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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.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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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** **Diagram (a):** Represents a foundation on a weaker soil layer underlain by a stronger sand layer. The diagram shows the structure's placement on two soil layers: the upper layer (weaker soil) and the lower layer (stronger soil). Key dimensions in the diagram are labeled, such as: - \(D_1\): Depth of weaker soil - \(D_2\): Depth of stronger soil - \(B\): Width of the foundation - \(H\): Height of the weaker soil layer - \(\phi_1\): Angle of friction of the weaker soil - \(\phi_2\): Angle of friction of the stronger soil - \(\gamma_1\): Unit weight of the weaker soil **Graph (b):** Illustrates the nature of variation of \(q_u\) with \(H/B\). The graph displays a curve that shows how the ultimate bearing capacity (\(q_u\)) changes as the ratio of the height of the weaker soil layer (\(H\)) to the width of the foundation (\(B\)) varies. Initially, the \(q_u\) value is relatively high when \(H/B\) is small, and it gradually decreases as \(H/B\) increases, showing an inverse relationship between \(q_u\) and \(H/B\).
Transcribed Image Text:**Figure 6.17** **Diagram (a):** Represents a foundation on a weaker soil layer underlain by a stronger sand layer. The diagram shows the structure's placement on two soil layers: the upper layer (weaker soil) and the lower layer (stronger soil). Key dimensions in the diagram are labeled, such as: - \(D_1\): Depth of weaker soil - \(D_2\): Depth of stronger soil - \(B\): Width of the foundation - \(H\): Height of the weaker soil layer - \(\phi_1\): Angle of friction of the weaker soil - \(\phi_2\): Angle of friction of the stronger soil - \(\gamma_1\): Unit weight of the weaker soil **Graph (b):** Illustrates the nature of variation of \(q_u\) with \(H/B\). The graph displays a curve that shows how the ultimate bearing capacity (\(q_u\)) changes as the ratio of the height of the weaker soil layer (\(H\)) to the width of the foundation (\(B\)) varies. Initially, the \(q_u\) value is relatively high when \(H/B\) is small, and it gradually decreases as \(H/B\) increases, showing an inverse relationship between \(q_u\) and \(H/B\).
### Problem 6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, the following parameters are given: - **Foundation Dimensions and Depth:** - Width (\( B \)) = 1.5 m - Depth of foundation (\( D_f \)) = 1 m - Thickness of clay layer (\( H \)) = 0.8 m - **Layer 1 Properties:** - Unit weight (\( \gamma_1 \)) = 16.5 kN/m³ - Cohesion (\( c_1 = c_{u(1)} \)) = 48 kN/m² - Angle of internal friction (\( \phi_1 \)) = 0 - **Layer 2 Properties:** - Unit weight (\( \gamma_2 \)) = 17.5 kN/m³ - Cohesion (\( c_2 = c_{u(2)} \)) = 96 kN/m² - Angle of internal friction (\( \phi_2 \)) = 0 Using Equation (6.46), determine the gross ultimate bearing capacity. ### Explanation: The problem involves calculating the bearing capacity of a foundation resting on a two-layered clay system. The necessary parameters like width, depth, unit weights, cohesion values, and angles of internal friction for each layer are specified. Since the angles of internal friction are zero, the soil behaves as a purely cohesive material. ### Diagram Explanation: The diagram (Figure 6.17) typically illustrates a cross-section of the layered foundation system, showing the different layers of clay with their respective properties, including the thickness (\( H \)), depth (\( D_f \)), and width (\( B \)). Each layer is labeled with its specific unit weight and cohesion.
Transcribed Image Text:### Problem 6.12 Refer to Figure 6.17. For a continuous foundation on layered clay, the following parameters are given: - **Foundation Dimensions and Depth:** - Width (\( B \)) = 1.5 m - Depth of foundation (\( D_f \)) = 1 m - Thickness of clay layer (\( H \)) = 0.8 m - **Layer 1 Properties:** - Unit weight (\( \gamma_1 \)) = 16.5 kN/m³ - Cohesion (\( c_1 = c_{u(1)} \)) = 48 kN/m² - Angle of internal friction (\( \phi_1 \)) = 0 - **Layer 2 Properties:** - Unit weight (\( \gamma_2 \)) = 17.5 kN/m³ - Cohesion (\( c_2 = c_{u(2)} \)) = 96 kN/m² - Angle of internal friction (\( \phi_2 \)) = 0 Using Equation (6.46), determine the gross ultimate bearing capacity. ### Explanation: The problem involves calculating the bearing capacity of a foundation resting on a two-layered clay system. The necessary parameters like width, depth, unit weights, cohesion values, and angles of internal friction for each layer are specified. Since the angles of internal friction are zero, the soil behaves as a purely cohesive material. ### Diagram Explanation: The diagram (Figure 6.17) typically illustrates a cross-section of the layered foundation system, showing the different layers of clay with their respective properties, including the thickness (\( H \)), depth (\( D_f \)), and width (\( B \)). Each layer is labeled with its specific unit weight and cohesion.
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