Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781337516877
Author: Das
Publisher: Cengage
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Chapter 11, Problem 11.21P

(a)

To determine

Calculate the average stress increase in the clay layer due to the applied load.

(a)

Expert Solution
Check Mark

Answer to Problem 11.21P

The average stress increase in the clay layer (Δσ) is 18.63kN/m2_.

Explanation of Solution

Given information:

The foundation load (P) is 478kN.

The length of the foundation (L) is 2.5m.

The breadth of the foundation (B) is 2.5m.

The depth of foundation (Df) is 1.5m.

The height of the clay layer 1 (H1) is 3.25m.

The height of the clay layer 2 (H2) is 3.5m.

The dry unit weight of sand (γd) is 16kN/m3.

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

The liquid limit (LL) is 37.

Calculation:

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

Calculate the distributed load (q) as shown below.

q=PBL

Substitute 478kN for P, 2.5m for B, and 2.5m for L.

q=4782.5×2.5=4786.25=76.48kN/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 the depth (z) of 1.75m:

Calculate the width (b) as shown below.

b=B2

Substitute 2.5m for B.

b=2.52=1.25m

Calculate the ratio (m1) as shown below.

m1=LB

Substitute 2.5m for L and 2.5m for B.

m1=2.52.5=1

Calculate the ratio (n1) as shown below.

n1=zb

Substitute 1.25m for b and 1.75m for z.

n1=1.751.25=1.4

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 0.522, for the values m1 of 1 and n1 of 1.4.

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

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

Substitute 76.48kN/m2 for q and 0.522 for I4 in Equation (1).

Δσz=76.48×0.522=39.92kN/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)
11.251.751.40.52239.92
11.253.52.80.21016.06
11.255.254.20.107.65

Table 1

Refer to table 1.

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

Δσ=Δσt+4Δσm+Δσb6

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 39.92kN/m2 for Δσt, 16.06kN/m2 for Δσm, and 7.65kN/m2 for Δσb.

Δσ=39.92+4×16.06+7.656=111.816=18.63kN/m2

Hence, the average stress increase in the clay layer (Δσ) is 18.63kN/m2_.

(b)

To determine

Calculate the primary consolidation settlement.

(b)

Expert Solution
Check Mark

Answer to Problem 11.21P

The primary consolidation settlement (SC) is 184mm_.

Explanation of Solution

Given information:

The foundation load (P) is 478kN.

The length of the foundation (L) is 2.5m.

The breadth of the foundation (B) is 2.5m.

The depth of foundation (Df) is 1.5m.

The height of the clay layer 1 (H1) is 3.25m.

The height of the clay layer 2 (H2) is 3.5m.

The dry unit weight of sand (γd) is 16kN/m3.

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

The liquid limit (LL) is 37.

The moisture content (w) is 19%.

The specific gravity of soil solids (Gs) is 2.71.

The preconsolidation pressure (σc) is 65kN/m2.

The swell index (CS) is 15CC.

Calculation:

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

The stress at the middle of the clay layer (Δσ=q) is 76.48kN/m2.

Calculate the compression index (CC) using the relation.

CC=0.009(LL10)

Substitute 37 for LL.

CC=0.009(3710)=0.243

Calculate the swell index (CS) as shown below.

CS=15CC

Substitute 0.243 for CC.

CS=15×0.243=0.0486

Calculate the saturated unit weight of clay layer ((γsat)clay) as shown below.

(γsat)clay=(1+w)γwGs1+wGs

Substitute 19% for w, 9.81kN/m3 for γw, and 2.71 for Gs.

(γsat)clay=(1+19100)×9.81×2.711+19100×2.71=1.19×26.58511.5149=20.88kN/m3

Calculate the void ratio (e0) as shown below.

e0=wGs

Substitute 19% for w and 2.71 for Gs.

e0=19100×2.71=0.515

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

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

Substitute 16kN/m3 for γd, 1.5m for Df, 18.8kN/m3 for (γsat), 9.81kN/m3 for γw, 20.88kN/m3 for (γsat)clay, 3.25m for H1, and 3.5m for H2.

σo=16×1.5+(18.89.81)(3.251.5)+(20.889.81)×3.52=24+15.7325+19.3725=59.1kN/m2

The effective stress σo=59.1kN/m2σc=65kN/m2. Hence, the clay is over consolidated.

