ1(H)2Ka

Substitute 18kN/m3 for γ1, 6.353 m for H, and 0.2994 for Ka.

Pa=12×18×6.3532×0.2994=108.76kN/m

Find the vertical component of the active earth pressure (Pv) using the equation:

Pv=Pasinα

Substitute 108.76kN/m for Pa and 10° for α.

Pv=108.76×sin10°=18.89kN/m

Find the horizontal component of the active earth pressure (Ph) using the equation:

Ph=Pacosα

Substitute 108.76kN/m for Pa and 10° for α.

Ph=108.76×cos10°=107.11kN/m

Find the weight of section 1 (W1).

W1=A1γc=(12x1H)γc

Here, A1 is the area of the section 1.

Substitute 1.5 m for x1, 6 m for H, and 24kN/m3 for γc.

W1=12×1.5×6×24=108kN/m

Find the moment arm or lever arm [(MA)1] of section 1 from point C.

(MA)1=23x1

Substitute 1.5 m for x1.

(MA)1=23×1.5m=1.0m

Find the moment about point C (MC,1) due to the self-weight of section 1:

MC,1=W1×(MA)1

Substitute 108kN/m for W1 and 1.0 m for (MA)1.

MC,1=108×1.0=108kN

Find the weight of section 2 (W2).

W2=A2γc=(x2H)γc

Here, A2 is the area of the section 2.

Substitute 0.5 m for x2, 6 m for H, and 24kN/m3 for γc.

W2=0.5×6×24=72kN/m

Find the moment arm or lever arm [(MA)2] of section 2 from point C.

(MA)2=x1+x22

Substitute 1.5 m for x1 and 0.5 m for x2.

(MA)2=1.5m+0.5m2=1.75m

Find the moment about point C (MC,2) due to the self-weight of section 2.

MC,2=W2×(MA)2

Substitute 72kN/m for W2 and 1.75 m for (MA)2.

MC,2=72×1.75=126kN

Find the weight of section 3 (W3).

W3=A3γc=(12x3H)γc

Here, A3 is the area of the section 3.

Substitute 2.0 m for x3, 6 m for H, and 24kN/m3 for γc.

W3=12×2.0×6×24=144kN/m

Find the moment arm or lever arm [(MA)3] of section 3 from point C.

(MA)3=x1+x2+x33

Substitute 0.5 m for x2, 1.5 m for x1, and 2.0 m for x3.

(MA)3=1.5+0.5+2.03=2.67m

Find the moment about point C (MC,3) due to the self-weight of section 3.

MC,3=W3×(MA)3

Substitute 144kN/m for W3 and 2.67 m for (MA)3.

MC,3=144kN/m×2.67m=384.5kN

Find the weight of section 4 (W4).

W4=A4γ1=(12x3H)γ1

Here, A4 is the area of the section 4.

Substitute 2.0 m for x3, 6 m for H, and 18kN/m3 for γ1.

W4=12×2.0×6×18=108kN/m

Find the moment arm or lever arm [(MA)4] of section 4 from point C.

(MA)4=x1+x2+23x3

Substitute 2.0 m for x3, 0.5 m for x2, and 1.5 m for x1.

(MA)4=1.5+0.5+23×2.0=3.33m

Find the moment about point C (MC,4) due to the self-weight of section 4.

MC,4=W4×(MA)4

Substitute 108 kN/m for W4 and 3.33 m for (MA)4.

MC,4=108×3.33=359.6kN

Find the weight of section 5 (W5).

W5=A5γ1=(12x3H1)γ1

Substitute 2 m for x3, 0.353 m for H, and 18kN/m3 for γ1.

W5=12×2.0×0.353×18=6.35kN/m

Find the moment arm or lever arm [(MA)5] of section 4 from point C.

[(MA)5]=x1+x2+23x3

Substitute 2.0 m for x3, 0.5 m for x2, and 1.5 m for x1.

(MA)4=1.5+0.5+23×2.0=3.33m

Find the moment about point C (MC,5) due to the self-weight of section 5.

MC,5=W5×(MA)5

Substitute 6.35kN/m for W5 and 3.33 m for (MA)4.

MC,5=6.35×3.33=21.15kN

Find the moment arm or lever arm [(MA)Pv] of vertical component of active earth pressure from point C.

(MA)Pv=x1+x2+x3

Substitute 0.5 m for x2, 1.5 m for x1, and 2.0 m for x3.

(MA)Pv=1.5+0.5+2.0=4.0m

Find the moment about point C (MC,Pv) due to the vertical component of active earth pressure.

MC,Pv=Pv×(MA)Pv

Substitute 18.89kN/m for Pv and 4.0 m for (MA)Pv.

MC,Pv=18.89×4=75.56kN

Find the total moment about the point C (MC).

MC=MC,1+MC,2+MC,3+MC,4+MC,5+MC,Pv

Substitute 108kN-m/m for MC,1, 126kN-m/m for MC,2, 384.5kN-m/m for MC,3, 359.6kN-m/m for MC,4, 21.15kN-m/m for MC,5, and 75.56kN-m/m for MC,Pv.

MC=(108+126+384.5+359.6+21.15+75.56)=1,074.8kN-m/m

Find the total vertical load (V) due to the retaining wall and the inclined backfill.

V=W1+W2+W3+W4+W5+Pv

Substitute 108kN/m for W1, 72kN/m for W2, 144kN/m for W3, 108kN/m for W4, 6.35kN/m for W5, and 18.89kN/m for Pv.

V=(108+72+144+108+6.35+18.89)=475.24kN/m

Summarize the values of weight, moment arm from C, and moment about C as shown in Table 1.

