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

Question

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Answer 1

Question

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

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Answer 2

Question

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

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Answer 3

Question

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

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Answer 4

Question

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

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Answer 5

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

Answer 6

Chapter 15, Problem 15.2P

To determine

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

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

Answer 7

Chapter 15, Problem 15.2P

To determine

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

Answer 8

Expert Solution & Answer

Answer to Problem 15.2P

The factor of safety with respect to overturning is $4.62_$ .

The factor of safety with respect to sliding is $2.11_$ .

The factor of safety with respect to bearing capacity failure is $7.4_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The cohesion $(c′1)$ of backfill soil is 0.

The unit weight $(γ1)$ of the backfill soil is $18.5 kN/m3$ .

The friction angle $(ϕ′1)$ of the backfill soil is $35°$ .

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

The backfill angle $(α)$ is $15°$ .

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 15.2P

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 $15°$ and effective soil friction angle $(ϕ′)$ of $35°$ is 0.2968.

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 $15°$ for $α$ .

$H1=2 ×tan15°=0.54 m$

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 0.54 m $H1$ .

$H′=5+1.0+0.54=6.54 m$

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

$Pa=12γ1(H′)2Ka$

Substitute $18.5 kN/m3$ for $γ1$ , 6.54 m for $H′$ , and 0.2968 for $Ka$ .

$Pa=12×18.5×(6.54)2×0.2968=117.4 kN/m$

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

$Pv=Pasinα$

Substitute $117.4 kN/m$ for $Pa$ and $15°$ for $α$ .

$Pv=117.4×sin15°=30.4kN/m$

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

$Ph=Pacosα$

Substitute $117.4kN/m$ for $Pa$ and $15°$ for $α$ .

$Ph=117.4×cos15°=113.4 kN/m$

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 $24 kN/m3$ for $γc$ .

$W1=12×1.5×6×24=108 kN/m$

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

$(MA)1=23x1$

Substitute 1.5 m for $x1$ .

$(MA)1=23×1.5=1.0 m$

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

$MC, 1=W1×(MA)1$

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

$MC, 1=108×1.0=108 kN$

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 $24 kN/m3$ for $γc$ .

$W2=0.5×6×24=72 kN/m$

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 $x2$ .

$(MA)2=1.5+0.52=1.75 m$

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

$MC, 2=W2×(MA)2$

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

$MC, 2=72×1.75=126 kN$

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 $24 kN/m3$ for $γc$ .

$W3=12×2.0×6×24=144 kN/m$

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 $x3$ .

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

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

$MC, 3=W3×(MA)3$

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

$MC, 3=144×2.67=384.5 kN$

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 $18.5 kN/m3$ for $γ1$ .

$W4=12×2.0×6×18.5 =111 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 4=W4×(MA)4$

Substitute $111 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 4=111×3.33=369.6 kN$

Find the weight of section 5 $(W5)$ .

$W5=A5γ1=(12x3H1)γ1$

Substitute 2.0 m for $x3$ , 0.54 m for H, and $18.5 kN/m3$ for $γ1$ .

$W5=12×2.0×0.54×18.5=10 kN/m$

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 $x1$ .

$(MA)4=1.5+0.5+23×2.0=3.33 m$

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

$MC, 5=W5×(MA)5$

Substitute $10 kN/m$ for $W4$ and 3.33 m for $(MA)4$ .

$MC, 5=10×3.33=33.33 kN$

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 $x3$ .

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

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

$MC, Pv=Pv×(MA)Pv$

Substitute $30.4 kN/m$ for $Pv$ and 4.0 m for $(MA)Pv$ .

$MC, Pv=30.4×4=121.6 kN$

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

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

Substitute $108 kN-m/m$ for $MC, 1$ , $126 kN-m/m$ for $MC, 2$ , $384.5 kN-m/m$ for $MC, 3$ , $369.6 kN-m/m$ for $MC, 4$ , $33.33 kN-m/m$ for $MC, 5$ , and $121.6 kN-m/m$ for $MC, Pv$ .

$∑MC=108+126+384.5+369.6+33.33+121.6=1,143 kN-m/m$

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

$∑V=W1+W2+W3+W4+W5+Pv$

Substitute $108 kN/m$ for $W1$ , $72 kN/m$ for $W2$ , $144 kN/m$ for $W3$ , $111 kN/m$ for $W4$ , $10 kN/m$ for $W5$ , and $30.4 kN/m$ for $Pv$ .

$∑V=108+72+144+111+10+30.4=475.4 kN/m$

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

Section	Weight (kN/m)	moment arm from C $(m)$	moment about C $(kN-m/m)$
1	108	1	108
2	72	1.75	126
3	144	2.67	384.5
4	111	3.33	369.6
5	10	3.33	33.33
	$pv=30.4$	4	121.6
	$∑V=475.4$		$∑MR=1,143$

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

$MO=PhH′3$

Substitute $113.4 kN/m$ for $Ph$ and 6.54 m for $H′$ .

