Concept explainers
(a)
Calculate the total consolidation settlement under the action of fill load.
(a)
Answer to Problem 11.1CTP
The total consolidation settlement (SC) is 0.188 m_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
The properties of clay and organic layers are given in the Table.
Calculation:
Consider the unit weight of water (γw) is 9.81 kN/m3.
Calculate the distributed load (q) as shown below.
q=tγfill
Substitute 1.75 m for t and 20.1 kN/m3 for γfill.
q=1.75×20.1=35.02 kN/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 3 m:
Calculate the width (b) as shown below.
b=B2
Substitute 8 m for B.
b=82=4 m
Calculate the ratio (m1) as shown below.
m1=LB
Substitute 8 m for L and 8 m for B.
m1=88=1
Calculate the ratio (n1) as shown below.
n1=zb
Substitute 4 m for b and 3 m 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.
0.8−0.750.8−0.6=0.8−I40.8−0.8920.25×(−0.092)=0.8−I4I4=0.823
Similarly calculate the remaining values and tabulate as in Table 1.
Calculate the increase in vertical stress (Δσz) as shown below.
Substitute 35.02 kN/m2 for q and 0.823 for I4 in Equation (1).
Δσz=35.02×0.823=28.82 kN/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=LB | b=B2 | Depth, z (m) | n1=zb | I4 | Δσz (kN/m2) |
1 | 4 | 3 | 0.75 | 0.823 | 28.82 |
1 | 4 | 5 | 1.25 | 0.599 | 20.98 |
1 | 4 | 7 | 1.75 | 0.403 | 14.11 |
Table 1
Refer to table 1.
Calculate the stress increase in the clay layer (Δσ′) using the relation.
Δσ′=Δσt′+4Δσm′+Δσb′6 (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 28.82 kN/m2 for Δσt′, 20.98 kN/m2 for Δσm′, and 14.11 kN/m2 for Δσb′ in Equation (2).
Δσ′=28.82+4×20.98+14.116=126.856=21.14 kN/m2
Calculate the average effective stress at the middle of the clay layer (σo′) as shown below.
σo′=γdDf+((γsat)sand−γw)(H1−Df)+((γsat)clay−γw)×H22
Substitute 17 kN/m3 for γd, 1.5 m for Df, 19.2 kN/m3 for (γsat)sand, 9.81 kN/m3 for γw, 18.8 kN/m3 for (γsat)clay, 3 m for H1, and 4 m for H2.
σo′=17×1.5+(19.2−9.81)(3−1.5)+(18.8−9.81)×42=25.5+14.085+17.98=57.565 kN/m2
Calculate the primary consolidation settlement (SC) using the relation.
SC=CCH1+e0log(σo′+Δσ′σo′) (3)
Substitute 0.31 for CC, 4 m for H, 1.08 for e0, 57.565 kN/m2 for σo′, and 21.14 kN/m2 for Δσ′ in Equation (3).
SC−clay=0.31×41+1.08log(57.565+21.1457.565)=0.59615×0.1358=0.081 m
For peat layer:
For the depth (z) of 7 m:
Calculate the ratio (n1) as shown below.
n1=zb
Substitute 4 m for b and 7 m 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.
1.8−1.751.8−1.6=0.388−I40.388−0.4490.25×(−0.061)=0.8−I4I4=0.403
Similarly calculate the remaining values and tabulate as in Table 2.
Calculate the increase in vertical stress (Δσz) as shown below.
Substitute 35.02 kN/m2 for q and 0.403 for I4 in Equation (1).
Δσz=35.02×0.403=14.11 kN/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=LB | b=B2 | Depth, z (m) | n1=zb | I4 | Δσz (kN/m2) |
1 | 4 | 7 | 1.75 | 0.403 | 14.11 |
1 | 4 | 7.9 | 1.975 | 0.342 | 11.98 |
1 | 4 | 8.8 | 2.2 | 0.302 | 10.58 |
Table 2
Refer to table 2.
Calculate the stress increase in the peat layer (Δσ′) using the relation.
Substitute 14.11 kN/m2 for Δσt′, 11.98 kN/m2 for Δσm′, and 10.58 kN/m2 for Δσb′ in Equation (2).
Δσ′=14.11+4×11.98+10.586=72.616=12.10 kN/m2
Calculate the average effective stress at the middle of the clay layer (σo′) as shown below.
σo′=γdDf+((γsat)sand−γw)(H1−Df)+((γsat)clay−γw)×H2+((γsat)peat−γw)×H32
Substitute 17 kN/m3 for γd, 1.5 m for Df, 19.2 kN/m3 for (γsat)sand, 9.81 kN/m3 for γw, 18.8 kN/m3 for (γsat)clay, 15 kN/m3 for (γsat)peat, 3 m for H1, 4 m for H2, and 1.8 m for H3.