Check for the condition σo+Δσ>σc as shown below.

Substitute 59.1kN/m2 for σo, 76.48kN/m2 for Δσ, and 65kN/m2 for σc.

59.1+76.48>65135.58kN/m2>65kN/m2

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

SC=CSH1+e0log(σcσo)+CCH1+e0log(σo+Δσσc)

Substitute 0.243 for CC, 0.0486 for CS, 3.5m for H, 0.515 for e0, 59.1kN/m2 for σo, 76.48kN/m2 for Δσ, and 65kN/m2 for σc.

SC=0.0486×3.51+0.515log(6559.1)+0.243×3.51+0.515log(59.1+76.4865)=0.1123×0.0413+0.5614×0.3193=0.184m×1,000mm1m=184mm

Therefore, the primary consolidation settlement (SC) is 184mm_.

(c)

To determine

Calculate the degree of consolidation after 2 years.

(c)

Expert Solution
Check Mark

Answer to Problem 11.21P

The degree of consolidation after 2 years (U) is 25%_.

Explanation of Solution

Given information:

The settlement after 2 years (SC(t)) is 46mm.

The foundation load (P) is 478kN.

The length of the foundation (L) is 2.5m.

The breadth of the foundation (B) is 2.5m.

The depth of foundation (Df) is 1.5m.

The height of the clay layer 1 (H1) is 3.25m.

The height of the clay layer 2 (H2) is 3.5m.

Calculation:

Refer to part (b).

The primary consolidation settlement (SC) is 184mm.

Calculate the degree of consolidation after 2 years (U) as shown below.

U=SC(t)SC

Substitute 46mm for SC(t) and 184mm for SC.

U=46184=0.25×100%=25%

Hence, the degree of consolidation (U) is 25%_.

(d)

To determine

Calculate the coefficient of consolidation for the pressure range.

(d)

Expert Solution
Check Mark

Answer to Problem 11.21P

The coefficient of consolidation of the clay (cv) is 0.075m2/year_.

Explanation of Solution

Given information:

The settlement after 2 years (SC(t)) is 46mm.

The foundation load (P) is 478kN.

The length of the foundation (L) is 2.5m.

The breadth of the foundation (B) is 2.5m.

The depth of foundation (Df) is 1.5m.

The height of the clay layer 1 (H1) is 3.25m.

The height of the clay layer 2 (H2) is 3.5m.

Calculation:

Refer to part (c).

The degree of consolidation (U) is 25%.

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 0.0491, for the value U of 25%.

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

Hdr=H32

Substitute 3.5m for H.

Hdr=3.52=1.75m

Calculate the coefficient of consolidation (cv) using the relation.

Tv=cvtHdr2

Substitute 0.0491 for Tv, 2year for t, and 1.75m for Hdr.

0.0491=cv×21.7522cv=0.1504cv=0.075m2/year

Hence, the coefficient of consolidation of the clay (cv) is 0.075m2/year_.

(e)

To determine

Calculate the settlement in 3 years.

(e)

Expert Solution
Check Mark

Answer to Problem 11.21P

The time settlement in 3 years (SC(3)) is 56mm_.

Explanation of Solution

Given information:

The foundation load (P) is 478kN.

The length of the foundation (L) is 2.5m.

The breadth of the foundation (B) is 2.5m.

The depth of foundation (Df) is 1.5m.

The height of the clay layer 1 (H1) is 3.25m.

The height of the clay layer 2 (H2) is 3.5m.

The time (t) is 3years.

Calculation:

Refer to part (b).

The primary consolidation settlement (SC) is 184mm.

Refer to part (d).

The coefficient of consolidation of the clay (cv) is 0.075m2/year.

Calculate the time factor (Tv) as shown below.

Tv=cvtHdr2

Substitute 3years for t, 0.075m2/year for cv, and 1.75m for Hdr.

Tv=0.075×31.752=0.073

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

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

Take the value of U as 30.5%, for the value Tv of 0.073.

Calculate the settlement in 3 years (SC(3)) as shown below.

U=SC(3)SC

Substitute 30.5% for U and 184mm for SC.

30.5100=SC(3)184SC(3)=56.12mmSC(3)56mm

Therefore, the time settlement in 3 years (SC(3)) is 56mm_.

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