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6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, given: . B = 1.5 m, D;= lm, H = 0.8 m 71 = 16.5 kN /m³, c₁ = Cu(1) = 48 kN/m²,0₁ = 0 72 = 17.5 kN/m³, C₂ = Cu(2) = 96 kN/m², 0₂ = 0 Using Eq. (6.46), determine the gross ultimate bearing capacity.

6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, given: . B = 1.5 m, D;= lm, H = 0.8 m 71 = 16.5 kN /m³, c₁ = Cu(1) = 48 kN/m²,0₁ = 0 72 = 17.5 kN/m³, C₂ = Cu(2) = 96 kN/m², 0₂ = 0 Using Eq. (6.46), determine the gross ultimate bearing capacity.

Principles of Foundation Engineering (MindTap Course List)
Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter9: Settlement Of Shallow Foundations
Section: Chapter Questions
Problem 9.5P
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This is Civil Engineering. Specifically Foundation Engineering. Please post thorough answers and work. Any plot needs to be computer generated. Thank you

Figure 6.17 ▸ Details Stronger soil (a) (b) DVB Weaker soil HB (a) Foundation on weaker soil layer underlain by stronger sand layer; (b) nature of variation of qu with H/B
Transcribed Image Text:Figure 6.17 ▸ Details Stronger soil (a) (b) DVB Weaker soil HB (a) Foundation on weaker soil layer underlain by stronger sand layer; (b) nature of variation of qu with H/B
6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, given: B = 1.5 m, D, 1m, H = 0.8 m 71 = 16.5 kN/m³, C₁=Cu(1) = 48 kN/m², ₁ = 0 72 = 17.5 kN/m³, C₂ = Cu(2) = 96 kN/m², 2 = 0 Using Eq. (6.46), determine the gross ultimate bearing capacity.
Transcribed Image Text:6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, given: B = 1.5 m, D, 1m, H = 0.8 m 71 = 16.5 kN/m³, C₁=Cu(1) = 48 kN/m², ₁ = 0 72 = 17.5 kN/m³, C₂ = Cu(2) = 96 kN/m², 2 = 0 Using Eq. (6.46), determine the gross ultimate bearing capacity.
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