σo′=17×1.5+(19.2−9.81)(3−1.5)+(18.8−9.81)×4+(15−9.81)×1.82=25.5+14.085+35.96+4.671=80.216 kN/m2
Calculate the primary consolidation settlement (SC) using the relation.
Substitute 7.2 for CC, 1.8 m for H, 6.4 for e0, 80.216 kN/m2 for σo′, and 12.10 kN/m2 for Δσ′ in Equation (3).
SC−Peat=7.2×1.81+6.4log(80.216+12.1080.216)=1.7514×0.061=0.107 m
Calculate the total consolidation settlement under the action of fill load (SC) as shown below.
SC=SC−clay+SC−peat
Substitute 0.081 m for SC−clay and 0.107 m for SC−peat.
SC=0.081+0.107=0.188
Hence, the total consolidation settlement (SC) is 0.188 m_.
(b)
Calculate the time for 99% primary consolidation for each layer.
(b)
Answer to Problem 11.1CTP
The time for 99% primary consolidation for clay (t99-clay) is 138 days_.
The time for 99% primary consolidation for peat (t99-peat) is 23 days_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
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 4 m for H2.
Hdr=42=2 m
Calculate the time for 99% consolidation (t99-clay) using the relation.
Tv=cvtH2dr (4)
Substitute 1.781 for Tv, 0.006 cm2/sec for cv, and 2 m for Hdr in Equation (4).
1.781=0.006 cm2/sec×t99-clay(2 m×100 cm1 m)20.006t99-clay=71,240t99-clay=11,873,333.333 sec×1 min60 sec×1 hr60 mins×1 day24 hrst99-clay=138 days
Hence, the time for 99% primary consolidation for clay (t99-clay) is 138 days_.
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 1.8 m for H3.
Hdr=1.8 m
Calculate the time for 99% consolidation (t99-peat) using the relation.
Substitute 1.781 for Tv, 0.029 cm2/sec for cv, and 1.8 m for Hdr in Equation (4).
1.781=0.029 cm2/sec×t99-peat(1.8 m×100 cm1 m)20.029t99-peat=57,704.4t99-peat=1,989,806.897 sec×1 min60 sec×1 hr60 mins×1 day24 hrst99-peat=23 days
Hence, the time for 99% primary consolidation for peat (t99-peat) is 23 days_.
(c)
Calculate the secondary compression in each layer up to end of 18 months.
(c)
Answer to Problem 11.1CTP
The secondary compression for clay (Ss−clay) is 0.055 m_.
The secondary compression for peat (Ss−peat) is 0.096 m_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
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 138 days.
The time for 99% primary consolidation for peat (t99-peat) is 23 days.
For clay:
Calculate the primary void ratio (Δeprimary) as shown below.
Δeprimary=CClog(σo′+Δσ′σo′) (5)
Substitute 0.31 for CC, 57.565 kN/m2 for σo′, and 21.14 kN/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=e0−Δeprimary (6)
Substitute 1.08 for e0 and 0.042 for Δeprimary in Equation (6).
ep=1.08−0.042=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=Cα′Hlog(t2t1) (8)
Substitute 0.0235 for Cα′, 4 m for H, 18 months for t2, and 138 days for t1 in Equation (8).
Ss−clay=0.0235×(4 m)log(18 months×30 days1 month138 days)=0.094×0.5925=0.055 m
Hence, the secondary compression for clay (Ss−clay) is 0.055 m_.
For peat:
Calculate the primary void ratio (Δeprimary) as shown below.
Substitute 7.2 for CC, 80.216 kN/m2 for σo′, and 12.10 kN/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=6.4−0.44=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′, 1.8 m for H, 18 months for t2, and 23 days for t1 in Equation (8).
Ss−clay=0.039×(1.8 m)log(18 months×30 days1 month23 days)=0.0702×1.3707=0.096 m
Hence, the secondary compression for peat (Ss−peat) is 0.096 m_.
(d)
Calculate the total settlement after 18 months.
(d)
Answer to Problem 11.1CTP
The total settlement after 18 months is 0.339 m_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
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 (Ss−clay) is 0.055 m.
The secondary compression for peat (Ss−peat) is 0.096 m.
Calculate the total settlement after 18 months as shown below.
Total settlement=SC+Ss−clay+Ss−peat
Substitute 0.188 m for SC, 0.055 m for Ss−clay, and 0.096 m for Ss−peat.
Total settlement=0.188+0.055+0.096=0.339 m
Hence, the total settlement after 18 months is 0.339 m_.
(e)
Calculate the excess pore water pressure at point A two months after the application of the fill load.
(e)
Answer to Problem 11.1CTP
The excess pore water pressure at point A (uz) is 2.114 kN/m2_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
The properties of clay and organic layers are given in the Table.
The depth (z) is 3.2 m.
Calculation:
Refer to part (a).
