**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.

Answered: Consolidation of Clay: Given Exercise…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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.

Expert Solution

Step 1

Given,

$\begin{array}{rcl} D e p t h o f c l a y l a y e r, H & = & 6 f t \\ P o r o s i t y o f c l a y, n & = & 0.629 \\ C o m p r e s s i o n i n d e x, C_{c} & = & 0.82 \\ L o a d a c t i n g a t t h e t o p s u r f a c e o f s o i l, P & = & 125 t o n s \\ D r y d e n s i t y o f s a n d, γ_{d} & = & 105 l b / f t^{3} \\ S a t u r a t e d d e n s i t y o f s i l t y s a n d, γ_{s a t} & = & 127 l b / f t^{3} \\ G r o u n d w a t e r t a b l e d e p t h f r o m s u r f a c e & = & 5 f t \end{array}$

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