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Contents:

## Introduction

There are two type of foundation settlements:

1. Immediate settlement – occurs immediately, usually within a week or two, small.
2. Consolidation settlement – long term, may be more than 10 years, occurs due to saturated weak clay layer under footing, can be very large.

Consolidation settlement normal occurs in saturated soft clay.  When the saturated clay is subjected to additional loading from structure above, it gradually release pore water and consolidated.  The amount of settlement depends on soil properties and the pressure apply to it.  There are many way to determine soil pressure under footing.  Two common methods are 2 to 1 method and Newmark's method

## Consolidation settlement equations Normally consolidated clay

DH = [Cc H /(1+eo)] log [(po+Dp)/po]

Overly consolidated clay:

When po+Dp < po

DH = [Cr H /(1+eo)] log [(po+Dp)/po]

When po+Dp > po

DH = [Cr H /(1+eo)] log [pc/po] + [Cc H /(1+eo)] log [(po+Dp)/pc]

Where

DH = Consolidation settlement,

H = Thickness of saturated clay layer,

Cc = Compression index of saturated clay,

Cr = Recompression index of saturated clay,

eo = initial void ratio of saturated clay,

po = Effective soil pressure before construction of footing.

pc = Pre-consolidation pressure.

## Consolidation, e- log p curves ## Example 12: Determine consolidation settlement of normally consolidated clay

Given:

• Column load: 50 kips including weight of footing.

• Footing type: 6 feet square footing, bottom of footing at 3 feet below ground surface.

• Soil profile:

• From 0 to -8 ft: clay sand, moist unit weight = 120 lbs/ft3.

• From -8 ft to -16 ft: saturated clay, effective unit weight = 80 lbs/ft3.

• Location of ground water table: -8 ft.

• Properties of saturated clay:

• Normally consolidated.

• Initial void ratio of saturated clay: 0.7

• Compression index of saturated clay: 0.5

RequirementDetermine consolidation settlement using 2:1 method for soil pressure.

Solution:

Thickness of clay layer H = 16-8 =8 ft

Overburden pressure Po= 120*8+80*8=1600 lbs/ft2.

Depth from bottom of footing to middle of clay layer = 8-3+8/2= 9 ft

Increase in soil pressure, Dp = 50000/(6+9)2 =222.2 lbs/ft2.

Consolidation settlement,

DH = [Cc H /(1+eo)] log [(po+Dp)/po]

= [0.5*8*12/(1+0.7)]log[(1600+ 222.2)/1600]

=1.6 inch

## Example 13: Determine consolidation settlement of overly consolidated clay (Po+DP < Pc).

Given:

• Column load: 40 kips including weight of footing.

• Footing type: 6 feet square footing, bottom of footing at 3 feet below ground surface.

• Soil profile:

• From 0 to -6 ft: clay sand, moist unit weight = 100 lbs/ft3.

• From -6 ft to -16 ft: saturated clay, effective unit weight = 70 lbs/ft3.

• Location of ground water table: -6 ft.

• Properties of saturated clay:

• Overly consolidated clay, preconsolidation pressure, Pc = 1600 lbs/ft2.

• Initial void ratio of saturated clay: 0.6

• Compression index of saturated clay: 0.5

• Recompression index of saturated clay: 0.1

RequirementDetermine consolidation settlement using 2:1 method for soil pressure.

Solution:

Thickness of clay layer H = 16-6 =10 ft

Overburden pressure Po= 100*6+70*10=1300 lbs/ft2

Depth from bottom of footing to middle of clay layer = 6-3+10/2=8 ft

Increase in soil pressure, Dp = 40000/(6+8)2 = 204 lbs/ft2.

Po+Dp = 1300+204=1504 lbs/ft2 <Pc=1600 lbs/ft2

Consolidation settlement,

DH = [Cr H /(1+eo)] log [(po+Dp)/po]

= [0.1*10*12/(1+0.6)]log[1504/1300]

= 0.5 inch

## Example 14: Determine consolidation settlement of overly consolidated clay (Po+DP > Pc).

Given:

• Column load: 80 kips including weight of footing.

• Footing type: 6 feet square footing, bottom of footing at 3 feet below ground surface.

• Soil profile:

• From 0 to -6 ft: clay sand, moist unit weight = 100 lbs/ft3.

• From -6 ft to -16 ft: saturated clay, effective unit weight = 70 lbs/ft3.

• Location of ground water table: -6 ft.

• Properties of saturated clay:

• Overly consolidated clay, preconsolidation pressure, Pc = 1600 lbs/ft2.

• Initial void ratio of saturated clay: 0.6

• Compression index of saturated clay: 0.5

• Recompression index of saturated clay: 0.1

Requirement

Determine consolidation settlement using 2:1 method for soil pressure.

Solution:

Thickness of clay layer H = 16-6 =10 ft

Overburden pressure Po= 100*6+70*10=1300 lbs/ft2

Depth from bottom of footing to middle of clay layer = 6-3+10/2=8 ft

Increase in soil pressure, Dp = 80000/(6+8)2 = 408 lbs/ft2.

Po+Dp = 1300+408=1708 lbs/ft2 <Pc=1600 lbs/ft2

Consolidation settlement,

DH = [Cr H /(1+eo)] log [pc/po] + [Cc H /(1+eo)] log [(po+Dp)/pc]

= [0.1*10*12/(1+0.6)]log[1600/1300]+[0.5*10*12/(1+0.6)]log[1708/1600]

= 1.7 inch

## Example 15: Determine consolidation settlement of normally consolidated clay

Given:

• Column load: 50 kips including weight of footing.

• Footing type: 6 feet square footing, bottom of footing at 3 feet below ground surface.

• Soil profile:

• From 0 to -8 ft: clay sand, moist unit weight = 120 lbs/ft3.

• From -8 ft to -16 ft: saturated clay, effective unit weight = 80 lbs/ft3.

• Location of ground water table: -8 ft.

• Properties of saturated clay:

• Normally consolidated.

• Initial void ratio of saturated clay: 0.7

• Compression index of saturated clay: 0.5

RequirementDetermine consolidation settlement at center and corner of footing using Newmark’s chart for vertical pressure

Solution:

Thickness of clay layer H = 16-8 =8 ft

Overburden pressure Po= 120*8+80*8=1600 lbs/ft2.

Depth from bottom of footing to middle of clay layer = 8-3+8/2= 9 ft

Use Newmark’s chart to determine increase in soil pressure:

Pressure at bottom of footing = 50000/36=1389 lbs/ft2. The number of units in the square with its center at center of chart is about 40

Dp = 1389x0.005x20=278 lbs/ft2.

Consolidation settlement at center of footing

DH = [Cc H /(1+eo)] log [(po+Dp)/po]

= [0.5*8*12/(1+0.7)]log[(1600+ 278)/1600]

= 2 inch

The number of units in the square with its corner at center of chart is about 24

Dp = 1389x0.005x24=167 lbs/ft2.

Consolidation settlement at center of footing

DH = [Cc H /(1+eo)] log [(po+Dp)/po]

= [0.5*8*12/(1+0.7)]log[(1600+ 167)/1600]

= 1.3 inch