The pore water pressure (uo=Δσ′) is 21.14 kN/m2.
Calculate the length of maximum drainage path (Hdr) using the relation.
Hdr=H22
Substitute 4 m for H.
Hdr=42=2 m
Calculate the time factor (Tv) as shown below.
Tv=cvtH2dr
Substitute 0.006 cm2/sec for cv, 2 months for t, and 2 m for Hdr.
Tv=0.006 cm2/sec×2 months×30 days1 month×24 hrs1 day×60 mins1 hr×60 sec1 min(2 m×100 cm1 m)2=0.7776
Calculate the ratio zHdr as shown below.
Substitute 3.2 m for z and 2 m 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=1−uzuo
Substitute 0.9 for U and 21.14 kN/m2 for uo.
0.9=1−uz21.14uz21.14=0.1uz=2.114 kN/m2
Hence, the excess pore water pressure at point A (uz) is 2.114 kN/m2_.
(f)
Calculate the effective stress at point A two months after the application of the fill load.
(f)
Answer to Problem 11.1CTP
The effective stress at point A is 87.38 kN/m2_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
The properties of clay and organic layers are given in the Table.
The depth (z) is 3.2 m.
Calculation:
Refer to part (a).
The pore water pressure (uo=Δσ′) is 21.14 kN/m2.
Refer to part (e)
The excess pore water pressure at point A (uz) is 2.114 kN/m2.
Calculate the increase in effective stress (Δu) after 2 months as shown below.
Δu=uo−uz
Substitute 21.14 kN/m2 for uo and 2.114 kN/m2 for uz.
Δu=21.14−2.114=19.026 kN/m2
Calculate the average effective stress at the point A (σo−A′) as shown below.
σo−A′=γdDf+((γsat)sand−γw)(H1−Df)+((γsat)clay−γw)×z
Substitute 17 kN/m3 for γd, 1.5 m for Df, 19.2 kN/m3 for (γsat)sand, 9.81 kN/m3 for γw, 18.8 kN/m3 for (γsat)clay, 3 m for H1, and 3.2 m for z.
σo−A′=17×1.5+(19.2−9.81)(3−1.5)+(18.8−9.81)×3.2=25.5+14.085+28.768=68.353 kN/m2
Calculate the final effective stress at point A as shown below.
Final effective stress=σo−A′+Δu
Substitute 68.353 kN/m2 for σo−A′ and 19.026 kN/m2 for Δu.
Final effective stress=68.353+19.026=87.38 kN/m2
Hence, the effective stress at point A is 87.38 kN/m2_.
(g)
Calculate the piezometer reading at point A two months after the application of the fill load.
(g)
Answer to Problem 11.1CTP
The piezometer reading at point A is 48.22 kN/m2_.
Explanation of Solution
Given information:
The thickness of fill material (t) is 1.75 m.
The compacted unit weight of fill material (γfill) is 20.1 kN/m3.
The length of the foundation (L) is 8 m.
The breadth of the foundation (B) is 8 m.
The depth of fill (Df) is 1.5 m.
The height of the layer silty sand (H1) is 3 m.
The height of the clay layer (H2) is 4 m.
The height of the peat layer (H3) is 1.8 m.
The dry unit weight of sand (γd) is 17 kN/m3.
The saturated unit weight of sand (γsat)sand is 19.2 kN/m3.
The saturated unit weight of clay (γsat)clay is 18.8 kN/m3.
The saturated unit weight of peat (γsat)peat is 15 kN/m3.
The time (t) is 8 months.
The properties of clay and organic layers are given in the Table.
The depth (z) is 3.2 m.
Calculation:
Refer to part (e)
The excess pore water pressure at point A (uz) is 2.114 kN/m2.
The piezometer reading is the total pore water pressure.
Calculate the piezometer reading (upiezometer) as shown below.
upiezometer=(H1−Df+z)γw+uz
Substitute 3 m for H1, 1.5 m for Df, 3.2 m for z, 9.81 kN/m3 for γw, and 2.114 kN/m2 for uz.
upiezometer=(3−1.5+3.2)9.81+2.114=48.221 kN/m2
Hence, the piezometer reading at point A is 48.22 kN/m2_.
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Chapter 11 Solutions
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- P.3.3 Oil of sp.gr. 0.9 flows through a vertical pipe (upwards). Two points A and B one above the other 40 cm apart in a pipe are connected by a U-tube carrying mercury. If the difference of pressure between A and B is 0.2 kg/cm², 1- Find the reading of the manometer. 2- If the oil flows through a horizontal pipe, find the reading in manometer for the same difference in pressure between A and B. Ans. 1- R= 0.12913 m, 2- R = 0.1575 m,arrow_forwardPlease solve the question by hand with a detailed explanation of the steps.arrow_forwardPlease solve the question by hand with a detailed explanation of the steps.arrow_forward
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