Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Question
**Consolidation of Clay:**

**Objective:**
Given Exercise Figure 7.4 and accompanying information, determine the ultimate settlement in the clay under the applied load. 

**Notes:**
- 1 ft = 0.3048 m
- 62.4 lb/ft³ = 1 Mg/m³

**Formula:**
Assuming the sand will not settle, the ultimate settlement in the clay can be calculated using the formula:

\[
\Delta H = \frac{H}{1 + e_0} C_c \log \left(1 + \frac{\Delta \sigma'}{\sigma'}\right)
\]

**Steps:**

a. **Effective Stress Determination:**
   - Determine the effective stress for the cross section prior to loading. Use an appropriate diagram to show stresses at depth.

b. **Calculate \(\Delta \sigma'\):**
   - Find the change in effective stress (\(\Delta \sigma'\)) on the clay layer caused by the imposed load. Calculate for the center of the clay layer at 5.8 m [19 ft] deep.

c. **Original Porosity (\(e_0\)):**
   - Determine \(e_0\) based on the original porosity of the clay.

d. **Settlement Calculation:**
   - Calculate how much settlement will occur in the clay.

**Exercise Figure 7.4: Explanation:**

The figure provides a cross-sectional view with the following layers:

- **0 to 5 ft:** Sand, dry unit weight \(\gamma_d = 105 \, \text{lb/ft}^3\)
- **5 to 16 ft:** Silty sand, saturated unit weight \(\gamma_{\text{sat}} = 127 \, \text{lb/ft}^3\), Groundwater Table (GWT) indicated at 5 ft
- **16 to 22 ft:** Clay, saturated unit weight \(\gamma_{\text{sat}} = 121 \, \text{lb/ft}^3\), with noted porosity (\(n = 0.629\)) and a compression index (\(C_c = 0.82\))

- **Load:** A point load of 125 tons is applied at the surface.

This setup is used to analyze the stress changes and resulting settlement in the clay layer.
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Transcribed Image Text:**Consolidation of Clay:** **Objective:** Given Exercise Figure 7.4 and accompanying information, determine the ultimate settlement in the clay under the applied load. **Notes:** - 1 ft = 0.3048 m - 62.4 lb/ft³ = 1 Mg/m³ **Formula:** Assuming the sand will not settle, the ultimate settlement in the clay can be calculated using the formula: \[ \Delta H = \frac{H}{1 + e_0} C_c \log \left(1 + \frac{\Delta \sigma'}{\sigma'}\right) \] **Steps:** a. **Effective Stress Determination:** - Determine the effective stress for the cross section prior to loading. Use an appropriate diagram to show stresses at depth. b. **Calculate \(\Delta \sigma'\):** - Find the change in effective stress (\(\Delta \sigma'\)) on the clay layer caused by the imposed load. Calculate for the center of the clay layer at 5.8 m [19 ft] deep. c. **Original Porosity (\(e_0\)):** - Determine \(e_0\) based on the original porosity of the clay. d. **Settlement Calculation:** - Calculate how much settlement will occur in the clay. **Exercise Figure 7.4: Explanation:** The figure provides a cross-sectional view with the following layers: - **0 to 5 ft:** Sand, dry unit weight \(\gamma_d = 105 \, \text{lb/ft}^3\) - **5 to 16 ft:** Silty sand, saturated unit weight \(\gamma_{\text{sat}} = 127 \, \text{lb/ft}^3\), Groundwater Table (GWT) indicated at 5 ft - **16 to 22 ft:** Clay, saturated unit weight \(\gamma_{\text{sat}} = 121 \, \text{lb/ft}^3\), with noted porosity (\(n = 0.629\)) and a compression index (\(C_c = 0.82\)) - **Load:** A point load of 125 tons is applied at the surface. This setup is used to analyze the stress changes and resulting settlement in the clay layer.
Expert Solution
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Step 1

Given,

Depth of clay layer, H= 6ft Porosity of clay,n=0.629Compression index, Cc=0.82Load acting at the top surface of soil,P =125tonsDry density of sand,γd=105lb/ft3Saturated density of silty sand,γsat=127lb/ft3Ground water table depth from surface = 5ft

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