$MO=113.4×6.543=247.2 kN-m/m$

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

$FS(overturning)=∑MRMO$

Substitute $1,143 kN-m/m$ for $∑MR$ and $247.2 kN-m/m$ for $MO$ .

$FS(overturning)=1,143247.2=4.62$

Therefore, the factor of safety with respect to overturning is $4.62_$ .

Check the stability with respect to sliding.

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

$Kp=tan2(45+ϕ′2)$

Substitute $35°$ for $ϕ′$ .

$Kp=tan2(45+35°2)=3.690$

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, $18.5 kN/m3$ for $γ2$ , and 3.690 for $Kp$ .

$Pp=12×18.5×(1)2×3.690=34.1 kN/m$

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

$δ′=23ϕ′$

Substitute $35°$ for $ϕ′1$ .

$δ′=23×35°=23.3°$

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

$FS(sliding)=(∑V)tanδ′+PpPh$

Substitute $475.4 kN/m$ for $∑V$ , $23.3°$ for $δ′$ , $34.1 kN/m$ for $Pp$ , and $113.4 kN/m$ for $Ph$ .

$FS(sliding)=475.4×tan(23.3°)+34.1113.4=2.11$

Therefore, the factor of safety with respect to sliding is $2.11_$ .

Check the stability with bearing capacity failure.

Find the eccentricity (e) using the equation:

$e=B2−∑MR−∑MO∑V$

Substitute 4 m for B, $1,143 kN-m/m$ for $∑MR$ , $247.2 kN-m/m$ for $∑MO$ , and $475.4 kN/m$ for $∑V$ .

$e=42−1,143−247.2475.4=2−1.884=0.116 m$

Check for eccentricity.

$e<B6$

Substitute 0.116 m for e and 4 m for B.

$0.116 m<460.116 m<0.67 m$

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 $475.4 kN/m$ for $∑V$ , 4 m for B, and 0.116 m for e.

$qtoe=475.44×(1+6×0.1164)=139.5 kN/m2$

Find the effective breadth $(B′)$ using the equation:

$B′=B−2e$

Substitute 4 m for B and 0.116 m for e.

$B′=4−(2×0.116)=3.768 m$

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 $35°$ is 33.30.

Take the value of bearing capacity factor, $Nc$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 46.12.

Take the value of bearing capacity factor, $Nγ$ with corresponding angle of friction $(ϕ′)$ of $35°$ is 48.03.

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

$q=γD$

Substitute 1.0 m for D and $18.5 kN/m3$ for $γ2$ .

$q=18.5×1.0=18.5 kN/m2$

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

$ψ=tan−1(Pacosα∑V)=tan−1(Ph∑V)$

Substitute $113.4 kN/m$ for $Pa$ and $475.4 kN/m$ for $∑V$ .

$ψ=tan−1(113.4475.4)=tan−1(0.238)=13.4°$

Find the inclination factor $Fqi$ and $Fci$ using the equation:

$Fqi=Fci=(1−ψ90)2$

Substitute $13.4°$ for $ψ°$ .

$Fqi=(1−13.4°90)2=0.72$

Find the depth factor $(Fqd)$ using the equation:

$Fqd=1+2tanϕ′(1−sinϕ′)2DfB$

Here, $Df$ is depth of flange.

Substitute $35°$ for $ϕ′$ , 1.0 m for $Df$ , and 4 m for B.

$Fqd=1+2tan(35°)×(1−sin35°)2×1.04=1+(1.4×0.182×0.25)=1.07$

The depth factor $Fγd$ is 1.0 for $ϕ′>0$ .

Find the inclination factor $Fγi$ using the equation:

$Fγi=(1−ψϕ′)2$

Substitute $35°$ for $ϕ′$ and $15°$ for $ψ$ .

$Fγi=(1−15°35°)2=0.33$

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

$qu=qNqFqdFqi+12γ2B′NγFγdFγi$

Substitute $18.5 kN/m2$ for q, 33.30 for $Nq$ , 1.07 for $Fqd$ , 0.72 for $Fqi$ , $18.5 kN/m3$ for $γ2$ , 3.768 m for $B′$ , 48.03 for $Nγ$ , 1 for $Fγd$ , and 0.33 for $Fγi$ .

$qu=[(18.5×33.30×1.07×0.72)+(12×18.5×3.768×48.03×1×0.33)]=474.60+552.43=1,027 kN/m2$

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

$FS(bearing)=quqmax$

Substitute $1,027 kN/m2$ for $qu$ and $139.5 kN/m2$ for $qmax$ .

$FS(bearing)=1,027139.5=7.4$

Therefore, the factor of safety with respect to bearing capacity failure is $7.4_$ .

Answer 12

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2P Ch. 15 - Prob. 15.3P Ch. 15 - Prob. 15.4P Ch. 15 - Prob. 15.5P Ch. 15 - Prob. 15.6P Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8P Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

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

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

Answer to Problem 15.2P

Explanation of Solution

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