Find: Bearing Capacity, qult [lb/ft2] infinite strip footing 3,200 4,800 C) 7,600 D) 8,200 4 [ft] 8 [ft] Clay Find the ultimate bearing capacity, q ‘ultimate’, in pounds per square feet. [pause] In this problem --- γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite strip footing 3,200 4,800 C) 7,600 D) 8,200 4 [ft] 8 [ft] Clay an infinite strip footing 8 feet wide, is buried to a depth of 4 feet beneath the soil surface. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite strip footing 3,200 4,800 C) 7,600 D) 8,200 4 [ft] 8 [ft] Clay The soil is a stiff clay with the given soil properties. [pause] Since the groundwater table is not mentioned, it is assumed to be very deep. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] 8 [ft] Clay We begin by writing out our equation for the ultimate bearing capacity of an infinite strip footing. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing cohesion term 4 [ft] 8 [ft] Clay This equation sums the resisting forces from cohesion, --- γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing cohesion term wedge 4 [ft] weight term 8 [ft] Clay the wedge weight, --- γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing cohesion term wedge overburden 4 [ft] weight term term 8 [ft] Clay and the overburden, of the soil. In this problem we are neglecting the footing thickness, assuming it to be zero. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] 8 [ft] Clay We begin by determining which variables we need to calculate. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc 8 [ft] Clay Let’s begin by solving for the bearing capacity factors, for the cohesion, N sub c, --- γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc 8 [ft] Nγ Clay for the wedge weight, N sub gamma, --- γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc 8 [ft] Nγ Clay and for the overburden weight, N sub q. In a drained condition, --- Nq γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay these bearing capacity factors are all a function of the friction angle, phi. Here, I’m using the wedge weight multiplier, N sub gamma, developed by Vesic. Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay We first solve for N sub q, the overburden multiplier, --- Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay by plugging in our phi value, of 16 degrees, --- Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay N sub q equals 4.34. The other two bearing capacity factors are solved --- Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] = 4.34 OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay by plugging in N sub q, and phi. [pause] N sub c equals --- Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] = 4.34 OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ = 11.63 8 [ft] Nγ = 2 * (Nq+1) * tan φ Clay 11.63, and N sub gamma equals, --- Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] = 4.34 OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? ? ? ? 4 [ft] Nc = (Nq-1) * cot φ = 11.63 8 [ft] Nγ = 2 * (Nq+1) * tan φ= 3.06 Clay 3.06. Nq = tan2(45+φ/2) * eπ * tanφ γT=120 [lb/ft3] = 4.34 OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] Nc = 11.63 8 [ft] Nγ = 3.06 Clay We have now solved our bearing capacity factors. Nq = 4.34 γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite Nq Nc Nγ φ 4 8 12 16 20 24 28 32 1.00 1.43 2.06 2.97 4.34 6.40 9.60 14.72 23.18 5.14 6.19 7.53 9.28 11.63 14.83 19.32 25.80 35.49 0.00 0.34 0.86 1.69 3.06 5.39 9.44 16.72 30.21 strip footing 4 [ft] 8 [ft] Clay As a time saving strategy, it may be helpful to print out a table of these bearing capacity factors at various values of phi. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite Nq Nc Nγ φ 4 8 12 16 20 24 28 32 1.00 1.43 2.06 2.97 4.34 6.40 9.60 14.72 23.18 5.14 6.19 7.53 9.28 11.63 14.83 19.32 25.80 35.49 0.00 0.34 0.86 1.69 3.06 5.39 9.44 16.72 30.21 strip footing 4 [ft] 8 [ft] Clay Here are the factors we just computed, for the friction angle of 16 degrees. γT=120 [lb/ft3] OCR>1 c=400[lb/ft2] φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] 8 [ft] Clay We next determine the effective unit weight, and the overburden stress, of the soil. Since the groundwater table is not mentioned, --- γT=120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT 8 [ft] Clay we’ll set the the effective unit weight equal to the total unit weight,which equals --- γT=120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay 120 pounds per cubic feet. [pause] Next, q, the overburden stress, is calculated. In the context of bearing capacity problems, --- γT=120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay q is the same as the vertical effective stress calculated in geostatic stress problems, --- γT=120 [lb/ft3] q= σ’v Nc = 11.63 Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay which is equal to the effective unit weight, times the thickness of the soil layer, for all soil layers above the footing base. γT=120 [lb/ft3] q= σ’v Nc = 11.63 q= γ’*h Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay This unit weight value corresponds to the 4 feet of soil above the footing base, which is 120 pounds per cubic foot, --- γT=120 [lb/ft3] q= σ’v Nc = 11.63 q= γ’*h Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ? ? 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay and the height of this soil layer is 4 feet. The overburden stress equals --- γT=120 [lb/ft3] γT=120 [lb/ft3] q= σ’v Nc = 11.63 q= γ’*h Nγ = 3.06 OCR>1 c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay 480 pounds per square feet. γT=120 [lb/ft3] q= σ’v Nc = 11.63 q= γ’*h Nγ = 3.06 OCR>1 q= 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] γ’= γT γ’= 120 [lb/ft3] 8 [ft] Clay Having solved for the effective unit weight and overburden stress, --- γT=120 [lb/ft3] q= σ’v Nc = 11.63 q= γ’*h Nγ = 3.06 OCR>1 q= 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] 8 [ft] Clay the ultimate bearing capacity due to general shear failure can be computed. The values are plugged in for the --- γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] 8 [ft] Clay cohesion term, --- γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] 8 [ft] Clay wedge weight term, --- γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing 4 [ft] 8 [ft] Clay and overburden term. γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ft2 lb ft2 lb qult=4,652 +1,469 4 [ft] ft2 lb +2,083 8 [ft] Clay Finally the, bearing capacity computes to, --- γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq strip footing ft2 lb ft2 lb qult=4,652 +1,469 4 [ft] ft2 lb +2,083 8 [ft] ft2 lb qult=8,204 Clay 8,204 pounds per square foot. γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq 3,200 4,800 C) 7,600 D) 8,200 strip footing ft2 lb ft2 lb qult=4,652 +1,469 4 [ft] ft2 lb +2,083 8 [ft] ft2 lb qult=8,204 Clay Looking at the possible solutions, --- γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
Find: Bearing Capacity, qult [lb/ft2] infinite qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq 3,200 4,800 C) 7,600 D) 8,200 strip footing ft2 lb ft2 lb qult=4,652 +1,469 4 [ft] ft2 lb +2,083 8 [ft] ft2 lb qult=8,204 AnswerD Clay the answer is D. γT=120 [lb/ft3] γ’= 120 [lb/ft3] Nc = 11.63 Nγ = 3.06 OCR>1 q = 480 [lb/ft2] c=400[lb/ft2] Nq = 4.34 φ=16˚
( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ γ d H*C σfinal ρcn= log Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