14.2 For the cantilever retaining wall shown in Figure 14.20, the wall dimensions are H = 24 ft, X1 = 1.2 ft, x2 = 1.8 ft, x3 = 4.5 ft, x4 = 10 ft, xs = 3 ft, D = 5.25 ft, and a = 10°. The soil properties are Y = 107 lb/ft', 'i = 32°, y2 = 112 lb/ft’, = 28°, and c = 630 Ib/ft. Calculate the factors of safety with respect to overturning, sliding, and bearing capacity. Use unit weight of concrete, y. = 150 lb/ft'.

Consider the following figure: The value of the active earth pressure coefficient, Ka from the table 11.2, values of Ka corresponding to the inclination of backfill α, in equation 11.25. The value of α = 10o and the effective soil friction angle in front of the heel and under the base, ϕ2' = 28o Ka = 0.321 Calculate the height of the inclined portion of the backfill by using the tangent rule as follows: Tan α = H1x4H1 = Tan α × x4Substitute the values as follows:H1 = Tan 10o × 10 ftH1 = 1.7630 ft Calculate the total height of the inclined backfill as follows: H' = H1 +H +x5Substitute the values as follows:H' = 1.7630 ft +24 ft +3 ftH' = 28.7630 ft Calculate the active earth pressure on the retaining wall due to the backfill as follows: Pa = γ1 H'2 Ka2Substitute the values as follows:Pa = 107× 28.7632 ×0.3212Pa = 14,208 lbft×1 kip 1,000 lbPa = 14.21 kipft Calculate the vertical component of the active earth pressure as follows: Pv = PaSinαSubstitute the vertical component of the active earth pressure as follows;Pv = 14.21 kipft×Sin10oPv = 2.468 kipft Calculate the horizontal component of the active earth pressure as follows; Ph = Pa CosαSubstitute the vertical component of the active earth pressure as follows;Ph = 14.21 kipft×Cos10oPh = 13.99 kipft Calculate the weight of the section-1 as follows: W1 = A1γcRewrite the equation as follows:W1 = x1×H×γcSubstitute the values as follows:W1 = 1.2 ft×24 ft×150 lbft3W1 = 4320 lbft3×1 kip1,000 lbW1 = 4.320 kipft3 Calculate the moment arm of section-1 as follows: MA1 = x3 + x2 - x12Substitute the values as follows:MA1 = 4.5 ft + 1.8ft - 1.2 ft2MA1 = 5.7 ft Calculate the moment about the point C due to the self weight of the section-1 as follows: MC1 =W1 × MA1Substitute the values as follows:MC1 =4.32 kipft × 5.7 ftMC1 =24.624 kip Calculate the weight of the section-2 as follows: W2 = A2×γcRewrite the equation as follows:W2 = X2×H2×γcSubstitute the values as follows:W2 = 06 ft×24 ft2×150 lbft3W2 = 1080 lbft ×kip1000 lbW2 = 1.080 kipft Calculate the moment arm of section-2 from point C as follows: MA2 = x3 + 23x6Substitute the values as follows:MA2 = 4.5 ft + 23×0.6 ftMA2 = 4.9 ft Calculate the moment about the point C due to the self weight of the section-2 as follows: MC2 =W2 × MA2Substitute the values as follows:MC2 =1.08 kipft × 4.9 ftMC2 =5.292 kip Calculate the weight of section-3 as follows: W3 = A3 γcRewrite the equation as follows:W3 = x3 + x2 + x4 x5 γcSubstitute the values as follows:W3 = 4.5 ft+ 1.8 ft + 10 ft × 3 ft 150 lbft3W3 = 7335 lbft × kip1000 lbW3 = 7.335 kipft Calculate the moment arm of section-3 from point C as follows: MA3 = x2 + x3 + x42Substitute the values as follows:MA3 = 1.8 ft + 4.5 ft + 10 ft2MA3 = 8.15 ft Calculate the moment about the point C due to the self weight of the section-3 as follows: MC3 =W3 × MA3Substitute the values as follows:MC3 =1.08 kipft × 8.15 ftMC3 =8.802 kip Calculate the weight of the section-4 as follows: W4 = A4 × γcRewrite the equation as follows:W4 = H + H +H12× x4× γcSubstitute teh values as follows:W4 = 24 ft + 24ft +1.763 ft2× 10 ft× 107 lbft3W4 = 26623 lbft×1 kip1000 lbW4 = 26.623 kipft Calculate the moment arm of section-4 as follows: MA4 = x2 + x3 + x42Substitute the values as follows:MA4 = 1.8 ft + 4.5 ft + 10 ft2MA4 = 11.3 ft Calculate the moment about the point C due to the self weight of the section-4 as follows: MC4 = W4 × MA4Substitute the values as follows:MC4 = 26.623 kipft × 11.3 ftMC4 = 300.81 kipCalculate the moment arm about the point C due to the vertical component of active earth pressure as follows: MAPv = x3 + x3 + x4Substitute the values as follows:MAPv = 4.5ft + 1.8ft + 10ftMAPv = 16.30 ft Calculate the moment about the point C due to the vertical component of active earth pressure as follows: MC, PV = Pv × MAPVSubstitute the values as follows:MC, PV = 2.468 kipft × 16.30 ftMC, PV = 40.228 kip Calculate the total moment about the point C as follows: ∑MC = MC1 + MC2 + MC3 + MC4 + MC PvSubstitute the values as follows:∑MC = 24.624 kip-ftft + 5.292 kip-ftft + 59.78 kip-ftft+ 300.80 kip-ftft + 40.228 kip-ftft∑MC = 430.734 kip-ftft Calculate the total vertical load due to the retaining wall and the inclined backfill as follows: ∑V = W1 + W2 + W3 + W4 + P vSubstitute the values as follows:∑V = 4.32 kipft + 1.08 kipft + 7.335 kipft+ 26.62 kipft + 2.468 kipft∑V = 41.823 kipft Calculate the overturning moment due to the horizontal component of the active earth pressure as follows: Mo = Ph ×H'3Substitute the values as follows:Mo = 13.99 kipft×28.736 ft3Mo = 134.13 kip-ftft Calculate the factor of safety about the toe of the retaining wall as follows: Factor of safetyOverturning = ∑MCMoSubstitute the valuesFactor of safetyOverturning = 430.734 kip-ftft134.13 kip-ftftFactor of safetyOverturning = 3.21 Therefore, the factor of safety with respect to overturning is 3.21. Calculate the factor of safety against sliding as follows: Factor of safetySliding = ∑V×tan23ϕ2' + B23c2'MoSubstitute the valuesFactor of safetySliding = 41.832 kipft×tan23×28o + 16.30ft ×23× 630 lbft214.21 kipft ×Cos 10oFactor of safetySliding = 1.5 Therefore, the factor of safety with respect to sliding is 1.50. Calculate the eccentricity as follows: e = B2 - ∑MR - ∑Mo∑VSubstitute the values as follows:e = 16.30 ft2 - 430.734 kip-ftft - 134.13 kip-ftft441.832 kip-ftfte = 8.15 ft - 7.09 ft e = 1.06 ft Calculate the maximum pressure occurring at the end of the toe base as follows: qtoe = ∑VB1+6eBSubstitute the values as follows:qtoe = 41.832kipft16.30 ft1+6 ×1.06 ft16.3 ftqtoe = 3.57kipft2 Calculate the value of the term B' as follows: B' = B -2eSubstitute the values as follows:B' = 16.3 ft -2×1.06 ftB' = 14.18 ft The respective values of the bearing capacity factors Nq, Nc, and Nγ are 14.72, 25.80, and 16.72 respectively for ϕ2' =28o. Calculate the value of Fcd as follows: Fcd = 1 +0.4DBSubstitute the valuesFcd = 1 +0.45.25 ft16.30 ftFcd = 1.129 Calculate the value of the load due to the soil in front of the heel as follows: q = γ2 DSubstitute the values as follows:q = 112lbft2 ×5.25 ftq = 588lbft2 ×kip1000 lbq = 0.588kipft2 Calculate the value of Ψ o as follows: Ψ o = tan-1 Pa cos α∑VSubstitute the values as follows:Ψ o = tan-1 14.21 kipft× cos 10o41.832 kipftΨ o = 18.5o Calculate the terms Fqi and Fci Fqi = Fci = 1-Ψo902Substitute the valuesFqi = Fci = 1-18.5o902Fqi = Fci = 0.63 Calculate the terms Fqd as follows: Fqd = 1+2 tanϕ2' 1-sinϕ2' DBSubstitute the values as follows:Fqd = 1+2 tan28o 1-sin28o 5.25 ft16.30 ftFqd = 1.096 Calculate the value of Fγi as follows: Fγi = 1-Ψ oϕ2'2Substitute the valuesFγi = 1-18.5 o28o2Fγi = 0.115 Assume the value of Fγd = 1 Calculate the ultimate bearing capacity as follows: qu = c2' Nc Fcd Fci + q Nq Fqd Fqi + 12γ2 B' Nγ Fγd FγiSubstitute the valuesqu =630lbft2×25.80×1.129×0.63+0.588kipft2×14.72×1.096×0.63 +12×112lbft2×14.18 ft×16.72×1×0.115qu = 11560.98 lbft2×kip1000 lb + 5.976 kipft2 + 1526.857 lbft2×kip1000 lb qu = 19.0632 kipft2 Calculate the factor of safety against the bearing capacity failure as follows: Factor of safetyBearing = quqtoeSubstitute the valuesFactor of safetyBearing = 19.063 kipft23.57 kipft2Factor of safetyBearing = 5.34 Therefore, the value of the factor of safety against the bearing capacity failure is 5.34.