Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚ 6 7 8 9
square footing 6 7 8 9 d t B [ft] Gravelly Sand Q=400 [k] Find the minimum base width, B, for the square footing. [pause] In this problem, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square footing 6 7 8 9 d t B [ft] Gravelly Sand Q=400 [k] a square footing supports an axial load, Q, and sits 3 feet beneath the surface of a gravelly sand. γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square footing 6 7 8 9 d t B [ft] Gravelly Sand Q=400 [k] A number of other variables are also given. [pause] γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚
square SF = 3 footing d t B [ft] Gravelly Sand Q=400 [k] Let’s begin with the safety factor, 3, which equals --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚ qult
square SF = 3 = footing qall d t B [ft] Gravelly Sand Q=400 [k] the ultimate bearing capacity divided by the allowable bearing capacity. Or more simply, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square SF = 3 = footing qall d qult = 3 * qall t B [ft] Gravelly Sand Q=400 [k] the ultimate bearing capacity is 3 times the allowable bearing capacity. As far as safety goes, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square SF = 3 = footing qall d qult > 3 * qall t B [ft] Gravelly Sand Q=400 [k] We should consider 3, to be the minimum allowable safety factor. Since both of these values are functions of --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square SF = 3 = footing qall d qult = 3 * qall t B [ft] qult = f(B) Gravelly Sand qall = f(B) Q=400 [k] the base width, B, our strategy will be to solve for both bearing capacities, in term of B, and then return to this equation and solve for B directly. γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square SF = 3 = footing qall d qult = 3 * qall t B [ft] qult = f(B) Gravelly Sand qall = f(B) Q=400 [k] Let’s begin by solving for the ultimate bearing capacity in terms of B. γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square footing d t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand Q=400 [k] For a square footing, our ultimate bearing capacity equals --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

square footing cohesion multiplier d t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand cohesion Q=400 [k] 1.2 times the cohesion, times the cohesion multiplier, plus --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

wedge weight square footing multiplier cohesion multiplier d t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand cohesion base width Q=400 [k] 0.4 times the effective unit weight times the base width times the wedge weight multiplier, plus --- effective unit γcon=150 [lb/ft3] weight γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

wedge weight square footing multiplier cohesion overburden multiplier d multiplier t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand cohesion base width Q=400 [k] the overburden stress times the overburden multiplier. effective unit γcon=150 [lb/ft3] overburden weight γT=120 [lb/ft3] stress d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

wedge weight square footing multiplier cohesion overburden multiplier d multiplier t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand cohesion base width Q=400 [k] The three multiplier terms are also called bearing capacity factors, --- effective unit γcon=150 [lb/ft3] overburden weight γT=120 [lb/ft3] stress d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

wedge weight square footing multiplier cohesion overburden multiplier d multiplier t qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq B [ft] Gravelly Sand cohesion base width Q=400 [k] which are all a functions of the angle of internal friction, phi. effective unit γcon=150 [lb/ft3] overburden weight γT=120 [lb/ft3] stress d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Gravelly Sand Q=400 [k] The bearing capacity factors are solved for by first solving --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Q=400 [k] N sub q, the overburden multiplier. Phi is --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Q=400 [k] plugged into the equation, and N sub q equals --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Nq = 37.75 Q=400 [k] 37.75 γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Nq = 37.75 Q=400 [k] The other two bearing capacity factors are solved for --- Nc = (Nq-1) * cot φ γcon=150 [lb/ft3] Nγ = 2 * (Nq+1) * tan φ γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Nq = 37.75 Q=400 [k] by plugging in phi and N sub q --- Nc = (Nq-1) * cot φ γcon=150 [lb/ft3] Nγ = 2 * (Nq+1) * tan φ γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion wedge overburden multiplier weight multiplier square d t multiplier footing B [ft] Nq = tan2(45+φ/2) * eπ * tanφ Gravelly Sand Nq = 37.75 Q=400 [k] [pause] Nc = (Nq-1) * cot φ=50.58 γcon=150 [lb/ft3] Nγ = 2 * (Nq+1) * tan φ=56.31 γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing B [ft] Gravelly Sand Q=400 [k] We’ll next solve for the cohesion, the effective unit weight and the overburden stress. γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing B [ft] Gravelly Sand Q=400 [k] Since the soil is a gravelly sand, --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] Gravelly Sand Q=400 [k] the cohesion equals 0. γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] γ’= γT-u Q
qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand Q=400 [k] Since the groundwater table is not mentioned, γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] γ’= γT-u Q
qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand Q=400 [k] the effective unit weight of the soil equals the total unit weight of the soil, γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] γ’= γT-u
Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand γ’= 120 [lb/ft3] Q=400 [k] which is 120 pounds per cubic feet. γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand γ’= 120 [lb/ft3] Q=400 [k] q= γ’*h Lastly, the overburden stress equals the effective unit weight of the soil, times the depth of the footing, --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand γ’= 120 [lb/ft3] Q=400 [k] q= γ’*h which is 120 pounds per cubic feet, times 3 feet, or, --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq cohesion overburden effective stress square unit weight d t footing c=0 [lb/ft2] B [ft] γ’= γT-u Gravelly Sand γ’= 120 [lb/ft3] Q=400 [k] q= γ’*h 360 pounds per square feet. γcon=150 [lb/ft3] q= 360 [lb/ft2] Nq = 37.75 γT=120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] Nγ = 56.31 φ=36˚ SF=3.0

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3]
Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq square d t footing B [ft] Gravelly Sand Q=400 [k] If we plug these values into the equation --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] c=0 [lb/ft2] γ’= 120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] q= 360 [lb/ft2] Nγ = 56.31 φ=36˚ SF=3.0

Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq qult=1.2 * 0 [lb/ft2] * 50.58 +0.4 * 120 [lb/ft3] * B * 56.31 square footing + 360 [lb/ft2] * 37.75 B [ft] Gravelly Sand Q=400 [k] we find the ultimate bearing capacity equals --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] c=0 [lb/ft2] γ’= 120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] q= 360 [lb/ft2] Nγ = 56.31 φ=36˚ SF=3.0

Q qult=1.2*c*Nc+0.4*γ’*B*Nγ+q*Nq qult=1.2 * 0 [lb/ft2] * 50.58 +0.4 * 120 [lb/ft3] * B * 56.31 square footing + 360 [lb/ft2] * 37.75 B [ft] qult=2,703 * B [lb/ft3] + 13,590 [lb/ft2] Q=400 [k] 2,703 times B pounds per cubic feet, plus 13,590 pounds per square foot. [pause] Returning to our initial equation, --- γcon=150 [lb/ft3] Nq = 37.75 γT=120 [lb/ft3] c=0 [lb/ft2] γ’= 120 [lb/ft3] Nc = 50.58 d=3 [ft] t=2 [ft] q= 360 [lb/ft2] Nγ = 56.31 φ=36˚ SF=3.0

SF = 3 = qall square d qult = 3 * qall t footing B [ft] qult=2,703 * B [lb/ft3] + 13,590 [lb/ft2] Q=400 [k] we have now solved for the ultimate bearing capacity, in terms of B. γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

SF = 3 = qall square d qult = 3 * qall t footing B [ft] qult=2,703 * B [lb/ft3] + 13,590 [lb/ft2] Q=400 [k] Next we’ll solve for the allowable stress, in terms of B. [pause] For a square footing, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

γcon*B2*t γT*B2*(d-t) Find: minimum B [ft] γcon=150 [lb/ft3]
Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square d t footing B [ft] Gravelly Sand Q=400 [k] the allowable bearing capacity equals the contribution from --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square d t axial footing stress B [ft] Gravelly Sand Q=400 [k] the axial stress, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square d t footing axial footing weight stress B [ft] Gravelly Sand Q=400 [k] the footing weight, --- γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square soil d t footing axial footing weight weight stress B [ft] Gravelly Sand Q=400 [k] and the weight of the soil, above the footing. γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square soil d t footing axial footing weight weight stress B [ft] Gravelly Sand Q + γcon*t + γT*(d-t) qall= Q=400 [k] B squared cancells out in the second and third terms, and the equation is simplified. Next, --- B2 γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square soil d t footing axial footing weight weight stress B [ft] Gravelly Sand Q + γcon*t + γT*(d-t) qall= Q=400 [k] the known values are plugged in the equation, and the allowable stress equals --- B2 γcon=150 [lb/ft3] γT=120 [lb/ft3] d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Q γcon*B2*t γT*B2*(d-t) Q + + qall= B2 B2 B2 square soil d t footing axial footing weight weight stress B [ft] Gravelly Sand Q + γcon*t + γT*(d-t) qall= Q=400 [k] 400,000 pounds divided by the base area, plus 420 pounds per square feet. B2 γcon=150 [lb/ft3] γT=120 [lb/ft3] 400,000 [lb] qall= +420 [lb/ft2] B2 d=3 [ft] t=2 [ft] φ=36˚ SF=3.0

Find: minimum B [ft] Q +420 [lb/ft2] qult qall qult = 3 * qall
SF = 3 = qall square d qult = 3 * qall t footing B [ft] qult=2,703 * B [lb/ft3] + 13,590 [lb/ft2] Having solved for the ultimate and allowable bearing capacities, in terms of the base width, B, we will now directly solve for B. 400,000 [lb] qall= +420 [lb/ft2] B2

Find: minimum B [ft] Q +420 [lb/ft2] qult qall qult = 3 * qall
SF = 3 = qall square d qult = 3 * qall t footing B [ft] qult=2,703 * B [lb/ft3] + 13,590 [lb/ft2] After plugging in these values and simplifying, --- 400,000 [lb] qall= +420 [lb/ft2] B2

Find: minimum B [ft] Q qult qall qult = 3 * qall
SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] We end up with a third degree polynomial. If we keep in mind that B is in units of feet, we can simplify the equation to read ---

SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] B in [ft] 0 equals, 901 times B cubed, plus 4,110 times B squared minus 400,000. Solving for the roots of B, the only non-imaginary solution is --- 0=901* B3 + 4,110 * B ,000

SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] B in [ft] B equals 6.37 feet. To ensure the safety factor is at least 3, --- 0=901* B3 + 4,110 * B ,000 B1=(imaginary) B2=(imaginary) B3=6.37 [ft]

SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] B in [ft] B should be rounded up to 7 feet. 0=901* B3 + 4,110 * B ,000 B1=(imaginary) B2=(imaginary) B=7 [ft] B3=6.37 [ft]

Find: minimum B [ft] Q 6 7 8 9 qult qall qult = 3 * qall
SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] B in [ft] Looking back at the possible solutions, --- 0=901* B3 + 4,110 * B ,000 B1=(imaginary) B2=(imaginary) B=7 [ft] B3=6.37 [ft]

Find: minimum B [ft] Q 6 7 8 9 AnswerB qult qall qult = 3 * qall
SF = 3 = qall square d qult = 3 * qall t footing B [ft] 0=901* B3 [lb/ft3] + 4,110 * B2 [lb/ft2] + 400,000 [lb] B in [ft] the answer is B. 0=901* B3 + 4,110 * B ,000 B1=(imaginary) AnswerB B2=(imaginary) B=7 [ft] B3=6.37 [ft]

( ) ? 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 ‘

Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚

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Find: minimum B [ft] γcon=150 [lb/ft3] γT=120 [lb/ft3] Q φ=36˚

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