PRINCIPLES OF GEOTECH.ENGINEERING >LL+M
PRINCIPLES OF GEOTECH.ENGINEERING >LL+M
9th Edition
ISBN: 9781337583879
Author: Das
Publisher: CENGAGE L
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Question
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Chapter 11, Problem 11.1CTP

(a)

To determine

Calculate the total consolidation settlement under the action of fill load.

(a)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The total consolidation settlement (SC) is 0.188m_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

Calculation:

Consider the unit weight of water (γw) is 9.81kN/m3.

Calculate the distributed load (q) as shown below.

q=tγfill

Substitute 1.75m for t and 20.1kN/m3 for γfill.

q=1.75×20.1=35.02kN/m2

Calculate the increase in vertical stress (Δσz) below the center of a rectangular area using the relation as follows.

Δσz=qI4 (1)

  Here, I4 is the influence factor and as a function of m1 and n1.

For clay layer:

For the depth (z) of 3m:

Calculate the width (b) as shown below.

b=B2

Substitute 8m for B.

b=82=4m

Calculate the ratio (m1) as shown below.

m1=LB

Substitute 8m for L and 8m for B.

m1=88=1

Calculate the ratio (n1) as shown below.

n1=zb

Substitute 4m for b and 3m for z.

n1=34=0.75

Similarly calculate the remaining values and tabulate as in Table 1.

Refer Table 10.11 “Variation of I4 with m1 and n1” in the Text Book.

Take the value of I4 as 0.892, for the values m1 of 1 and n1 of 0.6.

Take the value of I4 as 0.800, for the values m1 of 1 and n1 of 0.8.

Calculate the value of I4 for the values m1 of 1 and n1 of 0.75 by interpolation as shown below.

0.80.750.80.6=0.8I40.80.8920.25×(0.092)=0.8I4I4=0.823

Similarly calculate the remaining values and tabulate as in Table 1.

Calculate the increase in vertical stress (Δσz) as shown below.

Substitute 35.02kN/m2 for q and 0.823 for I4 in Equation (1).

Δσz=35.02×0.823=28.82kN/m2

Similarly calculate the increase in vertical stress values and tabulate as in Table 1.

Show the increase in vertical stress for each depth below the center of the loaded area as in Table 1.

m1=LBb=B2Depth, z(m)n1=zbI4Δσz(kN/m2)
1430.750.82328.82
1451.250.59920.98
1471.750.40314.11

Table 1

Refer to table 1.

Calculate the stress increase in the clay layer (Δσ) using the relation.

Δσ=Δσt+4Δσm+Δσb6 (2)

Here, Δσt is the increase in effective stress at the top layer, Δσm is the increase in effective stress at middle layer, and Δσb is the increase in effective stress at the bottom layer.

Substitute 28.82kN/m2 for Δσt, 20.98kN/m2 for Δσm, and 14.11kN/m2 for Δσb in Equation (2).

Δσ=28.82+4×20.98+14.116=126.856=21.14kN/m2

Calculate the average effective stress at the middle of the clay layer (σo) as shown below.

σo=γdDf+((γsat)sandγw)(H1Df)+((γsat)clayγw)×H22

Substitute 17kN/m3 for γd, 1.5m for Df, 19.2kN/m3 for (γsat)sand, 9.81kN/m3 for γw, 18.8kN/m3 for (γsat)clay, 3m for H1, and 4m for H2.

σo=17×1.5+(19.29.81)(31.5)+(18.89.81)×42=25.5+14.085+17.98=57.565kN/m2

Calculate the primary consolidation settlement (SC) using the relation.

SC=CCH1+e0log(σo+Δσσo) (3)

Substitute 0.31 for CC, 4m for H, 1.08 for e0, 57.565kN/m2 for σo, and 21.14kN/m2 for Δσ in Equation (3).

SCclay=0.31×41+1.08log(57.565+21.1457.565)=0.59615×0.1358=0.081m

For peat layer:

For the depth (z) of 7m:

Calculate the ratio (n1) as shown below.

n1=zb

Substitute 4m for b and 7m for z.

n1=74=1.75

Similarly calculate the remaining values and tabulate as in Table 2.