Sectionweightmoment arm from Cmoment about C
11081108
2721.75126
31442.67384.5
41083.33359.6
56.353.3321.15
 pv=18.89475.56
 V=475.24 MR=1,074.8

Find the overturning moment (MO) due the horizontal component of the active earth pressure using the equation:

MO=PhH3

Substitute 107.11kN/m for Ph and 6.353 m for H.

MO=107.11×6.3533=226.82kN-m/m

Find the factor of safety [FS(overturning)] about the toe (point C) of the retaining wall using the equation:

FS(overturning)=MRMO

Substitute 1,074.8kN-m/m for MR and 226.82kN-m/m for MO.

FS(overturning)=1,074.8226.82=4.74

Therefore, the factor of safety with respect to overturning is 4.74_.

Check the stability with respect to sliding.

Find the coefficient of passive earth pressure (Kp) using the equation:

Kp=tan2(45+ϕ22)

Substitute 24° for ϕ2.

Kp=tan2(45+24°2)=tan2(57°)=2.37

Find the passive earth pressure (Pp) using the equation:

Pp=12γ2D2Kp

Here, γ2 is the unit weigh of soil in front of the heel and under the base.

Substitute 1 m for D, 19.0kN/m3 for γ2, and 2.37 for Kp.

Pp=12×19.0×(1)2×2.37=22.51kN/m

Find the angle of friction (δ) between the soil and the base slab using the equation:

δ=23ϕ2

Substitute 24° for ϕ2.

δ=23×24°=16°

Find the factor of safety against sliding [FS(sliding)] using the equation:

FS(sliding)=(V)tanδ+PpPh

Substitute 475.24kN/m for V, 16° for δ, 22.51kN/m for Pp, and 107.11kN/m for Ph.

FS(sliding)=475.24×tan(16°)+22.51107.11=1.48

Therefore, the factor of safety with respect to sliding is 1.48_.

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

e=B2MRMOV

Substitute 4 m for B, 1,074.8kN-m/m for MR, 226.82kN-m/m for MO, and 475.24kN/m for V.

e=421,074.8226.82475.24=21.78=0.22m

Check for eccentricity.

e<B6

Substitute 0.22 m for e and 4 m for B.

0.22m<460.22m<0.67m

The eccentricity is within the limit. Therefore, there is no tensile stress produced at the end of the steel section.

Find the maximum pressure (qtoe) occurring at the end of the toe base using the equation:

qmax=qtoeqtoe=VB(1+6eB)

Substitute 475.24kN/m for V, 4 m for B, and 0.22 m for e.

qtoe=475.244×(1+6×0.224)=158kN/m2

Find the effective breadth (B) using the equation:

B=B2e

Substitute 4 m for B and 0.22 m for e.

B=4(2×0.22)=3.56m

Refer Table 16.2, “Bearing Capacity Factors” in the textbook.

Take the value of bearing capacity factor, Nq with corresponding angle of friction (ϕ) of 24° is 9.60.

Take the value of bearing capacity factor, Nc with corresponding angle of friction (ϕ) of 24° is 19.32.

Take the value of bearing capacity factor, Nγ with corresponding angle of friction (ϕ) of 24° is 9.44.

Find the depth factor Fcd using the equation:

Fcd=1+0.4(DB)

Substitute 1 m for D and 4 m for B.

Fcd=1+0.4×(1.0m4m)=1.10

Find the load (q) due the soil in front of heel using the equation:

q=γ2D

Substitute 19kN/m3 for γ2 and 1.0 m for D.

q=19×1.0=19kN/m2

Find the inclination angle of vertical load (ψ) using the equation:

ψ=tan1(PacosαV)=tan1(PhV)

Substitute 107.11kN/m for Pa and 475.24kN/m for V.

ψ=tan1(107.11475.24)=tan1(0.238)=12.7°

Find the inclination factor Fqi and Fci using the equation:

Fqi=Fci=(1ψ90)2

Substitute 12.7° for ψ°.

Fqi=(112.7°90)2=0.74

Find the depth factor (Fqd) using the equation:

Fqd=1+2tanϕ2(1sinϕ2)2DfB

Here, Df is depth of flange.

Substitute 24° for ϕ2, 1.0 m for Df, and 4 m for B.

Fqd=1+2tan(24°)×(1sin24°)2×1.04=1+(0.890×0.352×0.25)=1.08

The depth factor Fγd is 1.0 for ϕ>0.

Find the inclination factor Fγi using the equation:

Fγi=(1ψϕ2)2

Substitute 24° for ϕ2 and 12.7° for ψ.

Fγi=(112.7°24°)2=0.22

Find the ultimate bearing capacity of the shallow foundation (qu) using the equation:

qu=cNcFcdFci+qNqFqdFqi+12γ2BNγFγdFγi

Substitute 10kN/m2 for c, 19.32 for Nc, 1.1 for Fcd, 0.74 for Fci, 19kN/m2 for q, 9.60 for Nq, 1.08 for Fqd, 0.74 for Fqi, 19kN/m3 for γ2, 3.56 m for B, 9.44 for Nγ, 1 for Fγd, and 0.22 for Fγi.

qu=[(10×19.32×1.1×0.74)+(19×9.60×1.08×0.74)+(12×19×3.56×9.44×1×0.22)]=157.26+145.77+70.23=373.26kN/m2

Find the factor of safety against bearing capacity failure [FS(bearing)] using the equation:

FS(bearing)=quqmax

Substitute 373.26kN/m2 for qu and 158kN/m2 for qtoe.

FS(bearing)=373.26158=2.36

Therefore, the factor of safety with respect to bearing capacity failure is 2.36_.

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