I need answer only for question NO 14.2 HFigure 14.20factor of safety with respect to overturning and sliding and bearing capacity. Usaactive pressure for the calculation.14.6 Repeat Problem 14.5. Use Coulomb's active pressure calculation and & = .14.7 Determine the factor of safety of the retaining wall shown in Figure 14.22 withfailure The uniL weight of the concrete is 14.1 State whether the following are true or false.a. Cantilever retaining walls do not have steel reinforcements.b. Cantilever retaining walls have smaller cross sections than gravity retaining walls.c. Adhesion at the base of the retaining wall is generally less than the cohesion of the soilbeneath the base.d. Provision of a key at the base of the wall reduces the passive resistance.e. The pressure distribution beneath the base of a retaining wall is never uniform.For Problems 14.2 through 14.6, use k = k2 = and P, = 0 while using Eq. (14.10).14.2 For the cantilever retaining wall shown in Figure 14.20, the wall dimensions are H = 24 ft,1.2 ft, x2 = 1.8 ft, x3 = 4.5 ft, x4 = 10 ft, x5 = 3 ft, D = 5.25 ft, and =10°. The soil= Ixproperties are YI = 107 lb/ft', i = 32°, y2 = 112 lb/ft, = 28°, and c = 630 lb/ft.Calculate the factors of safety with respect to overturning, sliding, and bearing capacity. Useunit weight of concrete, y, = 150 lb/ft'.14.3 Repeat Problem 14.2 with the following:Wall dimensions: H = 18 ft, x, = 18 in., x230 in., x, = 4 ft, x = 6 ft, xg = 2.75 ft,%3DD = 4 ft, a = 10°110 Ib/ft", ; = 18", c = 800 lb/ft²Soil properties: YI = 117 lb/ft', i = 34°, y2 =Use unit weight of concrete, y = 150 lb/ft'.14.4 Repeat Problem 14.2 with the following:Wall dimensions: H = 22 ft, x = 12 in., x2 = 27 in., x3 = 4.5 ft, x = 8 ft, xs = 2.75 ft,%3DD = 4 ft, a = 5°Soil properties: yI = 110 lb/ft', 6i = 36°, y2= 120 lb/ft', = 15°, ch = 1000 lb/ft150 lb/ft.14.5 The gravity retaining wall shown in Figure 14.21 has wall dimensions of H = 15 ft, x = 1.5 ft,Use unit weight of concrete, Ye =W59 F Mostly cle

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.