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Concept explainers
Figure 16.21 shows a continuous foundation with a width of 1.8 m constructed at a depth of 1.2 m in a granular soil. The footing is subjected to an eccentrically inclined loading with e = 0.3 m, and α = 10°. Determine the gross ultimate load, Qu(ei), that the footing can support using:
- a. Meyerhof (1963) method [Eq. (16.52)]
- b. Saran and Agarwal (1991) method [Eq. (16.53)]
- c. Patra et al. (2012) reduction factor method [Eq. (16.54)]
(a)
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The gross ultimate load Qu(ei) that the footing can support by using Meyerhof method.
Answer to Problem 16.19P
The gross ultimate load Qu(ei) that the footing can support by using Meyerhof method is 984 kN/m_.
Explanation of Solution
Given information:
The unit weight of the soil γ is 17 kN/m3.
The value of cohesion c′ is 0.
The soil friction angle ϕ′ is 34°.
The location of depth of footing base Df is 1.2 m.
The width of the footing B is 1.8 m.
The value of eccentricity e is 0.3 m.
The inclined angle α is 10°.
Calculation:
Determine the effective width of the footing using the relation.
B′=B−2e
Substitute 1.8 m for B and 0.3 for e.
B′=1.8−2×0.3=1.2 m
For the continuous foundation, all shape factors are equal to one (λcs=λqs=λγs=1.0).
Determine the depth factor λcd using the relation.
λcd=1+0.4tan−1(DfB′)
Substitute 1.2 m for Df and 1.2 m for B′.
λcd=1+0.4(1.21.2)=1.4
Determine the depth factor λqd using the relation.
λqd=1+2tanϕ′(1−sinϕ′)2(DfB′)
Substitute 34° for ϕ′, 1.2 m for Df, and 1.2 m for B′.
λqd=1+2tan34°(1−sin34°)2(1.21.2)=1.262
Determine the inclination factor λqi using the relation.
λqi=(1−α90°)2
Substitute 10° for α.
λqi=(1−10°90°)2=0.79
Determine the inclination factor λγi using the relation.
λγi=(1−αϕ′)2
Substitute 10° for α and 34° for ϕ′.
λγi=(1−10°34°)2=0.498
Determine the ultimate bearing capacity of the soil (q′u) using the relation.
q′u=c′Ncλcsλcd+qNqλqsλqd+12γB′Nγλγsλγd=c′Ncλcsλcd+γDfNqλqsλqd+12γB′Nγλγsλγd (1)
Here, λγd is the depth factor and Nc, Nq, and Nγ are bearing-capacity factors.
Refer Table 16.2, “Bearing-capacity factors Nc, Nq, and Nγ” in the textbook.
For ϕ′=34°;
The values of Nc is 42.16, Nq is 29.44, and Nγ is 41.06.
Substitute 0 for c′, 42.16 for Nc, 1.0 for λcs, 1.4 for λcd, 17 kN/m3 for γ, 1.2 m for Df, 29.44 for Nq, 1.0 for λqs, 1.262 for λqd, 0.79 for λqi, 1.2 m for B′, 41.06 for Nγ, 1.0 for λγs, 1.0 for λγd, and 0.498 for λγi in Equation (1).
q′u={(0×42.16×1.0×1.4)+(17×1.2×29.44×1.0×1.262×0.79)+(12×17×1.2×41.06×1×1×0.498)}=807.33 kN/m2
Determine the gross ultimate load Qu(ei) using the relation.
Qu(ei)=q′uB′cosα
Substitute 807.33 kN/m2 for q′u, 1.2 m for B′, and 10° for α.
Qu(ei)=807.33×1.2cos10°=984 kN/m
Therefore, the gross ultimate load Qu(ei) that the footing can support by using Meyerhof method is 984 kN/m_.
(b)
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The gross ultimate load Qu(ei) that the footing can support by using Saran and Agarwal method.
Answer to Problem 16.19P
The gross ultimate load Qu(ei) that the footing can support by using Saran and Agarwal method is 1,237 kN/m_.
Explanation of Solution
Given information:
The unit weight of the soil γ is 17 kN/m3.
The value of cohesion c′ is 0.
The soil friction angle ϕ′ is 34°.
The location of depth of footing base Df is 1.2 m.
The width of the footing B is 1.8 m.
The value of eccentricity e is 0.3 m.
The inclined angle α is 10°.