Refer Table 10.11 “Variation of I4 with m1 and n1” in the Text Book.

Take the value of I4 as 0.449, for the values m1 of 1 and n1 of 1.6.

Take the value of I4 as 0.388, for the values m1 of 1 and n1 of 1.8.

Calculate the value of I4 for the values m1 of 1 and n1 of 1.75 by interpolation as shown below.

1.81.751.81.6=0.388I40.3880.4490.25×(0.061)=0.8I4I4=0.403

Similarly calculate the remaining values and tabulate as in Table 2.

Calculate the increase in vertical stress (Δσz) as shown below.

Substitute 35.02kN/m2 for q and 0.403 for I4 in Equation (1).

Δσz=35.02×0.403=14.11kN/m2

Similarly calculate the increase in vertical stress values and tabulate as in Table 2.

Show the increase in vertical stress for each depth below the center of the loaded area as in Table 2.

m1=LBb=B2Depth, z(m)n1=zbI4Δσz(kN/m2)
1471.750.40314.11
147.91.9750.34211.98
148.82.20.30210.58

Table 2

Refer to table 2.

Calculate the stress increase in the peat layer (Δσ) using the relation.

Substitute 14.11kN/m2 for Δσt, 11.98kN/m2 for Δσm, and 10.58kN/m2 for Δσb in Equation (2).

Δσ=14.11+4×11.98+10.586=72.616=12.10kN/m2

Calculate the average effective stress at the middle of the clay layer (σo) as shown below.

σo=γdDf+((γsat)sandγw)(H1Df)+((γsat)clayγw)×H2+((γsat)peatγw)×H32

Substitute 17kN/m3 for γd, 1.5m for Df, 19.2kN/m3 for (γsat)sand, 9.81kN/m3 for γw, 18.8kN/m3 for (γsat)clay, 15kN/m3 for (γsat)peat, 3m for H1, 4m for H2, and 1.8m for H3.

σo=17×1.5+(19.29.81)(31.5)+(18.89.81)×4+(159.81)×1.82=25.5+14.085+35.96+4.671=80.216kN/m2

Calculate the primary consolidation settlement (SC) using the relation.

Substitute 7.2 for CC, 1.8m for H, 6.4 for e0, 80.216kN/m2 for σo, and 12.10kN/m2 for Δσ in Equation (3).

SCPeat=7.2×1.81+6.4log(80.216+12.1080.216)=1.7514×0.061=0.107m

Calculate the total consolidation settlement under the action of fill load (SC) as shown below.

SC=SCclay+SCpeat

Substitute 0.081m for SCclay and 0.107m for SCpeat.

SC=0.081+0.107=0.188

Hence, the total consolidation settlement (SC) is 0.188m_.

(b)

To determine

Calculate the time for 99% primary consolidation for each layer.

(b)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The time for 99% primary consolidation for clay (t99-clay) is 138days_.

The time for 99% primary consolidation for peat (t99-peat) is 23days_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

Calculation:

The degree of consolidation (U) is 99%.

The clay layer is permeable and having less void ratio compared to peat layer. Hence, double drainage condition is assumed for the clay layer.

Calculate the time factor (Tv) as shown below.

Refer Table 11.7 “Variation of Tv with U” in the Text Book.

Take the value of Tv as 1.781, for the value U of 99%.

Calculate the length of maximum drainage path (Hdr) for clay layer using the relation.

Hdr=H22

Substitute 4m for H2.

Hdr=42=2m

Calculate the time for 99% consolidation (t99-clay) using the relation.

Tv=cvtH2dr (4)

Substitute 1.781 for Tv, 0.006cm2/sec for cv, and 2m for Hdr in Equation (4).

1.781=0.006cm2/sec×t99-clay(2m×100cm1m)20.006t99-clay=71,240t99-clay=11,873,333.333sec×1min60sec×1hr60mins×1day24hrst99-clay=138days

Hence, the time for 99% primary consolidation for clay (t99-clay) is 138days_.

The peat layer is low permeable and having high void ratio compared to clay layer. Hence, single drainage condition is assumed for the peat layer.

Calculate the length of maximum drainage path (Hdr) for clay layer using the relation.

Hdr=H3

Substitute 1.8m for H3.