Calculation:
Determine the ratio of (eB).
Substitute 0.3 for e and 1.8 m for B.
(eB)=0.31.8=0.167
Determine the gross ultimate load Qu(ei) using the relation.
Qu(ei)=B[c′Nc(ei)+qNq(ei)+12γBNγ(ei)]=B[c′Nc(ei)+γDfNq(ei)+12γBNγ(ei)] (2)
Here, Nc(ei), Nq(ei), and Nγ(ei) are the bearing capacity factors.
Refer Figure 16.14, “Variation of Nc(ei)-(b)” in the textbook.
Take the Nc(ei) value as 24.0 for ϕ′ value of 34° and the (eB) value of 0.167.
Refer Figure 16.15, “Variation of Nq(ei)-(b)” in the textbook.
Take the Nq(ei) value as 22.8 for ϕ′ value of 34° and the (eB) value of 0.167.
Refer Figure 16.16, “Variation of Nγ(ei)-(b)” in the textbook.
Take the Nγ(ei) value as 14.5 for ϕ′ value of 34° and the (eB) value of 0.167.
Substitute 0 for c′, 24.0 for Nc(ei), 17 kN/m3 for γ, 1.2 m for Df, 22.8 for Nq(ei), 1.8 m for B, and 14.5 for Nγ(ei), in Equation (2).
Qu(ei)=1.8(0×24.0+17×1.2×22.8+12×17×1.8×14.5)=1,237 kN/m
Therefore, the gross ultimate load Qu(ei) that the footing can support by using Saran and Agarwal method is 1,237 kN/m_.
(c)
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The gross ultimate load Qu(ei) that the footing can support by using Patra et al. reduction factor method.
Answer to Problem 16.19P
The gross ultimate load Qu(ei) that the footing can support by using Patra et al. reduction factor method is 1,006 kN/m_.
Explanation of Solution
Given information:
The unit weight of the soil γ is 17 kN/m3.
The value of cohesion c′ is 0.
The soil friction angle ϕ′ is 34°.
The location of depth of footing base Df is 1.2 m.
The width of the footing B is 1.8 m.
The value of eccentricity e is 0.3 m.
The inclined angle α is 10°.
Calculation:
For the continuous foundation, all shape factors are equal to one (λcs=λqs=λγs=1.0).
Determine the depth factor λcd using the relation.
λcd=1+0.4tan−1(DfB)
Substitute 1.2 m for Df and 1.8 m for B.
λcd=1+0.4(1.21.8)=1.26
Determine the depth factor λqd using the relation.
λqd=1+2tanϕ′(1−sinϕ′)2(DfB)
Substitute 34° for ϕ′, 1.2 m for Df, and 1.8 m for B.
λqd=1+2tan34°(1−sin34°)2(1.21.8)=1.175
Determine the ultimate bearing capacity of the soil (qu) using the relation.
qu=c′Ncλcsλcd+qNqλqsλqd+12γBNγλγsλγd=c′Ncλcsλcd+γDfNqλqsλqd+12γBNγλγsλγd (3)
Refer Table 16.2, “Bearing-capacity factors Nc, Nq, and Nγ” in the textbook.
Take the Nc as 42.16, Nq as 29.44, and Nγ as 41.06 for the ϕ′ value of 34°.
Substitute 0 for c′, 42.16 for Nc, 1.0 for λcs, 1.26 for λcd, 17 kN/m3 for γ, 1.2 m for Df, 29.44 for Nq, 1.0 for λqs, 1.175 for λqd, 1.8 m for B, 41.06 for Nγ, 1.0 for λγs, and 1.0 for λγd, in Equation (3).
qu={0×42.16×1.0×1.26+17×1.2×29.44×1.0×1.175+12×17×1.8×41.06×1×1}=1,333.89 kN/m2
Determine the gross ultimate load Qu(ei) using the relation.
Qu(ei)=Bqu[1−2(eB)](1−αϕ′)2−(Df/B)
Substitute 1.8 m for B, 1,333.89 kN/m2 for q′u, 0.3 m for e, 10° for α, 34° for ϕ′, and 1.2 m for Df.
Qu(ei)=1.8×1,333.89[1−2(0.31.8)](1−10°34°)2−(1.2/1.8)=1,006 kN/m
Therefore, the gross ultimate load Qu(ei) that the footing can support by using Patra et al. reduction factor method is 1,006 kN/m_.
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