Hdr=1.8m

Calculate the time for 99% consolidation (t99-peat) using the relation.

Substitute 1.781 for Tv, 0.029cm2/sec for cv, and 1.8m for Hdr in Equation (4).

1.781=0.029cm2/sec×t99-peat(1.8m×100cm1m)20.029t99-peat=57,704.4t99-peat=1,989,806.897sec×1min60sec×1hr60mins×1day24hrst99-peat=23days

Hence, the time for 99% primary consolidation for peat (t99-peat) is 23days_.

(c)

To determine

Calculate the secondary compression in each layer up to end of 18 months.

(c)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The secondary compression for clay (Ssclay) is 0.055m_.

The secondary compression for peat (Sspeat) is 0.096m_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

Calculation:

Refer to part (b).

The time for 99% primary consolidation for clay (t99-clay) is 138days.

The time for 99% primary consolidation for peat (t99-peat) is 23days.

For clay:

Calculate the primary void ratio (Δeprimary) as shown below.

Δeprimary=CClog(σo+Δσσo) (5)

Substitute 0.31 for CC, 57.565kN/m2 for σo, and 21.14kN/m2 for Δσ in Equation (5).

Δeprimary=0.31log(57.565+21.1457.565)=0.31×0.1358=0.042

Calculate the void ratio at the end of primary consolidation (ep) as shown below.

ep=e0Δeprimary (6)

Substitute 1.08 for e0 and 0.042 for Δeprimary in Equation (6).

ep=1.080.042=1.038

Calculate the magnitude of secondary compression index (Cα) as shown below.

Cα=Cα1+ep (7)

Here, Cα is the secondary compression index.

Substitute 0.048 for Cα and 1.038 for ep.

Cα=0.0481+1.038=0.0235

Calculate the secondary compression (Ss) as shown below.

Ss=CαHlog(t2t1) (8)

Substitute 0.0235 for Cα, 4m for H, 18months for t2, and 138days for t1 in Equation (8).

Ssclay=0.0235×(4m)log(18months×30days1month138days)=0.094×0.5925=0.055m

Hence, the secondary compression for clay (Ssclay) is 0.055m_.

For peat:

Calculate the primary void ratio (Δeprimary) as shown below.

Substitute 7.2 for CC, 80.216kN/m2 for σo, and 12.10kN/m2 for Δσ in Equation (5).

Δeprimary=7.2log(80.216+12.180.216)=7.2×0.061=0.44

Calculate the void ratio at the end of primary consolidation (ep) as shown below.

Substitute 6.4 for e0 and 0.44 for Δeprimary in Equation (6).

ep=6.40.44=5.96

Calculate the magnitude of secondary compression index (Cα) as shown below.

Substitute 0.273 for Cα and 5.96 for ep.

Cα=0.2731+5.96=0.039

Calculate the secondary compression (Ss) as shown below.

Substitute 0.039 for Cα, 1.8m for H, 18months for t2, and 23days for t1 in Equation (8).

Ssclay=0.039×(1.8m)log(18months×30days1month23days)=0.0702×1.3707=0.096m

Hence, the secondary compression for peat (Sspeat) is 0.096m_.

(d)

To determine

Calculate the total settlement after 18 months.

(d)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The total settlement after 18 months is 0.339m_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

Calculation:

Refer to part (a).

The total consolidation settlement (SC) is 0.188.

Refer to part (c).

The secondary compression for clay (Ssclay) is 0.055m.

The secondary compression for peat (Sspeat) is 0.096m.

Calculate the total settlement after 18 months as shown below.

Totalsettlement=SC+Ssclay+Sspeat

Substitute 0.188m for SC, 0.055m for Ssclay, and 0.096m for Sspeat.

Totalsettlement=0.188+0.055+0.096=0.339m

Hence, the total settlement after 18 months is 0.339m_.

(e)

To determine

Calculate the excess pore water pressure at point A two months after the application of the fill load.

(e)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The excess pore water pressure at point A (uz) is 2.114kN/m2_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

The depth (z) is 3.2m.

Calculation:

Refer to part (a).

The pore water pressure (uo=Δσ) is 21.14kN/m2.

Calculate the length of maximum drainage path (Hdr) using the relation.

Hdr=H22

Substitute 4m for H.

Hdr=42=2m

Calculate the time factor (Tv) as shown below.

Tv=cvtH2dr

Substitute 0.006cm2/sec for cv, 2months for t, and 2m for Hdr.

Tv=0.006cm2/sec×2months×30days1month×24hrs1day×60mins1hr×60sec1min(2m×100cm1m)2=0.7776

Calculate the ratio zHdr as shown below.

Substitute 3.2m for z and 2m for Hdr.

zHdr=3.22=1.6

Calculate the degree of consolidation (U) as shown below.

Refer Figure 11.29 “Variation of Uz with Tv and zHdr” in the Text Book.

Take the value of U as 0.9, for the values Tv of 0.7776 and zHdr of 1.6.

Calculate the excess pore water pressure after 2 months (uz) as shown below.

U=1uzuo

Substitute 0.9 for U and 21.14kN/m2 for uo.

0.9=1uz21.14uz21.14=0.1uz=2.114kN/m2

Hence, the excess pore water pressure at point A (uz) is 2.114kN/m2_.

(f)

To determine

Calculate the effective stress at point A two months after the application of the fill load.

(f)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The effective stress at point A is 87.38kN/m2_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

The depth (z) is 3.2m.

Calculation:

Refer to part (a).

The pore water pressure (uo=Δσ) is 21.14kN/m2.

Refer to part (e)

The excess pore water pressure at point A (uz) is 2.114kN/m2.

Calculate the increase in effective stress (Δu) after 2 months as shown below.

Δu=uouz

Substitute 21.14kN/m2 for uo and 2.114kN/m2 for uz.

Δu=21.142.114=19.026kN/m2

Calculate the average effective stress at the point A (σoA) as shown below.

σoA=γdDf+((γsat)sandγw)(H1Df)+((γsat)clayγw)×z

Substitute 17kN/m3 for γd, 1.5m for Df, 19.2kN/m3 for (γsat)sand, 9.81kN/m3 for γw, 18.8kN/m3 for (γsat)clay, 3m for H1, and 3.2m for z.

σoA=17×1.5+(19.29.81)(31.5)+(18.89.81)×3.2=25.5+14.085+28.768=68.353kN/m2

Calculate the final effective stress at point A as shown below.

Finaleffectivestress=σoA+Δu

Substitute 68.353kN/m2 for σoA and 19.026kN/m2 for Δu.

Finaleffectivestress=68.353+19.026=87.38kN/m2

Hence, the effective stress at point A is 87.38kN/m2_.

(g)

To determine

Calculate the piezometer reading at point A two months after the application of the fill load.

(g)

Expert Solution
Check Mark

Answer to Problem 11.1CTP

The piezometer reading at point A is 48.22kN/m2_.

Explanation of Solution

Given information:

The thickness of fill material (t) is 1.75m.

The compacted unit weight of fill material (γfill) is 20.1kN/m3.

The length of the foundation (L) is 8m.

The breadth of the foundation (B) is 8m.

The depth of fill (Df) is 1.5m.

The height of the layer silty sand (H1) is 3m.

The height of the clay layer (H2) is 4m.

The height of the peat layer (H3) is 1.8m.

The dry unit weight of sand (γd) is 17kN/m3.

The saturated unit weight of sand (γsat)sand is 19.2kN/m3.

The saturated unit weight of clay (γsat)clay is 18.8kN/m3.

The saturated unit weight of peat (γsat)peat is 15kN/m3.

The time (t) is 8months.

The properties of clay and organic layers are given in the Table.

The depth (z) is 3.2m.

Calculation:

Refer to part (e)

The excess pore water pressure at point A (uz) is 2.114kN/m2.

The piezometer reading is the total pore water pressure.

Calculate the piezometer reading (upiezometer) as shown below.

upiezometer=(H1Df+z)γw+uz

Substitute 3m for H1, 1.5m for Df, 3.2m for z, 9.81kN/m3 for γw, and 2.114kN/m2 for uz.

upiezometer=(31.5+3.2)9.81+2.114=48.221kN/m2

Hence, the piezometer reading at point A is 48.22kN/m2_.

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